Solucionário - Fundamentos da Termodinâmica - Borgnakke Sonntag - 8a Edição

Updated June 2013 SOLUTION MANUAL CHAPTER 1 Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CONTENT CHAPTER 1 SUBSECTION PROB NO Concept Problems 121 Properties Units and Force 2237 Specific Volume 3844 Pressure 4561 Manometers and Barometers 6283 Energy and Temperature 8495 Review problems 96101 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful InText Concept Questions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1a Make a control volume around the turbine in the steam power plant in Fig 12 and list the flows of mass and energy that are there Solution We see hot high pressure steam flowing in at state 1 from the steam drum through a flow control not shown The steam leaves at a lower pressure to the condenser heat exchanger at state 2 A rotating shaft gives a rate of energy power to the electric generator set WT 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1b Take a control volume around your kitchen refrigerator and indicate where the components shown in Figure 13 are located and show all flows of energy transfers Solution The valve and the cold line the evaporator is inside close to the inside wall and usually a small blower distributes cold air from the freezer box to the refrigerator room cb W Q Q leak The black grille in the back or at the bottom is the condenser that gives heat to the room air The compressor sits at the bottom Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1c Why do people float high in the water when swimming in the Dead Sea as compared with swimming in a fresh water lake As the dead sea is very salty its density is higher than fresh water density The buoyancy effect gives a force up that equals the weight of the displaced water Since density is higher the displaced volume is smaller for the same force Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1d Density of liquid water is ρ 1008 T2 kgm3 with T in oC If the temperature increases what happens to the density and specific volume Solution The density is seen to decrease as the temperature increases ρ T2 Since the specific volume is the inverse of the density v 1ρ it will increase 1e A car tire gauge indicates 195 kPa what is the air pressure inside The pressure you read on the gauge is a gauge pressure P so the absolute pressure is found as P Po P 101 195 296 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1f Can I always neglect P in the fluid above location A in figure 113 What does that depend on If the fluid density above A is low relative to the manometer fluid then you neglect the pressure variation above position A say the fluid is a gas like air and the manometer fluid is like liquid water However if the fluid above A has a density of the same order of magnitude as the manometer fluid then the pressure variation with elevation is as large as in the manometer fluid and it must be accounted for 1g A U tube manometer has the left branch connected to a box with a pressure of 110 kPa and the right branch open Which side has a higher column of fluid Solution Since the left branch fluid surface feels 110 kPa and the right branch surface is at 100 kPa you must go further down to match the 110 kPa The right branch has a higher column of fluid P o Box H Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Concept Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11 Make a control volume around the whole power plant in Fig 11 and with the help of Fig 12 list what flows of mass and energy are in or out and any storage of energy Make sure you know what is inside and what is outside your chosen CV Solution Smoke stack Boiler building Coal conveyor system Dock Turbine house Storage gypsum Coal storage flue gas cb Underground power cable W electrical Hot water District heating m Coal m m Flue gas Storage for later Gypsum fly ash slag transport out Cold return m m Combustion air Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12 Make a control volume around the refrigerator in Fig 13 Identify the mass flow of external air and show where you have significant heat transfer and where storage changes The valve and the cold line the evaporator is inside close to the inside wall and usually a small blower distributes cold air from the freezer box to the refrigerator room cb W Q Q leak The black grille in the back or at the bottom is the condenser that gives heat to the room air The compressor sits at the bottom The storage changes inside the box which is outside of the refrigeration cycle components of Fig 13 when you put some warmer mass inside the refrigerator it is being cooled by the evaporator and the heat is leaving in the condenser The condenser warms outside air so the air flow over the condenser line carries away some energy If natural convection is not enough to do this a small fan is used to blow air in over the condenser forced convection Likewise the air being cooled by the evaporator is redistributed inside the refrigerator by a small fan and some ducts Since the room is warmer than the inside of the refrigerator heat is transferred into the cold space through the sides and the seal around the door Also when the door is opened warm air is pulled in and cold air comes out from the refrigerator giving a net energy transfer similar to a heat transfer Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13 Separate the list P F V v ρ T a m L t and V into intensive extensive and non properties Solution Intensive properties are independent upon mass P v ρ T Extensive properties scales with mass V m Nonproperties F a L t V Comment You could claim that acceleration a and velocity V are physical properties for the dynamic motion of the mass but not thermal properties Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14 A tray of liquid water is placed in a freezer where it cools from 20oC to 5oC Show the energy flows and storage and explain what changes Inside the freezer box the walls are very cold as they are the outside of the evaporator or the air is cooled and a small fan moves the air around to redistribute the cold air to all the items stored in the freezer box The fluid in the evaporator absorbs the energy and the fluid flows over to the compressor on its way around the cycle see Fig 13 As the water is cooled it eventually reaches the freezing point and ice starts to form After a significant amount of energy is removed from the water it is turned completely into ice at 0oC and then cooled a little more to 5oC The water has a negative energy storage and the energy is moved by the refrigerant fluid out of the evaporator into the compressor and then finally out of the condenser into the outside room air C Borgnakke Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 15 The overall density of fibers rock wool insulation foams and cotton is fairly low Why is that Solution All these materials consist of some solid substance and mainly air or other gas The volume of fibers clothes and rockwool that is solid substance is low relative to the total volume that includes air The overall density is ρ m V msolid mair Vsolid Vair where most of the mass is the solid and most of the volume is air If you talk about the density of the solid only it is high Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 16 Is density a unique measure of mass distribution in a volume Does it vary If so on what kind of scale distance Solution Density is an average of mass per unit volume and we sense if it is not evenly distributed by holding a mass that is more heavy in one side than the other Through the volume of the same substance say air in a room density varies only little from one location to another on scales of meter cm or mm If the volume you look at has different substances air and the furniture in the room then it can change abruptly as you look at a small volume of air next to a volume of hardwood Finally if we look at very small scales on the order of the size of atoms the density can vary infinitely since the mass electrons neutrons and positrons occupy very little volume relative to all the empty space between them Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 17 Water in nature exists in different phases such as solid liquid and vapor gas Indicate the relative magnitude of density and specific volume for the three phases Solution Values are indicated in Figure 18 as density for common substances More accurate values are found in Tables A3 A4 and A5 Water as solid ice has density of around 900 kgm3 Water as liquid has density of around 1000 kgm3 Water as vapor has density of around 1 kgm3 sensitive to P and T Ice cube Liquid drops falling Cloud Steam water vapor cannot be seen what you see are tiny drops suspended in air from which we infer that there was some water vapor before it condensed Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 18 What is the approximate mass of 1 L of gasoline Of helium in a balloon at To Po Solution Gasoline is a liquid slightly lighter than liquid water so its density is smaller than 1000 kgm3 1 L is 0001 m3 which is a common volume used for food items A more accurate density is listed in Table A3 as 750 kgm3 so the mass becomes m ρ V 750 kgm3 0001 m3 075 kg The helium is a gas highly sensitive to P and T so its density is listed at the standard conditions 100 kPa 25C in Table A5 as ρ 01615 kgm3 m ρ V 01615 kgm3 0001 m3 1615 104 kg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 19 Can you carry 1 m3 of liquid water Solution The density of liquid water is about 1000 kgm3 from Figure 17 see also Table A3 Therefore the mass in one cubic meter is m ρV 1000 kgm3 1 m3 1000 kg and we can not carry that in the standard gravitational field Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 110 A heavy refrigerator has four heightadjustable feet What feature of the feet will ensure that they do not make dents in the floor Answer The area that is in contact with the floor supports the total mass in the gravitational field F PA mg so for a given mass the smaller the area is the larger the pressure becomes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 111 A swimming pool has an evenly distributed pressure at the bottom Consider a stiff steel plate lying on the ground Is the pressure below it just as evenly distributed Solution The pressure is force per unit area from page 13 P FA mgA The steel plate can be reasonable plane and flat but it is stiff and rigid However the ground is usually uneven so the contact between the plate and the ground is made over an area much smaller than the actual plate area Thus the local pressure at the contact locations is much larger than the average indicated above The pressure at the bottom of the swimming pool is very even due to the ability of the fluid water to have full contact with the bottom by deforming itself This is the main difference between a fluid behavior and a solid behavior Steel plate Ground 112 What physically determines the variation of the atmospheric pressure with elevation The total mass of the column of air over a unit area and the gravitation gives the force which per unit area is pressure This is an integral of the density times gravitation over elevation as in Eq14 To perform the integral the density and gravitation as a function of height elevation should be known Later we will learn that air density is a function of temperature and pressure and compositions if it varies Standard curve fits are known that describes this variation and you can find tables with the information about a standard atmosphere See problems 128 164 and 195 for some examples Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 113 Two divers swim at 20 m depth One of them swims right in under a supertanker the other stays away from the tanker Who feels a greater pressure Solution Each one feels the local pressure which is the static pressure only a function of depth Pocean P0 P P0 ρgH So they feel exactly the same pressure Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 114 A manometer with water shows a P of Po20 what is the column height difference Solution P Po20 ρHg H Po20 ρ g 1013 1000 Pa 20 997 kgm3 980665 ms2 0502 m Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 115 Does the pressure have to be uniform for equilibrium to exist No It depends on what causes a pressure difference Think about the pressure increasing as you move down into the ocean the water at different levels are in equilibrium However if the pressure is different at nearby locations at same elevation in the water or in air that difference induces a motion of the fluid from the higher towards the lower pressure The motion will persist as long as the pressure difference exist Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 116 A water skier does not sink too far down in the water if the speed is high enough What makes that situation different from our static pressure calculations The water pressure right under the ski is not a static pressure but a static plus dynamic pressure that pushes the water away from the ski The faster you go the smaller amount of water is displaced but at a higher velocity Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 117 What is the lowest temperature in degrees Celsuis In degrees Kelvin Solution The lowest temperature is absolute zero which is at zero degrees Kelvin at which point the temperature in Celsius is negative TK 0 K 27315 oC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 118 Convert the formula for water density in Intext Concept Question d to be for T in degrees Kelvin Solution ρ 1008 TC2 kgm3 We need to express degrees Celsius in degrees Kelvin TC TK 27315 and substitute into formula ρ 1008 TC2 1008 TK 273152 11446 TK2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 119 A thermometer that indicates the temperature with a liquid column has a bulb with a larger volume of liquid why is that The expansion of the liquid volume with temperature is rather small so by having a larger volume expand with all the volume increase showing in the very small diameter column of fluid greatly increases the signal that can be read Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 120 What is the main difference between the macroscopic kinetic energy in a motion like the blowing of wind versus the microscopic kinetic energy of individual molecules Which one can you sense with your hand Answer The microscopic kinetic energy of individual molecules is too small for us to sense however when the combined action of billions actually more like in the order of 1 E19 are added we get to the macroscopic magnitude we can sense The wind velocity is the magnitude and direction of the averaged velocity over many molecules which we sense The individual molecules are moving in a random motion with zero average on top of this mean or average motion A characteristic velocity of this random motion is the speed of sound around 340 ms for atmospheric air and it changes with temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 121 How can you illustrate the binding energy between the three atoms in water as they sit in a triatomic water molecule Hint imagine what must happen to create three separate atoms Answer If you want to separate the atoms you must pull them apart Since they are bound together with strong forces like nonlinear springs you apply a force over a distance which is work energy in transfer to the system and you could end up with two hydrogen atoms and one oxygen atom far apart so they no longer have strong forces between them If you do not do anything else the atoms will sooner or later recombine and release all the energy you put in and the energy will come out as radiation or given to other molecules by collision interactions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Properties Units and Force Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 122 An apple weighs 60 g and has a volume of 75 cm3 in a refrigerator at 8oC What is the apple density List three intensive and two extensive properties of the apple Solution ρ EA kg mA3 AE A 800 EA kg mA3 AE m V 006 0000 075 Intensive ρ 800 EA kg mA3 AE A v A1 ρE A 0001 25 EAmA3 A EkgE A T 8C P 101 kPa Extensive m 60 g 006 kg V 75 cmA3E A 0075 L 0000 075 mA3E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 123 One kilopond 1 kp is the weight of 1 kg in the standard gravitational field How many Newtons N is that F ma mg 1 kp 1 kg 9807 msA2E A 9807 N Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 124 A stainless steel storage tank contains 5 kg of oxygen gas and 7 kg of nitrogen gas How many kmoles are in the tank Table A2 MAO2E A 31999 MAN2E A 28013 nAO2E A mAO2E A MAO2E A A 5 31999E A 015625 kmol nAO2E A mAN2E A MAN2E A A 7 28013E A 024988 kmol nAtotE A nAO2E A nAN2E A 015625 024988 040613 kmol Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 125 A steel cylinder of mass 4 kg contains 4 L of liquid water at 25AoE AC at 100 kPa Find the total mass and volume of the system List two extensive and three intensive properties of the water Solution Density of steel in Table A3 ρ 7820 kgmA3E Volume of steel V mρ A 4 kg 7820 kgm3 E A 0000 512 mA3E Density of water in Table A4 ρ 997 kgmA3E Mass of water m ρV 997 kgmA3E A 0004 mA3E A 3988 kg Total mass m msteel mwater 4 3988 7988 kg Total volume V Vsteel Vwater 0000 512 0004 0004 512 mA3E A 451 L Extensive properties m V Intensive properties ρ or v 1ρ T P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 126 The standard acceleration at sea level and 45 latitude due to gravity is 980665 msA2E A What is the force needed to hold a mass of 2 kg at rest in this gravitational field How much mass can a force of 1 N support Solution ma 0 F F mg F mg 2 kg 980665 msA2E A 19613 N F mg m AF gE A A 1 N 980665 ms2 E A 0102 kg m F g Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 127 An aluminum piston of 25 kg is in the standard gravitational field where a force of 25 N is applied vertically up Find the acceleration of the piston Solution Fup ma F mg a AF mg mE A AF mE A g A 25 N 25 kgE A 9807 msA2E 0193 msA2 E g F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 128 When you move up from the surface of the earth the gravitation is reduced as g 9807 332 106 z with z as the elevation in meters How many percent is the weight of an airplane reduced when it cruises at 11 000 m Solution go 9807 ms 2 gH 9807 332 106 11 000 97705 ms2 Wo m go WH m gH WH Wo gH go 97705 9807 09963 Reduction 1 09963 00037 or 037 ie we can neglect that for most applications Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 129 A car rolls down a hill with a slope so the gravitational pull in the direction of motion is one tenth of the standard gravitational force see Problem 126 If the car has a mass of 2500 kg find the acceleration Solution ma F mg 10 a mg 10m g10 980665 ms2 10 0981 ms2 g This acceleration does not depend on the mass of the car Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 130 A van is driven at 60 kmh and is brought to a full stop with constant deceleration in 5 seconds If the total car and driver mass is 2075 kg find the necessary force Solution Acceleration is the time rate of change of velocity a dV dt 60 1000 3600 5 3333 ms2 ma F Fnet ma 2075 kg 3333 ms2 6916 N Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 131 A 1500kg car moving at 20 kmh is accelerated at a constant rate of 4 ms2 up to a speed of 75 kmh What are the force and total time required Solution a dV dt V t t V a 75 20 kmh 1000 mkm 3600 sh 4 ms2 382 sec F ma 1500 kg 4 ms2 6000 N Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 132 On the moon the gravitational acceleration is approximately onesixth that on the surface of the earth A 5kg mass is weighed with a beam balance on the surface on the moon What is the expected reading If this mass is weighed with a spring scale that reads correctly for standard gravity on earth see Problem 126 what is the reading Solution Moon gravitation is g gearth6 m m m Beam Balance Reading is 5 kg Spring Balance Reading is in kg units This is mass comparison Force comparison length F g Reading will be 5 6 kg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 133 The elevator in a hotel has a mass of 750 kg and it carries six people with a total mass of 450 kg How much force should the cable pull up with to have an acceleration of 1 ms2 in the upwards direction Solution The total mass moves upwards with an acceleration plus the gravitations acts with a force pointing down ma F F mg F ma mg ma g 750 450 kg 1 981 ms2 12 972 N F g Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 134 One of the people in the previous problem weighs 80 kg standing still How much weight does this person feel when the elevator starts moving Solution The equation of motion is ma F F mg so the force from the floor becomes F ma mg ma g 80 kg 1 981 ms2 8648 N x kg 981 ms2 Solve for x x 8648 N 981 ms2 8815 kg The person then feels like having a mass of 88 kg instead of 80 kg The weight is really force so to compare to standard mass we should use kp So in this example the person is experiencing a force of 88 kp instead of the normal 80 kp Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 135 A bottle of 12 kg steel has 175 kmole of liquid propane It accelerates horizontal with 3 ms2 what is the needed force Solution The molecular weight for propane is M 44094 from Table A2 The force must accelerate both the container mass and the propane mass m msteel mpropane 12 175 44094 90645 kg ma F F ma 90645 kg 3 ms2 2719 N Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 136 Some steel beams with a total mass of 700 kg are raised by a crane with an acceleration of 2 ms2 relative to the ground at a location where the local gravitational acceleration is 95 ms2 Find the required force Solution F ma Fup mg Fup ma mg 700 kg 2 95 ms2 80 500 N Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Specific Volume Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 137 A 1 m3 container is filled with 400 kg of granite stone 200 kg dry sand and 02 m3 of liquid 25C water Use properties from tables A3 and A4 Find the average specific volume and density of the masses when you exclude air mass and volume Solution Specific volume and density are ratios of total mass and total volume mliq Vliqvliq Vliq ρliq 02 m3 997 kgm3 1994 kg mTOT mstone msand mliq 400 200 1994 7994 kg Vstone mv mρ 400 kg 2750 kgm3 01455 m3 Vsand mv mρ 200 1500 01333 m 3 VTOT Vstone Vsand Vliq 01455 01333 02 04788 m3 v VTOT mTOT 047887994 0000599 m3kg ρ 1v mTOTVTOT 799404788 16696 kgm3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 138 A power plant that separates carbondioxide from the exhaust gases compresses it to a density of 110 kgm3 and stores it in an unminable coal seam with a porous volume of 100 000 m3 Find the mass they can store Solution m ρ V 110 kgm3 100 000 m3 11 10 6 kg Comment Just to put this in perspective a power plant that generates 2000 MW by burning coal would make about 20 million tons of carbondioxide a year That is 2000 times the above mass so it is nearly impossible to store all the carbondioxide being produced Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 139 A 15kg steel gas tank holds 300 L of liquid gasoline having a density of 800 kgm3 If the system is decelerated with 2g what is the needed force Solution m mtank mgasoline 15 kg 03 m3 800 kgm3 255 kg F ma 255 kg 2 981 ms2 5003 N cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 140 A 5 m3 container is filled with 900 kg of granite density 2400 kgm3 and the rest of the volume is air with density 115 kgm3 Find the mass of air and the overall average specific volume Solution mair ρ V ρair Vtot mgranite ρ 115 5 900 2400 115 4625 532 kg v V m 5 900 532 0005 52 m3kg Comment Because the air and the granite are not mixed or evenly distributed in the container the overall specific volume or density does not have much meaning Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 141 A tank has two rooms separated by a membrane Room A has 1 kg air and volume 05 m3 room B has 075 m3 air with density 08 kgm3 The membrane is broken and the air comes to a uniform state Find the final density of the air Solution Density is mass per unit volume m mA mB mA ρBVB 1 08 075 16 kg V VA VB 05 075 125 m3 ρ m V 16 125 kg m3 128 kgm 3 A B cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 142 One kilogram of diatomic oxygen O2 molecular weight 32 is contained in a 500 L tank Find the specific volume on both a mass and mole basis v and Solution From the definition of the specific volume v V m 05 1 m3 kg 05 m3kg V n V mM M v 32 kgkmol 05 m3kg 16 m3kmol v v Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Pressure Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 143 A 5000kg elephant has a cross sectional area of 002 m2 on each foot Assuming an even distribution what is the pressure under its feet Force balance ma 0 PA mg P mgA 5000 kg 981 ms2 4 002 m2 613 125 Pa 613 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 144 A valve in a cylinder has a cross sectional area of 11 cm2 with a pressure of 735 kPa inside the cylinder and 99 kPa outside How large a force is needed to open the valve Fnet PinA PoutA 735 99 kPa 11 cm2 6996 kPa cm2 6996 EAkN mA2 AE A 10A4E A mA2E 700 N cb P cyl Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 145 The hydraulic lift in an autorepair shop has a cylinder diameter of 02 m To what pressure should the hydraulic fluid be pumped to lift 40 kg of pistonarms and 700 kg of a car Solution Force acting on the mass by the gravitational field F ma mg 740 980665 72569 N 7257 kN Force balance F P PA0E A A F P P0 F A A π D2 1 4 0031416 m2 P 101 kPa EA 7257 kN 0031416 mA2 AE A 332 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 146 A hydraulic lift has a maximum fluid pressure of 500 kPa What should the pistoncylinder diameter be so it can lift a mass of 850 kg Solution With the piston at rest the static force balance is F P A F mg A π rA2E A π DA2E A4 PA P π DA2E A4 mg DA2E A A4mg P πE A D 2A mg Pπ EA 2A 850 kg 9807 ms2 E500 kPa π 1000 PakPa EA 0146 m Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 147 A laboratory room keeps a vacuum of 01 kPa What net force does that put on the door of size 2 m by 1 m Solution The net force on the door is the difference between the forces on the two sides as the pressure times the area F Poutside A Pinside A P A 01 kPa 2 m 1 m 200 N Remember that kPa is kNmA2E A Pabs Po P P 01 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 148 A vertical hydraulic cylinder has a 125mm diameter piston with hydraulic fluid inside the cylinder and an ambient pressure of 1 bar Assuming standard gravity find the piston mass that will create a pressure inside of 1500 kPa Solution Force balance F PA F PA0E AA mApE Ag PA0E A 1 bar 100 kPa A π4 DA2E A π4 0125A2E A 001227 mA2E cb g P o mp P PA0E A AA gE A 1500 100 kPa 1000 PakPa A001227 980665E A A m2 Ems2 E A 1752 kg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 149 A 75kg human footprint is 005 mA2E A when the human is wearing boots Suppose you want to walk on snow that can at most support an extra 3 kPa what should the total snowshoe area be Force balance ma 0 PA mg A Amg PE A A75 kg 981 ms2 E 3 kPaE A 0245 mA2E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 150 A pistoncylinder with cross sectional area of 001 mA2E A has a piston mass of 100 kg resting on the stops as shown in Fig P150 With an outside atmospheric pressure of 100 kPa what should the water pressure be to lift the piston Solution The force acting down on the piston comes from gravitation and the outside atmospheric pressure acting over the top surface Force balance F F PA mApE Ag PA0E AA Now solve for P divide by 1000 to convert to kPa for 2nd term P PA0E A A mpg EAE A 100 kPa A100 980665 001 1000E A kPa 100 kPa 9807 kPa 198 kPa Water cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 151 A large exhaust fan in a laboratory room keeps the pressure inside at 10 cm water vacuum relative to the hallway What is the net force on the door measuring 19 m by 11 m Solution The net force on the door is the difference between the forces on the two sides as the pressure times the area F Poutside A Pinside A P A 10 cm H2O 19 m 11 m 010 980638 kPa 209 mA2E A 2049 N Table A1 1 m H2O is 980638 kPa and kPa is kNmA2E A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 152 A tornado rips off a 100 mA2E A roof with a mass of 1000 kg What is the minimum vacuum pressure needed to do that if we neglect the anchoring forces Solution The net force on the roof is the difference between the forces on the two sides as the pressure times the area F Pinside A PoutsideA P A That force must overcome the gravitation mg so the balance is P A mg P mgA 1000 kg 9807 msA2E A 100 mA2E A 98 Pa 0098 kPa Remember that kPa is kNmA2E A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 153 A 5kg cannonball acts as a piston in a cylinder with a diameter of 015 m As the gunpowder is burned a pressure of 7 MPa is created in the gas behind the ball What is the acceleration of the ball if the cylinder cannon is pointing horizontally Solution The cannon ball has 101 kPa on the side facing the atmosphere ma F P1 A P0 A P1 P0 A 7000 101 kPa π 0152 4 m2 1219 kN a AF mE A A1219 kN 5 kgE A 24 380 ms2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 154 Repeat the previous problem for a cylinder cannon pointing 40 degrees up relative to the horizontal direction Solution ma F P1 P0 A mg sin 400 ma 7000 101 kPa π 0152 4 m2 5 9807 06428 N 1219 kN 3152 N 12187 kN a AF mE A A12187 kN 5 kgE A 24 374 ms2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 155 A 25 m tall steel cylinder has a cross sectional area of 15 mA2E A At the bottom with a height of 05 m is liquid water on top of which is a 1 m high layer of gasoline This is shown in Fig P155 The gasoline surface is exposed to atmospheric air at 101 kPa What is the highest pressure in the water Solution The pressure in the fluid goes up with the depth as P PAtopE A P PAtopE A ρgh and since we have two fluid layers we get P PAtopE A ρhAgasolineE A ρhAwaterE A g Air Water 1 m 05 m Gasoline The densities from Table A4 are ρAgasolineE A 750 kgmA3E A ρAwaterE A 997 kgmA3E P 101 kPa 750 1 997 05 kgm2 A9807 1000E A ms2 kPaPa 1132 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 156 An underwater buoy is anchored at the seabed with a cable and it contains a total mass of 250 kg What should the volume be so that the cable holds it down with a force of 1000 N Solution We need to do a force balance on the system at rest and the combined pressure over the buoy surface is the buoyancy lift equal to the weight of the displaced water volume ma 0 mH2Og mg F ρH2OVg mg F V mg F ρH2Og m Fg ρH2O 250 kg 1000 N981 msA2E A 997 kgmA3E A 0353 mA3E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 157 At the beach atmospheric pressure is 1025 mbar You dive 15 m down in the ocean and you later climb a hill up to 250 m elevation Assume the density of water is about 1000 kgmA3E A and the density of air is 118 kgmA3E A What pressure do you feel at each place Solution P ρgh Units from A1 1 mbar 100 Pa 1 bar 100 kPa PAoceanE A PA0E A P 1025 100 Pa 1000 kgmA3E A 981 msA2E A 15 m 24965 10A5E A Pa 250 kPa PAhillE A PA0E A P 1025 100 Pa 118 kgmA3E A 981 msA2E A 250 m 099606 10A5E A Pa 9961 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 158 What is the pressure at the bottom of a 5 m tall column of fluid with atmospheric pressure 101 kPa on the top surface if the fluid is a water at 20C b glycerine 25C or c gasoline 25C Solution Table A4 ρH2O 997 kgmA3E A ρGlyc 1260 kgmA3E A ρgasoline 750 kgmA3E P ρgh P PAtopE A P a P ρgh 997 9807 5 48 888 Pa P 101 4899 1499 kPa b P ρgh 1260 9807 5 61 784 Pa P 101 618 1628 kPa c P ρgh 750 9807 5 36 776 Pa P 101 368 1378 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 159 A steel tank of cross sectional area 3 mA2E A and 16 m tall weighs 10 000 kg and it is open at the top as shown in Fig P159 We want to float it in the ocean so it sticks 10 m straight down by pouring concrete into the bottom of it How much concrete should I put in Solution The force up on the tank is from the water pressure at the bottom times its area The force down is the gravitation times mass and the atmospheric pressure F PA ρoceangh P0A F mtank mconcreteg P0A The force balance becomes Air Ocean Concrete 10 m F F ρoceangh P0A mtank mconcreteg P0A Solve for the mass of concrete mconcrete ρoceanhA mtank 997 10 3 10 000 19 910 kg Notice The first term is the mass of the displaced ocean water The force up is the weight mg of this mass called buoyancy which balances with gravitation and the force from P0 cancel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 160 A piston mp 5 kg is fitted in a cylinder A 15 cm2 that contains a gas The setup is in a centrifuge that creates an acceleration of 25 ms2 in the direction of piston motion towards the gas Assuming standard atmospheric pressure outside the cylinder find the gas pressure Solution Force balance F F PA0E AA mApE Ag PA P PA0E A A mpg EAE A 101325 A 5 25 1000 00015E A AkPa kg ms2 EPa m2 E 1847 kPa gas g Po Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 161 Liquid water with density ρ is filled on top of a thin piston in a cylinder with crosssectional area A and total height H as shown in Fig P161 Air is let in under the piston so it pushes up spilling the water over the edge Derive the formula for the air pressure as a function of piston elevation from the bottom h Solution Force balance H h P 0 Piston F F PA P0A mAH2OE Ag P P0 mAH2OE AgA P P0 H hρg h V air P P 0 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Manometers and Barometers Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 162 A probe is lowered 16 m into a lake Find the absolute pressure there Solution The pressure difference for a column is from Eq12 and the density of water is from Table A4 P ρgH 997 kgmA3E A 981 ms2 16 m 156 489 Pa 156489 kPa PAoceanE A PA0E A P 101325 156489 2578 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 163 The density of atmospheric air is about 115 kgm3 which we assume is constant How large an absolute pressure will a pilot see when flying 2000 m above ground level where the pressure is 101 kPa Solution Assume g and ρ are constant then the pressure difference to carry a column of height 2000 m is from Fig210 P ρgh 115 kgm3 9807 msA2E A 2000 m 22 556 Pa 226 kPa The pressure on top of the column of air is then P PA0E A P 101 226 784 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 164 The standard pressure in the atmosphere with elevation H above sea level can be correlated as P PA0E A 1 HLA526E A with L 44 300 m With the local sea level pressure PA0E A at 101 kPa what is the pressure at 10 000 m elevation P PA0E A 1 HLA526E 101 kPa 1 10 00044 300A526E 263 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 165 A barometer to measure absolute pressure shows a mercury column height of 725 mm The temperature is such that the density of the mercury is 13 550 kgmA3E A Find the ambient pressure Solution Hg L 725 mm 0725 m ρ 13 550 kgmA3E The external pressure P balances the column of height L so from Fig 114 P ρ L g 13 550 kgmA3E A 980665 ms2 0725 m 10A3E A kPaPa 9634 kPa This is a more common type that does not involve mercury as an older wall mounted unit Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 166 A differential pressure gauge mounted on a vessel shows 125 MPa and a local barometer gives atmospheric pressure as 096 bar Find the absolute pressure inside the vessel Solution Convert all pressures to units of kPa PAgaugeE A 125 MPa 1250 kPa PA0E A 096 bar 96 kPa P PAgaugeE A PA0E A 1250 96 1346 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 167 A manometer shows a pressure difference of 1 m of liquid mercury Find P in kPa Solution Hg L 1 m ρ 13 580 kgmA3E A from Table A4 or read Fig 18 The pressure difference P balances the column of height L so from Eq12 P ρ g L 13 580 kgmA3E A 980665 ms2 10 m 10A3E A kPaPa 1332 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 168 Blue manometer fluid of density 925 kgmA3E A shows a column height difference of 3 cm vacuum with one end attached to a pipe and the other open to PA0E A 101 kPa What is the absolute pressure in the pipe Solution Since the manometer shows a vacuum we have PAPIPEE A PA0E A P P ρgh 925 kgmA3E A 9807 ms2 003 m 2721 Pa 0272 kPa PAPIPEE A 101 0272 10073 kPa cb P o Pipe Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 169 What pressure difference does a 10 m column of atmospheric air show Solution The pressure difference for a column is from Eq12 P ρgH So we need density of air from Fig 28 or Table A5 ρ 12 kgmA3E P 12 kgmA3E A 981 msA2E A 10 m 1177 Pa 012 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 170 A barometer measures 760 mmHg at street level and 735 mmHg on top of a building How tall is the building if we assume air density of 115 kgmA3E A Solution P ρgH H Pρg A 760 735 115 9807E A A mmHg kgm2s2 E A A13332 Pa mmHgE A 295 m Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 171 The pressure gauge on an air tank shows 75 kPa when the diver is 10 m down in the ocean At what depth will the gauge pressure be zero What does that mean Ocean HA2E A0 pressure at 10 m depth is PAwaterE A PA0E A ρLg 1013 A997 10 980665 1000E A 199 kPa Air Pressure absolute in tank PAtankE A 199 75 274 kPa Tank Pressure gauge reads zero at HA2E A0 local pressure 274 1013 A997 980665 1000E A L L 1766 m At this depth you will have to suck the air in it can no longer push itself through a valve Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 172 An exploration submarine should be able to go 1200 m down in the ocean If the ocean density is 1020 kgmA3E A what is the maximum pressure on the submarine hull Solution Assume we have atmospheric pressure inside the submarine then the pressure difference to the outside water is P ρLg 1020 kgmA3E A 1200 m 9807 ms2 1000 PakPa 12 007 kPa 12 MPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 173 A submarine maintains 101 kPa inside it and it dives 240 m down in the ocean having an average density of 1030 kgmA3E A What is the pressure difference between the inside and the outside of the submarine hull Solution Assume the atmosphere over the ocean is at 101 kPa then P is from the 240 m column water P ρLg 1030 kgmA3E A 240 m 9807 ms2 1000 2424 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 174 Assume we use a pressure gauge to measure the air pressure at street level and at the roof of a tall building If the pressure difference can be determined with an accuracy of 1 mbar 0001 bar what uncertainty in the height estimate does that corresponds to Solution ρair 1169 kgmA3E A from Table A5 P 0001 bar 100 Pa L AP ρgE A A 100 1169 9807E A 872 m As you can see that is not really accurate enough for many purposes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 175 The absolute pressure in a tank is 115 kPa and the local ambient absolute pressure is 97 kPa If a Utube with mercury density 13550 kgmA3E A is attached to the tank to measure the gage pressure what column height difference would it show Solution P PAtankE A PA0E A ρg H H PAtankE A PA0E Aρg 115 97 1000 Pa 13550 kgmA3E A 981 ms2 0135 m 135 cm H Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 176 An absolute pressure gauge attached to a steel cylinder shows 135 kPa We want to attach a manometer using liquid water a day that Patm 101 kPa How high a fluid level difference must we plan for Solution Since the manometer shows a pressure difference we have P PACYLE A PAatmE A ρ L g L P ρg A 135 101 kPa E997 kg m3 10 9807 ms2E A A1000 Pa kPaE A 3467 m H Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 177 A Utube manometer filled with water density 1000 kgm3 shows a height difference of 25 cm What is the gauge pressure If the right branch is tilted to make an angle of 30 with the horizontal as shown in Fig P177 what should the length of the column in the tilted tube be relative to the Utube Solution Same height in the two sides in the direction of g P FA mgA VρgA hρg 025 m 1000 kgmA3E A 9807ms2 24525 Pa 245 kPa h H sin 30 H hsin 30 2h 50 cm H h 30o Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 178 A pipe flowing light oil has a manometer attached as shown in Fig P178 What is the absolute pressure in the pipe flow Solution Table A3 ρAoilE A 910 kgmA3E A ρAwaterE A 997 kgmA3E PABOTE A PA0E A ρAwaterE A g HAtotE A PA0E A 997 kgmA3E A 9807 ms2 08 m PAoE A 7822 Pa PAPIPEE A PABOTE A ρAwaterE A g HA1E A ρAoilE A g HA2E PABOTE A 997 9807 01 910 9807 02 PABOTE A 9777 Pa 17849 Pa PAPIPEE A PAoE A 7822 9777 17849 Pa PAoE A 50594 Pa 101325 506 1064 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 179 The difference in height between the columns of a manometer is 200 mm with a fluid of density 900 kgm3 What is the pressure difference What is the height difference if the same pressure difference is measured using mercury density 13600 kg m3 as manometer fluid Solution P ρ1gh1 900 kgmA3E A 9807 ms2 02 m 176526 Pa 177 kPa hAHgE A P ρhg g ρ1 gh1 ρhg g A 900 13600E A 02 m 00132 m 132 mm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 180 Two cylinders are filled with liquid water ρ 1000 kgmA3E A and connected by a line with a closed valve as shown in Fig P180 A has 100 kg and B has 500 kg of water their crosssectional areas are AAAE A 01 m2 and AABE A 025 m2 and the height h is 1 m Find the pressure on each side of the valve The valve is opened and water flows to an equilibrium Find the final pressure at the valve location Solution VAAE A vAH2OE AmAAE A mAAE Aρ 01 AAAE AhAAE A hAAE A 1 m VABE A vAH2OE AmABE A mABE Aρ 05 AABE AhABE A hABE A 2 m PAVBE A PA0E A ρghABE AH 101325 1000 981 3 130 755 Pa PAVAE A PA0E A ρghAAE A 101325 1000 981 1 111 135 Pa Equilibrium same height over valve in both VAtotE A VAAE A VABE A hA2E AAAAE A hA2E A HAABE A hA2E A A hAAA hBHAB EAA AB E A 243 m PAV2E A PA0E A ρghA2E A 101325 1000 981 2431000 1252 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 181 Two pistoncylinder arrangements A and B have their gas chambers connected by a pipe Crosssectional areas are AA 75 cm2 and AB 25 cm2 with the piston mass in A being mA 25 kg Outside pressure is 100 kPa and standard gravitation Find the mass mB so that none of the pistons have to rest on the bottom Solution P Po o cb Force balance for both pistons F F A mAPAE Ag PA0E AAAAE A PAAAE B mAPBE Ag PA0E AAABE A PAABE Same P in A and B gives no flow between them A mPAg EAA E A PA0E A A mPBg EAB E A PA0E A mAPBE A mAPA E AAAAE A AABE A 25 2575 833 kg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 182 Two hydraulic pistoncylinders are of same size and setup as in Problem 181 but with negligible piston masses A single point force of 250 N presses down on piston A Find the needed extra force on piston B so that none of the pistons have to move Solution AAAE A 75 cmA2E A AABE A 25 cmA2E No motion in connecting pipe PAAE A PABE Forces on pistons balance Po Po cb A B FB FA PAAE A P0 FAAE A AAAE A PABE A P0 FABE A AABE A FABE A FAAE A A AB EAA E A 250 N A25 75E A 8333 N Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 183 A piece of experimental apparatus is located where g 95 ms2 and the temperature is 5C An air flow inside the apparatus is determined by measuring the pressure drop across an orifice with a mercury manometer see Problem 191 for density showing a height difference of 200 mm What is the pressure drop in kPa Solution P ρgh ρAHgE A 13600 kgmA3E P 13 600 kgmA3E A 95 ms2 02 m 25840 Pa 2584 kPa g Air Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Energy and Temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 184 An escalator brings four people whose total mass is 300 kg 25 m up in a building Explain what happens with respect to energy transfer and stored energy The four people 300 kg have their potential energy raised which is how the energy is stored The energy is supplied as electrical power to the motor that pulls the escalator with a cable Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 185 A car moves at 75 kmh its mass including people is 3200 kg How much kinetic energy does the car have KE ½ m VA2E A ½ 3200 kg A 75 1000 3600 2E A mA2E AsA2E A 694 444 J 694 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 186 A 52kg package is lifted up to the top shelf in a storage bin that is 4 m above the ground floor How much increase in potential energy does the package get The potential energy is from Eq15 pe gz so for a certain mass we get PE mgH 52 kg 981 msA2E A 4 m 2040 J 204 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 187 A car of mass 1775 kg travels with a velocity of 100 kmh Find the kinetic energy How high should it be lifted in the standard gravitational field to have a potential energy that equals the kinetic energy Solution Standard kinetic energy of the mass is KE ½ m VA2E A ½ 1775 kg A 100 1000 3600 2E A mA2E AsA2E ½ 1775 27778 Nm 684 800 J 6848 kJ Standard potential energy is POT mgh h ½ m VA2E A mg A 684 800 Nm 1775 kg 9807 ms2 E A 393 m Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 188 An oxygen molecule with mass m M mo 32 166 10A27E A kg moves with a velocity of 240 ms What is the kinetic energy of the molecule What temperature does that corresponds to if it has to equal 32 kT where k is Boltzmans constant and T is the absolute temperature in Kelvin KE ½ m VA2E A ½ 32 166 10A27E A kg 240A2E A mA2E AsA2E 153 E21 J So if the KE equals 15 kT then we get KE 15 kT 15 138065 10A23E A JK T T 23 KEk 23 153 E21 138065 10A23E A 739 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 189 What is a temperature of 5AoE AC in degrees Kelvin Solution The offset from Celsius to Kelvin is 27315 K so we get TK TC 27315 5 27315 26815 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 190 The human comfort zone is between 18 and 24AoE AC what is the range in Kelvin What is the maximum relative change from the low to the high temperature Solution TK TC 27315 The range in K becomes from 29115 to 29715 K The relative change is 6 degrees up from 29115 K which is Relative change A 6 29115E A 100 206 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 191 The density of mercury changes approximately linearly with temperature as ρHg 13595 25 T kg mA3E A T in Celsius so the same pressure difference will result in a manometer reading that is influenced by temperature If a pressure difference of 100 kPa is measured in the summer at 35C and in the winter at 15C what is the difference in column height between the two measurements Solution The manometer reading h relates to the pressure difference as P ρ L g L AP ρgE A The manometer fluid density from the given formula gives ρAsuE A 13595 25 35 135075 kgmA3E A ρAwE A 13595 25 15 136325 kgmA3E The two different heights that we will measure become LAsuE A A 100 103 E135075 9807E A AkPa PakPa kgm3 ms2 E A 07549 m LAwE A A 100 103 E136325 9807E A AkPa PakPa kgm3 ms2 E A 07480 m L LAsuE A LAwE A 00069 m 69 mm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 192 A mercury thermometer measures temperature by measuring the volume expansion of a fixed mass of liquid Hg due to a change in the density see problem 191 Find the relative change in volume for a change in temperature from 10C to 20C Solution From 10C to 20C At 10C ρHg 13595 25 10 13570 kgmA3E At 20C ρHg 13595 25 20 13545 kgmA3E The volume from the mass and density is V mρ Relative Change V20 V10 V10 mρ20 mρ10 mρ10 ρ10 ρ20 1 A13570 13545E A 1 00018 018 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 193 Density of liquid water is ρ 1008 T2 kgmA3E A with T in AoE AC If the temperature increases 10AoE AC how much deeper does a 1 m layer of water become Solution The density change for a change in temperature of 10AoE AC becomes ρ T2 5 kgmA3E A from an ambient density of ρ 1008 T2 1008 252 9955 kgmA3E A Assume the area is the same and the mass is the same m ρV ρAH then we have m 0 Vρ ρV V Vρρ and the change in the height is H AV AE A AHV VE A AHρ ρE A A1 5 E9955E A 0005 m barely measurable Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 194 Using the freezing and boiling point temperatures for water in both Celsius and Fahrenheit scales develop a conversion formula between the scales Find the conversion formula between Kelvin and Rankine temperature scales Solution TAFreezingE A 0 AoE AC 32 F TABoilingE A 100 AoE AC 212 F T 100 AoE AC 180 F TAoCE A TAFE A 3218 or TAFE A 18 TAoCE A 32 For the absolute K R scales both are zero at absolute zero TARE A 18 TAKE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 195 The atmosphere becomes colder at higher elevation As an average the standard atmospheric absolute temperature can be expressed as Tatm 288 65 10A3E A z where z is the elevation in meters How cold is it outside an airplane cruising at 12 000 m expressed in Kelvin and in Celsius Solution For an elevation of z 12 000 m we get Tatm 288 65 10A3E A z 210 K To express that in degrees Celsius we get TACE A T 27315 6315AoE AC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 196 Repeat problem 183 if the flow inside the apparatus is liquid water ρ 1000 kgm3 instead of air Find the pressure difference between the two holes flush with the bottom of the channel You cannot neglect the two unequal water columns Solution Balance forces in the manometer P P 1 2 h h 1 2 H H h2 H h1 hHg h1 h2 P1A ρAH2OE Ah1gA ρAHgE AH h1gA P2A ρAH2OE Ah2gA ρAHgE AH h2gA P1 P2 ρAH2OE Ah2 h1g ρAHgE Ah1 h2g P1 P2 ρAHgE AhHgg ρAH2OE AhHgg 13 600 02 95 1000 02 95 kgm3 m ms2 25 840 1900 Pa 23 940 Pa 2394 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 197 A dam retains a lake 6 m deep To construct a gate in the dam we need to know the net horizontal force on a 5 m wide and 6 m tall port section that then replaces a 5 m section of the dam Find the net horizontal force from the water on one side and air on the other side of the port Solution Pbot PA0E A P P ρgh 997 kgm3 9807 ms2 6 m 58 665 Pa 5866 kPa Neglect P in air Fnet Fright Fleft Pavg A PA0E AA Pavg PA0E A 05 P Since a linear pressure variation with depth Fnet PA0E A 05 PA PA0E AA 05 P A 05 5866 5 6 880 kN F F left right Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 198 In the city water tower water is pumped up to a level 25 m above ground in a pressurized tank with air at 125 kPa over the water surface This is illustrated in Fig P198 Assuming the water density is 1000 kgm3 and standard gravity find the pressure required to pump more water in at ground level Solution P ρ L g 1000 kgmA3E A 25 m 9807 ms2 245 175 Pa 2452 kPa PAbottomE A PAtopE A P 125 2452 370 kPa cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 199 The main waterline into a tall building has a pressure of 600 kPa at 5 m elevation below ground level How much extra pressure does a pump need to add to ensure a water line pressure of 200 kPa at the top floor 150 m above ground Solution The pump exit pressure must balance the top pressure plus the column P The pump inlet pressure provides part of the absolute pressure Pafter pump Ptop P P ρgh 997 kgmA3E A 9807 ms2 150 5 m 1 515 525 Pa 1516 kPa Pafter pump 200 1516 1716 kPa Ppump 1716 600 1116 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1100 Two cylinders are connected by a piston as shown in Fig P1100 Cylinder A is used as a hydraulic lift and pumped up to 500 kPa The piston mass is 25 kg and there is standard gravity What is the gas pressure in cylinder B Solution Force balance for the piston PBAB mpg P0AA AB PAAA AA π401A2E A 000785 mA2E A AB π40025A2E A 0000 491 mA2E PBAB PAAA mpg P0AA AB 500 000785 25 98071000 100 000785 0000 491 2944 kN PB 29440000 491 5996 kPa 60 MPa P B GAS A Oil Po cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1101 A 5kg piston in a cylinder with diameter of 100 mm is loaded with a linear spring and the outside atmospheric pressure of 100 kPa as shown in Fig P1101 The spring exerts no force on the piston when it is at the bottom of the cylinder and for the state shown the pressure is 400 kPa with volume 04 L The valve is opened to let some air in causing the piston to rise 2 cm Find the new pressure Solution A linear spring has a force linear proportional to displacement F k x so the equilibrium pressure then varies linearly with volume P a bV with an intersect a and a slope b dPdV Look at the balancing pressure at zero volume V 0 when there is no spring force F PA PAoE AA mApE Ag and the initial state These two points determine the straight line shown in the PV diagram Piston area AAPE A π4 01A2E A 000785 mA2E a PA0E A A mpg EAp E A 100 kPa A5 980665 000785E A Pa 1062 kPa intersect for zero volume VA2E A 04 000785 20 0557 L PA2E A PA1E A AdP dVE A V 400 A4001062 E04 0E A 0557 04 5153 kPa 400 1062 2 1 0 04 P V 0557 2 P UPDATED JULY 2013 SOLUTION MANUAL CHAPTER 1 English Units Fundamentals of Thermodynamics Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CHAPTER 1 SUBSECTION PROB NO ConceptStudy Guide Problems 102108 Properties and Units 109 Force Energy and Specific Volume 110115 Pressure Manometers and Barometers 116124 Temperature 125127 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Concept Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1102E A mass of 2 lbm has acceleration of 5 fts2 what is the needed force in lbf Solution Newtons 2nd law F ma F ma 2 lbm 5 fts2 10 lbm fts2 10 32174 lbf 031 lbf Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1103E How much mass is in 1 gallon of gasoline If helium in a balloon at atmospheric P and T Solution A volume of 1 gal equals 231 in3 see Table A1 From Table F3 the density is 468 lbmft3 so we get m ρV 468 lbmft3 1 231123 ft3 6256 lbm A more accurate value from Table F3 is ρ 848 lbmft3 For the helium we see Table F4 that density is 1008 103 lbmft3 so we get m ρV 1008 103 lbmft3 1 231123 ft3 000135 lbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1104E Can you easily carry a one gallon bar of solid gold Solution The density of solid gold is about 1205 lbmft3 from Table F2 we could also have read Figure 17 and converted the units V 1 gal 231 in3 231 123 ft3 013368 ft 3 Therefore the mass in one gallon is m ρV 1205 lbmft3 013368 ft3 161 lbm and some people can just about carry that in the standard gravitational field Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1105E What is the temperature of 5F in degrees Rankine Solution The offset from Fahrenheit to Rankine is 45967 R so we get TR TF 45967 5 45967 4547 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1106E What is the smallest temperature in degrees Fahrenheit you can have Rankine Solution The lowest temperature is absolute zero which is at zero degrees Rankine at which point the temperature in Fahrenheit is negative TR 0 R 45967 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1107E What is the relative magnitude of degree Rankine to degree Kelvin Look in Table A1 p 757 1 K 1 oC 18 R 18 F 1 R 5 9 K 05556 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1108E Chemical reaction rates genrally double for a 10 K increase in temperature How large an increase is that in Fahrenheit From the Conversion Table A1 1 K 1 oC 18 R 18 F So the 10 K increase becomes 10 K 18 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Properties and Units Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1109E An apple weighs 02 lbm and has a volume of 6 in3 in a refrigerator at 38 F What is the apple density List three intensive and two extensive properties for the apple Solution ρ EAlbm inA3 AE A 00333 EAlbm inA3 AE A 576 EAlbm ftA3 AE m V 02 6 Intensive ρ 576 EAlbm ftA3 AE A v A1 ρE A 00174 A ft3 ElbmE A T 38 F P 14696 lbfinA2E Extensive m 02 lbm V 6 inA3E A 0026 gal 000347 ftA3E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1110E A steel piston of 10 lbm is in the standard gravitational field where a force of 10 lbf is applied vertically up Find the acceleration of the piston Solution Fup ma F mg a AF mg mE A AF mE A g A 10 lbf 10 lbmE A 32174 ftsA2E 1 32174 32174 ftsA2E 0 ftsA2 E The mass does not move it is held stationary g F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Force Energy Density Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1111E A 2500lbm car moving at 25 mih is accelerated at a constant rate of 15 fts2 up to a speed of 50 mih What are the force and total time required Solution a AdV dtE A AV tE A t AV aE A t A50 25 mih 160934 mmi 328084 ftm 3600 sh 15 fts2 E A 244 sec F ma 2500 lbm 15 fts2 32174 lbm ft lbfs2 1165 lbf Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1112E An escalator brings four people of total 600 lbm and a 1000 lbm cage up with an acceleration of 3 ftsA2E A what is the needed force in the cable Solution The total mass moves upwards with an acceleration plus the gravitations acts with a force pointing down ma F F mg F ma mg ma g 1000 600 lbm 3 32174 ftsA2E 56 278 lbm ftsA2E A 56 278 lbf F g Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1113E One poundmass of diatomic oxygen O2 molecular weight 32 is contained in a 100gal tank Find the specific volume on both a mass and mole basis v and v Solution V 100 231 inA3E A 23 10012A3E A ftA3E A 1337 ftA3E A conversion seen in Table A1 This is based on the definition of the specific volume v Vm 1337 ftA3E A1 lbm 1337 ftA3E Albm Av E A Vn A V mME A Mv 32 1337 4278 ftA3E Albmol Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1114E A 30lbm steel gas tank holds 10 ftA3E A of liquid gasoline having a density of 50 lbmftA3E A What force is needed to accelerate this combined system at a rate of 15 fts2 Solution m mAtankE A mAgasolineE A 30 lbm 10 ftA3E A 50 lbmftA3E A 530 lbm cb F ma 530 lbm 15 ftsA2E A 32174 lbm ftsA2E A lbf 2471 lbf Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1115E A powerplant that separates carbondioxide from the exhaust gases compresses it to a density of 8 lbmftA3E A and stores it in an unminable coal seam with a porous volume of 3 500 000 ftA3E A Find the mass they can store Solution m ρ V 8 lbmftA3E A 3 500 000 ftA3E A 28 10A 7E A lbm Just to put this in perspective a power plant that generates 2000 MW by burning coal would make about 20 million tons of carbondioxide a year That is 2000 times the above mass so it is nearly impossible to store all the carbondioxide being produced Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1116E A laboratory room keeps a vacuum of 4 in of water due to the exhaust fan What is the net force on a door of size 6 ft by 3 ft Solution The net force on the door is the difference between the forces on the two sides as the pressure times the area F Poutside A Pinside A P A 4 in H2O 6 ft 3 ft 4 0036126 lbfinA2E A 18 ftA2E A 144 inA2E AftA2E 3746 lbf Table A1 1 in H2O is 0036 126 lbfinA2E A a unit also often listed as psi Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1117E A 150lbm human total footprint is 05 ft when the person is wearing boots If snow can support an extra 1 psi what should the total snow shoe area be Force balance ma 0 PA mg A Amg PE A A150 lbm 32174 fts2 E 1 lbfin2 E A A150 lbm 32174 fts2 E 32174 lbmfts2in2E 150 inA2E A 104 ftA2E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1118E A tornado rips off a 1000 ftA2E A roof with a mass of 2000 lbm What is the minimum vacuum pressure needed to do that if we neglect the anchoring forces Solution The net force on the roof is the difference between the forces on the two sides as the pressure times the area F Pinside A PoutsideA P A That force must overcome the gravitation mg so the balance is P A mg P mgA 2000 lbm 32174 ftsA2E A 1000 ftA2E A 2000 1000 144 psi 00139 psi Remember that psi lbfinA2E A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1119E A manometer shows a pressure difference of 35 in of liquid mercury Find P in psi Solution Hg L 35 in ρ 848 lbmftA3E A from Table F3 Pressure 1 psi 1 lbf in2 The pressure difference P balances the column of height L so from Eq22 P ρ g L 848 lbmftA3E A 32174 fts2 3512 ft 2473 lbfft2 2473 144 lbfin2 172 psi Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1120E A 7 ft m tall steel cylinder has a cross sectional area of 15 ftA2E A At the bottom with a height of 2 ft m is liquid water on top of which is a 4 ft high layer of gasoline The gasoline surface is exposed to atmospheric air at 147 psia What is the highest pressure in the water Solution The pressure in the fluid goes up with the depth as P PAtopE A P PAtopE A ρgh and since we have two fluid layers we get P PAtopE A ρhAgasolineE A ρhAwaterE Ag The densities from Table F3 are Air Water 4 ft 2 ft Gasoline ρAgasolineE A 468 lbmftA3E A ρAwaterE A 622 lbmftA3E P 147 468 4 622 2 A 32174 144 32174E A 1686 lbfinA2E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1121E A Utube manometer filled with water density 623 lbmftA3E A shows a height difference of 10 in What is the gauge pressure If the right branch is tilted to make an angle of 30 with the horizontal as shown in Fig P177 what should the length of the column in the tilted tube be relative to the Utube Solution h H 30 P FA mgA hρg A1012 623 32174 E32174 144E PAgaugeE A 036 lbfinA2E h H sin 30 H hsin 30 2h 20 in 0833 ft Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1122E A pistoncylinder with crosssectional area of 01 ftA2E A has a piston mass of 200 lbm resting on the stops as shown in Fig P150 With an outside atmospheric pressure of 1 atm what should the water pressure be to lift the piston Solution The force acting down on the piston comes from gravitation and the outside atmospheric pressure acting over the top surface Force balance F F PA mApE Ag PA0E AA Now solve for P multiply by 144 to convert from ftA2E A to inA2E A P PA0E A A mpg EAE A 14696 A 200 32174 01 144 32174E 14696 psia 1388 psia 2858 lbfinA2E Water cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1123E The main waterline into a tall building has a pressure of 90 psia at 16 ft elevation below ground level How much extra pressure does a pump need to add to ensure a waterline pressure of 30 psia at the top floor 450 ft above ground Solution The pump exit pressure must balance the top pressure plus the column P The pump inlet pressure provides part of the absolute pressure Pafter pump Ptop P P ρgh 622 lbmftA3E A 32174 ftsA2E A 450 16 ft A 1 lbf s2 E32174 lbm ftE 28 985 lbfftA2E A 2013 lbfinA2E Pafter pump 30 2013 2313 psia Ppump 2313 90 1413 psi Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1124E A piston mApE A 10 lbm is fitted in a cylinder A 25 inA2E A that contains a gas The setup is in a centrifuge that creates an acceleration of 75 ftsA2E A Assuming standard atmospheric pressure outside the cylinder find the gas pressure Solution Force balance F F PA0E AA mApE Ag PA P PA0E A A mpg EAE A 14696 A 10 75 25 32174E A Albm fts2 in2E A Albfs2 lbmftE 14696 9324 2402 lbfinA2E gas g Po Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1125E The human comfort zone is between 18 and 24C What is that range in Fahrenheit T 18C 32 18 18 644 F T 24C 32 18 24 752 F So the range is like 64 to 75 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1126E The atmosphere becomes colder at higher elevation As an average the standard atmospheric absolute temperature can be expressed as Tatm 518 384 10A3E A z where z is the elevation in feet How cold is it outside an airplane cruising at 32 000 ft expressed in Rankine and in Fahrenheit Solution For an elevation of z 32 000 ft we get Tatm 518 384 10A3E A z 3951 R To express that in degrees Fahrenheit we get TAFE A T 45967 6455 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1127E The density of mercury changes approximately linearly with temperature as ρAHgE A 8515 0086 T lbmftA3E A T in degrees Fahrenheit so the same pressure difference will result in a manometer reading that is influenced by temperature If a pressure difference of 147 lbfinA2E A is measured in the summer at 95 F and in the winter at 5 F what is the difference in column height between the two measurements Solution P ρgh h Pρg ρAsuE A 84333 lbmftA3E A ρAwE A 85107 lbmftA3E hAsuE A A147 144 32174 84333 32174E A 251 ft 3012 in hAwE A A147 144 32174 85107 32174E A 2487 ft 2984 in h hAsuE A hAwE A 0023 ft 028 in Updated June 2013 SOLUTION MANUAL CHAPTER 2 Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CONTENT CHAPTER 2 SUBSECTION PROB NO Concept problems 115 Phase diagrams triple and critical points 1624 General tables 2564 Ideal gas 6584 Compressibility factor 8597 Equations of state 98106 Review problems 107125 Linear interpolation 126130 Computer tables 131135 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful InText Concept Questions 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2a If the pressure is smaller than the smallest Psat at a given T what is the phase Refer to the phase diagrams in Figures 24 and 25 For a lower P you are below the vaporization curve or the sublimation curve and that is the superheated vapor region You have the gas phase ln P T Vapor L S Critical Point S 2b An external water tap has the valve activated by a long spindle so the closing mechanism is located well inside the wall Why is that Solution By having the spindle inside the wall the coldest location with water when the valve is closed is kept at a temperature above the freezing point If the valve spindle was outside there would be some amount of water that could freeze while it is trapped inside the pipe section potentially rupturing the pipe 5 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2c What is the lowest temperature approximately at which water can be liquid Look at the phase diagram in Fig 24 At the border between ice I ice III and the liquid region is a triple point which is the lowest T where you can have liquid From the figure it is estimated to be about 255 K ie at 18oC T 255 K 18C ln P T V L S CRP lowest T liquid 6 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2d Some tools should be cleaned in water at a least 150oC How high a P is needed Solution If I need liquid water at 150oC I must have a pressure that is at least the saturation pressure for this temperature Table B11 150oC Psat 4759 kPa 7 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2e Water at 200 kPa has a quality of 50 Is the volume fraction VgVtot 50 or 50 This is a twophase state at a given pressure and without looking in the table we know that vf is much smaller than vg From the definition of quality we get the masses from total mass m as mf 1 x m mg x m The volumes are Vf mf vf 1 x m vf Vg mg vg x m v g So when half the mass is liquid and the other half is vapor the liquid volume is much smaller that the vapor volume The vapor volume is thus much more than 50 of the total volume Only right at the critical point is vf vg for all other states vg vf and the difference is larger for smaller pressures 8 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2f Why are most of the compressed liquid or solid regions not included in the printed tables For the compressed liquid and the solid phases the specific volume and thus density is nearly constant These surfaces are very steep nearly constant v and there is then no reason to fill up a table with the same value of v for different P and T 2g Why is it not typical to find tables for Ar He Ne or air like an Appendix B table The temperature at which these substances are close to the twophase region is very low For technical applications with temperatures around atmospheric or higher they are ideal gases Look in Table A2 and we can see the critical temperatures as Ar 1508 K He 519 K Ne 444 K It requires a special refrigerator in a laboratory to bring a substance down to these cryogenic temperatures 9 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2h What is the percent change in volume as liquid water freezes Mention some effects the volume change can have in nature and in our households The density of water in the different phases can be found in Tables A3 and A4 and in Table B1 From Table B11 vf 000100 m3kg From Table B15 vi 00010908 m3kg Percent change 100 vi vf vf 100 00010908 0001 0001 91 increase Liquid water that seeps into cracks or other confined spaces and then freezes will expand and widen the cracks This is what destroys any porous material exposed to the weather on buildings roads and mountains It can burst water pipes and crack engine blocks that is why you put antifreeze in it 10 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2i How accurate is it to assume that methane is an ideal gas at room conditions From Table A2 Tc 1904 K Pc 460 MPa So at room conditions we have much higher T Tc and P Pc so this is the ideal gas region To confirm look in Table B72 100 kPa 300 K v 155215 m3kg Find the compressibility factor R from Table A5 as Z PvRT 100 kPa 155215 m3kg 05183 kJkgK 300 K 099823 so Z is 1 with an accuracy of 02 better than most measurements can be done 11 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2j I want to determine a state of some substance and I know that P 200 kPa is it helpful to write PV mRT to find the second property NO You need a second property Notice that two properties are needed to determine a state The EOS can give you a third property if you know two like PT gives v just as you would get by entering a table with a set PT This EOS substitutes for a table when it is applicable 12 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2k A bottle at 298 K should have liquid propane how high a pressure is needed use Fig D1 To have a liquid the pressure must be higher than or equal to the saturation pressure There is no printed propane table so we use the compressibility chart and Table A2 Propane Table A2 Tc 3698 K Pc 425 MPa The reduced temperature is Tr T Tc 298 3698 0806 for which we find in Fig D1 Pr sat 025 P Pr sat Pc 025 425 MPa 106 MPa 13 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2l A bottle at 298 K should have liquid propane how high a pressure is needed use the software To have a liquid the pressure must be higher than or equal to the saturation pressure There is no printed propane table but the software has propane included Start CATT3 select cryogenic substances propane select calculator select case 4 T x 25oC 0 P 09518 MPa 14 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Concept Problems 15 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 21 Are the pressures in the tables absolute or gauge pressures Solution The behavior of a pure substance depends on the absolute pressure so P in the tables is absolute 16 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 22 What is the minimum pressure for which I can have liquid carbon dioxide Look at the phase diagram in Fig 25 The minimum P in the liquid phase is at the triple point From Table 22 this is at 520 kPa a similar value around 4500 kPa is seen in Fig 25 The 100 kPa is below the triple point ln P T V L S 100 kPa 17 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 23 When you skate on ice a thin liquid film forms under the skate How can that be The ice is at some temperature below the freezing temperature for the atmospheric pressure of 100 kPa 01 MPa and thus to the left of the fusion line in the solid ice I region of Fig 24 As the skate comes over the ice the pressure is increased dramatically right under the blade so it brings the state straight up in the diagram crossing the fusion line and brings it into a liquid state at same temperature The very thin liquid film under the skate changes the friction to be viscous rather than a solid to solid contact friction Friction is thus significantly reduced Comment The latest research has shown that the pressure may not be enough to generate the liquid but that such a liquid layer always exist on an ice surface maybe only a few molecules thick dependent upon temperature At really low T say 40oC no such liquid layer exists which is why your finger can stick to such a surface 18 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 24 At a higher elevation like in mountains the pressure is lower what effect does that have for cooking food A lower pressure means that water will boil at a lower temperature see the vaporization line in Fig 24 or in Table B12 showing the saturated temperature as a function of the pressure You therefore must increase the cooking time a little 19 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 25 Water at room temperature and room pressure has v 1 10n m3kg what is n See Table B11 or B12 to determine it is in the liquid phase you should know this already Table A4 or from B11 at 20oC n 3 v 000100 m3kg 26 Can a vapor exist below the triple point temperature Look at the phase diagrams in Figs 24 and 25 Below the triple point the sublimation curve has very small pressures but not zero So for pressures below the saturation pressure the substance is a vapor If the phase diagram is plotted in linear coordinates the small vapor region is nearly not visible 27 In Example 21 b is there any mass at the indicated specific volume Explain This state is a twophase mixture of liquid and vapor There is no mass at the indicated state the v value is an average for all the mass so there is some mass at the saturated vapor state fraction is the quality x and the remainder of the mass is saturated liquid fraction 1x 20 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 28 Sketch two constantpressure curves 500 kPa and 30 000 kPa in a Tv diagram and indicate on the curves where in the water tables you see the properties P 05 MPa 30 CP v T v 500 kPa 30 MPa B 1 4 B13 B 1 4 B13 B13 B12 B11 B15 B15 B13 The 30 MPa line in Table B14 starts at 0oC and table ends at 380oC the line is continued in Table B13 starting at 375oC and table ends at 1300oC The 500 kPa line in Table B14 starts at 001oC and table ends at the saturated liquid state 15186oC The line is continued in Table B13 starting at the saturated vapor state 15186oC continuing up to 1300oC 21 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 29 If I have 1 L of R410A at 1 MPa 20oC how much mass is that R410A Tables B4 B41 Psat 14442 kPa at 20oC so superheated vapor B42 v 002838 m3kg under subheading 1000 kPa m V v 002838 m3kg 0001 m3 00352 kg 352 g T CP v 1444 kPa P CP v T 1000 1444 725 20 725 C 20 C 1000 kPa The Pv loglog diagram from CATT3 P in MPa and v in m3kg 22 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 210 Locate the state of ammonia at 200 kPa 10oC Indicate in both the Pv and the Tv diagrams the location of the nearest states listed in the printed Table B2 T CP v 200 kPa P CP v T 200 2909 189 10 0 189 C 10 C 150 kPa 23 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 211 Why are most of the compressed liquid or solid regions not included in the printed tables For the compressed liquid and the solid phases the specific volume and thus density is nearly constant These surfaces are very steep nearly constant v and there is then no reason to fill up a table with the same value of v for different P and T 24 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 212 How does a constant vprocess look like for an ideal gas in a PT diagram For an ideal gas Pv RT so then P Rv T Constant v is a straight line with slope Rv in the PT diagram P T Vapor Liquid Critical point S P RTv 25 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 213 If v RTP for an ideal gas what is the similar equation for a liquid The equation for a liquid is v Constant vo If you include that v increases a little with T then v vo C T To where C is a small constant with units m3kgK 26 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 214 To solve for v given P T in Eq214 what is the mathematical problem To solve for v in Eq 214 multiply with the denumerators so we get the form P f1v RT f2v a v b This is a 3rd order polynomial a cubic function in v The problem then is to find the roots or zero points in this cubic equation The mathematical subject to study is to find zero points of functions or roots Typically you will do it by some iteration technique Successive substitutions bisection NewtonRaphson are some of the methods you should learn 27 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 215 As the pressure of a gas becomes larger Z becomes larger than 1 What does that imply Pv Z RT So for a given P the specific volume v is then larger than predicted by the ideal gas law The molecules are pressed so close together that they have repulsive forces between them the electron clouds are getting closer The ideal gas law assumes the atoms molecules are point masses with no interactions between them and thus has a limit of zero specific volume as P goes to infinity Real molecules occupy some volume and the outer shell has a number of electrons with negative charges which can interact with one another if they are close enough 28 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Phase Diagrams Triple and Critical Points 29 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 216 Carbon dioxide at 280 K can be in different phases Indicate the pressure range you have for each of the three phases vapor liquid and solid Look at the PT phase diagram in Fig 25 at 280 K P 4000 kPa vapor 4000 kPa P 400 MPa liquid 400 MPa P solid ln P T V L S 280 K 30 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 217 Modern extraction techniques can be based on dissolving material in supercritical fluids such as carbon dioxide How high are pressure and density of carbon dioxide when the pressure and temperature are around the critical point Repeat for ethyl alcohol Solution CO2 Table A2 Pc 738 MPa Tc 304 K vc 000212 m3kg ρc 1vc 1000212 472 kgm3 C2H5OH Table A2 Pc 614 MPa Tc 514 K vc 000363 m3kg ρc 1vc 1000363 275 kgm3 31 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 218 The ice cap on the North Pole could be 1000 m thick with a density of 920 kgm3 Find the pressure at the bottom and the corresponding melting temperature Solution ρICE 920 kgm3 P ρgH 920 kgm3 980665 ms2 1000 m 9 022 118 Pa P Po P 101325 9022 9123 kPa See figure 37 liquid solid interphase TLS 1C 32 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 219 Find the lowest temperature at which it is possible to have water in the liquid phase At what pressure must the liquid exist Solution There is no liquid at lower temperatures than on the fusion line see Fig 37 saturated ice III to liquid phase boundary is at T 263K 10C and P 210 MPa ln P T V L S CRP lowest T liquid 33 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 220 Water at 27C can exist in different phases dependent upon the pressure Give the approximate pressure range in kPa for water being in each one of the three phases vapor liquid or solid Solution The phases can be seen in Fig 24 a sketch of which is shown to the right T 27 C 300 Κ From Fig 24 PVL 4 103 MPa 4 kPa PLS 103 MPa ln P T V L S CRP S 0 P 4 kPa VAPOR 0004 MPa P 1000 MPa LIQUID P 1000 MPa SOLID ICE 34 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 221 Dry ice is the name of solid carbon dioxide How cold must it be at atmospheric 100 kPa pressure If it is heated at 100 kPa what eventually happens Solution The phase boundaries are shown in Figure 25 At 100 kPa the carbon dioxide is solid if T 190 K It goes directly to a vapor state without becoming a liquid hence its name The 100 kPa is below the triple point ln P T V L S 100 kPa 35 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 222 What is the lowest temperature in Kelvin for which you can see metal as a liquid if the metal is a mercury b zinc Solution Assume the two substances have a phase diagram similar to Fig 25 then the triple point is the lowest T with liquid possible the data is from Table 21 Ta 39oC 234 K Tb 419oC 692 K 36 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 223 A substance is at 2 MPa 17C in a rigid tank Using only the critical properties can the phase of the mass be determined if the substance is oxygen water or propane Solution Find state relative to critical point properties which are from Table A2 a Oxygen O2 504 MPa 1546 K b Water H2O 2212 MPa 6473 K c Propane C3H8 425 MPa 3698 K State is at 17 C 290 K and 2 MPa Pc for all cases O2 T Tc Superheated vapor P Pc H2O T Tc P Pc you cannot say C3H8 T Tc P Pc you cannot say ln P T Vapor Liquid CrP a c b 37 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 224 Give the phase for the following states Solution a CO2 T 40C P 05 MPa Table A2 T Tc also P Pc superheated vapor assume ideal gas Table A5 b Air T 20C P 200 kPa Table A2 superheated vapor assume ideal gas Table A5 c NH3 T 170C P 600 kPa Table B22 or A2 T Tc superheated vapor P CP v T v T abc a b c P const 38 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful General Tables 39 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 225 Give the phase for the following states Solution a H2O T 260C P 5 MPa Table B11 or B12 B11 For given T read Psat 4689 MPa P Psat compressed liquid B12 For given P read Tsat 264C T Tsat compressed liquid b H2O T 2C P 100 kPa Table B11 T T triple point Table B15 at 2C read Psat 0518 kPa since P Psat compressed solid Note state b in Pv see the 3D figure is up on the solid face P CP v T CP v T a b a b P const P T v V L S CP a b 40 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 226 Determine the phase of the substance at the given state using Appendix B tables a Water 100C 500 kPa b Ammonia 10C 150 kPa c R410A 0C 350 kPa Solution a From Table B11 Psat100C 1013 kPa 500 kPa Psat then it is compressed liquid OR from Table B12 Tsat500 kPa 152C 100C Tsat then it is subcooled liquid compressed liquid b Ammonia NH3 Table B21 P Psat10 C 291 kPa Superheated vapor c R410A Table B41 P Psat0 C 799 kPa Superheated vapor The SL fusion line goes slightly to the left for water It tilts slightly to the right for most other substances ln P T Vapor L CrP a bc S 41 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 227 Give the missing property of PvT and x for water at a P 10 MPa v 0003 m3kg b 1 MPa 190C c 200C 01 m3kg d 10 kPa 10C Solution For all states start search in table B11 if T given or B12 if P given a P 10 MPa v 0003 m3kg so look in B12 at 10 MPa T 311C vf 0001452 v vg 001803 m3kg so LV x v vf vfg 0003 0001452001657 0093 b 1 MPa 190C Only one of the two lookups is needed B11 P Psat 12544 kPa so it is superheated vapor B12 T Tsat 17991C so it is superheated vapor B13 v 019444 020596 019444 190 17991 200 17991 02002 m3kg c 200C 01 m3kg look in B11 P Psat 15538 kPa vf 0001156 m3kg v vg 012736 m3kg so LV x v vf vfg 01 000115601262 07832 d 10 kPa 10C Only one of the two lookups is needed From B11 P Pg 12276 kPa so compressed liquid From B12 T Tsat 458C so compressed liquid From B11 v vf 0001 m3kg at given T not given P States shown are placed relative to the twophase region not to each other P CP v T CP v T a d c b d a c b P const 42 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 228 For water at 200 kPa with a quality of 10 find the volume fraction of vapor This is a twophase state at a given pressure Table B12 vf 0001 061 m3kg vg 088573 m3kg From the definition of quality we get the masses from total mass m as mf 1 x m mg x m The volumes are Vf mf vf 1 x m vf Vg mg vg x m v g So the volume fraction of vapor is Fraction Vg V Vg Vg Vf x m vg x m vg 1 xm vf 01 088573 01 088573 09 0001061 0088573 00895279 09893 Notice that the liquid volume is only about 1 of the total We could also have found the overall v vf xvfg and then V m v 43 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 229 Determine whether refrigerant R410A in each of the following states is a compressed liquid a superheated vapor or a mixture of saturated liquid and vapor Solution All cases are seen in Table B41 a 50C 005 m3kg From table B41 at 50C vg 000707 m3kg since v vg we have superheated vapor b 10 MPa 20C From table B41 at 20C Pg 9099 kPa since P Pg we have superheated vapor c 01 MPa 01 m3kg From table B41 at 01 MPa use 101 kPa vf 000074 and vg 02395 m3kg as vf v vg we have a mixture of liquid vapor d 20C 200 kPa superheated vapor P Pg 400 kPa at 20C 230 Show the states in Problem 229 in a sketch of the Pv diagram States shown are placed relative to the twophase region not to each other P CP v T CP v T a b d c d a b c P const 44 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 231 How great is the change in liquid specific volume for water at 20oC as you move up from state i towards state j in Fig 214 reaching 15 000 kPa State i here a is saturated liquid and up is then compressed liquid states a Table B11 vf 0001 002 m3kg at 234 kPa b Table B14 vf 0001 002 m3kg at 500 kPa c Table B14 vf 0001 001 m3kg at 2000 kPa d Table B14 vf 0001 000 m3kg at 5000 kPa e Table B14 vf 0000 995 m3kg at 15 000 kPa f Table B14 vf 0000 980 m3kg at 50 000 kPa Notice how small the changes in v are for very large changes in P v T v P a b c T 20 C o d e f fa P T v V L S CP a f 45 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 232 Fill out the following table for substance ammonia Solution P kPa T oC v m3kg x a 1003 25 01185 09238 b 1195 30 04824 05 a B21 vf v vg twophase mix Look in B21 x v vf vfg 01185 0001658012647 09238 b B21 P Psat 1195 kPa v vf x vfg 0001476 05 096192 04824 m3kg 233 Place the two states ab listed in Problem 232 as labeled dots in a sketch of the P v and Tv diagrams Solution T CP v a b P const P CP v T a b 1003 120 46 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 234 Give the missing property of P T v and x for R410A at a T 20oC P 450 kPa b P 300 kPa v 0092 m3kg Solution a B41 P Psat 3996 kPa compressed liquid v vf 0000803 m3kg x undefined b B42 v vg at 300 kPa superheated vapor T 20 0 20 0092 008916 009845 008916 1389C x undefined T CP v a b P const P v a T b 47 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 235 Fill out the following table for substance water Solution P kPa T oC v m3kg x a 500 20 0001002 Undefined b 500 15186 020 0532 c 1400 200 014302 Undefined d 8581 300 001762 08 a Table B11 P Psat so it is compressed liquid Table B14 b Table B12 vf v vg so two phase L V x v vf vfg 02 0001093 03738 0532 T Tsat 15186oC c Only one of the two lookup is needed Table B11 200oC P Psat superheated vapor Table B12 1400 kPa T Tsat 195oC Table B13 subtable for 1400 kPa gives the state properties d Table B11 since quality is given it is twophase v vf x vfg 0001404 08 002027 001762 m3kg 236 Place the four states ad listed in Problem 235 as labeled dots in a sketch of the P v and Tv diagrams Solution T CP v d a c b P const P CP v T a d c b 500 1400 8581 20 152 200 300 48 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 237 Determine the specific volume for R410A at these states a 15oC 400 kPa b 20oC 1500 kPa c 20oC quality 25 a Table B41 P Psat 4804 kPa so superheated vapor B42 v 006475 007227 006475 15 1998 0 1998 006662 m3kg b Table B41 P Psat 1444 kPa so compressed liquid v vf 0000923 m3kg c Table B41 vf 0000923 m3kg vfg 001666 m3kg so v vf x vfg 0000923 025 001666 000509 m3kg States shown are placed relative to the twophase region not to each other P CP v T CP v T c a b a c b P const 49 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 238 Give the missing property of P T v and x for CH4 at a T 155 K v 004 m3kg b T 350 K v 025 m3kg a B71 v vg 004892 m3kg 2phase x v vf vfg 004 0002877 004605 0806 P Psat 1296 kPa b B71 T Tc and v vc superheated vapor B72 located between 600 800 kPa P 600 200 025 030067 02251 030067 734 kPa T CP v P const P v T a b a b 50 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 239 Give the specific volume of carbondioxide at 20oC for 2000 kPa and repeat for 1400 kPa Table B31 20oC Psat 1969 kPa at 2000 kPa state is compressed liquid v vf 0000969 m3kg at 1400 kPa state is superheated vapor v 00296 m3kg The 2000 kPa is above and the 1400 kPa is below the vaporization line ln P T V L S b a 51 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 240 Calculate the following specific volumes a CO2 10C 80 quality b Water 4 MPa 90 quality c Nitrogen 120 K 60 quality Solution All states are twophase with quality given The overall specific volume is given by Eq21 or 22 v vf x vfg 1xvf x v g a CO2 10C 80 quality in Table B31 v 0001161 x 000624 0006153 m3kg b Water 4 MPa 90 quality in Table B12 v 00012521x x 004978 004493 m3kg c Nitrogen 120 K 60 quality in Table B61 v 0001915 x 000608 0005563 m3kg 52 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 241 Give the missing property of P T v and x for a R410A at T 25oC v 001 m3kg b R410A at 400 kPa v 0075 m3kg c NH3 at 10oC v 01 m3kg Solution a Table B41 v vg 001514 m3kg so state is twophase L V P Psat 16536 kPa x v vf vfg 001 0000944001420 06377 b Table B41 v vg so go to B42 superheated vapor 400 kPa Make linear interpolation T 0 20 0 0075 007227 007916 007227 792 x is undefined c Table B21 vf v vg 001514 m3kg so state is twophase L V P Psat 6152 kPa x v vf vfg 01 00016020381 04828 Each state is positioned relative to the twophase region not to each other T cp v P const P v T a c b a c b 53 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 242 You want a pot of water to boil at 105oC How heavy a lid should you put on the 15 cm diameter pot when Patm 101 kPa Solution Table B11 at 105oC Psat 1208 kPa A π 4 D2 π 4 0152 001767 m2 Fnet Psat Patm A 1208 101 kPa 001767 m2 03498 kN 350 N Fnet mlid g mlid Fnetg 350 9807 N ms2 357 kg Some lids are clamped on the problem deals with one that stays on due to its weight 54 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 243 Water at 400 kPa with a quality of 25 has its pressure raised 50 kPa in a constant volume process What is the new quality and temperature Solution State 1 from Table B12 at 400 kPa v vf x vfg 0001084 025 046138 011643 m3kg State 2 has same v at P 450 kPa also from Table B12 x v vf vfg 011643 0001088 041289 0279 T Tsat 14793 C T CP v P CP v T 400 450 143 148 143 C 148 C 55 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 244 A sealed rigid vessel has volume of 1 m3 and contains 2 kg of water at 100C The vessel is now heated If a safety pressure valve is installed at what pressure should the valve be set to have a maximum temperature of 200C Solution Process v Vm constant State 1 v1 12 05 m3kg from Table B11 it is 2phase State 2 200C 05 m3kg Table B13 between 400 and 500 kPa so interpolate CP T v 100 C 500 kPa 400 kPa P 400 05 053422 042492 053422 500 400 4313 kPa 56 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 245 Saturated water vapor at 200 kPa is in a constant pressure piston cylinder At this state the piston is 01 m from the cylinder bottom How much is this distance and the temperature if the water is cooled to occupy half the original volume Solution State 1 B 12 v1 vg 200 kPa 08857 m3kg T1 1202C Process P constant 200 kPa State 2 P v2 v12 044285 m3kg Table B12 v2 vg so two phase T2 Tsat 1202C Height is proportional to volume h2 h1 v2v1 01 m 05 005m 57 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 246 Saturated liquid water at 60C is put under pressure to decrease the volume by 1 keeping the temperature constant To what pressure should it be compressed Solution State 1 T 60C x 00 Table B11 v 0001017 m3kg Process T constant 60C State 2 T v 099 vf 60C 0990001017 00010068 m3kg Between 20 30 MPa in Table B14 P 238 MPa P CP v T v 20 MPa 1 2 1 2 30 MPa 58 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 247 Water at 400 kPa with a quality of 25 has its temperature raised 20oC in a constant pressure process What is the new quality and specific volume Solution State 1 from Table B12 at 400 kPa v vf x vfg 0001084 025 046138 0126185 m3kg State 2 has same P from Table B12 at 400 kPa T2 Tsat 20 14363 20 16363oC so state is superheated vapor look in B13 and interpolate between 150 and 200 C in the 400 kPa superheated vapor table x undefined v2 047084 053422 047084 16363 150 200 150 04881 m3kg T CP v 400 kPa P CP v T 400 144 164 1436 C 164 C 59 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 248 In your refrigerator the working substance evaporates from liquid to vapor at 20 oC inside a pipe around the cold section Outside on the back or below is a black grille inside which the working substance condenses from vapor to liquid at 45 oC For each location find the pressure and the change in specific volume v if the substance is ammonia Solution The properties come from the saturated tables where each phase change takes place at constant pressure and constant temperature Substance TABLE T Psat kPa v vfg m3kg Ammonia B21 45 oC 1782 00707 Ammonia B21 20 oC 190 0622 T CP 1 2 P CP v T 2 20 1 45 45C 3 4 3 4 60 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 249 Repeat the previous problem with the substances a R134a b R410A In your refrigerator the working substance evaporates from liquid to vapor at 20 oC inside a pipe around the cold section Outside on the back or below is a black grille inside which the working substance condenses from vapor to liquid at 45 oC For each location find the pressure and the change in specific volume v Solution The properties come from the saturated tables where each phase change takes place at constant pressure and constant temperature Substance TABLE T Psat kPa v vfg m3kg R134a B51 45 oC 1160 00165 R134a B51 20 oC 134 0146 R410A B41 45 oC 2728 000723 R410A B41 20 oC 400 0064 T CP 1 2 P CP v T 2 20 1 45 45C 3 4 3 4 61 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 250 Repeat Problem 248 with CO2 condenser at 20oC and evaporator at 30oC Solution The properties come from the saturated tables where each phase change takes place at constant pressure and constant temperature Substance TABLE T Psat kPa v vfg m3kg CO2 B31 20 oC 5729 000386 CO2 B31 30 oC 1428 0026 T CP 1 2 P CP v T 2 30 1 20 20C 3 4 3 4 62 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 251 A glass jar is filled with saturated water at 500 kPa quality 25 and a tight lid is put on Now it is cooled to 10C What is the mass fraction of solid at this temperature Solution Constant volume and mass v1 v2 Vm From Table B12 v1 0001093 025 03738 0094543 m3kg From Table B15 v2 00010891 x2 446756 v1 0094543 m3kg x2 00002 mass fraction vapor xsolid 1 x2 09998 or 9998 P CP v T CP v T 1 2 1 2 P T v V L S CP 1 2 63 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 252 Two tanks are connected as shown in Fig P252 both containing water Tank A is at 200 kPa v 05 m3kg VA 1 m3 and tank B contains 35 kg at 05 MPa 400C The valve is now opened and the two come to a uniform state Find the final specific volume Solution Control volume both tanks Constant total volume and mass process A B sup vapor State A1 P v mA VAvA 105 2 kg State B1 P T Table B13 vB 06173 m3kg VB mBvB 35 kg 06173 m3kg 21606 m3 Final state mtot mA mB 55 kg Vtot VA VB 31606 m3 v2 Vtotmtot 05746 m3kg 64 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 253 Saturated vapor R134a at 60oC changes volume at constant temperature Find the new pressure and quality if saturated if the volume doubles Repeat the question for the case the volume is reduced to half the original volume Solution 1 T x B41 v1 vg 001146 m3kg P1 Psat 16818 kPa 2 v2 2v1 002292 m3kg superheated vapor Interpolate between 1000 kPa and 1200 kPa P2 1000 200 002292 002311 001844 002311 1008 kPa 3 v3 v12 000573 m3kg vg two phase x3 vfg v3 vf 000573 0000951 001051 04547 P3 Psat 16818 kPa T CP v 1 2 P 1682 kPa P CP v T 2 1682 60 1 3 3 65 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 254 A steel tank contains 6 kg of propane liquid vapor at 20C with a volume of 0015 m3 The tank is now slowly heated Will the liquid level inside eventually rise to the top or drop to the bottom of the tank What if the initial mass is 1 kg instead of 6 kg Solution Constant volume and mass v2 v1 V m 6 kg 0015 m3 00025 m3kg v T 20C CP v c a b Vapor Liq A2 vc 000454 m3kg v1 eventually reaches sat liquid level rises to top If m 1 kg v1 0015 m3kg vc then it will reach saturated vapor level falls 66 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 255 Saturated water vapor at 60C has its pressure decreased to increase the volume by 10 keeping the temperature constant To what pressure should it be expanded Solution Initial state v 76707 m3kg from table B11 Final state v 110 vg 11 76707 84378 m3kg Interpolate at 60C between saturated P 1994 kPa and superheated vapor P 10 kPa in Tables B11 and B13 P 19941 10 19941 84378 76707 153345 76707 189 kPa P CP v T CP v T P 10 kPa 60 C o 10 kPa Comment Tv P 18 kPa software v is not linear in P more like 1P so the linear interpolation in P is not very accurate 67 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 256 Ammonia at 20oC with a quality of 50 and total mass 2 kg is in a rigid tank with an outlet valve at the top How much vapor mass can you take out through the valve assuming the temperature stays constant Solution The top has saturated vapor at 20oC as long as there is a two phase inside When there is no more liquid and vapor will be taken out pressure will drop for the remaining vapor so we can take it out until we reach P0 V m1v1 2 kg 05 0001638 014922 m3kg 015086 m 3 v2 14153 m3kg from Table B22 at 100 kPa m2 Vv2 015086 m3 14153 m3kg 01066 kg m m1 m2 2 01066 1893 kg P CP v T CP v T P 857 kPa 20 C o 68 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 257 A sealed rigid vessel of 2 m3 contains a saturated mixture of liquid and vapor R 134a at 10C If it is heated to 50C the liquid phase disappears Find the pressure at 50C and the initial mass of the liquid Solution Process constant volume and constant mass P v 2 1 State 2 is saturated vapor from table B51 P2 Psat50C 1318 MPa State 1 same specific volume as state 2 v1 v2 0015124 m3kg v1 0000794 x1 0048658 x1 02945 m Vv1 2 m30015124 m3kg 13224 kg mliq 1 x1 m 93295 kg 69 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 258 A storage tank holds methane at 120 K with a quality of 25 and it warms up by 5C per hour due to a failure in the refrigeration system How long time will it take before the methane becomes single phase and what is the pressure then Solution Use Table B71 Assume rigid tank v constant v1 v1 0002439 025 030367 0078366 m3kg We then also see that v1 vc 000615 m3kg All single phase when v vg T 145 K P 2 1 v t T 5Ch 145 120 5 5 hours P Psat 824 kPa 70 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 259 Ammonia at 10oC and mass 10 kg is in a piston cylinder with an initial volume of 1 m3 The piston initially resting on the stops has a mass such that a pressure of 900 kPa will float it Now the ammonia is slowly heated to 50oC Find the final pressure and volume Solution CV Ammonia constant mass Process V constant unless P Pfloat State 1 T 10 oC v1 V m 1 10 01 m3kg From Table B21 vf v v g x1 vfg v vf 01 00016 020381 04828 v P 2 1 1a P P 1 2 50 C 215 C State 1a P 900 kPa v v1 01 m3kg vg at 900 kPa This state is twophase T1a 2152oC Since T2 T1a then v2 v 1a State 2 50oC and on lines means P2 900 kPa which is superheated vapor Table B22 v2 018465 0144992 016482 m3kg V2 mv2 16482 m3 71 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 260 A 400m3 storage tank is being constructed to hold LNG liquified natural gas which may be assumed to be essentially pure methane If the tank is to contain 90 liquid and 10 vapor by volume at 100 kPa what mass of LNG kg will the tank hold What is the quality in the tank Solution CH4 is in the section B tables From Table B71 vf 0002366 m3kg interpolated From Table B72 vg 055665 m3kg first entry 100 kPa mliq vf Vliq 09 400 0002366 152 1555 kg mvap vg Vvap 01 400 055665 7186 kg mtot 152 227 kg x mvap mtot 472 10 4 72 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 261 A boiler feed pump delivers 005 m3s of water at 240C 20 MPa What is the mass flowrate kgs What would be the percent error if the properties of saturated liquid at 240C were used in the calculation What if the properties of saturated liquid at 20 MPa were used Solution State 1 T P compressed liquid seen in B14 v 0001205 m3kg m V v 0050001205 415 kgs vf 240C 0001229 m3kg m 4068 kgs error 2 vf 20 MPa 0002036 m3kg m 2456 kgs error 41 P CP v T CP v P 20 MPa 240 C o 20 MPa 240 The constant T line is nearly vertical for the liquid phase in the Pv diagram The state is at so high P T that the saturated liquid line is not extremely steep 73 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 262 A pistoncylinder arrangement is loaded with a linear spring and the outside atmosphere It contains water at 5 MPa 400C with the volume being 01 m3 If the piston is at the bottom the spring exerts a force such that Plift 200 kPa The system now cools until the pressure reaches 1200 kPa Find the mass of water the final state T2 v2 and plot the Pv diagram for the process Solution 1 Table B13 v1 005781 m3kg m Vv1 01005781 173 kg Straight line P Pa C v v2 v1 P2 Pa P1 Pa 001204 m3kg v2 vg1200 kPa so twophase T2 188C x2 v2 000113901622 00672 P v 5000 1200 200 1 2 a 005781 0 74 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 263 A pressure cooker closed tank contains water at 100C with the liquid volume being 120 of the vapor volume It is heated until the pressure reaches 20 MPa Find the final temperature Has the final state more or less vapor than the initial state Solution State 1 Vf mf vf Vg20 mgvg20 Table B11 vf 0001044 m3kg vg 16729 m3kg x1 mg mf mg mf 20 mfvf vg 20 mfvf vg 20 vf vg 20 vf 002088 002088 16729 001233 v1 0001044 001233167185 002166 m3kg State 2 v2 v1 002166 m3kg vg2MPa from B12 so twophase P v 2 1 At state 2 v2 vf x2 v fg 002166 0001177 x2 009845 x2 0208 More vapor at final state T2 Tsat2MPa 2124C 75 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 264 A pressure cooker has the lid screwed on tight A small opening with A 5 mm2 is covered with a petcock that can be lifted to let steam escape How much mass should the petcock have to allow boiling at 120oC with an outside atmosphere at 1013 kPa Table B11 Psat 1985 kPa F mg P A m P Ag 1985101310005106 9807 N ms2 00496 kg 50 g 76 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Ideal Gas Law 77 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 265 What is the relative change in P if we double the absolute temperature of an ideal gas keeping mass and volume constant What will it be if we double V having m T constant Ideal gas law PV mRT State 2 P2V mRT2 mR2T1 2P1V P2 2P 1 Relative change PP1 P1P1 1 100 State 3 P3V3 mRT1 P1V1 P3 P1V1V3 P12 Relative change PP1 P12P1 05 50 T P V T 3 2 1 T 1 2 V 2 1 3 78 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 266 A 1m3 tank is filled with a gas at room temperature 20C and pressure 200 kPa How much mass is there if the gas is a air b neon or c propane Solution Use Table A2 to compare T and P to the critical T and P with T 20C 29315 K P 200 kPa Pc for all Air T TCN2 TCO2 1546 K so ideal gas R 0287 kJkg K Neon T Tc 444 K so ideal gas R 041195 kJkg K Propane T Tc 370 K but P Pc 425 MPa so gas R 018855 kJkg K All states are ideal gas states so the ideal gas law applies PV mRT a m PV RT 0287 kJkgK 29315 K 200 kPa 1 m3 2377 kg b m PV RT 041195 kJkgK 29315 K 200 kPa 1 m3 1656 kg c m PV RT 018855 kJkgK 29315 K 200 kPa 1 m3 3618 kg 79 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 267 Calculate the ideal gas constant for argon and hydrogen based on table A2 and verify the value with Table A5 argon R R M 83145 39948 02081 kJkgK same as Table A5 hydrogen R R M 83145 2016 4124256 kJkgK same as Table A5 80 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 268 A pneumatic cylinder a piston cylinder with air must close a door with a force of 500 N The cylinder crosssectional area is 5 cm2 With V 50 cm3 T 20C what is the air pressure and its mass F PA PoA P Po FA 100 kPa 500 N 00005 m2 1000 NkN 1100 kPa m P1V1 RT1 1100 kPa 000005 m3 0287 kJkgK 293 K 000065 kg 065 g Comment Dependent upon your understanding of the problem you could also have neglected the atmospheric pressure to get 1000 kPa and 059 g for answers 81 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 269 Is it reasonable to assume that at the given states the substance behaves as an ideal gas Solution a Oxygen O2 at 30C 3 MPa Ideal Gas T Tc 155 K from A2 b Methane CH4 at 30C 3 MPa Ideal Gas T Tc 190 K from A2 c Water H2O at 30C 3 MPa NO compressed liquid P Psat B11 d R134a at 30C 3 MPa NO compressed liquid P Psat B51 e R134a at 30C 100 kPa Ideal Gas P is low Psat B51 ln P T Vapor Liq CrP a b c d e 82 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 270 Helium in a steel tank is at 250 kPa 300 K with a volume of 01 m3 It is used to fill a balloon and when the pressure drops to 125 kPa the flow of helium stops by itself If all the helium still is at 300 K how big a balloon did I get Solution State 1 m Vv assume ideal gas so m P1V1 RT1 250 kPa 01 m3 20771 kJkgK 300 K 00401 kg State 2 Same mass so then T2 T1 c i r c u s t h e r m o cb V2 mRT2 P2 P1V1 RT2 RT1 P2 V1 P1 P2 01 m3 250 125 02 m3 The balloon volume is Vballoon V2 V1 02 01 01 m3 83 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 271 A hollow metal sphere of 150mm inside diameter is weighed on a precision beam balance when evacuated and again after being filled to 875 kPa with an unknown gas The difference in mass is 00025 kg and the temperature is 25C What is the gas assuming it is a pure substance listed in Table A5 Solution Assume an ideal gas with total volume V π 60153 0001767 m3 M mR T PV 00025 kg 83145 kJkmolK 2982 K 875 kPa 0001767 m3 4009 kgkmol MHe Helium Gas 84 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 272 A spherical helium balloon of 10 m in diameter is at ambient T and P 15oC and 100 kPa How much helium does it contain It can lift a total mass that equals the mass of displaced atmospheric air How much mass of the balloon fabric and cage can then be lifted We need to find the masses and the balloon volume V π 6 D3 π 6 103 5236 m3 mHe ρV V v PV RT 20771 kJkgK 288 K 100 kPa 5236 m3 875 kg mair PV RT 100 kPa 5236 m3 0287 kJkgK 288 K 633 kg mlift mair mHe 633875 5455 kg 85 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 273 A glass is cleaned in 45oC hot water and placed on the table bottom up The room air at 20oC that was trapped in the glass gets heated up to 40oC and some of it leaks out so the net resulting pressure inside is 2 kPa above ambient pressure of 101 kPa Now the glass and the air inside cools down to room temperature What is the pressure inside the glass Solution 1 air 40oC 103 kPa 2 air 20oC Constant Volume V1 V2 AIR Slight amount of liquid water seals to table top Constant Mass m1 m2 Ideal Gas P1V1 m1RT1 and P2V2 m1RT2 Take Ratio P2 P1 T1 T2 103 kPa 20 273 40 273 964 kPa This is a vacuum relative to atm pressure so the glass is pressed against table 86 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 274 Air in an internal combustion engine has 227oC 1000 kPa with a volume of 01 m3 Now combustion heats it to 1800 K in a constant volume process What is the mass of air and how high does the pressure become The mass comes from knowledge of state 1 and ideal gas law m P1V1 RT1 1000 kPa 01 m3 0287 kJkgK 227 273 K 0697 kg The final pressure is found from the ideal gas law written for state 1 and state 2 and then eliminate the mass gas constant and volume V2 V1 between the equations P1 V1 m RT1 and P2 V2 m RT 2 P2 P1 T2T1 1000 kPa 1800 500 3600 kPa 87 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 275 Air in an automobile tire is initially at 10C and 190 kPa After the automobile is driven awhile the temperature gets up to 10C Find the new pressure You must make one assumption on your own Solution Assume constant volume V2 V1 and that air is an ideal gas P1V1 mRT1 and P2V2 mRT 2 so P2 P1 T2T1 190 kPa 28315 26315 2044 kPa 88 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 276 A rigid tank of 1 m3 contains nitrogen gas at 600 kPa 400 K By mistake someone lets 05 kg flow out If the final temperature is 375 K what is the final pressure Solution m PV RT 02968 kJkgK 400 K 600 kPa 1 m3 5054 kg m2 m 05 4554 kg P2 m2RT2 V 4554 kg 02968 kJkgK 375 K 1 m3 5069 kPa 89 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 277 Assume we have 3 states of saturated vapor R134a at 40 oC 0 oC and 40 oC Calculate the specific volume at the set of temperatures and corresponding saturated pressure assuming ideal gas behavior Find the percent relative error 100v vgvg with vg from the saturated R134a table Solution R134a Table values from Table B51 Psat vgT Ideal gas constant from Table A5 RR134a 008149 kJkg K T Psat kPa vg m3kg vIDG RT P sat error 40 oC 518 035696 036678 275 0 oC 294 006919 007571 94 40 oC 1017 002002 002509 253 v P v T 1 2 1 2 3 3 90 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 278 Do Problem 277 for R410A Solution R410A Table values from Table B41 Psat vgT Ideal gas constant from Table A5 RR410a 01146 kJkg K T Psat kPa vg m3kg vIDG RT P sat error 40 oC 1750 014291 01527 68 0 oC 7987 003267 003919 20 40 oC 24207 000967 001483 533 v P v T 1 2 1 2 3 3 91 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 279 Do Problem 277 for the substance ammonia Solution NH3 Table values from Table B21 Psat vgT Ideal gas constant from Table A5 Rammonia 04882 kJkg K T Psat kPa vg m3kg vIDG RT P sat error 40 oC 717 15526 15875 225 0 oC 4296 028929 03104 73 40 oC 1555 008313 009832 183 v P v T 1 2 1 2 3 3 92 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 280 A 1 m3 rigid tank has propane at 100 kPa 300 K and connected by a valve to another tank of 05 m3 with propane at 250 kPa 400 K The valve is opened and the two tanks come to a uniform state at 325 K What is the final pressure Solution Propane is an ideal gas P Pc with R 01886 kJkgK from Tbl A5 mA PAVA RTA 01886 kJkgK 300 K 100 kPa 1 m3 17674 kg mB PBVB RTB 01886 kJkgK 400 K 250 kPa 05 m3 16564 kg V2 VA VB 15 m3 m2 mA mB 34243 kg P2 m2RT2 V2 34243 kg 01886 kJkgK 325 K 15 m3 1399 kPa 93 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 281 A vacuum pump is used to evacuate a chamber where some specimens are dried at 50C The pump rate of volume displacement is 05 m3s with an inlet pressure of 01 kPa and temperature 50C How much water vapor has been removed over a 30 min period Solution Use ideal gas since P lowest P in steam tables From table A5 we get R 046152 kJkg K m m t with mass flow rate as m V v PV RT ideal gas m PV tRT 046152 kJkgK 32315 K 01 kPa 05 m3s 30 min 60 smin 0603 kg 94 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 282 A 1m3 rigid tank with air at 1 MPa 400 K is connected to an air line as shown in Fig P282 The valve is opened and air flows into the tank until the pressure reaches 5 MPa at which point the valve is closed and the temperature inside is 450K a What is the mass of air in the tank before and after the process b The tank eventually cools to room temperature 300 K What is the pressure inside the tank then Solution P T known at both states and assume the air behaves as an ideal gas mair1 RT1 P1V 0287 kJkgK 400 K 1000 kPa 1 m3 8711 kg mair2 RT2 P2V 0287 kJkgK 450 K 5000 kPa 1 m3 38715 kg Process 2 3 is constant V constant mass cooling to T3 P3 P2 T3T2 5000 kPa 300450 333 MPa 95 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 283 A cylindrical gas tank 1 m long inside diameter of 20 cm is evacuated and then filled with carbon dioxide gas at 20C To what pressure should it be charged if there should be 12 kg of carbon dioxide Solution Assume CO2 is an ideal gas table A5 R 01889 kJkg K Vcyl A L π 4022 1 0031416 m 3 P V mRT P mRT V P 12 kg 01889 kJkg Κ 27315 20 K 0031416 m3 2115 kPa 96 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 284 Ammonia in a pistoncylinder arrangement is at 700 kPa 80C It is now cooled at constant pressure to saturated vapor state 2 at which point the piston is locked with a pin The cooling continues to 10C state 3 Show the processes 1 to 2 and 2 to 3 on both a Pv and Tv diagram Solution State 1 T P from table B22 this is superheated vapor State 2 T x from table B21 State 3 T v twophase 700 290 P v 2 3 1 v T 2 3 1 80 14 10 97 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Compressibility Factor 98 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 285 Find the compressibility factor Z for saturated vapor ammonia at 100 kPa and at 2000 kPa Table B22 v1 11381 m3kg T1 336oC P1 100 kPa v2 006444 m3kg T2 4937oC P2 2000 kPa Table A5 R 04882 kJkg K Extended gas law Pv ZRT so we can calculate Z from this Z1 RT1 P1v1 04882 kJkgK 27315 336 K 100 kPa 11381 m3kg 0973 Z2 RT2 P2v2 04882 kJkgK 27315 4937 K 2000 kPa 006444 m3kg 08185 So state 1 is close to ideal gas and state 2 is not so close ln P r Z T 20 r T 07 r T 07 r 01 1 T 12 r 1 2 99 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 286 Find the compressibility for nitrogen at a 2000 kPa 120 K b 2000 kPa 300 K c 120 K v 0005 m3kg Solution Table B6 has the properties for nitrogen a B62 v 001260 m3kg Z PvRT 2000 kPa 001260 m3kg 02968 kJkgK 120 K 0707 b B62 v 004440 m3kg Z PvRT 2000 kPa 004440 m3kg 02968 kJkgK 300 K 0997 c B61 vf v vg so this is a twophase state Z PvRT 2000 kPa 0005 m3kg 02968 kJkgK 120 K 028 The liquid in this state is incompressible with a low volume and the vapor is very close to the critical point If you calculate a quality it is x 0507 The compressibility for the saturated vapor alone is 044 100 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 287 Find the compressibility for carbon dioxide at 60oC and 10 MPa using Fig D1 Solution Table A2 CO2 Tc 3041 K Pc 738 MPa Tr TTc 3333041 1095 Pr PPc 10738 1355 From Figure D1 Z 045 Compare with table B32 v 000345 m3kg Z PvRT 10 000 kPa 000345 m3kg 01889 kJkgK 333 K 055 ln P r Z T 20 r T 07 r T 07 r 01 1 T 11 r 101 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 288 What is the percent error in specific volume if the ideal gas model is used to represent the behavior of superheated ammonia at 40C 500 kPa What if the generalized compressibility chart Fig D1 is used instead Solution NH3 T 40C 31315 K Tc 4055 K Pc 1135 MPa from Table A1 Table B22 v 02923 m3kg Ideal gas v RT P 048819 kJkgK 313 K 500 kPa 03056 m3kg 45 error Figure D1 Tr 31315 4055 0772 Pr 05 1135 0044 Z 097 v ZRT P 02964 m3kg 14 error 102 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 289 A cylinder fitted with a frictionless piston contains butane at 25C 500 kPa Can the butane reasonably be assumed to behave as an ideal gas at this state Solution Butane 25C 500 kPa Table A2 Tc 425 K Pc 38 MPa Tr 25 273 425 0701 Pr 05 38 013 Look at generalized chart in Figure D1 Actual Pr Pr sat 01 liquid not a gas The pressure should be less than 380 kPa to have a gas at that T 103 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 290 Estimate the saturation pressure of chlorine at 300 K Solution We do not have a table in the B section for Chlorine so we must use the generalized chart Table A2 Pc 798 MPa Tc 4169 K Tr TTc 300 4169 07196 Figure D1 Pr sat 013 same estimation from Table D4 P Pc Pr sat 798 013 104 MPa If you use the CATT3 program then you will find Pr sat 0122 and P 973 kPa 104 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 291 A bottle with a volume of 01 m3 contains butane with a quality of 75 and a temperature of 300 K Estimate the total butane mass in the bottle using the generalized compressibility chart Solution We need to find the property v the mass is m Vv so find v given T1 and x as v vf x v fg Table A2 Butane Tc 4252 K Pc 38 MPa 3800 kPa Tr 3004252 0705 From Fig D1 or table D4 Zf 002 Zg 09 Pr sat 01 ln P r Z T 20 r f T 07 r T 07 r 01 1 g P Psat Pr sat Pc 01 380 MPa 1000 kPaMPa 380 kPa vf ZfRTP 002 014304 kJkgK 300 K380 kPa 000226 m3kg vg ZgRTP 09 014304 kJkgK 300 K380 kPa 01016 m3kg v 000226 075 01016 000226 0076765 m3kg m V v 01 0076765 1303 kg 105 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 292 Find the volume of 2 kg of ethylene at 270 K 2500 kPa using Z from Fig D1 Ethylene Table A2 Tc 2824 K Pc 504 MPa Table A5 R 02964 kJkg K The reduced temperature and pressure are Tr T Tc 270 2824 0956 Pr P Pc 25 504 0496 Enter the chart with these coordinates and read Z 076 V mZRT P 2 kg 076 02964 kJkgK 270 K 2500 kPa 00487 m3 ln P r Z T 20 r T 07 r T 07 r 01 1 T 12 r T 096 r 05 106 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 293 For Tr 07 what is the ratio vgvf using Fig D1 compared to Table D3 For the saturated states we can use Fig D1 with the estimates Zf 002 Zg 09 so vgvf ZRTPg ZRTPf ZgZf 002 09 00222 Table D3 list the entries more accurately than we can read the figure Zf 0017 Zg 0897 so vgvf ZRTPg ZRTPf ZgZf 0017 0897 00189 ln P r Z T 20 r T 07 r T 07 r 01 1 T 12 r 05 107 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 294 Argon is kept in a rigid 5 m3 tank at 30C 3 MPa Determine the mass using the compressibility factor What is the error if the ideal gas model is used Solution No Argon table so we use generalized chart Fig D1 Tr 243151508 1612 Pr 30004870 0616 Z 096 m PV ZRT 096 02081 kJkgK 2432 K 3000 kPa 5 m3 30875 kg Ideal gas Z 1 m PV RT 2964 kg 4 error 108 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 295 Refrigerant R32 is at 10 oC with a quality of 15 Find the pressure and specific volume Solution For R32 there is no section B table printed We will use compressibility chart From Table A2 Tc 3513 K Pc 578 MPa From Table A5 R 01598 kJkg K Tr TTc 2633513 0749 From Table D4 or Figure D1 Zf 0029 Zg 086 Pr sat 016 P Pr sat Pc 016 5780 925 kPa v vf x vfg Zf x Zfg RTP 0029 015 086 0029 01598 kJkgK 263 K 925 kPa 0007 m3kg ln P r Z T 20 r T 07 r T 07 r 01 1 109 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 296 To plan a commercial refrigeration system using R123 we would like to know how much more volume saturated vapor R123 occupies per kg at 30 oC compared to the saturated liquid state Solution For R123 there is no section B table printed We will use compressibility chart From Table A2 Tc 4569 K Pc 366 MPa M 15293 Tr TTc 2434569 053 R R M 831451 kJkmolK 15293 kgkmol 00544 kJkgK The value of Tr is below the range in Fig D1 so use the table D4 Table D4 Zg 0979 Zf 000222 Zfg 0979 00022 09768 Pr Pr sat 00116 P Pr Pc 425 kPa vfg Zfg RTP 09768 00544 kJkgK 243 K 425 kPa 0304 m3kg Comment If you check with the software the solution is off by a factor of 6 The linear interpolation is poor and so is the approximation for Pr sat so the real saturation pressure should be 675 kPa Also the very small value of Zf is inaccurate by itself minute changes in the curve gives large relative variations 110 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 297 A new refrigerant R125 is stored as a liquid at 20 oC with a small amount of vapor For a total of 15 kg R125 find the pressure and the volume Solution As there is no section B table use compressibility chart Table A2 R125 Tc 3392 K Pc 362 MPa Tr T Tc 25315 3392 0746 We can read from Figure D1 or a little more accurately interpolate from table D4 entries Pr sat 016 Zg 086 Zf 0029 P Pr sat Pc 016 3620 kPa 579 kPa Vliq Zf mliq RTP 0029 15 kg 006927 kJkgK 25315 K 579 kPa 00013 m3 ln P r Z T 20 r T 07 r T 07 r 01 1 sat vapor sat liq 111 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Equations of State For these problems see appendix D for the equation of state EOS and chapter 12 112 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 298 Determine the pressure of nitrogen at 160 K v 000291 m3kg using ideal gas van der Waal Equation of State and the nitrogen table Nitrogen from table A2 Tc 1262 K Pc 3390 kPa Ideal gas P RT v 02968 160 000291 kJkgK K m3kg 16 319 kPa For van der Waal equation of state from Table D1 we have b 1 8 RTc Pc 0125 02968 1262 3390 0001 381 m3kg a 27 b2 Pc 27 0001 3812 3390 0174 562 kPa m3kg 2 The EOS is P RT v b a v2 02968 160 000291 0001 381 0174 562 0002912 10 444 kPa Table B62 P 10 000 kPa 113 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 299 Determine the pressure of nitrogen at 160 K v 000291 m3kg using Redlich Kwong Equation of State and the nitrogen table Nitrogen from table A2 Tc 1262 K Pc 3390 kPa Tr TTc 1601262 126783 For RedlichKwong EOS we have the parameters from Table D1 b 008664 RTc Pc 008664 02968 1262 3390 0000 957 3 m3kg a 042748 T E1267812 3390E A 0157 122 kPa mA3E AkgA2E 12 r 042748 R2T2 c Pc 029682 12622 The equation is P A RT v bE A A a v2 bvE A A 02968 160 000291 0000 9573E A A 0157122 0002912 0000 9573 000291E A 10 357 kPa Table B62 P 10 000 kPa 114 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2100 Determine the pressure of nitrogen at 160 K v 000291 mA3E Akg using Soave Equation of State and the nitrogen table Nitrogen from table A2 Tc 1262 K Pc 3390 kPa Tr TTc 1601262 126783 For Soave EOS see Appendix D very close to RedlichKwong P A RT v bE A A a v2 bvE where the parameters are from Table D1 and D4 ω 0039 f 048 1574ω 0176ωA2E A 054112 ao 042748 1 f 1 TA12 rE A A2E A 0371184 b 008664 RTc Pc 008664 A02968 1262 3390E A 0000 957 3 mA3E Akg a 0371184 A R2 AET2 c Pc E 0371184 A029682 12622 E3390E A 0153 616 kPa mA3E AkgA2E P A RT v bE A A a v2 bvE A A 02968 160 000291 0000 9573E A A 0153 616 0002912 0000 9573 000291E A 10 669 kPa Nitrogen Table B62 P 10 000 kPa 115 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2101 Carbon dioxide at 60AoE AC is pumped at a very high pressure 10 MPa into an oil well to reduce the oil viscosity for better oil flow We want to find its specific volume from the CO2 table ideal gas and van der Waals equation of state by iteration Table B32 v 000345 mA3E Akg Ideal gas v ART PE A A01889 60 27315 E10 000E A AkJkgK K kPaE A 0006 293 mA3E Akg Carbon dioxide from table A2 Tc 3041 K Pc 7380 kPa For van der Waal equation of state from Table D1 we have b A1 8E A RTc Pc 0125 A01889 3041 7380E A 0000 972 98 mA3E Akg a 27 bA2E A Pc 27 0000 972 98A2E A 7380 0188 6375 kPa mA3E AkgA2E The EOS is P A RT v bE A A a v2 E A Since it is nonlinear in v we use trial and error starting with the Table entry v 000345 P A 01889 33315 000345 0000 972 98E A A0188 6375 0003452 E A 95578 kPa low v 0003 P A 01889 33315 0003 0000 972 98E A A0188 6375 00032 E A 10 0868 kPa high v 000307 P A 01889 33315 000307 0000 972 98E A A0188 6375 0003072 E A 99954 OK v 000307 mA3E Akg 116 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2102 Solve the previous problem using the RedlichKwong equation of state Notice this becomes trial and error Carbon dioxide from table A2 Tc 3041 K Pc 7380 kPa Tr TTc 333153041 109553 For RedlichKwong EOS we have the parameters from Table D1 b 008664 RTc Pc 008664 A01889 3041 7380E A 0000 6744 mA3E Akg a 042748 TA12 rE A A R2 AET2 c Pc E 042748 A 018892 30412 E10955312 7380E A 018262 kPa mA3E AkgA2E The equation is P A RT v bE A A a v2 bvE A 10 000 A01889 33315 v 0000 6744E A A 018262 v2 0000 6744 vE A Trial and error on v start guided by Table B32 v 00035 mA3E Akg P 97728 kPa so v smaller v 00033 mA3E Akg P 100447 kPa so v larger v 00034 mA3E Akg P 99065 kPa linear interpolation gives v 000333 mA3E Akg P 100027 kPa OK 117 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2103 Solve Problem 2101 using the Soave equation of state ω 0239 Notice this becomes trial and error Carbon dioxide from table A2 Tc 3041 K Pc 7380 kPa Tr TTc 333153041 109553 For Soave EOS see Appendix D very close to RedlichKwong P A RT v bE A A a v2 bvE where the parameters are from Table D1 and D4 ω 0239 f 048 1574ω 0176ωA2E A 084613 ao 042748 1 f 1 TA12 rE A A2E A 0394381 b 008664 RTc Pc 008664 A01889 3041 7380E A 0000 6744 mA3E Akg a 0394381 A R2 AET2 c Pc E 0394381 A018892 30412 E7380E A 0176 342 kPa mA3E AkgA2E The equation is P A RT v bE A A a v2 bvE A 10 000 A01889 33315 v 0000 6744E A A 0176 342 v2 0000 6744 vE A Trial and error on v start guided by Table B32 v 00035 mA3E Akg P 10 202 kPa so v larger v 00036 mA3E Akg P 10 051 kPa so v close v 000363 mA3E Akg P 10 006 kPa OK 118 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2104 A tank contains 835 kg methane in 01 mA3E A at 250 K Find the pressure using ideal gas van der Waal EOS and the methane table The state is given by T 250 K v Vm 01835 0011976 mA3E Akg Table A2 or B72 Tc 1904 K Pc 4600 kPa Tr T Tc A 250 1904E A 1313 P T Vapor Liquid CritP CH4 state S Ideal gas model P ART vE A A05183 250 0011976E A 10 820 kPa For van der Waal equation of state from Table D1 we have b A1 8E A RTc Pc 0125 A05183 1904 4600E A 0002 681 64 mA3E Akg a 27 bA2E A Pc 27 0002 681 64A2E A 4600 0893 15 kPa mA3E AkgA2E The EOS is P A RT v bE A A a v2 E A A 05183 250 0011976 0002 681 64E A A 089315 00119762 E A 7714 kPa Locating the state in Table B72 P 8000 kPa very close 119 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2105 Do the previous problem using the RedlichKwong equation of state The state is given by T 250 K v Vm 01835 0011976 mA3E Akg Table A2 or B72 Tc 1904 K Pc 4600 kPa Tr T Tc A 250 1904E A 1313 P T Vapor Liquid CritP CH4 state S Ideal gas model P ART vE A A05183 250 0011976E A 10 820 kPa For RedlichKwong equation of state we have the parameters from Table D1 b 008664 RTc Pc 008664 A05183 1904 4600E A 0001 858 7 mA3E Akg a 042748 TA12 rE A A R2 AET2 c Pc E 042748 A051832 19042 E131312 4600E A 0789809 kPa mA3E AkgA2E The equation is P A RT v bE A A a v2 bvE A A 05183 250 0011976 00018587E A A 0789809 00119762 00018587 0011976E A 8040 kPa Locating the state in Table B72 P 8000 kPa very close 120 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2106 Do Problem 2104 using the Soave EOS A tank contains 835 kg methane in 01 mA3E A at 250 K Find the pressure using ideal gas van der Waal EOS and the methane table The state is given by T 250 K v Vm 01835 0011976 mA3E Akg Table A2 or B72 Tc 1904 K Pc 4600 kPa Tr T Tc A 250 1904E A 1313 P T Vapor Liquid CritP CH4 state S Ideal gas model P ART vE A A05183 250 0011976E A 10 820 kPa For Soave EOS we have the parameters from Table D1 and D4 ω 0011 f 048 1574ω 0176ωA2E A 049729 ao 042748 1 f 1 TA12 rE A A2E A 0367714 b 008664 RTc Pc 008664 A05183 1904 4600E A 0001 8587 mA3E Akg a 0367714 A R2 AET2 c Pc E 0367714 A051832 19042 E4600E A 0778 482 kPa mA3E AkgA2E The equation is P A RT v bE A A a v2 bvE A A 05183 250 0011976 0001 8587E A A 0778 482 00119762 00018587 0011976E 81087 kPa Locating the state in Table B72 P 8000 kPa 121 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems 122 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2107 Determine the quality if saturated or temperature if superheated of the following substances at the given two states Solution a Water H2O use Table B11 or B12 1 120AE AC 1 mA3E Akg v vAgE A superheated vapor T 120 AE AC 2 10 MPa 001 mA3E Akg twophase v vAgE x 001 0001452 001657 0516 b Nitrogen NA2E A table B6 1 1 MPa 003 mA3E Akg superheated vapor since v vAgE Interpolate between sat vapor and superheated vapor B62 T 10373 12010373 A 003002416 003117002416E A 117 K 2 100 K 003 mA3E Akg sat liquid vapor as twophase v vAgE v 003 0001452 x 0029764 x 0959 States shown are placed relative to the twophase region not to each other 123 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2108 Give the phase and the missing properties of P T v and x Solution a R410A T 10AE AC v 001 mA3E Akg Table B41 v vAgE A 002383 mA3E Akg sat liquid vapor P Psat 10857 kPa x v vAfE AvAfgE A 001 0000886002295 02713 b H2O T 350AE AC v 02 mA3E Akg Table B11 at given T v vAgE A 000881 mA3E Akg sup vapor P 140 MPa x undefined c R410A T 5 AE AC P 600 kPa sup vapor P PAgE A 6789 kPa at 5AE AC Table B42 v 004351 mA3E Akg at 867AE AC v 004595 mA3E Akg at 0AE AC v 004454 mA3E Akg at 5AE AC d R134a P 294 kPa v 005 mA3E Akg Table B51 v vAgE A 006919 mA3E Akg twophase T TAsatE A 0AE AC x v vAfE AvAfgE A 005 0000773006842 07195 States shown are placed relative to the twophase region not to each other P CP v T CP v T bc a d bc P const a d 124 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2109 Find the phase quality x if applicable and the missing property P or T Solution a H2O T 120AE AC v 05 mA3E Akg Table B11 at given T v vAgE A 089186 sat liq vap P Psat 1985 kPa x v vfvfg 05 00010608908 056 b H2O P 100 kPa v 18 m3kg Table B12 at given P v vAgE A 1694 sup vap interpolate in Table B13 T A 18 1694 193636 1694E A 150 9962 9962 12165 C c H2O T 263 K v 02 mA3E Akg Table B15 at given T 10 AE AC v vAgE A 466757 sat solid vap P Psat 026 kPa x v vAiE AvAigE A 200 0001466756 04285 States shown are placed relative to the twophase region not to each other 125 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2110 Find the phase quality x if applicable and the missing property P or T Solution a NH3 P 800 kPa v 02 mA3E Akg Superheated Vapor v vAgE A at 800 kPa Table B 22 interpolate between 70AE AC and 80AE AC T 714AE AC b NH3 T 20AE AC v 01 mA3E Akg Table B21 at given T v vAgE A 014922 sat liq vap P Psat 8575 kPa x v vAfE AvAfgE A 01 000164014758 0666 126 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2111 Give the phase and the missing properties of P T v and x These may be a little more difficult if the appendix tables are used instead of the software a R410A at T 10C v 002 m3kg Table B41 v vg at 10C sup vap Table B42 interpolate between sat and sup both at 10C P 6807 600 6807 A 0036003471 004018003471E A 6617 kPa b H2O v 02 m3kg x 05 Table B11 sat liq vap v 1x vf x vg vf vg 04 m3kg since vf is so small we find it approximately where vg 04 m3kg vf vg 039387 at 150C vf vg 04474 at 145C An interpolation gives T 1494C P 4682 kPa c H2O T 60C v 0001016 m3kg Table B11 v vf 0001017 compr liq see Table B14 v 0001015 at 5 MPa so P 055000 199 251 MPa d NH3 T 30C P 60 kPa Table B21 P Psat sup vapor interpolate in Table B22 50 kPa to 100 kPa v 29458 14657 29458 A 60 50 100 50E A 265 mA3E Akg v is not linearly proportional to P more like 1P so the computer table gives a more accurate value of 245 m3kg e R134a v 0005 m3kg x 05 sat liq vap Table B51 v 1x vf x vg vf vg 001 m3kg vf vg 0010946 at 65C vf vg 0009665 at 70C An interpolation gives T 687C P 206 MPa States shown are placed relative to the twophase region not to each other P CP v T CP v T a b e a P const d b e c c d 127 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2112 Refrigerant410a in a pistoncylinder arrangement is initially at 50C x 1 It is then expanded in a process so that P Cv1 to a pressure of 100 kPa Find the final temperature and specific volume Solution State 1 50C x 1 Table B41 PA1E A 30652 kPa vA1E A 000707 mA3E Akg Process Pv C PA1E Av1 P2 Cv2 P1v1v2 State 2 100 kPa and vA2E A vA1E APA1E APA2E A 02167 mA3E Akg vA2E A vAgE A at 100 kPa TA2E A 5165C from Table B42 Notice T is not constant v P v T 1 2 1 2 The first part of the process may be in the sup vapor region 128 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2113 Consider two tanks A and B connected by a valve as shown in Fig P2113 Each has a volume of 200 L and tank A has R410A at 25C 10 liquid and 90 vapor by volume while tank B is evacuated The valve is now opened and saturated vapor flows from A to B until the pressure in B has reached that in A at which point the valve is closed This process occurs slowly such that all temperatures stay at 25C throughout the process How much has the quality changed in tank A during the process Solution A B vacuum State A1 Table B41 vAfE A 0000944 mA3E Akg vAgE A 001514 mA3E Akg mAA1E A A Vliq1 Evf 25C E A A Vvap1 Evg 25C E A A01 02 0000944E A A09 02 001514E A 21186 11889 33075 kg xAA1E A A11889 33075E A 03594 State B2 Assume A still twophase so saturated P for given T mAB2E A A VB Evg 25C E A A 02 001514E A 13210 kg State A2 mass left is mAA2E A 33075 13210 19865 kg vAA2E A A 02 19865E A 0010068 0000944 xA2 001420 xAA2E A 06425 x 0283 129 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2114 Water in a pistoncylinder is at 90C 100 kPa and the piston loading is such that pressure is proportional to volume P CV Heat is now added until the temperature reaches 200C Find the final pressure and also the quality if in the twophase region Solution Final state 200C on process line P CV State 1 Table B11 vA1E A 0001036 mA3E Akg PA2E A PA1E AvA2E AvA1E A from process equation Check state 2 in Table B11 vAgE ATA2E A 012736 PAgE ATA2E A 15538 MPa If vA2E A vAgE ATA2E A PA2E A 123 MPa PAgE A not OK If sat PA2E A PAgE ATA2E A 15538 kPa vA2E A 00161 mA3E Akg vAgE A sat OK PA2E A 15538 kPa xA2E A 00161 0001156 01262 0118 1 2 v P 130 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2115 A tank contains 2 kg of nitrogen at 100 K with a quality of 50 Through a volume flowmeter and valve 05 kg is now removed while the temperature remains constant Find the final state inside the tank and the volume of nitrogen removed if the valvemeter is located at a The top of the tank b The bottom of the tank Solution Table B61 v1 0001452 x1 0029764 0016334 mA3E Akg Vtank m1v1 00327 mA3E m2 m1 05 15 kg v2 Vtankm2 00218 vgT x2 002180001452 00312160001452 06836 Top flow out is sat vap vg 0031216 mA3E Akg Vout moutvg 00156 mA3E Bottom flow out is sat liq vf 0001452 Vout moutvf 0000726 mA3E 131 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2116 A springloaded pistoncylinder contains water at 500C 3 MPa The setup is such that pressure is proportional to volume P CV It is now cooled until the water becomes saturated vapor Sketch the Pv diagram and find the final pressure Solution State 1 Table B13 vA1E A 011619 mA3E Akg Process m is constant and P CA0E AV CA0E Am v C v P Cv C PA1E AvA1E A 3000011619 25820 kPa kgmA3E State 2 xA2E A 1 PA2E A CvA2E A on process line P v 1 2 Trial error on TA2satE A or PA2satE A Here from B12 at 2 MPa vAgE A 009963 C 20074 low 25 MPa vAgE A 007998 C 31258 high 225 MPa vAgE A 008875 C 25352 low Interpolate to get the right C PA2E A 2270 kPa 132 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2117 A container with liquid nitrogen at 100 K has a cross sectional area of 05 m2 Due to heat transfer some of the liquid evaporates and in one hour the liquid level drops 30 mm The vapor leaving the container passes through a valve and a heater and exits at 500 kPa 260 K Calculate the volume rate of flow of nitrogen gas exiting the heater Solution Properties from table B61 for volume change exit flow from table B62 V A h 05 003 0015 mA3E mAliqE A VvAfE A 00150001452 103306 kg mAvapE A VvAgE A 001500312 04808 kg mAoutE A 103306 04808 985 kg vAexitE A 015385 m3kg AV E A AmE AvAexitE A 985 1 h 015385 m3kg 15015 m3h 002526 m3min 133 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2118 For a certain experiment R410A vapor is contained in a sealed glass tube at 20C It is desired to know the pressure at this condition but there is no means of measuring it since the tube is sealed However if the tube is cooled to 20C small droplets of liquid are observed on the glass walls What is the initial pressure Solution Control volume R410A fixed volume V mass m at 20C Process cool to 20C at constant v so we assume saturated vapor State 2 vA2E A vAg at 20CE A 006480 mA3E Akg State 1 20C vA1E A vA2E A 006480 mA3E Akg interpolate between 400 and 500 kPa in Table B42 PA1E A 485 kPa T CP v P CP v 20 C 400 1444 20 20 500 kPa 500 kPa 400 kPa 2 P 1 134 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2119 A cylinderpiston arrangement contains water at 105C 85 quality with a volume of 1 L The system is heated causing the piston to rise and encounter a linear spring as shown in Fig P2119 At this point the volume is 15 L piston diameter is 150 mm and the spring constant is 100 Nmm The heating continues so the piston compresses the spring What is the cylinder temperature when the pressure reaches 200 kPa Solution PA1E A 1208 kPa vA1E A vf x vfg 0001047 085141831 120661 m VA1E A vA1E A A 0001 120661E A 828810A4E A kg vA2E A vA1E A VA2E A VA1E A 120661 15 18099 P PA1E A 1208 kPa TA2E A 2035C PA3E A PA2E A AksAp 2E A mvA3E A vA2E A linear spring AApE A π4 015A2E A 001767 mA2E A ks 100 kNm matches P in kPa 200 1208 100001767 2 828810A4E AvA3E A 18099 200 1208 265446 vA3E A 18099 vA3E A 21083 m3kg TA3E A 600 100 21083 20129722443201297 641C P v 1 2 3 200 1 15 liters 135 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2120 Determine the mass of methane gas stored in a 2 m3 tank at 30C 2 MPa Estimate the percent error in the mass determination if the ideal gas model is used Solution Table B7 Methane Table B71 at 30C 24315 K Tc 1906 K so superheated vapor in Table B72 Linear interpolation between 225 and 250 K at 2 MPa v 005289 A24315225 250225E A 006059 005289 005848 mA3E Akg m Vv 2005848 342 kg Ideal gas assumption v ART PE A A051835 24315 2000E A 006302 mA3E Akg m AV vE A A 2 006302E A 3174 kg Error m 246 kg ideal gas 72 too small Comment The compressibility of the methane Z 093 136 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2121 A cylinder containing ammonia is fitted with a piston restrained by an external force that is proportional to cylinder volume squared Initial conditions are 10C 90 quality and a volume of 5 L A valve on the cylinder is opened and additional ammonia flows into the cylinder until the mass inside has doubled If at this point the pressure is 12 MPa what is the final temperature Solution State 1 Table B21 vA1E A 00016 090205525 00016 018513 m3kg PA1E A 615 kPa VA1E A 5 L 0005 m3 mA1E A Vv 0005018513 0027 kg State 2 PA2E A 12 MPa Flow in so mA2E A 2 mA1E A 0054 kg Process Piston FAextE A KVA2E A PA P CVA2E A PA2E A PA1 E AVA2E AVA1E A2 From the process equation we then get VA2E A VA1E A AP2P1E AA 12E A 0005 A1200 615E AA 12E A 0006984 mA3E vA2E A Vm A0006984 0054E A 012934 mA3E Akg At PA2E A vA2E A TA2E A 709C 137 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2122 A cylinder has a thick piston initially held by a pin as shown in Fig P2122 The cylinder contains carbon dioxide at 200 kPa and ambient temperature of 290 K The metal piston has a density of 8000 kgm3 and the atmospheric pressure is 101 kPa The pin is now removed allowing the piston to move and after a while the gas returns to ambient temperature Is the piston against the stops Solution Force balance on piston determines equilibrium float pressure Piston mApE A AApE A l ρ ρApistonE A 8000 kgm3 PAext on CO2 E A PA0E A A mpg EAp E A 101 A Ap 01 9807 8000 EAp 1000E A 1088 kPa Pin released as PA1E A PAextE A piston moves up TA2E A To if piston at stops then VA2E A VA1E A ΗA2E AΗA1E A VA1E A 150 100 Ideal gas with TA2E A TA1E A then gives PA2E A PA1E A VA1E A VA2E A 200 A100 150E A 133 kPa PAextE piston is at stops and PA2E A 133 kPa 138 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2123 What is the percent error in pressure if the ideal gas model is used to represent the behavior of superheated vapor R410A at 60C 003470 m3kg What if the generalized compressibility chart Fig D1 is used instead iterations needed Solution Real gas behavior P 1000 kPa from Table B42 Ideal gas constant R AR E AM 83145172585 01146 kJkg K P RTv 01146 27315 60 00347 1100 kPa which is 10 too high Generalized chart Fig D1 and critical properties from A2 TArE A 333227315 713 0967 PAcE A 4901 kPa Assume P 1000 kPa PArE A 0204 Z 092 v ZRTP 092 01146 kJkgK 33315 K 1000 kPa 003512 m3kg too high Assume P 1050 kPa PArE A 0214 Z 0915 v ZRTP 0915 01146 kJkgK 33315 K 1050 kPa 003327 m3kg too low P 1000 1050 1000 A003470 003512 003327 003512E A 1011 kPa 11 high 139 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2124 An initially deflated and flat balloon is connected by a valve to a 12 m3 storage tank containing helium gas at 2 MPa and ambient temperature 20C The valve is opened and the balloon is inflated at constant pressure Po 100 kPa equal to ambient pressure until it becomes spherical at D1 1 m If the balloon is larger than this the balloon material is stretched giving a pressure inside as P PA0E A C 1 D1 D D1 D The balloon is inflated to a final diameter of 4 m at which point the pressure inside is 400 kPa The temperature remains constant at 20C What is the maximum pressure inside the balloon at any time during this inflation process What is the pressure inside the helium storage tank at this time Solution At the end of the process we have D 4 m so we can get the constant C as P 400 PA0E A C 1 A1 4E A A1 4E A 100 C 316 C 1600 kPa The pressure is P 100 1600 1 X 1 X 1 X D DA1E Differentiate to find max AdP dDE A C X 2 2 X 3 DA1E A 0 X 2 2 X 3 0 X 2 at max P D 2DA1E A 2 m V Aπ 6E A D3 418 m3 Pmax 100 1600 1 A1 2E A A1 2E A 500 kPa Helium is ideal gas A5 m APV RTE A A 500 4189 20771 29315E A 344 kg mTANK 1 APV RTE A A 2000 12 20771 29315E A 39416 kg mTANK 2 39416 344 35976 kg PT2 mTANK 2 RTV mTANK 1 mTANK 2 P1 18255 kPa 140 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2125 A pistoncylinder arrangement shown in Fig P2125 contains air at 250 kPa 300C The 50kg piston has a diameter of 01 m and initially pushes against the stops The atmosphere is at 100 kPa and 20C The cylinder now cools as heat is transferred to the ambient a At what temperature does the piston begin to move down b How far has the piston dropped when the temperature reaches ambient Solution Piston AApE A Aπ 4E A 01A2E A 000785 mA2E Balance forces when piston floats PAfloatE A Po A mpg EAp E A 100 A 50 9807 000785 1000E A 1625 kPa PA2E A PA3E A To find temperature at 2 assume ideal gas TA2E A TA1E A A P2 EP1 E A 57315 A1625 250E A 3725 K b Process 2 3 is constant pressure as piston floats to TA3E A To 29315 K VA2E A VA1E A AApE A H 000785 025 000196 mA3E A 196 L Ideal gas and PA2E A PA3E A VA3E A VA2E A A T3 ET2 E A 196 A29315 3725E A 154 L H V2 V3A 196154 0001000785 0053 m 53 cm 141 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Linear Interpolation 142 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2126 Find the pressure and temperature for saturated vapor R410A with v 01 mA3E Akg Solution Table B41 Look at the saturated vapor column vAgE A and it is found between 35AE AC and 30AE AC We must then do a linear interpolation between these values T 35 30 35 A 01 011582 009470 011582E A 35 5 0749 313AE AC P 2184 2696 2184 0749 2567 kPa v T 2 1 35 30 009470 011582 01 v P 2 1 009470 011582 01 2184 2696 To understand the interpolation equation look at the smaller and larger triangles formed in the figure The ratio of the side of the small triangle in v as 011582 01 to the side of the large triangle 011582 009470 is equal to 0749 This fraction of the total P 2696 2184 or T 30 35 is added to the lower value to get the desired interpolated result 143 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2127 Use a linear interpolation to estimate properties of ammonia to fill out the table below P kPa T AE AC v m3kg x a 550 075 b 80 20 c 10 04 Solution a Find the pressures in Table B21 that brackets the given pressure T 5 10 5 A 550 5159 6152 5159E A 5 5 0341 67 AE AC vAfE A 0001583 00016 0001583 0341 0001589 mA3E Akg vAgE A 024299 020541 024299 0341 0230175 mA3E Akg v vAfE A xvAfgE A 0001589 0750230175 0001589 01729 mA3E Akg b Interpolate between 50 and 100 kPa to get properties at 80 kPa v 28466 14153 28466 A 80 50 100 50E A 28466 14313 06 19878 mA3E Akg x Undefined c Table B21 v vAgE A so the state is superheated vapor Table B22 locate state between 300 and 400 kPa P 300 400 300 A 04 044251 032701 044251E 300 100 0368 3368 kPa x Undefined 144 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2128 Use a linear interpolation to estimate Tsat at 900 kPa for nitrogen Sketch by hand the curve PsatT by using a few table entries around 900 kPa from table B61 Is your linear interpolation over or below the actual curve Solution The 900 kPa in Table B61 is located between 100 and 105 K T 100 105 100 A 900 7792 10846 7792E A 100 5 03955 102 K The actual curve has a positive second derivative it curves up so T is slightly underestimated by use of the chord between the 100 K and the 105 K points as the chord is above the curve P T 100 105 110 7792 10846 14676 900 145 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2129 Use a double linear interpolation to find the pressure for superheated R134a at 13AE AC with v 03 mA3E Akg Solution Table B52 Superheated vapor At 10AE AC 03 mA3E Akg P 50 100 50 A 03 045608 022527 045608E A 838 kPa At 20AE AC 03 mA3E Akg P 50 100 50 A 03 047287 023392 047287E A 862 kPa Interpolating to get 13AE AC between the 10AE AC and 20AE AC Ps above P 838 310 862 838 845 kPa This could also be interpolated as following At 13AE AC 50 kPa v 045608 310 00168 04611 mA3E Akg At 13AE AC 100 kPa v 022527 310 00087 02279 mA3E Akg Interpolating at 03 mA3E Akg P 50 100 50 A 03 04611 02279 04611E A 845 kPa 146 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2130 Find the specific volume for COA2E A at 0AoE AC and 625 kPa Solution The state is superheated vapor in Table B32 between 400 and 800 kPa v 012552 006094 012552 A625 400 800 400E 012552 006458 05625 00892 mA3E Akg v P 400 625 800 012552 006094 147 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Computer Tables 148 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2131 Use the computer software to find the properties for water at the 4 states in Problem 235 Start the software click the tab for water as the substance and click the small calculator icon Select the proper CASE for the given properties CASE RESULT a 1 T P Compressed liquid x undefined v 0001002 m3kg b 5 P v Twophase T 1519C x 05321 c 1 T P Sup vapor x undefined v 0143 m3kg d 4 T x P Psat 8581 kPa v 001762 m3kg 149 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2132 Use the computer software to find the properties for ammonia at the 2 states listed in Problem 232 Start the software click the tab for cryogenic substances and click the tab for the substance ammonia Then click the small calculator icon and select the proper CASE for the given properties CASE RESULT a 2 T v Sup vapor x undefined P 1200 kPa b 4 T x Twophase P 2033 kPa v 003257 m3kg 150 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2133 Use the computer software to find the properties for ammonia at the 3 states listed in Problem 2127 Start the software click the tab for cryogenic substances select ammonia and click the small calculator icon Select the proper CASE for the given properties CASE RESULT a 8 P x T 6795C v 01719 m3kg b 1 T P Sup vapor x undefined v 1773 m3kg c 2 T v Sup vapor x undefined P 3304 kPa 151 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2134 Find the value of the saturated temperature for nitrogen by linear interpolation in table B61 for a pressure of 900 kPa Compare this to the value given by the computer software The 900 kPa in Table B61 is located between 100 and 105 K T 100 105 100 900 7792 10846 7792 100 5 03955 10198 K The actual curve has a positive second derivative it curves up so T is slightly underestimated by use of the chord between the 100 K and the 105 K points as the chord is above the curve From the computer software CASE 8 Px T 171C 10215 K So we notice that the curvature has only a minor effect P T 100 105 110 7792 10846 14676 900 152 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2135 Use the computer software to sketch the variation of pressure with temperature in Problem 244 Extend the curve a little into the singlephase region P was found for a number of temperatures A small table of P T values were entered into a spreadsheet and a graph made as shown below The superheated vapor region is reached at about 140C and the graph shows a small kink at that point T P 80 130 180 230 280 330 380 430 100 110 120 130 140 150 160 ENGLISH UNIT PROBLEMS SOLUTION MANUAL CHAPTER 2 English Units Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CHAPTER 2 SUBSECTION PROB NO ConceptStudy Guide Problems 136140 Phase diagrams 141142 General Tables 143158 Ideal Gas 159166 Review Problems 167169 Compressibility Factor 170171 Equations of state 172173 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Concept Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2136E Cabbage needs to be cooked boiled at 250 F What pressure should the pressure cooker be set for Solution If I need liquid water at 250 F I must have a pressure that is at least the saturation pressure for this temperature Table F71 250 F Psat 29823 psia The pot must have a lid that can be fastened to hold the higher pressure which is a pressure cooker Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2137E If I have 1 ft3 of ammonia at 15 psia 60 F how much mass is that Ammonia Tables F8 F81 Psat 10764 psia at 60 F so superheated vapor F82 v 215641 ft3lbm under subheading 15 psia m V v 215641 ft3lbm 1 ft3 00464 lbm T CP v 1076 psia P CP v T 15 108 273 60 273 F 60 F 15 psia The Pv loglog diagram from CATT3 P in psi and v in ft3lbm Crosshair indicates the state Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2138E For water at 1 atm with a quality of 10 find the volume fraction of vapor This is a twophase state at a given pressure Table F72 vf 001 672 ft3lbm vg 268032 ft3lbm From the definition of quality we get the masses from total mass m as mf 1 x m mg x m The volumes are Vf mf vf 1 x m vf Vg mg vg x m v g So the volume fraction of vapor is Fraction Vg V Vg Vg Vf x m vg x m vg 1 xm vf 01 268032 01 268032 09 0016 72 268032 269537 09944 Notice that the liquid volume is only about 05 of the total We could also have found the overall v vf xvfg and then V m v Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2139E Locate the state of R410A at 30 psia 20 F Indicate in both the Pv and the Tv diagrams the location of the nearest states listed in the printed table F9 From F9 F91 at 20 F Psat 40923 psi so we have superheated vapor F92 at 30 psi we find the state for 20 F T CP v 409 psia P CP v T 30 409 33 20 33 F 20 F 30 20 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2140E Calculate the ideal gas constant for argon and hydrogen based on Table F1 and verify the value with Table F4 The gas constant for a substance can be found from the universal gas constant from table A1 and the molecular weight from Table F1 Argon R R M 198589 39948 004971 Btu lbm R 38683 lbfft lbm R Hydrogen R R M 198589 2016 098506 Btu lbm R 7665 lbfft lbm R Recall from Table A1 1 Btu 7781693 lbfft Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Phase Diagrams Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2141E Water at 80 F can exist in different phases dependent on the pressure Give the approximate pressure range in lbfin2 for water being in each one of the three phases vapor liquid or solid Solution The phases can be seen in Fig 24 a sketch of which is shown to the right T 80 F 540 R 300 K From Fig 24 PVL 4 103 MPa 4 kPa 058 psia PLS 103 MPa 145 038 psia ln P T V L S CRP S 0 P 058 psia VAPOR 058 psia P 145 038 psia LIQUID P 145 038 psia SOLIDICE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2142E A substance is at 300 lbfin2 65 F in a rigid tank Using only the critical properties can the phase of the mass be determined if the substance is oxygen water or propane Solution Find state relative to the critical point properties Table F1 a Oxygen 731 lbfin2 2783 R b Water 3208 lbfin2 11651 R c Propane 616 lbfin2 6656 R P Pc for all and T 65 F 65 45967 525 R a O2 T Tc Yes gas and P Pc b H2O T Tc P Pc so you cannot say c C3H8 T Tc P Pc you cannot say ln P T Vapor Liquid CrP a c b Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful General Tables Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2143E Determine the missing property of P T v and x if apllicable for water at a 680 lbfin2 003 ft3lbm b 150 lbfin2 320 F c 400 F 3 ft3lbm Solution All cases can be seen from Table F71 a 680 lbfin2 003 ft3lbm vg 02183 vf 002472 ft3lbm so liquid vapor mixture b 150 lbfin2 320 F compressed liquid P PsatT 896 lbfin2 c 400 F 3 ft3lbm sup vapor v vgT 2339 ft3lbm States shown are placed relative to the twophase region not to each other P CP v T CP v T a c b a c b P const P T v V L S CP a b c Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2144E Determine the phase of the substance at the given state using Appendix F Tables Solution a water 200 F 70 psia F71 at 200 F P PsatT 1153 psia compressed liquid b Ammonia 10 F 20 lbfin2 F81 at 10 F P PsatT 38508 lbfin2 sup vapor F82 Superheated by 20 1663 F 366 F v 147635 ft3lbm c R410A 30 F 50 ft3lbm F91 at 30 F P PsatT 1118 lbfin2 sup vapor F92 Interpolate to find v The SL fusion line goes slightly to the left for water It tilts slightly to the right for most other substances ln P T Vapor L CrP a bc S States shown are placed relative to the twophase region not to each other P CP v T CP v T a bc a b c P const Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2145E Give the phase and the missing property of P T v and x for R134a at a T 10 F P 18 psia b P 40 psia v 13 ft3lbm Solution a Look in Table F101 at 10 F P Psat 1676 psia This state is compressed liquid so x is undefined and v vf 001173 ft3lbm b Look in Table F101 close to 50 psia there we see v vg 095 ft3lbm so superheated vapor Look then in Table F102 under 40 psia and interpolate between the 60 F and 80 F 40 psia 13 ft3lbm T 666 F For a better accuracy use the computer software CATT3 which gives T 674 F T CP v a b P const P v a T b Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2146E Give the phase and the missing property of P T v and x for ammonia at a T 120 F v 1876 ft3lbm b T 120 F x 05 Solution a Look in Table F81 at 120 F v vg 10456 ft3lbm so sup vap x undefined F82 P 175 psia b Look in Table F81 at 120 F v vf x vfg 002836 05 10172 053696 ft3lbm T CP v a b P const P v a T b 120 286 175 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2147E Give the phase and the specific volume Solution a R410A T 25 F P 30 lbfin2 Table F92 T Tsat 3324 F supvap v 19534 253324 203324 20347 19534 2004 ft3lbm b R410A T 25 F P 40 lbfin2 Table F92 T Tsat 210 F P Psat compresssed Liquid F91 v vf 001246 ft3lbm c H2O T 280 F P 35 lbfin2 Table F71 P Psat 492 psia superheated vapor v 21734 10711 21734 1520 10669 ft3lbm d NH3 T 60 F P 15 lbfin2 Table F81 Psat 1076 psia P Psat superheated vapor v 21564 ft3lbm States shown are placed relative to the twophase region not to each other P CP v T CP v T a c d b a c d b P const Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2148E Determine the specific volume for R410A at these states a 20 F 70 psia b 70 F 150 psia c 70 F quality 25 a F92 P Psat 9313 psia so sup vapor v 10783 70 60 75 60 08393 10783 09190 ft3lbm b F91 P Psat 21595 psia so superheated vapor F92 v 04236 70 60 80 60 04545 04236 043905 ft3lbm c F91 vf 001486 ft3lbm vfg 02576 ft3lbm v vf x vfg 001486 025 02576 007926 ft3lbm States shown are placed relative to the twophase region not to each other P CP v T CP v T c ab c ab P const Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2149E Give the missing property of P T v and x for a R410A at 80 F v 02 ft3lbm b R410A at 60 psia v 11 ft3lbm c Ammonia at 60 F v 32 ft3lbm a Table F91 at 80 F v vg 02308 ft3lbm so we have 2 phase LV x v vf vfg 02 001525 02156 08569 P Psat 250665 psia b Table F92 at 60 psia v vg 10038 ft3lbm so we have superheated vapor so x is undefined F92 between 20 and 40 F T 20 20 11 10783 11405 10783 2698 F c Table F81 at 60 F v vg 27481 ft3lbm so we have superheated vapor F82 between 90 and 100 psia P 90 10 32 33503 29831 33503 9409 psia States shown are placed relative to the twophase region not to each other P CP v T CP v T a bc a bc P const Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2150E Saturated liquid water at 150 F is put under pressure to decrease the volume by 1 while keeping the temperature constant To what pressure should it be compressed F71 v vf 001634 ft3lbm New v v 099 vf 00161766 ft3lbm look in F73 close to 2000 psia interpolate between 2000 and 8000 psia P 2610 psia P CP v T v 2000 psia 1 2 1 2 8000 psia Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2151E A sealed rigid vessel has volume of 35 ft3 and contains 2 lbm of water at 200 F The vessel is now heated If a safety pressure valve is installed at what pressure should the valve be set to have a maximum temperature of 400 F Solution Process v Vm constant State 1 v1 352 175 ft3lbm from Table F71 it is 2phase State 2 400F 175 ft3lbm Table F72 between 20 and 40 lbfin2 so interpolate CP T v 200 F 40 lbfin 20 lbfin2 2 P 20 40 20 175 25427 12623 25427 324 lbfin2 The result is 2897 psia if found by the software ie linear interpolation is not so accurate Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2152E You want a pot of water to boil at 220 F How heavy a lid should you put on the 6 inch diameter pot when Patm 147 psia Solution Table F71 at 220 F Psat 17189 psia A π 4 D2 π 4 62 28274 in2 Fnet Psat Patm A 17189 147 lbf in2 28274 in2 70374 lbf Fnet mlid g mlid Fnetg 70374 lbf 32174 fts2 32174 fts2 70374 32174 lbm fts2 70374 lbm Some lids are clamped on the problem deals with one that stays on due to its weight Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2153E Saturated water vapor at 240 F has its pressure decreased to increase the volume by 10 keeping the temperature constant To what pressure should it be expanded Solution Initial state v 163257 ft3lbm from table F71 Final state v 11 vg 11 163257 179583 ft3lbm Interpolate between sat at 240 F P 24968 lbfin2 and sup vapor in Table F72 at 240 F 20 lbfin2 P 24968 20 24968 179583 163257 20475 163257 230 lbfin2 P CP v T CP v T 240 F 2497 psi 20 psi Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2154E A glass jar is filled with saturated water at 300 F and quality 25 and a tight lid is put on Now it is cooled to 10 F What is the mass fraction of solid at this temperature Solution Constant volume and mass v1 v2 Vm From Table F71 v1 001745 025 64537 1630875 ft3lbm From Table F74 v2 001744 x2 9043 v1 1630875 ft3lbm x2 0000178 mass fraction vapor xsolid 1 x2 09998 or 9998 P CP v T CP v T 1 2 1 2 P T v V L S CP 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2155E A boiler feed pump delivers 100 ft3min of water at 400 F 3000 lbfin2 What is the mass flowrate lbms What would be the percent error if the properties of saturated liquid at 400 F were used in the calculation What if the properties of saturated liquid at 3000 lbfin2 were used Solution Table F73 v 00183 ft3lbm interpolate 20008000 psia m v V 100 60 0018334 9107 lbms vf 400 F 001864 ft3lbm m 8941 lbms error 18 vf 3000 lbfin2 003475 ft3lbm m 4796 lbms error 47 P CP v T CP v P 3000 psia 400 F 3000 400 247 695 The constant T line is nearly vertical for the liquid phase in the Pv diagram The state is at so high P T that the saturated liquid line is not extremely steep Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2156E A pressure cooker has the lid screwed on tight A small opening with A 00075 in2 is covered with a petcock that can be lifted to let steam escape How much mass should the petcock have to allow boiling at 250 F with an outside atmosphere at 15 psia Solution Table F71 at 250 F Psat 29823 psia Fnet Psat Patm A 29823 15 psia 00075 in2 0111 lbf Fnet mpetcock g mpetcock Fnetg 0111 lbf 32174 fts2 32174 fts2 0111 32174 lbm fts2 0111 lbm Some petcocks are held down by a spring the problem deals with one that stays on due to its weight Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2157E Two tanks are connected together as shown in Fig P252 both containing water Tank A is at 30 lbfin2 v 8 ft3lbm V 40 ft3 and tank B contains 8 lbm at 80 lbfin 2 750 F The valve is now opened and the two come to a uniform state Find the final specific volume Solution Control volume both tanks Constant total volume and mass process A B sup vapor State A1 P v twophase mA VAvA 40 ft3 8 ft3lbm 5 lbm State B1 P T Table F72 vB 8561 93222 89415 ft3lbm VB mBvB 8 lbm 89415 ft3lbm 71532 ft3 Final state mtot mA mB 5 8 13 lbm Vtot VA VB 111532 ft3 v2 Vtotmtot 111532 ft313 lbm 8579 ft3lbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2158E Ammonia at 70 F with a quality of 50 and total mass 45 lbm is in a rigid tank with an outlet valve at the bottom How much liquid mass can you take out through the valve assuming the temperature stays constant Solution The bottom has liquid until the state inside becomes saturated vapor V m1v1 45 05 002631 23098 525625 ft3 m2 Vv2 525625 ft3 23098 ft3lbm 22756 lbm m m1 m2 45 22756 2224 lbm P CP v T CP v T P 129 psia 70 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Ideal Gas Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2159E Give the phase and the specific volume for each of the following Solution a CO2 T 510 F P 75 lbfin2 Table F4 superheated vapor ideal gas v RTP 351 ftlbflbmR 510 4597 R 75 lbfin2 144 ft in 2 3152 ft3lbm b Air T 68 F P 2 atm Table F4 superheated vapor ideal gas v RTP 5334 ftlbflbmR 68 4597 R 2 146 lbfin2 144 ft in 2 66504 ft3lbm c Ar T 300 F P 30 lbfin2 Table F4 Ideal gas v RTP 3868 ftlbflbmR 300 4597 R 30 lbfin2 144 ft in 2 6802 ft3lbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2160E A cylindrical gas tank 3 ft long inside diameter of 8 in is evacuated and then filled with carbon dioxide gas at 77 F To what pressure should it be charged if there should be 26 lbm of carbon dioxide Solution Assume CO2 is an ideal gas table F4 P mRTV Vcyl A L π 4 8 in2 3 12 in 18096 in 3 P 26 lbm 351 ftlbmlbmR 77 45967 R 12 inft 18096 in3 3248 lbfin2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2161E A spherical helium balloon of 30 ft in diameter is at ambient T and P 60 F and 1469 psia How much helium does it contain It can lift a total mass that equals the mass of displaced atmospheric air How much mass of the balloon fabric and cage can then be lifted We need to find the masses and the balloon volume V π 6 D3 π 6 303 14 137 ft3 mHe ρV V v PV RT 1469 psi 14 137 ft3 144 inft2 3860 ftlbflbmR 520 R 14899 lbm mair PV RT 1469 psi 14 137 ft3 144 inft2 5334 ftlbflbmR 520 R 1078 lbm mlift mair mHe 1078 149 929 lbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2162E Helium in a steel tank is at 36 psia 540 R with a volume of 4 ft3 It is used to fill a balloon When the pressure drops to 18 psia the flow of helium stops by itself If all the helium still is at 540 R how big a balloon is produced Solution State 1 m Vv assume ideal gas so m P1V1 RT1 36 psi 4 ft3 144 inft2 386 ftlbflbmR 540 R 00995 lbm State 2 Same mass so then T2 T1 V2 mRT2 P2 P1V1 RT2 RT1 P2 V1 P1 P2 4 36 18 8 ft 3 c i r c u s t h e r m o cb The balloon volume is Vballoon V2 V1 8 4 4 ft3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2163E A 35 ft3 rigid tank has propane at 15 psia 540 R and connected by a valve to another tank of 20 ft3 with propane at 40 psia 720 R The valve is opened and the two tanks come to a uniform state at 600 R What is the final pressure Solution Propane is an ideal gas P Pc with R 3504 ftlbflbm R from Tbl F4 mA PAVA RTA 15 35 144 3504 540 3995 lbm m PBVB RTB 40 20 144 3504 720 4566 lbm V2 VA VB 55 ft3 m2 mA mB 8561 lbm P2 m2RT2 V2 8561 3504 600 55 144 22726 psia Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2164E What is the percent error in specific volume if the ideal gas model is used to represent the behavior of superheated ammonia at 100 F 80 lbfin2 What if the generalized compressibility chart Fig D1 is used instead Solution Ammonia Table F82 v 4186 ft3lbm Ideal gas v RT P 9072 5597 80 144 44076 ft3lbm 53 error Generalized compressibility chart and Table D4 Tr 55977299 0767 Pr 801646 00486 Z 096 v ZRT P 096 44076 ft3lbm 4231 ft3lbm 10 error Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2165E Air in an internal combustion engine has 440 F 150 psia with a volume of 2 ft3 Now combustion heats it to 3000 R in a constant volume process What is the mass of air and how high does the pressure become The mass comes from knowledge of state 1 and ideal gas law m P1V1 RT1 150 psia 2 ft3 144 inft 2 5334 ftlbflbmR 440 45967 R 090 lbm The final pressure is found from the ideal gas law written for state 1 and state 2 and then eliminate the mass gas constant and volume V2 V1 between the equations P1 V1 m RT1 and P2 V2 m RT 2 P2 P1 T2T1 150 psia 3000 440 45967 5002 psia Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2166E A 35 ft3 rigid tank has air at 225 psia and ambient 600 R connected by a valve to a piston cylinder The piston of area 1 ft2 requires 40 psia below it to float Fig P2166 The valve is opened and the piston moves slowly 7 ft up and the valve is closed During the process air temperature remains at 600 R What is the final pressure in the tank mA RTA PAVA 5334 ftlbflbmR 600 R 225 psia 35 ft3 144 inft 2 35433 lbm mB2 mB1 VA vB VBPB RT 40 psia 1 7 ft3 144 inft 2 5334 ftlbflbmR 600 R 126 lbm mA2 mA mB2 mB1 35433 126 34173 lbm PA2 VA mA2RT 34173 5334 600 35 144 217 psia Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2167E Give the phase and the missing properties of P T v and x These may be a little more difficult if the appendix tables are used instead of the software Solution a R410A at T 50 F v 04 ft3lbm Table F91 v vg 03818 ft3lbm sup vap F92 interpolate between sat and sup vap at 50 F Find v at 150 psia 50 F v 04066 Now interpolate between the sat vap 157473 psia and the 150 psia P 150 7473 04 0406603818 04066 152 lbfin2 b H2O v 2 ft3lbm x 05 Table F71 since vf is so small we find it approximately where vg 4 ft3lbm vf vg 43293 at 330 F vf vg 380997 at 340 F linear interpolation T 336 F P 113 lbfin2 c H2O T 150 F v 001632 ft3lbm Table F71 v vf compr liquid P 500 lbfin2 d NH3 T 80 F P 13 lbfin2 Table F81 P Psat sup vap interpolate between 10 and 15 psia v 2697 ft3lbm v is not linear in P more like 1P so computer table is more accurate e R134a v 008 ft3lbm x 05 Table F101 since vf is so small we find it approximately where vg 016 ft3lbm vf vg 01729 at 150 F vf vg 01505 at 160 F linear interpolation T 156 F P 300 lbfin2 States shown are placed relative to the twophase region not to each other P CP v T CP v T a b e a P const d b e c c d Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2168E A pressure cooker closed tank contains water at 200 F with the liquid volume being 110 of the vapor volume It is heated until the pressure reaches 300 lbfin2 Find the final temperature Has the final state more or less vapor than the initial state Solution Process Constant volume and mass Vf mf vf Vg10 mgvg10 Table F71 vf 001663 ft3lbm vg 33631 ft3lbm x1 mg mf mg mf 10 mfvf vg 10 mfvf vg 10 vf vg 10 vf 01663 01663 33631 000492 v2 v1 001663 x1 33615 01820 ft3lbm P2 v2 T2 Tsat300 psia 41743 F P v 2 1 At state 2 v2 vf x2 v fg 01820 001890 x2 15286 x2 0107 More vapor at final state Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2169E Refrigerant410A in a pistoncylinder arrangement is initially at 60 F x 1 It is then expanded in a process so that P Cv1 to a pressure of 30 lbfin2 Find the final temperature and specific volume Solution State 1 P1 18498 lbfin2 v1 03221 ft3lbm Process Pv C P1v1 P2v 2 State 2 P2 30 lbfin2 and on process line equation v2 P2 v1P1 0 3221 18498 30 19861 ft3lbm Table F92 between saturated at 3324 F and 20 F T2 279 F Notice T is not constant v P v T 1 2 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Compressiblity Factor Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2170E A substance is at 70 F 300 lbfin2 in a 10 ft3 tank Estimate the mass from the compressibility chart if the substance is a air b butane or c propane Solution Use Fig D1 for compressibility Z and table F1 for critical properties m PV ZRT 300 144 10 530 ZR 81509 ZR Air use nitrogen Pc 492 lbfin2 Tc 2272 R Pr 061 Tr 233 Z 098 m PV ZRT 81509 ZR 81509 098 5515 1508 lbm Butane Pc 551 lbfin2 Tc 7654 R Pr 0544 Tr 0692 Z 009 m PV ZRT 81509 ZR 81509 009 2658 3407 lbm Propane Pc 616 lbfin2 Tc 6656 R Pr 0487 Tr 0796 Z 008 m PV ZRT 81509 ZR 81509 008 3504 2908 lbm ln Pr Z T 20 r a b c T 07 r T 07 r 01 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2171E Determine the mass of an ethane gas stored in a 25 ft3 tank at 250 F 440 lbfin2 using the compressibility chart Estimate the error if the ideal gas model is used Solution Table F1 Tr 250 460 5497 129 and Pr 440708 0621 Figure D1 Z 09 m PV ZRT 09 5138 ftlbflbmR 710 R 440 psia 25 ft3 144 inft 2 4825 lbm Ideal gas Z 1 m PVRT 4321 lbm 10 error Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Equations of State Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2172E Determine the pressure of R410A at 100 F v 02 ft3lbm using ideal gas and the van der Waal Equation of State From F1 for R410A M 72585 Tc 6201 R Pc 711 psia For ideal gas we get R R M 154536 72585 2129 lbfftlbmR P RTv 2129 100 45967 02 144 lbfft lbmR R ft3lbm inft2 41373 psi For van der Waal equation of state from Table D1 we have b 1 8 RTc Pc 0125 2129 6201 711 144 0016 118 ft3lbm a 27 b2 Pc 27 0016 1182 711 4987 283 psi ft3lbm 2 The EOS is P RT v b a v2 2129 55967144 02 0016 118 4987 283 022 3253 psi From Table F92 we see it is around 300 psia Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2173E Determine the pressure of R410A at 100 F v 02 ft3lbm using ideal gas and the RedlichKwong Equation of State From F1 for R410A M 72585 Tc 6201 R Pc 711 psia For ideal gas we get R R M 154536 72585 2129 lbfftlbmR P RTv 2129 100 45967 02 144 lbfft lbmR R ft3lbm inft2 41373 psi For RedlichKwong EOS we have the parameters from Table D1 Tr TTc 100 45967 6201 090255 b 008664 RTc Pc 008664 2129 6201 711 144 0011 172 ft3lbm a 042748 T E09025512 711 1442 E A 12 r 042748 R2T2 c Pc 21292 62012 5319 372 psi ftA3E AlbmA2E The equation is P A RT v bE A A a v2 bvE A A2129 55967 144 02 0011 172E A A 5319 372 022 0011 172 02E A 312 psia From Table F92 we see it is around 300 psia Updated June 2013 SOLUTION MANUAL CHAPTER 3 Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CONTENT CHAPTER 3 SUBSECTION PROB NO Concept problems 127 Kinetic and potential energy 2836 Force displacement work 3744 Boundary work 4557 Heat transfer 5869 Properties u h from general tables 7081 Problem analysis 8288 Simple processes 89115 Specific heats solids and liquids 116126 Properties u h Cv Cp Ideal gas 127138 Specific heats ideal gas 139151 Polytropic process 152168 Multistep process all subtances 169184 Energy equation rate form 185199 General work 200209 More complex devices 210217 Review problems 218241 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful InText Concept Questions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3a In a complete cycle what is the net change in energy and in volume For a complete cycle the substance has no change in energy and therefore no storage so the net change in energy is zero For a complete cycle the substance returns to its beginning state so it has no change in specific volume and therefore no change in total volume 3b Explain in words what happens with the energy terms for the stone in Example 33 What would happen if it were a bouncing ball falling to a hard surface In the beginning all the energy is potential energy associated with the gravitational force As the stone falls the potential energy is turned into kinetic energy and in the impact the kinetic energy is turned into internal energy of the stone and the water Finally the higher temperature of the stone and water causes a heat transfer to the ambient until ambient temperature is reached With a hard ball instead of the stone the impact would be close to elastic transforming the kinetic energy into potential energy the material acts as a spring that is then turned into kinetic energy again as the ball bounces back up Then the ball rises up transforming the kinetic energy into potential energy mgZ until zero velocity is reached and it starts to fall down again The collision with the floor is not perfectly elastic so the ball does not rise exactly up to the original height losing a little energy into internal energy higher temperature due to internal friction with every bounce and finally the motion will die out All the energy eventually is lost by heat transfer to the ambient or sits in lasting deformation internal energy of the substance Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3c Make a list of at least 5 systems that store energy explaining which form of energy A spring that is compressed Potential energy 12 kx E 2 A battery that is charged Electrical potential energy V Amp h A raised mass could be water pumped up higher Potential energy mgH A cylinder with compressed air Potential internal energy like a spring A tank with hot water Internal energy mu A flywheel Kinetic energy rotation 12 IωAA2E A mass in motion Kinetic energy 12 mVA2E 3d A constant mass goes through a process where 100 J of heat transfer comes in and 100 J of work leaves Does the mass change state Yes it does As work leaves a control mass its volume must go up v increases As heat transfer comes in an amount equal to the work out means u is constant if there are no changes in kinetic or potential energy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3e The electric company charges the customers per kWhour What is that in SI units Solution The unit kWhour is a rate multiplied with time For the standard SI units the rate of energy is in W and the time is in seconds The integration in Eq34 and on page 135 becomes 1 kW hour 1000 W 60 Amin hourE A hour 60 A s minE A 3 600 000 Ws 3 600 000 J 36 MJ 3f Torque and energy and work have the same units N m Explain the difference Solution Work force displacement so units are N m Energy in transfer Energy is stored could be from work input 1 J 1 N m Torque force arm static no displacement needed Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3g What is roughly the relative magnitude of the work in the process 12c versus the process 12a shown in figure 315 By visual inspection the area below the curve 12c is roughly 50 of the rectangular area below the curve 12a To see this better draw a straight line from state 1 to point f on the axis This curve has exactly 50 of the area below it 3h Helium gas expands from 125 kPa 350 K and 025 mA3E A to 100 kPa in a polytropic process with n 1667 Is the work positive negative or zero The boundary work is W AP dVEA P drops but does V go up or down The process equation is PVAnE A C so we can solve for P to show it in a PV diagram P CVAnE as n 1667 the curve drops as V goes up we see VA2E A VA1E A giving dV 0 and the work is then positive P V W 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3i An ideal gas goes through an expansion process where the volume doubles Which process will lead to the larger work output an isothermal process or a polytropic process with n 125 The process equation is PVAnE A C The polytropic process with n 125 drops the pressure faster than the isothermal process with n 1 and the area below the curve is then smaller P V W 1 2 n 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3j Water is heated from 100 kPa 20AoE AC to 1000 kPa 200AoE AC In one case pressure is raised at T C then T is raised at P C In a second case the opposite order is done Does that make a difference for A1E AQA2E A and A1E AWA2E A Yes it does Both A1E AQA2E A and A1E AWA2E A are process dependent We can illustrate the work term in a Pv diagram P T V L CrP S 1000 a 20 200 1 2 100 T CP v a P v a 180 C 2 2 20 C 20 200 100 1 1000 200 C 100 1553 kPa 1000 1 b b In one case the process proceeds from 1 to state a along constant T then from a to state 2 along constant P The other case proceeds from 1 to state b along constant P and then from b to state 2 along constant T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3k A rigid insulated tank A contains water at 400 kPa 800AoE AC A pipe and valve connect this to another rigid insulated tank B of equal volume having saturated water vapor at 100 kPa The valve is opened and stays open while the water in the two tanks comes to a uniform final state Which two properties determine the final state Continuity eq mA2E A mA1AE A mA1BE A 0 mA2E A mA1AE A mA1BE Energy eq mA2E AuA2E A mA1AE AuA1AE A mA1BE AuA1BE A 0 0 Process Insulated A1E AQA2E A 0 Rigid VA2E A C VAAE A VABE A A1E AWA2E A 0 From continuity eq and process vA2E A VA2E AmA2E A A m1A Em2 E A vA1AE A A m1B Em2 E A vA1BE From energy eq uA2E A A m1A Em2 E A uA1AE A A m1B Em2 E A uA1BE Final state 2 vA2E A uA2E A both are the mass weighted average of the initial values Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3l To determine v or u for some liquid or solid is it more important that I know P or T T is more important v and u are nearly independent of P in the liquid and solid phases 3m To determine v or u for an ideal gas is it more important that I know P or T For v they are equally important v RTP but for u only T is important For an ideal gas u is a function of T only independent of P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3n I heat 1 kg of substance at a constant pressure 200 kPa 1 degree How much heat is needed if the substance is water at 10AoE AC steel at 25AoE AC air at 325 K or ice at 10AoE AC Heating at constant pressure gives recall the analysis in Section 39 page 109 A1E AQA2E A HA2E A HA1E A mhA2E A hA1E A m CApE A TA2E A TA1E A For all cases A1E AQA2E A 1 kg C 1 K Water 10AoE AC 200 kPa liquid so A4 C 418 kJkgK A1E AQA2E A 418 kJ Steel 25AoE AC 200 kPa solid so A3 C 046 kJkgK A1E AQA2E A 046 kJ Air 325 K 200 kPa gas so A5 CApE A 1004 kJkgK A1E AQA2E A 1004 kJ Ice 10AoE AC 200 kPa solid so A3 C 204 kJkgK A1E AQA2E A 204 kJ Comment For liquid water we could have interpolated hA2E A hA1E A from Table B11 and for ice we could have used Table B15 For air we could have used Table A7 If the temperature is very different from those given the tables will provide a more accurate answer Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Concept Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 31 What is 1 cal in SI units and what is the name given to 1 Nm Look in the conversion factor table A1 under energy 1 cal Int 41868 J 41868 Nm 41868 kg mA2E AsA2E This was historically defined as the heat transfer needed to bring 1 g of liquid water from 145AoE AC to 155AoE AC notice the value of the heat capacity of water in Table A4 1 Nm 1 J or Force times displacement energy in Joule Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 32 A car engine is rated at 110 kW What is the power in hp Solution The horsepower is an older unit for power usually used for car engines The conversion to standard SI units is given in Table A1 1 hp 07355 kW 7355 W 1 hp 07457 kW for the UK horsepower 110 kW 110 kW 07457 kWhp 1475 hp Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 33 Why do we write E or EA2E A EA1E A whereas we write A1E AQA2E A and A1E AWA2E A E or EA2E A EA1E A is the change in the stored energy from state 1 to state 2 and depends only on states 1 and 2 not upon the process between 1 and 2 A1E AQA2E A and A1E AWA2E A are amounts of energy transferred during the process between 1 and 2 and depend on the process path The quantities are associated with the process and they are not state properties Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 34 If a process in a control mass increases energy EA2E A EA1E A 0 can you say anything about the sign for A1E AQA2E A and A1E AWA2E A No The net balance of the heat transfer and work terms from the energy equation is EA2E A EA1E A A1E AQA2E A A1E AWA2E A 0 but that does not separate the effect of the two terms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 35 In Fig P35 CV A is the mass inside a piston cylinder CV B is that plus the piston outside which is the standard atmosphere Write the energy equation and work term for the two CVs assuming we have a nonzero Q between state 1 and state 2 CV A EA2E A EA1E A mAAE AeA2E A eA1E A mAAE AuA2E A uA1E A A1E AQA2E A A1E AWAA2E A1E AWAA2E A P dV PVA2E A VA1E A CV B EA2E A EA1E A mAAE AeA2E A eA1E A mApistE AeA2E A eA1E A mAAE AuA2E A uA1E A mApistE AgZA2E A ZA1E A A1E AQA2E A A1E AWBA2E A1E AWBA2E A PAoE A dV PAoE AVA2E A VA1E A Notice how the P inside CV A is P PAoE A mApistE Ag AAcylE A ie the first work term is larger than the second The difference between the work terms is exactly equal to the potential energy of the piston sitting on the left hand side in the CV B energy Eq The two equations are mathematically identical A1E AWAA2E A PVA2E A VA1E A PAoE A mApistE Ag AAcylE A VA2E A VA1E A A1E AWBA2E A mApistE AgVA2E A VA1E AAAcylE A1E AWBA2E A mApistE AgZA2E A ZA1E A P o g p A m m Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 36 A 500 W electric space heater with a small fan inside heats air by blowing it over a hot electrical wire For each control volume a wire only b all the room air and c total room plus the heater specify the storage work and heat transfer terms as 500W or 500W or 0 W neglect any E QA through the room walls or windows Storage Work Heat transfer Wire 0 W 500 W 500 W Room air 500 W 0 W 500 W Tot room 500 W 500 W 0 W Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 37 Two engines provide the same amount of work to lift a hoist One engine can provide 3 F in a cable and the other 1 F What can you say about the motion of the point where the force F acts in the two engines Since the two work terms are the same we get W F dx 3 F xA1E A 1 F xA2E A xA2E A 3 xA1E so the lower force has a larger displacement Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 38 Two hydraulic pistoncylinders are connected through a hydraulic line so they have roughly the same pressure If they have diameters of DA1E A and DA2E A 2DA1E A respectively what can you say about the piston forces FA1E A and FA2E A For each cylinder we have the total force as F PAAcylE A P π DA2E A4 FA1E A PAAcyl 1E A P π DA2 1E A4 FA2E A PAAcyl 2E A P π DA2 2E A4 P π 4 DA2 1E A4 4 FA1E cb 1 2 F 2 F 1 The forces are the total force acting up due to the cylinder pressure There must be other forces on each piston to have a force balance so the pistons do not move Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 39 Assume a physical setup as in Fig P35 We now heat the cylinder What happens to P T and v up down or constant What transfers do we have for Q and W pos neg or zero Solution Process P Po mpgAcyl C Heat in so T increases v increases and Q is positive As the volume increases the work is positive 1W2 P dV P o m p Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 310 A drag force on an object moving through a medium like a car through air or a submarine through water is F E A 0225 A ρVA2E A Verify the unit becomes Newton d Solution FAdE A 0225 A ρVA2E Units mA2E A kgmA3E A mA2E A sA2E A kg m sA2E A N Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 311 The sketch shows three physical situations show the possible process in a Pv diagram a b c V P 1 P 1 V stop V P 1 P 1 V 1 V P 1 P 1 V stop V 1 R 4 1 0 A P o m p Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 312 For the indicated physical setup in ab and c above write a process equation and the expression for work a P P1 and V Vstop or V Vstop and P P1 1W2 P1V2 V1 P1 Pfloat b P A BV 1W2 E 1A P1 P2V2 V1 2 c P P1 and V Vstop or V Vstop and P P1 1W2 P1V2 V1 P1 Pfloat a b c V P 1 P 1 V stop V P 1 P 1 V 1 V P 1 P 1 V stop V 1 R410A P o m p Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 313 Assume the physical situation as in Fig P311b what is the work term a b c or d a 1w2 P1v2 v1 b 1w2 v1P2 P1 c 1w2 E A P1 P2v2 v1 d 1w2 A1 2E A P1 P2v2 v1 1 2 Solution work term is formula c the area under the process curve in a Pv diagram The avg height is A1 2E A P1 P2 The base is v2 v1 v P 1 P 1 v 1 2 v 2 P 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 314 The sketch in Fig P314 shows a physical situation show the possible process in a Pv diagram a b c Solution v P 1 P 1 v stop v P 1 P 1 v 1 v P 1 P 1 v stop R 4 0 A P o m p Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 315 What can you say about the beginning state of the R410A in Fig P311 versus the case in Fig P314 for the same pistoncylinder For the case where the piston floats as in Fig P311 the pressure of the R410A must equal the equilibrium pressure that floats balance forces on the piston The situation in Fig P314 is possible if the R410A pressure equals or exceeds the float pressure Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 316 A piece of steel has a conductivity of k 15 WmK and a brick has k 1 WmK How thick a steel wall will provide the same insulation as a 10 cm thick brick The heat transfer due to conduction is from Eq 323 EA kA AdT dxE A kA AT xE A Q For the same area and temperature difference the heat transfers become the same for equal values of k x so A k xE A brick A k xE A steel xsteel xbrick ksteel kbrick 01 m A15 1E A 15 m Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 317 A thermopane window see Fig 338 traps some gas between the two glass panes Why is this beneficial The gas has a very low conductivity relative to a liquid or solid so the heat transfer for a given thickness becomes smaller The gap is furthermore made so small that possible natural convection motion is reduced to a minimum It becomes a trade off to minimize the overall heat transfer due to conduction and convection Typically these windows can be manufactured with an Eglaze to reduce radiation loss winter or gain summer multiple glazings lowE coating Gas filled space Spacer and sealer Window structural frame Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 318 On a chilly 10AoE AC fall day a house 20AoE AC inside loses 6 kW by heat transfer What transfer happens on a 30AoE AC warm summer day assuming everything else is the same The heat transfer is A QEA CA T where the details of the heat transfer is in the factor C Assuming those details are the same then it is the temperature difference that changes the heat transfer so A QEA CA T 6 kW CA 20 10 K CA 06 AkW KE Then A QEA CA T 06 AkW KE A 20 30 K 6 kW it goes in Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 319 Verify that a surface tension S with units Nm also can be called a surface energy with units JmA2E A The latter is useful for consideration of a liquid drop or liquid in small pores capillary Units Nm NmmA2E A JmA2E This is like a potential energy associated with the surface For water in small pores it tends to keep the water in the pores rather than in a drop on the surface Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 320 Some liquid water is heated so it becomes superheated vapor Do I use u or h in the energy equation Explain The energy equation for a control mass is muA2E A uA1E A A1E AQA2E A A1E AWA2E A The storage of energy is a change in u when we neglect kinetic and potential energy changes and that is always so To solve for the heat transfer we must know the work in the process and it is for a certain process P C that the work term combines with the change in u to give a change in h To avoid confusion you should always write the energy equation as shown above and substitute the appropriate expression for the work term when you know the process equation that allows you to evaluate work Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 321 Some liquid water is heated so it becomes superheated vapor Can I use specific heat to find the heat transfer Explain NO The specific heat cannot give any information about the energy required to do the phase change The specific heat is useful for single phase state changes only Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 322 Look at the R410A value for uAfE A at 50AoE AC Can the energy really be negative Explain The absolute value of u and h are arbitrary A constant can be added to all u and h values and the table is still valid It is customary to select the reference such that u for saturated liquid water at the triple point is zero The standard for refrigerants like R410A is that h is set to zero as saturated liquid at 40AoE AC other substances as cryogenic substances like nitrogen methane etc may have different states at which h is set to zero The ideal gas tables use a zero point for h as 25AoE AC or at absolute zero 0 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 323 A rigid tank with pressurized air is used to a increase the volume of a linear spring loaded piston cylinder cylindrical geometry arrangement and b to blow up a spherical balloon Assume that in both cases P A BV with the same A and B What is the expression for the work term in each situation The expression is exactly the same the geometry does not matter as long as we have the same relation between P and V then A1E AWA2E A P dV A BV dV AVA2E A VA1E A 05 B VA2 2E A VA2 1E A AVA2E A VA1E A 05 B VA2E A VA1E A VA2E A VA1E A 05 A B VA2E A A B VA1E A VA2E A VA1E A 05 PA1E A PA2E A VA2E A VA1E A Notice the last expression directly gives the area below the curve in the PV diagram P P 2 P V 1 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 324 An ideal gas in a pistoncylinder is heated with 2 kJ during an isothermal process How much work is involved Energy Eq u E A uA1E A A1E AqA2E A A1E AwA2E A 0 since uA2E A uA1E A isothermal 2 Then A1E AWA2E A m A1E AwA2E A A1E AQA2E A m A1E AqA2E A 2 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 325 An ideal gas in a pistoncylinder is heated with 2 kJ during an isobaric process Is the work pos neg or zero As the gas is heated u and T increase and since PV mRT it follows that the volume increase and thus work goes out w 0 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 326 You heat a gas 10 K at P C Which one in Table A5 requires most energy Why A constant pressure process in a control mass gives recall Section 39 and Eq344 A1E AqA2E A uA2E A uA1E A A1E AwA2E A uA2E A uA1E A PA1E AvA2E A vA1E A hA2E A hA1E A CApE A T The one with the highest specific heat is hydrogen HA2E A The hydrogen has the smallest mass but the same kinetic energy per mol as other molecules and thus the most energy per unit mass is needed to increase the temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 327 You mix 20AoE AC water with 50AoE AC water in an open container What do you need to know to determine the final temperature The process will take place at constant pressure atmospheric and you can assume there will be minimal heat transfer if the process is reasonably fast The energy equation then becomes UA2E A UA1E A 0 A1E AWA2E A PVA2E A VA1E A Which we can write as HA2E A HA1E A 0 mA2E AhA2E A mA1 20CE AhA1 20CE A mA1 50CE AhA1 50CE A You need the amount of mass at each temperature mA1 20CE A and mA1 50CE A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Kinetic and Potential Energy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 328 A piston motion moves a 25 kg hammerhead vertically down 1 m from rest to a velocity of 50 ms in a stamping machine What is the change in total energy of the hammerhead Solution CV Hammerhead The hammerhead does not change internal energy ie same P T but it does have a change in kinetic and potential energy EA2E A EA1E A muA2E A uA1E A m12VA2E A 2 0 mg Z2 0 0 25 kg 12 50 ms2 25 kg 980665 msA2E A 1 m 31 250 J 24517 J 31 005 J 31 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 329 A 1200 kg car is accelerated from 30 to 50 kmh in 5 s How much work is that If you continue from 50 to 70 kmh in 5 s is that the same The work input is the increase in kinetic energy EA2E A EA1E A 12mVA2 2E A VA2 1E A A1E AWA2E A 05 1200 kg 50A2E A 30A2E A A km h 2E 600 kg 2500 900 A 1000 m 3600 s 2E A 74 074 J 741 kJ The second set of conditions does not become the same EA2E A EA1E A 12mVA2 2E A VA2 1E A 600 kg 70A2E A 50A2E A A 1000 m 3600 s 2E A 111 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 330 The rolling resistance of a car depends on its weight as F 0006 mg How far will a car of 1200 kg roll if the gear is put in neutral when it drives at 90 kmh on a level road without air resistance Solution The car decreases its kinetic energy to zero due to the force constant acting over the distance m 12VA2 2E A 12VA2 1E A A1E AWA2E A F dx FL VA2E A 0 VA1E A 90 Akm hE A A90 1000 3600E A msA1E A 25 msA1E 12 mVA2 1E A FL 0006 mgL L EAEA 05 VA2 1 A E0006 gE A EAEA 0525A2 A E00069807 E A EAEAmA2 AsA2 A EmsA2 AE A 5311 m Remark Over 5 km The air resistance is much higher than the rolling resistance so this is not a realistic number by itself Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 331 A piston of mass 2 kg is lowered 05 m in the standard gravitational field Find the required force and work involved in the process Solution F ma 2 kg 980665 msA2E A 1961 N W F dx F dx F x 1961 N 05 m 9805 J Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 332 A 1200 kg car accelerates from zero to 100 kmh over a distance of 400 m The road at the end of the 400 m is at 10 m higher elevation What is the total increase in the car kinetic and potential energy Solution KE ½ m VA2 2E A VA2 1E A VA2E A 100 kmh A100 1000 3600E A ms 2778 ms KE ½ 1200 kg 2778A2E A 0A2E A msA2E A 463 037 J 463 kJ PE mgZA2E A ZA1E A 1200 kg 9807 msA2E A 10 0 m 117684 J 1177 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 333 A hydraulic hoist raises a 1750 kg car 18 m in an auto repair shop The hydraulic pump has a constant pressure of 800 kPa on its piston What is the increase in potential energy of the car and how much volume should the pump displace to deliver that amount of work Solution CV Car No change in kinetic or internal energy of the car neglect hoist mass EA2E A EA1E A PEA2E A PEA1E A mg ZA2E A ZA1E A 1750 kg 980665 msA2E A 18 m 30 891 J The increase in potential energy is work into car from pump at constant P W EA2E A EA1E A P dV P V V A E2 E1 EPE A A 30891 J 800 1000 PaE A 00386 m3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 334 Airplane takeoff from an aircraft carrier is assisted by a steam driven pistoncylinder device with an average pressure of 1250 kPa A 17500 kg airplane should be accelerated from zero to a speed of 30 ms with 30 of the energy coming from the steam piston Find the needed piston displacement volume Solution CV Airplane No change in internal or potential energy only kinetic energy is changed EA2E A EA1E A m 12 VA2 2E A 0 17500 kg 12 30A2E A msA2E 7 875 000 J 7875 kJ The work supplied by the piston is 30 of the energy increase W P dV PAavgE A V 030 EA2E A EA1E A 030 7875 kJ 23625 kJ V A W Pavg E A A23625 1250E A A kJ kPaE A 189 m3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 335 Solve Problem 334 but assume the steam pressure in the cylinder starts at 1000 kPa dropping linearly with volume to reach 100 kPa at the end of the process Solution CV Airplane EA2E A EA1E A m 12 V2 2 0 17 500 kg 12 302 msA2E 7875 000 J 7875 kJ W 030EA2E A EA1E A 030 7875 23625 kJ W P dV 12Pbeg Pend V V A W Pavg E A A 23625 kJ 121000 100 kPaE A 429 m3 P 1 2 V W 1000 100 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 336 A steel ball weighing 5 kg rolls horizontal with 10 ms If it rolls up an incline how high up will it be when it comes to rest assuming standard gravitation CV Steel ball Energy Eq EA2E A EA1E A A1E AQA2E A A1E AWA2E A 0 0 0 EA1E A muA1E A mgZA1E A 05 mVA2E A EA2E A muA2E A mgZA2E A 0 We assume the steel ball does not change temperature uA2E A uA1E A so then the energy equation gives muA2E A mgZA2E A muA1E A mgZA1E A 05 mVA2E A 0 mg ZA2E A ZA1E A 05 mVA2E A ZA2E A ZA1E A 05 VA2E Ag 05 10A2E A mA2E AsA2E A 981 msA2E A 51 m Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 337 A hydraulic cylinder of area 001 mA2E A must push a 1000 kg arm and shovel 05 m straight up What pressure is needed and how much work is done F mg 1000 kg 981 msA2E A 9810 N PA P FA 9810 N 001 mA2E 981 000 Pa 981 kPa W AF dxEA F x 9810 N 05 m 4905 J Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 338 A hydraulic cylinder has a piston of cross sectional area 10 cm2 and a fluid pressure of 2 MPa If the piston is moved 025 m how much work is done Solution The work is a force with a displacement and force is constant F PA W F dx PA dx PA x 2000 kPa 10 10A4E A m2 025 m 05 kJ Units kPa mA2E A m kN mA2E A mA2E A m kN m kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 339 Two hydraulic pistoncylinders are connected with a line The master cylinder has an area of 5 cmA2E A creating a pressure of 1000 kPa The slave cylinder has an area of 3 cmA2E A If 25 J is the work input to the master cylinder what is the force and displacement of each piston and the work output of the slave cylinder piston Solution W Fx dx P dv P A dx P A x xAmasterE A A W PAE A A 25 1000 5 104 E A A J kPa m2 E A 005 m Ax V 5 10A4E A 005 25 10A5E A mA3E A VAslaveE A A x xAslaveE A VA 25 10A5E A mA3E A 3 10A4E A mA2E A 00083 33 m FAmasterE A P A 1000 kPa 5 10A4E A mA2E A 10A3E A PakPa 500 N FAslaveE A P A 1000 kPa 10A3E A PakPa 3 10A4E A mA2E A 300 N WAslaveE A F x 300 N 008333 m 25 J Master Slave Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 340 The air drag force on a car is 0225 A ρVA2E A Assume air at 290 K 100 kPa and a car frontal area of 4 mA2E A driving at 90 kmh How much energy is used to overcome the air drag driving for 30 minutes The formula involves density and velocity and work involves distance so ρ A1 vE A A P RTE A A 100 0287 290E A 12015 EAEAkg mA3 AE V 90 Akm hE A 90 A1000 3600E A Am sE A 25 ms x V t 25 ms 30 min 60 smin 45 000 m Now F 0225 A ρ VA2E A 0225 4 mA2E A 12015 EAEAkg mA3 AE A 25A2E A EAEAmA2 EsA2 AE 67E58 mA2E A EAEAkg mA3 AE A EAEAmA2 EsA2 AE A 676 N WE E F x 676 N 45 000 m 30 420 000 J 3042 MJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 341 A bulldozer pushes 800 kg of dirt 100 m with a force of 1500 N It then lifts the dirt 3 m up to put it in a dump truck How much work did it do in each situation Solution W F dx F x 1500 N 100 m 150 000 J 150 kJ W F dz mg dz mg Z 800 kg 9807 msA2E A 3 m 23 537 J 235 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 342 Two hydraulic cylinders maintain a pressure of 1200 kPa One has a cross sectional area of 001 mA2E A the other 003 mA2E A To deliver a work of 1 kJ to the piston how large a displacement V and piston motion H is needed for each cylinder Neglect PAatmE A Solution W F dx P dV PA dx PA H PV V AW PE A A 1 kJ 1200 kPaE A 0000 833 mA3E Both cases the height is H VA HA1E A A0000833 001E A m 00833 m HA2E A A0000833 003E A m 00278 m cb 1 2 F 2 F 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 343 A linear spring F kAsE Ax xA0E A with spring constant kAsE A 500 Nm is stretched until it is 100 mm longer Find the required force and work input Solution F kAsE Ax xA0E A 500 01 50 N W F dx A ksx x0dx x0EA kAsE Ax xA0E AA2E A2 500 AN mE A 01A2E A2 mA2E A 25 J Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 344 A piston of 2 kg is accelerated to 20 ms from rest What constant gas pressure is required if the area is 10 cmA2E A the travel 10 cm and the outside pressure is 100 kPa CV Piston EA2E A E1PIST mu2 u1 m12VA2 2E A 0 mg 0 0 12 m VA2 2E A 05 2 kg 20A2E A msA2E A 400 J Energy equation for the piston is EA2E A E1 PIST Wgas Watm Pavg Vgas Po Vgas Vgas A L 10 cmA2E A 10 cm 00001 mA3E Pavg Vgas EA2E A E1PIST Po Vgas Pavg EA2E A E1PIST Vgas Po 400 J 00001 mA3E A 100 kPa 4000 kPa 100 kPa 4100 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Boundary work Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 345 A 25 kg piston is above a gas in a long vertical cylinder Now the piston is released from rest and accelerates up in the cylinder reaching the end 5 m higher at a velocity of 25 ms The gas pressure drops during the process so the average is 600 kPa with an outside atmosphere at 100 kPa Neglect the change in gas kinetic and potential energy and find the needed change in the gas volume Solution CV Piston EA2E A E1PIST mu2 u1 m12VA2 2E A 0 mg H2 0 0 25 kg 12 252msA2E A 25 kg 980665 msA2E A 5 m 78125 J 12258 J 90383 J 9038 kJ Energy equation for the piston is EA2E A E1 Wgas Watm Pavg Vgas Po Vgas remark Vatm Vgas so the two work terms are of opposite sign Vgas A 9038 600 100E A A kJ kPaE A 0018 m3 P 1 2 V Pavg V H P o g Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 346 The R410A in Problem 314 c is at 1000 kPa 50AoE AC with mass 01 kg It is cooled so the volume is reduced to half the initial volume The piston mass and gravitation is such that a pressure of 400 kPa will float the piston Find the work in the process If the volume is reduced the piston must drop and thus float with P 400 kPa The process therefore follows a process curve shown in the PV diagram Table B42 vA1E A 003320 mA3E Akg A1E AWA2E A APdVEA area PAfloatE A VA2E A VA1E A PAfloatE A VA1E A2 400 kPa 01 kg 00332 mA3E Akg 2 0664 kJ v P 1 P 1 v stop 400 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 347 A 400L tank A see figure P347 contains argon gas at 250 kPa 30AoE AC Cylinder B having a frictionless piston of such mass that a pressure of 150 kPa will float it is initially empty The valve is opened and argon flows into B and eventually reaches a uniform state of 150 kPa 30AoE AC throughout What is the work done by the argon Solution Take CV as all the argon in both A and B Boundary movement work done in cylinder B against constant external pressure of 150 kPa Argon is an ideal gas so write out that the mass and temperature at state 1 and 2 are the same PAA1E AVAAE A mAAE ARTAA1E A mAAE ARTA2E A PA2E A VAAE A VAB2E A VAB2E A PAA1E A PA2E A VAAE A VAAE A A250 04 150E A 04 02667 mA3E A A1E AWA2E A A 1 2 PextdVEA PAextE AVAB2E A VAB1E A 150 kPa 02667 0 mA3E A 40 kJ Notice there is a pressure loss in the valve so the pressure in B is always 150 kPa while the piston floats V P 1 2 B B B A Argon P o g Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 348 A piston cylinder contains 2 kg of liquid water at 20oC and 300 kPa as shown in Fig P348 There is a linear spring mounted on the piston such that when the water is heated the pressure reaches 3 MPa with a volume of 01 m3 a Find the final temperature b Plot the process in a Pv diagram c Find the work in the process Solution Take CV as the water This is a constant mass m2 m1 m State 1 Compressed liquid take saturated liquid at same temperature B11 v1 vf20 0001002 mA3E Akg State 2 v2 V2m 012 005 mA3E Akg and P 3000 kPa from B12 Twophase T2 2339oC Work is done while piston moves at linearly varying pressure so we get 1W2 P dV area Pavg V2 V1 A1 2E A PA1E A PA2E AVA2E A VA1E A 05 300 3000 kPa 01 0002 mA3E A 1617 kJ T CP v 2 1 300 kPa P CP v T 300 20 2 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful The process shown by CATT3 in a loglog diagram The linear relation P A Bv has B 551 MPamA3E Akg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 349 Air in a spring loaded pistoncylinder has a pressure that is linear with volume P A BV With an initial state of P 150 kPa V 1 L and a final state of 800 kPa and volume 15 L it is similar to the setup in Problem 348 Find the work done by the air Solution Knowing the process equation P A BV giving a linear variation of pressure versus volume the straight line in the PV diagram is fixed by the two points as state 1 and state 2 The work as the integral of PdV equals the area under the process curve in the PV diagram State 1 PA1E A 150 kPa VA1E A 1 L 0001 m3 State 2 PA2E A 800 kPa VA2E A 15 L 00015 m3 Process P A BV linear in V A1E AWA2E A A 1 2 PdVEA A P1 P2 E2E AVA2E A VA1E A P V W 1 2 A1 2E A 150 800 kPa 15 1 L 0001 m3L 02375 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 350 Heat transfer to a block of 15 kg ice at 10AoE AC melts it to liquid at 10AoE AC in a kitchen How much work does the water gives out Work is done against the atmosphere due to volume change in the process The pressure is 101 kPa so we approximate the states as saturated State 1 Compressed solid B15 v1 00010891 mA3E Akg State 2 Compressed liquid B11 v2 0001000 mA3E Akg A1E AWA2E A A PdVEA Po V2 V1 Po m vA2E A vA1E A 101325 kPa 15 kg 0001 00010891 mA3E Akg 00135 kJ Notice the work is negative the volume is reduced Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 351 A cylinder fitted with a frictionless piston contains 5 kg of superheated refrigerant R134a vapor at 1000 kPa 140C The setup is cooled at constant pressure until the R134a reaches a quality of 25 Calculate the work done in the process Solution Constant pressure process boundary work State properties from Table B52 State 1 v 003150 mA3E Akg State 2 v 0000871 025 001956 000576 mA3E Akg Interpolated to be at 1000 kPa numbers at 1017 kPa could have been used in which case v 000566 mA3E Akg A1E AWA2E A P dV P VA2E AVA1E A mP vA2E AvA1E A 5 kg 1000 kPa 000576 003150 mA3E Akg 1287 kJ T CP v 1 2 P 1000 kPa P CP v T 1000 39 140 2 1 cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 352 A pistoncylinder contains 2 kg water at 20AoE AC with volume 01 mA3E A By mistake someone locks the piston preventing it from moving while we heat the water to saturated vapor Find the final temperature volume and the process work Solution 1 vA1E A Vm 01 mA3E A2 kg 005 mA3E Akg twophase state 2 Constant volume vA2E A vAgE A vA1E A VA2E A VA1E A 01 mA3E A1E AWA2E A P dV 0 State 2 vA2E A xA2E A 1 TA2E A TAsE AaE A E AtE A 250 5 A 005 005013 004598 005013E A 2502C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Polytropic process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 353 A nitrogen gas goes through a polytropic process with n 13 in a pistoncylinder It starts out at 600 K 600 kPa and ends at 800 K Is the work positive negative or zero The work is a boundary work so it is W A PdVEA A Pm dvEA AREA so the sign depends on the sign for dV or dv The process looks like the following The actual process is on a steeper curve than n 1 As the temperature increases we notice the volume decreases so dv 0 W 0 Work is negative and goes into the nitrogen gas P v 2 1 T v 2 1 T T 1 2 T C v 03 P C v 13 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 354 Helium gas expands from 125 kPa 350 K and 025 m E A to 100 kPa in a polytropic process with n 1667 How much work does it give out 3 Solution Process equation PVAnE A constant PA1E AVAn 1E A PA2E AVAn 2E Solve for the volume at state 2 VA2E A VA1E A PA1E APA2E AA1nE A 025 A 125 100 06E A 02852 mA3E Work from Eq321 A1E AWA2E A EAEA PA2 AVA2 A PA1 A VA1 A E 1nE A A100 02852 125 025 1 1667E A kPa mA3E A 409 kJ The actual process is on a steeper curve than n 1 P V W 1 2 n 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 355 Air goes through a polytropic process from 125 kPa 325 K to 300 kPa and 500 K Find the polytropic exponent n and the specific work in the process Solution Process Pv E A Const PA1E AvAn 1E A PA2E A vAn 2E n Ideal gas Pv RT so vA1E A ART PE A A0287 325 125E A 07462 mA3E Akg vA2E A ART PE A A0287 500 300E A 047833 mA3E Akg From the process equation PA2E A PA1E A vA1E A vA2E AAnE A lnPA2E A PA1E A n lnvA1E A vA2E A n lnPA2E A PA1E A lnvA1E A vA2E A A ln 24 ln 156E A 1969 The work is now from Eq321 per unit mass and ideal gas law A1E AwA2E A A P2v2 P1v1 E1nE A RT2 T1 1n A0287500 325 E1 1969E A kJkgK K 518 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 356 A balloon behaves so the pressure is P C2 VA13E A C2 100 kPam The balloon is blown up with air from a starting volume of 1 mA3E A to a volume of 4 mA3E A Find the final mass of air assuming it is at 25AoE AC and the work done by the air Solution The process is polytropic with exponent n 13 PA1E A C2 VA13E A 100 1A13E A 100 kPa P2 C2 VA13E A 100 4A13E A 15874 kPa A1E AWA2E A P dV A P2V2 P1V1 E1 nE A Equation 318 and 321 A15874 4 100 1 1 13E A kPamA3E A 4012 kJ mA2E A A P2V2 E RT2 E A A 15874 4 0287 298E A AkPam3 EkJkgE A 7424 kg P V W 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 357 Consider a piston cylinder with 05 kg of R134a as saturated vapor at 10C It is now compressed to a pressure of 500 kPa in a polytropic process with n 15 Find the final volume and temperature and determine the work done during the process Solution Take CV as the R134a which is a control mass m2 m1 m Process PvA15E A constant until P 500 kPa 1 T x vA1E A 009921 mA3E Akg P PAsatE A 2017 kPa from Table B51 2 P process vA2E A vA1E A PA1E APA2E A A115E A 009921 2017500A23E A 005416 mA3E Akg Given P v at state 2 from B52 it is superheated vapor at T2 79C Process gives P C v A15E A which is integrated for the work term Eq321 A1E AWA2E A P dV A m 1 15E A PA2E AvA2E A PA1E AvA1E A A 2 05E A kg 500 005416 2017 009921 kPamA3E Akg 707 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Heat Transfer rates Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 358 The brake shoe and steel drum on a car continuously absorbs 75 W as the car slows down Assume a total outside surface area of 01 mA2E A with a convective heat transfer coefficient of 10 WmA2E A K to the air at 20C How hot does the outside brake and drum surface become when steady conditions are reached Solution Convection heat transfer Eq324 A QEA hAΤ Τ A Q EhAE A T ΤBRAKE 20 A 75 W 10 Wm2K1 01 m2 E A 75 C TBRAKE 20 75 95C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 359 A waterheater is covered up with insulation boards over a total surface area of 3 mA2E A The inside board surface is at 75C and the outside surface is at 18C and the board material has a conductivity of 008 Wm K How thick a board should it be to limit the heat transfer loss to 200 W Solution Steady state conduction through a single layer board Eq323 A QEA cond k A AT xE A x k Α Τ A QE x 008 A W m KE A 3 mA2E A A75 18 200E A AK WE 0068 m Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 360 Find the rate of conduction heat transfer through a 15 cm thick hardwood board k 016 Wm K with a temperature difference between the two sides of 20AoE AC One dimensional heat transfer by conduction we do not know the area so we can find the flux heat transfer per unit area WmA2E A A qEA A QEAA k AT xE A 016 A W m KE A A 20 0015E A AK mE A 213 WmA2E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 361 A 2 mA2E A window has a surface temperature of 15AoE AC and the outside wind is blowing air at 2AoE AC across it with a convection heat transfer coefficient of h 125 WmA2E AK What is the total heat transfer loss Solution Convection heat transfer Eq324 A QEA h A T 125 WmA2E AK 2 mA2E A 15 2 K 3250 W as a rate of heat transfer out 2 C o 15 C o Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 362 Due to a faulty door contact the small light bulb 25 W inside a refrigerator is kept on and limited insulation lets 50 W of energy from the outside seep into the refrigerated space How much of a temperature difference to the ambient at 20C must the refrigerator have in its heat exchanger with an area of 1 mA2E A and an average heat transfer coefficient of 15 WmA2E A K to reject the leaks of energy Solution A QEA tot 25 50 75 W to go out Convection heat transfer Eq324 A QEA hA T 15 1 T 75 W T A Q EhAE A A 75 W 15 Wm2K 1 m2 E A 5 C so T must be at least 25 C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 363 A large condenser heat exchanger in a power plant must transfer a total of 100 MW from steam running in a pipe to sea water being pumped through the heat exchanger Assume the wall separating the steam and seawater is 4 mm of steel conductivity 15 Wm K and that a maximum of 5C difference between the two fluids is allowed in the design Find the required minimum area for the heat transfer neglecting any convective heat transfer in the flows Solution Steady conduction through the 4 mm steel wall Eq323 A QEA k A AT xE A Α A QEA x kΤ A 100 106 W 0004 m 15 WmK 5 K 480 mA2E Condensing water Sea water cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 364 The black grille on the back of a refrigerator has a surface temperature of 35C with a total surface area of 1 mA2E A Heat transfer to the room air at 20C takes place with an average convective heat transfer coefficient of 15 WmA2E A K How much energy can be removed during 15 minutes of operation Solution Convection heat transfer Eq324 A QEA hA T Q A QEA t hA T t Q 15 WmA2E A K 1 mA2E A 3520 Κ 15 min 60 smin 202 500 J 2025 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 365 A pot of steel conductivity 50 Wm K with a 5 mm thick bottom is filled with 15C liquid water The pot has a diameter of 20 cm and is now placed on an electric stove that delivers 500 W as heat transfer Find the temperature on the outer pot bottom surface assuming the inner surface is at 15C Solution Steady conduction Eq323 through the bottom of the steel pot Assume the inside surface is at the liquid water temperature A QEA k A AT xE A Τ A QEA x kΑ T 500 W 0005 m 50 WmK Aπ 4E A 02A2E A mA2E A 159 K T 15 159 166C cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 366 A log of burning wood in the fireplace has a surface temperature of 450C Assume the emissivity is 1 perfect black body and find the radiant emission of energy per unit surface area Solution Radiation heat transfer Eq325 A QEA A 1 σ T4 567 10 8 WmA2E AKA4E A 27315 4504 KA4E 15 505 Wm2 155 kWm2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 367 A wall surface on a house is at 30C with an emissivity of ε 07 The surrounding ambient to the house is at 15C average emissivity of 09 Find the rate of radiation energy from each of those surfaces per unit area Solution Radiation heat transfer Eq325 A QEA A εσATA4E A σ 567 10 8 WmA2E AKA4E a A QEAA 07 567 108 WmA2E AKA4E A 27315 30A4E A KA4E A 335 Wm2 b A QEAA 09 567 108 WmA2E AKA4E A 28815A4E A KA4E A 352 Wm2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 368 A radiant heat lamp is a rod 05 m long and 05 cm in diameter through which 400 W of electric energy is deposited Assume the surface has an emissivity of 09 and neglect incoming radiation What will the rod surface temperature be Solution For constant surface temperature outgoing power equals electric power Radiation heat transfer Eq325 A QradE A εσAT4 A QelE A T4 A QelE A εσA 400 W 09 567 10 8 WmA2E AKA4E A 05 π 0005 mA2E A 99803 1011 K4 T 1000 K OR 725 C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 369 A radiant heating lamp has a surface temperature of 1000 K with ε 08 How large a surface area is needed to provide 250 W of radiation heat transfer Radiation heat transfer Eq325 We do not know the ambient so let us find the area for an emitted radiation of 250 W from the surface A QEA εσATA4E A A A Q εσT4 E A A 250 08 567 108 10004 E A A W Wm2 E 00055 mA2E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Properties u h from General Tables Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 370 Determine the phase of the following substances and find the values of the unknown quantities a Nitrogen P 2000 kPa 120 K v Z b Nitrogen 120 K v 00050 m3kg Z c Air T 100 C v 0500 m3kg P d R410A T 25C v 001 m3kg P h Solution a B62 at 2000 kPa 120 K v 00126 m3kg A5 R 02968 kJkgK Z APv RTE A A2000 00126 02968 120E A A kPa m3kg EkJkgK KE A 07075 b Table B61 vAfE A v vAgE A 000799 mA3E Akg so twophase L V P PAsatE A 2513 kPa x v vAfE AvAfgE A A0005 0001915 000608E A 05074 Z APv RTE A A2513 0005 02968 120E A A kPa m3kg EkJkgK KE A 0353 c Ideal gas P RTv 0287 kJkgK 37315 K 05 m3kg 214 kPa d B41 at 25C vAfE A 0000944 vg 001514 vAfE A v vg saturated P 16536 kPa x v vf vfg A001 0000944 00142E A 063775 h hAfE A x hAfgE A 9759 063775 18643 216486 kJkg States shown are placed relative to the twophase region not to each other P CP v T CP v T c bd bd a c P const a Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 371 Find the phase and the missing properties of T P v u and x for water at a 500 kPa 100AoE AC b 5000 kPa u 800 kJkg c 5000 kPa v 006 mA3E Akg d 6AoE AC v 1 mA3E Akg Solution a Look in Table B12 at 500 kPa T Tsat 151AoE AC compressed liquid Table B14 v 0001043 mA3E Akg u 4188 kJkg b Look in Table B12 at 5000 kPa u uf 114778 kJkg compressed liquid Table B14 between 180oC and 200oC T 180 200 180 A 800 75962 84808 75962E A 180 2004567 1891AoE AC v 0001124 04567 0001153 0001124 0001137 mA3E Akg c Look in Table B12 at 5000 kPa v vg 003944 mA3E Akg superheated vapor Table B13 between 400oC and 450oC T 400 50A 006 005781 00633 005781E A 400 5003989 41995oC u 290658 03989 299964 290658 29437 kJkg d B15 vi v vg 33414 mA3E Akg 2phase P Psat 8876 kPa x v vi vfg 1 00010898334138 00029895 u ui x ufg 34591 0002989527129 3378 kJkg 372 States shown are placed relative to the twophase region not to each other P CP v T CP v T b c c a b P const d d a Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 373 Find the missing properties and give the phase of the ammonia NH3 a T 65AoE AC P 600 kPa u v b T 20AoE AC P 100 kPa u v x c T 50AoE AC v 01185 mA3E Akg u P x Solution a Table B21 P Psat superheated vapor Table B22 v 05 025981 05 026888 02645 mA3E Akg u 05 14257 05 14443 1435 kJkg b Table B21 P Psat x undefined superheated vapor from B22 v 14153 mA3E Akg u 13745 kJkg c Sup vap v vAgE A Table B22 P 1200 kPa x undefined u 1383 kJkg States shown are placed relative to the twophase region not to each other P CP v T CP v T a c b a c b 1200 kPa 600 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 374 Find the missing properties of P T v u h and x and indicate the states in a Pv and Tv diagram for a Water at 5000 kPa u 1000 kJkg Table B1 reference b R134a at 20AoE AC u 300 kJkg c Nitrogen at 250 K 200 kPa Solution a Compressed liquid B14 interpolate between 220oC and 240oC T 2333oC v 0001213 mA3E Akg x undefined b Table B51 u uAgE A twophase liquid and vapor x u uAfE AuAfgE A 300 2270316216 0449988 045 v 0000817 045 003524 001667 m3kg c Table B61 T TAsatE A 200 kPa so superheated vapor in Table B62 x undefined v 05035546 038535 03704 m3kg u 0517723 19214 1847 kJkg States shown are placed relative to the twophase region not to each other P CP v T CP v T b c a b c a P const Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 375 Determine the phase and the missing properties a H2O 20C v 0001000 m3kg P u b R410A 400 kPa v 0075 m3kg T u c NH3 10C v 01 m3kg P u d N2 1013 kPa h 60 kJkg T v a Enter Table B11 with T and we see at 20C v vf so compressed liquid and Table B14 P 5000 kPa and u 8364 kJkg b Table B42 P 400 kPa v vg so superheated interpolate T 0 20 A 0075 007227 007916 07227E A 20 03962 79C u 26151 27644 26151 03962 26743 kJkg c Table B21 at 10C vf 00016 m3kg vg 020541 m3kg so twophase P Psat 6152 kPa x v vf vfg A01 00016 020381E A 04828 u uf x ufg 22599 x 10997 75693 kJkg d Table B61 shows that at 1013 kPa hf h hg 7669 kJkg so saturated twophase T 773 K x h hf hfg 60 1221519884 0916 v vf x vfg 0001240 0916 021515 01983 m3kg States shown are placed relative to the two phase region not to each other P CP v T CP v T a b a b P const cd cd Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 376 Find the missing properties of u h and x a HA2E AO T 120C v 05 mA3E Akg b HA2E AO T 100C P 10 MPa c NA2E A T 100 K x 075 d NA2E A T 200 K P 200 kPa e NHA3E A T 100C v 01 mA3E Akg Solution a Table B11 vAfE A v vAgE A LV mixture P 1985 kPa x 05 00010608908 056 u 50348 056 202576 16379 kJkg b Table B14 compressed liquid v 0001039 mA3E Akg u 4161 kJkg c Table B61 100 K x 075 v 0001452 075 002975 0023765 mA3E Akg u 7433 075 1375 288 kJkg d Table B62 200 K 200 kPa v 029551 mA3E Akg u 14737 kJkg e Table B21 v vAgE A superheated vapor x undefined B22 P 1600 400 A 01 010539 008248010539E A 1694 kPa States shown are placed relative to the twophase region not to each other P CP v T CP v T a d e c b e d a c b P const Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 377 Determine the phase of the following substances and find the values of the unknown quantities a R410A T 20C u 220 kJkg P x b Ammonia T 20C v 035 m3kg P u c Water P 400 kPa h 2800 kJkg T v a At 20C in B41 u uAgE A 24599 kJkg so x u uAfE AuAfgE A 220 279221807 08808 P Psat 3996 kPa b At 20C vAfE A 0001504 vg 062334 m3kg vAfE A v vg saturated P 1902 kPa x Av vf vfg E A A035 0001504 062184E A 056043 u uAfE A x uAfgE A 8876 056043 12107 76727 kJkg c B12 at 400 kPa h hg 27385 kJkg so superheated vapor we locate it between 150 and 200C and interpolate y 2800 275282286051 275282 043811 T 150 y 200 150 1719C v 047084 y 053422 047084 04986 m3kg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 378 Find the missing properties for COA2E A at a 20AoE AC 2 MPa v and h b 10AoE AC x 05 P u c 1 MPa v 005 mA3E Akg T h Solution a Table B31 P PAsatE A 5729 kPa so superheated vapor Table B32 v 00245 mA3E Akg h 36842 kJkg b Table B31 since x given it is twophase P PAsatE A 4502 kPa u uAfE A x uAfgE A 1076 05 16907 19214 kJkg c Table B31 v vAgE A 00383 mA3E Akg so superheated vapor Table B32 Between 0 and 20AoE AC so interpolate T 0 20 A 005 0048 00524 0048E A 20 04545 909AoE AC h 36114 37963 36114 04545 36954 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 379 Find the missing properties among T P v u h and x if applicable and indicate the states in a Pv and a Tv diagram for a R410A P 500 kPa h 300 kJkg b R410A T 10AoE AC u 200 kJkg c R134a T 40AoE AC h 400 kJkg Solution a Table B41 h hAgE A superheated vapor look in section 500 kPa and interpolate T 0 20 A 300 28784 30618 28784E A 20 066303 1326C v 005651 066303 006231005651 006036 mA3E Akg u 25959 066303 27502 25959 26982 kJkg b Table B41 u uAgE A 2559 kJkg LV mixture P 10857 kPa x A u uf Eufg E A A200 7224 18366E A 06956 v 0000886 06956 002295 001685 mA3E Akg h 7321 06956 20857 2183 kJkg c Table B51 h hAgE A twophase L V look in B51 at 40C x A h hf Ehfg E A A400 2565 1633E A 087875 P PAsatE A 1017 kPa v 0000 873 087875 001915 00177 mA3E Akg u 2557 087875 1438 3821 kJkg States shown are placed relative to the twophase region not to each other P CP v T CP v T b c a P C b c a Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 380 Saturated liquid water at 20AoE AC is compressed to a higher pressure with constant temperature Find the changes in u and h from the initial state when the final pressure is a 500 kPa b 2000 kPa Solution State 1 is located in Table B11 and the states ac are from Table B14 State u kJkg h kJkg u u uA1E h h hA1E Pv 1 8394 8394 a 8391 8441 003 047 05 b 8382 8582 012 188 2 For these states u stays nearly constant dropping slightly as P goes up h varies with Pv changes v P v T a b 1 ba1 T 20 C o P T v V L S CP 1 a b cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 381 Determine the phase of the following substances and find the values of the unknown quantities a Water P 500 kPa u 2850 kJkg T v b R134a T 10C v 008 m3kg P u c Ammonia T 20C u 1000 kJkg P x a B12 at 500 kPa u ug 2561 kJkg so superheated vapor we locate it between 300 and 350C and interpolate x 2850 2802928826 28029 0590966 T 300 x 350 300 32955C v 052256 x 057012 052256 05507 m3kg b B51 at 10C vAfE A 0000755 m3kg vg 009921 m3kg vAfE A v vg saturated P 2017 kPa x v vAfE A vAfgE A 0080000755009845 0805 u uAfE A x uAfgE A 18657 0805 1857 336 kJkg c B21 at 20C u uAgE A 12995 kJkg so P Psat 1902 kPa x u uAfE AuAfgE A 1000 887612107 07526 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Problem Analysis Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 382 Consider Problem 3101 Take the whole room as a CV and write both conservation of mass and energy equations Write some equations for the process two are needed and use those in the conservation equations Now specify the four properties that determines initial 2 and final state 2 do you have them all Count unknowns and match with equations to determine those CV Containment room and reactor Mass mA2E A mA1E A 0 mA2E A mA1E A VAreactorE AvA1E Energy muA2E A uA1E A A1E AQA2E A A1E AWA2E A Process Room volume constant V C A1E AWA2E A 0 Room insulated A1E AQA2E A 0 Using these in the equation for mass and energy gives mA2E A VA2E AvA2E A mA1E A muA2E A uA1E A 0 0 0 State 1 PA1E A TA1E A so Table B14 gives vA1E A uA1E A mA1E State 2 PA2E A We do not know one state 2 property and the total room volume Energy equation then gives uA2E A uA1E A a state 2 property State 2 PA2E A uA2E A vA2E Now we have the room volume as Continuity Eq mA2E A VvA2E A mA1E A so V mA1E A vA2E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 383 Consider a steel bottle as a CV It contains carbon dioxide at 20 oC quality 20 It has a safety valve that opens at 6 MPa The bottle is now accidentially heated until the safety valve opens Write the process equation that is valid until the valve opens and plot the Pv diagram for the process Solution CV carbon dioxide which is a control mass of constant volume Energy Eq35 326 uA2E A uA1E A A1E AqA2E A A1E AwA2E A Process V constant and m constant v constant A1E AwA2E A 0 State 1 Table B31 PA1E A PAsatE A 1970 kPa vA1E A vAfE A x vAfgE A State 2 6000 kPa vA1E A vA2E A Table B31 or B32 TA2E A uA2E A A1E AqA2E A uA2E A uA1E A State 2 could also be in the twophase region if vA1E A vA2E A vAgE A at 6 MPa v P 1 2 1970 6000 T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 384 A pistoncylinder contains water with quality 75 at 200 kPa Slow expansion is performed while there is heat transfer and the water is at constant pressure The process stops when the volume has doubled How do you determine the final state and the heat transfer CV Water this is a control mass we do not know size so do all per unit mass Energy Eq35 u E A uA1E A A1E AqA2E A A1E AwA2E 2 Process P C A1E AwA2E A Pdv PvA2E A vA1E A State 1 xA1E A PA1E A Table B12 gives TA1E A uA1E A State 2 vA2E A 2vA1E A PA2E A PA1E Compare vA2E A to vAgE A to determine if sup vapor or not Either find xA2E A or interpolate to get TA2E A and uA2E A in B13 Process P C so the work term integral is A1E AwA2E A Pdv PvA2E A vA1E A From the energy equation A1E AqA2E A uA2E A uA1E A A1E AwA2E A T CP v 2 1 200 kPa P CP v T 200 120 2 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 385 Consider Problem 3173 with the final state given but that you were not told the piston hits the stops and only told VAstopE A 2 VA1E A Sketch the possible Pv diagram for the process and determine which numbers you need to uniquely place state 2 in the diagram There is a kink in the process curve what are the coordinates for that state Write an expression for the work term CV R410A Control mass goes through process 1 2 3 As piston floats pressure is constant 1 2 and the volume is constant for the second part 2 3 So we have vA3E A vA2E A 2 vA1E A State 2 VA2E A VAstopE A vA2E A 2 vA1E A vA3E A and PA2E A PA1E A TA2E A State 3 Table B42 PT vA3E A 002249 mA3E Akg uA3E A 28791 kJkg Now we can find state 1 TA1E A vA1E A vA3E A2 0011245 mA3E Akg twophase Then state 2 the kink vA2E A vA3E A 002249 mA3E Akg PA2E A PA1E A 10857 kPa PAsatE A W P dV PVA2E A VA1E A Pm vA2E A vA1E A v P 1 2 3 R410A P o cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 386 Take problem 3210 and write the left hand side storage change of the conservation equations for mass and energy How do you write mA1E A and Eq 35 CV Both rooms A and B in tank B A Continuity Eq mA2E A mAA1E A mAB1E A 0 Energy Eq mA2E AuA2E A mAA1E AuAA1E A mAB1E AuAB1E A A1E AQA2E A A1E AWA2E A Notice how the state 1 term split into two terms mA1E A ρ dV 1v dV VAAE AvAA1E A VABE AvAB1E A mAA1E A mAB1E and for energy as mA1E AuA1E A ρu dV uv dV uAA1E AvAA1E AVAAE A uAB1E AvAB1E AVABE mAA1E AuAA1E A mAB1E AuAB1E A Formulation continues as Process constant total volume VAtotE A VAAE A VABE A and A1E AWA2E A A0E mA2E A mAA1E A mAB1E A vA2E A VAtotE AmA2E A etc Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 387 Two rigid insulated tanks are connected with a pipe and valve One tank has 05 kg air at 200 kPa 300 K and the other has 075 kg air at 100 kPa 400 K The valve is opened and the air comes to a single uniform state without any heat transfer How do you determine the final temperature and pressure Solution CV Total tank Control mass of constant volume Mass and volume mA2E A mAAE A mABE A V VAAE A VABE A Energy Eq UA2E A UA1E A mA2E A uA2E A mAAE AuAA1E A mABE AuAB1E A A1E AQA2E A A1E AWA2E A 0 Process Eq V constant A1E AWA2E A 0 Insulated A1E AQA2E A 0 Ideal gas at A1 VAAE A mAAE ARTAA1E APAAE A1E A uAA1E A from Table A7 Ideal gas at B1 VABE A mABE ARTAB1E A PAB1E A uAB1E A from Table A7 State 2 mA2E A mAAE A mABE A vA2E A V mA2E A Energy Eq uA2E A A mAuA1 mBuB1 Em2 E A Table A71 TA2E A PA2E A mA2E A RTA2E A V B A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 388 Look at problem 3183 and plot the Pv diagram for the process Only TA2E A is given how do you determine the 2nd property of the final state What do you need to check and does it have an influence on the work term Process P constant FA PA1E A if V Vmin V constant VA1aE A VAminE A if P PA1E State 1 P T VA1E A mRTA1E APA1E A 05 0287 10002000 007175 mA3E The only possible PV combinations for this system are shown in the diagram so both state 1 and 2 must be on the two lines For state 2 we need to know if it is on the horizontal P line segment or the vertical V segment Let us check state 1a State 1a PA1aE A PA1E A VA1aE A VAminE A Ideal gas so TA1aE A TA1E A A V1a EV1 E A We see if TA2E A TA1aE A then state 2 must have VA2E A VA1aE A VAminE A 003 mA3E A So state 2 is known by TA2E A vA2E A and P2 PA1E A T2 T1 V1 V2 If it was that TA2E A TA1aE A then we know state 2 as T2 P2 PA1E A and we then have V2 VA1E A T2 T1 The work is the area under the process curve in the PV diagram and so it does make a difference where state 2 is relative to state 1a For the part of the process that proceeds along the constant volume VAminE A the work is zero there is only work when the volume changes A1E AWA2E A A1 2 E PEA dV PA1E A VA1aE A VA1E A V P 1 2 1a P P 2 1 V T 1 2 1a T T 2 1a T 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Simple processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 389 A 100L rigid tank contains nitrogen NA2E A at 900 K 3 MPa The tank is now cooled to 100 K What are the work and heat transfer for this process Solution CV Nitrogen in tank mA2E A mA1E A Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E Process V constant vA2E A vA1E A Vm A1E AWA2E A A0E Table B62 State 1 vA1E A 00900 mA3E Akg m VvA1E A 1111 kg uA1E A 6917 kJkg State 2 100 K vA2E A vA1E A Vm look in Table B62 at 100 K 200 kPa v 01425 mA3E Akg u 717 kJkg 400 kPa v 00681 mA3E Akg u 693 kJkg so a linear interpolation gives PA2E A 200 200 009 0142500681 01425 341 kPa uA2E A 717 693 717 A 009 01425 00681 01425E A 700 kJkg A1E AQA2E A muA2E A uA1E A 1111 kg 700 6917 kJkg 6907 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 390 A constant pressure pistoncylinder aasembly contains 02 kg water as saturated vapor at 400 kPa It is now cooled so the water occupies half the original volume Find the work and the heat transfer in the process Solution Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E Process P C A1E AWA2E A PdV PVA2E A VA1E A Table B12 State 1 vA1E A 046246 mA3E Akg uA1E A 255355 kJkg State 2 vA2E A vA1E A 2 023123 mA3E Akg vAfE A x vAfgE A xA2E A A v2 vf2 Evfg2 E A A023123 0001084 046138E A 04988 uA2E A uAf2E A xA2E A uAfE AgE A2E A 60429 xA2E A 194926 157658 kJkg Process P C so the work term integral is A1E AWA2E A PVA2E AVA1E A 400 kPa 02 023123 046246 mA3E A 185 kJ From the energy equation A1E AQA2E A muA2E A uA1E A A1E AWA2E A 02 157658 255355 185 2139 kJ T CP v 1 2 P 400 kPa P CP v T 400 144 2 1 cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 391 Saturated vapor R410A at 0AoE AC in a rigid tank is cooled to 20AoE AC Find the specific heat transfer Solution CV R410A in tank mA2E A mA1E A Energy Eq35 uA2E A uA1E A A1E AqA2E A A1E AwA2E Process V constant vA2E A vA1E A Vm A1E AwA2E A A0E Table B41 State 1 uA1E A 2530 kJkg State 2 20AoE AC vA2E A vA1E A Vm look in Table B41 at 20AoE AC xA2E A A v2 vf2 Evfg2 E A A003267 0000803 006400E A 04979 uA2E A uAf2E A xA2E A uAfE AgE A2E A 2792 xA2E A 21807 1365 kJkg From the energy equation A1E AqA2E A uA2E A uA1E A 1365 2530 1165 kJkg V P 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 392 Ammonia at 0C quality 60 is contained in a rigid 200L tank The tank and ammonia is now heated to a final pressure of 1 MPa Determine the heat transfer for the process Solution CV NHA3E V P 1 2 Continuity Eq mA2E A mA1E A m Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E Process Constant volume vA2E A vA1E A A1E AWA2E A 0 State 1 Table B21 twophase state vA1E A 0001566 xA1E A 028763 017414 mA3E Akg uA1E A 17969 06 11383 86267 kJkg m VvA1E A 02 mA3E A017414 mA3E Akg 1148 kg State 2 PA2E A vA2E A vA1E A superheated vapor Table B22 TA2E A 100C uA2E A 14905 kJkg So solve for heat transfer in the energy equation A1E AQA2E A muA2E A uA1E A 1148 kg 14905 86267 kJkg 72075 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 393 A rigid tank contains 15 kg of R134a at 40C 500 kPa The tank is placed in a refrigerator that brings it to 20C Find the process heat transfer and show the process in a Pv diagram CV the R134a Energy Eq UA2E A UA1E A A1E AQA2E A A1E AWA2E A A1E AQA2E A Process Rigid tank V C v constant A1E AWA2E A A1 2 PdVEA 0 State 1 vA1E A 004656 m3kg uA1E A 40744 kJkg State 2 T v twophase straight down in Pv diagram from state 1 xA2E A v vAfE AvAfgE A 004656 0000738014576 031437 uA2E A uAfE A xA2E A uAfgE A 17365 031437 19285 23428 kJkg From the energy equation 1QA2E A muA2E A uA1E A 15 kg 23428 40744 kJkg 2597 kJ V P 2 125 C 1 450 134 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 394 A piston cylinder contains air at 600 kPa 290 K and a volume of 001 m E A A constant pressure process gives 54 kJ of work out Find the final volume the temperature of the air and the heat transfer 3 CV AIR control mass Continuity Eq mA2E A mA1E A 0 Energy Eq m uA2E A uA1E A A1E AQA2E A A1E AWA2E Process P C so A1E AWA2E A P dV PVA2E A VA1E A State 1 PA1E A TA1E AVA1E A State 2 PA1E A PA2E A mA1E A PA1E AVA1E ARTA1E A A 600 kPa 001 m3 E0287 kJkgK 290 KE A 00721 kg A1E AWA2E A PVA2E A VA1E A 54 kJ VA2E A VA1E A A1E AWA2E A P 54 kJ 600 kPa 009 mA3E VA2E A VA1E A A1E AWA2E A P 001 009 01 mA3E Ideal gas law PA2E AVA2E A mRTA2E TA2E A PA2E AVA2E A mR EAEA PA2 AVA2 A E PA1 AVA1 AE A TA1E A A010 001E A 290 K 2900 K Energy equation with us from table A71 A1E AQA2E A m uA2E A uA1E A A1E AWA2E 00721 kg 25638 20719 kJkg 54 kJ 2239 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 395 Two kg water at 120AoE AC with a quality of 25 has its temperature raised 20AoE AC in a constant volume process as in Fig P395 What are the heat transfer and work in the process Solution CV Water This is a control mass Energy Eq m uA2E A uA1E A A1E AQA2E A A1E AWA2E Process V constant A1E AWA2E A P dV 0 State 1 T xA1E A from Table B11 vA1E A vAfE A xA1E A vAfgE A 000106 025 08908 022376 mA3E Akg uA1E A uAfE A xA1E A uAfE AgE A 50348 025 202576 100992 kJkg State 2 TA2E A vA2E A vA1E A vAgE A2E A 050885 mA3E Akg so twophase xA2E A A v2 vf2 Evfg2 E A A022376 000108 050777E A 043855 uA2E A uAf2E A xA2E A uAfE AgE A2E A 58872 xA2E A 19613 144884 kJkg From the energy equation A1E AQA2E A muA2E A uA1E A 2 kg 144884 100992 kJkg 8778 kJ T CP v P CP v T 1985 3613 120 140 120 C 140 C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 396 A cylinder fitted with a frictionless piston contains 2 kg of superheated refrigerant R134a vapor at 350 kPa 100AoE AC The cylinder is now cooled so the R134a remains at constant pressure until it reaches a quality of 75 Calculate the heat transfer in the process Solution CV R134a mA2E A mA1E A m Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E Process P const A1E AWA2E A APdVEA PV PVA2E A VA1E A PmvA2E A vA1E A V P 1 2 V T 1 2 State 1 Table B52 hA1E A 49048 489522 490 kJkg State 2 Table B51 hA2E A 20675 075 19457 3527 kJkg 3509 kPa A1E AQA2E A muA2E A uA1E A A1E AWA2E A muA2E A uA1E A PmvA2E A vA1E A mhA2E A hA1E A A1E AQA2E A 2 kg 3527 490 kJkg 2746 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 397 A piston cylinder contains 15 kg water at 200 kPa 150AoE AC It is now heated in a process where pressure is linearly related to volume to a state of 600 kPa 350AoE AC Find the final volume the work and the heat transfer in the process Take as CV the 15 kg of water mA2E A mA1E A m Process Eq P A BV linearly in V State 1 P T vA1E A 095964 mA3E Akg uA1E A 257687 kJkg State 2 P T vA2E A 047424 mA3E Akg uA2E A 288112 kJkg VA2E A mvA2E A 07114 mA3E From process eq A1E AWA2E A P dV area Am 2E A PA1E A PA2E Av2 v1 A15 2E A kg 200 600 kPa 047424 095964 mA3E Akg 29124 kJ Notice volume is reduced so work is negative Energy Eq muA2E A uA1E A A1E AQA2E A A1E AWA2E A1E AQA2E A muA2E A uA1E A A1E AWA2E A 15 kg 288112 257687 kJkg 29124 kJ 1651 kJ P CP v T 2 200 1 6 00 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 398 A pistoncylinder contains 50 kg of water at 200 kPa with a volume of 01 m3 Stops in the cylinder are placed to restrict the enclosed volume to a maximum of 05 m3 The water is now heated until the piston reaches the stops Find the necessary heat transfer Solution CV H E AO m constant 2 Energy Eq35 meA2E A eA1E A muA2E A uA1E A A1E AQA2E A A1E AWA2E Process P constant forces on piston constant A1E AWA2E A P dV PA1E A VA2E A VA1E A Properties from Table B11 State 1 vA1E A 0150 0002 mA3E Akg 2phase as vA1E A vAgE xA1E A v1 vf vfg A0002 0001061 088467E A 0001061 hA1E A 50468 0001061 220196 50702 kJkg State 2 vA2E A 0550 001 mA3E Akg also 2phase same P xA2E A v2 vf vfg A001 0001061 088467E A 001010 hA2E A 50468 001010 220196 52692 kJkg Find the heat transfer from the energy equation as A1E AQA2E A muA2E A uA1E A A1E AWA2E A mhA2E A hA1E A A1E AQA2E A 50 kg 52692 50702 kJkg 995 kJ Notice that A1E AWA2E A PA1E A VA2E A VA1E A 200 kPa 05 01 m3 80 kJ V P 1 2 01 05 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 399 Ammonia 05 kg is in a piston cylinder at 200 kPa 10AoE AC is heated in a process where the pressure varies linear with the volume to a state of 120AoE AC 300 kPa Find the work and the heat transfer for the ammonia in the process Solution Take CV as the Ammonia constant mass Continuity Eq m2 m1 m Process P A BV linear in V State 1 Superheated vapor v1 06193 mA3E Akg uA1E A 13167 kJkg State 2 Superheated vapor v2 063276 mA3E Akg uA2E A 15420 kJkg Work is done while piston moves at increasing pressure so we get 1W2 P dV area Pavg V2 V1 A1 2E A PA1E A PA2E AmvA2E A vA1E A ½200 300 kPa 05 kg 063276 06193 mA3E Akg 1683 kJ Energy Eq muA2E A uA1E A A1E AQA2E A A1E AWA2E A1E AQA2E A muA2E A uA1E A A1E AWA2E A 05 kg 15420 13167 kJkg 1683 kJ 1143 kJ P P 2 P v 1 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3100 A pistoncylinder contains 1 kg water at 20 E AC with volume 01 mA3E A By mistake someone locks the piston preventing it from moving while we heat the water to saturated vapor Find the final temperature and the amount of heat transfer in the process o Solution CV Water This is a control mass Energy Eq m uA2E A uA1E A A1E AQA2E A A1E AWA2E Process V constant A1E AWA2E A 0 State 1 T vA1E A VA1E Am 01 mA3E Akg vAfE A so twophase xA1E A A v1 vf Evfg E A A010001002 577887E A 00017131 uA1E A uAfE A xA1E A uAfE AgE A 8394 xA1E A 231898 87913 kJkg State 2 vA2E A vA1E A 01 xA2E A 1 found in Table B11 between 210C and 215 C TA2E A 210 5 A 01010441 009479010441E A 210 5 04584 2123C uA2E A 259944 04584 260106 259944 26002 kJkg From the energy equation A1E AQA2E A muA2E A uA1E A 1 kg 26002 87913 kJkg 25123 kJ V P V T 2 1 2 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3101 A waterfilled reactor with volume of 1 m E A is at 20 MPa 360C and placed inside a containment room as shown in Fig P3101 The room is well insulated and initially evacuated Due to a failure the reactor ruptures and the water fills the containment room Find the minimum room volume so the final pressure does not exceed 200 kPa 3 Solution CV Containment room and reactor Mass mA2E A mA1E A VAreactorE AvA1E A 10001823 5485 kg Energy muA2E A uA1E A A1E AQA2E A A1E AWA2E A 0 0 0 State 1 Table B14 vA1E A 0001823 mA3E Akg uA1E A 17028 kJkg Energy equation then gives uA2E A uA1E A 17028 kJkg State 2 PA2E A 200 kPa uA2E A uAgE A Twophase Table B12 xA2E A uA2E A uAfE A uAfgE A 17028 50447202502 059176 vA2E A 0001061 059176 088467 052457 mA3E Akg VA2E A mA2E A vA2E A 5485 kg 052457 mA3E Akg 2877 mA3E P v 1 T v 1 2 200 kPa 200 2 u const P T v L CP 1 2 200 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3102 A rigid tank holds 075 kg ammonia at 70C as saturated vapor The tank is now cooled to 20C by heat transfer to the ambient Which two properties determine the final state Determine the amount of work and heat transfer during the process CV The ammonia this is a control mass Process Rigid tank V C v constant A1E AWA2E A A 1 2 PdVE A 0 Energy Eq UA2E A UA1E A A1E AQA2E A A1E AWA2E A A1E AQA2E A State 1 vA1E A 003787 mA3E Akg uA1E A 13389 kJkg State 2 T v twophase straight down in Pv diagram from state 1 xA2E A v vAfE AvAfgE A 003787 0001638014758 02455 uA2E A uAfE A xA2E A uAfgE A 27289 02455 10593 53295 kJkg A1E AQA2E A muA2E A uA1E A 075 kg 53295 13389 kJkg 6045 kJ V P 2 70 C 1 3312 858 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3103 Water in a 150L closed rigid tank is at 100C 90 quality The tank is then cooled to 10C Calculate the heat transfer during the process Solution CV Water in tank m E A mA1E A 2 Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E Process V constant vA2E A vA1E A A1E AWA2E A 0 State 1 Twophase L V look in Table B11 vA1E A 0001044 09 16719 15057 mA3E Akg uA1E A 41894 09 20876 22978 kJkg State 2 TA2E A vA2E A vA1E A mix of saturated solid vapor Table B15 vA2E A 15057 00010891 xA2E A 4667 xA2E A 0003224 uA2E A 35409 0003224 27155 34534 kJkg m VvA1E A 01515057 009962 kg A1E AQA2E A muA2E A uA1E A 009962 kg 34534 22978 kJkg 2633 kJ P CP v T CP v T 1 1 P const 2 2 P T v S V L V V L S CP 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3104 A 25 kg mass moves with 25 ms Now a brake system brings the mass to a complete stop with a constant deceleration over a period of 5 seconds The brake energy is absorbed by 05 kg water initially at 20AoE AC 100 kPa Assume the mass is at constant P and T Find the energy the brake removes from the mass and the temperature increase of the water assuming P C Solution CV The mass in motion EA2E A EA1E A E 05 mVA2E A 05 25 25A2E A1000 78125 kJ CV The mass of water muA2E A uA1E A HA2E AO E 78125 kJ uA2E A uA1E A 78125 kJ 05 kg 1563 kJkg uA2E A uA1E A 1563 8394 1563 99565 kJkg Assume uA2E A uAfE A then from Table B11 TA2E A 237AoE AC T 37AoE AC We could have used uA2E A uA1E A CT with C from Table A4 C 418 kJkg K giving T 1563418 37AoE AC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3105 A cylinder having a piston restrained by a linear spring of spring constant 15 kNm contains 05 kg of saturated vapor water at 120C as shown in Fig P3105 Heat is transferred to the water causing the piston to rise If the piston crosssectional area is 005 m2 and the pressure varies linearly with volume until a final pressure of 500 kPa is reached Find the final temperature in the cylinder and the heat transfer for the process Solution CV Water in cylinder Continuity mA2E A mA1E A m Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E A State 1 T x Table B11 vA1E A 089186 mA3E Akg uA1E A 25292 kJkg Process PA2E A PA1E A A ksm EAp 2 E A vA2E A vA1E A 1985 A15 05 0052 E A vA2E A 089186 State 2 PA2E A 500 kPa and on the process curve see above equation vA2E A 089186 500 1985 005275 09924 mA3E Akg P v Table B13 TA2E A 803C uA2E A 3668 kJkg The process equation allows us to evaluate the work A1E AWA2E A A PdVEA A P1 P2 E2 E A mvA2E A vA1E A A 1985 500 2 E A kPa 05 kg 09924 089186 mA3E Akg 1756 kJ Substitute the work into the energy equation and solve for the heat transfer A1E AQA2E A muA2E A uA1E A A1E AWA2E A 05 kg 3668 25292 kJkg 1756 kJ 587 kJ P v 1 2 T v 1 2 k m s Ap2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3106 A piston cylinder arrangement with a linear spring similar to Fig P3105 contains R134a at 15oC x 04 and a volume of 002 mA3E A It is heated to 60oC at which point the specific volume is 003002 mA3E Akg Find the final pressure the work and the heat transfer in the process Take CV as the R134a m2 m1 m mu2 u1 1Q2 1W2 State 1 T1 x1 Two phase so Table B51 P1 Psat 4895 kPa v1 vf x1 vfg 0000805 04 004133 001734 mA3E Akg u1 uf x1 ufg 2201 04 16635 28664 kJkg m V1v1 002 mA3E A 001734 mA3E Akg 11534 kg State 2 T v Superheated vapor Table B52 P2 800 kPa v2 003002 mA3E Akg u2 4212 kJkg V2 m v2 11534 kg 003002 mA3E Akg 003463 mA3E Work is done while piston moves at linearly varying pressure so we get 1W2 P dV area Pavg V2 V1 05P2 P1 V2 V1 05 4895 800 kPa 003463 002 mA3E A 9433 kJ Heat transfer is found from the energy equation 1Q2 mu2 u1 1W2 11534 4212 28664 943 1646 kJ R134a P P 2 P v 1 1 2 cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3107 A 10m high open cylinder Acyl 01 m2 contains 20C water above and 2 kg of 20C water below a 1985kg thin insulated floating piston shown in Fig P3107 Assume standard g Po Now heat is added to the water below the piston so that it expands pushing the piston up causing the water on top to spill over the edge This process continues until the piston reaches the top of the cylinder Find the final state of the water below the piston T P v and the heat added during the process Solution CV Water below the piston Piston force balance at initial state F F P E AA mApE Ag mABE Ag PA0E AA A State 1AAB E A Comp Liq v vAfE A 0001002 mA3E Akg uA1AE A 8395 kJkg VAA1E A mAAE AvAA1E A 0002 mA3E A mAtotE A VAtotE Av 10001002 998 kg mass above the piston mAB1E A mAtotE A mAAE A 996 kg PAA1E A PA0E A mApE A mABE AgA 101325 A1985 996 9807 E 01 1000E A 2185 kPa State 2AAE A PAA2E A PA0E A A mpg EAE A 1208 kPa vAA2E A VAtotE A mAAE A 05 mA3E Akg xAA2E A 05 000104714183 0352 TA2E A 105C uAA2E A 4400 0352 207234 11695 kJkg Continuity eq in A mAA2E A mAA1E Energy mAAE AuA2E A uA1E A A1E AQA2E A A1E AWA2E Process P linear in V as mABE A is linear with V A1E AWA2E A APdVEA A1 2E A2185 12082 kPa 1 0002 mA3E 16932 kJ P 1 2 V W cb A1E AQA2E A mAAE AuA2E A uA1E A A1E AWA2E A 21701 1693 23404 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3108 Assume the same setup as in Problem 3101 but the room has a volume of 100 mA3E A Show that the final state is twophase and find the final pressure by trial and error CV Containment room and reactor Mass mA2E A mA1E A VAreactorE AvA1E A 10001823 5485 kg Energy muA2E A uA1E A A1E AQA2E A A1E AWA2E A 0 0 0 uA2E A uA1E A 17028 kJkg Total volume and mass vA2E A VAroomE AmA2E A 01823 mA3E Akg State 2 uA2E A vA2E A Table B11 see Figure Note that in the vicinity of v 01823 mA3E Akg crossing the saturated vapor line the internal energy is about 2585 kJkg However at the actual state 2 u 17028 kJkg Therefore state 2 must be in the twophase region Trial error v vAfE A xvAfgE A u uAfE A xuAfgE uA2E A 17028 uAf E A A v2 vf Evfg E A uAfgE Compute RHS for a guessed pressure PA2E A PA2E A 600 kPa RHS 66988 A018230001101 031457E A 189752 17629 too large PA2E A 550 kPa RHS 65530 A018230001097 034159E A 190917 16681 too small Linear interpolation to match u 17028 gives PA2E A 5685 kPa v T 0184 u2585 1060 kPa sat vap 1060 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3109 A piston cylinder contains carbon dioxide at 20oC and quality 75 It is compressed in a process where pressure is linear in volume to a state of 3 MPa and 20oC Find the specific heat transfer CV Carbon dioxide out to the source both A1E AQA2E A and A1E AWA2E A Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E A Process P A BV A1E AWA2E A P dV ½ mPA1E A PA2E A vA2E A vA1E A State 1 Table B31 P 19696 kPa vA1E A 0000969 075 001837 001475 mA3E Akg uA1E A 3964 075 24625 22433 kJkg State 2 Table B3 v2 001512 mA3E Akg uA2E A 31021 kJkg A1E AwA2E A ½ PA1E A PA2E A vA2E A vA1E A ½ 19696 3000 kPa 001512 001475 mA3E Akg 092 kJkg A1E AqA2E A uA2E A uA1E A A1E AwA2E A 31021 22433 092 868 kJkg P v 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3110 A rigid steel tank of mass 25 kg contains 05 kg R410A at 0C with specific volume 001mkg The whole system is now heated to to a room temperature of 25C a Find the volume of the tank b Find the final P c Find the process heat transfer CV R410A and steel tank Control mass goes through process 1 2 Continuity Eq mA2E A mAR410aE A mAstE A 0 Energy Eq mAR410aE AuA2E A uA1E A mAstE AuA2E A uA1E A A1E AQA2E A A1E AWA2E A Process V C so A1E AWA2E A 0 State 1 TA1E A 0C vA1E A 001 m3kg V mvA1E A 0005 m3 xA1E A v vAfE A vAfgE A 001 0000855003182 028758 uA1E A uAfE A xA1E A uAfgE A 5707 xA1E A 19595 11342 kJkg State 2 T v supvapor straight up in Tv diagram from state 1 B41 at 25C vAfE A 0000944 m3kg vg 001514 m3kg vAfE A v vg saturated P 16536 kPa x v vf vfg A001 0000944 001420E A 063775 uA2E A uAfE A xA2E A uAfgE A 9603 xA2E A 16295 19995 kJkg From the energy Eq A1E AQA2E A mAR410aE AuA2E A uA1E A mAstE A CAstE ATA2E A TA1E A 05 kg 19995 11342 kJkg 25 kg 046 kJkgK 250 K 720 kJ v T 2 16536 kPa 1 0 25 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3111 The pistoncylinder in FigP3111 contains 01 kg water at 500C 1000 kPa The piston has a stop at half the original volume The water now cools to room temperature 25C a Sketch the possible water states in a Pv diagram b Find the final pressure and volume c Find the heat transfer and work in the process Energy Eq mu E A uA1E A A1E AQA2E A A1E AWA2E 2 Process Eq P C if v vAstopE A V C if P PAfloatE A State 1 vA1E A 035411 m3kg uA1E A 312434 kJkg State a vAaE A vA1E A2 0177055 mA3E Akg vAg 1000 kPaE A so TAaE A TAsat 1000 kPaE A 1799C The possible state 2 PV combinations are shown State a is 1000 kPa vAaE A so it is twophase with TAaE A 180C TA2E PA2E A PAsat 25 CE A 3169 kPa and vA2E A vAaE xA2E A vA2E A vAfE A vAfgE A 0177 000100343358 00040604 uA2E A uAfE A xA2E A uAfgE A 10486 xA2E A 23049 114219 kJkg VA2E A mvA2E A 01 kg 0177055 m3kg 00177 m3 A1E AWA2E A m P dv m PA1E A vA2E A vA1E A see area below process curve in figure 01 kg 1000 kPa 0177055 035411 m3kg 17706 kJ A1E AQA2E A muA2E A uA1E A A1E AWA2E A 01 kg 114219 312434 kJkg 17706 kJ 31872 kJ Water P o m p V P 2 500 C a 317 1000 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3112 A spring loaded pistoncylinder assembly contains 1 kg water at 500 E AC 3 MPa The setup is such that the pressure is proportional to volume P CV It is now cooled until the water becomes saturated vapor Sketch the Pv diagram and find the final state the work and heat transfer in the process o Solution State 1 Table B13 vA1E A 011619 mA3E Akg uA1E A 310792 kJkg Process m is constant and P CA0E AV CA0E Am v C v polytropic process with n 1 P Cv C PA1E AvA1E A 3000011619 25 820 kPa kgmA3E State 2 xA2E A 1 PA2E A CvA2E A on process line 2 1 P v C Trial error on TA2satE A or PA2satE A Here from B12 at 2 MPa vAgE A 009963 C PvAgE A 20074 low 25 MPa vAgE A 007998 C PvAgE A 31258 high 225 MPa vAgE A 008875 C PvAgE A 25352 low Now interpolate to match the right slope C PA2E A 2250 250 A25 820 25 352 31 258 25 352E A 2270 kPa vA2E A PA2E AC 227025820 00879 mA3E Akg uA2E A 260207 kJkg P is linear in V so the work becomes area in Pv diagram A1E AWA2E A P dv m A1 2E APA1E A PA2E AvA2E A vA1E A 1 kg A1 2E A 3000 2270 kPa 00879 011619 mA3E A 745 kJ From the energy Eq A1E AQA2E A muA2E A uA1E A A1E AWA2E A 1 260207 310792 745 125085 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3113 A piston cylinder contains 15 kg water at 600 kPa 350oC It is now cooled in a process where pressure is linearly related to volume to a state of 200 kPa 150oC Plot the Pv diagram for the process and find both the work and the heat transfer in the process Take as CV the 15 kg of water mA2E A mA1E A m Energy Eq35 muA2E A uA1E A A1E AQ2 A1E AW2 Process Eq P A BV linearly in V State 1 P T vA1E A 047424 mA3E Akg uA1E A 288112 kJkg State 2 P T vA2E A 095964 mA3E Akg uA2E A 257687 kJkg From process eq A1E AWA2E A P dV area Am 2E A PA1E A PA2E Av2 v1 A15 2E A kg 200 600 kPa 095964 047424 mA3E Akg 29124 kJ From energy eq A1E AQA2E A muA2E A uA1E A A1E AWA2E A 15 kg 257687 288112 kJkg 29124 kJ 16514 kJ P CP v T 1 200 2 6 00 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3114 Superheated refrigerant R134a at 20C 05 MPa is cooled in a pistoncylinder arrangement at constant temperature to a final twophase state with quality of 50 The refrigerant mass is 5 kg and during this process 500 kJ of heat is removed Find the initial and final volumes and the necessary work Solution CV R134a this is a control mass Continuity m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 500 kJ 1W2 State 1 T1 P1 Table B52 v1 004226 m3kg u1 39052 kJkg V1 mv1 0211 m3 State 2 T2 x2 Table B51 u2 22703 05 16216 30811 kJkg v2 0000817 05 003524 0018437 m3kg V2 mv2 00922 m3 1W2 500 kJ mu2 u1 500 kJ 5 kg 30811 39052 kJkg 879 kJ v P v T 1 2 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3115 Two kilograms of nitrogen at 100 K x 05 is heated in a constant pressure process to 300 K in a pistoncylinder arrangement Find the initial and final volumes and the total heat transfer required Solution Take CV as the nitrogen Continuity Eq m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Process P constant 1W2 PdV Pmv2 v1 State 1 Table B61 v1 0001452 05 002975 001633 m3kg V1 00327 m 3 h1 7320 05 16068 714 kJkg State 2 P 7792 kPa 300 K sup vapor interpolate in Table B62 v2 014824 011115014824 1792200 0115 m3kg V2 023 m 3 h2 31006 3096231006 1792200 30966 kJkg Now solve for the heat transfer from the energy equation 1Q2 mu2 u1 1W2 mh2 h1 2 kg 30966 714 kJkg 605 kJ V P 1 2 V T 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Energy Equation Solids and Liquids Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3116 In a sink 5 liters of water at 70oC is combined with 1 kg aluminum pots 1 kg of flatware steel and 1 kg of glass all put in at 20oC What is the final uniform temperature neglecting any heat loss and work Energy Eq U2 U1 miu2 u1i 1Q2 1W2 0 For the water vf 0001023 m3kg V 5 L 0005 m3 m Vv 48876 kg For the liquid and the metal masses we will use the specific heats Tbl A3 A4 so miu2 u1i miCv i T2 T1i T2miCv i miCv iT1 i noticing that all masses have the same T2 but not same initial T miCv i 48876 418 1 09 1 046 1 08 2259 kJK Energy Eq 2259 T2 48876 418 70 1 09 1 046 1 08 20 143011 432 14733 kJ T2 652oC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3117 A computer CPU chip consists of 50 g silicon 20 g copper 50 g polyvinyl chloride plastic It heats from 15oC to 70oC as the computer is turned on How much energy does the heating require Energy Eq U2 U1 mi u2 u1i 1Q2 1W2 For the solid masses we will use the specific heats Table A3 and they all have the same temperature so miu2 u1i miCv i T2 T1i T2 T1miCv i miCv i 005 07 002 042 005 096 00914 kJK U2 U1 00914 kJK 70 15 K 503 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3118 A copper block of volume 1 L is heat treated at 500C and now cooled in a 200L oil bath initially at 20C shown in Fig P3118 Assuming no heat transfer with the surroundings what is the final temperature Solution CV Copper block and the oil bath Also assume no change in volume so the work will be zero Energy Eq U2 U1 mmetu2 u1met moilu2 u1oil 1Q2 1W2 0 Properties from Table A3 and A4 mmet Vρ 0001 m3 8300 kgm3 83 kg moil Vρ 02 m3 910 kgm3 182 kg Solid and liquid Eq332 u Cv T Table A3 and A4 Cv met 042 kJ kg K Cv oil 18 kJ kg K The energy equation for the CV becomes mmetCv metT2 T1met moilCv oilT2 T1oil 0 83 kg 042 kJ kg K T2 500 C 182 kg 18 kJ kg K T2 20 C 0 33109 T2 1743 6552 0 T2 25 C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3119 A 1 kg steel pot contains 1 kg liquid water both at 15oC It is now put on the stove where it is heated to the boiling point of the water Neglect any air being heated and find the total amount of energy needed Solution Energy Eq U2 U1 1Q2 1W 2 The steel does not change volume and the change for the liquid is minimal so 1W2 0 State 2 T2 Tsat 1atm 100oC Tbl B11 u1 6298 kJkg u2 41891 kJkg Tbl A3 Cst 046 kJkg K Solve for the heat transfer from the energy equation 1Q2 U2 U1 mst u2 u1st mH2O u2 u1 H2O mstCst T2 T1 mH2O u2 u1 H2O 1Q2 1 kg 046 kJ kg K 100 15 K 1 kg 41891 6298 kJkg 391 35593 395 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3120 I have 2 kg of liquid water at 20oC 100 kPa I now add 20 kJ of energy at a constant pressure How hot does it get if it is heated How fast does it move if it is pushed by a constant horizontal force How high does it go if it is raised straight up a Heat at 100 kPa Energy equation E2 E1 1Q2 1W2 1Q2 PV2 V1 H2 H1 mh2 h1 h2 h1 1Q2m 8394 202 9404 kJkg Back interpolate in Table B11 T2 225oC We could also have used T 1Q2mC 20 2418 24oC b Push at constant P It gains kinetic energy 05 m V2 2 1W2 V2 2 1W2m 2 20 1000 J2 kg 1414 ms c Raised in gravitational field m g Z2 1W 2 Z2 1W2m g 20 000 J 2 kg 9807 ms2 1019 m Comment Notice how fast 500 kmh and how high it should be to have the same energy as raising the temperature just 2 degrees Ie in most applications we can disregard the kinetic and potential energies unless we have very high V or Z Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3121 A house is being designed to use a thick concrete floor mass as thermal storage material for solar energy heating The concrete is 30 cm thick and the area exposed to the sun during the daytime is 4 m 6 m It is expected that this mass will undergo an average temperature rise of about 3C during the day How much energy will be available for heating during the nighttime hours Solution CV Control mass concrete V 4 m 6 m 03 m 72 m3 Concrete is a solid with some properties listed in Table A3 m ρV 2200 kgm3 72 m3 15 840 kg Energy Eq mu2 u1 1Q2 1W2 1Q 2 The available heat transfer is the change in U From Eq333 and C from table A3 U m C T 15 840 kg 088 kJ kg K 3 K 41 818 kJ 4182 MJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3122 Because a hot water supply must also heat some pipe mass as it is turned on so it does not come out hot right away Assume 80oC liquid water at 100 kPa is cooled to 45oC as it heats 15 kg of copper pipe from 20 to 45oC How much mass kg of water is needed Solution CV Water and copper pipe No external heat transfer no work Energy Eq35 U2 U1 Ucu UH2O 0 0 From Eq333 and Table A3 Ucu mC Τ 15 kg 042 kJ kg K 45 20 K 1575 kJ From the energy equation mH2O Ucu uH2O mH2O Ucu CH2O ΤH2O 1575 418 35 1076 kg or using Table B11 for water mH2O Ucu u1 u2 1575 33484 18841 kJ kJkg 1076 kg Water Cu pipe The real problem involves a flow and is not analyzed by this simple process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3123 A car with mass 1275 kg drives at 60 kmh when the brakes are applied quickly to decrease its speed to 20 kmh Assume the brake pads are 05 kg mass with heat capacity of 11 kJkg K and the brake discsdrums are 40 kg steel Further assume both masses are heated uniformly Find the temperature increase in the brake assembly Solution CV Car Car loses kinetic energy and brake system gains internal u No heat transfer short time and no work term m constant Energy Eq35 E2 E1 0 0 mcar 1 2V2 2 V2 1 mbrakeu2 u1 The brake system mass is two different kinds so split it also use Cv from Table A3 since we do not have a u table for steel or brake pad material msteel Cv T mpad Cv T mcar 05 602 202 1000 3600 2 m2s2 4 046 05 11 kJ K T 1275 kg 05 3200 0077 16 m2s2 157 406 J 1574 kJ T 659 C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3124 A piston cylinder 05 kg steel altogether maintaining a constant pressure has 02 kg R134a as saturated vapor at 150 kPa It is heated to 40oC and the steel is at the same temperature as the R134a at any time Find the work and heat transfer for the process CV The R134a plus the steel Constant total mass m2 m1 m U2 U1 mR134au2 u1R134a msteelu2 u1 1Q2 1W2 State 1 B52 sat vapor v1 013139 m3kg u1 36806 kJkg State 2 B52 sup vapor v2 016592 m3kg u2 41159 kJkg V1 mv1 02 013139 002628 m3 V2 mv2 02 016592 003318 m3 Steel A3 Csteel 046 kJkgK Process P C for the R134a and constant volume for the steel 1W2 P dV P1V2 V1 150 kPa 003318 002628 m3 1035 kJ 1Q2 mR134a u2 u1 msteel u2 u1 1W2 mR134a u2 u1 msteelCsteel T2 T1 1W2 02 41159 36806 05 046 40 1729 1035 8706 13177 1035 2292 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3125 A 25 kg steel tank initially at 10oC is filled up with 100 kg of milk assume properties as water at 30oC The milk and the steel come to a uniform temperature of 5 oC in a storage room How much heat transfer is needed for this process Solution CV Steel Milk This is a control mass Energy Eq35 U2 U1 1Q2 1W2 1Q 2 Process V constant so there is no work 1W2 0 Use Eq333 and values from A3 and A4 to evaluate changes in u 1Q2 msteel u2 u1steel mmilku2 u1milk 25 kg 046 kJ kg K 5 10 Κ 100 kg 418 kJ kg K 5 30 Κ 1725 10450 10277 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3126 An engine consists of a 100 kg cast iron block with a 20 kg aluminum head 20 kg steel parts 5 kg engine oil and 6 kg glycerine antifreeze Everything begins at 5o C and as the engine starts we want to know how hot it becomes if it absorbs a net of 7000 kJ before it reaches a steady uniform temperature Energy Eq U2 U1 1Q2 1W 2 Process The steel does not change volume and the change for the liquid is minimal so 1W2 0 So sum over the various parts of the left hand side in the energy equation mFe u2 u1 mAl u2 u1Al mst u u1s t moil u2 u1oil mgly u2 u1gly 1Q 2 Table A3 CFe 042 CAl 09 Cst 046 all units of kJkg K Table A4 Coil 19 Cgly 242 all units of kJkg K So now we factor out T2 T1 as u2 u1 CT2 T1 for each term mFeCFe mAlCAl mstCst moilCoil mglyCgly T2 T1 1Q 2 T2 T1 1Q2 Σmi C i 7000 100 042 20 09 20 046 5 19 6 242 kJ kJK 7000 9322 K 75 K T2 T1 75oC 5 75 80oC Exhaust flow Air intake filter Coolant flow Atm air Shaft Fan power Radiator Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Properties u h Cv and Cp Ideal Gas Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3127 An ideal gas is heated from 500 to 1500 K Find the change in enthalpy using constant specific heat from Table A5 room temperature value and discuss the accuracy of the result if the gas is a Argon b Oxygen c Carbon dioxide Solution T1 500 K T2 1500 K h CP0T2T1 a Ar h 0520 kJkgK 1500500 K 520 kJkg Monatomic inert gas very good approximation b O2 h 0922 kJkgK 1500500 K 922 kJkg Diatomic gas approximation is OK with some error c CO2 h 0842 kJkgK 1500500 K 842 kJkg Polyatomic gas heat capacity changes see figure 326 See also appendix C for more explanation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3128 Use the ideal gas air table A7 to evaluate the heat capacity Cp at 300 K as a slope of the curve hT by hT How much larger is it at 1000 K and 1500 K Solution From Eq339 Cp dh dT h T h320 h290 320 290 1005 kJkg K 1000 K Cp h T h1050 h950 1050 950 110348 98944 100 1140 kJkg K 1500 K Cp h T h1550 h1450 1550 1450 169645 15754 100 121 kJkg K Notice an increase of 14 21 respectively h T 300 1000 1500 Cp 300 Cp 1500 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3129 Estimate the constant specific heats for R134a from Table B52 at 100 kPa and 125oC Compare this to table A5 and explain the difference Solution Using values at 100 kPa for h and u at 120oC and 130oC from Table B52 the approximate specific heats at 125oC are Cp h T 52198 51195 130 120 1003 kJkg K compared with 0852 kJkg K for the idealgas value at 25oC from Table A5 Cv u T 48936 48016 130 120 0920 kJkg K compared with 0771 kJkg K for the idealgas value at 25oC from Table A5 There are two reasons for the differences First R134a is not exactly an ideal gas at the given state 125oC and 100 kPa Second and by far the biggest reason for the differences is that R134a chemically CF3CH2 is a polyatomic molecule with multiple vibrational mode contributions to the specific heats see Appendix C such that they are strongly dependent on temperature Note that if we repeat the above approximation for Cp in Table B52 at 25oC the resulting value is 0851 kJkg K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3130 We want to find the change in u for carbon dioxide between 600 K and 1200 K a Find it from a constant Cvo from table A5 b Find it from a Cvo evaluated from equation in A6 at the average T c Find it from the values of u listed in table A8 Solution a u Cvo T 0653 kJkgK 1200 600 K 3918 kJkg b Tavg 1 2 1200 600 900 θ T 1000 900 1000 09 Cpo 045 167 09 127 092 039 093 12086 kJkg K Cvo Cpo R 12086 01889 10197 kJkg K u 10197 1200 600 6118 kJkg c u 99664 39272 60392 kJkg u T 300 600 1200 u 600 u 1200 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3131 Nitrogen at 300 K 3 MPa is heated to 500 K Find the change in enthalpy using a Table B6 b Table A8 and c Table A5 B62 h2 h1 51929 30494 21435 kJkg A8 h2 h1 52075 31167 20908 kJkg A5 h2 h1 Cpo T2 T1 1042 kJkgK 500 300 K 2084 kJkg Comment The results are listed in order of accuracy B62 best Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3132 We want to find the change in u for carbon dioxide between 50oC and 200oC at a pressure of 10 MPa Find it using ideal gas and Table A5 and repeat using the B section table Solution Using the value of Cvo for CO2 from Table A5 u Cvo T 0653 kJkgK 200 50 K 9795 kJkg Using values of u from Table B32 at 10 000 kPa with linear interpolation between 40oC and 60oC for the 50oC value u u200 u50 4376 2309 2067 kJkg Note Since the state 50oC 10 000 kPa is in the densefluid supercritical region a linear interpolation is quite inaccurate The proper value for u at this state is found from the CATT software to be 2451 instead of 2309 This results is u u200 u50 4376 2451 1925 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3133 We want to find the change in u for oxygen gas between 600 K and 1200 K a Find it from a constant Cvo from table A5 b Find it from a Cvo evaluated from equation in A6 at the average T c Find it from the values of u listed in table A8 Solution a u Cvo T 0662 kJkgK 1200 600 K 3972 kJkg b Tavg 1 2 1200 600 900 K θ T 1000 900 1000 09 Cpo 088 00001 09 054 092 033 093 10767 kJkgK Cvo Cpo R 10767 02598 08169 kJkgK u 08169 1200 600 4901 kJkg c u 88972 40446 4853 kJkg u T 300 600 1200 u 600 u 1200 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3134 For a special application we need to evaluate the change in enthalpy for carbon dioxide from 30oC to 1500oC at 100 kPa Do this using constant specific heat from Table A5 and repeat using Table A8 Which is the more accurate one Solution Using constant specific heat h CpoT 0842 1500 30 12377 kJkg Using Table A8 30C 30315 K h 21438 315 50 2579 21438 21712 kJkg 1500C 177315 K h 188243 7315 100 201767 188243 198136 kJkg h 198136 21712 17642 kJkg The result from A8 is best For large T or small T at high Tavg constant specific heat is poor approximation or it must be evaluated at a higher T A5 is at 25oC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3135 Water at 150C 400 kPa is brought to 1200C in a constant pressure process Find the change in the specific internal energy using a the steam tables b the ideal gas water table A8 and c the specific heat from A5 Solution a State 1 Table B13 Superheated vapor u1 256448 kJkg State 2 Table B13 u2 446723 kJkg u2 u1 446723 256448 190275 kJkg b Table A8 at 42315 K u1 59141 kJkg Table A8 at 147315 K u2 247425 kJkg u2 u1 247425 59141 18828 kJkg c Table A5 Cvo 141 kJkgK u2 u1 141 kJkgK 1200 150 K 14805 kJkg Notice how the average slope from 150C to 1200 C is higher than the one at 25C Cvo u T 25 150 1200 u 150 u 1200 Slope at 25 C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3136 Repeat the previous problem but use a constant specific heat at the average temperature from equation in Table A6 and also integrate the equation in Table A6 to get the change in enthalpy Tave 1 2 30 1500 27315 103815 K θ T1000 10382 Table A6 Cpo 12513 kJkgK h Cpoave T 12513 kJkgK 1470 K 1839 kJkg For the entry to Table A6 30C 30315 K θ1 030315 1500C 177315 K θ2 177315 h h2 h1 Cpo dT 045 θ2 θ1 167 1 2 θ2 2 θ1 2 127 1 3 θ2 3 θ1 3 039 1 4 θ2 4 θ1 4 17628 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3137 Water at 20C 100 kPa is brought to 200 kPa 1500C Find the change in the specific internal energy using the water table and the ideal gas water table in combination Solution State 1 Table B11 u1 uf 8395 kJkg State 2 Highest T in Table B13 is 1300C Using a u from the ideal gas tables A8 we get u1500 3139 kJkg u1300 269072 kJkg u1500 u1300 44826 kJkg We now add the ideal gas change at low P to the steam tables B13 ux 468323 kJkg as the reference u2 u1 u2 uxIDG ux u1 44828 468323 8395 5048 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3138 Reconsider Problem 3134 and examine if also using Table B3 would be more accurate and explain Table B3 does include nonideal gas effects however at 100 kPa these effects are extremely small so the answer from Table A8 is accurate Table B3 does not cover the 100 kPa superheated vapor states as the saturation pressure is below the triple point pressure Secondly Table B3 does not go up to the high temperatures covered by Table A8 and A9 at which states you do have ideal gas behavior Table B3 covers the region of states where the carbon dioxide is close to the twophase region and above the critical point dense fluid which are all states where you cannot assume ideal gas Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Specific Heats Ideal Gas Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3139 Air is heated from 300 to 350 K at V C Find 1q2 What if from 1300 to 1350 K Process V C 1W2 Ø Energy Eq u2 u1 1q2 0 1q2 u2 u1 Read the uvalues from Table A71 a 1q2 u2 u1 25032 21436 360 kJkg b 1q2 u2 u1 106794 102275 452 kJkg case a Cv 3650 072 kJkg K see A5 case b Cv 45250 0904 kJkg K 25 higher Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3140 A rigid container has 2 kg of carbon dioxide gas at 100 kPa 1200 K that is heated to 1400 K Solve for the heat transfer using a the heat capacity from Table A5 and b properties from Table A8 Solution CV Carbon dioxide which is a control mass Energy Eq35 U2 U1 m u2 u1 1Q2 1W 2 Process V 0 1W2 0 a For constant heat capacity we have u2 u1 Cvo T2 T1 so 1Q2 mCvo T2 T1 2 kg 0653 kJkgK 1400 1200 K 2612 kJ b Taking the u values from Table A8 we get 1Q2 m u2 u1 2 kg 121838 99664 kJkg 4435 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3141 Do the previous problem for nitrogen N2 gas A rigid container has 2 kg of carbon dioxide gas at 100 kPa 1200 K that is heated to 1400 K Solve for the heat transfer using a the heat capacity from Table A5 and b properties from Table A8 Solution CV Nitrogen gas which is a control mass Energy Eq35 U2 U1 m u2 u1 1Q2 1W 2 Process V 0 1W2 0 a For constant heat capacity we have u2 u1 Cvo T2 T1 so 1Q2 mCvo T2 T1 2 kg 0745 kJkgK 1400 1200 K 298 kJ b Taking the u values from Table A8 we get 1Q2 m u2 u1 2 kg 114135 957 kJkg 3687 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3142 Air 3 kg is in a pistoncylinder similar to Fig P35 at 27oC 300 kPa It is now heated to 500 K Plot the process path in a Pv diagram and find the work and heat transfer in the process Solution CV Air so this is a control mass Energy Eq35 U2 U1 m u2 u1 1Q2 1W 2 Process P C so 1W2 PdV P1V2 V1 State 1 T1 P1 ideal gas so P1V1 mRT1 V1 mR T1 P1 3 kg 0287 kJkgK 30015 K 300 kPa 086143 m3 State 2 T2 P2 P1 and ideal gas so P2V2 mRT 2 V2 mR T2 P2 3 0287 500300 1435 m3 From the process 1W2 PdV P V2 V1 300 kPa1435 086143 m3 1721 kJ From the energy equation 1Q2 m u2 u1 1W2 mCvo T2 T1 1W2 3 kg 0717 kJkgK 500 300 K 1721 kJ 6023 kJ T v 2 1 300 kPa P v T 300 300 2 1 T 1 2 500 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3143 A closed rigid container is filled with 15 kg water at 100 kPa 55oC 1 kg of stainless steel and 05 kg of PVC polyvinyl chloride both at 20oC and 01 kg of air at 400 K 100 kPa It is now left alone with no external heat transfer and no water vaporizes Find the final temperature and air pressure CV Container Process V constant 1W2 0 and also given 1Q2 0 Energy Eq U2 U1 miu2 u1i 1Q2 1W2 0 For the liquid and the metal masses we will use the specific heats Tbl A3 A4 so miu2 u1i miCv i T2 T1i T2 miCv i miCv iT1 i noticing that all masses have the same T2 but not same initial T miCv i 15 418 1 046 05 096 01 0717 7282 kJK The T for air must be converted to oC like the others Energy Eq T2 miCv i miCv iT1 i 7282 T2 15 418 55 1 046 05 096 20 01 0717 40027315 372745 kJ T2 512oC The volume of the air is constant so from PV mRT it follows that P varies with T P2 P1 T2T1 air 100 kPa 32434 K 400 K 81 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3144 A 250 L rigid tank contains methane at 500 K 1500 kPa It is now cooled down to 300 K Find the mass of methane and the heat transfer using ideal gas Solution CV Methane gas which is a control mass Energy Eq35 U2 U1 m u2 u1 1Q2 1W 2 Process V 0 1W2 0 State 1 Ideal gas so m P1VRT1 1500 025 05183 500 kPa m3 kJkgK K 1447 kg For constant heat capacity A5 we have u2 u1 Cvo T2 T1 so u2 u1 Cv T2 T1 1736 kJkgK 300 500 K 3472 kJkg 1Q2 mu2 u1 1447 kg 3472 kJkg 5024 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3145 A 10m high cylinder crosssectional area 01 m2 has a massless piston at the bottom with water at 20C on top of it shown in Fig P3145 Air at 300 K volume 03 m3 under the piston is heated so that the piston moves up spilling the water out over the side Find the total heat transfer to the air when all the water has been pushed out Solution H2O Po cb air V P 2 1 P P1 0 V V 1 max The water on top is compressed liquid and has volume and mass VH2O Vtot Vair 10 01 03 07 m 3 mH2O VH2Ovf 07 m3 0001002 m3kg 6986 kg The initial air pressure is then P1 P0 mH2OgA 101325 6986 9807 01 1000 16984 kPa and then mair PVRT 16984 03 0287 300 kPa m3 kJkgK K 0592 kg State 2 No liquid over piston P2 P0 101325 kPa V2 1001 1 m 3 State 2 P2 V2 T2 P1V1 T1P2V2 3001013251 1698403 59659 K The process line shows the work as an area 1W2 PdV 1 2P1 P2V2 V1 1 216984 1013251 03 9491 kJ The energy equation solved for the heat transfer becomes 1Q2 mu2 u1 1W2 mCvT2 T1 1W2 0592 kg 0717 kJkgK 59659 300 K 9491 kJ 2207 kJ Remark we could have used u values from Table A7 u2 u1 4325 21436 21814 kJkg versus 2125 kJkg with Cv Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3146 A cylinder with a piston restrained by a linear spring contains 2 kg of carbon dioxide at 500 kPa 400C It is cooled to 40C at which point the pressure is 300 kPa Calculate the heat transfer for the process Solution CV The carbon dioxide which is a control mass Continuity Eq m2 m1 0 Energy Eq m u2 u1 1Q2 1W 2 Process Eq P A BV linear spring 1W2 PdV 1 2P1 P2V2 V1 Equation of state PV mRT ideal gas State 1 V1 mRT1P1 2 018892 67315 500 05087 m 3 State 2 V2 mRT2P2 2 018892 31315 300 03944 m 3 1W2 1 2500 30003944 05087 4572 kJ To evaluate u2 u1 we will use the specific heat at the average temperature From Figure 326 CpoTavg 54 R 102 Cvo 083 Cpo R For comparison the value from Table A5 at 300 K is Cvo 0653 kJkg K 1Q2 mu2 u1 1W2 mCvoT2 T1 1W2 2 08340 400 4572 6433 kJ CO 2 Remark We could also have used the ideal gas table in A8 to get u2 u1 P v 2 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3147 Water at 100 kPa 400 K is heated electrically adding 700 kJkg in a constant pressure process Find the final temperature using a The water tables B1 b The ideal gas tables A8 c Constant specific heat from A5 Solution Energy Eq35 u2 u1 1q2 1w 2 Process P constant 1w2 P v2 v1 Substitute this into the energy equation to get 1q2 h2 h 1 Table B1 h1 267546 12685 9962 150 9962 277638 267546 27300 kJkg h2 h1 1q2 2730 700 3430 kJkg T2 400 500 400 3430 327811 348809 327811 4723C Table A8 h2 h1 1q2 7424 700 14424 kJkg T2 700 750 700 14424 133856 144343 133856 7495 K 4763C Table A5 h2 h1 Cpo T2 T1 T2 T1 1q2 Cpo 400 700 1872 7739K 5008C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3148 A constant pressure container is filled with 15 kg water at 100 kPa 55oC 1 kg of stainless steel and 05 kg of PVC polyvinyl chloride both at 20oC and 01 kg of air at 400 K 100 kPa It is now left alone with no external heat transfer and no water vaporizes Find the final temperature and process work Energy Eq U2 U1 miu2 u1i 1Q2 1W2 0 Process P constant 1W2 PV2 V1 and given 1Q2 0 For the liquid and the metal masses we will use the specific heats Tbl A3 A4 so miu2 u1i miCv i T2 T1i T2miCv i miCv iT1 i noticing that all masses have the same T2 but not same initial T Since it is only the air that changes volume then the work term and internal energy change for air combines to give changes in enthalpy as mair u2 u1 1W2 mair h2 h1 mairCp T2 T1 miCv i 15 418 1 046 05 096 01 1004 73104 kJK Energy Eq 73104 T2 15 418 55 1 046 05 096 20 01 1004 400 27315 376386 kJ T2 5148oC V1 mRT1P1 01 kg 0287 kJkgK 400 K 100 kPa 01148 m3 V2 V1 T2T1air 01148 32463 400 009317 m3 1W2 PV2 V1 100 kPa 009317 01148 m3 216 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3149 A spring loaded pistoncylinder contains 15 kg of air at 27C and 160 kPa It is now heated to 900 K in a process where the pressure is linear in volume to a final volume of twice the initial volume Plot the process in a Pv diagram and find the work and heat transfer Take CV as the air m2 m1 m mu2 u1 1Q2 1W2 Process P A BV 1W2 P dV area 05P1 P2V2 V1 State 1 Ideal gas V1 mRT1P1 15 0287 300160 08072 m3 Table A7 u1 u300 21436 kJkg State 2 P2V2 mRT2 so ratio it to the initial state properties P2V2 P1V1 P22 P1 mRT2 mRT1 T2 T1 P2 P1 T2 T1 12 160 900300 12 240 kPa Work is done while piston moves at linearly varying pressure so we get 1W2 05P1 P2V2 V1 05160 240 kPa 08072 m3 1614 kJ Heat transfer is found from energy equation 1Q2 mu2 u1 1W2 15674824 21436 1614 8521 kJ P V W 1 2 T V 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3150 A constant pressure piston cylinder contains 05 kg air at 300 K 400 kPa Assume the piston cylinder has a total mass of 1 kg steel and is at the same temperature as the air at any time The system is now heated to 1600 K by heat transfer a Find the heat transfer using constant specific heats for air b Find the heat transfer NOT using constant specific heats for air CV Air and Steel Energy Eq U2 U1 mairu2 u1 mstu2 u1 1Q2 1W2 Process P C 1W2 1 2 PdV P V2 V1 P mair v2 v1 1Q2 mairu2 u1air mstu2 u1st 1W2 mairh2 h1air mstu2 u1 st Use A3 u2 u1st C T2 T1 046 kJkgK 1600 300 K 598 kJkg Use A5 h2 h1air CpT2 T1 1004 kJkgK 1600 300 K 13052 kJkg 1Q2 mairh2 h1air mstu2 u1st 05 kg 1305 kJkg 1 kg 598 kJkg 12506 kJ Use air tables A7 h2 h1air 175733 30047 145686 kJkg 1Q2 mairh2 h1air mstu2 u1st 05 kg 145686 kJkg 1 kg 598 kJkg 132643 kJ Comment we could also have computed the work explicitly 1W2 P mair v2 v1 mair RT2 T1 05 kg 0287 kJkgK 1600 300 K 18655 kJ u2 u1air 129808 21436 108372 kJkg 1Q2 mairu2 u1air mstu2 u1st 1W 2 05 108372 598 18655 113986 18655 13264 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3151 An insulated cylinder is divided into two parts of 1 m3 each by an initially locked piston as shown in Fig P3151 Side A has air at 200 kPa 300 K and side B has air at 10 MPa 1000 K The piston is now unlocked so it is free to move and it conducts heat so the air comes to a uniform temperature TA TB Find the mass in both A and B and the final T and P CV A B Force balance on piston PAA PBA So the final state in A and B is the same State 1A Table A7 uA1 214364 kJkg mA PA1VA1RTA1 200 kPa 1 m30287 kJkgK 300 K 2323 kg State 1B Table A7 uB1 759189 kJkg mB PB1VB1RTB1 1000 kPa 1 m30287 kJkgK 1000 K 3484 kg For chosen CV 1Q2 0 1W2 0 so the energy equation becomes mAu2 u1A mBu2 u1B 0 mA mBu2 mAuA1 mBuB1 2323 214364 3484 759189 3143 kJ u2 3143 kJ 3484 2323 kg 54124 kJkg From interpolation in Table A7 T2 736 K P mA mBRT2Vtot 5807 kg 0287 kJ kg K 736 K 2 m3 613 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3152 Air in a piston cylinder is at 1800 K 7 MPa and expands in a polytropic process with n 15 to a volume 8 times larger Find the specific work and the specific heat transfer in the process and draw the Pv diagram Use constant specific heat to solve the problem CV Air of constant mass m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Process Pv150 constant v2v1 8 State 1 P1 7 MPa T1 1800 K State 2 v2 8v1 Must be on process curve so P2 P1 v1v2n 7000 18150 30936 kPa T2 T1 P2v2 P1v1 T1v1v2n 1 1800 1805 6364 K Work from the process expressed in Eq321 1w2 Pdv 1 n P2v2 P1v1 R 1 n T2 T1 1 15 0287 6364 1800 6679 kJkg Heat transfer from the energy equation 1q2 u2 u1 1w2 0717 6364 1800 6679 1664 kJkg Notice n 15 k 14 n k P v 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3153 Do the previous problem but do not use constant specific heats Air in a piston cylinder is at 1800 K 7 MPa and expands in a polytropic process with n 15 to a volume 8 times larger Find the specific work and the specific heat transfer in the process and draw the Pv diagram CV Air of constant mass m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Process Pv150 constant v2v1 8 State 1 P1 7 MPa T1 1800 K State 2 v2 8v1 Must be on process curve so P2 P1 v1v2n 7000 18150 30936 kPa T2 T1 P2v2 P1v1 T1v1v2n 1 1800 1805 6364 K Table A7 u1 148633 kJkg and interpolate u2 46306 kJkg Work from the process expressed in Eq321 1w2 Pdv 1 n P2v2 P1v1 R 1 n T2 T1 1 15 0287 6364 1800 6679 kJkg Heat transfer from the energy equation 1q2 u2 u1 1w2 46306 148633 6679 3554 kJkg Notice n 15 k 14 n k P v 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3154 Helium gas expands from 125 kPa 350 K and 025 m3 to 100 kPa in a polytropic process with n 1667 How much heat transfer is involved Solution CV Helium gas this is a control mass Energy equation mu2 u1 1Q2 1W 2 Process equation PVn constant P1Vn 1 P2Vn 2 Ideal gas A5 m PVRT 125 025 20771 350 0043 kg Solve for the volume at state 2 V2 V1 P1P21n 025 125 100 06 02852 m3 T2 T1 P2V2P1V1 350 K 100 02852 125 025 3194 K Work from Eq321 EA PA2 AVA2 A PA1 A VA1 A E 1nE A A100 02852 125 025 1 1667E A kPa mA3E A 409 kJ 1W2 Use specific heat from Table A5 to evaluate uA2E A uA1E A Cv 3116 kJkg K A1E AQA2E A muA2E A uA1E A A1E AWA2E A m Cv TA2E A TA1E A A1E AWA2E A 0043 kg 3116 kJkgK 3194 350 K 409 kJ 001 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3155 A piston cylinder contains 01 kg air at 300 K and 100 kPa The air is now slowly compressed in an isothermal T C process to a final pressure of 250 kPa Show the process in a PV diagram and find both the work and heat transfer in the process Solution Process T C ideal gas PV mRT constant The work was found as in Eq322 A1E AWA2E A PdV AmRT V EA dV mRT ln A V2 EV1 E A mRT ln A P1 EP2 E A 01 kg 0287 kJkgK 300 K ln 100 250 789 kJ since TA1E A TA2E A uA2E A uA1E The energy equation thus becomes A1E AQA2E A m uA2E A uA1E A A1E AWA2E A A1E AWA2E A 789 kJ P v 2 1 T v 2 1 T C P C v 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3156 A gasoline engine has a pistoncylinder with 01 kg air at 4 MPa 1527AoE AC after combustion and this is expanded in a polytropic process with n 15 to a volume 10 times larger Find the expansion work and heat transfer using Table A5 heat capacity Take CV as the air mA2E A mA1E A m Energy Eq35 muA2E A uA1E A A1E AQ2 A1E AW2 Process Eq PvAnE A Constant polytropic From the ideal gas law and the process equation we can get State 2 PA2E A PA1E A vA2E A vA1E AA nE A 4000 10A 15E A 1265 kPa TA2E A TA1E A PA2E AvA2E A PA1E AvA1E A 1527 273 A1265 10 4000E A 5693 K From process eq A1E AWA2E A P dV A m 1nE A PA2E Av2 PA1E Av1 AmR 1nE A TA2E A TA1E A A01 0287 1 15E A 5693 1800 7064 kJ From energy eq A1E AQA2E A muA2E A uA1E A A1E AWA2E A mCAvE ATA2E A TA1E A A1E AWA2E 01 kg 0717 kJkgK 5693 1800 K 7064 kJ 176 kJ P 1 2 T P C T 3 P v 1 2 T v 1 2 T T 2 1 T C v 05 P C v 15 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3157 Solve the previous problem using Table A7 Take CV as the air m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Process Eq Pvn Constant polytropic From the ideal gas law and the process equation we can get State 2 P2 P1 v2 v1 n 4000 10 15 1265 kPa T2 T1 P2v2 P1v1 1527 273 1265 10 4000 5693 K From process eq 1W2 P dV m 1n P2v2 P1v1 mR 1n T2 T1 01 0287 1 15 5693 1800 7064 kJ From energy eq 1Q2 mu2 u1 1W2 01 kg 41178 148633 kJkg 7064 kJ 368 kJ The only place where Table A7 comes in is for values of u1 and u 2 P 1 2 T P C T 3 P v 1 2 T v 1 2 T T 2 1 T C v 05 P C v 15 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3158 Find the specific heat transfer in Problem 355 Air goes through a polytropic process from 125 kPa 325 K to 300 kPa and 500 K Find the polytropic exponent n and the specific work in the process Solution Energy Eq u2 u1 1q2 1w 2 Process Pvn Const P1vn 1 P2 vn 2 Ideal gas Pv RT so v1 RT P 0287 325 125 07462 m3kg v2 RT P 0287 500 300 047833 m3kg From the process equation P2 P1 v1 v2n lnP2 P1 n lnv1 v2 n lnP2 P1 lnv1 v2 ln 24 ln 156 1969 The work is now from Eq321 per unit mass and ideal gas law 1w2 P2v2P1v1 1n RT2 T1 1n 0287500 325 11969 kJkgK K 518 kJkg From the energy equation 1q2 u2 u1 1w2 Cv T2 T1 1w2 0717 kJkgK 500 325 K 515 kJkg 7398 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3159 A pistoncylinder has nitrogen gas at 750 K and 1500 kPa Now it is expanded in a polytropic process with n 12 to P 750 kPa Find the final temperature the specific work and specific heat transfer in the process CV Nitrogen This is a control mass going through a polytropic process Continuty m2 m1 Energy Eq35 mu2 u1 1Q2 1W2 Process Pvn constant Substance ideal gas Pv RT T2 T1 P2P1 n1 n 750 12 750 1500 02 750 08909 668 K The work is integrated as in Eq321 1w2 Pdv 1 1 n P2v2 P1v1 R 1 n T2 T1 02968 1 12 kJkgK 668 750 K 1217 kJkg The energy equation with values of u from Table A8 is 1q2 u2 u1 1w2 5028 56845 1217 560 kJkg If constant specific heat is used from Table A5 1q2 CvT2 T1 1w2 0745668 750 1217 606 kJkg The actual process is on a steeper curve than n 1 P v w 1 2 n 1 750 1 P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3160 A pistoncylinder has 1 kg propane gas at 700 kPa 40C The piston cross sectional area is 05 m2 and the total external force restraining the piston is directly proportional to the cylinder volume squared Heat is transferred to the propane until its temperature reaches 700C Determine the final pressure inside the cylinder the work done by the propane and the heat transfer during the process Solution CV The 1 kg of propane Energy Eq35 mu2 u1 1Q2 1W 2 Process P Pext CV2 PV2 constant polytropic n 2 Ideal gas PV mRT and process yields P2 P1T2T1 n n1 700 70027315 4027315 23 14907 kPa The work is integrated as Eq321 1W2 1 2 PdV 1 n P2V2 P1V1 1 n mRT2 T1 1 018855 700 40 1 2 4148 kJ The energy equation with specific heat from Table A5 becomes 1Q2 mu2 u1 1W2 mCvT2 T1 1W2 1 kg 1490 kJkgΚ 700 40 K 4148 kJ 10249 kJ P V 2 1 P C V2 T V 2 1 T C V3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3161 A pistoncylinder arrangement of initial volume 0025 m3 contains saturated water vapor at 180C The steam now expands in a polytropic process with exponent n 1 to a final pressure of 200 kPa while it does work against the piston Determine the heat transfer in this process Solution CV Water This is a control mass State 1 Table B11 P 10022 kPa v1 019405 m3kg u1 25837 kJkg m Vv1 0025019405 0129 kg Process Pv const P1v1 P2v2 polytropic process n 1 v2 v1P1P2 019405 10021200 09723 m3kg State 2 P2 v2 Table B13 T2 155C u2 2585 kJkg 1W2 PdV P1V1 ln v1 v2 10022 kPa 0025 m3 ln 09723 019405 4037 kJ 1Q2 mu2 u1 1W2 01292585 25837 4037 4054 kJ P v 2 1 T v 2 1 P C v 1 T C Sat vapor line Notice T drops it is not an ideal gas Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3162 A piston cylinder contains pure oxygen at ambient conditions 20C 100 kPa The piston is moved to a volume that is 7 times smaller than the initial volume in a polytropic process with exponent n 125 Use constant heat capacity to find the final pressure and temperature the specific work and the specific heat transfer Energy Eq u2 u1 1q2 1w 2 Process Eq Pvn C P2 P1 v1v2 n 100 7 125 11386 kPa From the ideal gas law and state 2 P v we get T2 T1 P2P1v1v2 293 11386 100 17 4768 K We could also combine process eq and gas law to give T2 T1 v1v2 n1 Polytropic work Eq 321 1w2 1 1n P2v2 P1v1 R 1n T2 T1 1w2 02598 1 125 kJ kg K 4768 2932 K 19088 kJkg 1q2 u2 u1 1w2 Cv T2 T1 1w2 0662 4768 2932 19088 693 kJkg The actual process is on a steeper curve than n 1 P v w 1 2 n 1 100 2 P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3163 A pistoncylinder in a car contains 02 L of air at 90 kPa 20C shown in Fig P3163 The air is compressed in a quasiequilibrium polytropic process with polytropic exponent n 125 to a final volume six times smaller Determine the final pressure temperature and the heat transfer for the process Solution CV Air This is a control mass going through a polytropic process Continuty m2 m1 Energy Eq35 mu2 u1 1Q2 1W2 Process Pvn const P1v1n P2v2n P2 P1v1v2n 90 kPa 6125 84515 kPa Substance ideal gas Pv RT T2 T1P2v2P1v1 29315 K 8451590 6 4588 K m PV RT 0287 29315 90 02103 214104 kg The work is integrated as in Eq321 1w2 Pdv 1 1 n P2v2 P1v1 R 1 n T2 T1 0287 1 125 kJkgK 4588 29315 K 19017 kJkg The energy equation with values of u from Table A7 is 1q2 u2 u1 1w2 3294 20803 19017 688 kJkg 1Q2 m 1q2 00147 kJ ie a heat loss P v 2 1 P C v125 T v 2 1 T C v025 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3164 An air pistol contains compressed air in a small cylinder shown in Fig P3164 Assume that the volume is 1 cm3 pressure is 1 MPa and the temperature is 27C when armed A bullet m 15 g acts as a piston initially held by a pin trigger when released the air expands in an isothermal process T constant If the air pressure is 01 MPa in the cylinder as the bullet leaves the gun find a The final volume and the mass of air b The work done by the air and work done on the atmosphere c The work to the bullet and the bullet exit velocity Solution CV Air Air ideal gas mair P1V1RT1 1000 1060287 300 117105 kg Process PV const P1V1 P2V2 V2 V1P1P2 10 cm3 1W2 PdV V P1V1 dV P1V1 ln V2V1 2303 J 1W2ATM P0V2 V1 101 10 1 106 kJ 0909 J Wbullet 1W2 1W2ATM 1394 J 1 2 mbulletVexit2 Vexit 2WbulletmB12 2 1394001512 1363 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3165 Air goes through a polytropic process with n 13 in a pistoncylinder setup It starts at 200 kPa 300 K and ends with a pressure of 2200 kPa Find the expansion ratio v2v1 the specific work and the specific heat transfer Take CV as the air m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Process Eq Pvn Constant polytropic From the ideal gas law and the process equation we can get State 2 v2v1 P2P1 1n and Pv RT T2T1 v2 v1 1 n v2v1 P2P1 1n 2200 200 113 01581 T2 T1 P2P1 n1n 300 2200 200 0313 5217 K From process eq 1w2 P dv area 1 1n P2v2 P1v1 R 1n T2 T1 0287 113 kJkgK 5217 300 K 21209 kJkg From the energy equation and constant specific heat from Table A5 1q2 u2 u1 1w2 Cv T2 T1 1w2 0717 kJkgK 5217 300 K 21209 kJkg 5313 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3166 Nitrogen gas goes through a polytropic process with n 13 in a pistoncylinder arrangement It starts out at 600 K and 600 kPa and ends with 800 K Find the final pressure the process specific work and heat transfer Take CV as the nitrogen m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Process Eq Pvn Constant polytropic From the ideal gas law and the process equation we can get State 2 P2 P1 v2 v1 n and Pv RT T2T1 v2 v1 1 n P2 P1 T2 T1 nn1 600 800 600 1303 2087 kPa From process eq 1w2 P dv area 1 1n P2v2 P1v1 R 1n T2 T1 02968 113 800 600 1979 kJkg From the energy equation and Table A8 1q2 u2 u1 1w2 60941 44916 1979 kJkg 3765 kJkg From the energy equation and constant specific heat from Table A5 1q2 u2 u1 1w2 Cv T2 T1 1w2 0745 kJkgK 800 600 K 1979 kJkg 489 kJkg P 1 2 T P C T 433 P v 1 2 T v 1 2 T T 2 1 T C v 03 P C v 13 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3167 A pistoncylinder contains pure oxygen at 500 K 600 kPa The piston is moved to a volume such that the final temperature is 700 K in a polytropic process with exponent n 125 Use ideal gas approximation and constant heat capacity to find the final pressure the specific work and heat transfer Energy Eq u2 u1 1q2 1w 2 Process Eq Pvn C and ideal gas Pv RT gives P2 P1v1v2 n P1T2T1 nn1 600 75 5 3227 kPa Reversible work Eq 321 1w2 1 1n P2v2 P1v1 R 1n T2 T1 1w2 02598 1 125 kJ kg K 700 500 K 20784 kJkg 1q2 u2 u1 1w2 Cv T2 T1 1w2 0662 kJkgK 700 500 K 20784 kJkg 693 kJkg The actual process is on a steeper curve than n 1 P v w 1 2 n 1 100 2 P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3168 Calculate the heat transfer for the process described in Problem 357 Consider a piston cylinder with 05 kg of R134a as saturated vapor at 10C It is now compressed to a pressure of 500 kPa in a polytropic process with n 15 Find the final volume and temperature and determine the work done during the process Solution Take CV as the R134a which is a control mass Continuity m2 m1 m Energy mu2 u1 1Q2 1W2 Process Pv15 constant Polytropic process with n 15 1 T x P Psat 2017 kPa from Table B51 v1 009921 m3kg u1 37227 kJkg 2 P process v2 v1 P1P2 115 009921 20175000667 005416 m3kg Table B52 superheated vapor T2 79C V2 mv2 0027 m3 u2 4409 kJkg Process gives P C v 15 which is integrated for the work term Eq321 1W2 P dV mP2v2 P1v1115 205 kg 500005416 2017009921 kJkg 707 kJ 1Q2 mu2 u1 1W2 054409 37227 707 2725 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3169 A piston cylinder shown in Fig P3169 contains 05 m3 of R410A at 2 MPa 150oC The piston mass and atmosphere gives a pressure of 450 kPa that will float the piston The whole setup cools in a freezer maintained at 20oC Find the heat transfer and show the Pv diagram for the process when T2 20oC CV R410A Control mass Continuity m constant Energy Eq35 mu2 u1 1Q2 1W2 Process F F P A PairA F stop if V Vstop Fstop 0 This is illustrated in the Pv diagram shown below R410A Po State 1 v1 002247 m3kg u1 37349 kJkg m Vv 22252 kg State 2 T2 and on line compressed liquid see figure below v2 vf 0000 803 m3kg V2 001787 m3 u2 uf 2792 kJkg 1W2 PdV PliftV2 V1 450 kPa 001787 05 m3 2170 kJ Energy eq 1Q2 22252 kg 2792 37349 kJkg 2179 kJ 79066 kJ 2 MPa 450 kPa P 2 MPa P 450 kPa v 1 T 20 2 T v 150 168 20 1 2 323 P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3170 A cylinder containing 1 kg of ammonia has an externally loaded piston Initially the ammonia is at 2 MPa 180C and is now cooled to saturated vapor at 40C and then further cooled to 20C at which point the quality is 50 Find the total work and the heat transfer for the process assuming a piecewise linear variation of P versus V Solution o C C C 2 1 P 3 180 40 20 857 1555 2000 v o o cb State 1 T P Table B22 v1 010571 m3kg u1 16306 kJkg State 2 T x Table B21 sat vap P2 1555 kPa v2 008313 m3kg u2 13410 kJkg State 3 T x P3 857 kPa v3 0001638 0149222 007543 m3kg u3 27289 133222 80255 kJkg Sum the work as two integrals each evaluated by the area in the Pv diagram 1W3 1 3 PdV 2 P1 P2 mv2 v1 2 P2 P3 mv3 v2 2000 1555 2 kPa 1 kg 008313 010571 m3kg 1555 857 2 kPa 1 kg 007543 008313 m3kg 494 kJ From the energy equation 1Q3 mu3 u1 1W3 1 kg 80255 16306 kJkg 494 kJ 8775 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3171 10 kg of water in a piston cylinder arrangement exists as saturated liquidvapor at 100 kPa with a quality of 50 It is now heated so the volume triples The mass of the piston is such that a cylinder pressure of 200 kPa will float it as in Fig 3171 Find the final temperature and the heat transfer in the process Solution Take CV as the water Continuity Eq m2 m1 m Energy Eq mu2 u1 1Q2 1W2 Process v constant until P Plift then P is constant State 1 Twophase so look in Table B12 at 100 kPa u1 41733 05 208872 14617 kJkg v1 0001043 05 169296 08475 m3kg State 2 v2 P2 Plift v2 3 08475 25425 m3kg Interpolate T2 829C u2 371876 kJkg V2 mv2 25425 m3 From the process equation see PV diagram we get the work as 1W2 PliftV2 V1 200 kPa 10 kg 25425 08475 m3kg 3390 kJ From the energy equation we solve for the heat transfer 1Q2 mu2 u1 1W2 10 kg 371876 14617 kJkg 3390 kJ 25 961 kJ Po H2O cb V P 2 1 P P 1 2 cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3172 A helium gas is heated at constant volume from a state of 100 kPa 300 K to 500 K A following process expands the gas at constant pressure to three times the initial volume What are the specific work and the specific heat transfer in the combined process The two processes are 1 2 Constant volume V2 V 1 2 3 Constant pressure P3 P 2 V P 3 1 2 P P 1 2 Use ideal gas approximation for helium State 1 T P v1 RT1P1 State 2 V2 V1 P2 P1 T2T1 State 3 P3 P2 V3 3V2 T3 T2 v3v2 500 3 1500 K We find the work by summing along the process path 1w3 1w2 2w3 2w3 P3v3 v2 RT3 T2 20771 kJkgK 1500 500 K 2077 kJkg 1q3 u3 u1 1w3 Cv T3 T1 1w 3 3116 kJkgK 1500 300 K 2077 kJkg 5816 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3173 A vertical cylinder fitted with a piston contains 5 kg of R410A at 10C shown in Fig P3173 Heat is transferred to the system causing the piston to rise until it reaches a set of stops at which point the volume has doubled Additional heat is transferred until the temperature inside reaches 50C at which point the pressure inside the cylinder is 14 MPa a What is the quality at the initial state b Calculate the heat transfer for the overall process Solution CV R410A Control mass goes through process 1 2 3 As piston floats pressure is constant 1 2 and the volume is constant for the second part 2 3 So we have v3 v2 2 v1 State 3 Table B42 PT v3 002249 m3kg u3 28791 kJkg v P 1 2 3 R410A P o cb So we can then determine state 1 and 2 Table B41 v1 0011245 0000886 x1 002295 x1 04514 b u1 7224 04514 18366 15514 kJkg State 2 v2 002249 m3kg P2 P1 1086 kPa this is still 2phase We get the work from the process equation see PV diagram 1W3 1W2 1 2 PdV P1V2 V1 1086 kPa 5 kg 0011245 m3kg 611 kJ The heat transfer from the energy equation becomes 1Q3 mu3u1 1W3 5 kg 28791 15514 kJkg 611 kJ 7250 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3174 Water in a pistoncylinder Fig P3174 is at 101 kPa 25C and mass 05 kg The piston rests on some stops and the pressure should be 1000 kPa to float the piston We now heat the water so the piston just reaches the end of the cylinder Find the total heat transfer Solution Take CV as the water Continuity Eq m2 m1 m Energy Eq mu2 u1 1Q2 1W2 Process v constant until P Pfloat then P is constant Volume does go up so we get v2 5 v 1 State 1 v1 0001003 m3kg u1 10486 kJkg State 2 v2 P2 Pfloat so v2 5 0001003 0005015 m3kg x2 v2 vf vfg 0005015 0001127019332 002011 u2 76167 x2 182197 79831 kJkg From the process equation see PV diagram we get the work as 1w2 Pfloatv2 v1 1000 kPa 0005015 0001003 m3kg 4012 kJkg From the energy equation we solve for the heat transfer 1Q2 mu2 u1 1w2 0579831 10486 4012 3487 kJ H O Po 2 cb V P 2 1 P P 1 2 cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3175 A setup as in Fig P3169 has the R410A initially at 1000 kPa 50oC of mass 01 kg The balancing equilibrium pressure is 400 kPa and it is now cooled so the volume is reduced to half the starting volume Find the work and heat transfer for the process Take as CV the 01 kg of R410A Continuity Eq m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Process Eq P Pfloat or v C v1 State 1 P T v1 00332 m3kg u1 292695 kJkg State 2 P v v2 v12 00166 m3kg vg so it is twophase x2 v2 vf vfg 00166 00008030064 02468 u2 uf x2 ufg 2792 x2 21807 81746 kJkg From process eq 1W2 P dV area mP2 v2 v1 01 kg 400 kPa 00166 00332 m3kg 0664 kJ From energy eq 1Q2 mu2 u1 1W2 01 81746 292695 0664 218 kJ P CP v T 1 400 2 1000 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3176 A pistoncylinder contains 1 kg of liquid water at 20C and 300 kPa Initially the piston floats similar to the setup in Problem 3173 with a maximum enclosed volume of 0002 m3 if the piston touches the stops Now heat is added so a final pressure of 600 kPa is reached Find the final volume and the heat transfer in the process Solution Take CV as the water which is a control mass m2 m1 m Energy Eq mu2 u1 1Q2 1W2 Table B11 20C Psat 234 kPa State 1 Compressed liquid v vf20 0001002 m3kg u 8394 kJkg State 1a vstop 0002 m3kg 300 kPa State 2 Since P2 600 kPa Plift then piston is pressed against the stops v2 vstop 0002 m3kg and V 0002 m 3 For the given P vf v vg so 2phase T Tsat 15885 C x v vv 0002 0001101031457 0002858 u 66988 x 189752 6753 kJkg Work is done while piston moves at Plift constant 300 kPa so we get 1W2 P dV m Pliftv2 v1 1 kg 300 kPa 0002 0001002 m 3 030 kJ The heat transfer is from the energy equation 1Q2 mu2 u1 1W2 1 kg 6753 8394 kJkg 030 kJ 5917 kJ V P 1 2 H O Po 2 cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3177 A cylinderpiston arrangement contains 5 kg of water at 100C with x 20 and the piston mP 75 kg resting on some stops similar to Fig P3171 The outside pressure is 100 kPa and the cylinder area is A 245 cm2 Heat is now added until the water reaches a saturated vapor state Find the initial volume final pressure work and heat transfer terms and show the Pv diagram Solution CV The 5 kg water Continuty m2 m1 m Energy mu2 u1 1Q2 1W2 Process V constant if P Plift otherwise P Plift see Pv diagram P3 P2 Plift P0 mp g Ap 100 75 9807 000245 1000 400 kPa P v 1 2 3 100 C 143 C o o cb H O Po 2 cb State 1 Tx Table B11 v1 0001044 02 16719 m3kg V1 mv1 5 03354 1677 m 3 u1 41891 02 208758 8364 kJkg State 3 P x 1 Table B12 v3 04625 v1 u3 25536 kJkg Work is seen in the PV diagram if volume changes then P Plift 1W3 2W3 Pextmv3 v2 400 kPa 5 kg 046246 03354 m3kg 2541 kJ Heat transfer is from the energy equation 1Q3 5 kg 25536 8364 kJkg 2541 kJ 8840 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3178 A piston cylinder setup similar to Problem 3171 contains 01 kg saturated liquid and vapor water at 100 kPa with quality 25 The mass of the piston is such that a pressure of 500 kPa will float it The water is heated to 300C Find the final pressure volume work 1W2 and 1Q2 Solution Take CV as the water m2 m1 m Process v constant until P Plift To locate state 1 Table B12 v1 0001043 025169296 042428 m3kg u1 41733 025208872 93951 kJkg P 1 P lift V P 1 2 1a cb 1a v1a v1 042428 m3kg vg at 500 kPa state 1a is superheated vapor T1a 200C State 2 is 300C so heating continues after state 1a to 2 at constant P 2 T2 P2 Plift 500 kPa Table B13 v2 052256 m3kg u2 280291 kJkg V2 mv2 005226 m3 1W2 Plift V2 V1 500 kPa 005226 004243 m3 491 kJ The heat transfer is from the energy equation 1Q2 mu2 u1 1W2 01 kg 280291 93951 kJkg 491 kJ 19125 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3179 A pistoncylinder contains 01 kg R410A at 600 kPa 60oC It is now cooled so the volume is reduced to half the initial volume The piston has upper stops mounted and the piston mass and gravitation is such that a floating pressure is 400 kPa a Find the final temperature b How much work is involved c What is the heat transfer in the process d Show the process path in a Pv diagram Take as CV the 01 kg of R410A m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Process Eq P Pfloat or v C v1 State 1 P T v1 006023 m3kg u1 30491 kJkg State 2 P v v2 v12 0030115 m3kg vg so it is twophase x2 v2 vf vfg 0030115 00008030064 0458 u2 uf x2 ufg 2792 x2 21807 1278 kJkg From process eq 1W2 P dV area mP2 v2 v1 01 kg 400 kPa 0030115 006023 m3kg 12046 kJ From energy eq 1Q2 mu2 u1 1W2 01 kg 1278 30491 kJkg 12046 kJ 189 kJ P CP v T 1 201 2 400 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3180 A piston cylinder contains air at 1000 kPa 800 K with a volume of 005 m3 The piston is pressed against the upper stops see Fig P314c and it will float at a pressure of 750 kPa Now the air is cooled to 400 K What is the process work and heat transfer CV Air this is a control mass Energy Eq35 mu2 u1 1Q2 1W2 Process Eq P Pfloat or v C v1 State 1 u 59258 kJkg m PVRT 1000 005 0287 800 02178 kg We need to find state 2 Let us see if we proceed past state 1a during the cooling T1a T1 Pfloat P1 800 750 100 600 K so we do cool below T1a That means the piston is floating Write the ideal gas law for state 1 and 2 to get V2 P2 mRT2 P2T1 P1V1T2 1000 005 400 750 800 m3 00333 m 3 1W2 1aW2 P dV P2 V2 V1 750 kPa 00333 005 m3 125 kJ From energy eq 1Q2 mu2 u1 1W2 02178 kg 28649 59258 kJkg 125 kJ 792 kJ Air P o m p 1a 2 1 P V P V stop 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3181 The pistoncylinder arrangement in Fig P3181 contains 10 g ammonia at 20C with a volume of 1 L There are some stops so if the piston is at the stops the volume is 14 L The ammonia is now heated to 200C The piston and cylinder is made of 05 kg aluminum Assume that the mass has the same temperature as the ammonia at any time Find the final volume and the total heat transfer and plot the PV diagram for the process CV NH3 Control mass goes through process 1 2 3 Energy Eq U3 U1 mNH3 u3 u1 mAlu u3 u1 1Q3 1W 3 As piston floats pressure is constant 1 2 and the volume is constant for the second part 2 3 if we go this far So we have at stop v3 v2 14 v1 State 1 B21 v1 Vm 0001 001 01 m3kg v so 2phase P 8575 kPa x1 v vf vfg 01 0001638 014758 06665 u1 uf x1 ufg 27289 x1 10593 97891 kJkg State 2 v2 14 v1 014 m3kg P 8575 kPa still 2phase so T2 20oC State 3 200oC v3 v2 014 m3kg P 1600 kPa V 14 L u3 16765 kJkg We get the work from the process equation see PV diagram 1W3 1W2 1 2 PdV P1V2V1 8575 kPa 00014 0001 m3 0343 kJ The heat transfer from the energy equation becomes 1Q3 mNH3 u3 u1 mAluCAlu T3 T1 1W3 001 kg 16765 97891 kJkg 05 kg 09 kJkgK 20020 K 0343 kJ 8832 kJ V P 1 2 3 NH P o cb 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3182 Air in a pistoncylinder at 200 kPa 600 K is expanded in a constantpressure process to twice the initial volume state 2 shown in Fig P3182 The piston is then locked with a pin and heat is transferred to a final temperature of 600 K Find P T and h for states 2 and 3 and find the work and heat transfer in both processes Solution CV Air Control mass m2 m3 m1 Energy Eq35 u2 u1 1q2 1w2 Process 1 to 2 P constant 1w2 P dv P1v2 v1 RT2 T1 Ideal gas Pv RT T2 T1v2v1 2T1 1200 K P2 P1 200 kPa 1w2 RT1 1722 kJkg Table A7 h2 12778 kJkg h3 h1 6073 kJkg 1q2 u2 u1 1w2 h2 h1 12778 6073 6705 kJkg Process 23 v3 v2 2v1 2w3 0 P3 P2T3T2 P1T12T1 P12 100 kPa 2q3 u3 u2 4351 9334 4983 kJkg Po Air cb T v 2 1 3 P v 2 1 3 1200 600 100 200 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3183 A pistoncylinder has 05 kg air at 2000 kPa 1000 K as shown in Fig P3183 The cylinder has stops so Vmin 003 m3 The air now cools to 400 K by heat transfer to the ambient Find the final volume and pressure of the air does it hit the stops and the work and heat transfer in the process We recognize this is a possible twostep process one of constant P and one of constant V This behavior is dictated by the construction of the device Continuity Eq m2 m1 0 Energy Eq35 mu2 u1 1Q2 1W 2 Process P constant FA P1 if V Vmin V constant V1a Vmin if P P1 State 1 P T V1 mRT1P1 05 0287 10002000 007175 m 3 The only possible PV combinations for this system is shown in the diagram so both state 1 and 2 must be on the two lines For state 2 we need to know if it is on the horizontal P line segment or the vertical V segment Let us check state 1a State 1a P1a P1 V1a Vmin Ideal gas so T1a T1 V1a V1 1000 003 007175 418 K We see that T2 T1a and state 2 must have V2 V1a Vmin 003 m3 P2 P1 T2 T1 V1 V2 2000 400 1000 007175 003 19133 kPa The work is the area under the process curve in the PV diagram 1W2 1 2 P dV P1 V1a V1 2000 kPa 003 007175 m3 835 kJ Now the heat transfer is found from the energy equation us from Table A71 1Q2 mu2 u1 1W2 05 28649 75919 835 31985 kJ V P 1 2 1a P P 2 1 V T 1 2 1a T T 2 1a T 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3184 Air in a rigid tank is at 100 kPa 300 K with a volume of 075 m3 The tank is heated to 400 K state 2 Now one side of the tank acts as a piston letting the air expand slowly at constant temperature to state 3 with a volume of 15 m3 Find the pressures at states 2 and 3 Find the total work and total heat transfer State 1 m P1V1 RT1 100 075 0287 300 kPa m3 kJkg 0871 kg Process 1 to 2 Constant volume heating dV 0 1W2 0 P2 P1 T2 T1 100 400 300 1333 kPa Process 2 to 3 Isothermal expansion dT 0 u3 u2 and P3 P2 V2 V3 1333 075 15 6667 kPa 2W3 2 P 3 dV P2V2 ln V2 V3 1333 075 ln2 693 kJ The overall process 1W3 1W2 2W3 2W3 693 kJ From the energy equation 1Q3 mu3 u1 1W3 m Cv T3 T1 1W3 0871 kg 0717 kJkgK 400 300 K 693 kJ 1318 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3185 A 100 hp car engine has a drive shaft rotating at 2000 RPM How much torque is on the shaft for 25 of full power Solution Power 025 100 hp 025 735 kW if SI hp 18375 kW Tω ω angular velocity rads RPM 2 π 60 T Powerω power 60 RPM 2π 18375 kW 60 smin 2000 2π radmin 8773 Nm We could also have used UK hp to get 025 746 kW then T 89 Nm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3186 A crane use 2 kW to raise a 100 kg box 20 m How much time does it take Power W FV mgV mg L t t mgL W 2000 W 100 kg 9807 ms2 20 m 981 s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3187 An escalator raises a 100 kg bucket of sand 10 m in 1 minute Determine the rate of work done during the process Solution The work is a force with a displacement and force is constant F mg W F dx F dx F x 100 kg 980665 ms2 10 m 9807 J The rate of work is work per unit time W W t 9807 J 60 s 163 W Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3188 A pistoncylinder of cross sectional area 001 m2 maintains constant pressure It contains 1 kg water with a quality of 5 at 150oC If we heat so 1 gs liquid turns into vapor what is the rate of work out Vvapor mvapor vg Vliq mliq v f mtot constant mvapor mliq Vtot Vvapor Vliq m tot 0 m vapor m liq m liq m vapor V tot V vapor V liq m vaporvg m liqv f m vapor vg vf m vapor vfg W PV P m vapor vfg 4759 kPa 0001 kgs 039169 m3kg 01864 kW 186 W Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3189 A pot of water is boiling on a stove supplying 325 W to the water What is the rate of mass kgs vaporizing assuming a constant pressure process To answer this we must assume all the power goes into the water and that the process takes place at atmospheric pressure 101 kPa so T 100oC Energy equation dQ dE dW dU PdV dH hfg dm dQ dt hfg dm dt E A 0144 gs dm dt Q hfg 325 W 2257 kJkg The volume rate of increase is AdV dtE A Adm dtE A vfg 0144 gs 167185 mA3E Akg 024 10A3E A mA3E As 024 Ls Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3190 The heaters in a spacecraft suddenly fail Heat is lost by radiation at the rate of 100 kJh and the electric instruments generate 75 kJh Initially the air is at 100 kPa 25C with a volume of 10 m3 How long will it take to reach an air temperature of 20C Solution A Continuity Eq dM dt 0 EEnergy Eq dE dt Q el Q rad E A EA W 0 EKE 0 PE 0E AE E A AU E A AQ E AelE A AQ E AradE A AQ E AnetE A UA2E A UA1E A muA2E A uA1E A AQ E AnetE AtA2E A tA1E A Ideal gas m A P1V1 ERT1 E A A 100 10 0287 29815E A 11688 kg muA2E A uA1E A m CAv0E ATA2E A TA1E A 11688 0717 20 25 3771 kJ tA2E A tA1E A m uA2E A uA1E AAQ E AnetE A 3771 kJ 25 kJh 1508 h CV CM Air el Q rad Q Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3191 As fresh poured concrete hardens the chemical transformation releases energy at a rate of 2 Wkg Assume the center of a poured layer does not have any heat loss and that it has an average specific heat of 09 kJkgK Find the temperature rise during 1 hour of the hardening curing process Solution AU E A Amu E E A mCvAT E A AQ E A mAq E AT E A Aq E ACv 210A3E A kW 09 kJkgK 2222 10A3E A Csec T AT E At 2222 10A3E A 3600 8 C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3192 A pot of 12 kg water at 20AoE AC is put on a stove supplying 1250 W to the water After how long time can I expect it to come to a boil 100AoE AC Energy Equation on a rate form dEwater dt dUwater dt AQ E A AW E A AQ E A PAV E A AQ E A dUwater dt PAV E A dHwater dt mwaterCp dTwater dt Integrate over time Q AQ E A t H mwater hA2E A hA1E A mwaterCp TA2E A TA1E A t mwater hA2E A hA1E AAQ E A mwaterCp TA2E A TA1E AAQ E A 12 41902 8394125 12 418 100 20125 3217 s 55 min Comment Notice how close the two results are ie use of constant Cp is OK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3193 A computer in a closed room of volume 200 m3 dissipates energy at a rate of 10 kW The room has 50 kg wood 25 kg steel and air with all material at 300 K 100 kPa Assuming all the mass heats up uniformly how long will it take to increase the temperature 10C Solution CV Air wood and steel mA2E A mA1E A no work Energy Eq35 UA2E A UA1E A A1E AQA2E A AQ E At The total volume is nearly all air but we can find volume of the solids VAwoodE A mρ 50510 0098 m3 VAsteelE A 257820 0003 m3 VAairE A 200 0098 0003 199899 m3 mAairE A PVRT 101325 1998990287 300 23525 kg We do not have a u table for steel or wood so use heat capacity from A3 U mAairE A Cv mAwoodE A Cv mAsteelE A Cv T 23525 0717 50 138 25 046 kJK 10 K 16867 690 115 2492 kJ AQ E A t 10 kW t t 2492 kJ10 kW 2492 sec 42 minutes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3194 The rate of heat transfer to the surroundings from a person at rest is about 400 kJh Suppose that the ventilation system fails in an auditorium containing 100 people Assume the energy goes into the air of volume 1500 m3 initially at 300 K and 101 kPa Find the rate degrees per minute of the air temperature change Solution AQ E A n Aq E A 100 400 40 000 kJh 6667 kJmin dEair dt AQ E A mairCv dTair dt mair PVRT 101 kPa 1500 m3 0287 kJkgK 300 K 17596 kg dTair dt AQ E A mCv 6667 kJmin 17596 kg 0717 kJkgK 053Cmin Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3195 A steam generating unit heats saturated liquid water at constant pressure of 800 kPa in a piston cylinder If 15 kW of power is added by heat transfer find the rate kgs of saturated vapor that is made Solution Energy equation on a rate form making saturated vapor from saturated liquid AU E A Amu E E A Am E Au AQ E A AW E A AQ E A PAV E A AQ E A P Am E Av Rearrange to solve for heat transfer rate AQ E A Am E Au vP Am E A h Am E A hfg So now Am E A AQ E A hfg 15 kW 204804 kJkg 0732 gs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3196 A 500 Watt heater is used to melt 2 kg of solid ice at 10AoE AC to liquid at 5AoE AC at a constant pressure of 150 kPa a Find the change in the total volume of the water b Find the energy the heater must provide to the water c Find the time the process will take assuming uniform T in the water Solution Take CV as the 2 kg of water mA2E A mA1E A m Energy Eq35 muA2E A uA1E A A1E AQ2 A1E AW2 State 1 Compressed solid take saturated solid at same temperature v vAiE A10 00010891 mA3E Akg h hAiE A 35409 kJkg State 2 Compressed liquid take saturated liquid at same temperature v vAfE A 0001 h hAfE A 2098 kJkg Change in volume VA2E A VA1E A mvA2E A vA1E A 20001 00010891 0000178 mA3E Work is done while piston moves at constant pressure so we get A1E AWA2E A P dV area PV2 V1 150 0000178 0027 kJ 27 J Heat transfer is found from energy equation A1E AQA2E A muA2E A uA1E A A1E AWA2E A mhA2E A hA1E A 2 2098 35409 750 kJ The elapsed time is found from the heat transfer and the rate of heat transfer t A1E AQA2E AAQ E A 750 kJ 500 W 1000 JkJ 1500 s 25 min P T v V L S CP 2 1 SV LV P CP v T CP v 1 2 1 2 P const Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3197 A drag force on a car with frontal area A 2 mA2E A driving at 80 kmh in air at 20AoE AC is FAdE A 0225 A ρairVA2E A How much power is needed and what is the traction force AW E A FV V 80 Akm hE A 80 A1000 3600E A msA1E A 2222 msA1E ρAAIRE A A P RTE A A 101 0287 293E A 120 kgmA3E FAdE A 0225 AρVA2E A 0225 2 12 2222A2E A 26661 N AW E A FV 26661 N 2222 ms 5924 W 592 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3198 A mass of 3 kg nitrogen gas at 2000 K V C cools with 500 W What is dTdt Process V C A1E AWA2E A 0 AdE dtE A AdU dtE A mAdU dtE A mCAvE A AdT dtE A AQ E A W AQ E A 500 W CAv 2000E A Adu dTE A Au TE A u2100 u1900 21001900 A181908 162166 200E A 0987 kJkg K AdT dtE A A Q EmCv E A A 500 W 3 0987 kJKE A 017 AK sE Remark Specific heat from Table A5 has CAv 300E A 0745 kJkg K which is nearly 25 lower and thus would overestimate the rate with 25 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3199 Consider the pot in Problem 3119 Assume the stove supplies 1 kW of heat How much time does the process take A 1 kg steel pot contains 1 kg liquid water both at 15AoE AC It is now put on the stove where it is heated to the boiling point of the water Neglect any air being heated and find the total amount of energy needed Solution Energy Eq UA2E A UA1E A A1E AQA2E A A1E AWA2E The steel does not change volume and the change for the liquid is minimal so A1E AWA2E A 0 State 2 TA2E A TAsatE A 1atm 100AoE AC Tbl B11 uA1E A 6298 kJkg uA2E A 41891 kJkg Tbl A3 CAstE A 046 kJkg K Solve for the heat transfer from the energy equation A1E AQA2E A UA2E A UA1E A mAstE A uA2E A uA1E AAstE A mAH2OE A uA2E A uA1E AAH2OE mAstE ACAstE A TA2E A TA1E A mAH2OE A uA2E A uA1E AAH2OE A1E AQA2E A 1 kg 046 A kJ kg KE A 100 15 K 1 kg 41891 6298 kJkg 391 35593 395 kJ To transfer that amount of heat with a rate of 1 kW we get the relation A1E AQA2E A AQ E A dt AQ E A t t A1E AQA2E A AQ E A 395 kJ 1 kJs 395 s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful General work Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3200 Electric power is volts times ampere P V i When a car battery at 12 V is charged with 6 amp for 3 hours how much energy is delivered Solution Work term integrated as W A W EA dt AWE A t V i t 12 V 6 Amp 3 3600 s 777 600 J 7776 kJ Remark Volt times ampere is also watts 1 W 1 V 1 Amp 1 Js Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3201 A copper wire of diameter 2 mm is 10 m long and stretched out between two posts The normal stress pressure σ EL LoLo depends on the length L versus the unstretched length Lo and Youngs modulus E 11 10A6E A kPa The force is F Aσ and measured to be 110 N How much longer is the wire and how much work was put in Solution F As AE L LAoE A and L FLAoE A AE A Aπ 4E ADA2E A Aπ 4E A 0002A2E A 3142 10A6E A mA2E L EA 110 N 10 m 3142106 m211 10A6 A 10A3 A PaE A 0318 m A1E AWA2E A F dx A s dx AE EA x LAo AE A dx EAAE LAo AE A ½ xA2E A where x L LAoE EA314210 A6 A 11 10A6 A 10A3 A E10E A ½ 0318A2E A 1747 J Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3202 A film of ethanol at 20C has a surface tension of 223 mNm and is maintained on a wire frame as shown in Fig P3202 Consider the film with two surfaces as a control mass and find the work done when the wire is moved 10 mm to make the film 20 40 mm Solution Assume a free surface on both sides of the frame ie there are two surfaces 20 30 mm W A S dAEA 22310A3E A Nm 2 800 600 10A6E A mA2E 89210A6E A J 892 µJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3203 A 10L rigid tank contains R410A at 10C 80 quality A 10A electric current from a 6V battery is passed through a resistor inside the tank for 10 min after which the R410A temperature is 40C What was the heat transfer to or from the tank during this process Solution CV R410A in tank Control mass at constant V Continuity Eq mA2E A mA1E A m Energy Eq muA2E A uA1E A A1E AQA2E A A1E AWA2E Process Constant V vA2E A vA1E A no boundary work but electrical work v P 1 2 State 1 from table B41 vA1E A 0000827 08 004470 003659 mA3E Akg uA1E A 4232 08 20736 20821 kJkg m Vv 0010 mA3E A 003659 mA3E Akg 02733 kg State 2 Table B42 at 40C and vA2E A vA1E A 003659 mA3E Akg superheated vapor so use linear interpolation to get PA2E A 800 200 003659 004074003170 004074 800 200 045907 892 kPa uA2E A 28683 045907 28435 28683 28569 kJkg A1E AWA2 elecE A power t Amp volts t A10 6 10 60 1000E A 36 kJ A1E AQA2E A muA2E A uA1E A A1E AWA2E A 02733 kg 28569 20821 kJkg 36 kJ 148 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3204 A battery is well insulated while being charged by 123 V at a current of 6 A Take the battery as a control mass and find the instantaneous rate of work and the total work done over 4 hours Solution Battery thermally insulated Q 0 For constant voltage E and current i Power E i 123 6 738 W Units V A W W power dt power t 738 W 4 h 3600 sh 1 062 720 J 10627 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3205 A sheet of rubber is stretched out over a ring of radius 025 m I pour liquid water at 20AoE AC on it as in Fig P3205 so the rubber forms a half sphere cup Neglect the rubber mass and find the surface tension near the ring Solution F F F SL The length is the perimeter 2πr and there is two surfaces S 2 2πr mAH2oE A g ρAH2oE A Vg ρAH2oE A A 1 12E A π 2r A3E Ag ρAH2oE A π A2 3E A r A3E S ρAH2oE A A1 6E A rA2E A g 997 kgmA3E A A1 6E A 025A2E A mA2E A 981 msA2E A 1019 Nm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3206 Assume we fill a spherical balloon from a bottle of helium gas The helium gas provides work PdV that stretches the balloon material S dA and pushes back the atmosphere Po dV Write the incremental balance for dWhelium dWstretch dWatm to establish the connection between the helium pressure the surface tension S and Po as a function of radius WAHeE A P dV S dA PAoE A dV dWAHeE A P dV S dA PAoE A dV dV d Aπ 6E A DA3E A Aπ 6E A 3DA2E A dD dA d 2 π DA2E A 2π 2D dD P Aπ 2E A DA2E A dD S 4πD dD PAoE A Aπ 2E A DA2E A dD Divide by Aπ 2E A DA2E A to recognize PAHeE A PAoE A 8 AS DE A PAoE A 4 AS rE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3207 Assume a balloon material with a constant surface tension of S 2 Nm What is the work required to stretch a spherical balloon up to a radius of r 05 m Neglect any effect from atmospheric pressure Assume the initial area is small and that we have 2 surfaces inside and out W S dA S AA2E A AA1E A SAA2E A S 2 π DA2 2E A 2 Nm 2 π 1 mA2E A 1257 J WAinE A W 1257 J Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3208 A soap bubble has a surface tension of S 3 10A4E A Ncm as it sits flat on a rigid ring of diameter 5 cm You now blow on the film to create a half sphere surface of diameter 5 cm How much work was done A1E AWA2E A F dx S dA S A 2 S Aπ 2E A DA2E A Aπ 4E A D A2E A 2 3 10A4E A Ncm 100 cmm Aπ 2E A 005A2E A mA2E A 1 05 118 10A4E A J Notice the bubble has 2 surfaces AA1E A Aπ 4E A D A2E A AA2E A ½ π DA2E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3209 A 05mlong steel rod with a 1cm diameter is stretched in a tensile test What is the required work to obtain a relative strain of 01 The modulus of elasticity of steel is 2 10A8E A kPa Solution A1E AWA2E A A AEL0 E2E A eA2E A A Aπ 4E A 001A2E A 7854 10A6E A mA2E A1E AWA2E A A1 2E A 7854 10A6E A mA2E A 210A8E A kPa 05 m 10A3E AA2E A 393 J Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful More Complex Devices Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3210 A rigid tank is divided into two rooms by a membrane both containing water shown in Fig P3210 Room A is at 200 kPa v 05 m3kg VA 1 m3 and room B contains 35 kg at 05 MPa 400C The membrane now ruptures and heat transfer takes place so the water comes to a uniform state at 100C Find the heat transfer during the process Solution CV Both rooms A and B in tank B A Continuity Eq mA2E A mAA1E A mAB1E A Energy Eq mA2E AuA2E A mAA1E AuAA1E A mAB1E AuAB1E A A1E AQA2E A A1E AWA2E A State 1A P v Table B12 mAA1E A VAAE AvAA1E A 105 2 kg xAA1E A A v vf Evfg E A A05 0001061 088467E A 0564 uAA1E A uAfE A x uAfgE A 50447 0564 202502 16466 kJkg State 1B Table B13 vAB1E A 06173 uAB1E A 29632 VABE A mAB1E AvAB1E A 216 mA3E Process constant total volume VAtotE A VAAE A VABE A 316 mA3E A and A1E AWA2E A A0E mA2E A mAA1E A mAB1E A 55 kg vA2E A VAtotE AmA2E A 05746 mA3E Akg State 2 TA2E A vA2E A Table B11 twophase as vA2E A vAgE xA2E A v2 vf vfg A05746 0001044 167185E A 0343 uA2E A uAfE A x uAfgE A 41891 0343 208758 113495 kJkg Heat transfer is from the energy equation A1E AQA2E A mA2E AuA2E A mAA1E AuAA1E A mAB1E AuAB1E A 55 113495 2 16466 35 29632 kg kJkg 7421 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3211 A piston cylinder has a water volume separated in VAAE A 02 mA3E A and VABE A 03 mA3E A by a stiff membrane The initial state in A is 1000 kPa x 075 and in B it is 1600 kPa and 250C Now the membrane ruptures and the water comes to a uniform state at 200C What is the final pressure Find the work and the heat transfer in the process Take the water in A and B as CV Continuity mA2E A m1A m1B 0 Energy mA2E AuA2E A mA1AE Au1A mA1BE Au1B A1E AQA2E A A1E AWA2E A Process PA2E A PAeqE A constant P1A as piston floats and mApE A PAoE A do not change State 1A Two phase Table B12 vA1AE A 0001127 075 019332 0146117 m3kg uA1AE A 76167 075 182197 212815 kJkg State 1B Table B13 vA1BE A 014184 m3kg uA1BE A 269226 kJkg mA1AE A VA1AE AvA1AE A 13688 kg mA1BE A VA1BE AvA1BE A 2115 kg State 2 1000 kPa 200AoE AC sup vapor vA2E A 020596 m3kg uA2E A 26219 kJkg mA2E A mA1AE A mA1BE A 34838 kg VA2E A mA2E AvA2E A 34838 020596 07175 m3 Piston moves at constant pressure A1E AWA2E A P dV PAeqE A VA2E A VA1E A 1000 kPa 07175 05 m3 2175 kJ A1E AQA2E A mA2E AuA2E A m1Au1A m1Bu1B A1E AWA2E A 34838 26219 13688 212815 2115 269226 2175 744 kJ AH2O P o cb BH2O g p m Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3212 The cylinder volume below the constant loaded piston has two compartments A and B filled with water A has 05 kg at 200 kPa 150oC and B has 400 kPa with a quality of 50 and a volume of 01 m3 The valve is opened and heat is transferred so the water comes to a uniform state with a total volume of 1006 m3 a Find the total mass of water and the total initial volume b Find the work in the process c Find the process heat transfer Solution Take the water in A and B as CV Continuity m2 m1A m1B 0 Energy m2u2 m1Au1A m1Bu1B 1Q2 1W2 Process P constant P1A if piston floats VA positive ie if V2 VB 01 m3 State A1 Sup vap Table B13 v 095964 m3kg u 25769 kJkg V mv 05 kg 095964 m3kg 047982 State B1 Table B12 v 1x 0001084 x 04625 02318 m3kg m Vv 04314 kg u 60429 05 19493 15789 kJkg State 2 200 kPa v2 V2m 100609314 10801 m3kg Table B13 close to T2 200oC and u2 26544 kJkg So now V1 047982 01 05798 m3 m1 05 04314 09314 kg Since volume at state 2 is larger than initial volume piston goes up and the pressure then is constant 200 kPa which floats piston 1W2 P dV Plift V2 V1 200 kPa 1006 057982 m3 8524 kJ 1Q2 m2u2 m1Au1A m1Bu1B 1W2 09314 26544 05 25769 04314 15789 8524 588 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3213 Water in a tank A is at 250 kPa with a quality of 10 and mass 05 kg It is connected to a piston cylinder holding constant pressure of 200 kPa initially with 05 kg water at 400C The valve is opened and enough heat transfer takes place to have a final uniform temperature of 150C Find the final P and V the process work and the process heat transfer CV Water in A and B Control mass goes through process 1 2 Continuity Eq m2 mA1 mB1 0 m2 mA1 mB1 05 05 1 kg Energy Eq U2 U1 1Q2 1W2 State A1 vA1 0001067 xA1 071765 0072832 VA1 mv 0036416 m 3 uA1 53508 01 200214 73522 kJkg State B1 vB1 15493 m3kg uB1 296669 kJkg VB1 mB1vB1 077465 m 3 State 2 If V2 VA1 then P2 200 kPa that is the piston floats For T2 P2 150C 200 kPa superheated vapor u2 257687 kJkg v2 095964 m3kg V2 m2v2 095964 m3 VA1 checks OK The possible state 2 PV combinations are shown State a is at 200 kPa and va m2 VA1 0036 m3kg and thus twophase Ta 120C T 2 Process 1W2 P2 V2 V1 200 kPa 095964 077465 0036416 m3 2972 kJ From the energy Eq 1Q2 m2u2 mA1uA1 mB1uB1 1W 2 1 257687 05 735222 05 296669 2972 7556 kJ V P 2 150 C a 467 200 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3214 Two rigid tanks are filled with water Tank A is 02 m3 at 100 kPa 150oC and tank B is 03 m3 at saturated vapor 300 kPa The tanks are connected by a pipe with a closed valve We open the valve and let all the water come to a single uniform state while we transfer enough heat to have a final pressure of 300 kPa Give the two property values that determine the final state and find the heat transfer State A1 u 258275 kJkg v 193636 m3kg mA1 Vv 02193636 01033 kg State B1 u 254355 kJkg v 060582 m3kg mB1 Vv 03 060582 04952 kg The total volume and mass is the sum of volumes mass for tanks A and B m2 mA1 mB1 01033 04952 05985 kg V2 VA1 VB1 02 03 05 m 3 v2 V2m2 05 05985 08354 m3kg State 2 P2 v2 300 kPa 08354 m3kg T2 27476C and u2 276732 kJkg The energy equation is neglecting kinetic and potential energy m2 u2 mAuA1 mBuB1 1Q2 1W2 1Q2 1Q2 05985 276732 01033 258275 04952 254355 kg kJkg 1299 kJ B A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3215 A tank has a volume of 1 m3 with oxygen at 15oC 300 kPa Another tank contains 4 kg oxygen at 60oC 500 kPa The two tanks are connected by a pipe and valve which is opened allowing the whole system to come to a single equilibrium state with the ambient at 20oC Find the final pressure and the heat transfer CV Both tanks of constant volume Continuity Eq m2 m1A m1B 0 Energy Eq m2u2 m1Au1A m1Bu1B 1Q2 1W 2 Process Eq V2 VA VB constant 1W2 0 State 1A m1A RT1A P1AVA 02598 kJkgK 28815 K 300 kPa 1 m3 4007 kg State 1B VB P1B m1BRT1B 4 kg 02598 kJkgK 33315 K 500 kPa 06924 m3 State 2 T2 v2 V2m2 V2 VA VB 1 06924 16924 m 3 m2 m1A m1B 4007 4 8007 kg P2 V2 m2RT2 8007 kg 02598 kJkgK 29315 K 16924 m3 3603 kPa Heat transfer from energy equation 1Q2 m2u2 m1Au1A m1Bu1B m1Au2 u1A m1Bu2 u1B m1ACv T2 T1A m1B Cv T2 T1B 4007 kg0662 kJkgK20 15 K 4 kg0662 kJkgK 20 60 K 9265 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3216 A rigid insulated tank is separated into two rooms by a stiff plate Room A of 05 m3 contains air at 250 kPa 300 K and room B of 1 m3 has air at 500 kPa 1000 K The plate is removed and the air comes to a uniform state without any heat transfer Find the final pressure and temperature Solution CV Total tank Control mass of constant volume Mass and volume m2 mA mB V VA VB 15 m 3 Energy Eq U2 U1 m2 u2 mAuA1 mBuB1 Q W 0 Process Eq V constant W 0 Insulated Q 0 Ideal gas at 1 mA PA1VARTA1 250 050287 300 1452 kg uA1 214364 kJkg from Table A7 Ideal gas at 2 mB PB1VBRTB1 500 kPa 1 m30287 kJkgK 1000 K 1742 kg uB1 759189 kJkg from Table A7 m2 mA mB 3194 kg u2 m2 mAuA1 mBuB1 1452 214364 1742 759189 3194 51151 kJkg Table A71 T2 6986 K P2 m2 RT2 V 3194 kg 0287 kJkgK 6986 K 15 m3 4269 kPa A B cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3217 A rigid tank A of volume 06 m3 contains 3 kg water at 120oC and the rigid tank B is 04 m3 with water at 600 kPa 200oC They are connected to a piston cylinder initially empty with closed valves The pressure in the cylinder should be 800 kPa to float the piston Now the valves are slowly opened and heat is transferred so the water reaches a uniform state at 250oC with the valves open Find the final volume and pressure and the work and heat transfer in the process CV A B C Only work in C total mass constant m2 m1 0 m2 mA1 m B1 U2 U1 1Q2 1W2 1W2 PdV Plift V2 V1 1A v 063 02 m3kg xA1 02 00010608908 0223327 u 50348 0223327 202576 95589 kJkg 1B v 035202 m3kg mB1 04035202 11363 kg u 263891 kJkg m2 3 11363 41363 kg and V1 VA VB 1 m3 V2 VA VB VC 1 m3 VC Locate state 2 Must be on PV lines shown State 1a 800 kPa v1a VAVB m 024176 m3kg 800 kPa v1a T 173C too low Assume 800 kPa 250C v 029314 m3kg v1a OK V2 m2v2 41363 kg 029314 m3kg 12125 m3 Final state is 800 kPa 250C u2 271546 kJkg 1W2 Plift V2 V1 800 kPa 12125 1 m3 170 kJ 1Q2 m2u2 m1u1 1W2 m2u2 mA1uA1 mB1uB1 1W2 41363 271546 3 95589 11363 263891 170 11 232 28677 29986 170 5536 kJ A B C V P 2 1a P2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3218 Ten kilograms of water in a pistoncylinder setup with constant pressure is at 450C and a volume of 0633 m3 It is now cooled to 20C Show the Pv diagram and find the work and heat transfer for the process Solution CV The 10 kg water Energy Eq35 mu2 u1 1Q2 1W 2 Process P C 1W2 mPv2 v1 State 1 T v1 063310 00633 m3kg Table B13 P1 5 MPa h1 33162 kJkg State 2 P P 5 MPa 20C Table B14 v2 0000 999 5 m3kg h2 8865 kJkg v P 1 2 v T 1 2 5 MPa The work from the process equation is found as 1W2 10 kg 5000 kPa 00009995 00633 m3kg 3115 kJ The heat transfer from the energy equation is 1Q2 mu2 u1 1W2 mh2 h1 1Q2 10 kg 8865 33162 kJkg 32 276 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3219 A pistoncylinder Fig P3171 contains 1 kg of water at 20C with a volume of 01 m3 Initially the piston rests on some stops with the top surface open to the atmosphere Po and a mass so a water pressure of 400 kPa will lift it To what temperature should the water be heated to lift the piston If it is heated to saturated vapor find the final temperature volume and the work 1W2 Solution a State to reach lift pressure of P 400 kPa v Vm 01 m3kg Table B12 vf v vg 04625 m3kg T T sat 14363C b State 2 is saturated vapor at 400 kPa since state 1a is twophase V P 1 2 H O Po 2 1a v2 vg 04625 m3kg V2 m v2 04625 m3 Pressure is constant as volume increase beyond initial volume 1W2 P dV P V2 V1 Plift V2 V1 400 kPa 04625 01 m3 145 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3220 Two kilograms of water is contained in a pistoncylinder Fig P3220 with a massless piston loaded with a linear spring and the outside atmosphere Initially the spring force is zero and P1 Po 100 kPa with a volume of 02 m3 If the piston just hits the upper stops the volume is 08 m3 and T 600C Heat is now added until the pressure reaches 12 MPa Find the final temperature show the P V diagram and find the work done during the process Solution V P 2 3 1 V 1 P 1 V stop 3 2 State 1 v1 Vm 02 2 01 m3kg Process 1 2 3 or 1 3 State at stops 2 or 2 v2 Vstopm 04 m3kg T2 600C Table B13 Pstop 1 MPa P 3 since Pstop P3 the process is as 1 2 3 State 3 P3 12 MPa v3 v2 04 m3kg T3 770C W13 W12 W23 1 2P1 P2V2 V1 0 1 2100 1000 kPa 08 02 m3 330 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3221 Two springs with same spring constant are installed in a massless pistoncylinder with the outside air at 100 kPa If the piston is at the bottom both springs are relaxed and the second spring comes in contact with the piston at V 2 m3 The cylinder Fig P3221 contains ammonia initially at 2C x 013 V 1 m3 which is then heated until the pressure finally reaches 1200 kPa At what pressure will the piston touch the second spring Find the final temperature the total work done by the ammonia and the heat transfer Solution P P 0 W 2 1 W 3 2 0 2 V 3 1 2 3 0 V 1 cb State 1 P 3997 kPa Table B21 v 000156 01303106 00419 m3kg u 17052 013114578 31947 kJkg m Vv 100419 23866 kg At bottom state 0 0 m3 100 kPa State 2 V 2 m3 and on line 012 Final state 3 1200 kPa on line segment 2 Slope of line 012 P V P1 P0V 39971001 2997 kPa m3 P2 P1 V2 V1PV 3997 212997 6994 kPa State 3 Last line segment has twice the slope P3 P2 V3 V22PV V3 V2 P3 P22PV V3 2 120069945994 2835 m 3 v3 v1V3V1 0041928351 01188 T3 51C u3 138439 kJkg 1383 140481383 01188 011846 012378 011846 from B22 1W3 1W2 2W3 1 2 P1 P2V2 V1 1 2 P3 P2V3 V2 5496 7930 13426 kJ The energy equation gives the heat transfer as 1Q3 mu3 u1 1W3 23866 kg 138439 31947 kJkg 13426 kJ 26 758 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3222 Ammonia NH3 is contained in a sealed rigid tank at 0C x 50 and is then heated to 100C Find the final state P2 u2 and the specific work and heat transfer Solution Continuity Eq m2 m1 Energy Eq35 E2 E1 1Q2 1W2 0 Process V2 V1 v2 v1 0001566 05 028783 014538 m3kg Table B22 v2 T2 between 1000 kPa and 1200 kPa P2 1000 200 014538 017389 014347 017389 1187 kPa V P 1 2 u2 14905 14858 14905 0935 148583 kJkg u1 17969 05 11383 74884 kJkg Process equation gives no displacement 1w2 0 The energy equation then gives the heat transfer as 1q2 u2 u1 148583 74884 737 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3223 A pistoncylinder system contains 50 L of air at 300C 100 kPa with the piston initially on a set of stops as shown A total external constant force acts on the piston so a balancing pressure inside should be 200 kPa The cylinder is made of 2 kg of steel initially at 1300C The system is insulated so that heat transfer only occurs between the steel cylinder and the air The system comes to equilibrium Find the final temperature and the work done by the air in the process and plot the process PV diagram CV Steel water out to ambient T0 This is a control mass Energy Eq U2 U1 mairu2 u1 mstu2 u1 1Q2 1W2 Entropy Eq S2 S1 mairs2 s1 msts2 s1 dQT 1S2 gen 1S 2 gen Process 1Q2 0 and must be on PV diagram shown mair P1V1 RT1 100 005 0287 57315 00304 kg Since V1a V1 then T1a T1PfloatP1 57315 200100 11463 K Use constant Cv for air at 900 K Cv uT 0833 kJkgK from A7 To reach state 1a Uair mCvT 00304 0833 1146 573 145 kJ Ust mCvT 2 046 1146 1573 3928 kJ Conclusion from this is T2 is higher than T1a 1146 K piston lifts P2 P float Write the work as 1W2 P2 V2 V1 and use constant Cv in the energy Eq as mair Cv T2 T1 mst Cst T2 T1 P2mairv2 P2 V1 now P2 v2 RT2 for the air so isolate T2 terms as mair Cv R mCst T2 mair CvT1 air mCst T1 st P2V1 00304 112 2 046 T2 00304 0833 57315 2 046 157315 200 005 Solution gives T2 15427 K V2 00304028715427200 00673 m3 1W2 P2 V2 V1 200 00673 005 346 kJ V P 2 1 1a 200 100 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3224 A pistoncylinder arrangement has the piston loaded with outside atmospheric pressure and the piston mass to a pressure of 150 kPa It contains water at 2C which is then heated until the water becomes saturated vapor Find the final temperature and specific work and heat transfer for the process Solution CV Water in the piston cylinder Continuity m2 m1 Energy Eq35 per unit mass u2 u1 1q2 1w 2 Process P constant P1 1w2 1 2 P dv P1v2 v1 State 1 T1 P1 Table B15 compressed solid take as saturated solid v1 109103 m3kg u1 33762 kJkg State 2 x 1 P2 P1 150 kPa due to process Table B12 v2 vgP2 11593 m3kg T2 1114C u2 25197 kJkg From the process equation 1w2 P1v2 v1 150 kPa 11593 109103 m3kg 1737 kJkg From the energy equation 1q2 u2 u1 1w2 25197 33762 1737 3031 kJkg P T v V L S CP 2 1 SV LV P CP v T CP v 1 2 1 2 P C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3225 A pistoncylinder contains 1 kg of ammonia at 20C with a volume of 01 m3 shown in Fig P3225 Initially the piston rests on some stops with the top surface open to the atmosphere Po so a pressure of 1400 kPa is required to lift it To what temperature should the ammonia be heated to lift the piston If it is heated to saturated vapor find the final temperature volume and the heat transfer Solution CV Ammonia which is a control mass m2 m1 m mu2 u1 1Q2 1W2 State 1 20C v1 010 vg x1 01 0001638014758 06665 u1 uf x1 ufg 27289 06665 10593 9789 kJkg Process Piston starts to lift at state 1a Plift v1 State 1a 1400 kPa v1 Table B22 superheated vapor Ta 50 60 50 01 009942 010423 009942 512 C 1400 1200 857 P v 1 2 1a T v 1 2 1a State 2 x 10 v2 v1 V2 mv2 01 m3 T2 30 01 011049 5009397 011049 332 C u2 13387 kJkg 1W2 0 1Q2 m1q2 mu2 u1 1 kg 13387 9789 kJkg 3598 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3226 A piston held by a pin in an insulated cylinder shown in Fig P3226 contains 2 kg water at 100C quality 98 The piston has a mass of 102 kg with cross sectional area of 100 cm2 and the ambient pressure is 100 kPa The pin is released which allows the piston to move Determine the final state of the water assuming the process to be adiabatic Solution CV The water This is a control mass Continuity Eq m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Process in cylinder P Pfloat if piston not supported by pin P2 Pfloat P0 mpgA 100 102 9807 100104 103 200 kPa We thus need one more property for state 2 and we have one equation namely the energy equation From the equilibrium pressure the work becomes 1W2 Pfloat dV P2 mv2 v1 With this work the energy equation gives per unit mass u2 u1 1q2 1w2 0 P2v2 v1 or with rearrangement to have the unknowns on the left hand side u2 P2v2 h2 u1 P2v1 h2 u1 P2v1 24648 kJkg 200 kPa 16395 m3kg 27927 kJkg State 2 P2 h2 Table B13 T2 16175C p g o P m Water cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3227 A vertical cylinder Fig P3227 has a 6118kg piston locked with a pin trapping 10 L of R410A at 10C 90 quality inside Atmospheric pressure is 100 kPa and the cylinder crosssectional area is 0006 m2 The pin is removed allowing the piston to move and come to rest with a final temperature of 10C for the R 410A Find the final pressure final volume and the work done and the heat transfer for the R410A Solution CV R410A this is a control mass Energy Eq mu2 u1 1Q2 1W2 Process in cylinder P Pfloat if piston not supported by pin State 1 T x from table B41 v1 0000886 09 002295 0021541 m3kg u1 7224 09 18366 13164 kJkg m V1v1 0010 m30021541 m3kg 0464 kg Force balance on piston gives the equilibrium pressure P2 P0 mPg AP 100 6118 9807 0006 1000 200 kPa State 2 TP in Table B42 v2 01507 0163222 015696 m3kg u2 26506 279132 272095 kJkg V2 mv2 0464 kg 015696 m3kg 00728 m3 728 L 1W2 Pequil dV P2V2V1 200 kPa 00728 0010 m3 1256 kJ 1Q2 mu2 u1 1W 2 0464 kg 272095 13164 kJkg 1256 kJ 7773 kJ p g o P m R410A cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3228 A cylinder having an initial volume of 3 m3 contains 01 kg of water at 40C The water is then compressed in an isothermal quasiequilibrium process until it has a quality of 50 Calculate the work done in the process splitting it into two steps Assume the water vapor is an ideal gas during the first step of the process Solution CV Water State 2 40C x 1 Tbl B11 PG 7384 kPa vG 1952 m3kg State 1 v1 V1m 3 01 30 m3kg vG so H2O ideal gas from 12 so since constant T P1 PG vG v1 7384 1952 30 48 kPa V2 mv2 01 1952 1952 m3 T CP v 1 2 P P CP v T 3 738 40 2 1 3 sat P1 Process T C and ideal gas gives work from Eq321 1W2 1 2 PdV P1V1ln V1 V2 48 30 ln 1952 3 619 kJ v3 0001008 05 19519 97605 V3 mv3 0976 m3 P C Pg This gives a work term as 2W3 2 3 PdV Pg V3V2 7384 kPa 0976 1952 m3 721 kJ Total work 1W3 1W2 2W3 619 721 134 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3229 A springloaded pistoncylinder arrangement contains R134a at 20C 24 quality with a volume 50 L The setup is heated and thus expands moving the piston It is noted that when the last drop of liquid disappears the temperature is 40C The heating is stopped when T 130C Verify the final pressure is about 1200 kPa by iteration and find the work done in the process Solution CV R134a This is a control mass State 1 Table B51 v1 0000817 024 003524 0009274 P1 5728 kPa m V v1 0050 0009274 5391 kg Process Linear Spring P A Bv State 2 x2 1 T2 P2 1017 MPa v2 002002 m3kg Now we have fixed two points on the process line so for final state 3 P3 P1 P2 P1 v2 v1 v3 v1 RHS Relation between P3 and v 3 State 3 T3 and on process line iterate on P3 given T3 at P3 12 MPa v3 002504 P3 RHS 00247 at P3 14 MPa v3 002112 P3 RHS 03376 Linear interpolation gives P3 1200 00247 03376 00247 1400 1200 1214 kPa v3 002504 00247 03376 00247 002112 002504 002478 m3kg W13 P dV 1 2 P1 P3V3 V1 1 2 P1 P3 m v3 v1 1 2 5391 kg 5728 1214 kPa 002478 0009274 m3kg 747 kJ v P P 1 P 2 P 3 1 2 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3230 Water in a pistoncylinder similar to Fig P3225 is at 100oC x 05 with mass 1 kg and the piston rests on the stops The equilibrium pressure that will float the piston is 300 kPa The water is heated to 300oC by an electrical heater At what temperature would all the liquid be gone Find the final Pv the work and heat transfer in the process CV The 1 kg water Continuty m2 m1 m Energy mu2 u1 1Q2 1W2 Process V constant if P Plift otherwise P Plift see Pv diagram State 1 Tx Table B11 v1 0001044 05 16719 083697 m3kg u1 41891 05 208758 14627 kJkg State 1a 300 kPa v v1 vg 300 kPa 06058 m3kg so superheated vapor Piston starts to move at state 1a 1W1a 0 u1a 276882 kJkg 1Q1a mu u 1 kg 276882 14627 kJkg 130612 kJ State 1b reached before state 1a so v v1 vg see this in B11 T1b 120 5 083697 08908076953 08908 1222oC State 2 T2 T1a Table B13 v2 087529 u2 280669 kJkg Work is seen in the PV diagram when volume changes P Plift 1W2 1aW2 P2 mv2 v1 300 1087529 083697 115 kJ Heat transfer is from the energy equation 1Q2 1 kg 280669 14627 kJkg 115 kJ 13555 kJ H O Po 2 cb P v 1 1a 2 100 C o cb 1b Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3231 A pistoncylinder arrangement has a linear spring and the outside atmosphere acting on the piston shown in Fig P3231 It contains water at 3 MPa 400C with the volume being 01 m3 If the piston is at the bottom the spring exerts a force such that a pressure of 200 kPa inside is required to balance the forces The system now cools until the pressure reaches 1 MPa Find the heat transfer for the process Solution CV Water Continuity Eq m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 v 2 v 1 0 200 kPa 3 MPa 1 MPa 2 1 P V v State 1 Table B13 v1 009936 m3kg u1 29328 kJkg m Vv1 01009936 1006 kg Process Linear spring so P linear in v P P0 P1 P0vv 1 v2 P1 P0 P2 P0v1 3000 200 1000 200009936 002839 m3kg State 2 P2 v2 x2 v2 0001127019332 0141 T2 17991C u2 76162 x2 182197 101858 kJkg Process 1W2 PdV 1 2 mP1 P2v2 v1 1 2 1006 kg 3000 1000 kPa 002839 009936 m3kg 14279 kJ Heat transfer from the energy equation 1Q2 mu2 u1 1W2 1006 kg 101858 29328 kJkg 14279 kJ 20685 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3232 A 1 m3 tank containing air at 25oC and 500 kPa is connected through a valve to another tank containing 4 kg of air at 60oC and 200 kPa Now the valve is opened and the entire system reaches thermal equilibrium with the surroundings at 20oC Assume constant specific heat at 25oC and determine the final pressure and the heat transfer Control volume all the air Assume air is an ideal gas Continuity Eq m2 mA1 mB1 0 Energy Eq U2 U1 m2u2 mA1uA1 mB1uB1 1Q2 1W2 Process Eq V constant 1W2 0 State 1 mA1 RTA1 PA1VA1 0287 kJkgK2982 K 500 kPa1m3 584 kg VB1 PB1 mB1RTB1 4 kg0287 kJkgK3332 K 200 kNm2 191 m3 State 2 T2 20C v2 V2m2 m2 mA1 mB1 4 584 984 kg V2 VA1 VB1 1 191 291 m3 P2 m2RT2 V2 984 kg0287 kJkgK2932 K 291 m3 2845 kPa Energy Eq55 or 511 1Q2 U2 U1 m2u2 mA1uA1 mB1uB1 mA1u2 uA1 mB1u2 uB1 mA1Cv0T2 TA1 mB1Cv0T2 TB1 584 0717 20 25 4 0717 20 60 1356 kJ The air gave energy out A B Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3233 A rigid container has two rooms filled with water each 1 m3 separated by a wall see Fig P3210 Room A has P 200 kPa with a quality x 080 Room B has P 2 MPa and T 400C The partition wall is removed and the water comes to a uniform state which after a while due to heat transfer has a temperature of 200C Find the final pressure and the heat transfer in the process Solution CV A B Constant total mass and constant total volume Continuity m2 mA1 mB1 0 V2 VA VB 2 m3 Energy Eq35 U2 U1 m2u2 mA1uA1 mA1uA1 1Q2 1W2 1Q 2 Process V VA VB constant 1W2 0 State 1A Table B12 uA1 50447 08 202502 212447 kJkg vA1 0001061 08 088467 070877 m3kg State 1B Table B13 u B1 29452 vB1 01512 mA1 1vA1 1411 kg mB1 1vB1 6614 kg State 2 T2 v2 V2m 2 21411 6614 024924 m3kg Table B13 superheated vapor 800 kPa P2 1 MPa Interpolate to get the proper v2 P2 800 02492402608 02059602608 200 842 kPa u2 26288 kJkg From the energy equation 1Q2 8025 26288 1411 212447 6614 29452 1381 kJ A B Q P A1 v P A1 PB1 B1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3234 Consider the pistoncylinder arrangement shown in Fig P3234 A frictionless piston is free to move between two sets of stops When the piston rests on the lower stops the enclosed volume is 400 L When the piston reaches the upper stops the volume is 600 L The cylinder initially contains water at 100 kPa 20 quality It is heated until the water eventually exists as saturated vapor The mass of the piston requires 300 kPa pressure to move it against the outside ambient pressure Determine the final pressure in the cylinder the heat transfer and the work for the overall process Solution CV Water Check to see if piston reaches upper stops Energy Eq35 mu4 u1 1Q4 1W 4 Process If P 300 kPa then V 400 L line 21 and below If P 300 kPa then V 600 L line 34 and above If P 300 kPa then 400 L V 600 L line 23 State 1 v1 0001043 021693 033964 m v1 V1 04 033964 1178 kg u1 41736 02 20887 8351 kJkg State 3 v3 06 1178 05095 vG 06058 at P3 300 kPa Piston does reach upper stops to reach sat vapor State 4 v4 v3 05095 m3kg vG at P4 From Table B12 P4 361 kPa u4 25500 kJkg 1W4 1W2 2W3 3W4 0 2W3 0 1W4 P2V3 V2 300 06 04 60 kJ 1Q4 mu4 u1 1W4 117825500 8351 60 2080 kJ The three lines for process parts are shown in the PV diagram and is dictated by the motion of the piston force balance 2 1 3 4 P 4 2 P P 300 3 P 1 T v Wate Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3235 Ammonia 2 kg in a pistoncylinder is at 100 kPa 20oC and is now heated in a polytropic process with n 13 to a pressure of 200 kPa Do not use ideal gas approximation and find T2 the work and heat transfer in the process Take CV as the Ammonia constant mass Energy Eq35 mu2 u1 1Q2 1W2 Process Pvn constant n 13 State 1 Superheated vapor table B22 v1 12101 m3kg u1 13078 kJkg Process gives v2 v1 P1P21n 12101 100200 113 0710 m3kg State 2 Table B22 at 200 kPa interpolate u2 137649 kJkg T2 24oC Work is done while piston moves at increasing pressure so we get 1W2 m 1 n P2v2 P1v1 2 1 13 kg 200071 10012101 kJkg 1399 kJ Heat transfer is found from the energy equation 1Q2 mu2 u1 1W 2 2 137649 13078 1399 252 kJ P P 2 P v 1 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3236 A small flexible bag contains 01 kg ammonia at 10oC and 300 kPa The bag material is such that the pressure inside varies linear with volume The bag is left in the sun with with an incident radiation of 75 W loosing energy with an average 25 W to the ambient ground and air After a while the bag is heated to 30oC at which time the pressure is 1000 kPa Find the work and heat transfer in the process and the elapsed time Take CV as the Ammonia constant mass Continuity Eq m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Process P A BV linear in V State 1 Compressed liquid P Psat take saturated liquid at same temperature v1 vf 10 0001534 m3kg u1 uf 13396 kJkg State 2 Table B21 at 30oC P Psat so superheated vapor v2 013206 m3kg u2 13471 kJkg V2 mv2 00132 m 3 Work is done while piston moves at increacing pressure so we get 1W2 ½300 1000 kPa01 kg013206 0001534 m3kg 8484 kJ Heat transfer is found from the energy equation 1Q2 mu2 u1 1W2 01 kg 13471 13396 kJkg 8484 kJ 121314 8484 1298 kJ Q net 75 25 50 Watts Assume the constant rate Q net dQdt 1Q2 t so the time becomes t 1Q2 Q net 129800 50 JW 2596 s 433 min NH3 P CP v 300 2 1 1000 T CP v 2 1 10 30 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3237 A cylinderpiston arrangement contains 01 kg R410A of quality x 02534 and at 20C Stops are mounted so Vstop 3V1 see Fig P3237 The system is now heated to the final temperature of 20C Find the work and heat transfer in the process and draw the Pv diagram CV The R410A mass Energy Eq511 mu2 u1 1Q2 1W2 Process P Constant if V Vstop V Vstop if P P1 State 1 u1 2792 x1 21807 8318 kJkg P1 Psat 3996 kPa v1 0000803 x1 0064 001702 m3kg State 1a vstop 3 v1 005106 m3kg vg at P 1 State 2 at 20C T1 vstop vg 001758 m3kg so superheated vapor Table B42 Find it at P2 600 kPa u2 27356 kJkg V1 mv1 00017 m3 V2 mv2 00051 m 3 1W2 PdV P1 V2 V1 3996 kPa 00051 00017 m3 136 kJ 1Q2 mu2 u1 1W2 01 kg 27356 8318 kJkg 136 kJ 20398 kJ See the work term from the process in the Pv diagram v P 1 1a 2 R410A P o cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3238 A pistoncylinder shown in Fig P3238 contains R410A at 20C x 20 The volume is 02 m3 It is known that Vstop 04 m3 and if the piston sits at the bottom the spring force balances the other loads on the piston It is now heated up to 20C Find the mass of the fluid and show the Pv diagram Find the work and heat transfer Solution CV R410A this is a control mass Properties in Table B4 Continuity Eq m2 m1 Energy Eq35 E2 E1 mu2 u1 1Q2 1W 2 Process P A BV V 04 m3 A 0 at V 0 P 0 State 1 v1 0000803 02 00640 00136 m3kg u1 2792 02 21807 715 kJkg m m1 V1v1 14706 kg System on line V Vstop P1 3996 kPa Pstop 2P1 7992 kPa vstop 2v1 00272 m3kg State stop Pv Tstop 0C TWOPHASE STATE Since T2 Tstop v2 vstop 00272 m3kg State 2 T2 v2 Table B42 Interpolate between 1000 and 1200 kPa P2 1035 kPa u2 3665 kJkg From the process curve see also area in PV diagram the work is 1W2 PdV 1 2 P1 PstopVstop V1 1 2 3996 799202 1198 kJ From the energy equation 1Q2 mu2 u1 1W2 147062665 715 1198 29875 kJ P 0 0 02 04 1 V P 1 2P 1 T stop 0 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3239 A spherical balloon contains 2 kg of R410A at 0C 30 quality This system is heated until the pressure in the balloon reaches 1 MPa For this process it can be assumed that the pressure in the balloon is directly proportional to the balloon diameter How does pressure vary with volume and what is the heat transfer for the process Solution CV R410A which is a control mass m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 State 1 0C x 03 Table B41 gives P1 7987 kPa v1 0000855 03 003182 001040 m3kg u1 5707 03 19595 11586 kJkg Process P D V D3 PV 13 constant polytropic n 13 V2 mv2 V1 P2 P1 3 mv1 P2 P1 3 v2 v1 P2 P1 3 001040 1000 79873 002041 m3kg State 2 P2 1 MPa process v2 002041 Table B42 T2 725C sat vf 0000877 vfg 002508 m3kg uf 6802 ufg 18718 kJkg x2 07787 u2 2137 kJkg 1W2 P dV m P2v2 P1v1 1 n 2 1000 002041 7987 00104 1 13 1816 kJ 1Q2 mu2 u1 1W2 2 2137 11586 1816 2138 kJ P V W 1 2 sat vapor line Notice The R410A is not an ideal gas at any state in this problem R410A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3240 A pistoncylinder arrangement B is connected to a 1m3 tank A by a line and valve shown in Fig P3240 Initially both contain water with A at 100 kPa saturated vapor and B at 400C 300 kPa 1 m3 The valve is now opened and the water in both A and B comes to a uniform state a Find the initial mass in A and B b If the process results in T2 200C find the heat transfer and work Solution CV A B This is a control mass Continuity equation m2 mA1 mB1 0 Energy m2u2 mA1uA1 mB1uB1 1Q2 1W 2 System if VB 0 piston floats PB PB1 const if VB 0 then P2 PB1 and v VAmtot see PV diagram 1W2 PBdVB PB1V2 V1B PB1 V2 V1 tot State A1 Table B11 x 1 vA1 1694 m3kg uA1 25061 kJkg mA1 VAvA1 05903 kg State B1 Table B12 sup vapor vB1 10315 m3kg uB1 29655 kJkg mB1 VB1vB1 09695 kg V P 2 a PB1 2 m2 mTOT 156 kg At T2 PB1 v2 07163 va VAmtot 0641 so VB2 0 so now state 2 P2 PB1 300 kPa T2 200 C u2 26507 kJkg and V2 m2 v2 156 07163 1117 m3 we could also have checked Ta at 300 kPa 0641 m3kg T 155 C 1W2 PB1V2 V1 26482 kJ 1Q2 m2u2 mA1uA1 mB1uB1 1W2 4847 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3241 Consider the system shown in Fig P3241 Tank A has a volume of 100 L and contains saturated vapor R134a at 30C When the valve is cracked open R134a flows slowly into cylinder B The piston mass requires a pressure of 200 kPa in cylinder B to raise the piston The process ends when the pressure in tank A has fallen to 200 kPa During this process heat is exchanged with the surroundings such that the R134a always remains at 30C Calculate the heat transfer for the process Solution CV The R134a This is a control mass Continuity Eq m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Process in B If VB 0 then P Pfloat piston must move 1W2 Pfloat dV Pfloatmv2 v1 Work done in B against constant external force equilibrium P in cyl B State 1 30C x 1 Table B51 v1 002671 m3kg u1 39448 kJkg m Vv1 01 002671 3744 kg State 2 30C 200 kPa superheated vapor Table B52 v2 011889 m3kg u2 4031 kJkg From the process equation 1W2 Pfloatmv2 v1 2003744011889 002671 6902 kJ From the energy equation 1Q2 mu2 u1 1W2 3744 4031 39448 6902 1013 kJ V P 1 2 B B B A Tank P o g ENGLISH UNIT PROBLEMS SOLUTION MANUAL CHAPTER 3 English units Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CHAPTER 3 SUBSECTION PROB NO ConceptStudy Guide Problems 242250 Kinetic potential energy and work 251257 Properties uh 258260 Simple processes 261271 Specific heats 272281 Polytropic process 282284 Multistep proccesses 285290 Energy Eq rate form 291296 Heat Transfer Rates 297300 Review Problems 301306 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Concept Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3242E What is 1 cal in english units what is 1 Btu in ftlbf Look in Table A1 for the conversion factors under energy 1 Btu 7781693 lbfft 1 cal 41868 J 41868 1055 Btu 000397 Btu 000397 7781693 lbfft 3088 lbfft Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3243E Work as F x has units of lbfft what is that in Btu Look in Table A1 for the conversion factors under energy 1 lbfft 1 lbfft 7781693 ftlbfBtu 128507 103 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3244E Work in the expression Eq 318 or Eq 322 involves PV For P in psia and V in ft3 how does that become Btu Units psia ft3 lbfin2 ft3 lbfft ftin2 144 lbfft 1447781693 Btu 144 128507 103 Btu 018509 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3245E Look at the R410A value for uf at 60 F Can the energy really be negative Explain The absolute value of u and h are arbitrary A constant can be added to all u and h values and the table is still valid It is customary to select the reference such that u for saturated liquid water at the triple point is zero The standard for refrigerants like R410A is that h is set to zero as saturated liquid at 40 F other substances as cryogenic substances like nitrogen methane etc may have different states at which h is set to zero The ideal gas tables use a zero point for h as 77 F or at absolute zero 0 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3246E An ideal gas in a pistoncylinder is heated with 2 Btu during an isothermal process How much work is involved Energy Eq u2 u1 1q2 1w2 0 since u2 u1 isothermal Then 1W2 1Q2 m 1w2 2 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3247E You heat a gas 20 R at P C Which one in table F4 requires most energy Why A constant pressure process in a control mass gives recall p 109 and Eq 339 1q2 u2 u1 1w2 h2 h1 Cp T The one with the highest specific heat is hydrogen H2 The hydrogen has the smallest mass but the same kinetic energy per mol as other molecules and thus the most energy per unit mass is needed to increase the temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3248E The air drag force on a car is 0225 A ρV2 Verify that the unit becomes lbf F E A 0225 A ρVA2E d Units ftA2E A lbmftA3E A ftsA2E A lbmft sA2E A 132174 lbf Recall the result from Newtons law p 9 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Kinetic and Potential Energy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3249E An escalator raises a 200 lbm bucket of sand 30 ft in 1 minute Determine the amount of work done during the process Solution W AFdxEA FAdxEA F x mgH 200 lbm 32174 ftsA2E A 30 ft 200 lbf 30 ft 6000 ft lbf 6000778 Btu 771 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3250E A hydraulic hoist raises a 3650 lbm car 6 ft in an auto repair shop The hydraulic pump has a constant pressure of 100 lbfin2 on its piston What is the increase in potential energy of the car and how much volume should the pump displace to deliver that amount of work Solution CV Car No change in kinetic or internal energy of the car neglect hoist mass E2 E1 PE2 PE1 mg Z2 Z1 32174 6 32174 3650 21 900 lbfft The increase in potential energy is work into car from pump at constant P W EA2E A EA1E A P dV P V V A E2 E1 EPE A A 21 900 100 144E A 152 ft3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3251E A piston motion moves a 50 lbm hammerhead vertically down 3 ft from rest to a velocity of 150 fts in a stamping machine What is the change in total energy of the hammerhead Solution CV Hammerhead The hammerhead does not change internal energy ie same PT E2 E1 muA2E A u1 mA1 2E AVA2 2E A 0 mg hA2E A 0 0 50 12 150A2E A 50 32174 3 lbmfts2 562 500 482632174 17 333 lbfft A17 333 778E A Btu 2228 Btu Recall that 1 lbfft 32174 lbmfts2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3252E Airplane takeoff from an aircraft carrier is assisted by a steam driven pistoncylinder with an average pressure of 200 psia A 38 500 lbm airplane should be accelerated from zero to a speed of 100 fts with 30 of the energy coming from the steam piston Find the needed piston displacement volume Solution CV Airplane No change in internal or potential energy only kinetic energy is changed EA2E A EA1E A m 12 VA2 2E A 0 38 500 lbm 12 100A2E A ftsA2E A 192 500 000 lbmftsA2E A 5 983 092 lbfft The work supplied by the piston is 30 of the energy increase W P dV PAavgE A V 030 EA2E A EA1E A 030 5 983 092 lbfft 1 794 928 lbfft V A W Pavg E A A1 794 928 200E A A lbfft 144 lbfft2 E A 623 ftA3E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3253E A bulldozer pushes 1000 lbm of dirt 300 ft with a force of 400 lbf It then lifts the dirt 10 ft up to put it in a dump truck How much work did it do in each situation Solution W F dx F x 400 lbf 300 ft 120 000 lbfft 154 Btu W F dz mg dz mg Z 1000 lbm 32174 ftsA2E A 10 ft 32174 lbmft sA2E Albf 10 000 lbfft 1285 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3254E Two hydraulic cylinders maintain a pressure of 175 psia One has a cross sectional area of 01 ftA2E A the other 03 ftA2E A To deliver a work of 1 Btu to the piston how large a displacement V and piston motion H is needed for each cylinder Neglect PAatmE Solution W F dx P dV PA dx PA H P V W 1 Btu 77817 lbfft V AW PE A A 77817 lbfft 175 144 lbfft2 E A 0030 873 ftA3E Both cases the height is H VA HA1E A A0030873 01E A 03087 ft HA2E A A0030873 03E A 01029 ft cb 1 2 F F2 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3255E A linear spring F kAsE Ax xA0E A with spring constant ks 35 lbfft is stretched until it is 25 in longer Find the required force and work input Solution F kAsE Ax xA0E A 35 lbfft 2512 ft 7292 lbf W AFdxEA Aksx x0dx x0EA A1 2E A kAsE Ax xA0E AA2E A1 2E A 35 lbfft 2512A2E A ftA2E A 076 lbfft 97610A4E A Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3256E A cylinder fitted with a frictionless piston contains 10 lbm of superheated refrigerant R134a vapor at 100 lbfin2 300 F The setup is cooled at constant pressure until the R134a reaches a quality of 25 Calculate the work done in the process Solution Constant pressure process boundary work State properties from Table F10 State 1 Table F102 vA1E A 076629 ftA3E Albm State 2 Table F101 vA2E A 0013331 025 046652 012996 ftA3E Albm Interpolated to be at 100 psia numbers at 1015 psia could have been used A1E AWA2E A P dV P VA2E AVA1E A mP vA2E AvA1E A 10 lbm 100 lbfin2 A144 778E A A inft2 ElbfftBtuE A 012996 076629 ftA3E Albm 11778 Btu T CP v 1 2 P 100 psia P CP v T 100 79 300 2 1 cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3257E A piston of 4 lbm is accelerated to 60 fts from rest What constant gas pressure is required if the area is 4 inA2E A the travel 4 in and the outside pressure is 15 psia CV Piston EA2E A E1PIST mu2 u1 m12VA2 2E A 0 mg 0 0 12 m VA2 2E A 05 4 lbm 60A2E A ftsA2E A 7200 lbmftA2E AsA2E A A 7200 32174E A ftlbf Energy equation for the piston is EA2E A E1 PIST Wgas Watm Pavg Vgas Po Vgas Vgas A L 4 inA2E A 4 in 16 inA3E Pavg Vgas EA2E A E1PIST Po Vgas Pavg EA2E A E1PIST Vgas Po A 7200 lbfft 32174 16 in3 E A 15 lbfinA2E 16784 psia 15 psia 1828 psia Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Properties General Tables Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3258E Find the missing properties and give the phase of the substance a H2O u 1000 Btulbm T 270 F h v x b H2O u 450 Btulbm P 1500 lbfin2 T x v c R410A T 30 F P 120 lbfin2 h x Solution a Table F71 uAfE A u ug 2phase mixture of liquid and vapor x u uAfE A ufg 1000 2388185414 08912 v vAfE A x vfg 001717 08912 100483 8972 ftA3E Albm h hAfE A x hfg 23895 08912 93195 10695 Btulbm h u Pv 1000 41848 8972 144778 b Table F71 u uf so compressed liquid B13 x undefined T 4718 F v 0019689 ftA3E Albm c Table F91 P Psat x undef compr liquid Approximate as saturated liquid at same T h hAfE A 2411 Btulbm States shown are placed relative to the twophase region not to each other P CP v T CP v T a c b a c b P const Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3259E Find the missing properties among P T v u h together with x if applicable and give the phase of the substance a R410A T 50 F u 85 Btulbm b H2O T 600 F h 1322 Btulbm c R410A P 150 lbfin2 h 135 Btulbm Solution a Table F91 u ug LV mixture P 157473 lbfinA2E x 85 3106 7896 06831 v 0014 06831 03676 02653 ftA3E Albm h 3147 06831 8967 9272 Btulbm b Table F71 h hg superheated vapor follow 600 F in F72 P 200 lbfinA2E A v 3058 ftA3E Albm u 12089 Btulbm c Table F91 h hg superheated vapor so in F92 T 100 F v 0483 ftA3E Albm u 12125 Btulbm States shown are placed relative to the twophase region not to each other P CP v T CP v T a b c b c a P const Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3260E Find the missing properties among P T v u h together qwith x if applicable and give the phase of the substance a R134a T 140 F h 185 Btulbm b NH3 T 170 F P 60 lbfin2 c R134a T 100 F u 175 Btulbm Solution a Table F101 h hAgE A x undef superheated vapor F102 find it at given T between saturated 2439 psi and 200 psi to match h v 01836 02459 01836 A 185 18363 1868218363E A 02104 ftA3E Albm P 24393 200 24393 A 185 18363 1868218363E A 225 lbfinA2E b Table F81 P Psat x undef superheated vapor F82 v 63456 65694 2 6457 ftA3E Albm u hPv 1269459 70564 60 64575 144778 700115 7171 628405 Btulbm c Table F101 u uAgE A sup vapor Interpolate between 40 and 60 psia tables in F102 P 40 40 20175 1755717485 17557 40 20 0791667 558 lbfinA2E A v 14015 09091 14015 0791667 10117 ftA3E Albm h 18595 18494 18595 0791667 18515 Btulbm States shown are placed relative to the twophase region not to each other P CP v T CP v T a b b a P const c c Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Simple Processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3261E Ammonia at 30 F quality 60 is contained in a rigid 8ft3 tank The tank and ammonia are now heated to a final pressure of 150 lbfin2 Determine the heat transfer for the process Solution CV NHA3E V P 1 2 Continuity Eq mA2E A mA1E A m Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E Process Constant volume vA2E A vA1E A A1E AWA2E A 0 State 1 Table F81 twophase state vA1E A 002502 06 47978 2904 ftA3E Albm uA1E A 7506 06 49117 36975 Btulbm m VvA1E A 82904 2755 lbm State 2 PA2E A vA2E A vA1E A superheated vapor Table F82 TA2E A 258 F uA2E A 66142 Btulbm So solve for heat transfer in the energy equation A1E AQA2E A 2755 lbm 66142 36975 Btulbm 8036 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3262E Saturated vapor R410A at 30 F in a rigid tank is cooled to 0 F Find the specific heat transfer Solution CV R410A in tank mA2E A mA1E A Energy Eq35 uA2E A uA1E A A1E AqA2E A A1E AwA2E Process V constant vA2E A vA1E A Vm A1E AwA2E A A0E Table F91 State 1 vA1E A 05426 ftA3E Albm uA1E A 10863 Btulbm State 2 0 F vA2E A vA1E A Vm look in Table F91 at 0 F xA2E A A v2 vf2 Evfg2 E A A05426 001295 09448E A 0560595 uA2E A uAf2E A xA2E A uAfE AgE A2E A 1337 xA2E A 9275 65365 Btulbm From the energy equation A1E AqA2E A uA2E A uA1E A 65365 10863 4326 Btulbm V P 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3263E Saturated vapor R410A at 100 psia in a constant pressure piston cylinder is heated to 70 F Find the specific heat transfer Solution CV R410A mA2E A mA1E A m Energy Eq35 uA2E A uA1E A A1E AqA2E A A1E AwA2E Process P const A1E AwA2E A APdvEA Pv PvA2E A vA1E A State 1 Table F92 or F91 hA1E A 11938 Btulbm State 2 Table F92 hA2E A 13044 Btulbm A1E AqA2E A uA2E A uA1E A A1E AwA2E A uA2E A uA1E A PvA2E A vA1E A hA2E A hA1E A A1E Aq A2E A 13044 11938 1106 Btulbm v P 1 2 v T 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3264E A rigid tank holds 15 lbm R410A at 100 F as saturated vapor The tank is now cooled to 60 F by heat transfer to the ambient Which two properties determine the final state Determine the amount of work and heat transfer during the process CV The R410A this is a control mass Process Rigid tank V C v constant A1E AWA2E A A 1 2 PdVEA 0 Energy Eq UA2E A UA1E A A1E AQA2E A A1E AWA2E A A1E AQA2E A State 1 vA1E A 01657 ftA3E Albm uA1E A 1117 Btulbm State 2 T v twophase straight down in Pv diagram from state 1 xA2E A v vAfE AvAfgE A 01657 00145103076 049151 uA2E A uAfE A xA2E A uAfgE A 3478 049151 7582 72046 Btulbm A1E AQA2E A muA2E A uA1E A 15 lbm 72046 1117 Btulbm 595 Btu V P 2 100 F 1 333 185 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3265E A cylinder fitted with a frictionless piston contains 4 lbm of superheated refrigerant R134a vapor at 400 lbfin2 200 F The cylinder is now cooled so the R134a remains at constant pressure until it reaches a quality of 75 Calculate the heat transfer in the process Solution CV R134a mA2E A mA1E A m Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E Process P const A1E AWA2E A APdVEA PV PVA2E A VA1E A PmvA2E A vA1E A V P 1 2 V T 1 2 State 1 Table F102 hA1E A 19292 Btulbm State 2 Table F101 hA2E A 14062 075 4374 173425 Btulbm A1E AQA2E A muA2E A uA1E A A1E AWA2E A muA2E A uA1E A PmvA2E A vA1E A mhA2E A hA1E A 4 lbm 173425 19292 Btulbm 7798 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3266E A waterfilled reactor with volume of 50 ft3 is at 2000 lbfin2 560 F and placed inside a containment room as shown in Fig P3101 The room is well insulated and initially evacuated Due to a failure the reactor ruptures and the water fills the containment room Find the minimum room volume so the final pressure does not exceed 30 lbfin2 CV Containment room and reactor Mass mA2E A mA1E A VAreactorE AvA1E A 50002172 22957 lbm Energy Eq muA2E A uA1E A A1E AQA2E A A1E AWA2E A A0E A uA2E A uA1E A 5525 Btulbm State 2 30 lbfin2 uA2E A ug 2 phase Table F71 u 5525 21848 xA2E A 86941 xA2E A 03842 vA2E A 0017 03842 13808 5322 ftA3E Albm VA2E A mvA2E A 22957 lbm 5322 ftA3E Albm 12 218 ftA3E P v 1 T v 1 2 30 psia 30 2 u const 2000 P T v L CP 1 2 30 psia Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3267E Ammonia 1 lbm is in a piston cylinder at 30 psia 20 F is heated in a process where the pressure varies linear with volume to a state of 240 F 40 psia Find the work and the heat transfer in the process Solution Take CV as the Ammonia constant mass Continuity Eq m2 m1 m Process P A BV linear in V State 1 Superheated vapor v1 97206 ftA3E Albm u1 62239 30 97206 A144 778E A 5684 Btulbm State 2 Superheated vapor v2 109061 ftA3E Albm u2 7414 40 109061 144778 66065 Btulbm Work is done while piston moves at increasing pressure so we get 1W2 P dV area Pavg V2 V1 A1 2E A PA1E A PA2E AmvA2E A vA1E A ½30 40 psi 1 lbm 109061 97206 ftA3E Albm 41493 psiftA3E A 41493 144 lbfft 597492 lbfft 768 Btu see conversions in A1 p 755 Heat transfer from the energy equation 1Q2 mu2 u1 1W2 1 66065 5684 768 9993 Btu P P 2 P v 1 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3268E A piston cylinder arrangement with a linear spring similar to Fig P3105 contains R134a at 60 F x 06 and a volume of 07 ftA3E A It is heated to 140 F at which point the specific volume is 04413 ftA3E Albm Find the final pressure the work and the heat transfer in the process Take CV as the R134a m2 m1 m mu2 u1 1Q2 1W2 State 1 T1 x1 Two phase so Table F101 P1 Psat 72271 psia v1 vf x1 vfg 001291 06 06503 040309 ftA3E Albm u1 uf x1 ufg 9827 06 6931 139856 Btulbm m V1v1 07 ftA3E A 040309 ftA3E Albm 17366 lbm State 2 T v Superheated vapor Table F102 P2 125 psia v2 04413 ftA3E Albm u2 18077 Btulbm V2 m v2 17366 04413 076636 ftA3E Work is done while piston moves at linearly varying pressure so we get 1W2 P dV area Pavg V2 V1 05P2 P1 V2 V1 05 72271 125 psia 076636 07 ftA3E A 654545 144778 Btu 1212 Btu Heat transfer is found from the energy equation 1Q2 mu2 u1 1W2 17366 18077 139856 1212 7226 Btu P P 2 P v 1 1 2 cb R134a Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3269E Water in a 6ft3 closed rigid tank is at 200 F 90 quality The tank is then cooled to 20 F Calculate the heat transfer during the process Solution CV Water in tank mA2E A mA1E A muA2E A uA1E A A1E AQA2E A A1E AWA2E Process V constant vA2E A vA1E A A1E AWA2E A 0 State 1 Twophase L V look in Table F71 vA1E A 001663 09 336146 3027 ftA3E Albm uA1E A 16803 09 90615 9836 Btulbm State 2 TA2E A vA2E A vA1E A mix of sat solid vap Table F74 vA2E A 3027 001744 xA2E A 5655 xA2E A 000535 uA2E A 14931 000535 11665 14307 Btulbm m VvA1E A 6 ft3 3027 ftA3E Albm 0198 lbm A1E AQA2E A muA2E A uA1E A 0198 lbm 14307 9836 Btulbm 223 Btu P CP v T CP v T 1 1 P const 2 2 P T v S V L V V L S CP 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3270E A constant pressure pistoncylinder has 2 lbm water at 1100 F and 226 ftA3E A It is now cooled to occupy 110 of the original volume Find the heat transfer in the process CV Water mA2E A mA1E A m Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E Process P const A1E AWA2E A APdVEA PV PVA2E A VA1E A PmvA2E A vA1E A State 1 Table F72 T vA1E A Vm 2262 113 ftA3E Albm PA1E A 800 psia hA1E A 156781 Btulbm State 2 Table F72 P vA2E A vA1E A10 0113 ftA3E Albm twophase state TA2E A 510 520 510800 7435381148 74353 5183 F xA2E A vA2E A vAfE AvAfgE A 0113 00208705488 01679 hA2E A hAfE A xA2E A hAfgE A 50963 xA2E A 68962 62542 Btulbm A1E AQA2E A muA2E A uA1E A A1E AWA2E A mhA2E A hA1E A 2 62542 156781 18848 Btu V P 1 2 V T 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Solids and Liquids Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3271E I have 4 lbm of liquid water at 70 F 15 psia I now add 20 Btu of energy at a constant pressure How hot does it get if it is heated How fast does it move if it is pushed by a constant horizontal force How high does it go if it is raised straight up a Heat at 15 psia Energy equation EA2E A EA1E A A1E AQA2E A A1E AWA2E A A1E AQA2E A PVA2E A VA1E A HA2E A HA1E A mhA2E A hA1E A hA2E A hA1E A A1E AQA2E Am 3809 204 4309 Btulbm Back interpolate in Table F71 TA2E A 75 F We could also have used T A1E AQA2E AmC 20 4100 5 F b Push at constant P It gains kinetic energy 05 m VA2 2E A A1E AWA2E A V2 2 1W2m A 2 20 77817 lbfft4 lbm EA A 2 20 77817 32174 lbmfts2 4 lbm EA 500 fts c Raised in gravitational field m g Z2 A1E AWA2E Z2 A1E AWA2E Am g A 20 77817 lbfft 4 lbm 32174 fts2 E A 32174 Albmfts2 ElbfE A 3891 ft Comment Notice how fast 500 fts and how high it should be to have the same energy as raising the temperature just 5 degrees Ie in most applications we can disregard the kinetic and potential energies unless we have very high V or Z Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3272E A computer cpu chip consists of 01 lbm silicon 005 lbm copper 01 lbm polyvinyl chloride plastic It now heats from 60 F to 160 F as the computer is turned on How much energy did the heating require Energy Eq UA2E A UA1E A Amiu2 u1i EA A1E AQA2E A A1E AWA2E A For the solid masses we will use the specific heats Table F2 and they all have the same temperature so Amiu2 u1i EA AmiCv i T2 T1i EA TA2E A TA1E AAmiCv i EA AmiCv i EA 01 0167 005 01 01 0229 00446 BtuR UA2E A UA1E A 00446 BtuR 160 60 R 446 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3273E A copper block of volume 60 in3 is heat treated at 900 F and now cooled in a 3 ft3 oil bath initially at 70 F Assuming no heat transfer with the surroundings what is the final temperature CV Copper block and the oil bath Also assume no change in volume so the work will be zero Energy Eq UA2E A UA1E A mAmetE AuA2E A uA1E AAmetE A mAoilE AuA2E A uA1E AAoilE A A1E AQA2E A A1E AWA2E A 0 Solid and liquid u CAvE A T CAvE A from Table F2 and F3 mAmetE A Vρ 60 12A3E A ft3 518 lbmft3 17986 lbm mAoilE A Vρ 30 ft3 57 lbmft3 171 lbm CAvE A AmetE A 010 BtulbmR CAvE A AoilE A 043 BtulbmR Energy equation becomes mAmetE A CAvE A AmetE A TA2E A TA1metE A mAoilE A CAvE A AoilE A TA2E A TA1oilE A 0 17986 lbm 010 BtulbmR TA2E A 900 F 171 lbm 043 BtulbmR TA2E A 70 F 0 TA2E A 898 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Ideal Gas Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3274E Estimate the constant specific heats for R134a from Table F102 at 15 psia and 150 F Compare this to table F4 and explain the difference Solution Using values at 15 psia for h and u at 140 F and 160 F from Table F102 the approximate specific heats at 150 F are CApE A A h TE A A19995 19559 160 140E A 0218 Btulbm R compared with 0203 BtulbmR for the idealgas value at 77 F from Table F4 CAvE A A u TE A A18803 18408 160 140E A 0198 Btulbm R compared with 0184 BtulbmR for the idealgas value at 77 F from Table F4 There are two reasons for the differences First R134a is not exactly an ideal gas at the given state 150 F and 15 psia Second and by far the biggest reason for the differences is that R134a chemically CFA3E ACHA2E A is a polyatomic molecule with multiple vibrational mode contributions to the specific heats see Appendix C such that they are strongly dependent on temperature Note that if we repeat the above approximation for CApE A in Table F102 at 77 F the resulting value is 0203 Btulbm R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3275E Air is heated from 540 R to 640 R at V C Find A1E AqA2E A What if from 2400 to 2500 R Process V C A1E AWA2E A Ø Energy Eq uA2E A uA1E A A1E AqA2E A 0 A1E AqA2E A uA2E A uA1E A Read the uvalues from Table F5 a A1E AqA2E A uA2E A uA1E A 10934 9216 1718 Btulbm b A1E AqA2E A uA2E A uA1E A 47433 45264 217 Btulbm case a CAvE A 1718100 0172 BtulbmR see F4 case b CAvE A 217100 0217 BtulbmR 26 higher Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3276E A 30ft high cylinder crosssectional area 1 ft2 has a massless piston at the bottom with water at 70 F on top of it as shown in Fig P3107 Air at 540 R volume 10 ft3 under the piston is heated so that the piston moves up spilling the water out over the side Find the total heat transfer to the air when all the water has been pushed out Solution H2O Po cb air V P 2 1 P P1 0 V V 1 max The water on top is compressed liquid and has mass VAH2OE A VAtotE A VAairE A 30 1 10 20 ftA3E mAH2OE A VAH2OE AvAfE A 200016051 1246 lbm Initial air pressure is PA1E A PA0E A mAH2OE AgA 147 A g 1 144E A 23353 psia and then mAairE A APV RTE A A23353 10 144 5334 540E A 11675 lbm State 2 PA2E A PA0E A 147 lbfinA2E A VA2E A 30 1 30 ftA3E A1E AWA2E A APdVEA A1 2E A PA1E A PA2E AVA2E A VA1E A A1 2E A 23353 14730 10 144 54 796 lbfft 7043 Btu State 2 PA2E A VA2E A TA2E A A T1P2V2 EP1V1 E A A54014730 2335310E A 10197 R A1E AQA2E A muA2E A uA1E A A1E AWA2E A 11675 0171 10197 540 7043 1662 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3277E A closed rigid container is filled with 3 lbm water at 1 atm 130 F 2 lbm of stainless steel and 1 lbm of PVC polyvinyl chloride both at 70 F and 02 lbm of air at 700 R 1 atm It is now left alone with no external heat transfer and no water vaporizes Find the final temperature and air pressure CV Container Process V constant A1E AWA2E A 0 and also given A1E AQA2E A 0 Energy Eq UA2E A UA1E A Amiu2 u1i EA A1E AQA2E A A1E AWA2E A 0 For the liquid and the metal masses we will use the specific heats Tbl F3 F4 so Amiu2 u1i EA AmiCv i T2 T1i EA TA2E AmiCv i EA AmiCv iT1 i EA noticing that all masses have the same TA2E A but not same initial T AmiCv i EA 3 10 2 011 1 0229 02 0171 34832 BtuR The T for air must be converted to F like the others Energy Eq 34832 TA2E A 3 10 130 2 011 1 0229 70 02 0171 700 45967 429649 Btu TA2E A 12335 F The volume of the air is constant so from PV mRT it follows that P varies with T PA2E A PA1E A TA2E ATA1 airE A 1 atm 12335 45967 700 0833 atm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3278E An engine consists of a 200 lbm cast iron block with a 40 lbm aluminum head 40 lbm steel parts 10 lbm engine oil and 12 lbm glycerine antifreeze Everything begins at 40 F and as the engine starts it absorbs a net of 7000 Btu before it reaches a steady uniform temperature We want to know how hot it becomes Energy Eq UA2E A UA1E A A1E AQA2E A A1E AWA2E Process The steel does not change volume and the change for the liquid is minimal so A1E AWA2E A 0 So sum over the various parts of the left hand side in the energy equation mAFE AeE A uA2E A uA1E A mAAlE A uA2E A uA1E AAAlE A mAsE AtE A uA E A uA1E AAsE AtE mAoilE A uA2E A uA1E AAoilE A mAglyE A uA2E A uA1E AAglyE A A1E AQA2E Tbl F2 CAFE AeE A 01 CAAlE A 0215 CAsE AtE A 011 all units of BtulbmR Tbl F3 CAoilE A 046 CAglyE A 058 all units of BtulbmR So now we factor out TA2E A TA1E A as uA2E A uA1E A CTA2E A TA1E A for each term mAFE AeE ACAFE AeE A mAAlE ACAAlE A mAsE AtE ACAsE AtE A mAoilE ACAoilE A mAglyE ACAglyE A TA2E A TA1E A A1E AQA2E TA2E A TA1E A A1E AQA2E A ΣmAiE A CAiE A 7000 200 01 40 0215 40 011 10 046 12 058E A7000 4456E A R 157 R TA2E A TA1E A 157 F 40 157 197 F Exhaust flow Air intake filter Coolant flow Atm air Shaft Fan power Radiator Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3279E A car with mass 3250 lbm drives with 60 mih when the brakes are applied to quickly decrease its speed to 20 mih Assume the brake pads are 1 lbm mass with heat capacity of 02 Btulbm R and the brake discsdrums are 8 lbm steel where both masses are heated uniformly Find the temperature increase in the brake assembly CV Car Car looses kinetic energy and brake system gains internal u No heat transfer short time and no work term m constant Energy Eq35 E2 E1 0 0 mcar A1 2V2 E2 V2 1 EA mbrakeu2 u1 The brake system mass is two different kinds so split it also use Cv since we do not have a u table for steel or brake pad material msteel Cv T mpad Cv T mcar A1 2V2 E2 V2 1 EA 8011 102 BtuR T 3250 05 3600 4001466672 32174778 Btu 4469 Btu T 414 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3280E Water at 60 psia is brought from 320 F to 1800 F Evaluate the change in specific internal energy using a the steam tables b the ideal gas Table F6 and the specific heat F4 Solution a State 1 Table F73 Superheated vapor uA1E A 110946 Btulbm State 2 Table F73 uA2E A 172669 Btulbm uA2E A uA1E A 172669 110946 61723 Btulbm b Table F6 at 780 R uA1E A 19787 198589 780 42971 Btulbmol Table F6 at 2260 R uA2E A 15894 198589 2260 11 406 Btulbmol uA2E A uA1E A 11 406 4297118013 60935 Btulbm c Table F4 CAvoE A 0337 BtulbmR uA2E A uA1E A 0337 BtulbmR 1800 320 R 4988 Btulbm Notice how the average slope from 320 F to 1800 F is higher than the one at 77 F CAvoE A u T 77 320 1800 u 320 u 1800 Slope at 77 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3281E A 65 gallons rigid tank contains methane gas at 900 R 200 psia It is now cooled down to 540 R Assume ideal gas and find the needed heat transfer Solution Ideal gas and recall from Table A1 that 1 gal 231 inA3E A m AP1VRT1E A A 200 psi 65 gal 231 in3gal 9635 lbfftlbmR 900 R 12 inftE A 2886 lbm Process V constant VA1E A A1E AWA2E A 0 Use specific heat from Table F4 uA2E A uA1E A CAvE A TA2E A TA1E A 0415 BtulbmR 900 540 R 1494 Btulbm Energy Equation A1E AQA2E A muA2E A uA1E A 2886 1494 4312 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Polytropic Processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3282E Oxygen at 50 lbfin2 200 F is in a pistoncylinder arrangement with a volume of 4 ft3 It is now compressed in a polytropic process with exponent n 12 to a final temperature of 400 F Calculate the heat transfer for the process Continuity mA2E A mA1E A Energy EA2E A EA1E A muA2E A uA1E A A1E AQA2E A A1W2E A State 1 T P and ideal gas small change in T so use Table F4 m A P1V1 ERT1 E A A 50 psi 4 ft3 144 in2ft2 4828 lbfftlbmR 65967 RE A 09043 lbm Process PVAnE A constant A1E AWA2E A A 1 1nE A PA2E AVA2E A PA1E AVA1E A AmR 1nE A TA2E A TA1E A A09043 4828 1 12E A A400 200 778E A 5612 Btu A1E AQA2E A muA2E A uA1E A A1E AWA2E A mCAvE ATA2E A TA1E A A1E AWA2E 09043 0158 400 200 5612 2754 Btu P v 2 1 T v 2 1 T T 1 2 T C v02 P C v 12 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3283E An air pistol contains compressed air in a small cylinder as shown in Fig P3164 Assume that the volume is 1 in3 pressure is 10 atm and the temperature is 80 F when armed A bullet m 004 lbm acts as a piston initially held by a pin trigger when released the air expands in an isothermal process T constant If the air pressure is 1 atm in the cylinder as the bullet leaves the gun find a The final volume and the mass of air b The work done by the air and work done on the atmosphere c The work to the bullet and the bullet exit velocity CV Air Air ideal gas mAairE A A P1V1 ERT1 E A A 10 atm 147 psiatm 1 in3 5334 lbfftlbmR 53967 R 12 inftE A 42610A5E A lbm Process PV const PA1E AVA1E A PA2E AVA2E A VA2E A VA1E APA1E APA2E A 10 inA3E A1E AWA2E A APdVEA AP1V1 E 1VEA dV PA1E AVA1E A ln AV2V1E A 147 psi 1 in3 ln10 282 lbfft 00362 Btu A1E AWA2ATME A PA0E AVA2E A VA1E A 147 psi 9 in3 1103 lbfft 00142 Btu WAbulletE A A1E AWA2E A A1E AWA2ATME A 0022 Btu A1 2E A mAbulletE AVAexE AA2E VAexE A 2WAbulletmBE AA12E A 2002277832174 004A12E A 1659 fts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3284E Helium gas expands from 20 psia 600 R and 9 ftA3E A to 15 psia in a polytropic process with n 1667 How much heat transfer is involved Solution CV Helium gas this is a control mass Energy equation muA2E A uA1E A A1E AQA2E A A1E AWA2E Process equation PVAnE A constant PA1E AVAn 1E A PA2E AVAn 2E Ideal gas F4 m PVRT A20 psi 9 ft3 144 in2ft2 386 lbfftlbmR 600 RE A 0112 lbm Solve for the volume at state 2 VA2E A VA1E A PA1E APA2E AA1nE A 9 A 20 15 06E A 10696 ftA3E TA2E A TA1E A P2V2PA1E AVA1E A 600 R A15 10696 20 9E A 5348 R Work from Eq321 A1E AWA2E A EA PA2 AVA2 A PA1 A VA1 A E 1nE A A15 10696 20 9 1 1667E A psi ftA3E A 2933 psia ftA3E 4223 lbfft 543 Btu Use specific heat from Table F4 to evaluate uA2E A uA1E A Cv 0744 Btulbm R A1E AQA2E A muA2E A uA1E A A1E AWA2E A m Cv TA2E A TA1E A A1E AWA2E A 0112 lbm 0744 BtulbmR 5348 600 R 543 Btu 0003 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful More Complex Devices Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3285E Water in a tank A is at 270 F with a quality of 10 and mass 1 lbm It is connected to a piston cylinder holding constant pressure of 40 psia initially with 1 lbm water at 700 F The valve is opened and enough heat transfer takes place to have a final uniform temperature of 280 F Find the final P and V the process work and the process heat transfer Solution CV Water in A and B Control mass goes through process 1 2 Continuity Eq mA2E A mAA1E A mAB1E A 0 mA2E A mAA1E A mAB1E A 10 10 2 lbm Energy Eq UA2E A UA1E A A1E AQA2E A A1E AWA2E A State A1 vAA1E A 001717 xAA1E A 100483 1022 VAA1E A mv 1022 ftA3E uAA1E A 23881 01 85414 32422 Btulbm State B1 vAB1E A 17196 ftA3E Albm uAB1E A 125514 Btulbm VAB1E A mAB1E AvAB1E A 17196 ftA3E State 2 If VA2E A VAA1E A then PA2E A 40 psia that is the piston floats For TA2E A PA2E A 280 F 40 psia superheated vapor uA2E A 109731 Btulbm vA2E A 10711 ftA3E Albm VA2E A mA2E AvA2E A 21422 ftA3E A VAA1E A checks OK The possible state 2 PV combinations are shown State a is 40 psia vAaE A VAA1E AmA2E A 0511 and thus twophase TAaE A 2673 F less than TA2E Process A1E AWA2E A PA2E A VA2E A VA1E A 40 21422 1022 17196A144 778E A 2372 Btu From the energy Eq A1E AQA2E A mA2E AuA2E A mAA1E AuAA1E A mAB1E AuAB1E A A1E AWA2E 2 109731 10 32422 10 125514 2372 63898 Btu V P 2 280 F a 492 40 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3286E Two rigid tanks are filled with water as shown in Fig P3214 Tank A is 7 ftA3E A at 1 atm 280 F and tank B is 11 ftA3E A at saturated vapor 40 psia The tanks are connected by a pipe with a closed valve We open the valve and let all the water come to a single uniform state while we transfer enough heat to have a final pressure of 40 psia Give the two property values that determine the final state and find the heat transfer Solution State A1 u 11024 Btulbm v 29687 ftA3E Albm mAA1E A Vv 7029687 0236 lbm State B1 u 10923 Btulbm v 19501 ftA3E Albm mAB1E A Vv 110 19501 0564 lbm The total volume and mass is the sum of volumes mass for tanks A and B mA2E A mAA1E A mAB1E A 0236 0564 0800 lbm VA2E A VAA1E A VAB1E A 70 110 180 ftA3E vA2E A VA2E AmA2E A 180 0800 225 ftA3E Albm State 2 PA2E A vA2E A 40 psia 225 ftA3E Albm TA2E A 10535 F and uA2E A 13951 Btulbm The energy equation is neglecting kinetic and potential energy mA2E A uA2E A mAAE AuAA1E A mABE AuAB1E A A1E AQA2E A A1E AWA2E A A1E AQA2E A A1E AQA2E A 0800 13951 0236 11024 0564 10923 2399 Btu B A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3287E A vertical cylinder fitted with a piston contains 10 lbm of R410A at 50 F shown in Fig P3173 Heat is transferred to the system causing the piston to rise until it reaches a set of stops at which point the volume has doubled Additional heat is transferred until the temperature inside reaches 120 F at which point the pressure inside the cylinder is 200 lbfin2 a What is the quality at the initial state b Calculate the heat transfer for the overall process Solution CV R410A Control mass goes through process 1 2 3 As piston floats pressure is constant 1 2 and the volume is constant for the second part 2 3 So we have vA3E A vA2E A 2 vA1E A State 3 Table F92 PT vA3E A 03652 ftA3E Albm uA3E A 1235 Btulbm So we can determine state 1 and 2 Table F91 vA1E A 01826 001420 xA1E A03636 xA1E A 0463 uA1E A 3106 04637896 676 Btulbm State 2 vA2E A 03652 ft3lbm PA2E A PA1E A 1575 psia this is still 2phase We get the work from the process equation see PV diagram A1E AWA3E A A1E AWA2E A A 1 2 PdVEA PA1E AVA2E A VA1E A 1575 psia 10 lbm 03652 01826 ft3lbm 144 inftA2E A 41 4137 lbfft 532 Btu The heat transfer from the energy equation becomes A1E AQA3E A muA3E A uA1E A A1E AWA3E A 101235 676 532 6122 Btu V P 1 2 3 R410A P o cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3288E A piston cylinder shown in Fig P3169 contains 18 ftA3E A of R410A at 300 psia 300 F The piston mass and atmosphere gives a pressure of 70 psia that will float the piston The whole setup cools in a freezer maintained at 0 F Find the heat transfer and show the Pv diagram for the process when TA2E A 0 F Solution CV R410A Control mass Continuity m constant Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E A Process F F P A PAairE AA FAstopE if V VAstopE A FAstopE A A0E This is illustrated in the Pv diagram shown below R410A Po State 1 vA1E A 03460 ftA3E Albm uA1E A 15995 Btulbm m Vv 52023 lbm State 2 TA2E A and on line compressed liquid see figure below vA2E A vAfE A 001295 ftA3E Albm VA2E A 0674 ftA3E A uA2E A uAfE A 1337 Btulbm A1E AWA2E A APdVEA PAliftE AVA2E A VA1E A 70 psi 0674 18 ftA3E A 144 in2ft2 174 646 lbfft 2245 Btu Energy eq A1E AQA2E A 52023 1337 15995 2245 7850 Btu 300 70 P 300 P 70 v 1 T 0 2 T v 300 52 0 1 2 926 P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3289E A setup as in Fig P3169 has the R410A initially at 150 psia 120 F of mass 02 lbm The balancing equilibrium pressure is 60 psia and it is now cooled so the volume is reduced to half the starting volume Find the heat transfer for the process Solution Take as CV the 02 lbm of R410A Continuity Eq mA2E A mA1E A m Energy Eq35 muA2E A uA1E A A1E AQ2 A1E AW2 Process Eq P PAfloatE A or v C vA1E A State 1 P T vA1E A 05099 ftA3E Albm uA1E A 12528 Btulbm State 2 P v vA2E A vA1E A2 02550 ftA3E Albm vAgE A so it is twophase xA2E A vA2E A vAfE A vAfgE A 0255 0012909911 02443 uA2E A uAfE A xA2E A uAfgE A 1257 xA2E A 9333 3537 Btulbm From process eq A1E AWA2E A P dV area mPA2E A v2 v1 02 lbm 60 psi 0255 05099 ftA3E Albm 144 in2ft2 4405 lbfft 057 Btu From energy eq A1E AQA2E A muA2E A uA1E A A1E AWA2E A 02 3537 12528 057 1855 Btu P CP v T 1 60 2 150 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3290E A piston cylinder contains air at 150 psia 1400 R with a volume of 175 ftA3E A The piston is pressed against the upper stops see Fig P314c and it will float at a pressure of 110 psia Now the air is cooled to 700 R What is the process work and heat transfer CV Air this is a control mass Energy Eq35 muA2E A uA1E A A1E AQ2 A1E AW2 Process Eq P PAfloatE A or v C vA1E A State 1 u 24704 Btulbm m PVRT 150 psi 175 ft35334 lbfftlbmR 1400 R 0506 lbm We need to find state 2 Let us see if we proceed past state 1a during the cooling TA1aE A TA1E A PAfloatE A PA1E A 1400 R 110 150 102667 R so we do cool below TA1aE A That means the piston is floating Write the ideal gas law for state 1 and 2 to get VA2E A A mRT2 EP2 E A A P1V1T2 EP2T1 E A A150 175 700 110 1400E A 11932 ftA3E A1E AWA2E A A1aE AWA2E A P dV PA2E A VA2E A VA1E A 110 psia 11932 175 ftA3E A 88197 lbfft 1134 Btu From the energy equation A1E AQA2E A muA2E A uA1E A A1E AWA2E A 0506 lbm 1197 24704 Btulbm 1134 Btu 758 kJ Air P o m p 1a 2 1 P V P V stop 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3291E A mass of 6 lbm nitrogen gas at 3600 R V C cools with 1 Btus What is dTdt Process V C A1E AWA2E A 0 AdE dtE A AdU dtE A mAdU dtE A mCAvE A AdT dtE A AQ E A W AQ E A 1 Btus CAp 3600E A Adh dTE A Ah TE A h3800 h3400 38003400 A25857 22421 400 28013E A 03066 BtulbmR CAv 3600E A CAp 3600E A R 03066 5515 778 02357 BtulbmR AdT dtE A A Q EmCv E A A 1 Btus 6 02357 BtuRE A 071 AR sE Remark Specific heat from Table F4 has CAv 300E A 0178 BtulbmR which is nearly 25 lower and thus would overestimate the rate with 25 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3292E A crane use 7000 Btuh to raise a 200 lbm box 60 ft How much time does it take Power AW E A FV mgV mgAL tE F mg 200 A32174 32174E A lbf 200 lbf t AFL W E A A200 lbf 60 ft 7000 BtuhE A A200 60 3600 7000 77817E A s 79 s Recall Eq on page 9 1 lbf 32174 lbm ftsA2E A 1 Btu 77817 lbfft A1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3293E A computer in a closed room of volume 5000 ft3 dissipates energy at a rate of 10 kW The room has 100 lbm of wood 50 lbm of steel and air with all material at 540 R 1 atm Assuming all the mass heats up uniformly how long time will it take to increase the temperature 20 F CV Air wood and steel mA2E A mA1E A no work Energy Eq35 UA2E A UA1E A A1E AQA2E A AQ E At The total volume is nearly all air but we can find volume of the solids VAwoodE A mρ 100449 223 ft3 VAsteelE A 50488 0102 ft3 VAairE A 5000 223 0102 49977 ft3 mAairE A PVRT 147499771445334540 3673 lbm We do not have a u table for steel or wood so use specific heat U mAairE A Cv mAwoodE A Cv mAsteelE A Cv T 3673 0171 100 03 50 011 BtuF 20 F 12562 600 110 1966 Btu AQ E A t 101055 t t 196610 1055 207 sec 35 minutes u T 77 320 1800 u 320 u 1800 Slope at 77 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3294E Water is in a piston cylinder maintaining constant P at 330 F quality 90 with a volume of 4 ftA3E A A heater is turned on heating the water with 10 000 Btuh What is the elapsed time to vaporize all the liquid Solution Control volume water Continuity Eq mAtotE A constant mAvaporE A mAliqE on a rate form Am E AtotE A 0 Am E AvaporE A Am E AliqE A Am E AliqE A Am E AvaporE Energy equation AU E A AQ E A AW E A Am E AvaporE A uAfgE A AQ E A P Am E AvaporE A vAfgE Rearrange to solve for Am E AvaporE Am E AvaporE A uAfgE A PvAfgE A Am E AvaporE A hAfgE A AQ E From table F71 hAfgE A 8875 Btlbm vA1E A 001776 09 42938 38822 ftA3E Albm mA1E A VA1E AvA1E A 438822 10303 lbm mAliqE A 1xA1E AmA1E A 010303 lbm Am E AvaporE A AQ E AhAfgE A A10 000 8875E A A Btuh BtulbmE A 112676 lbmh 000313 lbms t mAliqE A Am E AvaporE A 010303 000313 329 s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3295E A pistoncylinder has 2 lbm of R134a at state 1 with 200 F 90 lbfin2 and is then brought to saturated vapor state 2 by cooling while the piston is locked with a pin Now the piston is balanced with an additional constant force and the pin is removed The cooling continues to a state 3 where the R134a is saturated liquid Show the processes in a PV diagram and find the work and heat transfer in each of the two steps 1 to 2 and 2 to 3 Solution CV R134a This is a control mass Properties from table F101 and F102 State 1 TP v 07239 ftA3E Albm u 194605 Btulbm State 2 given by fixed volume and x2 10 v2 v1 vg 1W2 0 TA2E A 50 10 0 7921 0 6632 0 7921 0 7239 553 F PA2E A 60311 72271 60311 05291 6664 psia u2 16495 16628 16495 05291 16565 Btulbm From the energy equation 1Q2 mu2 u1 1W2 mu2 u1 2 16565 194605 5791 Btu State 3 reached at constant P F constant state 3 P3 PA2E A and vA3E A vf 001271 001291 001271 05291 001282 ftA3E Albm uA3E A uf 9168 9495 9168 05291 9341 Btulbm 1W3 1W2 2W3 0 2W3 P dV PV3 V2 mPv3 v2 2 6664 001282 07239 A144 778E A 1754 Btu From the energy equation 2Q3 mu3 u2 2W3 29341 16565 1754 16202 Btu P V 1 2 3 cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Rates of Work Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3296E A force of 300 lbf moves a truck with 40 mih up a hill What is the power Solution AWE A F V 300 lbf 40 mih 12 000 A16093 328084 3600E A Albfft sE 17 600 Albfft sE A 2262 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Heat Transfer Rates Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3297E Find the rate of conduction heat transfer through a 15 cm thick hardwood board k 009 BtuhftR with a temperature difference between the two sides of 40 F One dimensional heat transfer by conduction we do not know the area so we can find the flux heat transfer per unit area BtuftA2E Ah t 15 cm 059 in 00492 ft A qEA A QEAA k AT xE A 009 A Btu hftRE A A 40 00492E A AR ftE A 732 BtuftA2E Ah Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3298E A waterheater is covered up with insulation boards over a total surface area of 30 ft2 The inside board surface is at 175 F and the outside surface is at 70 F and the board material has a conductivity of 005 Btuh ft F How thick a board should it be to limit the heat transfer loss to 720 Btuh Solution Steady state conduction through a single layer board A QEA cond k A AT xE A x k Α Τ A QE x 005 A Btu hftRE A 30 ft2 17570 R 720 Btuh 0219 ft 26 in Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3299E The sun shines on a 1500 ft2 road surface so it is at 115 F Below the 2 inch thick asphalt average conductivity of 0035 BtuhftF is a layer of compacted rubbles at a temperature of 60 F Find the rate of heat transfer to the rubbles Solution A QEA k A AT xE A 0035 A Btu hftRE A 1500 ft2 A115 60 212E A AR ftE A 17325 Βtuh Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3300E A cylinder having an initial volume of 100 ft3 contains 02 lbm of water at 100 F The water is then compressed in an isothermal quasiequilibrium process until it has a quality of 50 Calculate the work done in the process assuming water vapor is an ideal gas Solution State 1 TA1E A vA1E A Vm A100 02E A 500 ftA3E Albm vAgE A since PAgE A 095 psia very low so water is an ideal gas from 1 to 2 PA1E A PAgE A A vg Ev1 E A 0950 A350 500E A 06652 lbfinA2E VA2E A mvA2E A 02 lbm 350 ftA3E Albm 70 ftA3E vA3E A 001613 05350 001613 1750 ftA3E Albm For ideal gas and constant T the work term follows Eq 321 A1E AWA2E A APdVEA PA1E AVA1E A ln A V2 EV1 E A 06652 A144 778E A 100 ln A 70 100E A 433 Btu For the constant pressure part of the process the work becomes A2E AWA3E A PA2E A mvA3E A vA2E A 095 psi 02 lbm 175 350 ftA3E Albm 144 in2ft2 4788 lbfft 615 Btu A1E AWA3E A 615 433 1048 Btu T CP v 1 2 P P CP v T 3 095 100 2 1 3 sat P1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3301E A pistoncylinder contains 2 lbm of water at 70 F with a volume of 01 ft3 shown in Fig P3225 Initially the piston rests on some stops with the top surface open to the atmosphere Po so a pressure of 40 lbfin2 is required to lift it To what temperature should the water be heated to lift the piston If it is heated to saturated vapor find the final temperature volume and the heat transfer Solution CV Water This is a control mass mA2E A mA1E A m muA2E A uA1E A A1E AQA2E A 1W2 State 1 20 C v1 Vm 012 005 ftA3E Albm x 005 001605867579 00003913 u1 3809 0000391399564 3813 Btulbm To find state 2 check on state 1a P 40 psia v vA1E A 005 ftA3E Albm Table F71 vf v vg 10501 x1a 0 V P 2 1 1a P P 1 2 State 2 is saturated vapor at 40 psia as state 1a is twophase TA2E A 2673 F vA2E A vg 10501 ftA3E Albm VA2E A m vA2E A 210 ftA3E A uA2E A ug 109227 Btulbm Pressure is constant as volume increase beyond initial volume 1WA2E A P dV PAliftE A VA2E AV1 40 210 01 144 778 15475 Btu 1QA2E A muA2E A u1 1WA2E A 2 109227 3813 15475 2263 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3302E A twenty poundmass of water in a pistoncylinder with constant pressure is at 1100 F and a volume of 226 ft3 It is now cooled to 100 F Show the Pv diagram and find the work and heat transfer for the process Solution CV Water Energy Eq35 1Q2 mu2 u1 1W2 mh2 hA1E A Process Eq Constant pressure 1W2 mPv2 vA1E A Properties from Table F72 and F73 State 1 T1 v1 22620 113 ftA3E Albm P1 800 lbfinA2E A h1 15678 Btulbm State 2 800 lbfinA2E A 100 F v2 0016092 ftA3E Albm h2 7015 Btulbm v P 1 2 v T 1 2 800 psia The work from the process equation is found as 1W2 20 lbm 800 psi 0016092 113 ftA3E Albm 144 inA2E AftA2E 2 566 444 lbfft 3299 Btu The heat transfer from the energy equation is 1Q2 20 lbm 7015 15678 Btulbm 29 953 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3303E A cylinder fitted with a frictionless piston contains R134a at 100 F 80 quality at which point the volume is 3 Gal The external force on the piston is now varied in such a manner that the R134a slowly expands in a polytropic process to 50 lbfin2 80 F Calculate the work and the heat transfer for this process Solution CV The mass of R134a Properties in Table F101 vA1E A vf xA1E A vfg 001387 08 03278 02761 ftA3E Albm uA1E A 10851 08 6277 15873 Btulbm PA1E A 138926 psia m VvA1E A 3 231 12A 3E A 02761 0401 02761 14525 lbm State 2 vA2E A 11035 ftA3E Albm supvap uA2E A 17132 Btulbm linear interpolation is not so accurate as v is more like 1P Process n ln P1 P2 ln V2 V1 ln A138926 50E A ln A11035 02761E A 07376 A1E AWA2E A P dV P2 V2 P1 V1 1 n A50 11035 138926 02761 1 07376E A 14525 A144 778E A 1723 Btu A1E AQA2E A muA2E A uA1E A A1E AWA2E A 14525 17132 15873 1723 355 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3304E Ammonia NH3 is contained in a sealed rigid tank at 30 F x 50 and is then heated to 200 F Find the final state PA2E A uA2E A and the specific work and heat transfer Solution Continuity Eq mA2E A mA1E A Energy Eq35 EA2E A EA1E A A1E AQA2E A A 1W2 0 E E Process VA2E A VA1E A vA2E A vA1E A 002502 05 47945 2422 ftA3E Albm State 1 Table F81 u1 7506 05 49117 32065 Btulbm Table F82 vA2E A TA2E A between 150 psia and 175 psia V P 1 2 PA2E A 163 lbfinA2E A uA2E A 6335 Btulbm linear interpolation Process equation gives no displacement A1E AwA2E A 0 The energy equation then gives the heat transfer as A1E AqA2E A uA2E A uA1E A 6335 32065 31285 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3305E Water in a pistoncylinder similar to Fig P3225 is at 212 F x 05 with mass 1 lbm and the piston rests on the stops The equilibrium pressure that will float the piston is 40 psia The water is heated to 500 F by an electrical heater At what temperature would all the liquid be gone Find the final Pv the work and heat transfer in the process CV The 1 lbm water Continuty mA2E A mA1E A m Energy muA2E A uA1E A A1E AQA2E A A1E AWA2E A Process V constant if P PAliftE A otherwise P PAliftE A see Pv diagram State 1 Tx Table F71 vA1E A 001672 05 267864 134099 ftA3E Albm uA1E A 18009 05 89751 628845 Btulbm State 1a 40 psia v vA1E A vAg 40 psiaE A 10501 ftA3E Albm so superheated vapor Piston starts to move at state 1a A1E AWA1aE A 0 State 1b reached before state 1a so v vA1E A vAgE A see this in F71 TA1bE A 250 10 1340992 138247117674 138247 252 F State 2 TA2E A TA1aE A Table F72 vA2E A 14164 uA2E A 118006 Btulbm Work is seen in the PV diagram when volume changes P PAliftE A A1E AWA2E A A1aE AWA2E A PA2E A mvA2E A vA1E A 40 psi 1 lbm 14164 134099 ftA3E Albm 144 in2ft2 43435 lbfft 558 Btu Heat transfer is from the energy equation A1E AQA2E A 1 lbm 118006 628845 Btulbm 558 Btu 5568 Btu H O Po 2 cb P v 1 1a 2 100 C o cb 1b Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 3306E An insulated cylinder is divided into two parts of 10 ft3 each by an initially locked piston Side A has air at 2 atm 600 R and side B has air at 10 atm 2000 R as shown in Fig P3151 The piston is now unlocked so it is free to move and it conducts heat so the air comes to a uniform temperature TA TB Find the mass in both A and B and also the final T and P CV A B Then A1E AQA2E A A0E A A1E AWA2E A A0E A Force balance on piston PAAE AA PABE AA so final state in A and B is the same State 1A uAA1E A 102457 Btulbm mAAE A APV RTE A A294 psi 10 ft3 144 in2ft2 5334 lbfftlbmR600 RE A 1323 lbm State 1B uAB1E A 367642 Btulbm mABE A APV RTE A A147 psi 10 ft3 144 in2ft2 5334 lbfftlbmR 2000 RE A 1984 lbm For chosen CV A1E AQA2E A 0 A1E AWA2E A 0 so the energy equation becomes mAAE AuA2E A uA1E AAAE A mABE AuA2E A uA1E AABE A 0 mAAE A mABE AuA2E A mAAE AuAA1E A mABE AuAB1E 1323 102457 1984 367642 86495 Btu uA2E A 86495 Btu3307 lbm 26155 Btulbm TA2E A 1475 R P mtotRTA2VtotE A A3307 lbm 5334 lbfftlbmR 1475 R 20 ft3 144 in2ft2 E A 9034 Albfin2E Updated June 2013 SOLUTION MANUAL CHAPTER 4 Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CONTENT CHAPTER 4 SUBSECTION PROB NO InText concept questions ak ConceptStudy guide problems 1 10 Continuity equation and flow rates 1119 Single flow singledevice processes Nozzles diffusers 2030 Throttle flow 3139 Turbines expanders 4049 Compressors fans 5061 Heaters coolers 6274 Pumps pipe and channel flows 7583 Multiple flow singledevice processes Turbines compressors expanders 8490 Heat exchangers 91103 Mixing processes 104114 Multiple devices cycle processes 115125 Transient processes 126141 Review Problems 142155 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful InText Concept Questions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4a A mass flow rate into a control volume requires a normal velocity component Why The tangential velocity component does not bring any substance across the control volume surface as it flows parallel to it the normal component of velocity brings substance in or out of the control volume according to its sign The normal component must be into the control volume to bring mass in just like when you enter a bus it does not help that you run parallel with the bus side V V normal V tangential 4b Can a steady state device have boundary work No Any change in size of the control volume would require either a change in mass inside or a change in state inside neither of which is possible in a steadystate process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4c Can you say something about changes in E A and AV E A through a steady flow device m The continuity equation expresses the conservation of mass so the total amount of AmE A entering must be equal to the total amount leaving For a single flow device the mass flow rate is constant through it so you have the same mass flow rate across any total crosssection of the device from the inlet to the exit The volume flow rate is related to the mass flow rate as AV E A v AmE so it can vary if the state changes then v changes for a constant mass flow rate This also means that the velocity can change influenced by the area as AV E A VA and the flow can experience an acceleration like in a nozzle or a deceleration as in a diffuser 4d In a multiple device flow system I want to determine a state property Where should I be looking for informationupstream or downstream Generally flow is affected more by what happened to it which is upstream than what is in front of it Only the pressure information can travel upstream and give rise to accelerations nozzle or decelerations stagnation type flow Heat transfer that can heat or cool a flow can not travel very fast and is easily overpowered by the convection If the flow velocity exceeds the speed of sound even the pressure information can not travel upstream Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4e How does a nozzle or sprayhead generate kinetic energy By accelerating the fluid from a high pressure towards the lower pressure which is outside the nozzle The higher pressure pushes harder than the lower pressure so there is a net force on any mass element to accelerate it 4f What is the difference between a nozzle flow and a throttle process In both processes a flow moves from a higher to a lower pressure In the nozzle the pressure drop generates kinetic energy whereas that does not take place in the throttle process The pressure drop in the throttle is due to a flow restriction and represents a loss Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4g If you throttle a saturated liquid what happens to the fluid state What if this is done to an ideal gas The throttle process is approximated as a constant enthalpy process Changing the state from saturated liquid to a lower pressure with the same h gives a twophase state so some of the liquid will vaporize and it becomes colder 1 2 2 P v 1 h C T h C If the same process happens in an ideal gas then same h gives the same temperature h a function of T only at the lower pressure Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4h A turbine at the bottom of a dam has a flow of liquid water through it How does that produce power Which terms in the energy equation are important if the CV is the turbine only If the CV is the turbine plus the upstream flow up to the top of the lake which terms in the energy equation are then important The water at the bottom of the dam in the turbine inlet is at a high pressure It runs through a nozzle generating kinetic energy as the pressure drops This high kinetic energy flow impacts a set of rotating blades or buckets which converts the kinetic energy to power on the shaft and the flow leaves at low pressure and low velocity T H DAM Lake CV Turbine only The high P in and the low P out shows up in the h u Pv flow terms of the energy equation giving the difference in the flow work terms Pv in and out CV Turbine plus upstream flow For this CV the pressures in and out are the same 1 atm so the difference is in the potential energy terms gz included in htot Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4i If you compress air the temperature goes up why When the hot air high P flows in long pipes it eventually cools to ambient T How does that change the flow As the air is compressed volume decreases so work is done on a mass element its energy and hence temperature goes up If it flows at nearly constant P and cools its density increases v decreases so it slows down for same mass flow rate Am E A ρAV and flow area Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4j A mixing chamber has all flows at the same P neglecting losses A heat exchanger has separate flows exchanging energy but they do not mix Why have both kinds You might allow mixing when you can use the resulting output mixture say it is the same substance You may also allow it if you definitely want the outgoing mixture like water out of a faucet where you mix hot and cold water Even if it is different substances it may be desirable say you add water to dry air to make it more moist typical for a winter time airconditioning setup In other cases it is different substances that flow at different pressures with one flow heating or cooling the other flow This could be hot combustion gases heating a flow of water or a primary fluid flow around a nuclear reactor heating a transfer fluid flow Here the fluid being heated should stay pure so it does not absorb gases or radioactive particles and becomes contaminated Even when the two flows have the same substance there may be a reason to keep them at separate pressures 1 2 3 MIXING CHAMBER cb 1 2 3 cb 4 An open mixing chamber A closed tube in shell heat exchanger Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4k An initially empty cylinder is filled with air from 20AoE AC 100 kPa until it is full Assuming no heat transfer is the final temperature larger equal to or smaller than 20AoE AC Does the final T depend on the size of the cylinder This is a transient problem with no heat transfer and no work The balance equations for the tank as CV become Continuity Eq mA2E A 0 mAiE A Energy Eq mA2E AuA2E A 0 mAiE AhAiE A Q W mAiE AhAiE A 0 0 Final state uA2E A hAiE A uAiE A PAiE AvAiE A PA2E A PAiE TA2E A TAiE A and it does not depend on V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful ConceptStudy Guide Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 41 A temperature difference drives a heat transfer Does a similar concept apply to AmE A Yes A pressure difference drives the flow The fluid is accelerated in the direction of a lower pressure as it is being pushed harder behind it than in front of it This also means a higher pressure in front can decelerate the flow to a lower velocity which happens at a stagnation point on a wall F P A 1 1 F P A 2 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 42 What kind of effect can be felt upstream in a flow Only the pressure can be felt upstream in a subsonic flow In a supersonic flow no information can travel upstream The temperature information travels by conduction and even small velocities overpowers the conduction with the convection of energy so the temperature at a given location is mainly given by the upstream conditions and influenced very little by the downstream conditions 43 Which one of the properties P v T can be controlled in a flow How Since the flow is not contained there is no direct control over the volume and thus no control of v The pressure can be controlled by installation of a pump or compressor if you want to increase it or use a turbine nozzle or valve through which the pressure will decrease The temperature can be controlled by heating or cooling the flow in a heat exchanger Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 44 Air at 500 kPa is expanded to 100 kPa in two steady flow cases Case one is a nozzle and case two is a turbine the exit state is the same for both cases What can you say about the specific turbine work relative to the specific kinetic energy in the exit flow of the nozzle For these single flow devices let us assume they are adiabatic and that the turbine does not have any exit kinetic energy then the energy equations become Energy Eq413 nozzle h1 A1 2E A VA2 1E A gZ1 h2 A1 2E A VA2 2E A gZ2 Energy Eq413 turbine h1 A1 2E A VA2 1E A gZ1 h2 gZ2 wT Comparing the two we get A1 2E A VA2 2E A wT so the result is that the nozzle delivers kinetic energy of the same amount as the turbine delivers shaft work Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 45 Pipes that carry a hot fluid like steam in a power plant exhaust pipe for a diesel engine in a ship etc are often insulated Is that to reduce the heat loss or is there another purpose You definitely want to insulate pipes that carry hot steam from the boiler to the turbines in a power plant and pipes that flows hot water from one location to another Even if the energy in the flow is unwanted the pipes should be insulated for safety Any place that people could touch a hot surface or very cold surface there is a risk of a burn and that should be avoided Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 46 A windmill takes a fraction of the wind kinetic energy out as power on a shaft In what manner does the temperature and wind velocity influence the power Hint write the power as mass flow rate times specific work The work as a fraction f of the flow of kinetic energy becomes AW E A Am E Aw Am E A f A1 2E A VA2 inE A ρAVAinE A f A1 2E A VA2 inE A so the power is proportional to the velocity cubed The temperature enters through the density so assuming air is ideal gas ρ 1v PRT and the power is inversely proportional to temperature A windmill farm west of Denmark in the North Sea Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 47 An underwater turbine extracts a fraction of the kinetic energy from the ocean current In what manner does the temperature and water velocity influence the power Hint write the power as mass flow rate times specific work The work as a fraction f of the flow of kinetic energy becomes AW E A Am E Aw Am E A f A1 2E A VA2 inE A ρAVAinE A f A1 2E A VA2 inE A so the power is proportional to the velocity cubed The temperature enters through the density so assuming water is incompressible density is constant and the power does not vary with the temperature A proposed underwater tidal flow turbine farm Each turbine is for 1 MW with a diameter of 115 m mounted on the seafloor Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 48 A liquid water turbine in the bottom of a dam takes energy out as power on a shaft Which terms in the energy equation are changing and important The water at the bottom of the dam in the turbine inlet is at a high pressure In a standard turbine it runs through blade passages like a propeller In this case the inlet high pressure is used directly to generate the force on the moving blades For a Pelton turbine the water runs through a nozzle generating kinetic energy as the pressure drops The high kinetic energy flow impacts a set of rotating buckets and converts the flow kinetic energy to power on the shaft so the flow leaves at low pressure and low velocity From Environmental Energy Technologies Division of the Lawrence Berkeley National Laboratory and the US Department of Energy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 49 You blow a balloon up with air What kind of work terms including flow work do you see in that case Where is the energy stored As the balloon is blown up mass flow in has flow work associated with it Also as the balloon grows there is a boundary work done by the inside gas and a smaller boundary work from the outside of the balloon to the atmosphere The difference between the latter two work terms goes into stretching the balloon material and thus becomes internal energy or you may call that potential energy of the balloon material The work term to the atmosphere is stored in the atmosphere and the part of the flow work that stays in the gas is stored as the gas internal energy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 410 A storage tank for natural gas NG has a top dome that can move up or down as gas is added or subtracted from the tank maintaining 110 kPa 290 K inside A pipeline at 110 kPa 290 K now supplies some NG to the tank Does it change state during the filling process What happens to the flow work As the pressure inside the storage tank is the same as in the pipeline the state does not change However the tank volume goes up and work is done on the moving boundary at the 110 kPa so this work equals the flow work The net effect is the flow work goes into raising the dome Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Continuity equation and flow rates Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 411 A large brewery has a pipe of cross sectional area a 02 mA2E A flowing carbon dioxide at 400 kPa 10AoE AC with a volume flow rate of 03 mA3E As Find the velocity and the mass flow rate Am E A AVv AV E Av v RTP 01889 kJkgK 28315 K 400 kPa 01337 mA3E Akg Am E A AV E A v 03 mA3E As 01337 mA3E Akg 224 kgs V AV E A A 01337 mA3E As 02 mA2E A 067 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 412 Air at 35C 105 kPa flows in a 100 mm 150 mm rectangular duct in a heating system The mass flow rate is 0015 kgs What are the velocity of the air flowing in the duct and the volume flow rate Solution Assume a constant velocity across the duct area with A 100 150 10A6E A mA2E A 0015 mA2E Ideal gas so v ART PE A A0287 kJkgK 3082 K 105 kPaE A 08424 mA3E Akg and the volumetric flow rate from Eq 43 AV E A Am E Av AV 0015 kgs 08424 mA3E Akg 001264 mA3E As V AV EAE A A001264 m3s E0015 m2 E A 084 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 413 A pool is to be filled with 60 mA3E A water from a garden hose of 25 cm diameter flowing water at 2 ms Find the mass flow rate of water and the time it takes to fill the pool Solution With constant velocity we have AV E A Am E Av AV π 0025 m A2E A 2 ms 0003927 mA3E As t V AV E A 60 mA3E A 0003927 mA3E As 15279 s 4 h 14 min 39 s From table A3 we get the water density Am E A AV E A v ρAV E A 997 kg mA3E A 0003927 mA3E As 39 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 414 An empty bathtub has its drain closed and is being filled with water from the faucet at a rate of 10 kgmin After 10 minutes the drain is opened and 4 kgmin flows out and at the same time the inlet flow is reduced to 2 kgmin Plot the mass of the water in the bathtub versus time and determine the time from the very beginning when the tub will be empty Solution During the first 10 minutes we have A dmcv EdtE A Am E AiE A 10 kgmin m Am E A tA1E A 10 10 100 kg So we end up with 100 kg after 10 min For the remaining period we have A dmcv EdtE A Am E AiE A Am E AeE A 2 4 2 kgmin mA2E A Am E AnetE A tA2E A tA2E A EAEA m Am Anet E A 1002 50 min So it will take an additional 50 min to empty tAtotE A tA1E A tA2E A 10 50 60 min 10 10 20 100 m 2 0 10 0 0 t min 0 m t min kg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 415 A flat channel of depth 1 m has a fully developed laminar flow of air at Po To with a velocity profile as V 4Vc xH xHA2E A where Vc is the velocity on the centerline and x is the distance across the channel as shown in Fig P415 Find the total mass flow rate and the average velocity both as functions of Vc and H Am E A AVv AV E Av Since the velocity is distributed we need to integrate over the area From Eq42 AV E A Vlocal dA A Vx W dxE where W is the depth Substituting the velocity we get AV E A 4Vc xH 1 xH W dx 4 VAcE A WH A 0 1 z 1 z dzEA z xH 4 VAcE A WH A1 2 z2 1 E3 z3E A A 1 0E A A2 3E A VAcE A WH A2 3E A VAcE A A Average velocity V AV E A A A2 3E A VAcE Mass flow rate Am E A AV E Av A2 3E A VAcE A WHv Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 416 Nitrogen gas flowing in a 50mm diameter pipe at 15C 200 kPa at the rate of 005 kgs encounters a partially closed valve If there is a pressure drop of 30 kPa across the valve and essentially no temperature change what are the velocities upstream and downstream of the valve Solution Same inlet and exit area A Aπ 4E A 0050A2E A 0001963 mA2E Ideal gas vAiE A A RTi EPi E A A02968 kJkgK 2882 K 200 kPaE A 04277 mA3E Akg From Eq63 VAiE A A m vi EAE A A005 kgs 04277 m3kg E0001963 m2 E A 109 ms Ideal gas vAeE A A RTe EPe E A A02968 kJkgK 2882 K 170 kPaE A 05032 mA3E Akg VAeE A A m ve EAE A A005 kgs 05032 m3kg E0001963 m2 E A 128 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 417 A boiler receives a constant flow of 5000 kgh liquid water at 5 MPa 20C and it heats the flow such that the exit state is 450C with a pressure of 45 MPa Determine the necessary minimum pipe flow area in both the inlet and exit pipes if there should be no velocities larger than 20 ms Solution Mass flow rate from Eq 43 both V 20 ms Am E AiE A Am E AeE A AVv AiE A AVv AeE A 5000 A 1 3600E A kgs Table B14 vAiE A 0001 mA3E Akg Table B13 vAeE A 008003 006332 007166 mA3E Akg AAiE A vAiE A Am E AVAiE A 0001 mA3E Akg A5000 3600E A kgs 20 ms 694 10A5E A mA2E A 069 cmA2E AAeE A vAeE A Am E AVAeE A 007166 mA3E Akg A5000 3600E A kgs 20 ms 498 10A3E A mA2E A 50 cmA2E Inlet liquid i e Q Q boiler Super heater vapor cb Exit Superheated vapor Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 418 A 06 m diameter household fan takes air in at 98 kPa 20AoE AC and delivers it at 105 kPa 21AoE AC with a velocity of 15 ms What are the mass flow rate kgs the inlet velocity and the outgoing volume flow rate in mA3E As Solution Continuity Eq Am E AiE A Am E AeE A AV v Ideal gas v RTP Area A Aπ 4E A D A2E A Aπ 4E A 06A2E A 02827 mA2E AV E AeE A AVAeE A 02827 15 04241 mA3E As vAeE A A RTe EPe E A A0287 21 273 E105E A 08036 mA3E Akg Am E AiE A AV E AeE AvAeE A 0424108036 0528 kgs AVAiE A vAiE A Am E AiE A AVAeE A vAeE VAiE A VAeE A vAiE AvAeE A VAeE A A RTi EPive E A 15 A0287 kJkgK 20 273 K E98 kPa 08036 m3kgE A 16 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 419 An airport ventilation system takes 25 mA3E As air at 100 kPa 17AoE AC into a furnace and heats it to 52AoE AC and delivers the flow to a duct with crosssectional area 04 mA2E A at 110 kPa Find the mass flow rate and the velocity in the duct Solution The inflate flow is given by a Am E AiE Continuity Eq Am E AiE A AV E AiE A vAiE A Am E AeE A AAeE AVAeE AvAeE Ideal gas vAiE A A RTi EPi E A A0287 290 100E A 08323 Am3 EkgE vAeE A A RTe EPe E A A0287 52 273 E110E 08479 mA3E A kg Am E AiE A AV E AiE AvAiE A 25 mA3E As 08323 mA3E Akg 3004 kgs VAeE A Am E A vAeE A AAeE A A3004 08479 04E A EAEAmA3 As EmA2 AE A 637 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Single flow single device processes Nozzles diffusers Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 420 Liquid water at 15AoE AC flows out of a nozzle straight up 15 m What is nozzle VAexitE A Energy Eq413 hexit A1 2E A VA2 exitE A gHexit h2 A1 2E A VA2 2E A gH2 If the water can flow 15 m up it has specific potential energy of gH2 which must equal the specific kinetic energy out of the nozzle VA 2 exitE A2 The water does not change P or T so h is the same VA 2 exitE A2 gH2 Hexit gH VAexitE A A 2gHE A A 2 9807 15 m2s2 EA 1715 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 421 A nozzle receives an ideal gas flow with a velocity of 25 ms and the exit is at 100 kPa 300 K with a velocity of 250 ms Determine the inlet temperature if the gas is argon helium or nitrogen Solution CV Nozzle Am E Ai Am E Ae assume no heat transfer Energy Eq413 hi A1 2E A VA2 iE A A1 2E A VA2 eE A he hi he A1 2E AVA2 eE A A1 2E A VA2 iE hi he CApE A Ti Te A1 2E A VA2 eE A VA2 iE A A1 2E A 250A2E A 25A2E A mA2E AsA2E 309375 Jkg 30938 kJkg Specific heats for ideal gases are from table A5 Argon CApE A 052 kJkg K T A30938 052E A 595 Ti 3595 K Helium CApE A 5913 kJkg K T A30938 5193E A 596 Ti 306 K Nitrogen CApE A 1042 kJkg K T A30938 1042E A 297 Ti 330 K Inlet Low V Exit Hi V Hi P A Low P A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 422 A diffuser receives 01 kgs steam at 500 kPa 350AoE AC The exit is at 1 MPa 400AoE AC with negligible kinetic energy and the flow is adiabatic Find the diffuser inlet velocity and the inlet area Solution Energy Eq413 h1 A1 2E A VA2 1E A gZ1 h2 A1 2E A VA2 2E A gZ2 Process Z1 Z2 State 1 Table B13 h1 316765 kJkg v1 057012 mA3E Akg State 2 V2 0 Table B13 h2 326388 kJkg Then from the energy equation A1 2E A VA2 1E A h2 h1 326388 316765 9623 kJkg V1 2h2 h1 A 2 9623 1000EA 4387 ms The mass flow rate from Eq43 AmE A ρAV AVv A AmE AvV 01 kgs 057012 mA3E Akg 4387 ms 000013 m2 13 cm2 Inlet Low V Exit Hi V Hi P A Low P A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 423 In a jet engine a flow of air at 1000 K 200 kPa and 30 ms enters a nozzle as shown in Fig P423 where the air exits at 850 K 90 kPa What is the exit velocity assuming no heat loss Solution CV nozzle No work no heat transfer Continuity AmE AiE A AmE AeE A AmE Energy AmE A hAiE A ½VAiE A2E A AmE AhAeE A ½VAeE A2E A Due to high T take h from table A71 ½VAeE A2E A ½ VAiE A2E A hAiE A hAeE A 1 2000E A 30A2E A 104622 8774 045 16882 16927 kJkg VAeE A 2000 16927A12E A 5818 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 424 In a jet engine a flow of air at 1000 K 200 kPa and 40 ms enters a nozzle where the air exits at 500 ms 90 kPa What is the exit temperature assuming no heat loss Solution CV nozzle no work no heat transfer Continuity AmE AiE A AmE AeE A AmE Energy AmE A hAiE A ½VAiE A2E A AmE AhAeE A ½VAeE A2E A Due to the high T we take the h value from Table A71 hAeE A hAiE A ½ VAiE A2E A ½VAeE A2E 104622 kJkg 05 40A2E A 500A2E A m2s2 11000 kJJ 104622 1242 92202 kJkg Interpolation in Table A71 TAeE A 850 50 A92202 8774 93315 8774E A 890 K 40 ms 200 kPa 500 ms 90 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 425 Superheated vapor ammonia enters an insulated nozzle at 30C 1000 kPa shown in Fig P425 with a low velocity and at the steady rate of 001 kgs The ammonia exits at 300 kPa with a velocity of 450 ms Determine the temperature or quality if saturated and the exit area of the nozzle Solution CV Nozzle steady state 1 inlet and 1 exit flow insulated so no heat transfer Energy Eq413 q hAiE A VA2 iE A2 hAeE A VA2 eE A2 Process q 0 VAiE A 0 Table B22 hAiE A 14791 hAeE A 450A2E A21000 hAeE A 137785 kJkg Table B21 PAeE A 300 kPa Sat state at 924C hAeE A 137785 13789 xAeE A 129382 xAeE A 09584 vAeE A 0001536 xAeE A 040566 03903 mA3E Akg AAeE A AmE AeE AvAeE AVAeE A 001 03903 450 867 106 m2 Inlet Low V Exit Hi V Hi P A Low P A cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 426 The wind is blowing horizontally at 30 ms in a storm at PA0E A 20C toward a wall where it comes to a stop stagnation and leaves with negligible velocity similar to a diffuser with a very large exit area Find the stagnation temperature from the energy equation Solution Energy Eq h1 A1 2E A VA2 1E A gZ1 h2 A1 2E A VA2 2E A gZ2 Process Z1 Z2 and V2 0 h2 h1 A1 2E A VA2 1E A T2 T1 A1 2E A VA2 1E A CApE A 293 K A1 2E A 30 msA2E A 1004 kJkgK 1000 JkJ 29345 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 427 A sluice gate dams water up 5 m There is a small hole at the bottom of the gate so liquid water at 20AoE AC comes out of a 1 cm diameter hole Neglect any changes in internal energy and find the exit velocity and mass flow rate Solution Energy Eq413 h1 A1 2E A VA2 1E A gZ1 h2 A1 2E A VA2 2E A gZ2 Process h1 h2 both at P 1 atm V1 0 Z1 Z2 5 m Water 5 m A1 2E A VA2 2E A g Z1 Z2 V2 2gZ1 Z2 A 2 9806 5EA 9902 ms AmE A ρΑV AVv Aπ 4E A DA2E A V2v Aπ 4E A 001A2E A mA2E A 9902 ms 0001002 mA3E Akg 0776 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 428 A diffuser shown in Fig P428 has air entering at 100 kPa 300 K with a velocity of 200 ms The inlet crosssectional area of the diffuser is 100 mmA2E A At the exit the area is 860 mmA2E A and the exit velocity is 20 ms Determine the exit pressure and temperature of the air Solution Continuity Eq43 Am E AiE A AAiE AVAiE AvAiE A Am E AeE A AAeE AVAeE AvAeE A Energy Eqper unit mass flow 413 hAiE A A1 2E AVAiE A2E A hAeE A A1 2E AVAeE A2E hAeE A hAiE A A1 2E A 200A2E A1000 A1 2E A 20A2E A1000 198 kJkg TAeE A TAiE A hAeE A hAiE ACApE A 300 A 198 kJkg 1004 kJkgKE A 31972 K Now use the continuity equation and the ideal gas law vAeE A vAiE A A AeVe EAiVi E A RTAiE APAiE A A AeVe EAiVi E A RTAeE APAeE PAeE A PAiE A A Te ETi E A A AiVi EAeVe E A 100 kPa A 31972 300 E A A 100 200 860 20 E A 12392 kPa Inlet Low V Exit Hi V Hi P A Low P A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 429 A meteorite hits the upper atmosphere at 3000 ms where the pressure is 01 atm and the temperature is 40C How hot does the air become right in front of the meteorite assuming no heat transfer in this adiabatic stagnation process Solution Energy Eq h1 A1 2E A VA2 1E A gZ1 h2 A1 2E A VA2 2E A gZ2 Process Z1 Z2 and V2 0 h2 h1 A1 2E A VA2 1E A T2 T1 A1 2E A VA2 1E ACApE A 233 A1 2E A A 30002 E1004 1000E A 4715 K At this high temperature we cannot assume constant specific heats so use A7 h2 h1 A1 2E A VA2 1E A 2333 A1 2E A A30002 E1000E A 47333 kJkg Table A7 is listed to 3000 K so we have to extrapolate to get T2 3000 50 A 47333 352536 352536 346073E A 3935 K The value of CApE A over 3000 K is 1293 kJkgK from the last two table entries At this temperature there will be some chemical reactions that should be considered to have a more realistic temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 430 The front of a jet engine acts similar to a diffuser receiving air at 900 kmh 5C 50 kPa bringing it to 80 ms relative to the engine before entering the compressor If the flow area is increased to 120 of the inlet area find the temperature and pressure in the compressor inlet Solution CV Diffuser Steady state 1 inlet 1 exit flow no q no w Continuity Eq43 Am E Ai Am E Ae AVv Energy Eq412 Am E A hi A1 2E A VA2 iE A Am E A A1 2E A VA2 eE A he he hi Cp Te Ti A1 2E A VA2 iE A A1 2E A VA2 eE A A1 2E A A900 1000 3600E AA 2E A A1 2E A 80A2E A A1 2E A 250A2E A A1 2E A 80A2E A 28050 Jkg 2805 kJkg T 28051004 279 Te 5 279 229C Now use the continuity eq AiVi vi AeVe ve vAeE A vAiE A A AeVe EAiVi E vAeE A vAiE A A12 80 1 250E A vi 0384 Ideal gas Pv RT ve RTePe RT i 0384Pi Pe Pi TeT i0384 50 kPa 296268 0384 1438 kPa Fan Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Throttle flow Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 431 R410A at 5C 700 kPa is throttled so it becomes cold at 40C What is the exit P CV Throttle Steady state Process with q w 0 and VAiE A VAeE A ZAiE A ZAeE Energy Eq413 hAiE A hAeE A Inlet state Table B41 hAiE A 5022 kJkg slightly compressed liquid Exit state Table B41 since h hg 26283 kJkg it is twophase P Psat 175 kPa 2 P v 1 h C T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 432 Carbon dioxide is throttled from 20AoE AC 2 MPa to 800 kPa Find the exit temperature assuming ideal gas behavior and repeat for realgas behavior CV Throttle valve restriction Steady flow 1 inlet and exit no q w Energy Eq413 hAiE A hAeE A Ideal gas same h gives TAiE A TAeE A 20AoE AC Real gas A hi he 36842 kJkg EPe 08 MPa E A A Table B32 Te 53C 278 K E Comment As you go towards lower P it becomes closer to ideal gas and the constant h curve bends to become horizontal h becomes fct of T only e T v i h C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 433 Saturated liquid R134a at 25AoE AC is throttled to 300 kPa in a refrigerator What is the exit temperature Find the percent increase in the volume flow rate Solution Steady throttle flow Assume no heat transfer and no change in kinetic or potential energy hAeE A hAiE A hAf 25oCE A 23459 kJkg hAf eE A xAeE A hAfg eE A at 300 kPa From table B51 we get TAeE A TAsE AaE AtE A 300 kPa 056AoE AC Lets use 0AoE AC for the following we could interpolate to get at 056AoE AC xAeE A A he hf e Ehfg e E A A23459 200 19836E A 01744 vAeE A vAfE A xAeE A vAfE AgE A 0000773 xAeE A 006842 00127 mA3E Akg vAiE A vAf 25oCE A 0000829 mA3E Akg AV E A Am E Av so the ratio becomes A V e EV i E A A m ve Em vi E A A ve Evi E A A 00127 0000829E A 1532 So the increase is 1432 times or 1432 e T v i h C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 434 A supply line has a steady flow of R410A at 1000 kPa 60AoE AC from which a flow is taken out through a throttle with an exit flow at 300 kPa Find the exit temperature CV Throttle Steady state Process with q w 0 and VAiE A VAeE A ZAiE A ZAeE Energy Eq413 hAiE A hAeE A Inlet state Table B42 hAiE A 33575 kJkg Exit state Table B42 since h hg Interpolate T 40 20 A33575 32722 34481 32722E A 497AoE AC 2 P v 1 h C T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 435 Carbon dioxide used as a natural refrigerant flows out of a cooler at 10 MPa 40AoE AC after which it is throttled to 14 MPa Find the state T x for the exit flow CV Throttle Steady state Process with q w 0 and VAiE A VAeE A ZAiE A ZAeE Energy Eq413 hAiE A hAeE A Inlet state Table B32 hAiE A 20014 kJkg Exit state Table B31 since h hg 32387 kJkg Interpolate T 306AoE AC hf 192 kJkg hfg 30467 kJkg x 20014 192 30467 0594 2 P v 1 h C T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 436 Liquid water at 180AoE AC 2000 kPa is throttled into a flash evaporator chamber having a pressure of 500 kPa Neglect any change in the kinetic energy What is the fraction of liquid and vapor in the chamber Solution Energy Eq413 h1 A1 2E A VA2 1E A gZ1 h2 A1 2E A VA2 2E A gZ2 Process Z1 Z2 and V2 V1 h2 h1 76371 kJkg from Table B14 State 2 P2 h2 2 phase h2 hf x2 hfg x2 h2 hf hfg A76371 64021 210847E A 00586 Fraction of Vapor x2 00586 586 Liquid 1 x2 0941 941 Twophase out of the valve The liquid drops to the bottom Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 437 Helium is throttled from 12 MPa 20C to a pressure of 100 kPa The diameter of the exit pipe is so much larger than the inlet pipe that the inlet and exit velocities are equal Find the exit temperature of the helium and the ratio of the pipe diameters Solution CV Throttle Steady state Process with q w 0 and VAiE A VAeE A ZAiE A ZAeE Energy Eq413 hAiE A hAeE A Ideal gas TAiE A TAeE A 20C Am E A A AV RTPE A But Am E A V T are constant PAiE AAAiE A PAeE AAAeE A De EDi E A A Pi EPe E A 12E A A 12 01 E A 12E A 3464 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 438 Methane at 1 MPa 300 K is throttled through a valve to 100 kPa Assume no change in the kinetic energy What is the exit temperature Solution Energy Eq 413 h1 A1 2E A VA2 1E A gZ1 h2 A1 2E A VA2 2E A gZ2 Process Z1 Z2 and V2 V1 State 1 Table B72 h1 61876 kJkg Use energy eq h2 h1 61876 kJkg State 2 P2 h2 look in Β72 Superheated vapor Τ2 275 25 A61876 57236 62758 57236E A 296 K Notice if it had been ideal gas then Τ2 T1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 439 R134a is throttled in a line flowing at 25AoE AC 750 kPa with negligible kinetic energy to a pressure of 165 kPa Find the exit temperature and the ratio of exit pipe diameter to that of the inlet pipe DAexE ADAinE A so the velocity stays constant Solution Energy Eq413 h1 A1 2E A VA2 1E A gZ1 h2 A1 2E A VA2 2E A gZ2 Process Z1 Z2 and V2 V1 State 1 Table B51 h1 23459 kJkg v1 vf 0000829 mA3E Akg Use energy eq h2 h1 23459 kJkg State 2 P2 h2 2 phase and T2 Tsat 165 kPa 15C h2 hf x2 hfg 23459 kJkg x2 h2 hf hfg 23459 18019 209 02603 v2 vf x2 vfg 0000746 02603 011932 00318 mA3E Akg Now the continuity equation with V2 V1 gives from Eq43 Am E A ρΑV AVv A1V1v1 A2 V1 v2 A2 A1 v2 v1 D2 D12 D2D1 v2 v105 00318 000082905 619 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Turbines Expanders Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 440 A steam turbine has an inlet of 3 kgs water at 1200 kPa 350AoE AC and velocity of 15 ms The exit is at 100 kPa 150AoE AC and very low velocity Find the specific work and the power produced Solution Energy Eq413 h1 A1 2E A VA2 1E A gZ1 h2 A1 2E A VA2 2E A gZ2 wT Process Z1 Z2 and V2 0 Table B13 h1 315359 kJkg h2 277638 kJkg wT h1 A1 2E A VA2 1E A h2 315359 A 152 E2000E A 277638 3773 kJkg remember to convert mA2E AsA2E A Jkg to kJkg by dividing with 1000 W T Am E A wT 3 kgs 3773 kJkg 1132 kW W T 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 441 Air at 20 ms 1500 K 875 kPa with 5 kgs flows into a turbine and it flows out at 25 ms 850 K 105 kPa Find the power output using constant specific heats Solution Energy Eq413 h1 A1 2E A VA2 1E A gZ1 h2 A1 2E A VA2 2E A gZ2 wT Process Z1 Z2 and V2 0 Table A5 CApE A 1004 kJkgK wT h1 A1 2E A VA2 1E A h2 A1 2E A VA2 2E A CApE AT1 T2 A1 2E A VA2 1E A A1 2E A VA2 2E 1004 kJkgK 1500 850 K A202 252 E2000E A mA2E AsA2E A kJJ 6525 kJkg remember to convert mA2E AsA2E A Jkg to kJkg by dividing with 1000 W T Am E A wT 5 kgs 6525 kJkg 3263 kW W T 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 442 Solve the previous problem using Table A7 Air at 20 ms 1500 K 875 kPa with 5 kgs flows into a turbine and it flows out at 25 ms 850 K 105 kPa Find the power output using constant specific heats Solution Energy Eq413 h1 A1 2E A VA2 1E A gZ1 h2 A1 2E A VA2 2E A gZ2 wT Process Z1 Z2 and V2 0 Table A71 h1 16358 kJkg h2 8774 kJkg wT h1 h2 A1 2E A VA2 1E A A1 2E A VA2 2E A 16358 8774 kJkg A202 252 E2000E A mA2E AsA2E A kJJ 7583 kJkg remember to convert mA2E AsA2E A Jkg to kJkg by dividing with 1000 W T Am E A wT 5 kgs 7583 kJkg 3792 kW W T 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 443 A windmill with rotor diameter of 20 m takes 40 of the kinetic energy out as shaft work on a day with 20AoE AC and wind speed of 35 kmh What power is produced Solution Continuity Eq Am E AiE A Am E AeE A Am E Energy Am E A hAiE A ½VAiE A2E A gZAiE A Am E AhAeE A ½VAeE A2E A gZAeE A AW E Process information AW E A Am E A½VAiE A2E A 04 Am E A ρAV AVAiE A vAiE A Aπ 4E A D A2E A Aπ 4E A 20A2E A 31416 mA2E vAiE A RTAiE APAiE A A0287 293 1013E A 08301 mA3E Akg VAiE A 35 kmh A35 1000 3600E A 97222 ms Am E A AVAiE A vAiE A A31416 97222 08301E A 36795 kgs ½ VAiE A2E A ½ 97222A2E A mA2E AsA2E A 4726 Jkg AW E A 04 Am E A½ VAiE A2E A 04 36795 kgs 4726 Jkg 69 557 W 6956 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 444 A liquid water turbine receives 2 kgs water at 2000 kPa 20AoE AC and velocity of 15 ms The exit is at 100 kPa 20AoE AC and very low velocity Find the specific work and the power produced Solution Energy Eq413 h1 A1 2E A VA2 1E A gZ1 h2 A1 2E A VA2 2E A gZ2 wT Process Z1 Z2 and V2 0 State 1 Table B14 h1 8582 kJkg State 2 Table B11 h2 8394 which is at 23 kPa so we should add Pv 977 0001 to this wT h1 A1 2E A VA2 1E A h2 8582 1522000 8394 19925 kJkg W T Am E A wT 2 kgs 19925 kJkg 3985 kW Notice how insignificant the specific kinetic energy is Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 445 What is the specific work one can expect from the dam in Problem 427 Solution Energy Eq413 h1 A1 2E A VA2 1E A gZ1 h2 A1 2E A VA2 2E A gZ2 wT Process V2 V1 P1 P2 P0 h1 h2 Same velocity if same flow area incompressible flow and the water temperature does not change significantly The specific potential enrgy difference can be extracted wT gZ1 gZ2 9806 msA2E A 5 0 m 49 mA2E AsA2E A 49 Jkg This is a very small amount of specific work Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 446 A small highspeed turbine operating on compressed air produces a power output of 100 W The inlet state is 400 kPa 50C and the exit state is 150 kPa 30C Assuming the velocities to be low and the process to be adiabatic find the required mass flow rate of air through the turbine Solution CV Turbine no heat transfer no KE no PE Energy Eq413 hAinE A hAexE A wT Ideal gas so use constant specific heat from Table A5 wATE A hAinE A hAexE A CApE ATAinE A TAexE A 1004 kJkgK 50 30 K 803 kJkg AW E A Am E AwATE A Am E A AW E AwATE A 01 kW 803 kJkg 000125 kgs The dentists drill has a small air flow and is not really adiabatic Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 447 Hoover Dam across the Colorado River dams up Lake Mead 200 m higher than the river downstream The electric generators driven by waterpowered turbines deliver 1300 MW of power If the water is 175C find the minimum amount of water running through the turbines Solution CV HA2E AO pipe turbines T H DAM Lake Mead Continuity Am E AinE A Am E AexE A Energy Eq413 h VA2E A2 gzAinE A h VA2E A2 gzAexE A wATE Water states hAinE A hAexE A vAinE A vAexE A Now the specific turbine work becomes wATE A gzAinE A gzAexE A 9807 2001000 1961 kJkg Am E A AW E ATE AwATE A A1300103 kW E1961 kJkgE A 663 10A5E A kgs AV E A Am E Av 663 10A5E A kgs 0001001 mA3E Akg 664 mA3E As Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 448 A small turbine shown in Fig P 448 is operated at part load by throttling a 025 kgs steam supply at 14 MPa 250C down to 11 MPa before it enters the turbine and the exhaust is at 10 kPa If the turbine produces 110 kW find the exhaust temperature and quality if saturated Solution CV Throttle Steady q 0 and w 0 No change in kinetic or potential energy The energy equation then reduces to Energy Eq413 hA1E A hA2E A 29272 kJkg from Table B13 CV Turbine Steady no heat transfer specific work w A110 025E A 440 kJkg Energy Eq hA1E A hA2E A hA3E A w 29272 kJkg hA3E A 29272 440 24872 kJkg State 3 P h Table B12 h hAgE 24872 19183 xA3E A 23928 T 458C xA3E A 0959 T v 1 2 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 449 A small expander a turbine with heat transfer has 005 kgs helium entering at 1000 kPa 550 K and it leaves at 250 kPa 300 K The power output on the shaft is measured to 55 kW Find the rate of heat transfer neglecting kinetic energies Solution CV Expander Steady operation Cont AmE AiE A AmE AeE A AmE Energy AmE AhAiE A AQ E A AmE AhAeE A AW E W T i e Q cb AQ E A AmE A hAeE A hAiE A AW E Use specific heat from Table A5 CAp HeE A 5193 kJkg K AQ E A AmE ACApE A TAeE A TAiE A AW E 005 kgs 5193 kJkgK 300 550 K 55 kW 6491 55 99 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Compressors fans Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 450 A compressor in a commercial refrigerator receives R410A at 25AoE AC x 1 The exit is at 1000 kPa 40AoE AC Neglect kinetic energies and find the specific work Solution CV Compressor steady state single inlet and exit flow For this device we also assume no heat transfer and ZAiE A ZAeE W C i e cb From Table B41 hAiE A 26977 kJkg From Table B42 hAeE A 31605 kJkg Energy Eq413 reduces to wAcE A hAiE A hAeE A 26977 31605 kJkg 4628 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 451 A compressor brings nitrogen from 100 kPa 290 K to 2000 kPa The process has a specific work input of 450 kJkg and the exit temperature is 450 K Find the specific heat transfer using constant specific heats Solution CV Compressor Not adiabatic neglect kinetic and potential energy changes Energy Eq413 q hAiE A AVi 2E2A gZAiE A hAeE A AVe 2E2A w gZAeE Process ZAiE A ZAeE A AVi 2E2A AVe 2E2A Solve for the heat transfer q hAeE A w hAiE A CApE A TAeE A TAiE A w 1042 kJkgK 450 290 K 450 kJkg 2833 kJkg Recall standard sign notation is work positive out Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 452 A portable fan blows 03 kgs room air with a velocity of 15 ms see Fig P418 What is the minimum power electric motor that can drive it Hint Are there any changes in P or T Solution CV Fan plus space out to near stagnant inlet room air Energy Eq413 q hAiE A AVi 2E2A hAeE A AVe 2E2A w Here q 0 Vi 0 and hAiE A hAeE A same P and T w AVe 2E2A 15A2E A2 msA2E A 1125 Jkg 01125 kJkg AW E A Am E Aw 03 kgs 01125 kJkg 0034 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 453 A refrigerator uses the natural refrigerant carbon dioxide where the compressor brings 002 kgs from 1 MPa 20AoE AC to 6 MPa using 2 kW of power Find the compressor exit temperature Solution CV Compressor steady state single inlet and exit flow For this device we also assume no heat transfer and ZAiE A ZAeE W C i e cb From Table B32 hAiE A 34231 kJkg Energy Eq413 reduces to AW E A AmE AwAcE A AmE A hAiE A hAeE A hAeE A hAiE A AW E AAmE A hAeE A 34231 kJkg 2 kW 002 kgs 44231 kJkg From Table B32 TAeE A 100 20 44231 4216944502 42169 1177 AoE AC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 454 An air compressor takes in air at 100 kPa 17C and delivers it at 1 MPa 600 K to a constantpressure cooler which it exits at 300 K Find the specific compressor work and the specific heat transfer in the cooler Solution CV air compressor q 0 Continuity Eq Am E A2E A Am E A1E Energy Eq413 0 hA1E A wAc inE A hA2E 1 3 2 Q cool Compressor W c Compressor section Cooler section Table A7 hA1E A 29019 kJkg hA2E A 60732 kJkg hA3E A 30047 kJkg wAc inE A hA2E A hA1E A 60732 29019 31713 kJkg CV cooler w A0E A Continuity Eq Am E A3E A Am E A1E A Energy Eq413 0 hA2E A qAoutE A hA3E qAoutE A hA2E A hA3E A 60732 30047 30685 kJkg Comment For thess temperatures we could have used constant specific heats from A5 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 455 A compressor brings R134a from 150 kPa 10AoE AC to 1200 kPa 50AoE AC It is water cooled with a heat loss estimated as 40 kW and the shaft work input is measured to be 150 kW How much is the mass flow rate through the compressor Solution CV Compressor Steady flow Neglect kinetic and potential energies Energy AmE A hAiE A AQ E A AmE AhAeE A AW E AmE A AQ E A AW E AhAeE A hAiE A 1 2 Q cool Compressor W c Look in table B52 hAiE A 39384 kJkg hAeE A 42684 kJkg AmE A A 40 150 E42684 39384E A A kW kJkgE A 3333 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 456 The compressor of a large gas turbine receives air from the ambient at 95 kPa 20C with a low velocity At the compressor discharge air exits at 152 MPa 430C with velocity of 90 ms The power input to the compressor is 5000 kW Determine the mass flow rate of air through the unit Solution CV Compressor steady state single inlet and exit flow Energy Eq413 q hAiE A AVi 2E2A hAeE A AVe 2E2A w Here we assume q 0 and Vi 0 so using constant CAPoE A from A5 w CAPoE ATAeE A TAiE A AVe 2E2A 1004430 20 A 902 E2 1000E A 4155 kJkg Notice the kinetic energy is 1 of the work and can be neglected in most cases The mass flow rate is then from the power and the specific work Am E A W c w A5000 4155E A A kW kJkgE A 120 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 457 How much power is needed to run the fan in Problem 418 A household fan of diameter 06 m takes air in at 98 kPa 20AoE AC and delivers it at 105 kPa 21AoE AC with a velocity of 15 ms What are the mass flow rate kgs the inlet velocity and the outgoing volume flow rate in mA3E As Solution Continuity Eq Am E AiE A Am E AeE A AV v Ideal gas v RTP Area A Aπ 4E A D A2E A Aπ 4E A 06A2E A 02827 mA2E AV E AeE A AVAeE A 02827 15 04241 mA3E As vAeE A A RTe EPe E A A0287 21 273 E105E A 08036 mA3E Akg Am E AiE A AV E AeE AvAeE A 0424108036 0528 kgs AVAiE A vAiE A Am E AiE A AVAeE A vAeE VAiE A VAeE A vAiE AvAeE A VAeE A A RTi EPive E A 15 A0287 kJkgK 20 273 K E98 kPa 08036 m3kgE A 16 ms AW E A Am E AhAiE A ½VAiE A2E A hAeE A ½VAeE A2E A Am E A CApE A TAiE A TAeE A ½ VAiE A2E A ½VAeE A2E A 0528 1004 1 EAEA16A2 A 15A2 A E2000E A 0528 1004 0000155 053 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 458 A compressor in an industrial airconditioner compresses ammonia from a state of saturated vapor at 200 kPa to a pressure of 1000 kPa At the exit the temperature is measured to be 100AoE AC and the mass flow rate is 05 kgs What is the required motor size kW for this compressor Solution CV Compressor Assume adiabatic and neglect kinetic energies Energy Eq413 wACE A hA1E A hA2E A States 1 B22 hA1E A 14196 kJkg 2 B22 hA2ACE A 16643 kJkg Energy equation wACE A hA2E A hA1E A 16643 14196 2447 kJkg AW E A 05 kgs 2447 kJkg 122 kW v P 2 ac 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 459 An exhaust fan in a building should be able to move 3 kgs air atmospheric pressure air at 25 E AC through a 05 m diameter vent hole How high a velocity must it generate and how much power is required to do that o Solution CV Fan and vent hole Steady state with uniform velocity out Continuity Eq Am E A constant ρΑV AV v AVPRT Ideal gas Pv RT and area is A Aπ 4E A DA2E Now the velocity is found V Am E A RTAπ 4E A DA2E AP 3 kgs 0287 kJkgK 29315 Κ Aπ 4E A 05A2E A mA2E A 101 kPa 127 ms The kinetic energy out is A1 2E A VA2 2E A A1 2E A 127A2E A 1000 008065 kJkg which is provided by the work only two terms in energy equation that does not cancel we assume VA1E A 0 AW E AinE A Am E A A1 2E A VA2 2E A 3 kgs 008065 kJkg 0242 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 460 A compressor receives R410A as saturated vapor R410A at 400 kPa and brings it to 2000 kPa 60AoE AC Then a cooler brings it to saturated liquid at 2000 kPa see Fig P454 Find the specific compressor work and the specific heat transfer in the cooler CV Compressor Assume adiabatic and neglect kinetic potential energies Energy Eq413 w hA1E A hA2E A States 1 B42 hA1E A 2719 kJkg 2 B42 hA2E A 32062 kJkg 3 B41 hA3E A 10614 A231 5E A 11495 10614 1102 kJkg wACE A hA2E A hA1E A 32062 2719 4872 kJkg CV Cooler No work neglect kinetic potential energies Energy Eq 413 q hA3E A hA2E A 1102 32062 2104 kJkg The compressor is inside the unit to the left that also houses a fan and the heat exchanger C Borgnakke Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 461 An air flow is brought from 20AoE AC 100 kPa to 1000 kPa 330AoE AC by an adiabatic compressor driven by a 50kW motor What are the mass flow rate and the exit volume flow rate of air CV Compressor Assume adiabatic and neglect kinetic potential energies Energy Eq413 w hA1E A hA2E A wACE A hA2E A hA1E A CAPE A TA2E A TA1E A 1004 kJkgK 330 20 K 3112 kJkg The mass flow rate scale the work term so AmE A AW E A wACE A A 50 kW 3112 kJkgE A 01606 kgs AV E A AmE A v AmE A ART PE A 01606 kgs A0287 330 273 kJkg E1000 kPaE A 00278 mA3E As Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful HeatersCoolers Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 462 The air conditioner in a house or a car has a cooler that brings atmospheric air from 30AoE AC to 10AoE AC both states at 101 kPa If the flow rate is 075 kgs find the rate of heat transfer Solution CV Cooler Steady state single flow with heat transfer Neglect changes in kinetic and potential energy and no work term Energy Eq 413 qAoutE A hAiE A hAeE Use constant specific heat from Table A5 T is around 300 K qAoutE A hAiE A hAeE A CAPE A TAiE A TAeE A 1004 A kJ kg KE A 30 10 K 201 kJkg AQ E AoutE A AmE A qAoutE A 075 201 15 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 463 A boiler section boils 3 kgs saturated liquid water at 2000 kPa to saturated vapor in a reversible constantpressure process Find the specific heat transfer in the process CV Boiler Steady state single inlet and exit flow neglect potential energy and since velocities are low we neglect kinetic energies Energy Eq 413 q hAiE A hAeE From Table B12 hAiE A 90877 kJkg Table B13 hAeE A 279951 kJkg q hAeE A hAiE A 279951 90877 189074 kJkg PAiE cb PAeE A PAiE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 464 A condenser cooler receives 005 kgs R410A at 2000 kPa 80AoE AC and cools it to 10AoE AC Assume the exit properties are as for saturated liquid with the same T What cooling capacity kW must the condenser have Solution CV R410A condenser Steady state single flow heat transfer out and no work Energy Eq 412 AmE A hA1E A AmE A hA2E A AQ E Aout Inlet state Table B42 hA1E A 34322 kJkg Exit state Table B41 hA2E A 7321 kJkg compressed liquid To be more accurate we could have added PPsatv to the hA2E A Process Neglect kinetic and potential energy changes Cooling capacity is taken as the heat transfer out ie positive out so AQ E Aout AmE A hA1E A hA2E A 005 kgs 34322 7321 kJkg 135 kW 1 2 Q cool Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 465 Carbon dioxide enters a steadystate steadyflow heater at 300 kPa 300 K and exits at 275 kPa 1500 K as shown in Fig P465 Changes in kinetic and potential energies are negligible Calculate the required heat transfer per kilogram of carbon dioxide flowing through the heater Solution CV Heater Steady state single inlet and exit flow Energy Eq 413 q hAiE A hAeE Q i e Table A8 q hAeE A hAiE A 16149 2144 14005 kJkg If we use CAp0E A from A5 then q 0842 kJkgK 1500 300K 10104 kJkg Too large T TAaveE A to use CAp0E A at room temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 466 Find the heat transfer in Problem 417 A boiler receives a constant flow of 5000 kgh liquid water at 5 MPa 20C and it heats the flow such that the exit state is 450C with a pressure of 45 MPa Determine the necessary minimum pipe flow area in both the inlet and exit pipes if there should be no velocities larger than 20 ms Solution CV Heater Steady state single inlet and exit flow neglect potential energy and since velocities are low we neglect kinetic energies Energy Eq 413 q hAiE A hAeE From Table B14 hAiE A 8864 kJkg Table B13 hAeE A 333023 3316152 332319 kJkg q hAeE A hAiE A 332319 8864 32346 kJkg PAiE cb PAeE A PAiE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 467 A chiller cools liquid water for airconditioning purposes Assume 25 kgs water at 20AoE AC 100 kPa is cooled to 5AoE AC in a chiller How much heat transfer kW is needed Solution CV Chiller Steady state single flow with heat transfer Neglect changes in kinetic and potential energy and no work term Energy Eq 413 qAoutE A hAiE A hAeE Properties from Table B11 hAiE A 8394 kJkg and hAeE A 2098 kJkg Now the energy equation gives qAoutE A 8394 2098 6296 kJkg AQ E AoutE A AmE A qAoutE A 25 6296 1574 kW Alternative property treatment since single phase and small T If we take constant specific heat for the liquid from Table A4 qAoutE A hAiE A hAeE A CApE A TAiE A TAeE A 418 kJkgK 20 5 K 627 kJkg AQ E AoutE A AmE A qAoutE A 25 kgs 627 kJkg 15675 kW 1 2 Q cool Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 468 Saturated liquid nitrogen at 600 kPa enters a boiler at a rate of 0008 kgs and exits as saturated vapor see Fig P468 It then flows into a super heater also at 600 kPa where it exits at 600 kPa 280 K Find the rate of heat transfer in the boiler and the super heater Solution CV boiler steady single inlet and exit flow neglect KE PE energies in flow Continuity Eq m 1 m 2 m 3 1 2 3 Q Q boiler Super heater vapor cb 600 P 1 2 3 v T 1 2 3 v Table B61 h1 81469 kJkg h2 8685 kJkg Table B62 h3 28905 kJkg Energy Eq 413 qboiler h2 h1 8685 81469 16832 kJkg AQ E Aboiler m 1qboiler 0008 kgs 16832 kJkg 1346 kW CV Superheater same approximations as for boiler Energy Eq 413 qsup heater h3 h2 28905 8685 2022 kJkg AQ E Asup heater m 2qsup heater 0008 kgs 2022 kJkg 162 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 469 Carbon dioxide used as a natural refrigerant flows through a cooler at 10 MPa which is supercritical so no condensation occurs The inlet is at 220AoE AC and the exit is at 50AoE AC Find the specific heat transfer CV Cooler Steady state single flow with heat transfer Neglect changes in kinetic and potential energy and no work term Energy Eq 413 0 hAiE A hAeE A q Properties from Table B32 hAiE A 54291 kJkg and hAeE A 20014 312112 25613 kJkg Now the energy equation gives q 25613 54291 28679 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 470 In a steam generator compressed liquid water at 10 MPa 30C enters a 30mm diameter tube at the rate of 3 Ls Steam at 9 MPa 400C exits the tube Find the rate of heat transfer to the water Solution CV Steam generator Steady state single inlet and exit flow Constant diameter tube AAiE A AAeE A Aπ 4E A 003A2E A 00007068 mA2E Table B14 AmE A AV E AiE AvAiE A 000300010003 30 kgs VAiE A AV E AiE AAAiE A 000300007068 424 ms Exit state properties from Table B13 VAeE A VAiE A vAeE AvAiE A 424 00299300010003 12686 ms The energy equation Eq 412 is solved for the heat transfer as AQ E A AmE A A he hi Ve 2 Vi 2 2E 30 kgs A 31178 13486 126862 4242 E2 1000 E A kJkg 8973 kW Typically hot combustion gas in Steam exit cb liquid water in gas out Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 471 An oven has five radiant heaters each one is rated at 15 kW It should heat some 2kg steel plates from 20C to 800 K How many of these plates per minute can it heat CV Oven steady state operation A flow of plates in represents an AmE A Energy Eq 0 AmE A hAiE A hAeE A AQ E AQ E A AmE A hAeE A hAiE A AmE A C TAeE A TAiE A AmE A AQ E A C TAeE A TAiE A A 5 15 kW 046 800 293 kJkgE A 0322 kgs N AmE Am 0322 2 1s 0161 1s 965 per min Front end of oven with steel rollers to bring the plates in a skirt hangs down to limit heat losses C Borgnakke Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 472 A cryogenic fluid as liquid nitrogen at 90 K 400 kPa flows into a probe used in cryogenic surgery In the return line the nitrogen is then at 160 K 400 kPa Find the specific heat transfer to the nitrogen If the return line has a cross sectional area 100 times larger than the inlet line what is the ratio of the return velocity to the inlet velocity Solution CV line with nitrogen No kinetic or potential energy changes Continuity Eq AmE A constant AmE AeE A AmE AiE A AAeE AVAeE AvAeE A AAiE AVAiE AvAiE Energy Eq 413 q hAeE A hAiE State i Table B61 hAiE A 9558 kJkg vAiE A 0001343 mA3E Akg State e Table B62 hAeE A 16296 kJkg vAeE A 011647 mA3E Akg From the energy equation q hAeE A hAiE A 16296 9558 2585 kJkg From the continuity equation VAeE AVAiE A AAiE AAAeE A vAeE AvAiE A A 1 100E A A 011647 0001343E A 0867 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 473 An evaporator has R410A at 20AoE AC and quality 20 flowing in with the exit flow being saturated vapor at 20AoE AC Knowing that there is no work find the specific heat transfer CV Heater Steady state single inlet and exit flow Energy Eq413 0 q hA1E A hA2E Table B4 hA1E A 2824 02 24365 7697 kJkg hA2E A 27189 kJkg q hA2E A hA1E A 27189 7697 19492 kJkg 1 2 Q evap Evaporator vapor cb 400 P 1 2 v T 1 2 v 20 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 474 A flow of liquid glycerine flows around an engine cooling it as it absorbs energy The glycerine enters the engine at 60AoE AC and receives 19 kW of heat transfer What is the required mass flow rate if the glycerine should come out at maximum 95AoE AC Solution CV Liquid flow glycerine is the coolant steady flow no work Energy Eq AmE AhAiE A AQ E A AmE AhAeE AmE A AQ E A hAeE A hAiE A EAEA AQ A ECAgly TAe A TAi A AE From table A4 CAglyE A 242 kJkgK AmE A A 19 242 95 60E A A kW kJkgE A 0224 kgs Exhaust flow Air intake filter Coolant flow Atm air Shaft Fan power Radiator cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Pumps pipe and channel flows Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 475 An irrigation pump takes water from a river at 10AoE AC 100 kPa and pumps it up to an open canal where it flows out 100 m higher at 10AoE AC The pipe diameter in and out of the pump is 01 m and the motor driving the unit is 5 hp What is the flow rate neglecting kinetic energy and losses CV Pump plus pipe to canal This is a single steady flow device with atmospheric pressure in and out Energy Eq 0 hAiE A 0 gZAiE A wAp inE A hAeE A 0 gZAeE A wAp inE A gZAeE A ZAiE A 981 msA2E A 100 msA2E A 0981 kJkg AW E Ap inE A 5 hp 5 hp 0746 kWhp 373 kW Find the flow rate from the work AmE A AW E Ap inE A wAp inE A 373 kW 0981 kJkg 38 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 476 A pipe flows water at 15AoE AC from one building to another In the winter time the pipe loses an estimated 500 W of heat transfer What is the minimum required mass flow rate that will ensure that the water does not freeze ie reach 0AoE AC Solution Energy Eq AmE AhAiE A AQ E A AmE AhAeE Assume saturated liquid at given T from table B11 AmE A A Q Ehe hi E A A500 103 E0 6298E A A kW kJkgE A A 05 6298E A kgs 0007 94 kgs 1 2 Q Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 477 A river flowing at 05 ms across a 1mhigh and 10mwide area has a dam that creates an elevation difference of 2 m How much energy could a turbine deliver per day if 80 of the potential energy can be extracted as work CV Dam from upstream to downstream 2 m lower elevation Continuity Eq AmE A constant AmE AeE A AmE AiE A AAeE AVAeE AvAeE A AAiE AVAiE AvAiE Find the mass flow rate AmE A AVAiE Av ρAVAiE A 997 kgmA3E A 1 10 mA2E A 05 ms 4985 kgs Energy Eq 0 AmE A hAiE A 05VAiE A2E A gZAiE A AmE A hAeE A 05VAeE A2E A gZAeE A AW E The velocity in and out is the same assuming same flow area and the two enthalpies are also the same since same T and P PA0E A AW E A 08 AmE A gZAiE A ZAeE A 08 4985 kgs 9807 msA2E A 2 m 78 221 Js 78 221 W 782 kW W AW E A t 782 kJs 24 60 60 s 676 GJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 478 A steam pipe for a 300m tall building receives superheated steam at 200 kPa at ground level At the top floor the pressure is 125 kPa and the heat loss in the pipe is 110 kJkg What should the inlet temperature be so that no water will condense inside the pipe Solution CV Pipe from 0 to 300 m no KE steady state single inlet and exit flow Neglect any changes in kinetic energy Energy Eq 413 q hAiE A hAeE A gZAeE No condensation means Table B12 hAeE A hAgE A at 125 kPa 26854 kJkg hAiE A hAeE A gZAeE A q 26854 A9807 300 1000E A 110 28101 kJkg At 200 kPa T 170AoE AC Table B13 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 479 Consider a water pump that receives liquid water at 25AoE AC 100 kPa and delivers it to a same diameter short pipe having a nozzle with exit diameter of 2 cm 002 m to the atmosphere 100 kPa Neglect the kinetic energy in the pipes and assume constant u for the water Find the exit velocity and the mass flow rate if the pump draws a power of 1 kW Solution Continuity Eq AmE AiE A AmE AeE A AVv A Aπ 4E A DA2 eE A Aπ 4E A 002A2E A 31416 10A4E A mA2E Energy Eq 413 hAiE A A1 2E AVA 2 iE A gZAiE A hAeE A A1 2E AVA 2 eE A gZAeE A w Properties hAiE A uAiE A PAiE AvAiE A hAeE A uAeE A PAeE AvAeE A PAiE A PAeE A vAiE A vAeE w A1 2E A VA 2 eE A AW E A AmE A A1 2E A VA 2 eE A A A1 2E A VA 3 eE AvAeE VAeE A 2 W ve A A 13E A A2 1000 0001003 31416 10 4 E A A 13E A 1855 ms AmE A AVAeE AvAeE A 31416 10A4E A 1855 0001003 581 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 480 A small water pump is used in an irrigation system The pump takes water in from a river at 10AoE AC 100 kPa at a rate of 5 kgs The exit line enters a pipe that goes up to an elevation 20 m above the pump and river where the water runs into an open channel Assume the process is adiabatic and that the water stays at 10AoE AC Find the required pump work Solution CV pump pipe Steady state 1 inlet 1 exit flow Assume same velocity in and out no heat transfer Continuity Eq Am E AinE A Am E AexE A Am E Energy Eq 412 Am E AhAinE A 12VAinE A2 gzAinE A Am E AhAexE A 12 VAex 2E A gzAexE A AW E States hAinE A hAexE A same T P i e H cb AW E A Am E A gzAinE A zAexE A 5 kgs 9807 msA2E A 0 20 m 1000 JkJ 098 kW IE 098 kW required input Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 481 A small stream with water at 15AoE AC runs out over a cliff creating a 50 m tall waterfall Estimate the downstream temperature when you neglect the horizontal flow velocities upstream and downstream from the waterfall How fast was the water dropping just before it splashed into the pool at the bottom of the waterfall Solution CV Waterfall steady state Assume no AQ E A nor AW E Energy Eq 413 h A1 2E AV2 gZ const State 1 At the top zero velocity ZA1E A 100 m State 2 At the bottom just before impact ZA2E A 0 State 3 At the bottom after impact in the pool hA1E A 0 gZA1E A hA2E A A1 2E A VA2 2E A 0 hA3E A 0 0 Properties hA1E A hA2E A same T P A1 2E A VA2 2E A gZA1E VA2E A 2gZ1 A 2 9806 ms2 50 mEA 313 ms Energy equation from state 1 to state 3 hA3E A hA1E A gZA1E use h CApE A T with value from Table A4 liquid water TA3E A TA1E A gZA1E A CApE 20 9806 msA2E A 50 m 4180 JkgK 2012 C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 482 A cutting tool uses a nozzle that generates a high speed jet of liquid water Assume an exit velocity of 500 ms of 20AoE AC liquid water with a jet diameter of 2 mm 0002 m How much mass flow rate is this What size power pump is needed to generate this from a steady supply of 20AoE AC liquid water at 200 kPa Solution CV Nozzle Steady state single flow Continuity equation with a uniform velocity across A AmE A AVv Aπ 4E A DA2E A V v Aπ 4E A 0002A2E A 500 0001002 1568 kgs Assume ZAiE A ZAeE A Ø uAeE A uAiE A and VAiE A 0 PAeE A 100 kPa atmospheric Energy Eq413 hAiE A Ø Ø hAeE A A1 2E AVA2 eE A Ø w w hAiE A hAeE A A1 2E AVA2 eE A uAiE A uAeE A PAiE A vAiE A PAeE A vAeE A A1 2E AVA2 eE PAiE A PAeE A vAiE A A1 2E AVA2 eE A 0001002 mA3E Akg 200 100 kPa 05 A5002 m2s2 E1000 JkJE A 01002 125 125 kJkg AW E A AmE Aw 1568 kgs 125 kJkg 196 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 483 The main waterline into a tall building has a pressure of 600 kPa at 5 m below ground level A pump brings the pressure up so the water can be delivered at 200 kPa at the top floor 100 m above ground level Assume a flow rate of 10 kgs liquid water at 10AoE AC and neglect any difference in kinetic energy and internal energy u Find the pump work Solution CV Pipe from inlet at 5 m up to exit at 150 m 200 kPa Energy Eq413 hAiE A A1 2E AVi2 gZAiE A hAeE A A1 2E AVe2 gZAeE A w With the same u the difference in hs are the Pv terms w hAiE A hAeE A A1 2E A Vi2 Ve2 g ZAiE A ZAeE A PAiE AvAiE A PAeE AvAeE A g ZAiE A ZAeE A 600 0001 200 0001 9806 5 1001000 04 103 063 kJkg AW E A AmE Aw 10 063 63 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Multiple flow single device processes Turbines Compressors Expanders Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 484 An adiabatic steam turbine in a power plant receives 5 kgs steam at 3000 kPa 500AoE AC Twenty percent of the flow is extracted at 1000 kPa 350AoE AC to a feedwater heater and the remainder flows out at 200 kPa 200AoE AC see Fig P484 Find the turbine power output CV Turbine Steady state 1 inlet and 2 exit flows Continuity Eq49 Am E A1E A Am E A2E A Am E A3E A Am E A3E A Am E A1E A Am E A2E A 150 lbms Energy Eq410 Am E A1E AhA1E A AW E ATE A Am E A2E AhA2E A Am E A3E AhA3E Table B12 hA1E A 345648 kJkg hA2E A 315765 kJkg Table B11 hA3E A 287046 kJkg From the energy equation Eq410 AW E ATE A Am E A1E AhA1E A Am E A2E AhA2E A Am E A3E AhA3E A 5 345648 1 315765 4 287046 kJs 2643 kW WT 1 2 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 485 A compressor receives 005 kgs R410A at 200 kPa 20AoE AC and 01 kgs R410A at 400 kPa 0AoE AC The exit flow is at 1000 kPa 60AoE AC as shown in Fig P485 Assume it is adiabatic neglect kinetic energies and find the required power input CV whole compressor steady 2 inlets 1 exit no heat transfer AQ E A 0 Continuity Eq49 AmE A1E A AmE A2E A AmE A3E A 005 01 015 kgs Energy Eq410 AmE A1E AhA1E A AmE A2E AhA2E A AW E ACE A AmE A3E AhA3E A Table B32 hA1E A 27872 kJkg h2 29042 kJkg Table B32 h3 33575 kJkg W C 3 2 1 AW E ACE A 005 27872 01 29042 015 33575 7385 kW Power input is 74 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 486 Cogeneration is often used where a steam supply is needed for industrial process energy Assume a supply of 5 kgs steam at 05 MPa is needed Rather than generating this from a pump and boiler the setup in Fig P486 is used so the supply is extracted from the highpressure turbine Find the power the turbine now cogenerates in this process Solution CV Turbine steady state 1 inlet and 2 exit flows assume adiabatic AQ E ACVE A 0 Continuity Eq49 Am E A1E A Am E A2E A Am E A3E A Energy Eq410 AQ E ACVE A Am E A1E AhA1E A Am E A2E AhA2E A Am E A3E AhA3E A AW E ATE A Supply state 1 20 kgs at 10 MPa 500AE AC Process steam 2 5 kgs 05 MPa 155AE AC Exit state 3 20 kPa x 09 Table B13 hA1E A 33737 hA2E A 27559 kJkg Table B12 hA3E A 2514 09 23583 23739 kJkg W T 1 2 3 HP LP AW E ATE A 20 33737 5 27559 15 23739 kgs kJkg 18 084 kW 18084 MW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 487 A steam turbine receives steam from two boilers One flow is 5 kgs at 3 MPa 700AE AC and the other flow is 10 kgs at 800 kPa 500AE AC The exit state is 10 kPa with a quality of 96 Find the total power out of the adiabatic turbine Solution CV whole turbine steady 2 inlets 1 exit no heat transfer AQ E A 0 Continuity Eq49 AmE A1E A AmE A2E A AmE A3E A 5 15 20 kgs Energy Eq410 AmE A1E AhA1E A AmE A2E AhA2E A AmE A3E AhA3E A AW E ATE Table B13 h1 39117 kJkg h2 34806 kJkg Table B12 h3 1918 096 23928 24889 kJkg W T 1 2 3 AW E AT 5 39117 10 34806 15 24889 17 031 kW 17 MW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 488 A compressor receives 01 kgs R134a at 150 kPa 10AoE AC and delivers it at 1000 kPa 40AoE AC The power input is measured to be 3 kW The compressor has heat transfer to air at 100 kPa coming in at 20AoE AC and leaving at 30AoE AC How much is the mass flow rate of air Solution CV Compressor steady state single inlet and exit flow For this device we also have an air flow outside the compressor housing no changes in kenetic or potential energy W C 1 2 cb 3 4 Air Air Continuity Eq Am E A2E A Am E A1E A Energy Eq 412 Am E A1E AhA1E A AW E AinE A Am E AairE AhA3E A Am E A2E AhA2E A Am E AairE AhA4E A Ideal gas for air and constant specific heat hA4E A hA3E A CApE A AairE A TA4E A TA3E A Am E AairE A Am E A1E A hA1E A hA2E A AW E AinE A CApE A AairE A TA4E A TA3E A A01 39384 42025 3 E1004 30 20E A A0359 10E A A kW kJkgE 00359 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 489 Two steady flows of air enters a control volume shown in Fig P489 One is 0025 kgs flow at 350 kPa 150C state 1 and the other enters at 450 kPa 15C state 2 A single flow of air exits at 100 kPa 40C state 3 The control volume rejects 1 kW heat to the surroundings and produces 4 kW of power Neglect kinetic energies and determine the mass flow rate at state 2 Solution CV Steady device with two inlet and one exit flows we neglect kinetic energies Notice here the Q is rejected so it goes out 1 2 3 Engine Q loss W Continuity Eq49 Am E A1E A Am E A2E A Am E A3E A 0025 Am E A2E Energy Eq410 AmE A1E AhA1E A AmE A2E AhA2E A AmE A3E AhA3E A AW E ACVE A AQ E AlossE Substitute the work and heat transfer into the energy equation and use constant specific heat 0025 1004 4232 AmE A2E A 1004 2882 0025 AmE A2E A 1004 2332 40 10 Now solve for AmE A2E A AmE A2E A A40 10 0025 1004 2332 4232 E1004 2882 2332E A kWkJkg 00042 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 490 A large expansion engine has two low velocity flows of water entering High pressure steam enters at point 1 with 20 kgs at 2 MPa 500C and 05 kgs cooling water at 120 kPa 30C enters at point 2 A single flow exits at point 3 with 150 kPa 80 quality through a 015 m diameter exhaust pipe There is a heat loss of 300 kW Find the exhaust velocity and the power output of the engine Solution CV Engine Steady state Constant rates of flow Q loss and AW E State 1 Table B13 hA1E A 34676 kJkg State 2 Table B11 hA2E A 12577 kJkg h3 4671 08 22265 22483 kJkg 1 2 3 Engine Q loss W v3 000105 08 115825 092765 mA3E Akg Continuity Eq49 Am E A1E A Am E A2E A Am E A3E A 2 05 25 kgs AVv π4DA2E AVv Energy Eq410 Am E A1E AhA1E A Am E A2E AhA2E A Am E A3E AhA3E A 05 VA2E A Q loss AW E V Am E A3E AvA3E A Aπ 4E A DA2E A 25 092765 07854 015A2E A 1312 ms 05 V2 05 1312A2E A mA2E AsA2E A1000 JkJ 86 kJkg AW E A 2 34676 05 12577 25 22483 86 300 1056 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Heat Exchangers Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 491 A condenser heat exchanger brings 1 kgs water flow at 10 kPa quality 95 to saturated liquid at 10 kPa as shown in Fig P491 The cooling is done by lake water at 20C that returns to the lake at 30C For an insulated condenser find the flow rate of cooling water Solution CV Heat exchanger 1 95 4 30C 3 20C m cool 1 kgs 2 sat liq Table B11 hA20E A 8394 kJkg hA30E A 12577 kJkg Table B12 hA95 10kPaE A 19181 095 239282 2465 kJkg hAf 10 kPaE A 19181 kJkg Energy Eq410 Am E AcoolE AhA20E A AmE AH2OE AhA300E A A m E AcoolE AhA30E A AmE AH2OE AhAf 10 kPaE Am E AcoolE A AmE AH2OE A A h95 hfE 10kPa h30 h20 E A 1 kgs A2465 19181 12577 8394E A 543 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 492 Air at 600 K flows with 3 kgs into a heat exchanger and out at 100AoE AC How much kgs water coming in at 100 kPa 20AoE AC can the air heat to the boiling point CV Total heat exchanger The flows are not mixed so the two flowrates are constant through the device No external heat transfer and no work Energy Eq410 Am E AairE AhAair inE A Am E AwaterE AhAwater inE A Am E AairE AhAair outE A Am E AwaterE AhAwater outE Am E AairE AhAair inE A hAair outE A Am E AwaterE AhAwater outE A hAwater inE A Table B12 hAwater outE A hAwater inE A 267546 8394 25915 kJkg Table A71 hAair inE A hAair outE A 60732 37414 23318 kJkg Solve for the flow rate of water from the energy equation Am E AwaterE A Am E AairE A hair in hair out hwater out hwater in 3 kgs A23318 25915E A 027 kgs Air in cb Air out Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 493 Steam at 500 kPa 300AoE AC is used to heat cold water at 15AoE AC to 75AoE AC for domestic hot water supply How much steam per kg liquid water is needed if the steam should not condense Solution CV Each line separately No work but there is heat transfer out of the steam flow and into the liquid water flow Water line energy Eq Am E AliqE AhAiE A AQ E A Am E AliqE AhAeE A AQ E A Am E AliqE AhAeE A hAiE A For the liquid water look in Table B11 hAliqE A hAeE A hAiE A 31391 6298 25093 kJkg CApE A T 418 75 15 2508 kJkg Steam line energy has the same heat transfer but it goes out Steam Energy Eq Am E AsteamE AhAiE A AQ E A Am E AsteamE AhAeE A AQ E A Am E AsteamE AhAiE A hAeE A For the steam look in Table B13 at 500 kPa hAsteamE A hAiE A hAeE A 30642 274867 31553 kJkg Now the heat transfer for the steam is substituted into the energy equation for the water to give Am E AsteamE A Am E AliqE A hAliqE A hAsteamE A A25093 31553E A 0795 cb Steam in Steam out Cold water in Hot water out Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 494 A dualfluid heat exchanger has 5 kgs water entering at 40C 150 kPa and leaving at 10C 150 kPa The other fluid is glycol entering at 10C 160 kPa and leaving at 10C 160 kPa Find the required mass flow rate of glycol and the rate of internal heat transfer CV Heat exchanger steady flow 1 inlet and 1 exit for glycol and water each The two flows exchange energy with no heat transfer tofrom the outside 3 glycol 1 water 4 2 Continuity Eqs Each line has a constant flow rate through it Energy Eq410 Am E AH2OE A hA1E A Am E AglycolE A hA3E A Am E AH2OE A hA2E A Am E AglycolE A hA4E Process Each line has a constant pressure Table B1 hA1E A 16754 kJkg hA2E A 4199 kJkg Table A4 CAP glyE A 242 kJkgK so hA4E A hA3E A CAP glyE A TA4E A TA3E A 242 10 10 484 kJkg Am E AglycolE A Am E AH2OE A A h1 h2 Eh4 h3 E A 5 A16754 4199 484E A 1297 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 495 A heat exchanger shown in Fig P495 is used to cool an air flow from 800 K to 360 K both states at 1 MPa The coolant is a water flow at 15C 01 MPa If the water leaves as saturated vapor find the ratio of the flow rates AmE AH2OE AAmE AairE Solution CV Heat exchanger steady flow 1 inlet and 1 exit for air and water each The two flows exchange energy with no heat transfer tofrom the outside 3 water 1 air 4 2 Continuity Eqs Each line has a constant flow rate through it Energy Eq410 Am E AairE AhA1E A Am E AH2OE AhA3E A Am E AairE AhA2E A Am E AH2OE AhA4E Process Each line has a constant pressure Air states Table A71 hA1E A 82220 kJkg hA2E A 36086 kJkg Water states Table B11 hA3E A 6298 kJkg at 15C Table B12 hA4E A 26755 kJkg at 100 kPa Am E AH2OE AAm E AairE A A h1 h2 Eh 4 h3 E A A82220 36086 26755 6299E A 01766 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 496 A superheater brings 25 kgs saturated water vapor at 2 MPa to 450AoE AC The energy is provided by hot air at 1200 K flowing outside the steam tube in the opposite direction as the water which is a counter flowing heat exchanger Find the smallest possible mass flow rate of the air so the air exit temperature is 20AoE AC larger than the incoming water temperature so it can heat it Solution CV Superheater Steady state with no external AQ E A or any AW E A the two flows exchanges energy inside the box Neglect kinetic and potential energies at all states 3 water 1 air 4 2 Energy Eq410 AmE AH2OE A hA3E A Am E AairE A hA1E A AmE AH2OE A hA4E A Am E AairE A hA2E Process Constant pressure in each line State 1 Table B12 TA3E A 21242C hA3E A 279951 kJkg State 2 Table B13 hA4E A 335748 kJkg State 3 Table A7 hA1E A 127781 kJkg State 4 TA2E A TA3E A 20 23242C 50557 K A7 hA2E A 50336 A557 20E A 52398 50336 5091 kJkg From the energy equation we get Am E AairE A AmE AH2OE A hA4E A hA3E AhA1E A hA2E A 25 335748 279951 127781 5091 1815 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 497 A two fluid heat exchanger has 2 kgs liquid ammonia at 20AoE AC 1003 kPa entering state 3 and exiting at state 4 It is heated by a flow of 1 kgs nitrogen at 1500 K state 1 leaving at 600 K state 2 similar to Fig P495 Find the total rate of heat transfer inside the heat exchanger Sketch the temperature versus distance for the ammonia and find state 4 T v of the ammonia Solution CV Nitrogen flow line steady rates of flow AQ E A out and AW E A 0 Continiuty Am E A1 Am E A2 1 kgs Energy Eq Am E A1h1 Am E A2h2 AQ E Aout Tbl A8 h1 16807 kJkg h2 62724 kJkg AQ E Aout Am E A1h1 h2 1 16807 62724 10535 kW If Tbl A5 is used Cp 1042 kJkg K AQ E Aout Am E A1 Cp T1 T2 11042 1500 600 9378 kW CV The whole heat exchanger No external AQ E A constant pressure in each line Am E A1h1 Am E A3h3 Am E A1h2 Am E A3h4 h4 h3 Am E A1h1 h2Am E A3 h4 2743 10535 2 801 kJkg hg 2phase x4 h4 hf hfg 801 29825 11652 043147 v4 vf x4 vfg 0001658 043147012647 005623 mA3E Akg T4 T3a 25AoE AC This is the boiling temperature for 1003 kPa T x 293 298 3 3a 4 3 NH 1 N 4 2 3 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 498 In a coflowing same direction heat exchanger 1 kgs air at 500 K flows into one channel and 2 kgs air flows into the neighboring channel at 300 K If it is infinitely long what is the exit temperature Sketch the variation of T in the two flows CV Heat Exchanger no AW E A Continuity Eq Am E A1E A Am E A3E A and Am E A2E A Am E A4E A Energy Eq410 Am E A1E AhA1E A Am E A2E AhA2E A Am E A1E AhA3E A Am E A2E AhA4E Same exit T hA3E A hA4E A Am E A1E AhA1E A Am E A2E AhA2E A Am E A1E A Am E A2E A Using conctant specific heat TA3E A TA4E A A m AE 1 m 1 m 2 E TA1E A A m AE 2 m 1 m 2 E TA2E A A1 3E A 500 A2 3E A 300 367 K x cb 3 4 1 2 T x 300 500 1 T 2 T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 499 An airwater counter flowing heat exchanger has one line with 2 kgs at 125 kPa 1000 K entering and the air is leaving at 100 kPa 400 K The other line has 05 kgs water entering at 200 kPa 20AoE AC and leaving at 200 kPa What is the exit temperature of the water CV Heat Exchanger no AW E A Continuity Eq Am E A1E A Am E A3E A Am E AH2OE A and Am E A2E A Am E A4E A Am E AairE A Energy Eq410 Am E AH2OE AhA1E A Am E AairE AhA4E A Am E AH2OE AhA3E A Am E AairE AhA2E Flowrates and properties known except for TA1E hA1E A hA3E A Am E AairE A hA2E A hA4E A Am E AH2OE 8394 2 104622 4013 05 26636 kJkg hA1E A hA1E A 27066 kJkg at 200 kPa so twophase T 12023AoE AC x cb 3 4 1 2 T x 393 1000 3 T 2 T The water is not completely evaporated at the exit Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4100 An automotive radiator has glycerine at 95AoE AC enter and return at 55AoE AC as shown in Fig P4100 Air flows in at 20AoE AC and leaves at 25AoE AC If the radiator should transfer 25 kW what is the mass flow rate of the glycerine and what is the volume flow rate of air in at 100 kPa Solution If we take a control volume around the whole radiator then there is no external heat transfer it is all between the glycerin and the air So we take a control volume around each flow separately Glycerine Am E AhAiE A AQ E A Am E AhAeE Table A4 Am E AglyE A A Q Ehe hi E A A Q ECglyTeTiE A A 25 24255 95E A 0258 kgs Air Am E AhAiE A AQ E A Am E AhAeE Table A5 Am E AairE A A Q Ehe hi E A A Q ECairTeTiE A A 25 100425 20E A 498 kgs AV E A Am E AvAiE A vAiE A A RTi EPi E A A0287 293 100E A AkJkgK K kPaE A 08409 mA3E Akg AV E AairE A Am E AvAiE A 498 08409 419 mA3E As Exhaust flow Air intake filter Coolant flow 55 C Atm air Shaft power 95 C o o cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4101 A cooler in an air conditioner brings 05 kgs air at 35AoE AC to 5AoE AC both at 101 kPa and it then mix the output with a flow of 025 kgs air at 20AoE AC 101 kPa sending the combined flow into a duct Find the total heat transfer in the cooler and the temperature in the duct flow Solution 1 2 3 4 Q cool Cooler section Mixing section CV Cooler section no AW E A Energy Eq412 Am E Ah1 Am E AhA2E A Q cool Q cool Am E Ah1 hA2E A Am E A CApE A TA1E A TA2E A 05 1004 355 1506 kW CV mixing section no AW E A AQ E A Continuity Eq Am E A2E A Am E A3E A Am E A4E A Energy Eq410 Am E A2E AhA2E A Am E A3E AhA3E A Am E A4E AhA4E Am E A4E A Am E A2E A Am E A3E A 05 025 075 kgs Am E A4E AhA4E A Am E A2E A Am E A3E AhA4E A Am E A2E AhA2E A Am E A3E AhA3E Am E A2E A hA4E A hA2E A Am E A3E A hA4E A hA3E A Ø Am E A2E A CApE A TA4E A TA2E A Am E A3E A CApE A TA4E A TA3E A Ø TA4E A Am E A2E A Am E A4E A TA2E A Am E A3E A Am E A4E A TA3E A 505075 20025075 10C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4102 A copper wire has been heat treated to 1000 K and is now pulled into a cooling chamber that has 15 kgs air coming in at 20AoE AC the air leaves the other end at 60AoE AC If the wire moves 025 kgs copper how hot is the copper as it comes out Solution CV Total chamber no external heat transfer Energy eq Am E AcE AuE A h AiE AcuE A Am E AairE A hAiE A AairE A Am E AcuE A hAe cuE A Am E AairE A hAe airE Am E AcuE A hAeE A hAiE A AcuE A Am E AairE A hAiE A hAeE A AairE A Am cuE A CAcuE A TAeE A TAiE A AcuE A Am airE A CAp airE A TAeE A TAiE A AairE Heat capacities from A3 for copper and A5 for air TAeE A TAiE AAcuE A A m airCp air Em cuCcu E A TAeE A TAiE AAairE A A15 1004 025 042E A 20 60 K 5737 K TAeE A TAiE A 5737 1000 5737 4263 K Air Air Cu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4103 A coflowing samedirection heat exchanger has one line with 025 kgs oxygen at 17C 200 kPa entering and the other line has 06 kgs nitrogen at 150 kPa 500 K entering The heat exchanger is very long so the two flows exit at the same temperature Use constant heat capacities and find the exit temperature CV Heat Exchanger no AW E A Continuity Eq Am E A1E A Am E A3E A and Am E A2E A Am E A4E A Energy Eq410 Am E A1E AhA1E A Am E A2E AhA2E A Am E A1E AhA3E A Am E A2E AhA4E Same exit T TA3E A TA4E A TAexE Using conctant specific heat Am E A1E A CAP O2E ATA1E A Am E A2E A CAP N2E ATA2E A Am E A1E A CAP O2E ATAexE A Am E A2E A CAP N2E ATAexE TAexE A A m AE 1CP O2 m 1CP O2 m 2CP N2 E TA1E A A m AE 2CP N2 m 1CP O2 m 2CP N2 E TA2E A A025 0922 290 06 1042 500 025 0922 06 1042E A 4434 K x cb 3 4 1 2 T x 290 500 1 T 2 T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Mixing processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4104 Two flows of air are both at 200 kPa one has 1 kgs at 400 K and the other has 2 kgs at 290 K The two flows are mixed together in an insulated box to produce a single exit flow at 200 kPa Find the exit temperature Solution Cont Am E A1E A Am E A2E A Am E A3E Energy Am E A1E AhA1E A Am E A2E AhA2E A Am E A3E AhA3E A Am E A1E A Am E A2E AhA3E 1 2 3 Mixing chamber Solve for the exit enthalpy hA3E A Am E A1E AhA1E A Am E A2E AhA2E A Am E A1E A Am E A2E A since the Ts are modest use constant specific heats TA3E A Am E A1E ATA1E A Am E A2E ATA2E A Am E A1E A Am E A2E A A1 3E A 400 A2 3E A 290 3267 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4105 Two air flows are combined to a single flow Flow one is 1 mA3E As at 20AoE AC and the other is 2 mA3E As at 200AoE AC both at 100 kPa They mix without any heat transfer to produce an exit flow at 100 kPa Neglect kinetic energies and find the exit temperature and volume flow rate Solution Cont Am E A1E A Am E A2E A Am E A3E Energy Am E A1E AhA1E A Am E A2E AhA2E A Am E A3E AhA3E A Am E A1E A Am E A2E AhA3E 1 2 3 Mixing section Am E A1E A hA3E A hA1E A Am E A2E A hA3E A hA2E A 0 Am E A1E ACApE A TA3E ATA1E A Am E A2E ACApE A TA3E ATA2E A 0 TA3E A Am E AiE AAm E A3E ATA1E A Am E A2E AAm E A3E ATA2E We need to find the mass flow rates vA1E A RTA1E APA1E A 0287 kJkgK 293 K100 kPa 08409 mA3E Akg vA2E A RTA2E APA2E A 0287 kJkgK 473 K100 kPa 13575 mA3E Akg Am E A1E A A V 1 Ev1 E A A 1 08409E A 11892 Akg sE A Am E A2E A A V 2 Ev2 E A A 2 13575E A 14733 Akg sE Am E A3E A Am E A1E A Am E A2E A 26625 kgs TA3E A A11892 26625E A 20 A14733 26625E A 200 1196AoE A C vA3E A A RT3 EP3 E A A0287 1196 273 E100E A 11268 mA3E Akg AV E A3E A Am E A3E A vA3E A 26625 kgs 11268 mA3E Akg 30 mA3E As Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4106 A flow of water at 2000 kPa 20AoE AC is mixed with a flow of 2 kgs water at 2000 kPa 180AoE AC What should the flowrate of the first flow be to produce an exit state of 200 kPa and 100AoE AC Solution CV Mixing chamber and valves Steady state no heat transfer or work terms Continuity Eq49 m 1 m 2 m 3 Energy Eq410 Am E A1E AhA1E A Am E A2E AhA2E A Am E A3E AhA3E A Am E A1E A Am E A2E AhA3E 1 2 3 MIXING CHAMBER Properties Table B11 hA1E A 858 kJkg hA3E A 4190 kJkg Table B14 hA2E A 7637 kJkg AmE A1E A AmE A2E A A h2 h3 Eh3 h1 E A 2 A7637 4190 4190 858E A 2069 kgs 2 P v 3 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4107 An open feedwater heater in a powerplant heats 4 kgs water at 45oC 100 kPa by mixing it with steam from the turbine at 100 kPa 250oC Assume the exit flow is saturated liquid at the given pressure and find the mass flow rate from the turbine Solution CV Feedwater heater No external Q or W 1 2 3 MIXING CHAMBER cb Continuity Eq49 m 1 m 2 m 3 Energy Eq410 m 1h1 m 2h2 m 3h3 m 1 m 2h 3 State 1 Table B11 h hf 18842 kJkg at 45oC State 2 Table B13 h2 297433 kJkg State 3 Table B12 h3 hf 41744 kJkg at 100 kPa m 2 m 1 h1 h3 h3 h2 4 18842 41744 41744 297433 0358 kgs T v 1 2 3 100 kPa 2 P v 1 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4108 Two flows are mixed to form a single flow Flow at state 1 is 15 kgs water at 400 kPa 200oC and flow at state 2 is 500 kPa 100oC Which mass flow rate at state 2 will produce an exit T3 150oC if the exit pressure is kept at 300 kPa Solution CV Mixing chamber and valves Steady state no heat transfer or work terms Continuity Eq49 m 1 m 2 m 3 Energy Eq410 m 1h1 m 2h2 m 3h3 m 1 m 2h 3 1 2 3 MIXING CHAMBER Properties Table B13 h1 286051 kJkg h3 276095 kJkg Table B14 h2 41932 kJkg m 2 m 1 h1 h3 h3 h2 15 286051 276095 276095 41932 00638 kgs 2 P v 3 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4109 A desuperheater has a flow of ammonia 15 kgs at 1000 kPa 100oC which is mixed with another flow of ammonia at 25oC and quality 50 in an adiabatic mixing chamber Find the flow rate of the second flow so the outgoing ammonia is saturated vapor at 1000 kPa CV Desuperheater No external Q or W Continuity Eq49 m 1 m 2 m 3 Energy Eq410 m 1h1 m 2h2 m 3h3 m 1 m 2h 3 State 1 Table B22 h1 16643 kJkg State 2 Table B21 h2 29825 05 11652 88085 kJkg State 3 Table B22 h3 14634 kJkg m 2 m 1 h1 h3 h3 h2 15 16643 14634 14634 88085 0517 kgs 1 2 3 MIXING CHAMBER cb T v 1 2 3 1000 kPa 2 P v 3 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4110 A mixing chamber with heat transfer receives 2 kgs of R410A at 1 MPa 40C in one line and 1 kgs of R410A at 15C quality 50 in a line with a valve The outgoing flow is at 1 MPa 60C Find the rate of heat transfer to the mixing chamber Solution CV Mixing chamber Steady with 2 flows in and 1 out heat transfer in 1 2 3 Heater Mixer Q Continuity Eq49 m 1 m 2 m 3 m 3 2 1 3 kgs Energy Eq410 m 1h1 m 2h2 Q m 3h3 Properties Table B42 h1 31605 kJkg h3 33575 kJkg Table B41 h2 8115 05 20164 18197 kJkg Energy equation then gives the heat transfer as Q 3 33575 2 31605 1 18197 19318 kW 2 P v 13 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4111 A geothermal supply of hot water at 500 kPa 150C is fed to an insulated flash evaporator at the rate of 15 kgs A stream of saturated liquid at 200 kPa is drained from the bottom of the chamber and a stream of saturated vapor at 200 kPa is drawn from the top and fed to a turbine Find the mass flow rate of the two exit flows CV Flash chamber Disregard any kinetic and potential energies Continuity Eq 0 m 1 m 2 m 3 Energy Eq 0 m 1h1 m 2h2 m 3h3 Process Q 0 W 0 h1 6322 kJkg from Table B14 B11 63218 State 2 and 3 Saturated vapor h2 270663 kJkg Saturated liquid h3 50468 kJkg Substitute m 3 from continuity eq into the energy eq to get 0 m 1h1 m 2h2 m 1 m 2 h3 m 2 m 1 h1 h3 h2 h3 m 1 h1 hf hg hf 15 6322 50468 270663 50468 00869 kgs m 3 m 1 m 2 15 00869 1413 kgs The fraction of the flow at state 2 equals the quality x in the flow right after the valve and the fraction going out at state 3 is 1 x 1 2 3 Twophase out of the valve The liquid drops to the bottom Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4112 An insulated mixing chamber receives 2 kgs R134a at 1 MPa 100C in a line with low velocity Another line with R134a as saturated liquid 60C flows through a valve to the mixing chamber at 1 MPa after the valve as shown in Fig P4110 The exit flow is saturated vapor at 1 MPa flowing at 20 ms Find the flow rate for the second line Solution CV Mixing chamber Steady state 2 inlets and 1 exit flow Insulated q 0 No shaft or boundary motion w 0 Continuity Eq49 m 1 m 2 m 3 Energy Eq410 m 1h1 m 2h2 m 3 h3 1 2 V2 3 m 2 h2 h3 1 2 V2 3 m 1 h3 1 2 V2 3 h1 1 Table B52 1 MPa 100C h1 48336 kJkg 2 Table B51 x 60C h2 28779 kJkg 3 Table B51 x 1 1 MPa 20 ms h3 41954 kJkg Now solve the energy equation for m 2 m 2 2 41954 1 2 202 1 1000 48336 28779 41954 1 2 202 1000 2 6382 02 13175 02 0964 kgs Notice how kinetic energy was insignificant 1 2 3 MIXING CHAMBER cb 2 P v 3 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4113 To keep a jet engine cool some intake air bypasses the combustion chamber Assume 2 kgs hot air at 2000 K 500 kPa is mixed with 15 kgs air 500 K 500 kPa without any external heat transfer Find the exit temperature by using constant specific heat from Table A5 Solution CV Mixing Section Continuity Eq49 m 1 m 2 m 3 m 3 2 15 35 kgs Energy Eq410 m 1h1 m 2h2 m 3h3 h3 m 1h1 m 2h2 m 3 For a constant specific heat divide the equation for h3 with Cp to get T3 m 1 m 3 T1 m 2 m 3 T2 2 35 2000 15 35 500 1357 K 1 2 3 Mixing section Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4114 Solve the previous problem using values from Table A7 To keep a jet engine cool some intake air bypasses the combustion chamber Assume 2 kgs hot air at 2000 K 500 kPa is mixed with 15 kgs air 500 K 500 kPa without any external heat transfer Find the exit temperature by using values from Table A7 Solution CV Mixing Section Continuity Eq49 m 1 m 2 m 3 m 3 2 15 35 kgs Energy Eq410 m 1h1 m 2h2 m 3h3 h3 m 1h1 m 2h2 m 3 Using A7 we look up the h at states 1 and 2 to calculate h3 h3 m 1 m 3 h1 m 2 m 3 h2 2 35 225158 15 35 50336 1502 kJkg Now we can backinterpolate to find at what temperature do we have that h T3 1350 50 1502 145543 151527 145543 1389 K This procedure is the most accurate 1 2 3 Mixing section Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Multiple Devices Cycle Processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4115 A flow of 5 kgs water at 100 kPa 20oC should be delivered as steam at 1000 kPa 350oC to some application Consider compressing it to 1000 kPa 20oC and then heat it at constant 1000 kPa to 350oC Which devices are needed and find the specific energy transfers in those devices To raise the pressure of a liquid flow requires a pump which delivers the difference between the flow work going out and in Constant pressure heating is a simple heater or heat exchanger Inlet state B11 h1 8394 kJkg v1 0001002 m3kg State 3 h3 315765 kJkg The pump delivers the difference between the flow work out and flow work in Pump wp P2 P1 v1 1000 100 kPa 0001002 m3kg 09 kJkg Heater q h3 h2 h3 h1 wp 315765 8394 09 30728 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4116 A twostage compressor takes nitrogen in at 20oC 150 kPa and compresses it to 600 kPa 450 K Then it flows through an intercooler where it cools to 320 K and the second stage compresses it to 3000 kPa 530 K Find the specific work in each of the two compressor stages and the specific heat transfer in the intercooler The setup is like this CV Stage 1 nitrogen Steady flow Process adiabatic q 0 Energy Eq 413 wC1 h2 h1 Assume constant CP0 1042 from A5 wC1 h1 h2 CP0T1 T2 1042 kJkgK 293 450 K 1636 kJkg CV Intercooler no work and no changes in kinetic or potential energy q23 h3 h2 CP0T3 T2 1042 320 450 1355 kJkg CV Stage 2 Analysis the same as stage 1 wC2 h3 h4 CP0T3 T4 1042 320 530 2188 kJkg C1 C2 1 2 3 4 2 W Q 1 W cooler inter Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4117 The intercooler in the previous problem uses cold liquid water to cool the nitrogen The nitrogen flow is 01 kgs and the liquid water inlet is 20oC and is set up to flow in the opposite direction from the nitrogen so the water leaves at 35oC Find the flow rate of the water Solution CV Heat exchanger steady 2 flows in and two flows out 1 3 2 4 Continuity eq m 1 m 2 m H2O m 3 m 4 m N2 Energy eq 0 m H2Oh3 h4 m N2 h2 h1 Due to the lower range of temperature we will use constant specific heats from A5 and A4 so the energy equation is approximated as 0 m H2OCpH2O T3 T4 m N2Cp T2 T1 Now solve for the water flow rate m H2O m N2 CpN2 T2 T1 CpH2O T4 T3 01 kgs 1042 450 320 418 35 20 0216 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4118 The following data are for a simple steam power plant as shown in Fig P4118 State 1 2 3 4 5 6 7 P MPa 62 61 59 57 55 001 0009 T C 45 175 500 490 40 h kJkg 194 744 3426 3404 168 State 6 has x6 092 and velocity of 200 ms The rate of steam flow is 25 kgs with 300 kW power input to the pump Piping diameters are 200 mm from steam generator to the turbine and 75 mm from the condenser to the steam generator Determine the velocity at state 5 and the power output of the turbine Solution Turbine A5 π4022 0031 42 m2 v5 006163 m3kg V5 m v5A5 25 kgs 0061 63 m3kg 0031 42 m2 49 ms h6 19183 092 23928 23932 kJkg wT h5 h6 1 2 V2 5 V2 6 3404 23932 492 2002 2 1000 992 kJkg W T m wT 25 kgs 992 kJkg 24 800 kW Remark Notice the kinetic energy change is small relative to enthalpy change Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4119 For the same steam power plant as shown in Fig P4118 assume the cooling water comes from a lake at 15C and is returned at 25C Determine the rate of heat transfer in the condenser and the mass flow rate of cooling water from the lake Solution Condenser A7 π400752 0004 418 m2 v7 0001 008 m3kg V7 m v7A7 25 0001 008 0004 418 57 ms h6 19183 092 23928 23932 kJkg qCOND h7 h6 1 2 V2 7 V2 6 168 23932 572 2002 21000 22452 kJkg Q COND 25 kgs 22452 kJkg 56 130 kW This rate of heat transfer is carried away by the cooling water so Q COND m H2Ohout hinH2O 56 130 kW m H2O 56 130 1049 630 kW kJkg 13396 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4120 For the same steam power plant as shown in Fig P4118 determine the rate of heat transfer in the economizer which is a low temperature heat exchanger Find also the rate of heat transfer needed in the steam generator Solution Economizer A7 πD2 74 0004 418 m2 v7 0001 008 m3kg V2 V7 m v7A7 25 0001 0080004 418 57 ms V3 v3v2V2 0001 118 0001 008 57 63 ms V2 so kinetic energy change unimportant qECON h3 h2 744 194 5500 kJkg Q ECON m qECON 25 kgs 5500 kJkg 13 750 kW Generator A4 πD2 44 0031 42 m2 v4 0060 23 m3kg V4 m v4A4 25 kgs 0060 23 m3kg 0031 42 m2 479 ms qGEN 3426 744 4792 63221000 2683 kJkg Q GEN m qGEN 25 kgs 2683 kJkg 67 075 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4121 A somewhat simplified flow diagram for a nuclear power plant is given in Fig P4121 Mass flow rates and the various states in the cycle are shown in the accompanying table The cycle includes a number of heaters in which heat is transferred from steam taken out of the turbine at some intermediate pressure to liquid water pumped from the condenser on its way to the steam drum The heat exchanger in the reactor supplies 157 MW and it may be assumed that there is no heat transfer in the turbines a Assume the moisture separator has no heat transfer between the two turbinesections determine the enthalpy and quality h4 x4 b Determine the power output of the lowpressure turbine c Determine the power output of the highpressure turbine d Find the ratio of the total power output of the two turbines to the total power delivered by the reactor Solution HP W 2 3 17 12 moisture separator W LP 9 4 5 8 a Moisture Separator steady state no heat transfer no work Continuity Eq m 3 m 4 m 9 m 4 m 3 m 9 62874 4662 58212 kgs Energy Eq m 3h3 m 4h4 m 9h9 From the energy equation we get h4 9h9 m 4 m 3h3 m 62874 2517 4662 558 58212 26739 kJkg h4 26739 56618 x4 21606 x4 09755 b Low Pressure Turbine steady state no heat transfer Continuity Eq m 4 m 5 m 8 Energy Eq m 4h4 m 5h5 m 8h8 W CVLP m 5 m 4 m 8 58212 2772 5544 kgs W CVLP m 4h4 m 5h5 m 8h8 58212 26739 5544 2279 2772 2459 22 489 kW 22489 MW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful c High Pressure Turbine steady state no heat transfer Energy Eq m 2h2 m 3h3 m 12h12 m 17h17 W CVHP W CVHP m 2h2 m 3h3 m 12h12 m 17h 17 756 2765 62874 2517 8064 2517 4662 2593 18 394 kW 18394 MW d η W HP W LPQ REACT 22489 18394 157 026 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4122 Consider the power plant as described in the previous problem a Determine the quality of the steam leaving the reactor b What is the power to the pump that feeds water to the reactor Solution a Reactor Cont m 20 m 21 Q CV 157 MW Energy Eq412 Q CV m 20h20 m 21h 21 21 19 20 Q 157 000 kW 1386 kgs 1221 kJkg 1386 kgs h21 h21 13343 kJkg 12824 x21 14583 kJkg x21 00349 b CV Reactor feedwater pump Cont m 19 m 20 Energy Eq m 19h19 m 19h20 W CvP Table B1 h19 h277C 7240 kPa 1220 kJkg h20 1221 kJkg W CvP m 19h19 h20 1386 kgs 1220 1221 kJkg 1386 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4123 A R410A heat pump cycle shown in Fig P4123 has a R410A flow rate of 005 kgs with 5 kW into the compressor The following data are given State 1 2 3 4 5 6 P kPa 3100 3050 3000 420 400 390 T C 120 110 45 10 5 h kJkg 377 367 134 280 284 Calculate the heat transfer from the compressor the heat transfer from the R 410A in the condenser and the heat transfer to the R410A in the evaporator Solution CV Compressor Q COMP m h1 h6 W COMP 005 377 284 50 035 kW CV Condenser Q COND m h3 h2 005 kgs 134 367 kJkg 1165 kW CV Valve h4 h3 134 kJkg CV Evaporator Q EVAP m h5 h4 005 kgs 280 134 kJkg 73 kW v P compressor evaporator condenser valve 2 1 3 4 5 6 Q H W C Q L cb Evaporator Condenser 3 4 5 6 1 2 Q loss Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4124 A modern jet engine has a temperature after combustion of about 1500 K at 3200 kPa as it enters the turbine setion see state 3 Fig P4124 The compressor inlet is 80 kPa 260 K state 1 and outlet state 2 is 3300 kPa 780 K the turbine outlet state 4 into the nozzle is 400 kPa 900 K and nozzle exit state 5 at 80 kPa 640 K Neglect any heat transfer and neglect kinetic energy except out of the nozzle Find the compressor and turbine specific work terms and the nozzle exit velocity Solution The compressor turbine and nozzle are all steady state single flow devices and they are adiabatic We will use air properties from table A71 h1 26032 h2 80028 h3 163580 h4 93315 h5 64953 kJkg Energy equation for the compressor gives wc in h2 h1 80028 26032 53936 kJkg Energy equation for the turbine gives wT h3 h4 163580 93315 70265 kJkg Energy equation for the nozzle gives h4 h5 ½ V2 5 ½ V2 5 h4 h5 93315 64953 28362 kJkg V5 2 h4 h5 12 2 28362 1000 12 753 ms cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4125 A proposal is made to use a geothermal supply of hot water to operate a steam turbine as shown in Fig P4125 The highpressure water at 15 MPa 180C is throttled into a flash evaporator chamber which forms liquid and vapor at a lower pressure of 400 kPa The liquid is discarded while the saturated vapor feeds the turbine and exits at 10 kPa 90 quality If the turbine should produce 1 MW find the required mass flow rate of hot geothermal water in kilograms per hour Solution Separation of phases in flashevaporator constant h in the valve flow so Table B13 h1 7635 kJkg h1 7635 60474 x 21338 x 007439 m 2m 1 Table B12 h2 27386 kJkg FLASH EVAP H O 2 Sat liquid out Sat vapor W Turb 1 2 3 4 h3 19183 09 23928 23454 kJkg Energy Eq412 for the turbine W m 2h2 h3 m 2 1000 27386 23454 kW kJkg 2543 kgs m 1 m 2x 3419 kgs 123 075 kgh Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Transient processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4126 An initially empty cylinder is filled with air from 20oC 100 kPa until it is full Assuming no heat transfer is the final temperature larger equal to or smaller than 20oC Does the final T depend on the size of the cylinder This is a transient problem with no heat transfer and no work The balance equations for the tank as CV become Continuity Eq m2 0 mi Energy Eq m2u2 0 mihi Q W mihi 0 0 Final state u2 hi P2 P i T2 Ti and it does not depend on V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4127 An initially empty canister of volume 02 m3 is filled with carbon dioxide from a line at 800 kPa 400 K Assume the process runs until it stops by itself and it is adiabatic Use constant specific heat to find the final temperature in the canister CV Canister and valve transient process with no heat transfer or work Continuity Eq415 m2 m1 min Energy Eq416 m2u2 m1u1 min hin 1Q2 1W 2 Process 1Q2 0 V constant so 1W2 0 State 1 m1 0 m2 min State 2 P2 Pline 800 kPa one more property Energy Eq u2 hin uin RTin Cvo T2 Cvo Tin RTin CPoTin T2 CPoCvo Tin k Tin 1289 400 K 5156 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4128 Repeat the previous problem but use the ideal gas Tables A8 to solve it CV Canister and valve transient process with no heat transfer or work Continuity Eq415 m2 m1 min Energy Eq416 m2u2 m1u1 min hin 1Q2 1W 2 Process 1Q2 0 V constant so 1W2 0 State 1 m1 0 m2 min State 2 P2 Pline 800 kPa one more property Energy Eq u2 hin 30376 kJkg from A8 back interpolate for u2 T2 4959 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4129 A tank contains 1 m3 air at 100 kPa 300 K A pipe flowing air at 1000 kPa 300 K is connected to the tank and it is filled slowly to 1000 kPa Find the heat transfer to reach a final temperature of 300 K CV The tank volume and the compressor This is a transient problem filling of tank Continuity Eq415 m2 m1 min Q i TANK 1 2 Energy Eq416 m2u2 m1u1 1Q2 1W2 minh in Process Constant volume 1W2 0 States u1 u2 uin u300 hin uin RTin m1 P1V1RT1 100 kPa 1 m3 0287 kJkgK 300 K 11614 kg m2 P2V2RT2 1000 kPa 1 m30287 kJkgK 300 K 116144 kg Heat transfer from the energy equation 1Q2 m2u2 m1u1 minhin m1 min u1 m1u1 minuin minRTin m1u1 m1u1 minu1 minuin minRTin minRT in 116144 11614 kg 0287 kJkgK 300 K 900 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4130 A 1m3 tank contains ammonia at 150 kPa 25C The tank is attached to a line flowing ammonia at 1200 kPa 60C The valve is opened and mass flows in until the tank is half full of liquid by volume at 25C Calculate the heat transferred from the tank during this process Solution CV Tank Transient process as flow comes in State 1 Table B22 interpolate between 20 C and 30C v1 09552 m3kg u1 13806 kJkg m1 Vv1 109552 1047 kg State 2 05 m3 liquid and 05 m3 vapor from Table B21 at 25C vf 0001658 m3kg vg 012813 m3kg mLIQ2 050001658 30157 kg mVAP2 05012813 3902 kg m2 30547 kg x2 mVAP2m2 001277 From continuity equation mi m2 m1 30442 kg Table B21 u2 2966 001277 10384 3099 kJkg State inlet Table B22 hi 15533 kJkg Energy Eq416 QCV mihi m2u2 m1u1 QCV 30547 3099 1047 13806 30442 15533 379 636 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4131 A 25L tank initially is empty and we want 10 g of ammonia in it The ammonia comes from a line with saturated vapor at 25C To end up with the desired amount we cool the can while we fill it in a slow process keeping the can and content at 30C Find the final pressure to reach before closing the valve and the heat transfer Solution CV Tank Continuity Eq415 mi m 2 Energy Eq416 m2u2 0 mihi 1Q2 State 2 30C v2 Vm2 00025 m30010 kg 025 m3kg From Table B22 we locate the state between 500 and 600 kPa P2 500 600 500 025 028103 023152 028103 5627 kPa u2 1370 kJkg State i Table B22 hi 14635 kJkg Now use the energy ewquation to solve for the heat transfer 1Q2 m2u2 mihi m2u2 hi 001 kg 1370 14635 kJkg 0935 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4132 An evacuated 150L tank is connected to a line flowing air at room temperature 25C and 8 MPa pressure The valve is opened allowing air to flow into the tank until the pressure inside is 6 MPa At this point the valve is closed This filling process occurs rapidly and is essentially adiabatic The tank is then placed in storage where it eventually returns to room temperature What is the final pressure Solution CV Tank Continuity Eq415 mi m 2 Energy Eq416 mihi m2u2 u2 h i Use constant specific heat CPo from table A5 then energy equation T2 CPCv Ti kTi 14 2982 4175 K Process constant volume cooling to T3 P3 P2 T3T2 60 MPa 29815 K 4175 K 429 MPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4133 An insulated 2m3 tank is to be charged with R 134a from a line flowing the refrigerant at 3 MPa 90C The tank is initially evacuated and the valve is closed when the pressure inside the tank reaches 3 MPa Find the mass in the tank and its final temperature CV Tank and valve transient process with no heat transfer or work Continuity Eq415 m2 m1 min Energy Eq416 m2u2 m1u1 min hin 1Q2 1W 2 Process 1Q2 0 V constant so 1W2 0 State 1 m1 0 m2 min State 2 P2 Pline 3 MPa one more property Energy Eq u2 hin 43619 kJkg State 2 P u interpolate B52 T2 100 10 43619 43377 44648 43377 1019C v2 000665 01904 000734 000665 0006781 m3kg m2 Vv2 2 m30006781 m3kg 2949 kg line R134a Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4134 Find the final state for the previous problem if the valve is closed when the tank reaches 2 MPa CV Tank and valve transient process with no heat transfer or work Continuity Eq415 m2 m1 min Energy Eq416 m2u2 m1u1 min hin 1Q2 1W 2 Process 1Q2 0 V constant so 1W2 0 State 1 m1 0 m2 min State 2 P2 2 MPa one more property Energy Eq u2 hin 43619 kJkg State 2 P u B52 T2 90C very close v2 001137 m3kg m2 Vv2 2 m3001137 m3kg 1759 kg line R134a Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4135 Helium in a steel tank is at 250 kPa 300 K with a volume of 01 m3 It is used to fill a balloon When the tank pressure drops to 150 kPa the flow of helium stops by itself If all the helium still is at 300 K how big a balloon did I get Assume the pressure in the balloon varies linearly with volume from 100 kPa V 0 to the final 150 kPa How much heat transfer did take place Solution Take a CV of all the helium This is a control mass the tank mass changes density and pressure Energy Eq U2 U1 1Q2 1W 2 Process Eq P 100 CV State 1 P1 T1 V1 State 2 P2 T2 V2 Ideal gas P2 V2 mRT2 mRT1 P1V1 c i r c u s t h e r m o cb V2 V1P1P2 01 250150 016667 m3 Vbal V2 V1 016667 01 006667 m 3 1W2 P dV AREA ½ P1 P2 V2 V1 ½ 250 150 kPa 006667 m3 13334 kJ U2 U1 1Q2 1W2 m u2 u1 mCv T2 T1 0 so 1Q2 1W2 13334 kJ Remark The process is transient but you only see the flow mass if you select the tank or the balloon as a control volume That analysis leads to more terms that must be elliminated between the tank control volume and the balloon control volume Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4136 A 25L tank shown in Fig P4136 that is initially evacuated is connected by a valve to an air supply line flowing air at 20C 800 kPa The valve is opened and air flows into the tank until the pressure reaches 600 kPaDetermine the final temperature and mass inside the tank assuming the process is adiabatic Develop an expression for the relation between the line temperature and the final temperature using constant specific heats Solution CV Tank Continuity Eq415 m2 mi Energy Eq416 m2u2 mihi Table A7 u2 hi 29364 kJkg T2 4100 K TANK m2 RT2 P2V 600 0025 0287 410 01275 kg Assuming constant specific heat hi ui RTi u2 RTi u2 ui CvoT2 Ti CvoT2 Cvo R Ti CPoTi T2 Cvo CPo Ti kTi For Ti 2932K constant CPo T2 1402932 4105 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4137 A nitrogen line 300 K and 05 MPa shown in Fig P4137 is connected to a turbine that exhausts to a closed initially empty tank of 50 m3 The turbine operates to a tank pressure of 05 MPa at which point the temperature is 250 K Assuming the entire process is adiabatic determine the turbine work Solution CV turbine tank Transient process Conservation of mass Eq415 mi m2 m Energy Eq416 mihi m2u2 WCV WCV mhi u2 Table B62 Pi 05 MPa Ti 300 K Nitrogen hi 31028 kJkg 2 P2 05 MPa T2 250 K u2 18389 kJkg v2 0154 m3kg m2 Vv2 50 m3 0154 m3kg 3247 kg WCV 3247 kg 31028 18389 kJkg 41 039 kJ 4104 MJ W Turb 1 2 TANK We could with good accuracy have solved using ideal gas and Table A5 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4138 A 1 m3 rigid tank contains 100 kg R410A at ambient temperature 15C A valve on top of the tank is opened and saturated vapor is throttled to ambient pressure 100 kPa and flows to a collector system During the process the temperature inside the tank remains at 15C The valve is closed when no more liquid remains inside Calculate the heat transfer to the tank CV Tank no work and neglect kinetic and potential energies Continuity Eq415 m2 m1 me Energy Eq416 m2u2 m1u1 1Q2 mehe State 1 has v1 Vm1 1100 001 m3kg vg so twophase B41 x1 001 0000904001955 046527 u1 uf x1ufg 8002 x1 1771 16242 kJkg State 2 is saturated vapor T2 15C v2 vg 002045 m3kg u2 ug 25712 kJkg m2 Vv2 1 m3 002045 m3kg 489 kg The exit state e he hg 28279 kJkg 1Q2 m2u2 m1u1 mehe 489 25712 100 16242 100 489 28279 10 782 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4139 A 200 liter tank initially contains water at 100 kPa and a quality of 1 Heat is transferred to the water thereby raising its pressure and temperature At a pressure of 2 MPa a safety valve opens and saturated vapor at 2 MPa flows out The process continues maintaining 2 MPa inside until the quality in the tank is 90 then stops Determine the total mass of water that flowed out and the total heat transfer Solution CV Tank no work but heat transfer in and flow out Denoting State 1 initial state State 2 valve opens State 3 final state Continuity Eq m3 m1 me Energy Eq m3u3 m1u1 mehe 1Q3 e Q sat vap cv State 1 Table B12 v1 vf x1vfg 0001043 001169296 001797 m3kg u1 uf x1ufg 41733 001208872 43822 kJkg m1 Vv1 02 m3001797 m3kg 1113 kg State 3 2MPa v3 vf x3vfg 0001177 09009845 08978 m3kg u3 uf x3ufg 90642 09169384 243088 kJkg m3 Vv3 02 m3008978 m3kg 223 kg Exit state 2MPa he hg 279951 kJkg Hence me m1 m3 1113 kg 223 kg 890 kg Applying the 1st law between state 1 and state 3 1Q3 m3u3 m1u1 mehe 223 243088 1113 43822 890 279951 25 459 kJ 2546 MJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4140 A 1L can of R134a is at room temperature 20C with a quality 50 A leak in the top valve allows vapor to escape and heat transfer from the room takes place so we reach a final state of 5C with a quality of 100 Find the mass that escaped and the heat transfer Solution CV The can of R134a not including the nozzlevalve out to ambient 20oC Continuity Eq m2 m1 me Energy Eq m2u2 m1u1 mehe 1Q2 1W 2 Process Eq V constant 1W2 PdV 0 State 1 Tx v1 vf x1vfg 0000817 05 003524 0018437 m3kg u1 uf x1ufg 22703 05 16216 30811 kJkg m1 V v1 0001 m3 0018437 m3kg 005424 kg State 2 Tx v2 vg 005833 m3kg u2 ug 38085 kJkg m2 Vv2 0001 m3 005833 m3kg 0017144 kg Exit state e Saturated vapor starting at 20oC ending at 5oC so we take an average he 05he1 he2 05 40984 40132 40558 kJkg me m1 m2 00371 kg The heat transfer from the energy equation becomes 1Q2 m2u2 m1u1 mehe 65293 167119 15047 4864 kJ Sat vapor out Liquid Vapor Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4141 A 2 m tall cylinder has a small hole in the bottom It is filled with liquid water 1 m high on top of which is 1 m high air column at atmospheric pressure of 100 kPa As the liquid water near the hole has a higher P than 100 kPa it runs out Assume a slow process with constant T Will the flow ever stop When Solution Pbot Pair ρgLf For the air PV mRT and the total height is H 2 m Pair mRTVair Vair A Lair A H Lf Pbot maRaTa AHLf ρfg Lf Pa1Va1 AHLf ρf gLf Pa1La1 HLf ρf gLf P o Solve for Lf ρf 1vf 100021002 998 kgm 3 Pa1 La1 ρf g Lf H Lf Po H Lf ρfgH Po Lf ρfgL2 f Po H Pa1 La1 0 Put in numbers and solve quadratic eq L EA PAo AH PAa1 ALAa1 A EρgE A 0 2 f H Poρg Lf PAoE Aρg EA 100 kPa mA3 A sA3 A E 998 9807 kg mE A 10217 m EA PAo AH PAa1 ALAa1 A EρgE A A 100 21 E9989807E A 10217 m LA2E AfE A 12217 LAfE A 10217 0 Air Water 1 m 1 m LAfE A A12217 2E A A 122172 E4 12217 4 EA 61085 52055 11314 or 0903 m so the second root is the solution Verify PAaE A2E A PAaE A1E A EA LAa1 A EHLAf AE A 100 A 1 2 0903E A 91158 kPa ρgLAfE A 998 9807 0903 8838 Pa 8838 kPa PAbotE A PAaE A2E A ρgLAfE A 91158 8838 99996 kPa OK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4142 A pipe of radius R has a fully developed laminar flow of air at Po To with a velocity profile as V Vc 1 rR A2E A where Vc is the velocity on the center line and r is the radius as shown in Fig P4142 Find the total mass flow rate and the average velocity both as functions of Vc and R Am E A AVv AV E Av Since the velocity is distributed we need to integrate over the area From Eq42 AV E A Vlocal dA A Vr 2πr drE where W is the depth Substituting the velocity we get AV E A A Vc 2πr 1 rR2 AE dr 2π VAcE A RA2E A A 0 1 z 1 z2 dzE 2π VAcE A RA2E A A1 2 z2 1 E4 z4E A A 1 0E A Aπ 2E A VAcE A RA2E A A1 2E A VAcE A A V AV E A A A1 2E A VAcE Am E A AV E Av Aπ 2E A VAcE A RA2E Av Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4143 Steam at 3 MPa 400C enters a turbine with a volume flow rate of 5 mA3E As An extraction of 15 of the inlet mass flow rate exits at 600 kPa 200C The rest exits the turbine at 20 kPa with a quality of 90 and a velocity of 20 ms Determine the volume flow rate of the extraction flow and the total turbine work Solution Inlet flow Table B13 v1 009936 mA3E Akg h1 323082 kJkg Am E A1 AV E Av1 5009936 5032 kgs Extraction flow v2 035202 mA3E Akg h2 285012 kJkg Am E A2 015 Am E A1 755 kg s AV E A2 Am E A2v 755 kgs 035202 mA3E Akg 2658 mA3E A s Exit flow AmE A3 085 Am E A1 4277 kg s Table B12 h3 25138 09 235833 2373 88 kJkg Energy Eq 0 Am E A1 h1 Am E A2h2 AmE A3 h3 AW E AT AW E AT Am E A1 h1 Am E A2h2 AmE A3 h3 5032 323082 755 285012 4277 237388 39 525 kW 395 MW W T 1 2 3 Exit flow Extraction flow Inlet flow HP LP section section Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4144 In a glass factory a 2 m wide sheet of glass at 1500 K comes out of the final rollers that fix the thickness at 5 mm with a speed of 05 ms Cooling air in the amount of 20 kgs comes in at 17AoE AC from a slot 2 m wide and flows parallel with the glass Suppose this setup is very long so the glass and air comes to nearly the same temperature a coflowing heat exchanger what is the exit temperature Solution Energy Eq m glasshAglass 1E A m airhair 2 m glasshAglass 3E A m airhair 4 AmE AglassE A ρAV E A ρAV 2500 kgmA3E A 2 m 0005 m 05 ms 125 kgs AmE AglassE ACAglassE A T3 T1 AmE Aair CAPaE A T4 T2 T4 T3 CAglassE A 080 kJkg K CAPaE A 1004 kJkg K T3 m glassCglass T1 m airCPa T2 m glassCglass m airCPa A1250801500 201004290 125080 201004E A K 6923 K We could use table A71 for air but then it will be trial and error 1 2 3 4 Air Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4145 Assume a setup similar to the previous problem but the air flows in the opposite direction of the glass it comes in where the glass goes out How much air flow at 17AoE AC is required to cool the glass to 450 K assuming the air must be at least 120 K cooler than the glass at any location Solution Energy Eq m 1h1 m 4h4 m 3h3 m 2h2 T4 290 K and T3 450 K AmE AglassE A ρAV E A ρAV 2500 kgmA3E A 2 m 0005 m 05 ms 125 kgs T2 T1 120 K 1380 K AmE A m 4 m 2 m 1 h1h3 h2h4 Let us check the limit and since T is high use table A71 for air h4 29043 kJkg h2 149133 kJkg AmE A m 4 m 2 m 1 h1h3 h2h4 m 1 E CAglass AT1T3 h2h4 AmE A 125 kgs A 08 1500450 E 149133 29043E A 8743 kgs 1 2 3 4 Air Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4146 Two kg of water at 500 kPa 20AoE AC is heated in a constant pressure process to 1700AoE AC Find the best estimate for the heat transfer Solution CV Heater steady state 1 inlet and exit no work term no KE PE Continuity Eq Am E AinE A Am E AexE A Am E A Energy Eq413 q hAinE A hAexE A q hAexE A hAinE steam tables only go up to 1300AoE AC so use an intermediate state at lowest pressure closest to ideal gas hAxE A1300AoE AC 10 kPa from Table B13 and table A8 for the high T change h from 1300AoE AC to 1700AoE AC hAexE A hAinE A hAexE A hAxE A hAxE A hAinE A 451555 341677 54097 8396 64245 kJkg Q mhAexE A hAinE A 2 64245 12 849 kJ Why is it done this way The two tables have different offset values for h Once differences in h between two states are taken from the same table the offset goes out Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4147 A 500L insulated tank contains air at 40C 2 MPa A valve on the tank is opened and air escapes until half the original mass is gone at which point the valve is closed What is the pressure inside then Solution State 1 ideal gas mA1E A PA1E AVRTA1E A A 2000 05 0287 3132E A 11125 kg Continuity eq415 mAeE A mA1E A mA2E A mA2E A mA1E A2 mAeE A mA2E A 55625 kg Energy Eq416 mA2E AuA2E A mA1E AuA1E A mAeE AhAe avgE Substitute constant specific heat from table A5 and evaluate the exit enthalpy as the average between the beginning and the end values hAe avgE A CAPE A TA1E A TA2E A2 556250717 TA2E A 1112507173132 556251004 3132 TA2E A2 Solving TA2E A 2394 K PA2E A A m2RT2 EVE A A55625 0287 2394 05E A 764 kPa cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4148 Three air flows all at 200 kPa are connected to the same exit duct and mix without external heat transfer Flow one has 1 kgs at 400 K flow two has 3 kgs at 290 K and flow three has 2 kgs at 700 K Neglect kinetic energies and find the volume flow rate in the exit flow Solution Continuity Eq m 1 m 2 m 3 m 4h4 Energy Eq m 1h1 m 2h2 m 3h3 m 4h4 AV E A4 AmE A v4 h4 m 1 m 4 h1 m 2 m 4 h2 m 3 m 4 h3 A1 6E A 4013 A3 6E A 29043 A2 6E A 71356 44995 kJkg T4 440 20 A44995 44193 46234 44193E A 44786 K v4 RT4 P4 0287 kJkgK 44786 K200 kPa 0643 mA3E Akg AV E A4 AmE A4v4 6 0643 3858 mA3E As 2 4 1 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4149 Consider the power plant as described in Problem 4121 a Determine the temperature of the water leaving the intermediate pressure heater TA13E A assuming no heat transfer to the surroundings b Determine the pump work between states 13 and 16 Solution a Intermediate Pressure Heater Continuity Eq AmE A11E A AmE A13E A AmE A12E A AmE A15E A AmE A14E A feedwater line closed Energy Eq410 AmE A11E AhA11E A AmE A12E AhA12E A AmE A15E AhA15E A AmE A13E AhA13E A AmE A14E AhA14E AmE A14E A AmE A12E A AmE A15E A 8064 4662 12726 kgs 7562846 80642517 4662584 756hA13E A 12726349 hA13E A 53035 kJkg TA13E A 1263C b The high pressure pump Energy Eq412 Am E A13E AhA13E A Am E A16E AhA16E A AW E ACvPE AW E ACvPE A Am E A13E AhA13E A hA16E A 756 kgs 53035 565 kJkg 2620 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4150 Consider the powerplant as described in Problem 4121 a Find the power removed in the condenser by the cooling water not shown b Find the power to the condensate pump c Do the energy terms balance for the low pressure heater or is there a heat transfer not shown Solution a Condenser Continuity Eq AmE A5E A AmE A10E A AmE A6E A also AmE A5E A AmE A3E A AmE A9E A AmE A8E Energy Eq410 AQ E ACVE A AmE A5E AhA5E A AmE A10E AhA10E A AmE A6E AhA6E AmE A10E A AmE A6E A AmE A5E A AmE A6E A AmE A3E A AmE A9E A AmE A8E A 756 62874 4662 2772 756 5544 2016 kgs AQ E ACVE A 5544 2279 2016 14251 756 1383 AQ E ACVE A 118 765 kW 11877 MW b The condensate pump AW E ACvPE A Am E A6E AhA6E A hA7E A 756 kgs 13831 140 kJkg 1278 kW c Low pressure heater Assume no heat transfer Am E A14E AhA14E A Am E A8E AhA8E A Am E A7E AhA7E A Am E A9E AhA9E A Am E A10E AhA10E A Am E A11E AhA11E LHS 12726349 27722459 756140 4662558 24 443 kW RHS 12726 2772 4662 14251 756 28487 24 409 kW A slight imbalance but OK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4151 A 1mA3E A 40kg rigid steel tank contains air at 500 kPa and both tank and air are at 20C The tank is connected to a line flowing air at 2 MPa 20C The valve is opened allowing air to flow into the tank until the pressure reaches 15 MPa and is then closed Assume the air and tank are always at the same temperature and the final temperature is 35C Find the final air mass and the heat transfer Solution Control volume Air and the steel tank Continuity Eq415 mA2E A mA1E A mAiE A Energy Eq416 mA2E AuA2E A mA1E AuA1E AAAIRE A mASTE AuA2E A uA1E AASTE A mAiE AhAiE A A1E AQA2E mA1 AIRE A A P1V ERT1 E A A 500 1 0287 2932E A 594 kg mA2 AIRE A A P2V ERT2 E A A 1500 1 0287 3082E A 1696 kg mAiE A mA2E A mA1E AAairE A 1696 594 1102 kg The energy equation now gives A1E AQA2E A mA2E AuA2E A mA1E AuA1E AAairE A mAstE AuA2E A uA1E AAstE A mAiE AhAiE A mA1E AuA2E A uA1E A mAiE AuA2E A uAiE A RTAiE A mAstE ACAstE ATA2E A TA1E A mA1E ACAvE ATA2E A TA1E A mAiE ACAvE ATA2E A TAiE A RTAiE A mAstE ACAstE ATA2E A TA1E A 594 071735 20 1102071735 20 0287 2932 40 04635 20 63885 808795 276 4689 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4152 A steam engine based on a turbine is shown in Fig P4152 The boiler tank has a volume of 100 L and initially contains saturated liquid with a very small amount of vapor at 100 kPa Heat is now added by the burner and the pressure regulator does not open before the boiler pressure reaches 700 kPa which it keeps constant The saturated vapor enters the turbine at 700 kPa and is discharged to the atmosphere as saturated vapor at 100 kPa The burner is turned off when no more liquid is present in the boiler Find the total turbine work and the total heat transfer to the boiler for this process Solution CV Boiler tank Heat transfer no work and flow out Continuity Eq415 mA2E A mA1E A mAeE Energy Eq416 mA2E AuA2E A mA1E AuA1E A QACVE A mAeE AhAeE State 1 Table B11 100 kPa vA1E A 0001 043 uA1E A 41736 kJkg mA1E A VvA1E A 010001 043 95877 kg State 2 Table B11 700 kPa vA2E A vAgE A 02729 uA2E A 25725 kJkg mA2E A VvAgE A 0102729 0366 kg Exit state Table B11 700 kPa hAeE A 27635 kJkg From continuity eq mAeE A mA1E A mA2E A 95511 kg QACVE A mA2E AuA2E A mA1E AuA1E A mAeE AhAeE 0366 25725 95877 41736 95511 27635 224 871 kJ 2249 MJ CV Turbine steady state inlet state is boiler tank exit state Turbine exit state Table B11 100 kPa hAeE A 26755 kJkg WAturbE A mAeE A hAinE A hAexE A 95511 27635 26755 8405 kJ BOILER Pressure regulator Adiabatic turbine W cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4153 An insulated springloaded pistoncylinder shown in Fig P4153 is connected to an air line flowing air at 600 kPa 700 K by a valve Initially the cylinder is empty and the spring force is zero The valve is then opened until the cylinder pressure reaches 300 kPa By noting that uA2E A uAlineE A CAVE ATA2E A TAlineE A and hAlineE A uAlineE A RTAlineE A find an expression for TA2E A as a function of PA2E A PAoE A TAlineE A With P 100 kPa find TA2E A Solution CV Air in cylinder insulated so A1E AQA2E A 0 Continuity Eq415 mA2E A mA1E A mAinE A Energy Eq416 mA2E AuA2E A mA1E AuA1E A mAinE AhAlineE A A1E AWA2E mA1E A 0 mAinE A mA2E A mA2E AuA2E A mA2E AhAlineE A A1 2E A PA0E A PA2E AmA2E AvA2E uA2E A A1 2E A PA0E A PA2E AvA2E A hAlineE Use constant specific heat in the energy equation CAvE ATA2E A TAlineE A uAlineE A A1 2E A PA0E A PA2E ARTA2E APA2E A hAlineE A Cv 1 E2 P0 P2 P2 RE ATA2E A R CAvE ATAlineE with s TA2E A A R Cv E 2 3 R Cv E A TAlineE A CAvE AR 1k1 k 14 TA2E A A k 1 1 2 3 k 2 E3 1E A TAlineE A A 3k 2k 1E A TAlineE A 1105 TAlineE A 7737 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4154 A massloaded pistoncylinder shown in Fig P4154 containing air is at 300 kPa 17C with a volume of 025 mA3E A while at the stops V 1 mA3E A An air line 500 kPa 600 K is connected by a valve that is then opened until a final inside pressure of 400 kPa is reached at which point T 350 K Find the air mass that enters the work and heat transfer Solution CV Cylinder volume Continuity Eq415 mA2E A mA1E A mAinE A Energy Eq416 mA2E AuA2E A mA1E AuA1E A mAinE AhAlineE A QACVE A A1E AWA2E Process PA1E A is constant to stops then constant V to state 2 at PA2E State 1 PA1E A TA1E A mA1E A A P1V ERT1 E A A 300 025 0287 2902E A 090 kg State 2 Open to PA2E A 400 kPa TA2E A 350 K mA2E A A 400 1 0287 350E A 3982 kg mAiE A 3982 090 3082 kg Only work while constant P A1E AWA2E A PA1E AVA2E A VA1E A 3001 025 225 kJ Energy Eq QACVE A mAiE AhAiE A mA2E AuA2E A mA1E AuA1E A A1E AWA2E QACVE A 3982 0717 350 090 0717 2902 225 3082 1004 600 8192 kJ We could also have used the air tables A71 for the us and hAiE A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4155 A 2mA3E A storage tank contains 95 liquid and 5 vapor by volume of liquified natural gas LNG at 160 K as shown in Fig P4155 It may be assumed that LNG has the same properties as pure methane Heat is transferred to the tank and saturated vapor at 160 K flows into a steady flow heater which it leaves at 300 K The process continues until all the liquid in the storage tank is gone Calculate the total amount of heat transfer to the tank and the total amount of heat transferred to the heater Solution CV Tank flow out transient Continuity Eq mA2E A mA1E A mAeE Energy Eq QATankE A mA2E AuA2E A mA1E AuA1E A mAeE AhAeE At 160 K from Table B7 LIQUID Q tank Q heater VAPOR mAfE A VAfE A vAfE A A095 2 000297E A 63973 kg mAgE A VAgE AvAgE A A005 2 003935E A 2541 kg mA1E A 642271 kg mA2E A VvAg2E A 2003935 50826 kg mA1E AuA1E A 6397310635 25412077 67507 kJ mAeE A mA1E A mA2E A 591445 kg QATankE A 50826 2077 67 507 591445 2703 237 931 kJ CV Heater steady flow P PAG 160 KE A 1593 kPa QAHeaterE A mAe TankE AhAeE A hAiE AAHeaterE 5914456129 2703 202 629 kJ ENGLISH UNIT PROBLEMS SOLUTION MANUAL CHAPTER 4 English units Fundamentals of Thermodynamics Borgnakke Sonntag 8e 8e 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CHAPTER 4 SUBSECTION PROB NO Continuity and Flow Rates 156159 Single Flow Devices 160187 Multiple Flow Devices 188197 Multiple Devices Cycle Processes 198203 Transient processes 204211 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Continuity and Flow Rates Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4156E Refrigerant R410A at 100 psia 60 F flows at 01 lbms in a 25 ft2 cross sectional area pipe Find the velocity and the volume flow rate m AVv V v Table F92 v 06848 ft3lbm V v m 06848 ft3lbm 01 lbms 006848 ft3s V V A 006848 ft3s 25 ft2 00274 fts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4157E Air at 95 F 16 lbfin2 flows in a 4 in 6 in rectangular duct in a heating system The volumetric flow rate is 30 cfm ft3min What is the velocity of the air flowing in the duct Solution Assume a constant velocity across the duct area with A 4 6 1 144 0167 ft 2 and the volumetric flow rate from Eq43 V m v AV V V A 30 60 0167 30 fts Ideal gas so note note ideal gas v RT P 5334 5547 16 144 12842 ft3lbm m V v 30 60 12842 00389 lbms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4158E A pool is to be filled with 2500 ft3 water from a garden hose of 1 in diameter flowing water at 6 fts Find the mass flow rate of water and the time it takes to fill the pool Solution With constant velocity we have V m v AV π 1122 ft2 6 fts 01309 ft3s t V V 2500 ft3 01309 ft3s 19 098 s 5 h 18 min 18 s From table F3 we get the water density m V v ρV 622 lbm ft3 01309 ft3s 814 lbms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4159E A hot air home heating system takes 500 ft3min cfm air at 147 psia 65 F into a furnace and heats it to 130 F and delivers the flow to a square duct 05 ft by 05 ft at 15 psia What is the velocity in the duct Solution The inflate flow is given by a m i Continuity Eq m i V i vi m e AeVev e Ideal gas vi Pi RTi 5334 525 147 144 1323 ft3 lbm ve Pe RTe 15 psi 144 in2ft2 5334 lbfftlbmR 130 460 R 1457 ft3 lbm m i V ivi 500 ft3min 60 smin 1323 ft3 lbm 063 lbms Ve m ve Ae 063 1457 05 05 ft3 ft s 2 367 fts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4160E Liquid water at 60 F flows out of a nozzle straight up 40 ft What is nozzle Vexit Energy Eq413 hexit 1 2 V2 exit gHexit h2 1 2 V2 2 gH2 If the water can flow 40 ft up it has specific potential energy of gH2 which must equal the specific kinetic energy out of the nozzle V 2 exit2 The water does not change P or T so h is the same V 2 exit2 gH2 Hexit gH Vexit 2gH 2 32174 40 ft2s2 507 fts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4161E A diffuser receives 02 lbms steam at 80 psia 600 F The exit is at 150 psia 700 F with negligible kinetic energy and the flow is adiabatic Find the diffuser inlet velocity and the inlet area Solution Energy Eq413 h1 1 2 V2 1 gZ1 h2 1 2 V2 2 gZ2 Process Z1 Z2 State 1 Table F72 h1 133066 Btulbm v1 7794 ft3lbm State 2 V2 0 Table F72 h2 137655 Btulbm Then from the energy equation 1 2 V2 1 h2 h1 137655 133066 4589 Btulbm V1 2h2 h1 2 32174 778 4589 15157 fts The mass flow rate from Eq43 m ρAV AVv A m vV 02 lbms 7794 ft3lbm 15157 fts 0000103 ft2 0148 in2 Inlet Low V Exit Hi V Hi P A Low P A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4162E Saturated vapor R134a leaves the evaporator in a heat pump at 50 F with a steady mass flow rate of 02 lbms What is the smallest diameter tubing that can be used at this location if the velocity of the refrigerant is not to exceed 20 fts Solution Mass flow rate Eq43 m V v AVv Exit state Table F101 T 50 F x 1 v vg 0792 ft3lbm The minimum area is associated with the maximum velocity for given m AMIN m vg VMAX 20 fts 02 lbms 0792 ft3lbm 000792 ft2 π 4 D 2 MIN DMIN 01004 ft 1205 in cb Exit Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Single Flow Devices Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4163E In a jet engine a flow of air at 1800 R 30 psia and 90 fts enters a nozzle where the air exits at 1500 R 13 psia as shown in Fig P423 What is the exit velocity assuming no heat loss Solution CV nozzle No work no heat transfer Continuity m i m e m Energy m hi ½Vi2 m he ½Ve2 Due to high T take h from table F5 ½Ve2 ½ Vi2 hi h e 902 2 32174 778 44979 36928 016 8051 8067 Btulbm Ve 2 32174 lbmfts2lbf 778 lbfftBtu 8067 Btulbm12 2010 fts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4164E A sluice gate dams water up 15 ft A 05 in diameter hole at the bottom of the gate allows liquid water at 70 F to come out Neglect any changes in internal energy and find the exit velocity and mass flow rate Solution Energy Eq413 h1 1 2 V2 1 gZ1 h2 1 2 V2 2 gZ2 Process h1 h2 both at P 1 atm V1 0 Z1 Z2 15 ft Water 15 ft 1 2 V2 2 g Z1 Z2 V2 2gZ1 Z2 2 32174 15 3107 fts m ρΑV AVv π 4 D2 V2v π 4 05122 ft2 3107 fts 001605 ft3lbm 2639 lbms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4165E A diffuser shown in Fig P428 has air entering at 147 lbfin2 540 R with a velocity of 600 fts The inlet crosssectional area of the diffuser is 02 in2 At the exit the area is 175 in2 and the exit velocity is 60 fts Determine the exit pressure and temperature of the air Solution Continuity Eq43 m i AiVivi m e AeVeve Energy Eqper unit mass flow 413 hi 1 2Vi 2 he 1 2Ve 2 he hi 126002 60232174 778 7119 Btulbm Te Ti he hiCp 540 R 7119 Btulbm 024 BtulbmR 5697 R Now use the continuity equation and the ideal gas law ve vi AeVe AiVi RTiPi AeVe AiVi RTeP e Pe Pi Te Ti AiVi AeVe 147 5697 540 02 600 175 60 1772 lbfin2 Inlet Low V Exit Hi V Hi P A Low P A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4166E Nitrogen gas flows into a convergent nozzle at 30 lbfin2 600 R and very low velocity It flows out of the nozzle at 15 lbfin2 500 R If the nozzle is insulated find the exit velocity Solution CV Nozzle steady state one inlet and exit flow insulated so it is adiabatic Inlet Low V Exit Hi V Hi P A Low P A cb Energy Eq413 h1 0 h2 1 2 V2 2 V2 2 2 h1 h2 2 CPN2 T1 T2 2 0249 BtulbmR 600 500 R 249 Btulbm V2 2 2 249 Btulbm 778 lbfftBtu 32174 lbmfts2lbf 1 246 562 ft2 s 2 V2 1116 fts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4167E A meteorite hits the upper atmosphere at 10 000 fts where the pressure is 01 atm and the temperature is 40 F How hot does the air become right in front of the meteorite assuming no heat transfer in this adiabatic stagnation process Solution Energy Eq h1 1 2 V2 1 gZ1 h2 1 2 V2 2 gZ2 Process Z1 Z2 and V2 0 h2 h1 1 2 V2 1 T2 T1 1 2V2 1Cp 41967 1 2 10 0002 024 778 32174 8742 R At this high temperature we cannot assume constant specific heats so use F5 h2 h1 1 2 V2 1 1004 1 2 778 32174 10 0002 20979 Btulbm Table F5 is listed to 5400 R so we have to extrapolate to get T2 5400 100 20979 1515632 1515632 1484762 7286 R The value of Cp over 5400 R is 0309 BtulbmR from the last two table entries At this temperature there will be some chemical reactions that should be considered to have a more realistic temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4168E Refrigerant R410A flows out of a cooler at 70 F 220 psia after which it is throttled to 77 psia Find the state T x for the exit flow CV Throttle Steady state Process with q w 0 and Vi Ve Zi Z e Energy Eq413 hi he Inlet state Table F91 hi 3917 Btulbm Exit state Table F91 since h hg 11821 Btulbm Interpolate T 10 F hf 170 Btulbm hfg 10122 Btulbm x 3917 170 10122 0219 2 P v 1 h C T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4169E R410A at 90 F 300 psia is throttled so it becomes cold at 10 F What is exit P State 1 is slightly compressed liquid so Table F91 h hf 4730 Btulbm At the lower temperature it becomes twophase since the throttle flow has constant h and at 10 F hg 11821 Btulbm P Psat 7693 psia 1 2 2 P v 1 h C T h C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4170E Saturated liquid R410A at 30 F is throttled to 40 psia in a refrigerator What is the exit temperature Find the percent increase in the volume flow rate Solution Steady throttle flow Assume no heat transfer and no change in kinetic or potential energy he hi hf 30 F 2411 Btulbm hf e xe hfg e at 40 psia From table F91 we get Te Tsat 40 psia 21 F xe hfg e he hf e 2411 6354 108776 01632 ve vf xe vfg 001253 xe 147584 025338 ft3lbm vi vf 30 F 001364 ft3lbm V m v so the ratio becomes V e V i m ve m vi ve vi 025338 001364 1858 So the increase is 1758 times or 1758 e T v i h C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4171E Saturated vapor R410A at 75 psia is throttled to 15 psia What is the exit temperature Repeat the question if you assumed it behaves like an ideal gas CV Throttle Steady state Process with q w 0 and Vi Ve Zi Z e Energy Eq413 hi he Inlet state Table F92 hi 11809 Btulbm Exit state Table F92 since h hg Exit state very close to T 20 F For an ideal gas h is only a function of T so Te Ti 871 F 2 P v 1 h C T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4172E Helium is throttled from 175 lbfin2 70 F to a pressure of 15 lbfin2 The diameter of the exit pipe is so much larger than the inlet pipe that the inlet and exit velocities are equal Find the exit temperature of the helium and the ratio of the pipe diameters CV Throttle Steady state Process with q w 0 and Vi Ve Zi Z e Energy Eq413 hi he Ideal gas Ti Te 70 F m AV RTP But m V T are constant PiAi PeAe De Di Pi Pe 12 175 15 12 3416 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4173E A liquid water turbine receives 4 lbms water at 300 psia 77 F with a velocity of 50 fts The exit is at 15 psia 77 F with very low velocity Find the specific work and the power produced Solution Energy Eq413 h1 1 2 V2 1 gZ1 h2 1 2 V2 2 gZ2 wT Process Z1 Z2 and V2 0 State 1 Table F71 h1 4509 Btulbm Add Pv 3000464 001606 32174144 1075 Btulbm State 2 Table F71 h2 4509 Add Pv 150464 001606 32174144 0052 Btulbm wT h1 1 2 V2 1 h2 4509 1075 1 2 778 32174 502 4509 0052 1073 Btulbm W T m wT 4 1073 429 Btus Notice how insignificant the specific kinetic energy is 005 Btulbm The added Pv terms are due to P being higher than saturated P see p134 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4174E Hoover Dam across the Colorado River dams up Lake Mead 600 ft higher than the river downstream as shown in Fig P447 The electric generators driven by waterpowered turbines deliver 12 106 Btus If the water is 65 F find the minimum amount of water running through the turbines Solution CV H2O pipe turbines T H DAM Lake Mead Continuity m in m ex Energy Eq413 h V22 gzin h V22 gzex wT Water states hin hex vin vex Now the specific turbine work becomes wT gzin gzex 32174 fts2 600 ft778 lbfftBtu 32174 lbmfts2lbf 0771 Btulbm m W TwT 12106Btus 0771 Btulbm 1556106 lbms V m v 1556106 0016043 24 963 ft3s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4175E A small expander a turbine with heat transfer has 01 lbms helium entering at 160 psia 1000 R and it leaves at 40 psia 540 R The power output on the shaft is measured to 55 Btus Find the rate of heat transfer neglecting kinetic energies Solution CV Expander Steady operation Continuity Eq m i m e m Energy Eq m hi Q m he W WT i e Q cb Q m hehi W Use heat capacity from Table F4 Cp He 124 Btulbm R Q m Cp TeTi W 01 lbms 124 BtulbmR 540 1000 R 55 Btus 5704 55 20 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4176E A small highspeed turbine operating on compressed air produces a power output of 01 hp The inlet state is 60 lbfin2 120 F and the exit state is 147 lbfin2 20 F Assuming the velocities to be low and the process to be adiabatic find the required mass flow rate of air through the turbine Solution CV Turbine no heat transfer no KE no PE Energy Eq413 hin hex wT Ideal gas so use constant specific heat from Table A5 wT hin hex CpTin Tex 024120 20 336 Btulbm W m wT m W wT 01 hp 550 lbfftshp 778 lbfftBtu 336 Btulbm 00021 lbms 757 lbmh The dentists drill has a small air flow and is not really adiabatic Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4177E A compressor in a commercial refrigerator receives R410A at 10 F and x 1 The exit is at 200 psia 120 F Neglect kinetic energies and find the specific work Solution CV Compressor steady state single inlet and exit flow For this device we also assume no heat transfer and Zi Z e W C i e cb From Table F91 hi 11629 Btulbm From Table F92 he 13702 Btulbm Energy Eq413 reduces to wc hi he 11629 13702 2073 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4178E A compressor in an industrial airconditioner compresses ammonia from a state of saturated vapor at 20 psia to a pressure of 125 psia At the exit the temperature is measured to be 200 F and the mass flow rate is 1 lbms What is the required power input to this compressor Solution CV Compressor Assume adiabatic and neglect kinetic energies Energy Eq413 wC h1 h2 States 1 F82 h1 60547 Btulbm 2 F82 h2AC 71044 Btulbm Energy equation wC h2 h1 71044 60547 10497 Btulbm W 1 lbms 10497 Btulbm 105 Btus v P 2 ac 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4179E An exhaust fan in a building should be able to move 6 lbms air at 144 psia 68 F through a 14 ft diameter vent hole How high a velocity must it generate and how much power is required to do that Solution CV Fan and vent hole Steady state with uniform velocity out Continuity Eq m constant ρΑV AV v AVPRT Ideal gas Pv RT and area is A π 4 D2 Now the velocity is found V m RTπ 4 D2 P 6 lbms 5334 lbfftlbmR 4597 68 R π 4 142 144 psi 144 in2ft2 52908 fts The kinetic energy out is 1 2 V2 2 1 2 529082 32174 43502 lbfftlbm which is provided by the work only two terms in energy equation that does not cancel we assume V1 0 W in m 1 2 V2 2 6 43502 261 lbffts 0335 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4180E An air flow is brought from 77 F 147 psia to 150 psia 620 F by an adiabatic compressor driven by a 50kW motor What are the mass flow rate and the exit volume flow rate of air CV Compressor Assume adiabatic and neglect kinetic potential energies Energy Eq413 w h1 h2 wC h2 h1 CP T2 T1 024 BtulbmR 620 77 R 13032 Btulbm The mass flow rate scale the work term so m W wC 13032 Btulbm 501055 Btus 03637 lbms V m v m RT P 03637 lbms 5334 620 45967 ftlbflbm 150 lbfin2 144 inft 2 2666 ft3s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4181E Carbon dioxide gas enters a steadystate steadyflow heater at 45 lbfin2 60 F and exits at 40 lbfin2 1800 F It is shown in Fig P465 here changes in kinetic and potential energies are negligible Calculate the required heat transfer per lbm of carbon dioxide flowing through the heater Solution CV Heater Steady state single inlet and exit flow Energy Eq413 q hi h e Q i e Table F6 q he hi 204708 1434 4401 4684 Btulbm If we use CP0 then q 0203 BtulbmR 1800 60 R 3532 Btulbm Too large T Tave to use CP0 at room temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4182E A condenser cooler receives 01 lbms of R410A at 300 psia 140 F and cools it to 70 F Assume the exit properties are as for saturated liquid same T What cooling capacity Btuh must the condenser have Solution CV R410A condenser Steady state single flow heat transfer out and no work Energy Eq412 m h1 m h2 Q out Inlet state Table F92 h1 13734 Btulbm Exit state Table F91 h2 3917 Btulbm compressed liquid Process Neglect kinetic and potential energy changes Cooling capacity is taken as the heat transfer out ie positive out so Q out m h1 h2 01 lbms 13734 3917 Btulbm 982 Btus 1 2 Q cool Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4183E In a steam generator compressed liquid water at 1500 lbfin2 100 F enters a 1in diameter tube at the rate of 5 ft3min Steam at 1250 lbfin2 750 F exits the tube Find the rate of heat transfer to the water Solution CV Steam generator Steady state single inlet and exit flow Constant diameter tube Ai Ae π 4 1 12 2 000545 ft2 Table B14 m V ivi 5 600016058 18 682 lbmh Vi V iAi 5000545 60 153 fts Exit state properties from Table F72 and inlet from F73 Ve Vi vevi 153 fts 0526001605 5014 fts The energy equation Eq412 is solved for the heat transfer as Q m he hi Ve 2 Vi 2 2 18 68213415 7199 501421532 232174778 2381107 Btuh Typically hot combustion gas in Steam exit cb liquid water in gas out Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4184E An oven has five radiant heaters each one is rated at 15 Btus It should heat some 4lbm steel plates from 77 F to 1400 R How many of these plates per minute can it heat CV Oven steady state operation A flow of plates in represents an m Energy Eq 0 m hi he Q Q m he hi m C Te Ti m Q C Te Ti 5 15 Btus 011 1400 5367 Btulbm 079 lbms N m m 079 4 1s 0197 per s 1185 per min Front end of oven with steel rollers to bring the plates in a skirt hangs down to limit heat losses C Borgnakke Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4185E An evaporator has R410A at 0 F and quality 20 flowing in with the exit flow being saturated vapor at 0 F Knowing that there is no work find the specific heat transfer CV Heater Steady state single inlet and exit flow Energy Eq413 0 q h1 h 2 Table F91 h1 1352 02 10376 3427 Btulbm h2 11728 Btulbm q h2 h1 11728 3427 8301 Btulbm 1 2 Q evap Evaporator vapor cb 63 P 1 2 v T 1 2 v 0 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4186E A flow of liquid glycerine flows around an engine cooling it as it absorbs energy The glycerine enters the engine at 140 F and receives 13 hp of heat transfer What is the required mass flow rate if the glycerine should come out at a maximum 200 F Solution CV Liquid flow glycerine is the coolant steady flow no work Energy Eq m hi Q m he EA AQ A ECAgly TAe A TAi A AE m Q he hi From table F3 CAglyE A 058 Btulbm R Am E A A13 hp 254443600 btushp E058 btulbmR 200 140 RE A 0264 lbms Exhaust flow Air intake filter Coolant flow Atm air Shaft Fan power Radiator cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4187E A small water pump is used in an irrigation system The pump takes water in from a river at 50 F 1 atm at a rate of 10 lbms The exit line enters a pipe that goes up to an elevation 100 ft above the pump and river where the water runs into an open channel Assume the process is adiabatic and that the water stays at 50 F Find the required pump work Solution CV pump pipe Steady state 1 inlet 1 exit flow Assume same velocity in and out no heat transfer Continuity Eq Am E AinE A Am E AexE A Am E Energy Eq412 Am E AhAinE A 12VAinE A2 gzAinE A Am E AhAexE A 12 VAex 2E A gzAexE A AW E States hAinE A hAexE A same T P i e H cb AW E A Am E AgzAinE A zAexE A 10 lbms A 32174 fts2 E32174 lbm fts2 lbfE A 100 ft 1000 lbffts 1285 Btus IE 1285 Btus required input Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Multiple Flow Devices Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4188E A steam turbine receives water at 2000 lbfin2 1200 F at a rate of 200 lbms as shown in Fig P484 In the middle section 50 lbms is withdrawn at 300 lbfin2 650 F and the rest exits the turbine at 10 lbfin2 95 quality Assuming no heat transfer and no changes in kinetic energy find the total turbine work CV Turbine Steady state 1 inlet and 2 exit flows Continuity Eq49 Am E A1E A Am E A2E A Am E A3E A Am E A3E A Am E A1E A Am E A2E A 150 lbms Energy Eq410 Am E A1E AhA1E A AW E ATE A Am E A2E AhA2E A Am E A3E AhA3E Table F72 hA1E A 15986 Btulbm hA2E A 13416 Btulbm Table F71 hA3E A hAfE A xA3E AhAfgE A 1612 095 9821 10942 Btulbm WT 1 2 3 From the energy equation Eq410 AW E ATE A Am E A1E AhA1E A Am E A2E AhA2E A Am E A3E AhA3E A 200 15986 50 13416 150 10942 885 10A4E A Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4189E A condenser as the heat exchanger shown in Fig P491 brings 1 lbms water flow at 1 lbfin2 from 500 F to saturated liquid at 1 lbfin2 The cooling is done by lake water at 70 F that returns to the lake at 90 F For an insulated condenser find the flow rate of cooling water Solution CV Heat exchanger Energy Eq410 Am E AcoolE AhA70E A AmE AH2OE AhA500E A A m E AcoolE AhA90E A AmE AH2OE AhAf 1E 1 500 F 4 90 F 3 70 F m cool 1 lbms 2 sat liq Table F71 hA70E A 3809 Btulbm hA90E A 5807 Btulbm hAf1E A 6974 Btulbm Table F72 hA5001E A 12885 btulbm Am E AcoolE A Am E AH2OE A A h500 hfE 1 h90 h70 E A 1 lbms A12885 6974 5807 3809E A 61 lbms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4190E A heat exchanger is used to cool an air flow from 1400 to 680 R both states at 150 lbfin2 The coolant is a water flow at 60 F 15 lbfin2 and it is shown in Fig P495 If the water leaves as saturated vapor find the ratio of the flow rates Am E AH2OE AAm E AairE A Solution CV Heat exchanger steady flow 1 inlet and 1 exit for air and water each The two flows exchange energy with no heat transfer tofrom the outside 3 water 1 air 4 2 Continuity Eqs Each line has a constant flow rate through it Energy Eq410 Am E AairE AhA1E A Am E AH2OE AhA3E A Am E AairE AhA2E A Am E AH2OE AhA4E Process Each line has a constant pressure Table F5 hA1E A 343016 Btulbm hA2E A 16286 Btulbm Table F7 hA3E A 2808 Btulbm hA4E A 11509 Btulbm at 15 psia Am E AH2OE AAm E AairE A A h1 h2 Eh4 h3 E A A343016 16286 11509 2808E A 01604 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4191E A dualfluid heat exchanger has 10 lbms water entering at 100 F 20 psia and leaving at 50 F 20 psia The other fluid is glycol entering at 10 F 22 psia and leaving at 50 F 22 psia Find the required mass flow rate of glycol and the rate of internal heat transfer CV Heat exchanger steady flow 1 inlet and 1 exit for glycol and water each The two flows exchange energy with no heat transfer tofrom the outside 3 glycol 1 water 4 2 Continuity Eqs Each line has a constant flow rate through it Energy Eq410 Am E AH2OE A hA1E A Am E AglycolE A hA3E A Am E AH2OE A hA2E A Am E AglycolE A hA4E Process Each line has a constant pressure Table F7 hA1E A 6804 Btulbm hA2E A 1805 Btulbm Table F3 CAP glyE A 058 BtulbmR so hA4E A hA3E A CAP glyE A TA4E A TA3E A 058 50 10 232 Btulbm Am E AglycolE A Am E AH2OE A A h1 h2 Eh4 h3 E A 10 lbms A6804 1805 232E A 2155 lbms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4192E An automotive radiator has glycerine at 200 F enter and return at 130 F as shown in Fig P4100 Air flows in at 68 F and leaves at 77 F If the radiator should transfer 33 hp what is the mass flow rate of the glycerine and what is the volume flow rate of air in at 15 psia Solution If we take a control volume around the whole radiator then there is no external heat transfer it is all between the glycerin and the air So we take a control volume around each flow separately Heat transfer AQ E A 33 hp 33 25444 3600 23324 Btus Glycerine energy Am E AhAiE A AQ E A Am E AhAeE Table F3 Am E AglyE A A Q Ehe hi E A A Q ECglyTeTiE A A 23324 058130 200E A 0574 lbms Air Am E AhAiE A AQ E A Am E AhAeE Table F4 Am E AairE A A Q Ehe hi E A A Q ECairTeTiE A A 23324 02477 68E A 8835 lbms AV E A Am E AvAiE A vAiE A A RTi EPi E A A5334 5277 15 144E A 1303 ftA3E Albm AV E AairE A Am E AvAiE A 8835 1303 115 ftA3E As Exhaust flow Air intake filter Coolant flow 130 F Atm air Shaft power 200 F cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4193E Steam at 80 psia 600 F is used to heat cold water at 60 F to 170 F for domestic hot water supply How much steam per lbm liquid water is needed if the steam should not condense Solution CV Each line separately No work but there is heat transfer out of the steam flow and into the liquid water flow Water line energy Eq Am E AliqE AhAiE A AQ E A Am E AliqE AhAeE A AQ E A Am E AliqE AhAeE A hAiE A For the liquid water look in Table F71 hAliqE A hAeE A hAiE A 13796 2808 10988 btulbm CApE A T 10 170 60 110 btulbm Steam line energy has the same heat transfer but it goes out Steam Energy Eq Am E AsteamE AhAiE A AQ E A Am E AsteamE AhAeE A AQ E A Am E AsteamE AhAiE A hAeE A For the steam look in Table F72 at 80 psia hAsteamE A hAiE A hAeE A 133066 118361 14705 btulbm Now the heat transfer for the steam is substituted into the energy equation for the water to give Am E AsteamE A Am E AliqE A hAliqE A hAsteamE A A10988 14705E A 0747 cb Steam in Steam out Cold water in Hot water out Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4194E A copper wire has been heat treated to 1800 R and is now pulled into a cooling chamber that has 3 lbms air coming in at 70 F the air leaves the other end at 120 F If the wire moves 05 lbms copper how hot is the copper as it comes out Solution CV Total chamber no external heat transfer Energy eq Am E AcE AuE A h AiE AcuE A Am E AairE A hAiE A AairE A Am E AcuE A hAe cuE A Am E AairE A hAe airE Am E AcuE A hAeE A hAiE A AcuE A Am E AairE A hAiE A hAeE A AairE A Am cuE A CAcuE A TAeE A TAiE A AcuE A Am airE A CAp airE A TAeE A TAiE A AairE Heat capacities from F2 for copper and F4 for air TAeE A TAiE A AcuE A A m airCp air Em cuCcu E A TAeE A TAiE A AairE A A 3 024 05 01E A 70 120 R 720 R TAeE A TAiE A 720 1800 720 1080 R Air Air Cu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4195E Two flows of air are both at 30 psia one has 1 lbms at 720 R and the other has 2 lbms at 520 R The two flows are mixed together in an insulated box to produce a single exit flow at 30 psia Find the exit temperature Solution Cont Am E A1E A Am E A2E A Am E A3E Energy Am E A1E AhA1E A Am E A2E AhA2E A Am E A3E AhA3E A Am E A1E A Am E A2E AhA3E 1 2 3 Mixing chamber Solve for the exit enthalpy hA3E A Am E A1E AhA1E A Am E A2E AhA2E A Am E A1E A Am E A2E A since the Ts are modest use constant specific heats TA3E A Am E A1E ATA1E A Am E A2E ATA2E A Am E A1E A Am E A2E A A1 3E A 720 A2 3E A 520 5867 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4196E A desuperheater has a flow of ammonia 3 lbms at 150 psia 200 F which is mixed with another flow of ammonia at 80 F and quality 25 in an adiabatic mixing chamber Find the flow rate of the second flow so the outgoing ammonia is saturated vapor at 150 psia CV Desuperheater No external AQ E A or AW E 1 2 3 MIXING CHAMBER cb Continuity Eq49 Am E A1E A Am E A2E A Am E A3E A Energy Eq410 Am E A1E AhA1E A Am E A2E AhA2E A Am E A3E AhA3E A Am E A1E A Am E A2E AhA3E State 1 Table F82 hA1E A 70796 Btulbm State 2 Table F81 h2 13168 025 49794 25617 Btulbm State 3 Table F82 h3 62945 Btulbm AmE A2E A AmE A1E A A h1 h3 Eh3 h2 E A 3 A70796 62945 62945 25617E A 0631 lbms T v 1 2 3 150 psia 2 P v 3 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4197E An insulated mixing chamber receives 4 lbms R134a at 150 lbfin2 220 F in a line with low velocity Another line with R134a as saturated liquid 130 F flows through a valve to the mixing chamber at 150 lbfin2 after the valve The exit flow is saturated vapor at 150 lbfin2 flowing at 60 fts Find the mass flow rate for the second line Solution CV Mixing chamber Steady state 2 inlets and 1 exit flow Insulated q 0 No shaft or boundary motion w 0 Continuity Eq49 m 1 m 2 m 3 Energy Eq410 m 1h1 m 2h2 m 3 h3 A1 2E A VA2 3E A m 2 h2 h3 A1 2E A VA2 3E A m 1 h3 A1 2E A VA2 3E A h1 State 1 Table F101 150 psia 220 F h1 20963 Btulbm State 2 Table F101 x 0 130 F h2 11988 Btulbm State 3 Table F102 x 1 150 psia h3 18061 Btulbm A1 2E A VA2 3E A A1 2E A 602 32174 778 0072 Btulbm Am E A2 Am E A1 h3 A1 2E A VA2 3E A h1 h2 h3 A1 2E A VA2 3E A 4 18061 0072 20963 11988 18061 0072 1904 lbms Notice how kinetic energy was insignificant 1 2 3 MIXING CHAMBER cb 2 P v 3 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Multiple Devices Cycle Processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4198E The following data are for a simple steam power plant as shown in Fig P4118 State 1 2 3 4 5 6 7 P psia 900 890 860 830 800 15 14 T F 115 350 920 900 110 h Btulbm 853 323 1468 1456 1029 78 State 6 has xA6E A 092 and velocity of 600 fts The rate of steam flow is 200 000 lbmh with 400 hp input to the pump Piping diameters are 8 in from steam generator to the turbine and 3 in from the condenser to the steam generator Determine the power output of the turbine and the heat transfer rate in the condenser Turbine AA5E A πDA2 5E A4 0349 ftA2E A vA5E A 0964 ftA3E Albm VA5E A AmE AvA5E AAA5E A A200 000 0964 3600 0349E A 153 fts w hA5E A 05VA2 5E A hA6E A 05VA2 6E A 1456 1029 A6002 1532 E2 25 037E A 4202 Btulbm Recall the conversion 1 Btulbm 25 037 ftA2E AsA2E A 1 hp 2544 Btuh AW E ATURBE A A4202 200 000 2544E A 33 000 hp Condenser AA7E A πDA2 7E A4 00491 ftA2E A vA7E A 001617 ftA3E Albm VA7E A AmE AvA7E AAA7E A A200000 001617 3600 00491E A 18 fts q 7802 10287 A182 6002 E2 25 037E A 9579 Btulbm AQ E ACONDE A 200 000 9579 191610A8E A Btuh Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4199E For the same steam power plant as shown in Fig P4118 and Problem 4198E determine the rate of heat transfer in the economizer which is a low temperature heat exchanger and the steam generator Determine also the flow rate of cooling water through the condenser if the cooling water increases from 55 to 75 F in the condenser Condenser AA7E A πDA2 7E A4 00491 ftA2E A vA7E A 001617 ftA3E Albm VA7E A AmE AvA7E AAA7E A A200000 001617 3600 00491E A 18 fts q 7802 10287 A182 6002 E2 25 037E A 9579 Btulbm AQ E ACONDE A 200 000 9579 191610A8E A Btuh Economizer VA3E A VA2E A since liquid v is constant vA3E A vA2E A and AA3E A AA2E A q hA3E A hA2E A 3230 853 2377 Btulbm AQ E AECONE A 200 000 2377 47510A7E A Btuh Generator AA4E A πDA2 4E A4 0349 ftA2E A vA4E A 09595 ftA3E Albm VA4E A AmE AvA4E AAA4E A A200 000 09505 3600 0349E A 151 fts AA3E A πDA2 3E A4 0349 ftA2E A vA3E A 00491 ftA3E Albm VA3E A AmE AvA3E AAA3E A A200 000 00179 3600 00491E A 20 fts q 14678 3230 A1512 202 E2 25 037E A 11452 Btulbm AQ E AGENE A 200 000 11452 229110A8E A Btuh Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4200E A twostage compressor takes nitrogen in at 80 F 20 psia and compresses it to 80 psia 800 R Then it flows through an intercooler where it cools to 580 R and the second stage compresses it to 400 psia 1000 R Find the specific work in each of the two compressor stages and the specific heat transfer in the intercooler The setup is like this CV Stage 1 nitrogen Steady flow Process adibatic q 0 reversible sAgenE A 0 Energy Eq 413 wAC1E A hA2E A hA1E A Assume constant CAP0E A 0249 from F4 wAC1E A hA1E A hA2E A CAP0E ATA1E A TA2E A 0249 BtulbmR 540 800 K 6474 Btulbm CV Intercooler no work and no changes in kinetic or potential energy qA2E A3E A hA3E A hA2E A CAP0E ATA3E A TA2E A 0249 800 580 1355 Btulbm CV Stage 2 Analysis the same as stage 1 wAC2E A hA3E A hA4E A CAP0E ATA3E A TA4E A 0249 580 1000 1046 Btulbm C1 C2 1 2 3 4 2 W Q 1 W cooler inter Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4201E The intercooler in the previous problem uses cold liquid water to cool the nitrogen The nitrogen flow is 01 lbms and the liquid water inlet is 77 F and is set up to flow in the opposite direction from the nitrogen so the water leaves at 105 F Find the flow rate of the water Solution CV Heat exchanger steady 2 flows in and two flows out 1 3 2 4 Continuity eq AmE A1E A AmE A2E A AmE AH2OE A AmE A3E A AmE A4E A AmE AN2E Energy eq 0 AmE AH2OE AhA3E A hA4E A AmE AN2E A hA2E A hA1E A Due to the lower range of temperature we will use constant specific heats from F4 and F3 so the energy equation is approximated as 0 AmE AH2OE ACApE AH2OE A TA3E A TA4E A AmE AN2E ACApE A TA2E A TA1E A Now solve for the water flow rate AmE AH2OE A AmE AN2E A CApE AN2E A TA2E A TA1E A CApE AH2OE A TA4E A TA3E A 01 lbms A0249 BtulbmR 800 580 R E 10 BtulbmR 105 77 RE A 0196 lbms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4202E A R410A heat pump cycle shown in Fig P4123 has a R410A flow rate of 01 lbms with 4 Btus into the compressor The following data are given State 1 2 3 4 5 6 P psia 410 405 400 62 60 58 T F 220 200 110 10 14 h Btulbm 154 150 56 120 122 Calculate the heat transfer from the compressor the heat transfer from the R410A in the condenser and the heat transfer to the R410A in the evaporator Solution CV Compressor AQ E ACOMPE A AmE AhA1E A hA6E A AW E ACOMPE A 01 154 122 40 08 Btus CV Condenser AQ E ACONDE A AmE AhA3E A hA2E A 01 lbms 56 150 Btulbm 94 Btus CV Valve hA4E A hA3E A 56 Btulbm CV Evaporator AQ E AEVAPE A AmE A hA5E A hA4E A 01 lbms 120 56 Btulbm 64 Btus v P compressor evaporator condenser valve 2 1 3 4 5 6 Q H W C Q L cb Evaporator Condenser 3 4 5 6 1 2 Q loss Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4203E A proposal is made to use a geothermal supply of hot water to operate a steam turbine as shown in Fig P4125 The high pressure water at 200 lbfin2 350 F is throttled into a flash evaporator chamber which forms liquid and vapor at a lower pressure of 60 lbfin2 The liquid is discarded while the saturated vapor feeds the turbine and exits at 1 lbfin2 90 quality If the turbine should produce 1000 hp find the required mass flow rate of hot geothermal water in poundmass per hour Solution Separation of phases in flashevaporator constant h in the valve flow so Table F73 hA1E A 3218 Btulbm hA1E A 3218 26225 x 9158 x 006503 AmE A2E AAmE A1E Table F72 hA2E A 11780 Btulbm FLASH EVAP H O 2 Sat liquid out Sat vapor W Turb 1 2 3 4 Table F71 hA3E A 6974 09 1036 10021 Btulbm AW E A AmE A2E AhA2E A hA3E A AmE A2E A A 1000 2545 11780 10021E A 14 472 lbmh AmE A1E A 222 539 lbmh Notice conversion 1 hp 2445 Btuh from Table A1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Transient Processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4204E A 1 Gallon tank initially is empty and we want to have 003 lbm of R410A in it The R410A comes from a line with saturated vapor at 20 F To end up with the desired amount we cool the can while we fill it in a slow process keeping the can and content at 20 F Find the final pressure to reach before closing the valve and the heat transfer Solution CV Tank Continuity Eq420 mA2E A 0 mAiE A Energy Eq421 mA2E AuA2E A 0 mAiE AhAiE A A1E AQA2E A State 2 20 F vA2E A VmA2E A 1 23112A3E A0030 4456 ftA3E Albm From Table F92 we locate the state between 15 and 20 psia P2 15 20 15 A 4456 46305 34479 46305E A 1574 psia u2 11268 Btulbm State i Table F91 hi 11907 Btulbm Now use the energy equation to solve for the heat transfer 1Q2 m2u2 mihi m2u2 hi 003 11268 11907 0192 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4205E An initially empty cylinder is filled with air from 70 F 15 psia until it is full Assuming no heat transfer is the final temperature larger equal to or smaller than 70 F Does the final T depends on the size of the cylinder This is a transient problem with no heat transfer and no work The balance equations for the tank as CV become Continuity Eq mA2E A 0 mAiE A Energy Eq mA2E AuA2E A 0 mAiE AhAiE A Q W mAiE AhAiE A 0 0 Final state uA2E A hAiE A PA2E A PAiE TA2E A TAiE A and it does not depend on V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4206E A tank contains 10 ftA3E A air at 15 psia 540 R A pipe flowing air at 150 psia 540 R is connected to the tank and it is filled slowly to 150 psia Find the heat transfer to reach a final temperature of 540 R CV The tank volume and the compressor This is a transient problem filling of tank Continuity Eq420 mA2E A mA1E A mAinE A Q i TANK 1 2 Energy Eq421 mA2E AuA2E A mA1E AuA1E A A1E AQA2E A A1E AWA2E A mAinE AhAinE Process Constant volume A1E AWA2E A 0 States uA1E A uA2E A uAinE A uA540E A hAinE A uAinE A RTAinE mA1E A PA1E AVA1E ARTA1E A 15 144 105334 540 075 lbm mA2E A PA2E AVA2E ARTA2E A 150 144 105334 540 7499 lbm Heat transfer from the energy equation A1E AQA2E A mA2E AuA2E A mA1E AuA1E A mAinE AhAinE A mA1E A mAinE A uA1E A mA1E AuA1E A mAinE AuAinE A mAinE ARTAinE mA1E AuA1E A mA1E AuA1E A mAinE AuA1E A mAinE AuAinE A mAinE ARTAinE A mAinE ARTAinE 7499 075 lbm 5334 ftlbflbmR 540 R 194 395 ftlbf 250 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4207E A 1ft3 tank shown in Fig P4136 that is initially evacuated is connected by a valve to an air supply line flowing air at 70 F 120 lbfin2 The valve is opened and air flows into the tank until the pressure reaches 90 lbfin2 Determine the final temperature and mass inside the tank assuming the process is adiabatic Develop an expression for the relation between the line temperature and the final temperature using constant specific heats Solution CV Tank Continuity Eq420 mA2E A mAiE A Energy Eq421 mA2E AuA2E A mAiE AhAiE A Table F5 uA2E A hAiE A 12678 Btulbm TA2E A 740 R TANK mA2E A A P2V ERT2 E A A90 144 1 5334 740E A 03283 lbm Assuming constant specific heat hAiE A uAiE A RTAiE A uA2E A RTAiE A uA2E A uAiE A CAVoE ATA2E A TAiE A CAVoE ATA2E A CAVoE A RTAiE A CAPoE ATAiE A TA2E A A CPo ECvo E A TAiE A kTAiE For TAiE A 5297 R constant CAPoE A TA2E A 140 5297 7416 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4208E Helium in a steel tank is at 40 psia 540 R with a volume of 4 ftA3E A It is used to fill a balloon When the tank pressure drops to 24 psia the flow of helium stops by itself If all the helium still is at 540 R how big a balloon did I get Assume the pressure in the balloon varies linearly with volume from 147 psia V 0 to the final 24 psia How much heat transfer did take place Solution Take a CV of all the helium This is a control mass the tank mass changes density and pressure Energy Eq UA2E A UA1E A A1E AQA2E A A1E AWA2E Process Eq P 147 CV State 1 PA1E A TA1E A VA1E State 2 PA2E A TA2E A VA2E A Ideal gas PA2E A VA2E A mRTA2E A mRTA1E A PA1E AVA1E A c i rc u s t her m o cb VA2E A VA1E APA1E APA2E A 4 4024 66667 ftA3E VAbalE A VA2E A VA1E A 66667 4 26667 ftA3E A1E AWA2E A P dV AREA ½ PA1E A PA2E A VA2E A VA1E A ½ 40 24 26667 144 12 288 lbfft 15791 Btu UA2E A UA1E A A1E AQA2E A A1E AWA2E A m uA2E A uA1E A mCAvE A TA2E A TA1E A 0 so A1E AQA2E A A1E AWA2E A 1579 Btu Remark The process is transient but you only see the flow mass if you select the tank or the balloon as a control volume That analysis leads to more terms that must be elliminated between the tank control volume and the balloon control volume Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4209E A 20ft3 tank contains ammonia at 20 lbfin2 80 F The tank is attached to a line flowing ammonia at 180 lbfin2 140 F The valve is opened and mass flows in until the tank is half full of liquid by volume at 80 F Calculate the heat transferred from the tank during this process Solution CV Tank Transient process as flow comes in mA1E A VvA1E A 2016765 1193 lbm mAf2E A VAf2E AvAf2E A 100026677 374855 lbm mAg2E A VAg2E AvAg2E A 1019531 5120 lbm mA2E A mAf2E A mAg2E A 379975 lbm xA2E A mAg2E A mA2E A 0013475 Table F81 uA2E A 1309 0013475 4434 1369 Btulbm uA1E A 5950 Btulbm hAiE A 6670 Btulbm Continuity Eq mAiE A mA2E A mA1E A 378782 lbm From continuity equation mAiE A mA2E A mA1E A 379975 1193 378782 lbm Energy eq QACVE A mAiE AhAiE A mA2E AuA2E A mA1E AuA1E QACVE A 379975 1369 1193 5950 378782 6670 201 339 Btu line Q Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4210E A nitrogen line 540 R and 75 lbfin2 is connected to a turbine that exhausts to a closed initially empty tank of 2000 ft3 as shown in Fig P4137 The turbine operates to a tank pressure of 75 lbfin2 at which point the temperature is 450 R Assuming the entire process is adiabatic determine the turbine work CV turbine tank Transient problem Conservation of mass mAiE A mA2E A m Energy Eq mAiE AhAiE A mA2E AuA2E A WACVE A WACVE A mhAiE A uA2E A Inlet state PAiE A 75 lbfinA2E A TAiE A 540 R Final state 2 PA2E A 75 lbfinA2E A TA2E A 450 R vA2E A RTA2E APA2E A 5515 45075 144 2298 ft3lbm mA2E A VvA2E A 20002298 87032 lbm hAiE A uA2E A uAiE A RTAiE A uA2E A RTAiE A CAvE A TAiE A TA2E A A 5515 77817E A 540 0178 540 450 3827 1602 5429 ABtu lbmE WACVE A 87032 5429 47 250 Btu W Turb 1 2 TANK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 4211E A massloaded pistoncylinder containing air is at 45 lbfin2 60 F with a volume of 9 ft3 while at the stops V 36 ft3 An air line 75 lbfin2 1100 R is connected by a valve as shown in Fig P4154 The valve is then opened until a final inside pressure of 60 lbfin2 is reached at which point T 630 R Find the air mass that enters the work and heat transfer Solution CV Cylinder volume Continuity Eq420 mA2E A mA1E A mAinE A Energy Eq421 mA2E AuA2E A mA1E AuA1E A mAinE AhAlineE A A1E AQA2E A A1E AWA2E Process PA1E A is constant to stops then constant V to state 2 at PA2E State 1 PA1E A TA1E A mA1E A A P1V ERT1 E A A 45 9 144 5334 5197E A 2104 lbm State 2 Open to PA2E A 60 lbfinA2E A TA2E A 630 R Table F5 hi 26613 btulbm u1 8868 Btulbm u2 10762 Btulbm AIR 1W2 APdVEA P1Vstop V1 45 36 9A144 778E A 2249 Btu m2 P2V2RT2 6036144 5334630 9256 lbm mi mA2E A mA1E A 9256 2104 7152 lbm A1E AQA2E A m2u2 m1u1 mi hi 1W2 9256 10762 2104 8868 7152 26613 2249 8689 Btu Updated June 2013 SOLUTION MANUAL CHAPTER 5 Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CONTENT CHAPTER 5 SUBSECTION PROB NO InText concept questions ag Concept problems 114 Heat engines and refrigerators 1536 Second law and processes 3743 Carnot cycles and absolute temperature 4471 Actual cycles 7280 Finite T heat transfer 8196 Ideal gas Carnot cycles 97100 Review problems 101120 The clipart used in the solution manual is from Microsoft Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful InText Concept Questions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5a Electrical appliances TV stereo use electric power as input What happens to the power Are those heat engines What does the second law say about those devices Most electric appliances such as TV VCR stereo and clocks dissipate power in electrical circuits into internal energy they get warm some power goes into light and some power into mechanical energy The light is absorbed by the room walls furniture etc and the mechanical energy is dissipated by friction so all the power eventually ends up as internal energy in the room mass of air and other substances These are not heat engines just the opposite happens namely electrical power is turned into internal energy and redistributed by heat transfer These are irreversible processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5b Geothermal underground hot water or steam can be used to generate electric power Does that violate the second law No Since the earth is not uniform we consider the hot water or steam supply as coming from one energy source the high T and we must reject heat to a low temperature reservoir as the ocean a lake or the atmosphere which is another energy reservoir Iceland uses a significant amount of steam to heat buildings and to generate electricity Microsoft clipart Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5c A windmill produces power on a shaft taking kinetic energy out of the wind Is it a heat engine Is it a perpetual machine Explain Since the wind is generated by a complex system driven by solar heat input and radiation out to space it is a kind of heat engine Microsoft clipart Within our lifetime it looks like it is perpetual However with a different time scale the climate will change the sun will grow to engulf the earth as it burns out of fuel There is a storage effect and a nonuniform distribution of states in the system that drives this Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5d heat engines and heat pumps refrigerators are energy conversion devices altering amounts of energy transfer between Q and W Which conversion direction Q W or W Q is limited and which is unlimited according to the second law The work output of a heat engine is limited Q to W You can transform W to Q unlimited a heat pump that does not work well or you may think about heat generated by friction 5e Ice cubes in a glass of liquid water will eventually melt and all the water approach room temperature Is this a reversible process Why There is heat transfer from the warmer ambient to the water as long as there is a temperature difference Eventually the temperatures approach each other and there is no more heat transfer This is irreversible as we cannot make icecubes out of the water unless we run a refrigerator and that requires a work from the surroundings which does not leave the surroundings unchanged Microsoft clipart Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5f Does a process become more or less reversible with respect to heat transfer if it is fast rather than slow Hint Recall from Chapter 3 that Q CA T If the higher heat transfer rate is caused by a larger T then the process is more irreversible so as the process would be slower due to a lower T then it approaches a reversible process If the rate of heat transfer is altered due to the factor CA with the same T then it is irreversible to the same degree Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5g If you generated hydrogen from say solar power which of these would be more efficient 1 transport it and then burn it in an engine or 2 convert the solar power to electricity and transport that What else would you need to know in order to give a definite answer Case 1 First there is a certain efficiency when converting solar power to hydrogen Then the transport and packaging of hydrogen has some energy expenditures associated with it The hydrogen could be compressed to a high pressure typically 70 MPa which is expensive in terms of work input and then stored in a tank One alternative would be to cool it down to become a liquid to have a much smaller volume but the temperature at which this happens is very low so the cooling and continued cooling under transport requires a significant work input also Certain materials like metalhydrides boron salt slurries and nanocarbon fibers allows storage of hydrogen at more modest pressures and are all currently under investigation as other alternative storage methods After the hydrogen is transported to an engine then the engine efficiency determines the work output Case 2 If the solar power is located where there is access to electrical transmission lines then it can be used in solar panels solar heating of water or other substance to run a heat engine cycle like a power plant to generate electricity All of these processes have a certain efficiency that must be evaluated to estimate the overall efficiency To make new transmission lines is costly and has an impact on the environment that must be considered You also need to look at the time of dayyear at which the power is required and when it is available The end use also presents some limitations like if the power should be used for a car then the energy must be stored temporarily like in a battery Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Concept Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 51 Two heat engines operate between the same two energy reservoirs and both receives the same QH One engine is reversible and the other is not What can you say about the two QLs The reversible heat engine can produce more work has a higher efficiency than the irreversible heat engine and due to the energy conservation it then gives out a smaller QL compared to the irreversible heat engine Wrev QH QL rev Wirrev QH QL irrev QL rev QL irrev Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 52 Compare two domestic heat pumps A and B running with the same work input If A is better than B which one heats the house most The statement that A is better means it has a higher COP and since QH A COPA W QH B COPB W it can thus provide more heat to the house The higher heat comes from the higher QL it is able to draw in Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 53 Suppose we forget the model for heat transfer as Q CA T can we draw some information about direction of Q from the second law One of the classical statements of the second law is the Clausius statement saying that you cannot have heat transfer from a lower temperature domain to a higher temperature domain without work input The opposite namely a transfer of heat from a high temperature domain towards a lower temperature domain can happen which is a heat engine with zero efficiency That is the only direction the heat transfer can have namely from the high T towards the low T environment Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 54 A combination of two heat engines is shown in Fig P54 Find the overall thermal efficiency as a function of the two individual efficiencies The overall efficiency ηTH W net Q H W 1 W 2 Q H η1 W 2 Q H For the second heat engine and the energy Eq for the first heat engine W 2 η2 Q M η2 1 η1 Q H so the final result is ηTH η1 η2 1 η1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 55 Compare two heat engines receiving the same Q one at 1200 K and the other at 1800 K they both reject heat at 500 K Which one is better The maximum efficiency for the engines are given by the Carnot heat engine efficiency as ηTH W net Q H 1 TL TH Since they have the same low temperature the one with the highest TH will have a higher efficiency and thus presumably better Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 56 A car engine takes atmospheric air in at 20oC no fuel and exhausts the air at 20oC producing work in the process What do the first and the second laws say about that Energy Eq W QH QL change in energy of air OK 2nd law Exchange energy with only one reservoir NOT OK This is a violation of the statement of KelvinPlanck Remark You cannot create and maintain your own energy reservoir Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 57 A combination of two refrigerator cycles is shown in Fig P57 Find the overall COP as a function of COP1 and COP2 The overall COP becomes COP β Q L W tot Q L W 1 W 1 W tot COP1 W 1 W tot COP1 1 1 W 2W 1 where we used W tot W 1 W 2 Use definition of COP2 and energy equation for refrigerator 1 to eliminate Q M as COP2 Q M W 2 and Q M W 1 Q L so we have W 2 Q M COP2 W 1 Q L COP2 and then W 2 W 1 1 Q LW 1 COP2 1 COP1 COP2 Finally substitute into the first equation and rearrange a little to get COP β COP1 COP2 COP1 COP2 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 58 After you have returned from a car trip the car engine has cooled down and is thus back to the state in which it started What happened to all the energy released in the burning of the gasoline What happened to all the work the engine gave out Solution All the energy from the fuel generates heat and work out of the engine The heat is directly dissipated in the atmosphere and the work is turned into kinetic energy and internal energy by all the frictional forces wind resistance rolling resistance brake action Eventually the kinetic energy is lost by braking the car so in the end all the energy is absorbed by the environment increasing its internal energy Microsoft clipart Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 59 Does a reversible heat engine burning coal which in practice cannot be done reversibly have impacts on our world other than depletion of the coal reserve Solution When you burn coal you form carbon dioxide CO2 which is a greenhouse gas It absorbs energy over a wide spectrum of wavelengths and thus traps energy in the atmosphere that otherwise would go out into space Coal from various locations also has sulfur and other substances like heavy metals in it The sulfur generates sulfuric acid resulting in acid rain in the atmosphere and can damage the forests Microsoft clipart Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 510 If the efficiency of a power plant goes up as the low temperature drops why do power plants not just reject energy at say 40oC In order to reject heat the ambient must be at the low temperature Only if we moved the plant to the North Pole would we see such a low T Remark You cannot create and maintain your own energy reservoir Microsoft clipart Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 511 If the efficiency of a power plant goes up as the low temperature drops why not let the heat rejection go to a refrigerator at say 10oC instead of ambient 20oC The refrigerator must pump the heat up to 20oC to reject it to the ambient The refrigerator must then have a work input that will exactly offset the increased work output of the power plant if they are both ideal As we can not build ideal devices the actual refrigerator will require more work than the power plant will produce extra Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 512 A coalfired power plant operates with a high T of 600oC whereas a jet engine has about 1400 K Does that mean we should replace all power plants with jet engines The thermal efficiency is limited by the Carnot heat engine efficiency That is the low temperature is also important Here the power plant has a much lower T in the condenser than the jet engine has in the exhaust flow so the jet engine does not necessarily have a higher efficiency than the power plant Gasturbines are used in power plants where they can cover peak power demands needed for shorter time periods and their high temperature exhaust can be used to boil additional water for the steam cycle W T Q H Q L W P in from coal to ambient Microsoft clipart Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 513 A heat transfer requires a temperature difference see chapter 3 to push the Q What implications does that have for a real heat engine A refrigerator This means that there are temperature differences between the source of energy and the working substance so TH is smaller than the source temperature This lowers the maximum possible efficiency As heat is rejected the working substance must have a higher temperature TL than the ambient receiving the Q L which lowers the efficiency further For a refrigerator the high temperature must be higher than the ambient to which the Q H is moved Likewise the low temperature must be lower than the cold space temperature in order to have heat transfer from the cold space to the cycle substance So the net effect is the cycle temperature difference is larger than the reservoir temperature difference and thus the COP is lower than that estimated from the cold space and ambient temperatures Both of these situations and statements are illustrated in Fig527 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 514 Hot combustion gases air at 1500 K are used as the heat source in a heat engine where the gas is cooled to 750 K and the ambient is at 300 K This is not a constant T source How does that affect the efficiency Solution If the efficiency is written as ηTH W net Q H 1 TL TH then TH is somewhere between 1500 K and 750 K and it is not a linear average H Q W L Q T L HE 1 2 cb After studying chapter 6 and 7 we can solve this problem and find the proper average high temperature based on properties at states 1 and 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Heat Engines and Refrigerators Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 515 A window mounted air conditioner removes 35 kJ from the inside of a home using 175 kJ work input How much energy is released outside and what is its coefficient of performance CV AC unit The energy QH goes into the outside air Energy Eq QH W QL 175 35 525 kJ COP β QL W 35 175 2 H Q L Q 35 kJ T L T amb REF W 175 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 516 A lawnmower tractor engine produces 18 hp using 40 kW of heat transfer from burning fuel Find the thermal efficiency and the rate of heat transfer rejected to the ambient Conversion Table A1 18 hp 18 07457 kW 13423 kW Efficiency ηTH W outQ H 13423 40 033 Energy equation Q L Q H W out 40 13423 266 kW Q H Microsoft clipart Q L W out Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 517 Calculate the thermal efficiency of the steam power plant cycle described in Example 49 Solution From solution to Example 49 wnet wt wp 6407 4 6367 kJkg qH qb 2831 kJkg ηTH wnetqH 6367 2831 0225 W T Q H Q L W P in Q 1 2 Notice we cannot write wnet qH qL as there is an extra heat transfer 1Q 2 as a loss in the line This needs to be accounted for in the overall energy equation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 518 Assume we have a refrigerator operating at steady state using 500 W of electric power with a COP of 25 What is the net effect on the kitchen air Take a CV around the whole kitchen The only energy term that crosses the control surface is the work input W apart from energy exchanged with the kitchen surroundings That is the kitchen is being heated with a rate of W Remark The two heat transfer rates are both internal to the kitchen Q H goes into the kitchen air and Q L actually leaks from the kitchen into the refrigerated space which is the reason we need to drive it out again Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 519 A room is heated with a 1500 W electric heater How much power can be saved if a heat pump with a COP of 25 is used instead Assume the heat pump has to deliver 1500 W as the Q H Heat pump β Q HW IN W IN Q Hβ 1500 25 600 W So the heat pump requires an input of 600 W thus saving the difference W saved 1500 W 600 W 900 W H Q W L Q T L HP Room in cb Microsoft clipart Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 520 Calculate the coefficient of performance of the R134a refrigerator given in Example 48 Solution From the definition β Q LW IN 1454 5 291 Notice we cannot write W IN Q H Q L as there is a small Q in the compressor This needs to be accounted for in the overall energy equation Q H W Q L Evaporato Condense Q loss Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 521 Calculate the thermal efficiency of the steam power plant cycle described in Problem 4118 From solution to Problem 4118 Turbine A5 π4022 003142 m2 V5 m v5A5 25 kgs 006163 m3kg 003142 m2 49 ms h6 19183 092 23928 23932 kJkg wT 3404 23932 2002 4922 1000 992 kJkg W T m wT 25 992 24 800 kW W NET 24800 300 24 500 kW From the solution to Problem 4120 Economizer A7 πD2 74 0004 418 m2 v7 0001 008 m3kg V2 V7 m vA7 25 0001008 0004418 57 ms V3 v3v2V2 0001 118 0001 008 57 63 ms V2 so kinetic energy change is unimportant qECON h3 h2 744 194 5500 kJkg Q ECON m qECON 25 5500 13 750 kW Generator A4 πD2 44 0031 42 m2 v4 0060 23 m3kg V4 m v4A4 25 0060 230031 42 479 ms qGEN 3426 744 4792 63221000 2683 kJkg Q GEN 25 2683 67 075 kW The total added heat transfer is Q H 13 758 67 075 80 833 kW ηTH W NETQ H 24500 80833 0303 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 522 A large coal fired power plant has an efficiency of 45 and produces net 1500 MW of electricity Coal releases 25 000 kJkg as it burns so how much coal is used per hour From the definition of the thermal efficiency and the energy release by the combustion called heating value HV we get W η Q H η m HV then m η HV W 1500 MW 045 25000 kJkg 1500 1000 kJs 045 25000 kJkg 13333 kgs 480 000 kgh Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 523 A window airconditioner discards 17 kW to the ambient with a power input of 500 W Find the rate of cooling and the coefficient of performance Solution In this case Q H 17 kW goes to the ambient so Energy Eq Q L Q H W 17 05 12 kW βREFRIG Q L W 12 05 24 H Q 17 kW W 05 kW L Q T L T amb REF Microsoft clipart Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 524 An industrial machine is being cooled by 04 kgs water at 15oC that is chilled from 35oC by a refrigeration unit with a COP of 3 Find the rate of cooling required and the power input to the unit Energy equation for heat exchanger Q L m h1 h2 m CP T1 T2 04 kgs 418 kJkgK 35 15 K 3344 kW β COP Q L W Q W L T H H Q REF 1 2 cb W Q L β 3344 3 1115 kW Comment An outside cooling tower is often used for this see Chapter 11 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 525 Calculate the COP of the R410A heat pump cycle described in Problem 4123 The cycle is given by the following states v P compressor evaporator condenser valve 2 1 3 4 5 6 Q H W C Q L cb Evaporator Condenser 3 4 5 6 1 2 Q loss Where Q H m h2 h3 005 kgs 367 134 kJk 1165 kW The COP is β COP Q H W IN 1165 kW 5 kW 233 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 526 A window airconditioner unit is placed on a laboratory bench and tested in cooling mode using 750 W of electric power with a COP of 175 What is the cooling power capacity and what is the net effect on the laboratory Definition of COP β Q L W Cooling capacity Q L β W 175 750 1313 W For steady state operation the Q L comes from the laboratory and Q H goes to the laboratory giving a net to the lab of W Q H Q L 750 W that is heating it Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 527 A farmer runs a heat pump with a 2 kW motor It should keep a chicken hatchery at 30oC which loses energy at a rate of 10 kW to the colder ambient Tamb What is the minimum coefficient of performance that will be acceptable for the heat pump Solution Power input W 2 kW Energy Eq for hatchery Q H Q Loss 10 kW Definition of COP β COP Q H W 10 2 5 Q leak Q Q H L W 2 kW HP cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 528 A sports car engine delivers 100 hp to the driveshaft with a thermal efficiency of 25 The fuel has a heating value of 40 000 kJkg Find the rate of fuel consumption and the combined power rejected through the radiator and exhaust Solution Heating value HV Q H m HV From the definition of the thermal efficiency W η Q H η m HV m ηHV W 100 07355 025 40 000 000736 kgs 736 gs Conversion of power from hp to kW in Table A1 Q L Q H W W η W 1 η 1 W 1 025 1 100 hp 07355 kWhp 221 kW Exhaust flow Air intake filter Shaft Fan power Fuel line cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 529 R410A enters the evaporator the cold heat exchanger in an AC unit at 20oC x 28 and leaves at 20oC x 1 The COP of the refrigerator is 15 and the mass flow rate is 0003 kgs Find the net work input to the cycle Energy equation for heat exchanger Q L m h2 h1 m hg hf x1 hfg m hfg x1 hfg m 1 x1hfg Q L 1 2 cb 0003 kgs 072 24365 kJkg 05263 kW β COP Q L W W Q L β 05263 15 035 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 530 In a Rankine cycle steam power plant 09 MW is taken out in the condenser 063 MW is taken out in the turbine and the pump work is 003 MW Find the plant thermal efficiency If everything could be reversed find the COP as a refrigerator Solution W T Q H Q L W P in CV Total plant Energy Eq Q H W Pin W T Q L Q H W T Q L W Pin 063 09 003 15 MW ηTH W T W Pin Q H 063 003 15 040 β Q L W T W Pin 09 063 003 15 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 531 An experimental power plant generates 130 MW of electrical power It uses a supply of 1200 MW from a geothermal source and rejects energy to the atmosphere Find the power to the air and how much air should be flowed to the cooling tower kgs if its temperature cannot be increased more than 12oC Solution CV Total power plant Energy equation gives the amount of heat rejection to the atmosphere as Q L Q H W 1200 130 1070 MW The energy equation for the air flow that absorbs the energy is Q L m air h m air Cp T m air Q L CpT 1070 1000 1004 12 MW kJkgK K 88 811 kgs This is too large to make so some cooling by liquid water or evaporative cooling should be used see chapter 11 Microsoft clipart H Q W T L L Q HE Air cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 532 A water cooler for drinking water should cool 25 Lh water from 18oC to 10oC while the water reservoir also gains 60 W from heat transfer Assume a small refrigeration unit with a COP of 25 does the cooling Find the total rate of cooling required and the power input to the unit The mass flow rate is m ρV 25 103 0001002 1 3600 kgs 693 gs Energy equation for heat exchanger Q L m h1 h2 Q H TR m CP T1 T2 Q H TR Q W L T H H Q REF 1 2 cb HTR Q 693 103 kgs 418 kJkgK 18 10 K 60 W 2918 W β COP Q L W W Q L β 2918 25 1167 W Comment The unit does not operate continuously so the instantaneous power is higher during the periods it does operate Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 533 A large stationary diesel engine produces 5 MW with a thermal efficiency of 40 The exhaust gas which we assume is air flows out at 800 K and the intake air is 290 K How large a mass flow rate is that assuming this is the only way we reject heat Can the exhaust flow energy be used Heat engine Q H W outηTH 5 04 125 MW Energy equation Q L Q H W out 125 5 75 MW Exhaust flow Q L m airh800 h290 m air Q L h800 h290 75 1000 8222 29043 kW kJkg 141 kgs The flow of hot gases can be used to heat a building or it can be used to heat water in a steam power plant since that operates at lower temperatures Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 534 For each of the cases below determine if the heat engine satisfies the first law energy equation and if it violates the second law a Q H 6 kW Q L 4 kW W 2 kW b Q H 6 kW Q L 0 kW W 6 kW c Q H 6 kW Q L 2 kW W 5 kW d Q H 6 kW Q L 6 kW W 0 kW Solution 1st law 2nd law a Yes Yes possible b Yes No impossible Kelvin Planck c No Yes but energy not conserved d Yes Yes Irreversible Q over T H Q W L Q T L T H HE cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 535 For each of the cases in problem 534 determine if a heat pump satisfies the first law energy equation and if it violates the second law a Q H 6 kW Q L 4 kW W 2 kW b Q H 6 kW Q L 0 kW W 6 kW c Q H 6 kW Q L 2 kW W 5 kW d Q H 6 kW Q L 6 kW W 0 kW Solution 1st law 2nd law a Satisfied Does not violate b Satisfied Does not violate c Violated Does not violate but 1st law d Satisfied Does violate Clausius H Q W L Q T L T H HP cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 536 Calculate the amount of work input a refrigerator needs to make ice cubes out of a tray of 025 kg liquid water at 10oC Assume the refrigerator has β 35 and a motorcompressor of 750 W How much time does it take if this is the only cooling load CV Water in tray We neglect tray mass Energy Eq mu2 u1 1Q2 1W2 Process P constant Po 1W2 P dV Pomv2 v1 1Q2 mu2 u1 1W2 mh2 h1 Tbl B11 h1 4199 kJkg Tbl B15 h2 3336 kJkg 1Q2 0253334 4199 93848 kJ Consider now refrigerator β QLW W QLβ 1Q2 β 9384835 2681 kJ For the motor to transfer that amount of energy the time is found as W W dt W t t WW 2681 1000J 750 W 3575 s Comment We neglected a baseload of the refrigerator so not all the 750 W are available to make ice also our coefficient of performance is very optimistic and finally the heat transfer is a transient process All this means that it will take much more time to make icecubes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Second Law and Processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 537 Prove that a cyclic device that violates the KelvinPlanck statement of the second law also violates the Clausius statement of the second law Solution Proof very similar to the proof in section 52 HE violating Kelvin receives QH from TH and produces net W QH This W input to HP receiving QL from TL HP discharges QH QL to TH Net Q to TH is QH QH QL QL Q W HE HP H Q L Q H Q L T H T L CV Total HE HP together transfers QL from TL to TH with no W thus violates Clausius Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 538 Discuss the factors that would make the power plant cycle described in Problem 4118 an irreversible cycle Solution General discussion but here are a few of the most significant factors 1 Combustion process that generates the hot source of energy 2 Heat transfer over finite temperature difference in boiler 3 Flow resistance and friction in turbine results in less work out 4 Flow friction and heat loss tofrom ambient in all pipes 5 Heat transfer over finite temperature difference in condenser Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 539 Discuss the factors that would make the heat pump described in Problem 4123 an irreversible cycle Solution General discussion but here are a few of the most significant factors 1 Unwanted heat transfer in the compressor 2 Pressure loss back flow leak in compressor 3 Heat transfer and pressure drop in line 1 2 4 Pressure drop in all lines 5 Throttle process 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 540 Assume a cyclic machine that exchanges 6 kW with a 250oC reservoir and has a Q L 0 kW W 6 kW b Q L 6 kW W 0 kW and Q L is exchanged with a 30oC ambient What can you say about the processes in the two cases a and b if the machine is a heat engine Repeat the question for the case of a heat pump Solution Heat engine a Since Q L 0 impossible Kelvin Planck b Possible irreversible ηeng 0 Ηeat pump a Possible irreversible like an electric heater b Impossible β Clausius Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 541 Consider a heat engine and heat pump connected as shown in figure P541 Assume TH1 TH2 Tamb and determine for each of the three cases if the setup satisfy the first law andor violates the 2nd law Q H1 Q L1 W 1 Q H2 Q L2 W 2 a 6 4 2 3 2 1 b 6 4 2 5 4 1 c 3 2 1 4 3 1 Solution 1st law 2nd law a Yes Yes possible b Yes No combine Kelvin Planck c Yes No combination clausius Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 542 Consider the four cases of a heat engine in problem 534 and determine if any of those are perpetual machines of the first or second kind a Q H 6 kW Q L 4 kW W 2 kW b Q H 6 kW Q L 0 kW W 6 kW c Q H 6 kW Q L 2 kW W 5 kW d Q H 6 kW Q L 6 kW W 0 kW H Q W L Q T L T H HE cb Solution 1st law 2nd law a Yes Yes possible b Yes No impossible Kelvin Planck Perpetual machine second kind It violates the 2nd law converts all Q to W c No Yes but energy not conserved Perpetual machine first kind It generates energy inside d Yes Yes Irreversible Q over T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 543 The simple refrigeration cycle is shown in Problem 523 and in Fig 56 Mention a few of the processes that are expected to be irreversible The throttling process is highly irreversible Both of the heat transfer processes are externally irreversible large T between the working substance and the source or sink energy reservoir Both of the heat transfer processes are also internally irreversible smaller T in the substance so there is not a single uniform T across the flow cross sectional area This is necessary to redistribute the energy uniformly in the working substance The compressor has friction and flow losses so not all of the shaft work goes into raising the substance pressure Such an effect is described by a device efficiency in chapter 7 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Carnot Cycles and Absolute Temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 544 Calculate the thermal efficiency of a Carnot cycle heat engine operating between reservoirs at 300oC and 45oC Compare the result to that of Example 47 Solution ηTH Wnet QH 1 TL TH 1 45 273 300 273 0445 Carnot ηEX 47 wnet qH 6408 4 28311 0225 efficiency about ½ of the Carnot Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 545 An ideal Carnot heat engine has an efficiency of 40 If the high temperature is raised 15 what is the new efficiency keeping the same low temperature Solution ηTH Wnet QH 1 TL TH 04 TL TH 06 so if TH is raised 15 the new ratio becomes TL TH new 06 115 05217 ηTH new 1 05217 0478 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 546 At a few places where the air is very cold in the winter like 30oC it is possible to find a temperature of 13oC down below ground What efficiency will a heat engine have operating between these two thermal reservoirs Solution ηTH 1 TL TH The ground becomes the hot source and the atmosphere becomes the cold side of the heat engine ηTH 1 273 30 273 13 1 243 286 015 This is low because of the modest temperature difference Microsoft clipart Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 547 Consider the combination of a heat engine and a heat pump as in Problem 541 with a low temperature of 400 K What should the high temperature be so that the heat engine is reversible For that temperature what is the COP for a reversible heat pump For all three cases of the heat engine the ratio between the heat transfers and the work term is the same as Q H Q L W 642 321 For a reversible heat engine we must have the heat transfer ratio equal to the temperature ratio so Q H Q L TH TL 3 2 400 K TH TH 32 400 K 600 K The COP is COPHP Q H W 3 1 3 TH TH TL 600 600 400 W L1 Q T H1 H1 Q 400 K HE L2 Q HP Q T H2 H2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 548 Find the power output and the low T heat rejection rate for a Carnot cycle heat engine that receives 6 kW at 250oC and rejects heat at 30oC as in Problem 540 Solution From the definition of the absolute temperature Eq 54 ηcarnot 1 TL TH 1 303 523 042 Definition of the heat engine efficiency gives the work as W η Q H 042 6 252 kW Apply the energy equation Q L Q H W 6 252 348 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 549 A large heat pump should upgrade 4 MW of heat at 65C to be delivered as heat at 145C What is the minimum amount of work power input that will drive this For the minimum work we assume a Carnot heat pump and Q L 4 MW βHP W in Q H TH TL TH 27315 145 145 65 5227 βREF βHP 1 Q L W in 4227 Now we can solve for the work W in Q LβREF 44227 0946 MW This is a domestic or small office building size AC unit much smaller than the 4 MW in this problem C Borgnakke Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 550 A temperature of about 001 K can be achieved by magnetic cooling In this process a strong magnetic field is imposed on a paramagnetic salt maintained at 1 K by transfer of energy to liquid helium boiling at low pressure The salt is then thermally isolated from the helium the magnetic field is removed and the salt temperature drops Assume that 1 mJ is removed at an average temperature of 01 K to the helium by a Carnotcycle heat pump Find the work input to the heat pump and the coefficient of performance with an ambient at 300 K Solution β QL Win TL TH TL 01 2999 000033 Win β QL 000033 1 103 3 J Remark This is an extremely large temperature difference for a heat pump A real one is built as a refrigerator within a refrigerator etc Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 551 The lowest temperature that has been achieved is about 1 106 K To achieve this an additional stage of cooling is required beyond that described in the previous problem namely nuclear cooling This process is similar to magnetic cooling but it involves the magnetic moment associated with the nucleus rather than that associated with certain ions in the paramagnetic salt Suppose that 10 µJ is to be removed from a specimen at an average temperature of 105 K ten micro joules is about the potential energy loss of a pin dropping 3 mm Find the work input to a Carnot heat pump and its coefficient of performance to do this assuming the ambient is at 300 K Solution The heat removed from the cold space is QL 10 µJ 10106 J at TL 105 K Carnot heat pump satisfies Eq74 QH QL TH TL 10 106 J 300 105 300 J From the energy equation for the heat pump Win QH QL 300 10 106 300 J β QL Win 10106 300 333108 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 552 Consider the setup with two stacked temperature wise heat engines as in Fig P54 Let TH 850 K TM 600 K and TL 350 K Find the two heat engine efficiencies and the combined overall efficiency assuming Carnot cycles The individual efficiencies η1 1 TM TH 1 600 850 0294 η2 1 TL TM 1 350 600 0417 The overall efficiency ηTH W net Q H W 1 W 2 Q H η1 W 2 Q H For the second heat engine and the energy Eq for the first heat engine W 2 η2 Q M η2 1 η1 Q H so the final result is ηTH η1 η2 1 η1 0294 0417 1 0294 0588 Comment It matches a single heat engine ηTH 1 TL TH 1 350 850 0588 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 553 Find the maximum coefficient of performance for the refrigerator in your kitchen assuming it runs in a Carnot cycle Solution The refrigerator coefficient of performance is β QLW QLQH QL TLTH TL Assuming TL 0C TH 35C β 27315 35 0 78 Actual working fluid temperatures must be such that TL Trefrigerator and TH T room A refrigerator does not operate in a Carnot cycle The actual vapor compression cycle is examined in Chapter 9 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 554 A car engine burns 5 kg fuel equivalent to addition of QH at 1500 K and rejects energy to the radiator and the exhaust at an average temperature of 750 K If the fuel provides 40 000 kJkg what is the maximum amount of work the engine can provide Solution A heat engine QH m qfuel 5 kg 40 000 kJkg 200 000 kJ Assume a Carnot efficiency maximum theoretical work η 1 TL TH 1 750 1500 05 W η QH 100 000 kJ Exhaust flow Air intake filter Coolant flow Atm air Shaft Fan power Radiator Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 555 An airconditioner provides 1 kgs of air at 15C cooled from outside atmospheric air at 35C Estimate the amount of power needed to operate the airconditioner Clearly state all assumptions made Solution Consider the cooling of air which needs a heat transfer as Q air m h m CpT 1 kgs 1004 kJkg K 20 K 20 kW Assume Carnot cycle refrigerator β Q L W Q L Q H Q L TL TH TL 273 15 35 15 144 W Q L β 200 144 139 kW This estimate is the theoretical maximum performance To do the required heat transfer TL 5C and TH 45C are more likely secondly β βcarnot H Q W L Q REF 35 C 15 C cb o o Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 556 A refrigerator should remove 400 kJ from some food Assume the refrigerator works in a Carnot cycle between 15oC and 45oC with a motorcompressor of 400 W How much time does it take if this is the only cooling load Assume Carnot cycle refrigerator β Q L W Q L Q H Q L TL TH TL 273 15 45 15 43 This gives the relation between the low T heat transfer and the work as Q L Q t β W 43 W t Q β W 400 1000 43 400 J W 233 s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 557 Calculate the amount of work input a refrigerator needs to make ice cubes out of a tray of 025 kg liquid water at 10oC Assume the refrigerator works in a Carnot cycle between 8oC and 35oC with a motorcompressor of 600 W How much time does it take if this is the only cooling load Solution CV Water in tray We neglect tray mass Energy Eq mu2 u1 1Q2 1W2 Process P constant Po 1W2 P dV Pomv2 v1 1Q2 mu2 u1 1W2 mh2 h1 Tbl B11 h1 4199 kJkg Tbl B15 h2 3336 kJkg 1Q2 025 kg 3334 4199 kJkg 93848 kJ Consider now refrigerator β QL W QL QH QL TL TH TL 273 8 35 8 616 W QL β 1Q2 β 93848 616 1524 kJ For the motor to transfer that amount of energy the time is found as W W dt W t t W W 1524 1000 600 J W 254 s Comment We neglected a baseload of the refrigerator so not all the 600 W are available to make ice also our coefficient of performance is very optimistic and finally the heat transfer is a transient process All this means that it will take much more time to make icecubes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 558 We propose to heat a house in the winter with a heat pump The house is to be maintained at 20C at all times When the ambient temperature outside drops to 10C the rate at which heat is lost from the house is estimated to be 25 kW What is the minimum electrical power required to drive the heat pump Solution Minimum power if we assume a Carnot cycle Q leak Q Q H L W HP Energy equation for the house steady state Q H Q leak 25 kW β Q H W IN TH THTL 2932 20 10 9773 W IN 25 9773 256 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 559 A household freezer operates in a room at 20C Heat must be transferred from the cold space at a rate of 2 kW to maintain its temperature at 30C What is the theoretically smallest power motor required to operate this freezer Solution Assume a Carnot cycle between TL 30C and TH 20C β Q L W in TL TH TL 27315 30 20 30 486 W in Q Lβ 2486 041 kW This is the theoretical minimum power input Any actual machine requires a larger input H Q W L Q T L T amb REF 2 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 560 A thermal storage is made with a rock granite bed of 2 m3 which is heated to 400 K using solar energy A heat engine receives a QH from the bed and rejects heat to the ambient at 290 K The rock bed therefore cools down and as it reaches 290 K the process stops Find the energy the rock bed can give out What is the heat engine efficiency at the beginning of the process and what is it at the end of the process Solution Assume the whole setup is reversible and that the heat engine operates in a Carnot cycle The total change in the energy of the rock bed is u2 u1 q C T 089 kJkgK 400 290 K 979 kJkg m ρV 2750 kgm3 2 m3 5500 kg Q mq 5500 kg 979 kJkg 538 450 kJ To get the efficiency use the CARNOT cycle result as η 1 ToTH 1 290400 0275 at the beginning of process η 1 ToTH 1 290290 00 at the end of process W Q Q H L HE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 561 It is proposed to build a 1000MW electric power plant with steam as the working fluid The condensers are to be cooled with river water see Fig P561 The maximum steam temperature is 550C and the pressure in the condensers will be 10 kPa Estimate the temperature rise of the river downstream from the power plant Solution W NET 106 kW TH 550C 8233 K PCOND 10 kPa TL TG P 10 kPa 458C 319 K ηTH CARNOT TH TH TL 8232 319 8232 06125 Q L MIN 106 1 06125 06125 06327 106 kW But m H2O 60 8 1060 0001 80 000 kgs having an energy flow of Q L MIN m H2O h m H2O CP LIQ H2O T H2O MIN TH2O MIN m H2OCP LIQ H2O Q L MIN 80000 4184 06327106 kW kgs kJkgK 19C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 562 A certain solarenergy collector produces a maximum temperature of 100C The energy is used in a cyclic heat engine that operates in a 10C environment What is the maximum thermal efficiency If the collector is redesigned to focus the incoming light what should the maximum temperature be to produce a 25 improvement in engine efficiency Solution For TH 100C 3732 K TL 2832 K ηth max TH TH TL 90 3732 0241 The improved efficiency is ηth max 0241 125 0301 With the Carnot cycle efficiency ηth max TH TH TL 1 TH TL 0301 Then TH TL 1 0301 405 K 132C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 563 A constant temperature of 125C must be obtained in a cryogenic experiment although it gains 120 W due to heat transfer What is the smallest motor you would need for a heat pump absorbing heat from the container and rejecting heat to the room at 20C Solution We do not know the actual device so find the work for a Carnot cycle βREF Q L W TL TH TL 14815 20 125 1022 W Q L βREF 120 W1022 117 4 W Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 564 Helium has the lowest normal boiling point of any of the elements at 42 K At this temperature the enthalpy of evaporation is 833 kJkmol A Carnot refrigeration cycle is analyzed for the production of 1 kmol of liquid helium at 42 K from saturated vapor at the same temperature What is the work input to the refrigerator and the coefficient of performance for the cycle with an ambient at 300 K Solution For the Carnot cycle the ratio of the heat transfers is the ratio of temperatures QL n h fg 1 kmol 833 kJkmol 833 kJ QH QL TH TL 833 300 42 5950 kJ WIN QH QL 5950 833 58867 kJ β QL WIN 833 58867 00142 TH TL TL Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 565 R134a fills a 01m3 capsule at 20C 200 kPa It is placed in a deep freezer where it is cooled to 10C The deep freezer sits in a room with ambient temperature of 20C and has an inside temperature of 10C Find the amount of energy the freezer must remove from the R134a and the extra amount of work input to the freezer to perform the process Solution CV R134a out to the 10C space Energy equation mu2 u1 1Q2 1W 2 Process V Const v2 v1 1W2 0 Table B52 v1 011436 m3kg u1 39527 kJkg m V v1 087443 kg State 2 v2 v1 vg 009921 m3kg Table B51 sup vap Tv interpolate between 150 kPa and 200 kPa in B52 P2 150 50 011436 013602 010013 013602 150 50 06035 180 kPa u2 37344 06035 37231 37344 37276 kJkg 1Q2 mu2 u1 087443 kg 37276 39527 kJkg 1968 kJ Consider the freezer and assume Carnot cycle β QL W QL QH QL TL TH TL 273 10 20 10 8767 Win QL β 1968 kJ 8767 2245 kJ 10 C o R 134a Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 566 A heat engine has a solar collector receiving 02 kW per square meter inside which a transfer media is heated to 450 K The collected energy powers a heat engine which rejects heat at 40oC If the heat engine should deliver 25 kW what is the minimum size area solar collector Solution TH 450 K TL 40oC 31315 K ηHE 1 TL TH 1 31315 450 0304 W η Q H Q H η W 25 0304 kW 8224 kW Q H 02 kWm2 A A Q H 02 41 m2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 567 A heat pump is driven by the work output of a heat engine as shown in figure P567 If we assume ideal devices find the ratio of the total power Q L1 Q H2 that heats the house to the power from the hot energy source Q H1 in terms of the temperatures βHP Q H2W Q H2Q H2 Q L2 Troom TroomTamb W ηHE Q H1 1 Troom TH Q H1 W Q H2βHP Troom TroomTamb Q H2 Q L1 Q H1 W 11 TH Troom Q H1 EA 1A Troom E TH A A TroomTamb Troom AE A A Troom E TH E A EA Troom TA2 Aroom ATAH A E TroomT Aamb AE A Q H2 Q L1 Q H1 11 Troom TH TAroomE A EA 1 TH E A EA 1 A Troom E TH A Troom TAamb AE A E A Troom E TH E A 1 A TH Troom E Troom Tamb E AE A Troom E TH E A A TH Tamb E Troom Tamb E A W L1 Q T H H1 Q House T room HE L2 Q HP H2 Q T amb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 568 Sixty kilograms per hour of water runs through a heat exchanger entering as saturated liquid at 200 kPa and leaving as saturated vapor The heat is supplied by a Carnot heat pump operating from a lowtemperature reservoir at 16C with a COP of half that of a similar Carnot unit Find the rate of work into the heat pump Solution CV Heat exchanger Am E A1E A Am E A2E A Am E A1E AhA1E A AQ E AHE A Am E A1E AhA2E Table B12 hA1E A 5047 kJkg hA2E A 27067 kJkg TAHE A TAsatE AP 12093 27315 39408 K AQ E AHE A A 60 3600E A27067 5047 367 kW H Q W L Q T L HP 1 2 cb First find the COP of a Carnot heat pump β AQ E AHE AAW E A TAHE A TAHE A TALE A 39408 39408 28915 376 Now we can do the actual one as βA HE A 3762 188 AW E A AQ E AHE AβA HE A 367 kW 188 1952 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 569 A power plant with a thermal efficiency of 40 is located on a river similar to Fig P561 With a total river mass flow rate of 1 10A5E A kgs at 15AoE AC find the maximum power production allowed if the river water should not be heated more than 1 degree The maximum heating allowed determines the maximum AQ E ALE A as AQ E ALE A AmE AH2OE A h AmE AH2OE A CAP LIQ H2O E ATAH2OE A 1 10A5E A kgs 418 kJkgK 1 K 418 MW AW E ANETE A1ηATH acE A 1 AW E ANETE A AQ E ALE A 1ηATH acE A 1 AQ E ALE A A ηTH ac E1 ηTH ac E 418 MW A 04 1 04E A 279 MW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 570 Liquid sodium leaves a nuclear reactor at 800C and is used as the energy source in a steam power plant The condenser cooling water comes from a cooling tower at 15C Determine the maximum thermal efficiency of the power plant Is it misleading to use the temperatures given to calculate this value Solution TAHE A 800C 10732 K TALE A 15C 2882 K ηATH MAXE A A TH TL ETH E A A10732 2882 10732E A 0731 It might be misleading to use 800C as the value for TAHE A since there is not a supply of energy available at a constant temperature of 800C liquid Na is cooled to a lower temperature in the heat exchanger The Na cannot be used to boil HA2E AO at 800C Similarly the HA2E AO leaves the cooling tower and enters the condenser at 15C and leaves the condenser at some higher temperature The water does not provide for condensing steam at a constant temperature of 15C LIQ Na 800 o C REACTOR ENERGY TO H O 2 COND COOLING TOWER ENERGY FROM STEAM POWER PLANT Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 571 The management of a large factory cannot decide which of two fuels to purchase The selected fuel will be used in a heat engine operating between the fuel burning temperature and a low exhaust temperature of Fuel A burns at 2200 K and exhausts at 450 K delivering 30 000 kJkg and costs 150kg Fuel B burns at 1200 K and exhausts at 350 K delivering 40 000 kJkg and costs 130kg Which fuel would you buy and why Solution Fuel A ηATHAE A 1 A TL ETH E A 1 A 450 2200E A 0795 WAAE A ηATHAE A QAAE A 0795 30 000 23 850 kJkg WAAE AAAE A 23 85015 15 900 kJ Fuel B ηATHBE A 1 A TL ETH E A 1 A 350 1200E A 0708 WABE A ηATHBE A QABE A 0708 40 000 28 320 kJkg WABE AABE A 28 32013 21 785 kJ Select fuel B for more work per dollar though it has a lower thermal efficiency Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 572 A sales person selling refrigerators and deep freezers will guarantee a minimum coefficient of performance of 45 year round How would you evaluate that Are they all the same Solution Assume a high temperature of 35C If a freezer compartment is included TALE A 20C deep freezer and fluid temperature is then TALE A 30C βAdeep freezerE A A TL ETH TL E A A27315 30 35 30E A 374 A hot summer day may require a higher TAHE A to push QAHE A out into the room so even lower β Claim is possible for a refrigerator but not for a deep freezer Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 573 A cyclic machine shown in Fig P573 receives 325 kJ from a 1000 K energy reservoir It rejects 125 kJ to a 400 K energy reservoir and the cycle produces 200 kJ of work as output Is this cycle reversible irreversible or impossible Solution ηACarnotE A 1 A TL ETH E A 1 A 400 1000E A 06 ηAengE A A W QH E A A200 325E A 0615 ηACarnotE This is impossible H Q 325 kJ W 200 kJ L Q 125 kJ T 1000 K H HE cb T 400 K L Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 574 Consider the previous problem and assume the temperatures and heat input are as given If the actual machine has an efficiency that is half that of the corresponding Carnot cycle find the work out and the rejected heat transfer ηACarnotE A 1 A TL ETH E A 1 A 400 1000E A 06 ηAengE A ηACarnotE A2 03 A W QH E A W 03 QAHE A 03 325 975 kJ H Q 325 kJ W L Q T 1000 K H HE cb T 400 K L QALE A QAHE A W 325 975 kJ 2275 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 575 Repeat problem 561 using a more realistic thermal efficiency of 45 AW E ANETE A 10A6E A kW ηATH acE A AQ E AHE A ηATH acE A 045 AQ E ALE A AQ E AHE A AW E ANETE A AW E ANETE A ηATH acE A AW E ANETE A AW E ANETE A1ηATH acE A 1 10A6E A kW A 1 045 045 E A 1222 10A6E A kW But AmE AH2OE A A60 8 1060 0001E A 80 000 kgs having an energy flow of AQ E ALE A AmE AH2OE A h AmE AH2OE A CAP LIQ H2O E ATAH2OE TAH2OE A A Q L Em H2OCP LIQ H2O E A A 1222 106 E80 000 418E A A kW kgs kJkgKE A 365C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 576 An inventor has developed a refrigeration unit that maintains the cold space at 10C while operating in a 25C room A coefficient of performance of 85 is claimed How do you evaluate this Solution βACarnotE A A QL EWin E A A TL ETH TL E A A 26315 25 10E A 752 85 βACarnotE A impossible claim H Q W L Q T 10C L T 25C H REF Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 577 A heat pump receives energy from a source at 80AoE AC and delivers energy to a boiler that operates at 350 kPa The boiler input is saturated liquid water and the exit is saturated vapor both at 350 kPa The heat pump is driven by a 25 MW motor and has a COP that is 60 of a Carnot heat pump COP What is the maximum mass flow rate of water the system can deliver TAHE A TAsatE A 13888AoE AC 412 K hAfgE A 21481 kJkg βAHP CarnotE A A Q H EW in E A A TH ETH TL E A A 412 13888 80E A 7 βAHP acE A 06 7 42 AQ HW E inE A AQ HE A 42 AW inE A 42 25 MW 105 MW AmE A hAfgE AmE A AQ HE A hAfgE A 10 500 kW 21481 kJkg 489 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 578 In a remote location you run a heat engine to provide power to run a refrigerator The input to the heat engine is 800 K and the low T is 400 K it has an actual efficiency equal to half of that of the corresponding Carnot unit The refrigerator has TALE A 10AoE AC and TAHE A 35AoE AC with a cop that is onethird that of the corresponding Carnot unit Assume a cooling capacity of 2 kW is needed and find the rate of heat input to the heat engine Heat engine ηACarnotE A 1 A TL ETH E A 1 A400 800E A 05 ηAacE A 025 Refrigerator βAref CarnotE A A TL ETH TL E A A 273 10 35 10E A 5844 βAref acE A 195 Cooling capacity AQ E ALE A 2 kW βAref acE A AW E A AW E A 2 kW195 1026 kW This work must be provided by the heat engine AW E A ηAacE A AQ E AHE AQ E AHE A AW E A ηAacE A 1026 kW 025 41 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 579 A car engine with a thermal efficiency of 33 drives the airconditioner unit a refrigerator besides powering the car and other auxiliary equipment On a hot 35AoE AC summer day the AC takes outside air in and cools it to 5AoE AC sending it into a duct using 2 kW of power input and it is assumed to be half as good as a Carnot refrigeration unit Find the rate of fuel kW being burned extra just to drive the AC unit and its COP Find the flow rate of cold air the AC unit can provide W extra η AQ E AH extraE A AQ E AH extraE A W extra η 2 kW 033 6 kW β A QL EWIN E A 05 βACarnotE A 05 A TL ETH TL E A 05 A5 27315 35 5E A 4636 AQ E ALE A β AW E A 4636 2 kW 9272 kW AmE AairE A CAP airE A TAairE AmE AairE A AQ E ALE A CAP airE A TAairE A A 9272 kW 1004 kJkgK 35 5 KE A 0308 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 580 A large heat pump should upgrade 5 MW of heat at 85C to be delivered as heat at 150C Suppose the actual heat pump has a COP of 25 how much power is required to drive the unit For the same COP how high a high temperature would a Carnot heat pump have assuming the same low T This is an actual COP for the heat pump as βAHPE A COP AQ E AHE AAW E AinE A 25 AQ E ALE AAW E AinE A 15 AW E AinE A AQ E ALE A 15 5 15 3333 MW The Carnot heat pump has a COP given by the temperatures as βAHPE A AQ E AHE AAW E AinE A A TH ETH TL E A 25 TAHE A 25 TAHE A 25 TALE A TAHE A A25 15E A TALE A A5 3E A 85 27315 597 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Finite T Heat Transfer Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 581 A refrigerator keeping 5AoE AC inside is located in a 30AoE AC room It must have a high temperature T above room temperature and a low temperature T below the refrigerated space in the cycle to actually transfer the heat For a T of 0 5 and 10AoE AC respectively calculate the COP assuming a Carnot cycle Solution From the definition of COP and assuming Carnot cycle β A QL EWIN E A A TL ETH TL E A when Ts are absolute temperatures T TAHE TAHE TALE TALE β AoE AC K AoE AC K a 0 30 303 5 278 111 b 5 35 308 0 273 78 c 10 40 313 5 268 596 Notice how the COP drops significantly with the increase in T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 582 The ocean near Havaii has 20AoE AC near the surface and 5AoE AC at some depth A power plant based on this temperature difference is being planned How large an efficiency could it have If the two heat transfer terms QH and QL both require a 2 degree difference to operate what is the maximum efficiency then Solution TAHE A 20C 2932 K TALE A 5C 2782 K ηATH MAXE A A TH TL ETH E A A2932 2782 2932E A 0051 ηATH modE A A TH TL ETH E A A2912 2802 2912E A 0038 This is a very low efficiency so it has to be done on a very large scale to be economically feasible and then it will have some environmental impact Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 583 A house is cooled by a heat pump driven by an electric motor using the inside as the lowtemperature reservoir The house gains energy in direct proportion to the temperature difference as Q gain KTH TL Determine the minimum electric power to drive the heat pump as a function of the two temperatures Solution Refrigerator COP β AQ E ALE AAW E AinE A TALE ATAHE A TALE A Heat gain must be removed AQ E ALE A Q gain KTAHE A TALE A Solve for required work and substitute in for β AW E AinE A AQ E ALE Aβ KTAHE A TALE A TAHE A TALE ATALE A AW E AinE A KTAHE A TALE AA2E ATALE Q KT T gain Q Q H L W HP cb H L Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 584 An air conditioner in a very hot region uses a power input of 25 kW to cool a 5C space with the high temperature in the cycle as 40C The QH is pushed to the ambient air at 30C in a heat exchanger where the transfer coefficient is 50 Wm2K Find the required minimum heat transfer area Solution AW E A 25 kW Q H βHP βHP TAHE ATAHE A TALE A A273 40 40 5E A 8943 Q H AW E A βHP 25 8943 2236 kW h A T A Q H h T A 2236 103 E50 40 30E A 4472 m2 T H Q W T L L Q amb REF 25 kW 30 C 40 C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 585 A small house is kept at 20C inside loses 12 kW to the outside ambient at 0C A heat pump is used to help heat the house together with possible electric heat The heat pump is driven by a motor of 25 kW and it has a COP that is ¼ of a Carnot heat pump unit Find the actual COP for the heat pump and the amount of electric heat that must be used if any to maintain the house temperature CV House Energy 0 AQ E AHE A AW E AelE A AQ E ALossE Definition of COP β COPAHPE A A Q H EW E A A1 4E A A TH ETH TL E A A1 4E A A29315 20 0E A 3664 AQ E AHE A COPAHPE A AW E AHPE A 3664 25 kW 916 kW AW E AelE A AQ E ALossE A AQ E AHE A 12 916 284 kW CV Total Energy AQ E ALE A AQ E AHE A AW E AHPE A 916 25 666 kW Entropy 0 AQ LE ATALE A AQ lossE ATALE A S gen S gen AQ lossE A AQ LE A TALE A A12 666 27315E A 00195 kWK Q loss Q Q H L W 25 kW HP cb W el Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 586 Consider a room at 20AoE AC that is cooled by an air conditioner with a COP of 32 using a power input of 2 kW and the outside is at 35AoE AC What is the constant in the heat transfer Eq 514 for the heat transfer from the outside into the room Definition of the coefficient of performance Eq52 and Eq514 AQ E ALE A βAACE AW E A 32 2 kW 64 kW AQ E Aleak inE A CA ΔT CA AQ E ALE A ΔT A 64 kW 35 20 KE A 0427 kWK Q leak Q Q H L W 2 kW AC cb T L Here TALE A TAhouseE A 20AoE AC TAHE A TAambE A 35AoE AC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 587 A car engine operates with a thermal efficiency of 35 Assume the air conditioner has a coefficient of performance of β 3 working as a refrigerator cooling the inside using engine shaft work to drive it How much fuel energy should be spend extra to remove 1 kJ from the inside Solution Car engine W ηAengE A QAfuelE A Air conditioner β A QL EWE W ηAengE A QAfuelE A A QL EβE A QAfuelE A QL ηAengE A β A 1 035 3E A 0952 kJ W L Q T H H Q T L REF Fuel Q HE L eng Q FUEL Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 588 Arctic explorers are unsure if they can use a 5kW motor driven heat pump to stay warm It should keep their shelter at 15C The shelter losses energy at a rate of 05 kW per degree difference to the colder ambient The heat pump has a COP that is 50 of a Carnot heat pump If the ambient temperature can fall to 25C at night would you recommend this heat pump to the explorers CV Heat pump The heat pump should deliver a rate of heating that equals the heat loss to the ambient for steady inside temperature COP β AQ E AHE AAW E A 05 βACarnotE A A1 2E A A TH ETH TL E A A1 2E A A273 15 15 25E A 36 The heat pump can then provide a heating capacity of AQ E AHE A β AW E A 36 5 kW 18 kW The heat loss is AQ E Aleak outE A CA ΔT 05 kWK 15 25 K 20 kW The heat pump is not sufficient to cover the loss and not recommended Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 589 Using the given heat pump in the previous problem how warm would it make the shelter in the arctic night The high is now an unknown so both the heat loss and the heat pump performance depends on that The energy balance around the shelter then gives AQ E AHE A β AW E A AQ E Aleak outE A CA ΔT Substitute the expression for β and CA ΔT to give A1 2E A A TH ETH TL E A AW E A 05 kWK TAHE A TALE A Multiply with the temperature difference factor 2 and divide by the work to get TAHE A A05 2 5 KE A TAHE A TALE AA2E A A02 KE A TAHE A TALE AA2E Solve this equation like 02 xA2E A x TALE A 0 with x TAHE A TALE A and TALE A 24815 K x TAHE A TALE A 3781 K negative root discarded TAHE A x TALE A 3781 25 128C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 590 An air conditioner cools a house at TALE A 20C with a maximum of 12 kW power input The house gains 06 kW per degree temperature difference to the ambient and the refrigeration COP is β 06 βACarnotE A Find the maximum outside temperature TAHE A for which the air conditioner provides sufficient cooling Solution Q leak Q Q H L W 12 kW HP cb T L Here TALE A TAhouseE TAHE A TAambE In this setup the low temperature space is the house and the high temperature space is the ambient The heat pump must remove the gain or leak heat transfer to keep it at a constant temperature Q leak 06 TAambE A TAhouseE A AQ E ALE A which must be removed by the heat pump β AQ E ALE A AW E A 06 βcarnot 06 TAhouseE A TAambE A TAhouseE A Substitute in for AQ E ALE A and multiply with TAambE A TAhouseE AAW E A 06 TAambE A TAhouseE A A2E A 06 TAhouseE A AW E A Since TAhouseE A 29315 K and AW E A 12 kW it follows TAambE A TAhouseE A A2E A TAhouseE A AW E A 29315 12 35178 KA2E Solving TAambE A TAhouseE A 1876 TAambE A 3119 K 388 C Comment We did assume here that β 06 βcarnot the statement could also have been understood as β 06 βcarnot which would lead to a slightly different result Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 591 A house is cooled by an electric heat pump using the outside as the high temperature reservoir For several different summer outdoor temperatures estimate the percent savings in electricity if the house is kept at 25C instead of 20C Assume that the house is gaining energy from the outside directly proportional to the temperature difference as in Eq 514 Solution Airconditioner Refrigerator AQ E ALEAKE A TAHE A TALE A AMax PerfE A A Q L EW IN E A A TL ETH TL E A A KTH TL EW IN E A AW E AINE A A KTH TL2 ETL E A TALA E A 20C 2932 K B TALB E A 25C 2982 K TAHE AC AW E AINA E AK AW E AINB E AK saving 45 2132 1341 371 40 1364 0755 446 35 0767 0335 563 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 592 A heat pump has a coefficient of performance that is 50 of the theoretical maximum It maintains a house at 20C which leaks energy of 06 kW per degree temperature difference to the ambient For a maximum of 10 kW power input find the minimum outside temperature for which the heat pump is a sufficient heat source Solution Q leak Q Q H L W 1 kW HP cb 06 TAHE A TL CV House For constant 20C the heat pump must provide AQ E Aleak 06 T AQ E AH AQ E Aleak 06 TAHE A TL β AW E CV Heat pump Definition of the coefficient of performance and the fact that the maximum is for a Carnot heat pump β A Q H EW E A A Q H EQ H Q L E A 05 βACarnotE A 05 A TH ETH TL E A Substitute into the first equation to get 06 TAHE A TL 05 TAHE A TAHE A TL 1 TAHE A TL 2 05 06 TAHE A 1 05 06 29315 24429 TAHE A TL 1563 TL 20 1563 44 C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 593 The room in problem 590 has a combined thermal mass of 2000 kg wood 250 kg steel and 500 kg drywall gypsum Estimate how quickly the room heats up if the air conditioner is turned off on a day when it is 35AoE AC outside Without the airconditioner the house gains heat and the energy equation for the house becomes m C AdT dtE A AQ E AinE A The gain is due to the temperature difference as AQ E AinE A 06 TAHE A TL 06 kWK 35 20 K 9 kW The combined mC is using an estimate C for gypsum as 1 kJkgK mC 2000 138 250 046 500 1 kJK 3375 kJK AdT dtE A AQ E AinE A mC 9 kW 3375 kJK 000267 Ks Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 594 A window air conditioner cools a room at TL 22AoE AC with a maximum of 12 kW power input possible The room gains 06 kW per degree temperature difference to the ambient and the refrigeration COP is β 06 βACarnotE A Find the actual power required on a day when the temperature is 30AoE AC outside COP refrigerator β A QL EWIN E A 06 βACarnotE A 06 A TL ETH TL E A A27315 22 30 22E A 221 Heat gain AQ E ALE A 06 TAHE A TL 06 kWK 30 22 K 48 kW AW E A AQ E ALE Aβ 48 kW221 0217 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 595 On a cold 10AoE AC winter day a heat pump provides 20 kW to heat a house maintained at 20AoE AC and it has a COPHP of 4 How much power does the heat pump require The next day a winter storm brings the outside to 15AoE AC assuming the same COP and the same house heat transfer coefficient for the heat loss to the outside air How much power does the heat pump require then If we look at the heat loss for the house we have AQ E AlossE A 20 kW CA T CA A 20 kW 20 10 KE A 0667 kWK So now with the new outdoor temperature we get AQ E AlossE A CA T 0667 kWK 20 15 K 233 kW AQ E AlossE A AQ E AHE A COP AW E A AW E A AQ E AlossE A COP A233 kW 4E A 583 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 596 In the previous problem it was assumed that the COP will be the same when the outside temperature drops Given the temperatures and the actual COP at the 10AoE AC winter day give an estimate for a more realistic COP for the outside at 15AoE AC case As the outside T drops the temperature in the low temperature heat exchanger drops and it will be harder for the heat pump A reasonable assumption is then that the reduced COP will follow the ideal Carnot cycle COP At 10AoE AC βACarnotE A A TH ETH TL E A A 29315 20 10E A 9772 COP 4 At 15AoE AC βACarnotE A A TH ETH TL E A A 29315 20 15E A 8376 COP A8376 9772E A 4 343 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Ideal Gas Carnot Cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 597 Hydrogen gas is used in a Carnot cycle having an efficiency of 60 with a low temperature of 300 K During the heat rejection the pressure changes from 90 kPa to 120 kPa Find the high and low temperature heat transfer and the net cycle work per unit mass of hydrogen Solution As the efficiency is known the high temperature is found as η 06 1 A TL ETH E A TAHE A TALE A 1 06 750 K Now the volume ratio needed for the heat transfer TA3E A TA4E A TALE A is vA3E A vA4E A RTA3E A PA3E A RTA4E A PA4E A PA4E A PA3E A 120 90 1333 so from Eq59 we have with R 41243 kJkgK from Table A5 qALE A RTALE A ln v3vA4E A 35595 kJkg Using the efficiency from Eq55 then qAHE A qALE A 1 06 8899 kJkg The net work equals the net heat transfer w qAHE A qALE A 5339 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 598 Carbon dioxide is used in an ideal gas refrigeration cycle reverse of Fig 524 Heat absorption is at 250 K and heat rejection is at 325 K where the pressure changes from 1200 kPa to 2400 kPa Find the refrigeration COP and the specific heat transfer at the low temperature The analysis is the same as for the heat engine except the signs are opposite so the heat transfers move in the opposite direction β AQ E ALE A AW E A βcarnot TALE A TAHE A TALE A A 250 325 250E A 333 qAHE A RTAHE A lnv2v1 RTAHE A ln A P1 EP2 E A 01889 kJkgK 325 K lnA2400 1200E A 4255 kJkg qALE A qAHE A TALE A TAHE A 4255 kJkg 250 325 3273 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 599 An ideal gas Carnot cycle with air in a piston cylinder has a high temperature of 1000 K and a heat rejection at 400 K During the heat addition the volume triples Find the two specific heat transfers q in the cycle and the overall cycle efficiency Solution The Pv diagram of the cycle is shown to the right From the integration along the process curves done in the main text we have Eq57 qAHE A R TAHE A lnv2v1 0287 kJkg 1000 ln3 3153 kJkg Since it is a Carnot cycle the knowledge of the temperatures gives the cycle efficiency as ηATHE A 1 A TL ETH E A 1 A 400 1000E A 06 from which we can get the other heat transfer from Eq54 qALE A qAHE A TALE A TAHE A 3153 400 1000 1261 kJkg P v 1 2 3 4 1200 K 400 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5100 Air in a pistoncylinder goes through a Carnot cycle with the Pv diagram shown in Fig 524 The high and low temperatures are 600 K and 300 K respectively The heat added at the high temperature is 250 kJkg and the lowest pressure in the cycle is 75 kPa Find the specific volume and pressure after heat rejection and the net work per unit mass Solution qAHE A 250 kJkg TH 600 K TL 300 K P3 75 kPa The states as shown in figure 524 1 600 K 2 600 K 3 75 kPa 300 K 4 300 K Since this is a Carnot cycle and we know the temperatures the efficiency is η 1 A TL ETH E A 1 A300 600E A 05 and the net work becomes wANETE A ηqAHE A 05 250 125 kJkg The heat rejected is qALE A qAHE A wANETE A 125 kJkg After heat rejection is state 4 From equation 59 34 Eq59 qALE A RTL ln v3v4 v3 RT3 P3 0287 kJkgK 300 K 75 kPa 1148 m3kg v4 v3 expqALE ARTALE A 1148 m3kg exp1250287 300 02688 m3kg P4 RT4 v4 0287 kJkgK 300 K 02688 m3kg 320 kPa P v 1 2 3 4 600 K 300 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5101 A 4L jug of milk at 25C is placed in your refrigerator where it is cooled down to 5C The high temperature in the Carnot refrigeration cycle is 45C the low temperature is 5C and the properties of milk are the same as for liquid water Find the amount of energy that must be removed from the milk and the additional work needed to drive the refrigerator Solution CV milk out to the 5 C refrigerator space Energy Eq muA2E A uA1E A A1E AQA2E A A1E AWA2E A Process P constant 1 atm A1E AWA2E A Pm v2 v1 State 1 Table B11 v1 vf 0001003 mA3E Akg h1 hf 10487 kJkg mA2E A mA1E A VA1E AvA1E A 0004 0001003 3988 kg State 2 Table B11 h2 hf 2098 kJkg A1E AQA2E A muA2E A uA1E A A1E AWA2E A muA2E A uA1E A Pm v2 v1 mhA2E A hA1E A A1E AQA2E A 3998 2098 10487 3988 8389 33455 kJ CV Refrigeration cycle TL 5 C TAHE A 45 C assume Carnot Ideal β QALE A W QALE A QAHE A QALE A TALE A TAHE A TALE A 27815 45 5 5563 W QALE A β 33455 kJ 5563 6014 kJ MILK 5 C AIR o Remark If you calculate the work term A1E AWA2E A you will find that it is very small the volume does not change liquid The heat transfer could then have been done as muA2E A uA1E A without any change in the numbers Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5102 Consider the combination of the two heat engines as in Fig P54 How should the intermediate temperature be selected so the two heat engines have the same efficiency assuming Carnot cycle heat engines Heat engine 1 ηATH 1E A 1 A TM ETH E Heat engine 2 ηATH 2E A 1 A TL ETM E ηATH 1E A ηATH 2E A 1 A TM ETH E A 1 A TL ETM E A A TM ETH E A A TL ETM E TAME A A TLTH E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5103 Consider a combination of a gas turbine power plant and a steam power plant as shown in Fig P54 The gas turbine operates at higher temperatures thus called a topping cycle than the steam power plant then called a bottom cycle Assume both cycles have a thermal efficiency of 32 What is the efficiency of the overall combination assuming QL in the gas turbine equals QH to the steam power plant Let the gas turbine be heat engine number 1 and the steam power plant the heat engine number 2 Then the overall efficiency ηTH W net Q H W 1 W 2 Q H η1 W 2 Q H For the second heat engine and the energy Eq for the first heat engine W 2 η2 Q M η2 1 η1 Q H so the final result is ηTH η1 η2 1 η1 032 0321 032 0538 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5104 We wish to produce refrigeration at 30C A reservoir shown in Fig P5104 is available at 200C and the ambient temperature is 30C Thus work can be done by a cyclic heat engine operating between the 200C reservoir and the ambient This work is used to drive the refrigerator Determine the ratio of the heat transferred from the 200C reservoir to the heat transferred from the 30C reservoir assuming all processes are reversible Solution Equate the work from the heat engine to the refrigerator AW E A Q H1 A TH T0 ETH E also AW E A AQ E AL2E A A T0 TL ETL E Q H1 AQ E AL2E A A To TL ETL E A A TH ETH To E A A 60 2432 E A A 4732 170 E A 0687 QH1 W QL1 HE QH2 QL2 REF T 200 C H T 30 C o T 30 C o T 30 C L Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5105 Redo the previous problem assuming the actual devices both have a performance that is 60 of the theoretical maximum For the heat engine this means AW E A 06 Q H1 A TH T0 ETH E For the refrigerator it means AW E A AQ E AL2E A A T0 TL ETL E A 06 The ratio of the two heat transfers becomes Q H1 AQ E AL2E A A To TL ETL E A A 1 06E A A TH ETH To E A A 1 06E A 60 2432 E A A 4732 170 E A A 1 036E A 191 As the heat engine delivers less work and the refrigerator requires more work energy from the high T source must increase significantly Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5106 A house should be heated by a heat pump β 22 and maintained at 20AoE AC at all times It is estimated that it loses 08 kW per degree the ambient is lower than the inside Assume an outside temperature of 10oC and find the needed power to drive the heat pump Solution Ambient TALE A 10oC Heat pump β AQ E AHE AAW E A House AQ E AHE A AQ E Aleak 08 TAHE A TALE A AW E A AQ E AHE Aβ AQ E Aleak β 08 TAHE A TALE A β 0820 10 22 1091 kW Q leak Q Q H L W HP cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5107 Give an estimate for the COP in the previous problem and the power needed to drive the heat pump when the outside temperature drops to 15AoE AC Solution Minimum power if we assume a Carnot cycle We assume the heat transfer coefficient stays the same AQ E AH AQ E Aleak 25 kW CA ΔT CA 20 10 CA A5 6E A kWK AQ E Aleak new CA ΔT A5 6E A 20 15 29167 kW β A Q H EW IN E A A TH ETHTL E A A29315 35E A 83757 AW E AINE A A29167 83757E A 348 kW Comment Leak heat transfer increases and COP is lower when T outside drops Q leak Q Q H L W HP Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5108 A farmer runs a heat pump with a motor of 2 kW It should keep a chicken hatchery at 30AoE AC which loses energy at a rate of 05 kW per degree difference to the colder ambient The heat pump has a coefficient of performance that is 50 of a Carnot heat pump What is the minimum ambient temperature for which the heat pump is sufficient Solution CV Hatchery steady state To have steady state at 30AoE AC for the hatchery Energy Eq AQ E AHE A AQ E ALossE A β AACE AW E A Process Eq AQ E ALossE A 05 TAHE A TAambE A β AACE A ½ βACARNOT E COP for the reference Carnot heat pump β ACARNOT E A A Q H EW E A A Q H EQ H Q L E A A TH ETH TL E A A TH E TH Tamb E Substitute the process equations and this β ACARNOT E A into the energy Eq 05 kWK TAHE A TAambE A ½ A TH E TH Tamb E A AW E TAHE A TAambE AA2E A ½ TAHE AW E A05 kWK TAHE AW E A 273 30 K 2 K 606 KA2E TAHE A TAambE A 2462 K TAambE A 30 2462 538AoE AC Comment That of course is not a very low temperature and the size of the system is not adequate for most locations Q leak Q Q H L W 2 kW HP cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5109 An airconditioner with a power input of 12 kW is working as a refrigerator β 3 or as a heat pump β 4 It maintains an office at 20C year round which exchanges 05 kW per degree temperature difference with the atmosphere Find the maximum and minimum outside temperature for which this unit is sufficient Solution Analyze the unit in heat pump mode Replacement heat transfer equals the loss AQ E A 05 kWK TAHE A Tamb AW E A A Q H EβE A 05 kWK A TH Tamb E4E A TAHE A Tamb 4 A W E05E A 96 K Heat pump mode Minimum Tamb 20 96 104 C The unit as a refrigerator must cool with rate AQ E A 05 Tamb Thouse AW E A A Q L EβE AError Bookmark not definedA 05 kWK Tamb Thouse 3 Tamb Thouse 3 A W E05E A 72 K Refrigerator mode Maximum Tamb 20 72 272 C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5110 An airconditioner on a hot summer day removes 8 kW of energy from a house at 21AoE AC and pushes energy to the outside which is at at 31AoE AC The house has 15 000 kg mass with an average specific heat of 095 kJkgK In order to do this the cold side of the airconditioner is at 5AoE AC and the hot side is at 40AoE AC The air conditioner refrigerator has a COP that is 60 of a corresponding Carnot refrigerator Find the actual COP of the airconditioner and the power required to run it A steady state refrigerator definition of COP COP βREF Q L AWE A Q L Q H Q L 06 βCarnot Carnot βCarnot A TL ETH TL E A A5 27315 40 5E A 795 βREF 06 795 477 AWE A Q L βREF 8 kW 477 168 kW H Q W L Q REF 5 C o 31 C amb 40 C 21 C o Q leak from atm 31 C House Any heat transfer must go from a higher to a lower T domain Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5111 The airconditioner in the previous problem is turned off How quickly does the house heat up in degrees per second AoE ACs Once the AC unit is turned off we do not cool the house so heat leaks in from the atm at a rate of 8 kW that is what we had to remove to keep steady state Energy Eq E CV AQ E AleakE A 8 kW mAhouseE A CAPE A AdT dtE AdT dtE A AQ E AleakE A mAhouseE A CAPE A A 8 kW 15 000 095 kJKE A 056 10A3E A Ks Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5112 Air in a rigid 1 m3 box is at 300 K 200 kPa It is heated to 600 K by heat transfer from a reversible heat pump that receives energy from the ambient at 300 K besides the work input Use constant specific heat at 300 K Since the coefficient of performance changes write dQ mAairE A CAvE A dT and find dW Integrate dW with temperature to find the required heat pump work Solution COP β A QH EWE A A QH EQH QL E A A TH ETH TL E mAairE A P1V1 RT1 200 1 0287 300 2322 kg dQAHE A mAairE A Cv dTH β dW A TH ETH TL E A dW dW mAairE A CAvE A A TH ETH TL E A dTH 1W2 mAairE A CAvE A 1 A TL ETE A dT mAairE A Cv 1 A TL ETE A dT mAairE A Cv T2 T1 TL ln T2 T1 2322 kg 0717 kJkgK 600 300 300 ln A600 300E A K 1531 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5113 A Carnot heat engine shown in Fig P5113 receives energy from a reservoir at Tres through a heat exchanger where the heat transferred is proportional to the temperature difference as AQ E AHE A KTAresE A TAHE A It rejects heat at a given low temperature TL To design the heat engine for maximum work output show that the high temperature TAHE A in the cycle should be selected as TH A TresTL E Solution W ηATHE AQAHE A A TH TL ETH E A KTAresE A TAHE A maximize WTAHE A AδW δTH E A 0 AδW δTH E A KTAresE A TAHE ATALE ATAHE A2E A K1 TALE ATAHE A 0 TAHE A A TresTL E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5114 A combination of a heat engine driving a heat pump see Fig P5114 takes waste energy at 50C as a source QAw1E A to the heat engine rejecting heat at 30C The remainder QAw2E A goes into the heat pump that delivers a QH at 150C If the total waste energy is 5 MW find the rate of energy delivered at the high temperature Solution Waste supply AQ E Aw1E A AQ E Aw2E A 5 MW Heat Engine AW E A η AQ E Aw1E A 1 TL1 TH1 AQ E Aw1E Heat pump AW E A AQ E AH βHP AQ E AW2 β AQ E Aw2E A TH1 TH TH1 Equate the two work terms 1 TL1 TH1 AQ E Aw1E A AQ E Aw2E A TH TH1 TH1 Substitute AQ E Aw1E A 5 MW AQ E Aw2E A 1 30315323155 AQ E Aw2E A AQ E Aw2E A 150 50 32315 20 5 AQ E Aw2E A AQ E Aw2E A 100 AQ E Aw2E A 08333 MW AQ E Aw1E A 5 08333 41667 MW AW E A η AQ E Aw1E A 006189 41667 0258 MW AQ E AH AQ E Aw2E A AW E A 109 MW For the heat pump β 42315 100 423 W QL Qw1 HE Qw2 QH HP Waste source Ambient 30 C Waste source HEAT 150 C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5115 A furnace shown in Fig P5115 can deliver heat QH1 at TH1 and it is proposed to use this to drive a heat engine with a rejection at Tatm instead of direct room heating The heat engine drives a heat pump that delivers QH2 at Troom using the atmosphere as the cold reservoir Find the ratio QH2QH1 as a function of the temperatures Is this a better setup than direct room heating from the furnace Solution CV Heat Eng AW E AHEE A ηAQ E AH1E A where η 1 TAatmE ATAH1E CV Heat Pump AW E AHPE A AQ E AH2E Aβ where β TArmE ATArmE A TAatmE A Work from heat engine goes into heat pump so we have AQ E AH2E A β AW E AHPE A β η AQ E AH1E A and we may substitute Ts for β η If furnace is used directly AQ E AH2E A AQ E AH1E A so if βη 1 this proposed setup is better Is it For TAH1E A TAatmE A formula shows that it is good for Carnot cycles In actual devices it depends whether βη 1 is obtained Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5116 Consider the rock bed thermal storage in Problem 560 Use the specific heat so you can write dQH in terms of dTArockE A and find the expression for dW out of the heat engine Integrate this expression over temperature and find the total heat engine work output Solution The rock provides the heat QAHE dQAHE A dUArockE A mC dTArockE dW ηdQAHE A 1 To Trock mC dTArockE m ρV 2750 kgmA3E A 2 mA3E A 5500 kg 1W2 1 To Trock mC dTArockE A mC T2 T1 To ln T2 T1 5500 kg 089 kJkgK 290 400 290 ln A290 400E A K 81 945 kJ W Q Q H L HE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5117 Consider a Carnot cycle heat engine operating in outer space Heat can be rejected from this engine only by thermal radiation which is proportional to the radiator area and the fourth power of absolute temperature Q rad KAT4 Show that for a given engine work output and given TH the radiator area will be minimum when the ratio TLTH 34 Solution W NET Q H TH TL TH Q L TH TL TL also Q L KAT4 L W NET KT4 H AT4 L T4 H TH TL 1 A TL TH 3 TL TH 4 const Differentiating dA TL TH 3 TL TH 4 A 3 TL TH 2 4 TL TH 3 d TL TH 0 dA dTLTH A TH 2 4 TL TH 3 3 TL TH TL 3 TL TH 4 0 TL TH 3 4 for min A Check that it is minimum and not maximum with the 2nd derivative 0 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5118 A Carnot heat engine operating between a high TH and low TL energy reservoirs has an efficiency given by the temperatures Compare this to two combined heat engines one operating between TH and an intermediate temperature TM giving out work WA and the other operating between TM and TL giving out WB The combination must have the same efficiency as the single heat engine so the heat transfer ratio QHQL ψTHTL QHQM QMQL The last two heat transfer ratios can be expressed by the same function ψ involving also the temperature TM Use this to show a condition the function ψ must satisfy The overall heat engine is a Carnot heat engine so Q H Q L TH TL ψTHTL The individual heat engines Q H Q M ψTHTM and Q M Q L ψTMTL Since an identity is Q H Q L Q H Q M Q M Q L ψTHTL it follows that we have ψTHTL ψTHTM ψTMTL Notice here that the product of the two functions must cancel the intermediate temperature TM this shows a condition the function ψ must satisfy The Kelvin and Rankine temperature scales are determined by the choice of the function ψTHTL TH TL Q H Q L satisfying the above restriction Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5119 On a cold 10oC winter day a heat pump provides 20 kW to heat a house maintained at 20oC and it has a COPHP of 4 using the maximum power available The next day a winter storm brings the outside to 15oC assuming that the COPHP changes by the same percentage as a Carnot unit and that the house loses heat to the outside air How cold is the house then If we look at the heat loss for the house we have Q loss 20 kW CA T CA 20 kW 20 10 K 0667 kWK Q loss Q H COP W W Q loss COP 20 kW 4 5 kW At 10oC β Carnot 1 TH TH TL1 29315 20 10 9772 With a changed COP we get COPHP 4 β Carnot 2β Carnot 1 49772 β Carnot 2 At the new unknown temperature of the house TH we have β Carnot 2 TH TL2 TH Q loss CA TH TL2 The energy equation for the house becomes Q loss Q H and substitution gives 0667 kWK TH TL2 COPHP W 49772 TH TL2 TH 5 kW or TH TL22 4 5 kW 9772 0667 kWK TH 307 K TH Solve with x TH TL2 x2 307 x 30727315 15 0 x 3072 30722 79252 12 29729 TH x TL2 29729 15 147oC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5120 A 10m3 tank of air at 500 kPa 600 K acts as the hightemperature reservoir for a Carnot heat engine that rejects heat at 300 K A temperature difference of 25C between the air tank and the Carnot cycle high temperature is needed to transfer the heat The heat engine runs until the air temperature has dropped to 400 K and then stops Assume constant specific heat capacities for air and find how much work is given out by the heat engine Solution Q H W Q L HE AIR 300 K TH Tair 25C TL 300 K mair RT1 P1V 500 10 0287 600 2904 kg dW ηdQH Tair 25 1 TL dQ H dQH mairdu mairCvdTair W dW mairCv Ta 25 dTa 1 TL mairCv Ta2 Ta1 TL ln Ta2 25 Ta1 25 2904 kg 0717 kJkgK 400 600 300 ln 375 575 K 14943 kJ UPDATED JULY 2013 SOLUTION MANUAL CHAPTER 5 English Units Fundamentals of Thermodynamics Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CHAPTER 5 SUBSECTION PROB NO Heat Engines and Refrigerators 121131 Carnot Cycles and Absolute Temperature 132145 Finite T Heat Transfer 146152 Ideal gas Carnot cycle 153154 Review Problems 155161 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Heat Engines and Refrigerators Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5121E A window mounted airconditioner removes 35 Btu from the inside of a home using 175 Btu work input How much energy is released outside and what is its coefficient of performance CV Refrigerator The energy QH goes into the kitchen air Energy Eq QH W QL 175 35 525 btu COP β QL W 35 175 2 H Q L Q 35 Btu T L T amb REF W 175 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5122E A lawnmower tractor engine produces 18 hp using 40 Btus of heat transfer from burning fuel Find the thermal efficiency and the rate of heat transfer rejected to the ambient Conversion Table A1 18 hp 18 254443600 Btus 12722 Btus Efficiency ηTH W outQ H 12722 40 0318 Energy equation Q L Q H W out 40 1272 273 Btus Q H Q L W out Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5123E Calculate the thermal efficiency of the steam power plant cycle described in Problem 4198 Solution From solution to problems 4198 199 W NET 33 000 400 32 600 hp 83 107 Btuh Q Htot Q econ Q gen 475 107 2291 108 2766 108 Btuh η W Q H 83 107 2766 108 030 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5124E A large coal fired power plant has an efficiency of 45 and produces net 1500 MW of electricity Coal releases 12 500 Btulbm as it burns so how much coal is used per hour From the definition of the thermal efficiency and the energy release by the combustion called heating value HV we get W η Q H η m HV then m η HV W 1500 MW 045 12 500 btulbm 1500 10001055 Btus 045 12 500 Btulbm 252765 lbms 909 950 lbmh Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5125E An industrial machine is being cooled by 08 lbms water at 60 F which is chilled from 95 F by a refrigeration unit with a COP of 3 Find the rate of cooling required and the power input to the unit Energy equation for heat exchanger Q L m h1 h2 m CP T1 T2 08 lbms 1 BtulbmR 95 60 R 28 Btus β COP Q L W Q W L T H H Q REF 1 2 cb W Q L β 28 3 933 Btus Comment An outside cooling tower is often used for this see Chapter 11 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5126E A water cooler for drinking water should cool 10 galh water from 65 F to 50 F using a small refrigeration unit with a COP of 25 Find the rate of cooling required and the power input to the unit The mass flow rate is m ρV 10 231 123 001603 1 3600 lbms 00232 lbms Energy equation for heat exchanger Q L m h1 h2 m CP T1 T2 Q W L T H H Q REF 1 2 cb 00232 10 65 50 0348 Btus β COP Q L W W Q L β 0348 Btus 25 0139 Btus Comment The unit does not operate continuously Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5127E A window airconditioner unit is placed on a laboratory bench and tested in cooling mode using 075 Btus of electric power with a COP of 175 What is the cooling power capacity and what is the net effect on the laboratory Definition of COP β Q L W Cooling capacity Q L β W 175 075 1313 Btus For steady state operation the Q L comes from the laboratory and Q H goes to the laboratory giving a net to the lab of W Q H Q L 075 Btus that is heating it Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5128E A farmer runs a heat pump with a 2 kW motor It should keep a chicken hatchery at 90 F which loses energy at a rate of 10 Btus to the colder ambient Tamb What is the minimum coefficient of performance that will be acceptable for the heat pump Power input W 2 kW 2 25444 3600 1414 Btus Energy Eq for hatchery Q H Q Loss 10 Btus Definition of COP β COP Q H W 10 1414 707 Q leak Q Q H L W 2 kW HP cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5129E R410A enters the evaporator the cold heat exchanger in an AC unit at 0 F x 28 and leaves at 0 F x 1 The COP of the refrigerator is 15 and the mass flow rate is 0006 lbms Find the net work input to the cycle Energy equation for heat exchanger Q L m h2 h1 m hg hf x1 hfg m hfg x1 hfg m 1 x1hfg Q L 1 2 cb 0006 lbms 072 10376 Btulbm 0448 Btus β COP Q L W W Q L β 0448 15 03 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5130E A large stationary diesel engine produces 2 000 hp with a thermal efficiency of 40 The exhaust gas which we assume is air flows out at 1400 R and the intake is 520 R How large a mass flow rate is that if that accounts for half the Q L Can the exhaust flow energy be used Power 2 000 hp 2 000 25444 3600 14136 Btus Heat engine Q H W outηTH 14 136 04 3534 Btus Energy equation Q L Q H W out 3534 14136 21204 Btus Exhaust flow 1 2Q L m airh1400 h520 m air 1 2 h1400 h520 Q L 1 2 21204 34302 12438 485 lbms The flow of hot gases can be used to heat a building or it can be used to heat water in a steam power plant since it operates at lower temperatures Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5131E Calculate the amount of work input a refrigerator needs to make ice cubes out of a tray of 05 lbm liquid water at 50 F Assume the refrigerator has β 35 and a motorcompressor of 750 W How much time does it take if this is the only cooling load Solution CV Water in tray We neglect tray mass Energy Eq mu2 u1 1Q2 1W2 Process P constant Po 1W2 P dV Pomv2 v1 1Q2 mu2 u1 1W2 mh2 h1 Tbl F71 h1 1805 Btulbm Tbl F74 h2 14334 Btulbm 1Q2 0514334 1805 80695 Btu Consider now refrigerator β QLW W QLβ 1Q2 β 8069535 2306 Btu For the motor to transfer that amount of energy the time is found as W W dt W t t WW 2306 1055750 324 s Comment We neglected a baseload of the refrigerator so not all the 750 W are available to make ice also our coefficient of performance is very optimistic and finally the heat transfer is a transient process All this means that it will take much more time to make icecubes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Carnot Cycles and Absolute T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7132E Calculate the thermal efficiency of a Carnotcycle heat engine operating between reservoirs at 920 F and 110 F Compare the result with that of Problem 5123 Solution TH 920 F TL 110 F ηCarnot 1 TH TL 1 110 45967 920 45967 0587 about twice 5123 03 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5133E A steam power plant has 1200 F in the boiler 630 Btus work out of turbine 900 Btus is taken out at 100 F in the condenser and the pump work is 30 Btus Find the plant thermal efficiency Assume the same pump work and heat transfer to the boiler how much is the turbine power if the plant is running in a Carnot cycle Solution W T Q H Q L W P in to ambient CV Total plant Energy Eq Q H W Pin W T Q L Q H 630 900 30 1500 Btus ηTH Pin Q H W T W Error Bookmark not defined 600 1500 040 ηcarnot W net Q H 1 TLTH 1 100 45967 1200 45967 0663 W T W Pin ηcarnotQ H 0663 1500 Btus 995 Btus W T 995 30 1025 Btu s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5134E A large heat pump should upgrade 4000 Btus of heat at 175 F to be delivered as heat at 280 F What is the minimum amount of work power input that will drive this For the minimum work we assume a Carnot heat pump and Q L 4000 Btus βHP W in Q H TH TL TH 4597 280 280 175 704 βREF βHP 1 Q L W in 604 Now we can solve for the work W in Q LβREF 4000604 662 Btus This is a domestic or small office building size AC unit much smaller than the 4000 Btus in this problem C Borgnakke Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5135E A car engine burns 10 lbm of fuel equivalent to addition of QH at 2600 R and rejects energy to the radiator and the exhaust at an average temperature of 1300 R If the fuel provides 17 200 Btulbm what is the maximum amount of work the engine can provide Solution A heat engine QH m qfuel 10 17200 170 200 Btu Assume a Carnot efficiency maximum theoretical work η 1 TL TH 1 1300 2600 05 W η QH 05 170 200 85 100 Btu Exhaust flow Air intake filter Shaft Fan power Fuel line cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5136E Consider the combination of a heat engine and a heat pump as given in problem 541 with a low temperature of 720 R What should the high temperature be so the heat engine is reversible For that temperature what is the COP for a reversible heat pump For all three cases of the heat engine the ratio between the heat transfers and the work term is the same as Q H Q L W 642 321 For a reversible heat engine we must have the heat transfer ratio equal to the temperature ratio so Q H Q L TH TL 3 2 720 R TH TH 32 720 R 1080 R The COP is COPHP Q H W 3 1 3 TH TH TL 1080 1080 720 W L1 Q T H1 H1 Q 720 R HE L2 Q HP Q T H2 H2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5137E An airconditioner provides 1 lbms of air at 60 F cooled from outside atmospheric air at 95 F Estimate the amount of power needed to operate the air conditioner Clearly state all assumptions made Solution Consider the cooling of air which needs a heat transfer as Q air m h m Cp T 1 lbms 024 BtulbmR 95 60 R 84 Btus Assume Carnot cycle refrigerator β Q L W Q L Q H Q L TL TH TL 60 45967 95 60 148 W Q L β 84 148 057 Btus This estimate is the theoretical maximum performance To do the required heat transfer TL 40 F and TH 110 F are more likely secondly β βcarnot H Q W L Q REF 95 F 60 F cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5138E We propose to heat a house in the winter with a heat pump The house is to be maintained at 68 F at all times When the ambient temperature outside drops to 15 F the rate at which heat is lost from the house is estimated to be 80 000 Btuh What is the minimum electrical power required to drive the heat pump Solution Minimum power if we assume a Carnot cycle Q H Q leak 80 000 Btuh Q leak Q Q H L W HP β Q H W IN TH TH TL 5277 68 15 9957 W IN 80 000 9957 8035 Btuh 2355 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5139E Consider the setup with two stacked temperature wise heat engines as in Fig P54 Let TH 1500 R TM 1000 R and TL 650 R Find the two heat engine efficiencies and the combined overall efficiency assuming Carnot cycles The individual efficiencies η1 1 TM TH 1 1000 1500 0333 η2 1 TL TM 1 650 1000 035 The overall efficiency ηTH W net Q H W 1 W 2 Q H η1 W 2 Q H For the second heat engine and the energy Eq for the first heat engine W 2 η2 Q M η2 1 η1 Q H so the final result is ηTH η1 η2 1 η1 0333 035 1 0333 0566 Comment It matches a single heat engine ηTH 1 TL TH 1 650 1500 0567 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5140E A thermal storage is made with a rock granite bed of 70 ft3 which is heated to 720 R using solar energy A heat engine receives a QH from the bed and rejects heat to the ambient at 520 R The rock bed therefore cools down and as it reaches 520 R the process stops Find the energy the rock bed can give out What is the heat engine efficiency at the beginning of the process and what is it at the end of the process Solution Assume the whole setup is reversible and that the heat engine operates in a Carnot cycle The total change in the energy of the rock bed is u2 u1 q C T 021 BtulbmR 720 520 R 42 Btulbm m ρV 172 lbmft3 70 ft3 12040 lbm Q mq 12040 lbm 42 Btulbm 505 680 Btu To get the efficiency assume a Carnot cycle device η 1 To TH 1 520720 028 at the beginning of process η 1 To TH 1 520520 0 at the end of process W Q Q H L HE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5141E A heat engine has a solar collector receiving 600 Btuh per square foot inside which a transfer media is heated to 800 R The collected energy powers a heat engine which rejects heat at 100 F If the heat engine should deliver 8500 Btuh what is the minimum size area solar collector Solution TH 800 R TL 100 45967 560 R ηHE 1 TL TH 1 560 800 030 W η Q H Q H η W 8500 030 28 333 Btuh Q H 600 A A Q H 600 47 ft2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5142E Sixhundred poundmass per hour of water runs through a heat exchanger entering as saturated liquid at 250 F and leaving as saturated vapor The heat is supplied by a Carnot heat pump operating from a lowtemperature reservoir at 60 F with a COP of half of the similar carnot unit Find the rate of work into the heat pump Solution CV Heat exchanger m 1 m 2 m 1h1 Q H m 1h 2 Table F71 h1 21858 Btulbm h2 116419 Btulbm H Q W L Q T L HP 1 2 cb Q H 600 3600 116419 21858 1576 Btus For the Carnot heat pump TH 250 F 710 R β Q HW TH TL TH 710 190 3737 βac 37372 187 W Q Hβ ac 1576187 843 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5143E A power plant with a thermal efficiency of 40 is located on a river similar to Fig P561 With a total river mass flow rate of 2 105 lbms at 60 F find the maximum power production allowed if the river water should not be heated more than 2 F The maximum heating allowed determines the maximum Q L as Q L m H2O h m H2O CP LIQ H2O TH2O 2 105 lbms 10 BtulbmR 2 R 4 105 Btus W NET1ηTH ac 1 W NET Q L 1ηTH ac 1 Q L 1 ηTH ac ηTH ac 4 105 Btus 04 1 04 267 105 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5144E A nuclear reactor provides a flow of liquid sodium at 1500 F which is used as the energy source in a steam power plant The condenser cooling water comes from a cooling tower at 60 F Determine the maximum thermal efficiency of the power plant Is it misleading to use the temperatures given to calculate this value Solution L I Q N a R E A C T O R E N E R G Y T OH O 2 1 5 0 0F C O N D C O O L I T O W E E N E R G Y F R O M S T E A M P O W E R P L A N T 6 0F L I Q H O 2 TH 1500 F 1960 R TL 60 F 520 R ηTH MAX TH TH TL 1960 520 19860 0735 It might be misleading to use 1500 F as the value for TH since there is not a supply of energy available at a constant temperature of 1500 F liquid Na is cooled to a lower temperature in the heat exchanger The Na cannot be used to boil H2O at 1500 F Similarly the H2O leaves the cooling tower and enters the condenser at 60 F and leaves the condenser at some higher temperature The water does not provide for condensing steam at a constant temperature of 60 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5145E An inventor has developed a refrigeration unit that maintains the cold space at 14 F while operating in a 77 F room A coefficient of performance of 85 is claimed How do you evaluate this Solution Assume Carnot cycle then βCarnot Win QL THTL TL 14 45967 77 14 75 85 βCarnot impossible claim H Q W L Q T 14 F L T 77 F H REF Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Finite T Heat Transfer Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5146E A car engine operates with a thermal efficiency of 35 Assume the air conditioner has a coefficient of performance that is one third of the theoretical maximum and it is mechanically pulled by the engine How much fuel energy should you spend extra to remove 1 Btu at 60 F when the ambient is at 95 F Solution Air conditioner β QL W TL TH TL 60 45967 95 60 148 βactual β 3 493 W QL β 1 493 0203 Btu Work from engine W ηeng Qfuel 0203 Btu Qfuel W ηeng 0203 035 058 Btu W L Q T H H Q T L REF Fuel Q HE L eng Q FUEL Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5147E In a remote location you run a heat engine to provide the power to run a refrigerator The input to the heat engine is at 1450 R and the low T is 700 R it has an actual efficiency equal to ½ of the corresponding Carnot unit The refrigerator has a TL 15 F and TH 95 F with a COP that is 13 of the corresponding Carnot unit Assume a cooling capacity of 7000 Btuh is needed and find the rate of heat input to the heat engine Heat engine ηCarnot 1 TH TL 1 700 1450 0517 ηac 0259 Refrigerator βref Carnot TH TL TL 4597 15 95 15 5934 βref ac 1978 Cooling capacity Q L 7000 Btuh βref ac W W 7000 Btuh 1978 35389 Btuh This work must be provided by the heat engine W ηac Q H Q H W ηac 35389 Btuh 0259 13 664 Btuh 3796 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5148E A heat pump cools a house at 70 F with a maximum of 4000 Btuh power input The house gains 2000 Btuh per degree temperature difference to the ambient and the heat pump coefficient of performance is 60 of the theoretical maximum Find the maximum outside temperature for which the heat pump provides sufficient cooling Solution Q leak Q Q H L W 4000 Btuh AC cb T L Here TL T house TH T amb In this setup the low temperature space is the house and the high temperature space is the ambient The heat pump must remove the gain or leak heat transfer to keep it at a constant temperature Q leak 2000 Tamb Thouse Q L which must be removed by the heat pump β Q H W 1 Q L W 06 βcarnot 06 Tamb Tamb Thouse Substitute in for Q L and multiply with Tamb Thouse Tamb Thouse 2000 Tamb Thouse 2 W 06 Tamb Since Thouse 5297 R and W 4000 Btuh it follows T2 amb 10586 Tamb 2795227 0 Solving Tamb 5545 R 948 F Comment We did assume here that β 06 βcarnot the statement could also have been understood as β 06 βcarnot which would lead to a slightly different result Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5149E A small house is kept at 77 F inside looses 12 Btus to the outside ambient at 32 F A heat pump is used to help heat the house together with possible electric heat The heat pump is driven by a motor of 25 kW and it has a COP that is ¼ of a Carnot heat pump unit Find the actual COP for the heat pump and the amount of electric heat that must be used if any to maintain the house temperature CV House Energy 0 Q H W el Q Loss Definition of COP β COPHP W Q H 1 4 TH TH TL 1 4 5367 77 32 2982 Q H COPHP W HP 2982 251055 Btus 7066 Btus W el Q Loss Q H 12 7066 4934 Btus Q loss Q Q H L W 25 kW HP cb W el Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5150E A house is cooled by an electric heat pump using the outside as the high temperature reservoir For several different summer outdoor temperatures estimate the percent savings in electricity if the house is kept at 77 F instead of 68 F Assume that the house is gaining energy from the outside directly proportional to the temperature difference Solution Airconditioner Refrigerator Q LEAK TH TL Max Perf W in Q L TH TL TL W in KTH TL W in TL KTH TL2 A TLA 68 F 5277 R B TLB 77 F 5367 R TH F W INAK W INBK saving 115 4186 2691 357 105 2594 1461 437 95 1381 0604 563 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5151E Arctic explorers are unsure if they can use a 5 kW motor driven heat pump to stay warm It should keep their shelter at 60 F which loses energy at a rate of 03 Btus per degree difference to the colder ambient The heat pump has a COP that is 50 of a Carnot heat pump If the ambient temperature can fall to 10 F at night would you recommend this heat pump to the explorers CV Heat pump The heat pump should deliver a rate of heating that equals the heat loss to the ambient for steady inside temperature COP β Q HW 05 βCarnot 1 2 TH TL TH 1 2 4597 60 60 10 37 The heat pump can then provide a heating capacity of Q H β W 37 5 kW 185 kW 1753 Btus The heat loss is Q leak out CA ΔT 03 BtusR 60 10 R 21 Btus The heat pump is not sufficient to cover the loss and not recommended Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5152E Using the given heat pump in the previous problem how warm could it make the shelter in the arctic night The high is now an unknown so both the heat loss and the heat pump performance depends on that The energy balance around the shelter then gives Q H β W Q leak out CA ΔT Substitute the expression for β and CA ΔT to give 1 2 TH TL TH W 03 BtusR TH TL Multiply with the temperature difference factor 2 and divide by the work 5 kW 51055 Btus so we get TH 03 2 51055 R TH TL2 01137 R TH TL2 Solve this equation like 01137 x2 x TL 0 with x TH TL and TL 4597 10 4497 R x TH TL 6744 R negative root discarded TH x TL 6744 10 574 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Ideal Gas Garnot Cycle Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5153E Carbon dioxide is used in an ideal gas refrigeration cycle reverse of Fig 524 Heat absorption is at 450 R and heat rejection is at 585 R where the pressure changes from 180 psia to 360 psia Find the refrigeration COP and the specific heat transfer at the low temperature The analysis is the same as for the heat engine except the signs are opposite so the heat transfers move in the opposite direction β Q L W βcarnot TL TH TL 450 585 450 333 qH RTH lnv2v1 RTH ln P2 P1 351 ftlbflbmR 585 R ln360 180 14 2327 ftlbflbm 1829 Btulbm qL qH TL TH 1829 450 585 1407 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5154E Air in a pistoncylinder goes through a Carnot cycle with the Pv diagram shown in Fig 524 The high and low temperatures are 1200 R and 600 R respectively The heat added at the high temperature is 100 Btulbm and the lowest pressure in the cycle is 10 lbfin2 Find the specific volume and pressure at all 4 states in the cycle assuming constant specific heats at 80 F Solution qH 100 Btulbm TH 1200 R TL 600 R P3 10 lbfin2 Cv 0171 Btulbm R R 5334 ftlbflbmR The states as shown in figure 524 1 1200 R 2 1200 R 3 10 psi 600 R 4 600 R As we know state 3 we can work backwards towards state 1 v3 RT3 P3 5334 600 10 144 22225 ft3lbm Process 23 from Eq58 Cv constant Cv ln TL TH R ln v3v2 0 ln v3v2 Cv R ln TL TH 01715334 ln 6001200 17288 v2 v3 exp 17288 2222556339 39449 ft3lbm Process 12 and Eq57 qH RTH ln v2 v1 ln v2 v1 qH RTH 100 7785334 1200 121547 v1 v2 exp 121547 11699 ft3lbm v4 v1 v3 v2 11699 2222539449 6591 ft3lbm P1 RT1 v1 5334 120011699144 3799 psia P2 RT2 v2 5334 120039449 144 1127 psia P4 RT4 v4 5334 6006591 144 337 psia Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5155E We wish to produce refrigeration at 20 F A reservoir is available at 400 F and the ambient temperature is 80 F as shown in Fig P5104 Thus work can be done by a cyclic heat engine operating between the 400 F reservoir and the ambient This work is used to drive the refrigerator Determine the ratio of the heat transferred from the 400 F reservoir to the heat transferred from the 20 F reservoir assuming all processes are reversible Solution Equate the work from the heat engine to the refrigerator W QL1 QH1 HE QL2 QH2 REF T 860 R H T 540 R o T 540 R o T 440 R L W Q H1 TH TH T0 also W Q L2 TL T0 TL Q H1 Q L2 TO TL TL TH TH TO 100 440 860 320 0611 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5156E An airconditioner on a hot summer day removes 8 Btus of energy from a house at 70 F and pushes energy to the outside which is at 88 F The house has 30 000 lbm mass with an average specific heat of 023 BtulbmR In order to do this the cold side of the airconditioner is at 40 F and the hot side is 100 F The air conditioner refrigerator has a COP that is 60 of a corresponding Carnot refrigerator Find the actual airconditioner COP and the required power to run it A steady state refrigerator definition of COP COP βREF Q L W Q L Q H Q L 06 βCarnot Carnot βCarnot TL TH TL 40 4597 100 40 8328 βREF 06 8328 5 W Q L βREF 8 Btus 5 16 Btus H Q W L Q REF 40 F 88 F amb 100 F 70 F Q leak from atm 88 F House Any heat transfer must go from a higher to a lower T domain Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5157E The air conditioner in the previous problem is turned off How fast does the house heat up in degrees per second Fs Once the AC unit is turned off we do not cool the house so heat leaks in from the atm at a rate of 8 Btus that is what we had to remove to keep steady state Energy Eq E CV Q leak 8 Btus mhouse CP dT dt dT dt Q leak mhouse CP 8 Btus 30 000 023 BtuR 116 103 Rs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5158E A window air conditioner cools a room at TL 68 F with a maximum of 12 kW power input The room gains 033 Btus per degree temperature difference to the ambient and the refrigeration COP is β 06 βCarnot Find the maximum outside temperature TH for which the air conditioner provides sufficient cooling Solution Q leak Q Q H L W 12 kW HP cb T L Here TL T house TH T amb In this setup the low temperature space is the house and the high temperature space is the ambient The heat pump must remove the gain or leak heat transfer to keep it at a constant temperature Q leak 033 Tamb Thouse Q L which must be removed by the heat pump β Q L W 06 βcarnot 06 Thouse Tamb Thouse Substitute in for Q L and multiply with Tamb ThouseW 033 Tamb Thouse 2 06 Thouse W Since Thouse 4597 68 5277 R and W 12 kW 11374 Btus it follows Tamb Thouse 2 06033 5277 11374 109128 R 2 Solving Tamb Thouse 3303 Tamb 56073 R 101 F Comment We did assume here that β 06 βcarnot the statement could also have been understood as β 06 βcarnot which would lead to a slightly different result Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5159E The room in problem 5158E has a combined thermal mass of 4 000 lbm wood 500 lbm steel and 1000 lbm plaster board Estimate how fast the room heats up if the airconditioner is turned off on a day it is 95 F outside Without the airconditioner the house gains heat and the energy equation for the house becomes m C dT dt Q in The gain is due to the temperature difference as Q in 033 TH TL 033 BtusR 95 68 R 891 Btus The combined mC is using an estimate C for gypsum as 024 BtulbmR mC 4000 033 500 011 1000 024 BtuR 1615 BtuR dT dt Q in mC 891 Btus 1615 BtuR 00055 Rs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5160E A 350ft3 tank of air at 80 lbfin2 1080 R acts as the hightemperature reservoir for a Carnot heat engine that rejects heat at 540 R A temperature difference of 45 F between the air tank and the Carnot cycle high temperature is needed to transfer the heat The heat engine runs until the air temperature has dropped to 700 R and then stops Assume constant specific heat capacities for air and find how much work is given out by the heat engine Solution Q H W Q L HE AIR 540 R TH Tair 45 TL 540 R mair RT1 P1V 80 350 144 5334 1080 69991 lbm dW ηdQH Tair 45 1 TL dQ H dQH mairdu mairCvdTair W dW mair Cv Ta 45 dTa 1 TL mair Cv Ta2 Ta1 TL ln Ta2 45 Ta1 45 69991 lbm 0171 BtulbmR 700 1080 540 ln 655 1035 R 1591 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 5161E Air in a rigid 40 ft3 box is at 540 R 30 lbfin2 It is heated to 1100 R by heat transfer from a reversible heat pump that receives energy from the ambient at 540 R besides the work input Use constant specific heat at 540 R Since the coefficient of performance changes write dQ mair Cv dT and find dW Integrate dW with temperature to find the required heat pump work Solution COP β QH W QH QH QL TH TH TL mair P1V1 RT1 30 40 144 540 5334 60 lbm dQH mair Cv dTH β dW TH TH TL dW dW mair Cv TH TH TL dTH 1W2 mair Cv 1 TL T dT mair Cv 1 T TL dT mair Cv T2 T1 TL ln T2 T1 60 lbm 0171 BtulbmR 1100 540 540 ln 1100 540 R 1804 Btu Updated June 2013 SOLUTION MANUAL CHAPTER 6 Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CONTENT CHAPTER 6 SUBSECTION PROB NO InText concept questions aj ConceptStudy Guide problems 116 Inequality of Clausius 1723 Entropy of a pure substance 2434 Reversible processes 3559 Entropy of a liquid or solid 6076 Entropy of ideal gases 7799 Polytropic processes 100114 Entropy generation 115161 Rates or fluxes of entropy 162172 Review 173189 Problem solution repeated but using the Pr and vr functions in Table A72 92 105 an additional air problem The clipart included in the solution manual is from Microsoft Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful InText Concept Questions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6a Does Clausius say anything about the sign for o dQ No The total net heat transfer can be coming in like in a heat engine Wout QH QL in which case it is positive It can also be net going out like in a refrigerator or heat pump Win QH QL in which case the sign is negative Finally if you look at a transmission gearbox there could be no heat transfer first approximation in which case the integral is zero 6b Does the statement of Clausius require a constant T for the heat transfer as in a Carnot cycle No The statement for a cycle involves an integral of dQT so T can vary which it does during most processes in actual devices This just means that you cannot that easily get a closed expression for the integral 6c How can you change s of a substance going through a reversible process From the definition of entropy ds dq T for a reversible process Thus only heat transfer gives a change in s expansioncompression involving work does not give such a contribution 6d A reversible process adds heat to a substance If T is varying does that influence the change in s Yes Reversible ds dq T So if T goes up it means that s changes less per unit of dq and the opposite if T decreases then s changes more per unit of dq Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6e Water at 100 kPa 150oC receives 75 kJkg in a reversible process by heat transfer Which process changes s the most constant T constant v or constant P ds dq T Look at the constant property lines in a Ts diagram Fig 65 The constant v line has a higher slope than the constant P line also at positive slope Thus both the constant P and v processes have an increase in T As T goes up the change in s is smaller for the same area heat transfer under the process curve in the Ts diagram as compared with the constant T process The constant T isothermal process therefore changes s the most In a reversible process the area below the process curve in the Ts diagram is the heat transfer 2 1 T s q Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6f A liquid is compressed in a reversible adiabatic process What is the change in T If the process is reversible then s is constant ds dq T 0 Change in s for a liquid an incompressible substance is Eq 610 ds C T dT From this it follows that if ds 0 then T is constant 6g An ideal gas goes through a constant T reversible heat addition process How do the properties v u h s P change up down or constant Ideal gas uT hT so they are both constant Eq 62 gives ds dqT 0 so s goes up by qT Eq 612 gives ds Rv dv so v increases Eq 614 gives ds RP dP so P decreases 2 1 P v T s 1 2 T q Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6h Carbon dioxide is compressed to a smaller volume in a polytropic process with n 12 How do the properties u h s P T change up down or constant For carbon dioxide Table A5 k 1289 so we have n k and the process curve can be recognized in Figure 813 From this we see a smaller volume means moving to the left in the Pv diagram and thus also up From Pv diagram P up T up From Ts diagram Since T is up then s down As T is up so is h and u 2 1 P v T s 1 2 T n 1 q n 12 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6i A substance has heat transfer out Can you say anything about changes in s if the process is reversible If it is irreversible Reversible ds dq T 0 since dq 0 Irreversible ds dq T dsgen dq 0 but dsgen 0 You cannot say ds depends on the magnitude of dqT versus dsgen 6j A substance is compressed adiabatically so P and T go up Does that change s If the process is reversible then s is constant ds dq T 0 If the process is irreversible then s goes up ds dq T dsgen dsgen 0 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Concept Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 61 When a substance has completed a cycle v u h and s are unchanged Did anything happen Explain Yes During various parts of the cycle work and heat transfer may be transferred That happens at different P and T The net work out equals the net heat transfer in energy conservation so dependent upon the sign it is a heat engine or a heat pump refrigerator The net effect is thus a conversion of energy from one storage location to another and it may also change nature some Q was changed to W or the opposite Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 62 Assume a heat engine with a given QH Can you say anything about QL if the engine is reversible If it is irreversible For a reversible heat engine it must be that o dQ T 0 QH TH QL TL or integrals if T not constant So as TL is lower than TH then QL must be correspondingly lower than QH to obtain the net zero integral For an irreversible heat engine we have o dQ T QH TH QL TL 0 This means that QL is larger than before given QH and the Ts The irreversible heat engine rejects more energy and thus gives less out as work Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 63 CV A is the mass inside a pistoncylinder CV B is that plus part of the wall out to a source of 1Q2 at Ts Write the entropy equation for the two control volumes assuming no change of state of the piston mass or walls Fig P63 The general entropy equation for a control mass is Eq637 S2 S1 1 2 dQ T 1S2 gen The left hand side is storage so that depends of what is inside the CV and the integral is summing the dQT that crosses the control volume surface while the process proceeds from 1 to 2 CV A mA s2 s1 1 2 dQ TA 1S2 gen CV A CV B mA s2 s1 1 2 dQ Ts 1S2 gen CV B In the first equation the temperature is that of mass mA which possibly changes from 1 to 2 whereas in the second equation it is the reservoir temperature Ts The two entropy generation terms are also different the second one includes the first one plus any s generated in the walls that separate the mass mA from the reservoir and there is a Q over a finite temperature difference When the storage effect in the walls are neglected the left hand sides of the two equations are equal P o T m p A m s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 64 Consider the previous setup with the mass mA and the piston cylinder of mass mp starting out at two different temperatures After a while the temperature becomes uniform without any external heat transfer Write the entropy equation storage term S2 S1 for the total mass In this case the storage effect must be summed up over all the mass inside the control volume S2 S1 mA s2 s1A mP s2 s1 P mA s2 s1A mP CP ln T2 T1 P The last equation assumed a constant specific heat for the solid material of the piston a common assumption There is only a single temperature T2 but there are two different temperatures for state 1 T1 A and T1 P The temperature T2 would be found from the energy equation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 65 Water at 100oC quality 50 in a rigid box is heated to 110oC How do the properties P v x u and s change increase stay about the same or decrease A fixed mass in a rigid box give a constant v process So P goes up in the twophase region P Psat at given T v stays constant x goes up we get closer to the saturated vapor state see Pv diagram u goes up Q in and no work s goes up Q in P v 1 2 T s 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 66 Liquid water at 20oC 100 kPa is compressed in a pistoncylinder without any heat transfer to a pressure of 200 kPa How do the properties T v u and s change increase stay about the same or decrease Adiabatic dq 0 dq T ds 0 Incompressible dv 0 dw P dv 0 T v u and s they are all constant Only the pressure and enthalpy goes up In the Ts diagram the two states are in the same location as T does not go up for v constant P v 1 2 T s 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 67 A reversible process in a pistoncylinder is shown in Fig P67 Indicate the storage change u2 u1 and transfers 1w2 and 1q2 as positive zero or negative 1w2 P dv 0 1q2 T ds 0 u2 u1 0 from general shape of the constant u curves Further out in the ideal gas region the constant u curve become horizontal u fctT only P v 1 2 T s 1 2 u C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 68 A reversible process in a pistoncylinder is shown in Fig P68 Indicate the storage change u2 u1 and transfers 1w2 and 1q2 as positive zero or negative P v 1 2 2 1 T s 1w2 P dv 0 1q2 T ds 0 u2 u1 1q2 1w2 0 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 69 Air at 290 K 100 kPa in a rigid box is heated to 325 K How do the properties P v u and s change increase stay about the same or decrease Rigid box v constant P u and s all increases P v 1 2 1 T s 2 P P 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 610 Air at 20oC 100 kPa is compressed in a pistoncylinder without any heat transfer to a pressure of 200 kPa How do the properties T v u and s change increase about the same or decrease T goes up v goes down u goes up work in q 0 s constant P v 1 2 1 T s 2 P P 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 611 Carbon dioxide is compressed to a smaller volume in a polytropic process with n 14 How do the properties u h s P T change up down or constant For carbon dioxide Table A5 k 1289 so we have n k and the process curve can be recognized in Figure 613 From this we see a smaller volume means moving to the left in the Pv diagram and thus also up P up T up As T is up so is h and u From the Ts diagram as n k then we move to larger T means s is up 2 1 P v T s 1 2 T n 1 q n 14 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 612 Process A Air at 300 K 100 kPa is heated to 310 K at constant pressure Process B Heat air at 1300 K to 1310 K at constant 100 kPa Use the table below to compare the property changes Property A B A B A B a v2 v1 b h2 h1 c s2 s1 a Ideal gas Pv RT so v goes with absolute T v RP T thus the same b Since dh CP dT and CP increases with T c At constant P ds CPT dT CP is only 15 higher at 1300 K compared to 300 K see Fig 311 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 613 Why do we write S or S2 S1 whereas we write dQT and 1S2 gen This is similar to the terms in the continuity equation m2 m1 versus mi me and the energy equation E2 E1 versus 1Q2 1W2 The first part is the change in the storage S2 S1 of entropy in the mass and the second part is the transfer or generation during the process from 1 to 2 The storage terms correspond to the left hand side of the balance equation for entropy The integral dQT represents a transfer of entropy across the control volume surface during the process from 1 to 2 and the 1S2 gen expresses the total amount of entropy being generated inside the control volume and both are on the right hand side of the balance equation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 614 A reversible heat pump has a flux of s entering as Q LTL What can you say about the exit flux of s at TH For the entropy equation 63 and 642 the rate of storage is zero and the generation term is zero Thus the entropy equation becomes 0 Q L TL Q H TH 0 So Q L TL Q H TH flux of s We have the same flux of s in as out matching the result in chapter 5 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 615 An electric baseboard heater receives 1500 W of electrical power that heats the room air which loses the same amount out through the walls and windows Specify exactly where entropy is generated in that process a Electrical heating wire electrical work turned into internal energy leaving as heat transfer b Heat transfer from hot wire to cooler room air ie in the wire coverings c Room air to walls d Inside walls and windows heat transfer over a finite T e from outside wall surface to ambient T The electric wire is inside the pipe and surrounded by sand to electrically isolate it The pipe has fins mounted to increase the surface area Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 616 A 500 W electric space heater with a small fan inside heats air by blowing it over a hot electrical wire For each control volume a wire at Twire only b all the room air at Troom and c total room plus the heater specify the storage entropy transfer terms and entropy generation as rates neglect any Q through the room walls or windows Storage Q T S gen CV Wire only 0 500 W Twire 500 W Twire CV Room air 500 W Troom 500 W Twire 500 W 1 Troom 1 Twire CV Total room 500 W Troom 0 500 W Troom Remark Room only receives the electrical power input of 500 W Some of the heaters can be radiant heaters in which case the fan is not needed Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Inequality of Clausius Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 617 Consider the steam power plant in Example 47 and assume an average T in the line between 1 and 2 Show that this cycle satisfies the inequality of Clausius Solution Show Clausius dQ T 0 For this problem we have three heat transfer terms qb 2831 kJkg qloss 21 kJkg qc 21733 kJkg dq T Tb qb Tavg 12 qloss Tc qc 2831 573 21 568 21733 318 193 kJkg K 0 OK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 618 A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC Examine each of three cases with respect to the inequality of Clausius a W 6 kW b W 0 kW c Carnot cycle Solution TH 250 273 523 K TL 30 273 303 K Case a dQ T 6000 523 0 303 1147 kWK 0 Impossible b dQ T 6000 523 6000 303 833 kWK 0 OK c dQ T 0 6000 523 303 Q L Q L 303 523 6 kW 3476 kW W Q H Q L 2529 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 619 Use the inequality of Clausius to show that heat transfer from a warm space towards a colder space without work is a possible process ie a heat engine with no work output Clausius dQ T 0 or T dQ 0 Take CV as the space separating the warm and cold space It is the same Q that crosses each surface from energy equation so dQ T Q Twarm Q Tcold Q 1 Twarm 1 Tcold 0 OK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 620 Use the inequality of Clausius to show that heat transfer from a cold space towards a warmer space without work is an impossible process ie a heat pump with no work input Clausius dQ T 0 or T dQ 0 Take CV as the space separating the warm and cold space It is the same Q that crosses each surface from energy equation so dQ T Q Tcold Q Twarm Q 1 Tcold 1 Twarm 0 Impossible Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 621 Assume the heat engine in Problem 534 has a high temperature of 1000 K and a low temperature of 400 K What does the inequality of Clausius say about each of the four cases Solution Cases a dQ T 6 1000 4 400 0004 kWK 0 OK b dQ T 6 1000 0 400 0006 kWK 0 Impossible c dQ T 6 1000 2 400 0001 kWK 0 Impossible d dQ T 6 1000 6 400 0009 kWK 0 OK H Q W L Q T 400 K L T 1000 K H HE cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 622 Let the steam power plant in Problem 530 have 700oC in the boiler and 40oC during the heat rejection in the condenser Does that satisfy the inequality of Clausius Repeat the question for the cycle operated in reverse as a refrigerator Solution Q H 1 MW Q L 058 MW dQ T 1000 973 580 313 082 kWK 0 OK Refrigerator dQ T 580 313 1000 973 082 0 Cannot be possible W T Q H Q L W P in from coal to ambient Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 623 Examine the heat engine given in Problem 574 to see if it satisfies the inequality of Clausius Solution QH 325 kJ at TH 1000 K QL 125 kJ at TL 400 K dQ T 325 1000 125 400 00125 kJK 0 Impossible H Q 325 kJ W 200 kJ L Q 125 kJ T 1000 K H HE cb T 400 K L Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Entropy of a pure substance Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 624 Determine the entropy for these states a Nitrogen P 2000 kPa 120 K b Nitrogen 120 K v 0005 m3kg c R410A T 25oC v 001 m3kg Solution a Table B61 P Psat 2513 kPa so superheated vapor B62 at 2000 kPa s 48116 kJkgK b Table B61 vf v vg 000799 m3kg so twophase L V x v vfvfg 0005 0001915 000608 05074 s sf x sfg 38536 x 07659 42422 kJkg K c Table B41 vf v vg 001514 m3kg so twophase L V x v vfvfg 001 0000944 001420 063775 s sf x sfg 03631 x 06253 07619 kJkg K P v T s a bc bc T P a Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 625 Determine the missing property among P T s and x for R410A at a T 20oC v 01377 m3kg b T 20oC v 001377 m3kg c P 200 kPa s 1409 kJkgK a B41 v vg 00648 m3kg B42 superheated vapor so x is undefined very close to 200 kPa s 11783 kJkgK b B41 0000923 vf v vg 001758 m3kg Twophase P Psat 14442 kPa x v vfvfg 001377 0000923 001666 077113 s sf x sfg 03357 077113 06627 08467 kJkgK c Table B42 at 200 kPa s sg so superheated vapor x is undefined and we find the state at T 60oC b P v T s b a a T P c c Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 626 Find the missing properties of P v s and x for ammonia NH3 at a T 65C P 600 kPa b T 20C u 800 kJkg c T 50C v 01185 m3kg a B22 average between 60C and 70C v 025981 0269992 026435 m3kg s 56383 570942 56739 kJkgK b B21 u ug 13322 kJkg P Psat 8575 kPa x u ufufg 800 27289 10593 049666 v 0001638 x 014758 007494 m3kg s 10408 x 40452 304989 kJkgK c B21 v vg 006337 m3kg B22 superheated vapor so x is undefined very close to 1200 kPa s 51497 kJkgK v P s T a b c a b c Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 627 Find the entropy for the following water states and indicate each state on a Ts diagram relative to the twophase region a 250oC v 002 m3kg b 250oC 2000 kPa c 2oC 100 kPa Solution a Table B11 0001251 vf v vg 005013 m3kg Twophase x 002 0001251 004887 038365 s sf x sfg 27927 038365 32802 405 kJkg K b Table B11 P Psat 3973 kPa superheated vapor B13 s 65452 kJkg K c Table B11 T Ttripple 001oC so goto B15 Table B15 P Psat 05177 kPa so compressed solid s 12369 kJkg K v P s T a b c a b c Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 628 Determine the missing property among P v s and x for CO2 and indicate each state on a Ts diagram relative to the twophase region a T 20oC P 2000 kPa b T 20oC s 149 kJkgK c T 10oC s 1 kJkgK a Table B31 at 20oC P Psat 19696 kPa Compressed liquid v vf 0000969 m3kg s sf 01672 kJkgK x is undefined b Table B31 s sg 10406 kJkgK superheated vapor Table B32 located between 1400 kPa and 2000 kPa P 1400 600 149 15283 14438 15283 1400 600 04532 1672 kPa v 003648 002453 003648 04532 003106 m3kg c Table B31 02501 sf s sg 12328 kJkgK Twophase x s sf sfg 1 02501 09828 0763 P Psat 26487 kPa v vf x vfg 0001017 0763 001303 001096 m3kg v P s T a b c a b c Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 629 Determine the missing property among T P v s a H2O 20oC v 001000 m3kg b R410A 400 kPa s 117 kJkgK c NH3 10oC v 01 m3kg d N2 1013 kPa s 35 kJkgK a Table B11 at 20oC v vf 0001002 m3kg Compr Liquid B14 at about 5000 kPa s 02955 kJkgK b Table B41 s sg 10779 kJkgK Sup vapor Table B42 T 0 20 117 11483 12108 11483 20 03472 694oC v 007227 007916 007227 03472 007466 m3kg c Table B21 vf v vg 020541 so twophase L V P 6152 kPa x v vf vfg 01 00016 020381 04828 s sf x sfg 08778 x 43266 29667 kJkgK d Table B61 sf s sg 54033 kJkgK so twophase T 773 K x s sf sfg 35 28326 25707 025962 v vf x vfg 000124 x 021515 00571 m3kg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 630 Two kg water at 120oC with a quality of 25 has its temperature raised 20oC in a constant volume process What are the new quality and specific entropy Solution State 1 from Table B11 at 120oC v vf x vfg 0001060 025 08908 022376 m3kg State 2 has same v at 140oC also from Table B11 x v vf vfg 022376 000108 050777 04385 s sf x sfg 1739 04385 51908 4015 kJkg K T CP v P CP v T 1985 3613 120 140 120 C 140 C T CP s 198 kPa 120 140 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 631 Two kg water at 400 kPa with a quality of 25 has its temperature raised 20oC in a constant pressure process What is the change in entropy Solution State 1 from Table B12 at 400 kPa s sf x sfg 17766 025 51193 30564 kJkg State 2 has same P from Table B12 at 400 kPa T2 Tsat 20 14363 20 16363oC so state is superheated vapor look in B13 and interpolate between 150 and 200 C in the 400 kPa superheated vapor table s2 69299 71706 69299 16363 150 200 150 69955 kJkgK s2 s1 69955 30564 39391 kJkgK T CP v 400 kPa P CP v T 400 144 164 1436 C 164 C T CP s 400 kPa 144 164 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 632 Saturated liquid water at 20oC is compressed to a higher pressure with constant temperature Find the changes in u and s when the final pressure is a 500 kPa b 2000 kPa c 20 000 kPa Solution kJkg kJkg K B11 u1 8394 s1 02966 B14 ua 8391 sa 02965 u 003 s 00001 B14 ub 8382 sb 02962 u 012 s 00004 B14 uc 8275 sc 02922 u 119 s 00044 Nearly constant u and s incompressible media v P s T a b c 1 cba1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 633 Saturated vapor water at 250oC is expanded to a lower pressure with constant temperature Find the changes in u and s when the final pressure is a 100 kPa b 50 kPa c 10 kPa Solution Table B11 for the first state then B13 for the a b and c states kJkg kJkg K kJkg kJkg K u1 260237 s1 60729 ua 273373 sa 80332 u 13136 s 19603 ub 273497 sb 83555 u 1326 s 22826 uc 273595 sc 91002 u 13358 s 30273 v P s T 1 a b c a b c 1 3973 kPa 100 50 10 Remark You approach ideal gas as P drops so u is uT but s is still sTP Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 634 Determine the missing property among P T s x for the following states a Ammonia 25oC v 010 m3kg b Ammonia 1000 kPa s 52 kJkg K c R410A 500 kPa s 14 kJkg K d R410A 50oC s 08 kJkg K Solution Table P kPa T oC s kJkg K x a B21 1003 25 41601 07776 b B22 1000 4253 52 c B42 500 100 14 d B41 3065 50 08 06973 a x 01 0001658012647 07776 s sf x sfg 1121 x 39083 41601 kJkg K b T 40 10 52 5177852654 51778 4253oC superheated vapor so x is undefined c s sg 10647 so superheated vapor found close to 100oC d sf s sg so twophase P Psat 30652 kPa x 08 0506704206 069734 v P s T d a b c a b c d Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Reversible processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 635 In a Carnot engine with ammonia as the working fluid the high temperature is 60C and as QH is received the ammonia changes from saturated liquid to saturated vapor The ammonia pressure at the low temperature is 190 kPa Find TL the cycle thermal efficiency the heat added per kilogram and the entropy s at the beginning of the heat rejection process Solution T s 1 2 4 3 Constant T constant P from 1 to 2 Table B21 qH Tds T s2 s1 T s fg h2 h1 hfg 9970 kJkg States 3 4 are twophase Table B21 TL T3 T4 TsatP 20C ηcycle 1 TH TL 1 2532 3332 024 Table B21 s3 s2 sg60C 46577 kJkg K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 636 Consider a Carnotcycle heat pump with R410A as the working fluid Heat is rejected from the R410A at 35C during which process the R410A changes from saturated vapor to saturated liquid The heat is transferred to the R410A at 0C a Show the cycle on a Ts diagram b Find the quality of the R410A at the beginning and end of the isothermal heat addition process at 0C c Determine the COP for the cycle Solution a 1 2 3 4 35 0 T s b From Table B41 state 3 is saturated liquid s4 s3 04189 kJkg K 02264 x408104 x4 02375 State 2 is saturated vapor so from Table B41 s1 s2 09671 kJkg K 02264 x108104 x1 0914 c β qH wIN TH TH TL 30815 35 88 Ts diagram from CATT3 for R410A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 637 Do Problem 636 using refrigerant R134a instead of R410A Consider a Carnotcycle heat pump with R410A as the working fluid Heat is rejected from the R410A at 35C during which process the R410A changes from saturated vapor to saturated liquid The heat is transferred to the R410A at 0C a Show the cycle on a Ts diagram b Find the quality of the R410A at the beginning and end of the isothermal heat addition process at 0C c Determine the coefficient of performance for the cycle Solution a 1 2 3 4 35 0 T s b From Table B51 state 3 is saturated liquid s4 s3 11673 kJkg K 10 x407262 x4 02303 State 2 is saturated vapor so from Table B51 s1 s2 17139 kJkg K 10 x107262 x1 0983 c β qH wIN TH TH TL 30815 35 88 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 638 Water is used as the working fluid in a Carnot cycle heat engine where it changes from saturated liquid to saturated vapor at 200C as heat is added Heat is rejected in a constant pressure process also constant T at 20 kPa The heat engine powers a Carnot cycle refrigerator that operates between 15C and 20C Find the heat added to the water per kg water How much heat should be added to the water in the heat engine so the refrigerator can remove 1 kJ from the cold space Solution Carnot cycle heat engine T s 1 2 4 3 Constant T constant P from 1 to 2 Table B21 qH Tds T s2 s1 T sfg hfg 47315 41014 1940 kJkg States 3 4 are twophase Table B21 TL T3 T4 TsatP 6006oC Carnot cycle refrigerator TL and TH are different from above βref QL W TL TH TL 273 15 20 15 258 35 737 W QL β 1 737 0136 kJ The needed work comes from the heat engine W ηHE QH H2O ηHE 1 TH TL 1 333 473 0296 QH H2O W ηHE 0136 0296 046 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 639 Water at 1 MPa 250C is expanded in a pistoncylinder to 200 kPa x 10 in a reversible process Find the sign for the work and the sign for the heat transfer Solution The process is not specified but the beginning and end states are and we assume a unidirectional process so v keeps decreasing State 1 Table B13 v1 023268 m3kg u1 27099 kJkg s1 69246 kJkg K State 2 Table B11 v2 08857 m3kg u2 25295 kJkg s2 71271 kJkg K Reversible process dw P dv dq T ds v2 v1 1w2 P dv 0 s2 s1 1q2 T ds 0 1 2 P v T s 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 640 R410A at 1 MPa and 60oC is expanded in a piston cylinder to 500 kPa 40oC in a reversible process Find the sign for both the work and the heat transfer for this process Solution The process is not specified but the beginning and end states are and we assume a unidirectional process so v keeps increasing 1w2 P dv so sign dv 1q2 T ds so sign ds State 1 B42 v1 003470 m3kg s1 12019 kJkg K State 2 B42 v2 006775 m3kg s2 12398 kJkg K dv 0 w is positive ds 0 q is positive 2 1 P v T s 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 641 A pistoncylinder compressor takes R410A as saturated vapor 500 kPa and compresses it in a reversible adiabatic process to 3000 kPa Find the final temperature and the specific compression work CV R410A this is a control mass Energy Eq35 u2 u1 1q2 1w2 Entropy Eq63 s2 s1 dqT 1q2 T Process Adiabatic and reversible 1q2 0 so then s2 s 1 State 1 P1 x 1 u1 24829 kJkg s1 10647 kJkgK State 2 P2 s2 s1 T2 60 20 10647 09933 10762 09933 60 2008613 772oC u2 27496 29838 27496 08613 29513 kJkg Now the work becomes 1w2 u1 u2 24829 29513 4684 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 642 A pistoncylinder receives R410A at 500 kPa and compresses it in a reversible adiabatic process to 1800 kPa 60oC Find the initial temperature CV R410A this is a control mass Energy Eq35 u2 u1 1q2 1w2 Entropy Eq63 s2 s1 dqT 1q2 T Process Adiabatic and reversible 1q2 0 so then s2 s 1 State 1 P1 s1 s2 11076 kJkgK T1 1389 1389 11076 10647 11155 10647 216oC 2 1 P v T s 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 643 Compression and heat transfer brings carbon dioxide in a pistoncylinder from 1400 kPa 20oC to saturated vapor in an isothermal process Find the specific heat transfer and the specific work Solution m constant Energy Eq35 u2 u1 1q2 1w2 Entropy Eq63 s2 s1 dqT 1q2 T Process T C and assume reversible 1q2 T s2 s1 State 1 Table B42 u1 25918 kJkg s1 10057 kJkg K State 2 Table B41 u2 25816 kJkg s2 09984 kJkg K P v 1 2 T 2 1 T s 1q2 273 20 09984 10057 214 kJkg 1w2 1q2 u1 u2 214 25918 25816 112 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 644 A piston cylinder maintaining constant pressure contains 01 kg saturated liquid water at 100C It is now boiled to become saturated vapor in a reversible process Find the work term and then the heat transfer from the energy equation Find the heat transfer from the entropy equation is it the same Energy Eq mu2 u1 1Q2 1W 2 Entropy Eq ms2 s1 dq T 0 T 1Q2 Process P C 1W2 m Pv2 v1 m P vfg 01 kg 1013 kPa 167185 m3kg 16936 kJ From the energy equation we get 1Q2 mu2 u1 1W2 m ufg 1W2 01 208758 16936 2257 kJ or mh2 h1 m hfg 01 225703 2257 kJ From the entropy equation we can get 1Q2 mTs2 s1 m T sfg 01 37315 6048 22568 kJ So they are equal to within round off errors Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 645 A piston cylinder contains 025 kg of R134a at 100 kPa It will be compressed in an adiabatic reversible process to 400 kPa and should be 70oC What should the initial temperature be CV R134a which is a control mass Entropy Eq63 ms2 s1 dQT 0 State 2 s2 s1 19051 kJkgK Work backwards from state 2 to state 1 State 1 100 kPa s1 T1 264C P v 1 2 2 1 T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 646 A pistoncylinder contains 05 kg of water at 200 kPa 300C and it now cools to 150C in an isobaric process The heat goes into a heat engine that rejects heat to the ambient at 25C shown in Fig P646 and the whole process is assumed to be reversible Find the heat transfer out of the water and the work given out by the heat engine CV H2O Energy Eq35 mu2 u1 1Q2 1W 2 Entropy Eq63 ms2 s1 dQT 0 Process P C W P dV PV2 V1 m P v2 v1 State 1 B13 s1 78926 kJkgK h1 307179 kJkg State 2 B13 s2 72795 kJkg K h2 27688 kJkg From the process equation and the energy equation 1Q2 mu2 u1 1W2 mh2 h1 0527688 307179 151495 kJ CV Total Energy Eq35 mu2 u1 QL 1W2 W HE Entropy Eq63 ms2 s1 QLTamb 0 QL mTambs1 s2 05 kg 29815 K 78926 72795 kJkgK 91398 kJ Now the energy equation for the heat engine gives WHE 1Q2 QL 151495 91398 601 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 647 A cylinder fitted with a piston contains ammonia at 50C 20 quality with a volume of 1 L The ammonia expands slowly and during this process heat is transferred to maintain a constant temperature The process continues until all the liquid is gone Determine the work and heat transfer for this process Solution CV Ammonia in the cylinder Energy Eq35 mu2 u1 1Q2 1W 2 Entropy Eq63 ms2 s1 dQT 0 Process T constant to x2 10 P constant 2033 MPa W P dV PV2 V1 m P v2 v1 dQT 1Q2 T 1 2 T s 50 C NH 3 o Table B21 T1 50C x1 020 V1 1 L v1 0001777 02 006159 0014095 m3kg s1 15121 02 32493 21620 kJkg K m V1v1 00010014095 0071 kg v2 vg 006336 m3kg s2 sg 47613 kJkg K 1W2 Pmv2 v1 2033 0071 006336 0014095 711 kJ From the entropy equation 1Q2 Tm s2 s1 3232 Κ 0071 kg 47613 21620 kJkgK 5965 kJ or 1Q2 mu2 u1 1W2 mh2 h1 h1 42148 02 105001 63148 kJkg h2 147149 kJkg 1Q2 0071 kg 147149 63148 kJkg 5965 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 648 Water in a pistoncylinder at 400oC 2000 kPa is expanded in a reversible adiabatic process The specific work is measured to be 41572 kJkg out Find the final P and T and show the Pv and the Ts diagram for the process Solution CV Water which is a control mass Adiabatic so 1q2 0 Energy Eq35 u2 u1 1q2 1w2 1w 2 Entropy Eq63 s2 s1 dqT 0 since reversible State 1 Table B13 u1 294521 kJkg s1 7127 kJkg K State 2 s u u2 u1 1w2 294521 41572 252949 kJkg sat vapor 200 kPa T 12023C v P s T 2 1 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 649 A pistoncylinder has 2 kg water at 1000 kPa 200C which is now cooled with a constant loading on the piston This isobaric process ends when the water has reached a state of saturated liquid Find the work and heat transfer and sketch the process in both a Pv and a Ts diagram Solution CV H2O Energy Eq35 mu2 u1 1Q2 1W 2 Entropy Eq63 ms2 s1 dQT Process P C W P dV PV2 V1 State 1 B13 v1 020596 m3kg s1 66939 kJkgK u1 262190 kJkg State 2 B12 v2 0001127 m3kg s2 21386 kJkg K u2 76167 kJkg From the process equation 1W2 m P v2 v1 2 1000 0001127 020596 4097 kJ From the energy equation we get 1Q2 mu2 u1 1W2 2 76167 262190 4097 41302 kJ 2 1 P v T s 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 650 One kilogram of water at 300C expands against a piston in a cylinder until it reaches ambient pressure 100 kPa at which point the water has a quality of 902 It may be assumed that the expansion is reversible and adiabatic What was the initial pressure in the cylinder and how much work is done by the water Solution CV Water Process Rev Q 0 Energy Eq35 mu2 u1 1Q2 1W2 1W 2 Entropy Eq63 ms2 s1 dQT Process Adiabatic Q 0 and reversible s2 s 1 State 2 P2 100 kPa x2 0902 from Table B12 s2 13026 0902 60568 67658 kJkg K u2 41736 0902 20887 23014 kJkg State 1 At T1 300C s1 67658 Find it in Table B13 P1 2000 kPa u1 27726 kJkg From the energy equation 1W2 mu1 u2 127726 23014 4712 kJ v P s T 2 1 1 2 T 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 651 Water at 1000 kPa 250C is brought to saturated vapor in a rigid container shown in Fig P854 Find the final T and the specific heat transfer in this isometric process Solution Energy Eq35 u2 u1 1q2 1w2 Entropy Eq63 s2 s1 dqT Process v constant 1w2 0 State 1 T P Table B13 u1 270991 kJkg v1 023268 m3kg State 2 x 1 and v2 v1 so from Table B11 we see P2 800 kPa T2 170 5 023268 02428302168 024283 170 5 038993 17195C u2 257646 038993 258019 257646 25779 kJkg From the energy equation 1q2 u2 u1 25779 270991 132 kJkg v P s T 2 1 1 2 v C Notice to get 1q2 T ds we must know the function Ts which we do not readily have for this process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 652 Estimate the specific heat transfer from the area in the Ts diagram and compare it to the correct value for the states and process in Problem 651 Solution Energy Eq35 u2 u1 1q2 1w2 Entropy Eq63 s2 s1 dqT Process v constant 1w2 0 State 1 T P Table B13 u1 270991 kJkg v1 023268 m3kg s1 69246 kJkg K State 2 x 1 and v2 v1 so from Table B11 we see P2 800 kPa T2 170 5 023268 02428302168 024283 170 5 038993 17195C u2 257646 038993 258019 257646 25779 kJkg s2 66663 038993 66256 66663 66504 kJkg K From the energy equation 1q2 actual u2 u1 25779 270991 132 kJkg Assume a linear variation of T versus s 1q2 T ds area 1 2 T1 T2s2 s1 1 2 17195 2 27315 250 K 66504 69246 kJkgK 13274 kJkg very close ie the v C curve is close to a straight line in the Ts diagram Look at the constant v curves in Fig E1 In the twophase region they curve slightly and more so in the region above the critical point v P s T 2 1 1 2 v C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 653 A closed tank V 10 L containing 5 kg of water initially at 25C is heated to 150C by a heat pump that is receiving heat from the surroundings at 25C Assume that this process is reversible Find the heat transfer to the water and the change in entropy Solution CV Water from state 1 to state 2 Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq63 ms2 s1 dQT Process constant volume reversible isometric so 1W2 0 State 1 v1 Vm 0002 from Table B11 x1 0002 000100343358 0000023 u1 10486 0000023 23049 10493 kJkg s1 03673 0000023 81905 036759 kJkg K Continuity eq same mass and V C fixes v2 State 2 T2 v2 v1 so from Table B11 x2 0002 0001090039169 00023233 u2 63166 00023233 192787 63614 kJkg s2 18417 00023233 49960 18533 kJkg K Energy eq has W 0 thus provides heat transfer as 1Q2 mu2 u1 265605 kJ The entropy change becomes ms2 s1 518533 036759 74286 kJK P v 1 2 T 2 1 T s Notice we do not perform the integration dQT to find change in s as the equation for the dQ as a function of T is not known Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 654 A pistoncylinder has 2 kg of R410A at 60C 100 kPa which is compressed to 1000 kPa The process happens so slowly that the temperature is constant Find the heat transfer and work for the process assuming it to be reversible Solution CV R410A Control Mass Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq63 ms2 s1 dQT Process T constant and assume reversible process 1 TP Table B42 v1 037833 m3kg u1 3094 kJkg s1 14910 kJkg K 2 TP Table B42 v2 003470 m3kg u2 30104 kJkg s2 12019 kJkg K 2 1 P v T s 1 2 T From the entropy equation 2nd law 1Q2 mTs2 s1 2 kg 33315 K 12019 14910 kJkgK 19263 kJ From the energy equation 1W2 1Q2 mu2 u1 19263 kJ 2 kg 30104 3094 kJkg 1759 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 655 A heavily insulated cylinderpiston contains ammonia at 1200 kPa 60C The piston is moved expanding the ammonia in a reversible process until the temperature is 20C During the process 200 kJ of work is given out by the ammonia What was the initial volume of the cylinder CV ammonia Control mass with no heat transfer State 1 Table B22 v1 01238 m3kg s1 52357 kJkg K u1 14048 kJkg Entropy Eq ms2 s1 dQT 1S 2 gen Process reversible 1S2 gen 0 and adiabatic dQ 0 s2 s 1 P v 1 2 2 1 T s State 2 T2 s2 x2 52357 0365752498 0928 u2 8876 092812107 121195 kJkg 1Q2 0 mu2 u1 1W2 m121195 14048 200 m 1037 kg V1 mv1 1037 01238 01284 m3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 656 Water at 1000 kPa 250C is brought to saturated vapor in a pistoncylinder with an isothermal process Find the specific work and heat transfer Estimate the specific work from the area in the Pv diagram and compare it to the correct value Solution Continuity Eq m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq63 ms2 s1 dQT Process T constant reversible State 1 Table B13 v1 023268 m3kg u1 270991 kJkg s1 69246 kJkg K State 2 T x Table B11 P2 3973 kPa v2 005013 m3kg u2 260237 kJkg s2 60729 kJkg K P v 1 2 T 2 1 T s From the entropy equation 1q2 T ds Ts2 s1 250 273 60729 69246 4456 kJkg From the energy equation 1w2 1q2 u1 u2 4456 270991 260237 338 kJkg Estimation of the work term from the area in the Pv diagram 1w2 area 1 2 P1P2v2 v1 1 21000 3973005013 023268 454 kJkg Not extremely accurate estimate Pv curve not linear more like Pv constant as curve has positive curvature the linear variation overestimates area Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 657 A rigid insulated vessel contains superheated vapor steam at 3 MPa 400C A valve on the vessel is opened allowing steam to escape The overall process is irreversible but the steam remaining inside the vessel goes through a reversible adiabatic expansion Determine the fraction of steam that has escaped when the final state inside is saturated vapor CV steam remaining inside tank Rev Adiabatic inside only Continuity Eq m2 m1 m Energy Eq mu2 u1 1Q2 1W2 Entropy Eq ms2 s1 dQT 1S 2 gen P v 1 2 2 1 T s CV m2 Rev 1S2 gen 0 Adiabatic Q 0 s2 s1 69212 sG at T 2 T2 141C v2 vg at T2 04972 m3kg me m1 m1m2 m1 1 m2 m1 1 v1 v2 1 009936 04972 080 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 658 Water at 100 kPa 25C is brought to the boiling point in a pistoncylinder with an isobaric process The heat is supplied by a heat pump with the cold side at the ambient temperature of 25C Assume that the whole process is reversible and find the work input to the heat pump per kg of water CV H2O Energy Eq35 u2 u1 1q2 1w 2 Entropy Eq63 s2 s1 dqT 0 Process P C w P dv Pv2 v1 State 1 B11 s1 03673 kJkgK h1 10487 kJkg State 2 B12 s2 13025 kJkg K h2 41744 kJkg From the process equation and the energy equation 1q2 u2 u1 1w2 h2 h1 41744 10487 31257 kJkg CV Total Energy Eq35 u2 u1 qL 1w2 w HP Entropy Eq63 s2 s1 qLTamb 0 qL Tambs2 s1 29815 K 13025 03673 kJkgK 27883 kJkg Now the energy equation for the heat engine gives wHP 1q2 qL 31257 27883 3374 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 659 Water at 1000 kPa 200C is brought to saturated vapor in a pistoncylinder with an isobaric process Find the specific work and heat transfer Estimate the specific heat transfer from the area in the Ts diagram and compare it to the correct value Solution CV H2O Energy Eq35 u2 u1 1q2 1w 2 Entropy Eq63 s2 s1 dqT Process P C w P dv Pv2 v1 State 1 B13 v1 020596 m3kg s1 66939 kJkgK u1 262190 kJkg State 2 B13 v2 019444 m3kg s2 65864 kJkg K u2 258364 kJkg T2 17991C From the process equation 1w2 P v2 v1 1000 kPa 019444 020596 m3kg 1152 kJkg From the energy equation 1q2 u2 u1 1w2 258364 262190 1152 4978 kJkg Now estimate the heat transfer from the Ts diagram 1q2 T ds AREA 1 2 T1 T2s2 s1 1 2 200 17991 2 27315 K 65864 66939 kJkgK 463105 01075 4978 kJkg very close approximation The P C curve in the Ts diagram is nearly a straight line Look at the constant P curves on FigE1 Up over the critical point they curve significantly 2 1 P v T s 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Entropy of a liquid or a solid Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 660 Two 5 kg blocks of steel one at 250oC the other at 25oC come in thermal contact Find the final temperature and the change in entropy of the steel CV Both blocks no external heat transfer C from Table A3 Energy Eq U2 U1 mAu2 u1A mBu2 u1B 0 0 mACT2 TA1 mBCT2 TB1 T2 mA mB mATA1 mBTB1 1 2 TA1 1 2 TB1 1375oC Entropy Eq637 S2 S1 mAs2 s1A mBs2 s1B 1S 2 gen Entropy changes from Eq611 S2 S1 mAC ln TA1 T2 mBC ln T2 TB1 5 046 ln 1375 27315 250 27315 5 046 ln 1375 27315 29815 05569 07363 01794 kJK A B Heat transfer over a finite temperature difference is an irreversible process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 661 A rigid tank of 12 kg steel contains 15 kg of R134a at 40oC 500 kPa The tank is placed in a refrigerator that brings it to 20oC Find the process heat transfer and the combined steel and R134a change in entropy CV The steel tank and the R134a The energy equation Eq 35 now becomes summing over the mass mst u2 u1st mR134a u2 u1R134a 1Q2 0 Process No change in volume so no work as used above Use specific heat from Table A3 for steel and Table B5 for R134a R134a v1 004656 m3kg u1 40744 kJkg s1 17971 kJkgK State 2 v2 v1 vg x2 v2 vfvfg 004656 0000738 014576 0314366 u2 uf x2 ufg 17365 x2 19285 234275 kJkg s2 sf x2 sfg 09007 x2 08388 116439 kJkgK Now the heat transfer from the energy equation 1Q2 mR134au2 u1R134a mst Cst T2 T1 15 234275 40744 12 046 20 40 29287 kJ Steel msts2 s1st mstCst ln T2T1 12046 ln 25315 31315 011741 kJK Entropy change for the total control volume steel and R134a S2 S1 mst s2 s1st mR134a s2 s1R134a 011741 15116439 17971 1066 kJK 20 C o Q 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 662 A large slab of concrete 5 8 03 m is used as a thermal storage mass in a solarheated house If the slab cools overnight from 23C to 18C in an 18C house what is the net entropy change associated with this process Solution CV Control mass concrete V 5 8 03 12 m3 m ρV 2200 12 26 400 kg Energy Eq mu2 u1 1Q2 1W 2 Entropy Eq ms2 s1 1Q2 T0 1S 2 gen Process V constant so 1W2 0 Use heat capacity Table A3 for change in u of the slab 1Q2 mCT 26400 kg 088 kJkgK 5 K 116 160 kJ We add all the storage changes as in Eq639 Sslab ms2 s1 m C ln T1 T2 26400 kg 088 kJkgK ln 2912 2962 3955 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 663 A foundry form box with 25 kg of 200C hot sand is dumped into a bucket with 50 L water at 15C Assuming no heat transfer with the surroundings and no boiling away of liquid water calculate the net entropy change for the mass Solution CV Sand and water constant pressure process msandu2 u1sand mH2Ou2 u1H2O PV2 V1 msandhsand mH2OhH2O 0 For this problem we could also have said that the work is nearly zero as the solid sand and the liquid water will not change volume to any measurable extent Now we get changes in us instead of hs For these phases CV CP C which is a consequence of the incompressibility Now the energy equation becomes msandCsandTsand mH2OCH2OTH2O 0 25 kg 08 kJkgK T2 200 Κ 50103 m3 0001001 m3kg 4184 kJkgΚ T2 15 K 0 T2 312C S2 S1 msand s2 s1 mH2O s2 s1 msand Csand lnT2T1 mH2O CH2O lnT2T1 25 08 ln 3043 47315 4995 4184 ln 3043 28815 257 kJK Box holds the sand for form of the cast part Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 664 Heat transfer to a block of 15 kg ice at 10oC melts it to liquid at 10oC in a kitchen Find the entropy change of the water Water changes state from nearly saturated solid to nearly saturated liquid The pressure is 101 kPa but we approximate the state properties with saturated state at the same temperature State 1 Compressed saturated solid B15 s1 12995 kJkgK State 2 Compressed saturated liquid B11 s2 01510 kJkgK The entropy change is s s2 s1 0151 12995 14505 kJkgK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 665 In a sink 5 liters of water at 70oC is combined with 1 kg aluminum pots 1 kg of flatware steel and 1 kg of glass all put in at 20oC What is the final uniform temperature and change in stored entropy neglecting any heat loss and work Energy Eq U2 U1 miu2 u1i 1Q2 1W2 0 Entropy Eq S2 S1 dQT 1S2 gen For the water vf 0001023 m3kg V 5 L 0005 m3 m Vv 48876 kg For the liquid and the metal masses we will use the specific heats Tbl A3 A4 so miu2 u1i miCv i T2 T1i T2miCv i miCv iT1 i noticing that all masses have the same T2 but not same initial T miCv i 48876 418 1 09 1 046 1 08 2259 kJK Energy Eq 2259 T2 48876 418 70 1 09 1 046 1 08 20 143011 432 T2 652oC S2 S1 mis2 s1i mi Ci ln T2 Ti1 48876 418 ln 6522 27315 70 27315 1 09 046 08 ln 6522 27315 20 27315 028659 030986 002327 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 666 A piston cylinder has constant pressure of 2000 kPa with water at 20oC It is now heated up to 100oC Find the heat transfer and the entropy change using the steam tables Repeat the calculation using constant heat capacity and incompressibility Solution CV Water Constant pressure heating Energy Eq35 u2 u1 1q2 1w 2 Entropy Eq637 s2 s1 1q2 TSOURCE 1s2 gen Process P P1 1w2 Pv2 v1 The energy equation then gives the heat transfer as 1q2 u2 u1 1w2 h2 h 1 Steam Tables B14 h1 8582 kJkg s1 02962 kJkg K h2 42045 kJkg s2 13053 kJkg K 1q2 h2 h1 8582 42045 33463 kJkg s2 s1 13053 02962 10091 kJkg K Now using values from Table A4 Liquid water Cp 418 kJkg K h2 h1 CpT2 T1 418 80 3344 kJkg s2 s1 Cp lnT2T1 418 ln 37315 29315 10086 kJkg K Approximations are very good Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 667 A 4 L jug of milk at 25C is placed in your refrigerator where it is cooled down to the refrigerators inside constant temperature of 5C Assume the milk has the property of liquid water and find the entropy change of the milk Solution CV Jug of milk Control mass at constant pressure Continuity Eq m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 dQT 1S 2 gen State 1 Table B11 v1 vf 0001003 m3kg sf 03673 kJkg K m Vv 0004 m3 0001003 m3kg 3988 kg State 2 Table B11 s sf 00761 kJkg K The change of entropy becomes S2 S1 ms2 s1 3988 kg 00761 03673 kJkgK 11613 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 668 A constant pressure container of 12 kg steel contains 15 kg of R134a at 40oC 500 kPa The container is placed in a refrigerator that brings it to 20oC Find the process heat transfer and the combined steel and R134a change in entropy CV The steel container and the R134a The energy equation Eq 35 now becomes summing over the mass mst u2 u1st mR134a u2 u1R134a 1Q2 1W 2 Process P C so 1W2 P V2 V1 P mR134a v2 v1 R134a Substitute the work into the energy eqution and we combine the R134a terms 1Q2 mR134ah2 h1R134a mst Cst T2 T1 Use specific heat from Table A3 for steel and Table B5 for R134a R134a v1 004656 m3kg h1 43072 kJkg s1 17971 kJkgK State 2 500 kPa 20oC compressed liquid v2 0000738 s2 sf 09007 kJkgK h2 hf ΔPv 17374 5001337 0000738 1740 kJkg notice how the correction for P higher than Psat is small Now the heat transfer from the energy equation 1Q2 mR134ah2 h1R134a mst Cst T2 T1 15 1740 43072 12 046 20 40 4182 kJ Steel msts2 s1st mstCst ln T2T1 12046 ln 25315 31315 011741 kJkgK Entropy change for the total control volume steel and R134a S2 S1 mst s2 s1st mR134a s2 s1R134a 011741 1509007 17971 1462 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 669 A 10kg steel container is cured at 500oC An amount of liquid water at 15oC 100 kPa is added to the container so a final uniform temperature of the steel and the water becomes 50oC Neglect any water that might evaporate during the process and any air in the container How much water should be added and how much was the entropy changed CV The steel and the water no external heat transfer nor any work Energy Eq mH2O u2 u1H2O mst u2 u1 0 mH2o 2093 6298 mstC T2 T1 0 mH2O 14632 kJkg 10 046 50 500 kJ 0 mH2O 207014632 14147 kg Entropy Eq 637 mH2O s2s1 mst s2 s1 1S 2 gen S2 S1 14147 kg 07037 02245 kJkgK 10kg 046 kJkgK ln 50 273 773 67792 40141 27651 kJK cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 670 A pan in an autoshop contains 5 L of engine oil at 20oC 100 kPa Now 3 L of hot 100oC oil is mixed into the pan Neglect any work term and find the final temperature and the entropy change Solution Since we have no information about the oil density we assume the same for both from Table A4 ρ 885 kgm3 Energy Eq m2u2 mAuA mBuB 0 0 u CvT so same Cv 19 kJkg K for all oil states T EA mAB A E mA2 AE A TABE A A5 8E A 20 A3 8E A 100 500AoE AC 32315 K 2 mA m2 TA SA2E A SA1E A mA2E AsA2E A mAAE AsAAE A mABE AsABE A mAAE AsA2E A sAAE A mABE AsA2E A sABE A 0005 885 19 ln A32315 29315E A 0003 885 19 ln A32315 37315E 08192 07257 00935 kJK Entropy generation is the total change in S recall Eq639 no external Q Oils shown before mixed to final uniform state Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 671 A computer CPU chip consists of 50 g silicon 20 g copper 50 g polyvinyl chloride plastic It heats from 15AoE AC to 75AoE AC as the computer is turned on How much did the entropy increase CV CPU chip The process has electrical work input and no heat transfer Entropy Eq SA2E A SA1E A Amis2 s1i EA dQT A1E ASA2 genE A A1E ASA2 genE For the solid masses we will use the specific heats Table A3 and they all have the same temperature so Amis2 s1i EA AmiCi lnT2 T1i EA ln TA2E ATA1E A AmiCi EA AmiCi EA 005 07 002 042 005 096 00914 kJK SA2E A SA1E A 00914 kJK ln 34815 28815 00173 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 672 A 5kg aluminum radiator holds 2 kg of liquid R134a at 10AoE AC The setup is brought indoors and heated with 220 kJ Find the final temperature and the change in entropy of all the mass Solution CV The aluminum radiator and the R134a Energy Eq35 mA2E AuA2E A mA1E AuA1E A A1E AQA2E A 0 Process No change in volume so no work as used above The energy equation now becomes summing over the mass mAalE A uA2E A uA1E AAalE A mAR134aE A uA2E A uA1E AAR134aE A A1E AQA2E Use specific heat from Table A3 and A4 mAalE ACAalE A TA2E A TA1E A m AR134aE AC AR134aE A ln TA2E A TA1E A A1E AQA2E TA2E A TA1E A A1E AQA2E A mAalE ACAalE A m AR134aE AC AR134aE A 220 kJ 5 09 2 143 kJK 2989oC TA2E A 10 2989 1989oC Entropy change for solid A3 and liquid A4 from Eq611 SA2E A SA1E A mAalE A sA2E A sA1E AAalE A mAR134aE A sA2E A sA1E AAR134aE mAalE ACAalE A ln TA2E ATA1E A m AR134aE AC AR134aE A ln TA2E ATA1E A 5 09 2 143 kJK ln A1989 27315 E10 27315E A 0792 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 673 A 12 kg steel container has 02 kg superheated water vapor at 1000 kPa both at 200oC The total mass is now cooled to ambient temperature 30AoE AC How much heat transfer was taken out and what is the steelwater entropy change Solution CV Steel and the water control mass of constant volume Energy Eq35 UA2E A UA1E A A1E AQA2E A A1E AWA2E Process V constant A1E AWA2E A 0 State 1 HA2E A0 Table B13 uA1E A 26219 kJkg vA1E A 020596 mA3E Akg sA1E A 66939 kJkg K State 2 HA2E A0 TA2E A vA2E A vA1E A from Table B11 xA2E A v vf vfg A020596 0001004 328922E A 0006231 uA2E A 12577 xA2E A 229081 14004 kJkg sA2E A 04369 xA2E A 80164 048685 kJkg K A1E AQA2E A muA2E A uA1E A mAsteelE ACAsteelE A TA2E A TA1E A mAH2OE A uA2E A uA1E A AH2OE 12 kg 046 kJkgK 30 200 K 02 kg14004 26219 kJkg 14348 kJ Entropy changes from Eq611 and the water tables SA2E A SA1E A mA2E A sA2E A mA1E AsA1E A mAsteelE ACAsteelE A ln T2 T1 mAH2OE A sA2E A sA1E AAH2OE 12 046 kJK ln A30315 47315E A 02 kg 048685 66939 kJkgK 24574 12414 3699 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 674 Find the total work the heat engine can give out as it receives energy from the rock bed as described in Problem 560 see FigP 674 Hint write the entropy balance equation for the control volume that is the combination of the rock bed and the heat engine Solution To get the work we must integrate over the process or do the 2nd law for a control volume around the whole setup out to TA0E CV Heat engine plus rock bed out to TA0E A W and QALE A goes out W Q H Q L HE CV Energy Eq35 UA2E A UA1E AArockE A QALE A W Entropy Eq63 11 SA2E A SA1E AArockE A A QL ET0 E A mC ln A T2 ET1 E A 5500 089 ln A290 400E A 157415 kJK QALE A TA0E A SA2E A SA1E AArockE A 290 157415 456 504 kJ The energy drop of the rock UA2E A UA1E AArockE A equals QAHE A into heat engine UA2E AUA1E AArockE A mC TA2E ATA1E A 5500 089 290 400 538 450 kJ W UA2E A UA1E AArockE A QALE A 538450 456504 81 946 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 675 Consider problem 660 if the two blocks of steel exchange energy through a heat engine similar to the setup in Problem 674 Find the work output of the heat engine CV Both blocks no external heat transfer C from Table A3 Energy Eq UA2E A UA1E A mAAE AuA2E A uA1E AAAE A mABE AuA2E A uA1E AABE A 0 W mAAE ACTA2E A TAA1E A mABE ACTA2E A TAB1E A Entropy Eq637 SA2E A SA1E A mAAE AsA2E A sA1E AAAE A mABE AsA2E A sA1E AABE A A1E ASA2 genE Process Assume reversible then A1E ASA2 genE A 0 Entropy changes from Eq611 SA2E A SA1E A mAAE AC ln A T2 ETA1 E A mABE AC ln A T2 ETB1 E A 0 Now solve for TA2E A by combining the ln terms as 0 ln A T2 ETA1 E A ln A T2 ETB1 E A mBmA ln A T2 ETA1 E A A T2 ETB1 E A mBmA So the factors inside the ln function equal to one The rewrite as A T2 ETA1 E A A T2 ETB1 E A mBmA 1 TA2E A T xA A1 T xB B1 xAAE A mAAE A mAAE A mABE A xABE A mABE A mAAE A mABE A In the actual case the two masses are the same so xAAE A xABE A ½ and the result is TA2E A A TA1 TB1 EA A 52315 29815EA 39494 K 1218AoE AC The work output comes from the energy equation W mAAE ACTAA1E A TA2E A mABE ACTAB1E A TA2E A 5 kg 046 kJkgK 250 1218 25 1218 K 722 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 676 Two kg of liquid lead initially at 400C is poured into a form It then cools at constant pressure down to room temperature of 20C as heat is transferred to the room The melting point of lead is 327C and the enthalpy change between the phases hAifE A is 246 kJkg The specific heats are in Tables A3 and A4 Calculate the net entropy change for the mass Solution CV Lead constant pressure process mAPbE AuA2E A uA1E AAPbE A A1E AQA2E A PVA2E A VA1E A We need to find changes in enthalpy u Pv for each phase separately and then add the enthalpy change for the phase change Consider the process in several steps Cooling liquid to the melting temperature Solidification of the liquid to solid recall sAifE A hAifE AT see page 268 Cooling of the solid to the final temperature A1E AQA2E A mAPbE AhA2E A hA1E A mAPbE AhA2E A hA327solE A hAifE A hA327fE A hA400E A 2 kg 0138 20 327 246 0155 327 400 kJkg 84732 492 2263 15656 kJ SA2E A SA1E A mAPbE ACp sollnT2600 hif600 CP liqln600T1E 2 A0138 ln 29315 600 246 E600 0155 ln 600 67315E A 0315 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Entropy of ideal gases Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 677 Air inside a rigid tank is heated from 300 to 350 K Find the entropy increase sA2E A sA1E A What if it is from 1300 to 1350 K Process V C A1E AWA2E A Ø Entropy change from Eq617 a sA2E A sA1E A CAvoE A ln A T2 ET1 E A 0717 ln A 350 300 E A 01105 kJkgK b sA2E A sA1E A CAvoE A ln A T2 ET1 E A 0717 ln A 1350 1300 E A 002706 kJkgK From A7 case a CAvE A u T 3650 072 kJkg K see A5 case b CAvE A u T 45250 0904 kJkg K 25 higher so result should have been 00341 kJkgK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 678 A rigid tank contains 1 kg methane at 500 K 1500 kPa It is now cooled down to 300 K Find the heat transfer and the change in entropy using ideal gas Ideal gas constant volume so there is no work Energy Eq 35 muA2E A uA1E A A1E AQA2E A A1E AWA2E A A1E AQA2E A Use specific heat from Table A5 uA2E A uA1E A CAvE A TA2E A TA1E A 1736 300 500 3472 kJkg A1E AQA2E A muA2E A uA1E A 1 3472 3472 kJ The change in s for an ideal gas Eqs61617 and vA2E A vA1E A gives msA2E A sA1E A m CAvoE A ln A T2 ET1 E A R ln A v2 Ev1 E A m CAvoE A ln A T2 ET1 E 1 kg 1736 kJkgK ln A300 500E A 08868 kJK P v 1 1 T s 1 P P 2 2 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 679 Three kg of air is in a pistoncylinder keeping constant pressure at 27AoE AC 300 kPa It is now heated to 500 K Plot the process path in a Ts diagram and find the heat transfer in the process Solution CV Air so this is a control mass Energy Eq35 UA2E A UA1E A m uA2E A uA1E A A1E AQA2E A A1E AWA2E Process P C so A1E AWA2E A APdVEA PA1E AVA2E A VA1E A m PvA2E A PvA1E A State 1 TA1E A PA1E A ideal gas so PA1E AVA1E A mRTA1E State 2 TA2E A PA2E A PA1E A and ideal gas so PA2E AVA2E A mRTA2E From the energy equation A1E AQA2E A m uA2E A uA1E A m PvA2E A PvA1E A m hA2E A hA1E A mCAPoE A TA2E A TA1E A 3 kg 1004 kJkgK 500 300 K 6024 kJ T s 2 1 300 kPa P v T 300 300 2 1 T 1 2 500 If we were to find the change in entropy we get sA2E A sA1E A CAPoE A lnTA2E A TA1E A 1004 ln 500300 05129 kJkgK The process curve in the Ts diagram is T TA1E A exps sA1E ACAPoE A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 680 A pistoncylinder setup contains air at 100 kPa 400 K which is compressed to a final pressure of 1000 kPa Consider two different processes i a reversible adiabatic process and ii a reversible isothermal process Show both processes in Pv and a Ts diagram Find the final temperature and the specific work for both processes Solution CV Air control mass of unknown size and mass Energy Eq35 uA2E A uA1E A A1E AqA2E A A1E AwA2E Entropy Eq63 sA2E A sA1E A dqT i dq 0 so A1E AqA2E A 0 ii T C so dqT A1E AqA2E AT i For this process the entropy equation reduces to sA2E A sA1E A 0 so we have constant s an isentropic process The relation for an ideal gas constant s and k becomes Eq623 TA2E A TA1E A PA2E A PA1E AA k1 k E A 400 A 1000 100 04 E14 E A 400 10A 028575E A 772 K From the energy equation we get the work term A1E AwA2E A uA1E A uA2E A CAvE ATA1E A TA2E A 0717400 772 2667 kJkg ii For this process TA2E A TA1E A so since ideal gas we get uA2E A uA1E A also sAο T2E A sAο T1E A Energy Eq A1E AwA2E A A1E AqA2E Now from the entropy equation we solve for A1E AqA2E A A1E AwA2E A A1E AqA2E A TsA2E A sA1E A TsA T2E A sA T1E A R ln P2 P1 RT ln P2 P1 0287 400 ln 10 264 kJkg P v 1 2ii 2ii 1 T s 1 P P 2 2i 2i Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 681 Prove that the two relations for changes in s Eqs616 and 617 are equivalent once we assume constant specific heat Hint recall the relation for specific heat in Eq 342 From Eq 342 CApoE A CAvoE A R Start with Eq616 sA2E A sA1E A CApoE A ln A T2 ET1 E A R ln P2 P1 Now substitute Eq342 to get sA2E A sA1E A CAvoE A R ln A T2 ET1 E A R ln P2 P1 CAvoE A ln A T2 ET1 E A R ln P2 P1 ln A T2 ET1 E A CAvoE A ln A T2 ET1 E A R ln P2 P1 A T1 ET2 E A Use the ideal gas law Pv RT for both states to get the ratio A P2v2 ERT2 P1v1 RT1 E A P2 P1 A T1 ET2 E A A v1 Ev2 E so then we get to Eq 617 as sA2E A sA1E A CAvoE A ln A T2 ET1 E A R ln A v1 Ev2 E A CAvoE A ln A T2 ET1 E A R ln A v2 Ev1 E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 682 A closed rigid container is filled with 15 kg water at 100 kPa 55AoE AC 1 kg of stainless steel and 05 kg of PVC polyvinyl chloride both at 20oC and 01 kg of hot air at 400 K 100 kPa It is now left alone with no external heat transfer and no water vaporizes Find the final temperature and the change in entropy of the masses Energy Eq UA2E A UA1E A Amiu2 u1i EA A1E AQA2E A A1E AWA2E A 0 Process A1E AQA2E A 0 A1E AWA2E A 0 For the liquid and the metal masses we will use the specific heats Tbl A3 A4 so Amiu2 u1i EA AmiCv i T2 T1i EA TA2E AmiCv i EA AmiCv iT1 i EA noticing that all masses have the same TA2E A but not same initial T AmiCv i EA 15 418 1 046 05 096 01 0717 7282 kJK Energy Eq 7282 TA2E A 15 418 55 1 046 05 096 20 01 0717 400 27315 372745 kJ TA2E A 512AoE AC 3243 K The change in entropy for the solids and liquid follow Eq611 and that for the ideal gas is from Eq616 or 617 or 619 we use here 617 since vA2E A vA1E SA2E A SA1E A Ami s2 s1i EA ACv i mi lnT2 T1i E 15 418 ln A 3243 32815E A 1 046 ln A 3243 29315E A 05 096 ln A 3243 29315E A 01 0717 ln A3243 400E 000588 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 683 Water at 150C 400 kPa is brought to 1200C in a constant pressure process Find the change in the specific entropy using a the steam tables b the ideal gas water Table A8 and c the specific heat from A5 Solution a State 1 Table B13 Superheated vapor sA1E A 69299 kJkgK State 2 Table B13 sA2E A 97059 kJkgK sA2E A sA1E A 97059 69299 2776 kJkgK b Table A8 at 42315 K sAo T1E A 1113891 kJkgK Table A8 at 147315 K sAo T2E A 1386383 kJkgK sA2E A sA1E A sAo T2E A sAo T1E A R ln P2 P1 sAo T2E A sAo T1E sAo T2E A sAo T1E A 1386383 1113891 272492 kJkgK c Table A5 CApoE A 1872 kJkgK sA2E A sA1E A CApoE A ln A T2 ET1 E A 1872 ln A147315 42315E A 23352 kJkgK Notice how the average slope from 150C to 1200C is higher than the one at 25C CApoE A h T 25 150 1200 h 150 h 1200 Slope at 25C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 684 R410A at 400 kPa is brought from 20oC to 120oC in a constant pressure process Evaluate the change in specific entropy using Table B4 and using ideal gas with Cp 081 kJkgK Table B42 s1 12108 kJkgK s2 14788 kJkgK s2 s1 14788 12108 0268 kJkgK Eq 616 s2 s1 Cpo ln T1 T2 081 ln 39315 29315 0238 kJkgK Two explanations for the difference are as the average temperature is higher than 25oC we could expect a higher value of the specific heat and secondly it is not an ideal gas if you calculate Z PvRT 094 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 685 R410A at 300 kPa 20oC is brought to 200oC in a constant volume process Evaluate the change in specific entropy using Table B4 and using ideal gas with Cv 0695 kJkgK Table B42 s1 12485 kJkgK v1 010720 m3kg s2 16413 kJkgK v2 010714 m3kg at 500 kPa very close we could have interpolated between 400 and 500 kPa s2 s1 16413 12485 03928 kJkgK Eq 617 s2 s1 Cvo ln T1 T2 0695 ln 47315 29315 0333 kJkgK Two explanations for the difference are as the average temperature is higher than 25oC we could expect a higher value of the specific heat and secondly it is not an ideal gas if you calculate Z PvRT 096 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 686 Consider a small air pistol with a cylinder volume of 1 cm3 at 250 kPa 27C The bullet acts as a piston initially held by a trigger The bullet is released so the air expands in an adiabatic process If the pressure should be 120 kPa as the bullet leaves the cylinder find the final volume and the work done by the air Solution CV Air Assume a reversible adiabatic process Energy Eq35 u2 u1 0 1w2 Entropy Eq637 s2 s1 dqT 1s2 gen 0 State 1 T1P1 State 2 P2 So we realize that one piece of information is needed to get state 2 Process Adiabatic 1q2 0 Reversible 1s2 gen 0 With these two terms zero we have a zero for the entropy change So this is a constant s isentropic expansion process giving s2 s1 From Eq623 T2 T1 P2 P1 k1 k 300 14 120 250 04 300 048 028575 24324 K The ideal gas law PV mRT at both states leads to V2 V1 P1 T2P2 T1 1 250 24324120 300 1689 cm3 The work term is from Eq629 or Eq44 with polytropic exponent n k 1W2 1 1 k P2V2 P1V1 1 1 14 120 1689 250 1 106 kPa m 3 0118 J Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 687 Oxygen gas in a piston cylinder at 300 K 100 kPa with volume 01 m3 is compressed in a reversible adiabatic process to a final temperature of 700 K Find the final pressure and volume using Table A5 and repeat the process with Table A8 Solution CV Air Assume a reversible adiabatic process Energy Eq35 u2 u1 0 1w2 Entropy Eq637 s2 s1 dqT 1s2 gen 0 Process Adiabatic 1q2 0 Reversible 1s2 gen 0 Properties Table A5 k 1393 With these two terms zero we have a zero for the entropy change So this is a constant s isentropic expansion process From Eq623 P2 P1 T2 T1 k k1 100 11393 700 300 1393 2015 kPa Using the ideal gas law to eliminate P from this equation leads to Eq624 V2 V1 T2 T1 1 1k 01 11393 700 300 1 00116 m 3 Using the ideal gas tables A8 we get s2 s1 so T2 so T1 R lnP2P1 0 or P2 P1 expso T2 so T1R P2 100 exp72336 6416802598 23195 kPa V2 V1 T2 T1P1P2 01 m3 700 300 100 23195 0010 m 3 P v 1 1 T s 1 P P 2 2 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 688 Air in a rigid tank is at 100 kPa 300 K with a volume of 075 m3 The tank is heated to 400 K state 2 Now one side of the tank acts as a piston letting the air expand slowly at constant temperature to state 3 with a volume of 15 m3 Find the entropy at states 1 2 and 3 Solution State 1 P1 T1 m P1V1 RT1 100 075 0287 300 kPa m3 kJkg 0871 kg Process 1 to 2 Constant volume heating dV 0 1W2 0 State 2 T2 P2 P1 T2 T1 100 400 300 1333 kPa Process 2 to 3 Isothermal expansion dT 0 u3 u2 and State 3 T3 P3 P2 V2 V3 1333 075 15 6667 kPa The entropy is generally used as a relative value with different reference states in different tables We can use here the reference as in Table A7 so s so T R lnPP0 where so T is absolute entropy from A7 at T and P0 100 kPa s1 686926 kJkgK which is at 300 K 100 kPa s2 715926 0287 ln1333100 707677 kJkgK s3 715926 0287 ln6667100 727561 kJkgK To get total entropy for the given mass multiply with m 0871 kg which will give 59831 61639 63371 all in kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 689 An insulated pistoncylinder setup contains carbon dioxide gas at 800 kPa 300 K which is then compressed to 6 MPa in a reversible adiabatic process Calculate the final temperature and the specific work using a ideal gas tables A8 and b using constant specific heats A5 Solution CV CO2 a control mass undergoing a reversible adiabatic process Energy Eq35 u2 u1 0 1w2 Entropy Eq637 s2 s1 dqT 1s2 gen 0 Process Adiabatic 1q2 0 Reversible 1s2 gen 0 State 1 300 K 800 kPa State 2 6000 kPa With two terms zero in the entropy equation we have a zero for the entropy change So this is a constant s isentropic expansion process s2 s1 a Table A8 for CO2 and Eq619 s2 s1 0 so T2 so T1 R lnP2P1 so T2 so T1 R ln P2 P1 48631 01889 ln 6000 800 52437 kJkgK Now interpolate in A8 to find T2 T2 450 50 52437 5232553375 52325 4553 K 1w2 u2 u1 271 1577 1133 kJkg b Table A5 k 1289 CVo 0653 kJkg K and now Eq623 T2 T1 P2 P1 k1 k 300 6000 800 0224 4711 K 1w2 CVoT2T1 0653 kJkgK 4711 300K 1117 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 690 Extend the previous problem to solve using Table B3 CV CO2 a control mass undergoing a reversible adiabatic process Energy Eq35 u2 u1 0 1w2 Entropy Eq637 s2 s1 dqT 1s2 gen 0 Process Adiabatic 1q2 0 Reversible 1s2 gen 0 State 1 300 K 800 kPa State 2 6000 kPa With two terms zero in the entropy equation we have a zero for the entropy change So this is a constant s isentropic expansion process s2 s1 From Table B32 s1 18240 kJkgK u1 3335 kJkg T2 2607oC 5339 K u2 5029 kJkg 1w2 u2 u1 5029 3335 1694 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 691 A handheld pump for a bicycle has a volume of 25 cm3 when fully extended You now press the plunger piston in while holding your thumb over the exit hole so that an air pressure of 300 kPa is obtained The outside atmosphere is at P0 T0 Consider two cases 1 it is done quickly 1 s and 2 it is done very slowly 1 h a State assumptions about the process for each case b Find the final volume and temperature for both cases Solution CV Air in pump Assume that both cases result in a reversible process State 1 P0 T0 State 2 300 kPa One piece of information must resolve the for a state 2 property Case I Quickly means no time for heat transfer Q 0 so a reversible adiabatic compression u2 u1 1w2 s2 s1 dqT 1s2 gen 0 With constant s and constant heat capacity we use Eq623 T2 T1 P2 P1 k1 k 298 14 300 101325 04 4053 K Use ideal gas law PV mRT at both states so ratio gives V2 P1V1T2T1P2 1148 cm3 Case II Slowly time for heat transfer so T constant T0 The process is then a reversible isothermal compression T2 T0 298 K V2 V1P1P2 844 cm3 P v 1 2ii 2ii 1 T s 1 P P 2 2i 2i Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 692 A pistoncylinder shown in Fig P692 contains air at 1380 K 15 MPa with V1 10 cm3 Acyl 5 cm2 The piston is released and just before the piston exits the end of the cylinder the pressure inside is 200 kPa If the cylinder is insulated what is its length How much work is done by the air inside Solution CV Air Cylinder is insulated so adiabatic Q 0 Continuity Eq m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 1W 2 Entropy Eq637 ms2 s1 dQT 1S2 gen 0 1S 2 gen State 1 T1 P1 State 2 P2 So one piece of information is needed for the assume reversible process 1S2 gen 0 s2 s1 0 which is also Eq63 State 1 Table A7 u1 10952 kJkg so T1 85115 kJkg K m P1V1RT1 0287 1380 15000 10106 0000379 kg State 2 P2 and from Entropy eq s2 s1 so from Eq619 s T2 s T1 R ln P2 P1 85115 0287 ln 200 15000 72724 kJkg K Now interpolate in Table A7 to get T2 T2 440 20 72724 725607730142 725607 4472 K u2 31564 33031 31564 036 32092 kJkg V2 V1 T2 P1 T1P2 10 4472 15000 1380 200 243 cm3 L2 V2 Acyl 2435 486 cm 1w2 u1 u2 7743 kJkg 1W2 m1w2 02935 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 693 Argon in a light bulb is at 90 kPa and 20oC when it is turned on and electric input now heats it to 60oC Find the entropy increase of the argon gas Solution CV Argon gas Neglect any heat transfer Energy Eq35 mu2 u1 1W2 electrical in Entropy Eq637 s2 s1 dqT 1s2 gen 1s 2 gen Process v constant and ideal gas P2 P1 T2T 1 Evaluate changes in s from Eq616 or 817 s2 s1 Cp ln T2T1 R ln P2 P1 Eq616 Cp ln T2T1 R ln T2 T1 Cv lnT2T1 Eq617 0312 kJkgK ln 60 273 20 273 004 kJkg K cb Since there was no heat transfer but work input all the change in s is generated by the process irreversible conversion of W to internal energy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 694 We wish to obtain a supply of cold helium gas by applying the following technique Helium contained in a cylinder at ambient conditions 100 kPa 20C is compressed in a reversible isothermal process to 600 kPa after which the gas is expanded back to 100 kPa in a reversible adiabatic process a Show the process on a Ts diagram b Calculate the final temperature and the net work per kilogram of helium Solution a 1 T 2 P 2 1 3 1 2 3 3 2 T s s s T P P 2 P v 1 600 100 3 b The adiabatic reversible expansion gives constant s from the entropy equation Eq637 With ideal gas and constant specific heat this gives relation in Eq623 T3 T2P3P2 k1 k 29315 10060004 14315 K The net work is summed up over the two processes The isothermal process has work as Eq631 1w2 RT1 lnP2P1 20771 kJkgK 29315 K ln600100 10910 kJkg The adiabatic process has a work term from energy equation with no q 2w3 CVoT2T3 3116 kJkgK 29315 14315 K 4674 kJkg The net work is the sum wNET 10910 4674 6236 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 695 A 1m3 insulated rigid tank contains air at 800 kPa 25C A valve on the tank is opened and the pressure inside quickly drops to 150 kPa at which point the valve is closed Assuming that the air remaining inside has undergone a reversible adiabatic expansion calculate the mass withdrawn during the process Solution CV Air remaining inside tank m2 ContEq m2 m Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 dQT 1S2 gen Process adiabatic 1Q2 0 and reversible 1S2 gen 0 P v 1 2 2 1 T s CV m 2 Entropy eq then gives s2 s1 and ideal gas gives the relation in Eq623 T2 T1P2P1 k1 k 2982 K 1508000286 1848 K m1 P1VRT1 800 kPa 1 m30287 kJkgK 2982 K 935 kg m2 P2VRT2 150 kPa 1 m30287 kJkgK 1848 K 283 kg me m1 m2 652 kg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 696 Two rigid insulated tanks are connected with a pipe and valve One tank has 05 kg air at 200 kPa 300 K and the other has 075 kg air at 100 kPa 400 K The valve is opened and the air comes to a single uniform state without any heat transfer Find the final temperature and the change in entropy of the air Solution CV Total tank Control mass of constant volume Mass and volume m2 mA mB V VA VB Energy Eq U2 U1 m2 u2 mAuA1 mBuB1 1Q2 1W2 0 Process Eq V constant 1W2 0 Insulated 1Q2 0 Ideal gas at A1 VA mARTA1PA1 05 0287 300 200 02153 m 3 Ideal gas at B1 VB mBRTB1 PB1 075 0287 400 100 0861 m3 State 2 m2 mA mB 125 kg V2 VA VB 10763 m3 Energy Eq u2 m2 mAuA1 mBuB1 and use constant specific heat T2 m2 mA TA1 mB m2 TB1 05 125 300 075 125 400 360 K P2 m2 RT2V 125 kg 0287 kJkgK 360 K 10763 m3 120 kPa S2 S1 mACP lnT2TA1 RlnP2PA1 mBCP lnT2TB1 RlnP2PB1 05 1004 ln360 300 0287 ln120 200 0751004 ln360 400 0287 ln120 100 05 13514 075 01581 05571 kJK B A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 697 Two rigid tanks shown in Fig P697 each contain 10 kg N2 gas at 1000 K 500 kPa They are now thermally connected to a reversible heat pump which heats one and cools the other with no heat transfer to the surroundings When one tank is heated to 1500 K the process stops Find the final P T in both tanks and the work input to the heat pump assuming constant heat capacities Solution Control volume of hot tank B Process constant volume mass so no work Energy equation Eq35 and specific heat in Eq520 gives U2 U1 mCvT2 T1 1Q2 10 kg 07448 kJkgK 500 K 3724 kJ P2 P1T2T1 15P1 750 kPa HP W HE 1 3 Q 1 2 Q A 1 3 B 1 2 State 1 initial 2 final hot 3 final cold To fix temperature in cold tank CV total For this CV only WHP cross the control surface no heat transfer The entropy equation Eq637 for a reversible process becomes S2 S1tot 0 mhot s2 s1 mcold s3 s1 Use specific heats to evaluate the changes in s from Eq616 and division by m Cphot lnT2 T1 R lnP2 P1 Cpcold lnT3 T1 R lnP3 P1 0 P3 P1T3T1 and P2 P1T2T 1 Now everything is in terms of T and Cp Cv R so Cvhot lnT2T1 Cvcold lnT3T1 0 same Cv T3 T1T1T2 667 K P3 333 kPa Qcold 1Q3 m Cv T3 T1 2480 kJ WHP 1Q2 Qcold 1Q2 1Q3 1244 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 698 A hydrogen gas in a pistoncylinder assembly at 300 K 100 kPa with a volume of 01 m3 is now slowly compressed to a volume of 001 m3 while cooling it in a reversible isothermal process What is the final pressure the heat transfer and the change in entropy Solution CV Hydrogen control mass Energy Eq 35 mu2 u1 1Q2 1W2 Process T constant so with ideal gas u2 u1 P v 1 2 2 1 T s 1 P P 2 From the process equation T2 T1 and ideal gas law we get P2 P1 V1V2 10 P1 1000 kPa we can calculate the work term as in Eq322 1Q2 1W2 PdV P1V1 ln V2V1 100 kPa 01 m3 ln 110 230 kJ The change of entropy from the entropy equation Eq63 is ms2 s1 1Q2T1 23 300 kJ K 007667 kJK If instead we use Eq617 we would get S ms2 s1 mCvo ln T1 T2 R ln v1 v2 m R ln v1 v2 P1V1T1 ln v1 v2 1Q2T 1 consistent with the above result Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 699 A rigid tank contains 4 kg air at 300oC 4 MPa which acts as the hot energy reservoir for a heat engine with its cold side at 20oC shown in Fig P699 Heat transfer to the heat engine cools the air down in a reversible process to a final 20oC and then stops Find the final air pressure and the work output of the heat engine W Q H Q L HE CV total Air Ambient Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq63 ms2 s1 dQT T not constant Process v constant 1W2 0 State 2 T2 and v2 v1 P2 P1T2T1 4000 2931557315 20459 kPa From the energy equation QH 1Q2 mu2 u1 m Cvo T2 T1 4 kg 0717 kJkgK 29315 57315 K 8030 kJ Take now CV total as the air plus heat engine out to ambient Entropy Eq63 ms2 s1 QL Tamb QL mTamb s2 s1 mTamb Cvo ln T2 T1 4 kg 29315 K 0717 kJkgK ln2931557315 5637 kJ Now the CV heat engine can give the engine work from the energy equation Energy HE WHE QH QL 8030 5637 2393 kJ Notice to get 1q2 T ds we must know the function Ts which we do not readily have for this process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Polytropic processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6100 An ideal gas having a constant specific heat undergoes a reversible polytropic expansion with exponent n 14 If the gas is carbon dioxide will the heat transfer for this process be positive negative or zero Solution T s n k P const n k 1 2 n k CO2 Table A5 k 1289 n Since n k and P2 P1 it follows that s2 s1 and thus Q flows out 1Q2 0 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6101 Repeat the previous problem for the gas carbon monoxide CO Solution T s n k P const n k 1 2 n k CO Table A5 k 1399 n Since n k and P2 P1 it follows that s2 s1 and thus adiabatic 1Q2 0 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6102 A nitrogen gas goes through a polytropic process with n 13 in a pistoncylinder arrangement It starts out at 600 K 600 kPa and ends at 800 K Is the heat transfer positive negative or zero T s n k P const n k 1 2 n k N2 Table A5 k 140 and n 13 k Since n k and T2 T1 process goes up on the n k curve s2 s1 and thus q must go out q is negative Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6103 A cylinderpiston contains 1 kg methane gas at 100 kPa 300 K The gas is compressed reversibly to a pressure of 800 kPa Calculate the work required if the process is adiabatic Solution CV Methane gas of constant mass m2 m1 m and reversible process Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq63 37 ms2 s1 dQT 1S2 gen dQT Process 1Q2 0 s2 s 1 thus isentropic process s constant and ideal gas gives the relation in Eq623 with k 1299 from Table A5 T2 T1 P2P1 k1 k 300 K 800 100 0230 48399 K 1W2 mCV0T2 T1 1 kg 1736 kJkgK 48399 29315 K 3313 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6104 Do the previous problem but assume the process is isothermal Solution CV Methane gas of constant mass m2 m1 m and reversible process Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 dQT 1S2 gen dQT Process T constant For ideal gas then u2 u1 and s T2 s T1 Energy eq gives 1W2 1Q2 and dQT 1Q2T with the entropy change found from Eq616 1W2 1Q2 mTs2 s1 mRT lnP2P1 05183 kJK 300 K ln 800 100 3233 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6105 A pistoncylinder contains air at 300 K 100 kPa It is now compressed in a reversible adiabatic process to a volume 7 times as small Use constant heat capacity and find the final pressure and temperature the specific work and specific heat transfer for the process Solution Expansion ratio v2 v1 17 Process eq Rev adiabatic and ideal gas gives Pvn C with n k P2 P1 v2v1k 714 15245 P2 P1 714 100 15245 15245 kPa T2 T1 v1v2k1 300 704 6534 K 1q2 0 kJkg Polytropic process work term from Eq629 1w2 R 1 k T2 T1 0287 04 kJkgK 6534 300 K 2536 kJkg Notice Cv Rk1 so the work term is also the change in u consistent with the energy equation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6106 A pistoncylinder contains pure oxygen at 500 K 600 kPa The piston is moved to a volume such that the final temperature is 700 K in a polytropic process with exponent n 125 Use ideal gas approximation and constant heat capacity to find the final pressure the specific work and the heat transfer Take CV as the oxygen m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Process Eq Pvn Constant polytropic From the ideal gas law and the process equation we can get State 2 P2 P1 v2 v1 n and Pv RT T2T1 v2 v1 1 n P2 P1 T2 T1 nn1 600 700 500 125025 32269 kPa From process eq 1w2 P dv area 1 1n P2v2 P1v1 R 1n T2 T1 02598 1125 700 500 2078 kJkg From the energy equation and Table A8 1q2 u2 u1 1w2 48018 33172 2078 kJkg 5934 kJkg From the energy equation and constant specific heat from Table A5 1q2 u2 u1 1w2 Cv T2 T1 1w2 0662 kJkgK 700 500 K 2078 kJkg 754 kJkg The function Ts can be done with constant CP a CP nRn1 0377 P 1 2 T P C T 5 P v 1 2 T s 1 2 T T 2 1 T C expsa P C v 125 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6107 Do Problem 6103 and assume the process is polytropic with n 115 Process Pvn constant with n 115 The TP relation is given in Eq628 T2 T1 P2P1 n1 n 300 800 100 0130 3931 K and the work term is given by Eq629 1W2 mP dv mP2v2 P1v11 n mR T2 T11 n 1 kg 05183 kJ kgK 3931 29315 1 115 K 3454 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6108 Hot combustion air at 2000 K expands in a polytropic process to a volume 6 times as large with n 13 Find the specific boundary work and the specific heat transfer Energy Eq u2 u1 1q2 1w 2 Reversible work Eq 629 1w2 1 1n P2v2 P1v1 R 1n T2 T1 Process Eq Pvn C T2 T1 v1v2 n1 2000 1 6 03 11684 K Properties from Table A71 u1 167752 kJkg u2 90546 kJkg 1w2 0287 1 13 kJkgK 11684 2000 K 7956 kJkg 1q2 u2 u1 1w2 90546 167752 7956 235 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6109 Air in a pistoncylinder is at 1800 K 7 MPa and expands in a polytropic process with n 15 to a volume eight times larger Find the specific work and specific heat transfer in the process using Table A7 and draw the Ts diagram CV Air of constant mass m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Process Pv150 constant v2v1 8 State 1 P1 7 MPa T1 1800 K State 2 v2 8v1 Must be on process curve so P2 P1 v1v2n 7000 18150 30936 kPa T2 T1 P2v2 P1v1 T1v1v2n 1 1800 1805 6364 K Table A7 u1 148633 kJkg and interpolate u2 46306 kJkg Work from the process expressed in Eq321 629 1w2 Pdv 1 n P2v2 P1v1 R 1 n T2 T1 1 15 0287 6364 1800 6679 kJkg Heat transfer from the energy equation 1q2 u2 u1 1w2 46306 148633 6679 3554 kJkg Notice n 15 k 14 n k P v 1 2 T s 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6110 Helium in a pistoncylinder at 20C 100 kPa is brought to 400 K in a reversible polytropic process with exponent n 125 You may assume helium is an ideal gas with constant specific heat Find the final pressure and both the specific heat transfer and specific work Solution CV Helium Continuity Eq m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Process Pvn C Pv RT Tvn1 C Table A5 Cv 3116 kJkg K R 20771 kJkg K From the process equation and T1 29315 T2 400 K T1 vn1 T2 vn1 v2 v1 T1 T2 1n1 02885 P2 P1 v1 v2n 473 P2 473 kPa The work is from Eq629 per unit mass 1w2 P dv C vn dv C 1n v2 1n v1 1n 1 1n P2 v2 P1 v1 R 1n T2 T1 20771 1 125 kJkgK 400 29315 K 8877 kJkg The heat transfer follows from the energy equation 1q2 u2 u1 1w2 Cv T2 T1 8877 5548 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6111 The power stroke in an internal combustion engine can be approximated with a polytropic expansion Consider air in a cylinder volume of 02 L at 7 MPa 1800 K shown in Fig P6111 It now expands in a reversible polytropic process with exponent n 15 through a volume ratio of 101 Show this process on Pv and Ts diagrams and calculate the work and heat transfer for the process Solution CV Air of constant mass m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq63 37 ms2 s1 dQT 1S2 gen dQT Process PV150 constant V2V1 10 State 1 P1 7 MPa T1 1800 K V1 02 L m1 RT1 P1V1 0287 1800 7000 02 103 271103 kg State 2 v V2m Must be on process curve so Eq624 gives T2 T1 V1V2n1 1800 11005 5692 K Table A7 u1 148633 kJkg and interpolate u2 411707 kJkg Notice n 15 k 14 n k Work from the process expressed in Eq629 1W2 PdV mRT2 T11 n 271103 02875692 1800 1 15 191 kJ Heat transfer from the energy equation 1Q2 mu2 u1 1W2 271103 kg 411707 148633 kJkg 191 kJ 100 kJ P v 1 2 T s 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6112 A cylinderpiston contains saturated vapor R410A at 10C the volume is 10 L The R410A is compressed to 2 MPa 60C in a reversible internally polytropic process Find the polytropic exponent n and calculate the work and heat transfer Solution CV R410A of constant mass m2 m1 m out to ambient Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 dQT 1S2 gen 1Q2Tamb 1S2 gen Process P1v1n P2v2n Eq636 State 1 T1 x1 Table B41 P1 10857 kPa v1 002383 m3kg m V1v1 001002383 04196 kg State 2 T2 P2 Table B42 v2 001536 m3kg From process eq P2P1 2000 10857 002383 001536 n n 139106 The work is from Eq629 1W2 PdV m P2v2 P1v1 1n 04196 2000 001536 10857 002383 1 139106 520 kJ Heat transfer from energy equation 1Q2 mu2 u1 1W2 04196 2899 2559 520 907 kJ P v 1 2 T s 1 2 10 60 1086 2000 LV Notice n 139 k 117 n k The two curves are shown below from the CATT3 process plot function Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6113 Air goes through a polytropic process with n 13 in a pistoncylinder setup It starts at 200 kPa 300 K and ends with a pressure of 2200 kPa Find the expansion ratio v2v1 the specific work and the specific heat transfer Take CV as the air m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Process Eq Pvn Constant polytropic From the ideal gas law and the process equation we can get State 2 v2v1 P2P1 1n and Pv RT T2T1 v2 v1 1 n v2v1 P2P1 1n 2200 200 113 01581 T2 T1 P2P1 n1n 300 2200 200 0313 5217 K From process eq 1w2 P dv area 1 1n P2v2 P1v1 R 1n T2 T1 0287 113 5217 300 21209 kJkg From the energy equation and constant specific heat from Table A5 1q2 u2 u1 1w2 Cv T2 T1 1w2 0717 kJkgK 5217 300 K 21209 kJkg 5313 kJkg From the energy equation and Table A7 1q2 u2 u1 1w2 3760 21436 21209 5045 kJkg process plotted from CATT3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6114 A cylinderpiston contains air at ambient conditions 100 kPa and 20C with a volume of 03 m3 The air is compressed to 800 kPa in a reversible polytropic process with exponent n 12 after which it is expanded back to 100 kPa in a reversible adiabatic process a Show the two processes in Pv and Ts diagrams b Determine the final temperature and the net work Solution a P T v s 1 2 3 1 2 3 P P 2 1 m P1V1RT1 100 03 0287 2932 03565 kg b The process equation is expressed in Eq628 T2 T1P2P1 n1 n 2932 800 100 0167 4149 K The work is from Eq629 1w2 1 2 Pdv 1n P2v2P1v1 1n RT2T1 1120 028741492932 1746 kJkg Isentropic relation is from Eq623 T3 T2 P3P2 k1 k 4149 100 800 0286 2289 K With zero heat transfer the energy equation gives the work 2w3 CV0T2 T3 07174149 2289 1333 kJkg wNET 035651746 1333 147 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Entropy generation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6115 Consider a heat transfer of 100 kJ from 1500 K hot gases to a steel container at 750 K that has a heat transfer of the 100 kJ out to some air at 375 K Determine the entropy generation in each of the control volumes indicated in Fig P6115 There is no change in energy or entropy in the indicated control volumes so no storage effect There is a transfer of energy in and out of each CV and an associated transfer of entropy Take CV1 Take CV2 Energy Eq Energy Eq 0 Q Q 0 Q Q Entropy Eq Entropy Eq 0 Q TH Q TM Sgen CV1 0 Q TM Q TL S gen CV2 Sgen CV1 Q TM Q TH Sgen CV2 Q TL Q TM 100 750 100 1500 00667 kJK 100 375 100 750 0133 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6116 A rigid tank has 01 kg saturated vapor R410A at 0oC that is cooled to 20oC by a 20oC heat sink Show the process in a Ts diagram find the change in entropy of the R410A the heat sink and the total entropy generation Solution CV R410A in tank out to the sink m2 m1 Energy Eq35 mu2 u1 1Q2 1W 2 Entropy Eq 637 ms2 s1 1Q2 Tsink 1S2 gen Process V constant v2 v1 Vm 1W2 0 Table B41 State 1 u1 25302 kJkg s1 10368 kJkgK State 2 20oC v2 v1 Vm look in Table B41 at 20oC x2 vfg2 v2 vf2 003267 0000803 006400 04979 u2 uf2 x2 ufg2 2792 x2 21807 1365 kJkg s2 sf2 x2 sfg2 01154 x2 09625 05946 kJkgK From the energy equation 1Q2 mu2 u1 01 kg 1365 2530 kJkg 1165 kJ S2 S1R410a ms2 s1 01 05946 10368 00442 kJK S2 S1sink 1Q2 Tsink 1165 kJ25315 K 0046 kJK 1S2 gen ms2 s1 1Q2 Tsink 00442 0046 00018 kJK 2 1 P v T s 1 2 v C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6117 One kg water at 500oC and 1 kg saturated water vapor both at 200 kPa are mixed in a constant pressure and adiabatic process Find the final temperature and the entropy generation for the process Solution Continuity Eq m2 mA mB 0 Energy Eq35 m2u2 mAuA mBuB 1W 2 Entropy Eq637 m2s2 mAsA mBsB dQT 1S 2 gen Process P Constant 1W2 PdV PV2 V1 Q 0 Substitute the work term into the energy equation and rearrange to get m2u2 P2V2 m2h2 mAuA mBuB PV1 mAhA mBhB where the last rewrite used PV1 PVA PVB State A1 Table B13 hA 348703 kJkg sA 85132 kJkg K State B1 Table B12 hB 270663 kJkg sB 71271 kJkg K Energy equation gives h EA mAB A E mA2 AE A hABE A A1 2E A 348703 A1 2E A 270663 309683 kJkg 2 mA m2 hA State 2 PA2E A hA2E A 309683 kJkg sA2E A 79328 kJkg K TA2E A 3122C With the zero heat transfer we have A1E ASA2 genE A mA2E AsA2E A mAAE AsAAE A mABE AsABE 2 79328 1 85132 1 71271 0225 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6118 A car uses an average power of 25 hp for a one hour round trip With a thermal efficiency of 35 how much fuel energy was used What happened to all the energy What change in entropy took place if we assume ambient at 20AoE AC Since it is a round trip there are no changes in storage of energy for the car after it has cooled down again All the energy is given out to the ambient in the form of exhaust flow hot air and heat transfer from the radiator and underhood air flow W A W dtEA 25 hp 07457 kWhp 3600 s 67 113 kJ η Q Fuel energy used to deliver the W Q E η 67 113 kJ 035 191 751 kJ S Q T 191 751 kJ 29315 K 6541 kJK All the energy Q ends up in the ambient at the ambient temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6119 A computer chip dissipates 2 kJ of electric work over time and rejects that as heat transfer from its 50AoE AC surface to 25AoE AC air How much entropy is generated in the chip How much if any is generated outside the chip CV1 Chip with surface at 50AoE AC we assume chip state is constant Energy UA2E A UA1E A 0 A1E AQA2E A A1E AWA2E A WAelectrical inE A QAout 1E Entropy SA2E A SA1E A 0 Qout 1 Tsurf A1E ASA2 gen1E A1E ASA2 gen1E A Qout 1 Tsurf Welectrical in Tsurf A 2 kJ 32315 KE A 619 JK CV2 From chip surface at 50AoE AC to air at 25AoE AC assume constant state Energy UA2E A UA1E A 0 A1E AQA2E A A1E AWA2E A QAout 1E A QAout 2E Entropy SA2E A SA1E A 0 Qout1 Tsurf Qout2 Tair A1E ASA2 gen2E A1E ASA2 gen2E A Qout2 Tair Qout1 Tsurf A 2 kJ 29815 KE A A 2 kJ 32315 KE A 0519 JK 25 C air o 50 C o Q air flow cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6120 An insulated cylinderpiston contains R134a at 1 MPa 50C with a volume of 100 L The R134a expands moving the piston until the pressure in the cylinder has dropped to 100 kPa It is claimed that the R134a does 190 kJ of work against the piston during the process Is that possible Solution CV R134a in cylinder Insulated so assume Q 0 State 1 Table B52 vA1E A 002185 mA3E Akg uA1E A 40939 kJkg sA1E A 17494 kJkg K m VA1E AvA1E A 01002185 4577 kg Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E A A0E A 190 uA2E A uA1E A A1E AWA2E Am 36789 kJkg State 2 PA2E A uA2E A Table B52 TA2E A 1925C sA2E A 17689 kJkg K Entropy Eq637 msA2E A sA1E A AdQTEA A1E ASA2genE A A1E ASA2genE A A1E ASA2genE A msA2E A sA1E A 00893 kJK This is possible since A1E ASA2genE A A0E 2 1 P v T s 1 2 s C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6121 A rigid tank holds 075 kg ammonia at 70C as saturated vapor The tank is now cooled to 20C by heat transfer to the ambient at 20C Determine the amount of entropy generation during the process CV The ammonia out to 20C this is a control mass Process Rigid tank V C v constant A1E AWA2E A A 1 2 PdVE A 0 Energy Eq muA2E A uA1E A A1E AQA2E A A1E AWA2E A A1E AQA2E A Entropy Eq637 msA2E A sA1E A AdQTEA A1E ASA2genE A A1E AQA2E ATA0E A A1E ASA2genE A State 1 vA1E A 003787 mA3E Akg uA1E A 13389 kJkg sA1E A 43533 kJkgK State 2 T v twophase straight down in Pv diagram from state 1 xA2E A v vAfE AvAfgE A 003787 0001638014758 02455 uA2E A uAfE A xA2E A uAfgE A 27289 02455 10593 53295 kJkg sA2E A sAfE A xA2E A sAfgE A 10408 02455 40452 20339 kJkgK A1E AQA2E A muA2E A uA1E A 075 kg 53295 13389 kJkg 6045 kJ A1E ASA2genE A msA2E A sA1E A A1E AQA2E ATA0E A 07520339 43533 604529315 0322 kJK V P 2 70 C 1 3312 858 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6122 The unrestrained expansion of the reactor water in Problem 3101 has a final state in the twophase region Find the entropy generated in the process A waterfilled reactor with volume of 1 m3 is at 20 MPa 360C and placed inside a containment room as shown in Fig P3101 The room is well insulated and initially evacuated Due to a failure the reactor ruptures and the water fills the containment room Find the minimum room volume so the final pressure does not exceed 200 kPa Solution CV Containment room and reactor Mass m2 m1 Vreactorv1 10001823 5485 kg Energy Eq35 mu2 u1 1Q2 1W2 0 0 0 Entropy Eq637 ms2 s1 dQT 1S 2 gen State 1 T P Table B14 u1 17028 kJkg s1 3877 Energy equation implies u2 u1 17028 kJkg State 2 P2 200 kPa u2 ug Twophase Table B12 x2 u2 uf ufg 17028 50447202502 059176 v2 0001061 059176 088467 052457 m3kg s2 sf x2sfg 153 059176 5597 48421 kJkg K V2 m2 v2 5485 052457 2877 m 3 From the entropy equation the generation is 1S2 gen ms2 s1 5485 48421 3877 5294 kJK P v 1 T s 1 2 200 kPa 200 2 u const Entropy is generated due to the unrestrained expansion No work was taken out as the volume goes up Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6123 Heat transfer from a 20oC kitchen to a block of 15 kg ice at 10oC melts it to liquid at 10oC How much entropy is generated Water changes state from nearly saturated solid to nearly saturated liquid The pressure is 101 kPa but we approximate the state properties with saturated state at the same temperature CV Ice out to the 20oC kitchen air Energy Eq35 mu2 u1 1Q2 1W2 1Q2 Entropy Eq637 ms2 s1 dQT 1S2 gen 1Q2T0 1S2 gen State 1 Compressed saturated solid B15 u1 35409 kJkg s1 12995 kJkgK State 2 Compressed saturated liquid B11 u2 4199 kJkg s2 01510 kJkgK Heat transfer from the energy Eq 1Q2 mu2 u1 15 4199 35409 59412 kJ From the entropy Eq 1S2 gen ms2 s1 1Q2T0 15 0151 12995 5941229315 0149 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6124 Ammonia is contained in a rigid sealed tank unknown quality at 0oC When heated in boiling water to 100oC its pressure reaches 1200 kPa Find the initial quality the heat transfer to the ammonia and the total entropy generation Solution CV Ammonia which is a control mass of constant volume Energy Eq35 u2 u1 1q2 1w2 Entropy Eq637 s2 s1 dqT 1s 2 gen State 2 1200 kPa 100oC Table B22 s2 55325 kJkg K v2 014347 m3kg u2 14858 kJkg State 1 v1 v2 Table B21 x1 014347 0001566028763 049336 u1 74128 kJkg s1 07114 x1 46195 29905 kJkg K Process V constant 1w2 0 1q2 u2 u1 14858 74128 74452 kJkg To get the total entropy generation take the CV out to the water at 100oC 1s2 gen s2 s1 1q2T 55325 29905 7445237315 0547 kJkg K v P s T 2 1 1 2 v C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6125 Water in a pistoncylinder is at 101 kPa 25C and mass 05 kg The piston rests on some stops and the pressure should be 1000 kPa to float the piston We now heat the water from a 200C reservoir so the volume becomes 5 times the initial volume Find the total heat transfer and the entropy generation Solution Take CV as the water out to the reservoir Continuity Eq m2 m1 m Energy Eq mu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 dQT 1S2 gen 1Q2Tres 1S2 gen Process v constant then P C Pfloat Volume does go up so we get v2 5 v 1 State 1 v1 0001003 m3kg u1 10486 kJkg s1 03673 kJkgK State 2 P2 Pfloat v2 5 0001003 0005015 m3kg T2 17991C x2 v2 vf vfg 0005015 0001127019332 002011 u2 76167 x2 182197 79831 kJkg s2 21386 x2 44478 22280 kJkgK From the process equation see PV diagram we get the work as 1w2 Pfloatv2 v1 1000 kPa 0005015 0001003 m3kg 4012 kJkg From the energy equation we solve for the heat transfer 1Q2 mu2 u1 1w2 0579831 10486 4012 3487 kJ 1S2 gen ms2 s1 1Q2Tres 0522280 03673 348747315 01934 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6126 Do Problem 6125 assuming the pistoncylinder is 15 kg of steel and has the the same temperature as the water at any time Solution Take CV as the water and steel out to the reservoir Continuity Eq m2 m1 mH2O msteel Energy Eq mH2O u2 u1H2O msteel u2 u1steel 1Q2 1W2 Entropy Eq637 mH2Os2 s1H2O msteels2 s1steel 1Q2Tres 1S2 gen Process v constant then P C Pfloat Volume does go up so we get v2 5 v 1 State 1 v1 0001003 m3kg u1 10486 kJkg s1 03673 kJkgK State 2 P2 Pfloat v2 5 0001003 0005015 m3kg T2 17991C x2 v2 vf vfg 0005015 0001127019332 002011 u2 76167 x2 182197 79831 kJkg s2 21386 x2 44478 22280 kJkgK There is only work when piston moves and then P Pfloat so the work is 1W2 PfloatV2 V1 1000 kPa 05 0005015 0001003 m3 2006 kJ From the energy equation we solve for the heat transfer 1Q2 U2 U1 1W2 0579831 10486 15 046 17991 25 2006 4556 kJ For the entropy change we use B11 for water and A5 and Eq611 for steel 1S2 gen S2 S1 1Q2Tres 0522280 03673 15 046 ln 45306 29815 4556 47315 0256 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6127 A cylinder fitted with a movable piston contains water at 3 MPa 50 quality at which point the volume is 20 L The water now expands to 12 MPa as a result of receiving 600 kJ of heat from a large source at 300C It is claimed that the water does 124 kJ of work during this process Is this possible Solution CV H2O in Cylinder Energy Eq mu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 1Q2 Tsource 1S2gen Tsource 300oC Process 1Q2 600 kJ 1W2 124 kJ State 1 3 MPa x1 05 Table B12 T1 2339oC v1 vf x1vfg 0001216 05006546 0033948 m3kg u1 uf x1ufg 18045 kJkg s1 sf x1sfg 44162 kJkgK m1 V1v1 002 m3 0033948 m3kg 0589 kg Now solve for u2 from the energy equation u2 u1 1Q2 1W2m1 18045 600 1240589 26126 kJkg State 2 P2 12 MPa u2 26126 kJkg Table B13 T2 200oC s2 65898 kJkgK From the entropy equation 1S2gen ms2 s1 1Q2 Tsource 0589 kg 65898 44162 kJkgK 600 300 273 kJK 02335 kJK 0 Process is possible P v 1 2 T 2 1 T s 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6128 A pistoncylinder device keeping a constant pressure has 1 kg water at 20oC and 1 kg of water at 100oC both at 500 kPa separated by a thin membrane The membrane is broken and the water comes to a uniform state with no external heat transfer Find the final temperature and the entropy generation for the process Solution Continuity Eq m2 mA mB 0 Energy Eq35 m2u2 mAuA mBuB 1W 2 Entropy Eq637 m2s2 mAsA mBsB dQT 1S 2 gen Process P Constant 1W2 PdV PV2 V1 Q 0 Substitute the work term into the energy equation and rearrange to get m2u2 P2V2 m2h2 mAuA mBuB PV1 mAhA mBhB where the last rewrite used PV1 PVA PVB State A1 Table B14 hA 8441 kJkg sA 02965 kJkg K State B1 Table B14 hB 41932 kJkg sB 13065 kJkg K Energy equation gives h EA mAB A E mA2 AE A hABE A A1 2E A 8441 A1 2E A 41932 251865 kJkg 2 mA m2 hA State 2 hA2E A 251865 kJkg PA2E A 500 kPa from Table B14 TA2E A 60085C sA2E A 083184 kJkg K With the zero heat transfer we have A1E ASA2 genE A mA2E AsA2E A mAAE AsAAE A mABE AsABE 2 083184 1 02965 1 13065 00607 kJK F Water cb 20 C 100 C Water Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6129 Reconsider Problem 3109 where COA2E A is compressed from 20AoE AC x 075 to a state of 3 MPa 20AoE AC in a pistoncylinder where pressure is linear in volume Assume heat transfer is from a reservoir at 100AoE AC and find the specific entropy generation in the process external to the COA2E A CV Carbon dioxide out to the source both A1E AQA2E A and A1E AWA2E A Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E A Entropy Eq637 SA2E A SA1E A AdQTEA A1E ASA2genE A A1E AQA2E ATAresE A A1E ASA2genE Process P A BV A1E AWA2E A P dV ½ mPA1E A PA2E A vA2E A vA1E A State 1 Table B31 P 19696 kPa vA1E A 0000969 075 001837 001475 mA3E Akg uA1E A 3964 075 24625 22433 kJkg sA1E A 01672 075 11157 1004 kJkgK State 2 Table B3 v2 001512 mA3E Akg uA2E A 31021 kJkg sA2E A 13344 kJkgK A1E AwA2E A ½ PA1E A PA2E A vA2E A vA1E A ½ 19696 3000 001512 001475 092 kJkg A1E AqA2E A uA2E A uA1E A A1E AwA2E A 31021 22433 092 868 kJkg A1E AsA2genE A sA2E A sA1E A A1E AqA2E ATAresE A 13344 1004 A 868 37315E A 0098 kJkgK P v 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6130 A pistoncylinder contains 1 kg water at 150 kPa 20C The piston is loaded so pressure is linear in volume Heat is added from a 600C source until the water is at 1 MPa 500C Find the heat transfer and the total change in entropy Solution CV HA2E AO out to the source both A1E AQA2E A and A1E AWA2E A Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E A Entropy Eq637 msA2E A sA1E A A1E AQA2E A TASOURCEE A A1E ASA2 genE Process P A BV A1E AWA2E A P dV ½ PA1E A PA2E A VA2E A VA1E A State 1 B11 Compressed liquid use saturated liquid at same T vA1E A 0001002 mA3E Akg uA1E A 8394 kJkg sA1E A 02966 kJkg K State 2 Table B13 sup vap vA2E A 035411 mA3E Akg uA2E A 31243 kJkg sA2E A 77621 kJkg K P v 1 2 2 1 T s P P 2 1 A1E AWA2E A ½ 1000 150 kPa 1 kg 035411 0001002 mA3E Akg 203 kJ A1E AQA2E A 131243 8394 203 32434 kJ msA2E A sA1E A 1 kg 77621 02968 kJkgK 74655 kJK A1E AQA2E A Tsource 37146 kJK for source Q A1E AQA2E A A1E ASA2 genE A msA2E A sA1E A A1E AQA2E A TASOURCEE A SAH2OE A SAsourceE A 74655 37146 3751 kJK Remark This is an external irreversible process delta T to the source Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6131 A closed rigid container is filled with 15 kg water at 100 kPa 55AoE AC 1 kg of stainless steel and 05 kg of PVC polyvinyl chloride both at 20oC and 01 kg of air at 400 K 100 kPa It is now left alone with no external heat transfer and no water vaporizes Find the final temperature and the entropy generation for the process CV Container Process V constant A1E AWA2E A 0 and also given A1E AQA2E A 0 Energy Eq UA2E A UA1E A Amiu2 u1i EA A1E AQA2E A A1E AWA2E A 0 Entropy Eq637 SA2E A SA1E A Amis2 s1i EA 0 A1E ASA2genE For the liquid and the metal mass we will use the specific heat Tbl A3 A4 so Amiu2 u1i EA AmiCv i T2 T1i EA TA2E AmiCv i EA AmiCv iT1 i EA Amis2 s1i EA AmiCv i lnT2T1i EA noticing that all masses have the same TA2E A but not same initial T AmiCv i EA 15 418 1 046 05 096 01 0717 7282 kJK Energy Eq 7282 TA2E A 15 418 55 1 046 05 096 20 01 0717 40027315 372745 kJ TA2E A 512AoE AC The volume of the air is constant so entropy change from Eq617 is the same expression as for the solids and liquids given above A1E ASA2genE A SA2E A SA1E A Amis2 s1i EA 15 418 ln3243532815 1 046 05 096 lnA32435 29315E A 01 0717 lnA32435 400E A 007304 kJK 009507 kJK 001503 kJK 0007 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6132 A cylinderpiston contains water at 200 kPa 200C with a volume of 20 L The piston is moved slowly compressing the water to a pressure of 800 kPa The loading on the piston is such that the product PV is a constant Assuming that the room temperature is 20C show that this process does not violate the second law Solution CV Water cylinder out to room at 20C Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E Entropy Eq637 msA2E A sA1E A A1E AQA2E A TAroomE A A1E ASA2 genE Process PV constant Pmv vA2E A PA1E AvA1E APA2E A1E AwA2E A APdvEA PA1E AvA1E A lnvA2E AvA1E A State 1 Table B13 vA1E A 10803 mA3E Akg uA1E A 26544 kJkg sA1E A 75066 kJkg K State 2 PA2E A vA2E A PA1E AvA1E APA2E A 200 10803800 02701 mA3E Akg Table B13 uA2E A 26550 kJkg sA2E A 68822 kJkg K A1E AwA2E A 200 10803 lnA 02701 10803 E A 2995 kJkg A1E AqA2E A uA2E A uA1E A A1E AwA2E A 26550 26544 2995 2989 kJkg A1E AsA2genE A sA2E A sA1E A A 1q2 ETroom E A 68822 75066 A 2989 29315E 0395 kJkg K 0 satisfies entropy eq Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6133 A rigid steel tank of mass 25 kg contains 05 kg R410A at 0C with specific volume 001mkg The whole system heats up to the room temperature 25C Find the process heat transfer and the entropy generation CV R410A and steel tank out to room T Control mass Continuity Eq mA2E A mAR410aE A mAstE A 0 Energy Eq mAR410aE AuA2E A uA1E A mAstE AuA2E A uA1E A A1E AQA2E A A1E AWA2E A Entropy Eq mAR410aE AsA2E A sA1E A mAstE AsA2E A sA1E A A1E AQA2E ATAroomE A A1E ASA2genE A Process V C so A1E AWA2E A 0 State 1 TA1E A 0C vA1E A 001 m3kg V mvA1E A 0005 m3 xA1E A v vAfE A vAfgE A 001 0000855003182 028758 uA1E A uAfE A xA1E A uAfgE A 5707 xA1E A 19595 11342 kJkg sA1E A 02264 xA1E A 08104 045945 kJkgK State 2 T v supvapor straight up in Tv diagram from state 1 B41 at 25C vAfE A 0000944 m3kg vg 001514 m3kg vAfE A v vg saturated P 16536 kPa x v vf vfg A001 0000944 001420E A 063775 uA2E A uAfE A xA2E A uAfgE A 9603 xA2E A 16295 19995 kJkg sA2E A sAfE A xA2E A sAfgE A 03631 xA2E A 06253 07619 kJkgK From the energy Eq A1E AQA2E A mAR410aE AuA2E A uA1E A mAstE A CAstE ATA2E A TA1E A 05 kg 19995 11342 kJkg 25 kg 046 kJkgK 250 K 720 kJ A1E ASA2genE A mAR410aE AsA2E A sA1E A mAstE A CAstE A lnTA2E ATA1E A A1E AQA2E ATAroomE 05 kg 07619 045945 A kJ kgKE A 25 kg 046 A kJ kgKE A ln A29815 27315E 7229815 kJK 01512 kJK 01007 kJK 02415 kJK 00104 kJK v T 2 16536 kPa 1 0 25 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6134 A pistoncylinder has ammonia at 2000 kPa 80oC with a volume of 01 m3 The piston is loaded with a linear spring and outside ambient is at 20oC shown in Fig P6134 The ammonia now cools down to 20oC at which point it has a quality of 15 Find the work the heat transfer and the total entropy generation in the process CV Ammonia out to the ambient both 1Q2 and 1W2 Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 1Q2 Tambient 1S2 gen Process P A BV 1W2 P dV ½ mP1 P2 v2 v1 State 1 Table B22 v1 007595 m3kg u1 14216 kJkg s1 50707 kJkg K m V1v1 01007595 131665 kg State 2 Table B21 v2 0001638 015 014758 0023775 m3kg u2 27289 015 10593 431785 kJkg s2 10408 015 40452 164758 kJkg K 1W2 ½ mP1 P2 v2 v1 ½ 131665 kg 2000 8575 kPa 0023775 007595 m3kg 9815 kJ 1Q2 mu2 u1 1W2 131665 431785 14216 9815 140139 kJ 1S2 gen ms2 s1 1Q2 Tamb 131665 164758 50707 140139 29315 4507051 578045 0273 kJk P v 1 2 2 1 T s P2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6135 One kilogram of ammonia NH3 is contained in a springloaded pistoncylinder Fig P6135 as saturated liquid at 20C Heat is added from a reservoir at 100C until a final condition of 800 kPa 70C is reached Find the work heat transfer and entropy generation assuming the process is internally reversible Solution CV NH3 out to the reservoir Continuity Eq m2 m1 m Energy Eq35 E2 E1 mu2 u1 1Q2 1W 2 Entropy Eq637 S2 S1 dQT 1S2gen 1Q2Tres 1S 2gen Process P A BV linear in V 1W2 PdV 1 2 P1 P2V2 V1 1 2 P1 P2mv2 v1 State 1 Table B21 P1 19008 kPa v1 0001504 m3kg u1 8876 kJkg s1 03657 kJkg K State 2 Table B22 sup vapor P v 1 2 2 1 T s P2 v2 0199 m3kg u2 14383 kJkg s2 55513 kJkg K 1W2 1 219008 800 kPa 1 kg 01990 0001504 m3kg 97768 kJ 1Q2 mu2 u1 1W2 114383 8876 97768 14473 kJ 1S2gen ms2 s1 1Q2Tres 155513 03657 14473 37315 1307 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6136 A 5 kg aluminum radiator holds 2 kg of liquid R134a both at 10oC The setup is brought indoors and heated with 220 kJ from a heat source at 100oC Find the total entropy generation for the process assuming the R134a remains a liquid Solution CV The aluminum radiator and the R134a Energy Eq35 m2u2 m1u1 1Q2 0 Process No change in volume so no work as used above The energy equation now becomes summing over the mass mal u2 u1al mR134a u2 u1R134a 1Q 2 Use specific heat from Table A3 and A4 malCal T2 T1 m R134aC R134a ln T2 T1 1Q 2 T2 T1 1Q2 malCal m R134aC R134a 220 5 09 2 143 2989oC T2 10 2989 1989oC Entropy generation from Eq637 1S2 gen ms2 s1 1Q2T malCal ln T2T1 m R134aC R134a ln T2T1 1Q2 Tamb 5 09 2 143 ln 1989 27315 10 27315 220 37315 07918 05896 0202 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6137 A pistoncylinder of total 1 kg steel contains 05 kg ammonia at 1600 kPa both masses at 120oC Some stops are placed so a minimum volume is 002 m3 shown in Fig P6137 Now the whole system is cooled down to 30oC by heat transfer to the ambient at 20oC and during the process the steel keeps same temperature as the ammonia Find the work the heat transfer and the total entropy generation in the process Energy Eq35 mu2 u1 mstu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 msts2 s1 1Q2Tamb 1S 2 gen State 1 v1 011265 m3kg u1 15166 kJkg s1 55018 kJkg K V1 mv1 005634 m3 Stop 1a vstop Vm 00205 004 m3kg Pstop P1 T 42oC State 2 30oC Tstop so v2 vstop 004 m3kg x2 v2vf vfg 004 000168 010881 035217 u2 32046 x2 10169 67858 kJkg s2 12005 x2 37734 25294 kJkg K 1W2 P dV P1m v2v1 1600 05 0004 011268 5814 kJ 1Q2 m u2 u1 mstu2 u1 1W2 05 67858 15166 104630 120 5814 41901 414 5814 51855 kJ 1S2 gen ms2 s1 mst s2 s1 1Q2T amb 05 25294 55018 1046 ln 27330 273120 5185 29315 14862 01196 16277 002186 kJK 2 1 P v T s 1 2 1a 42 30 1a NH P o 3 T o Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6138 A pistoncylinder contains 01 kg water at 500C 1000 kPa The piston has a stop at half the original volume similar to Fig 6137 The water now cools to room temperature 25C Find the heat transfer and the entropy generation Energy Eq mu2 u1 1Q2 1W 2 Entropy Eq ms2 s1 1Q2Troom 1S2gen Process Eq P C if v vstop V C if P Pfloat State 1 v1 035411 m3kg u1 312434 kJkg s1 77621 kJkgK State a va v12 0177055 m3kg vg 1000 kPa so Ta Tsat 1000 kPa 1799C The possible state 2 PV combinations are shown State a is 1000 kPa va so it is twophase with Ta 180C T 2 P2 Psat 25 C 3169 kPa and v2 v a x2 v2 vf vfg 0177 000100343358 00040604 u2 uf x2 ufg 10486 x2 23049 114219 kJkg s2 sf x2 sfg 03673 x2 81905 040056 kJkgK 1W2 m P dv m P1 v2 v1 see area below process curve in figure 01 kg 1000 kPa 0177055 035411 m3kg 17706 kJ 1Q2 mu2 u1 1W2 01 kg 114219 312434 kJkg 17706 kJ 31872 kJ 1S2gen m s2 s1 1Q2Troom 01 kg040056 77621 kJkgK 31872 kJ29815 K 073615 kJK 106899 kJK 0333 kJK Water P o m p V P 2 500 C a 317 1000 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6139 A hollow steel sphere with a 05m inside diameter and a 2mm thick wall contains water at 2 MPa 250C The system steel plus water cools to the ambient temperature 30C Calculate the net entropy change of the system and surroundings for this process CV Steel water out to ambient T0 This is a control mass Energy Eq U2 U1 mH2Ou2 u1 msteelu2 u1 1Q2 1W2 Entropy Eq S2 S1 dQT 1S2 gen 1Q2T0 1S2 gen Process V constant 1W2 0 msteel ρVsteel 8050 π605043 053 12746 kg VH2O π6053 mH2O Vv 6545102011144 0587 kg v2 v1 011144 0001004 x2 32889 x2 3358103 u2 12578 3358103 22908 1335 kJkg s2 04639 3358103 80164 04638 kJkg K 1Q2 mH2Ou2 u1 msteelu2 u1 05871335 26796 12746 04830 250 14946 1346 28406 kJ S2 S1 mH2Os2 s1 msteels2 s1 058704638 6545 12746 048 ln 30315 52315 6908 kJK SSURR 1Q2T0 284063032 9370 kJK 1S2 gen S2 S1 1Q2T0 6908 9370 2462 kJK Water Ambient Steel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6140 A cylinderpiston arrangement contains 10 g ammonia at 20C with a volume of 1 L There are some stops so if the piston is at the stops the volume is 14 L The ammonia is now heated to 200C by a 240C source The piston and cylinder is made of 05 kg aluminum and assume that the mass has the same temperature as the ammonia at any time Find the total heat transfer and the total entropy generation CV NH3 Control mass goes out to source Energy Eq U3 U1 mNH3 u3 u1 mAlu u3 u1 1Q3 1W 3 Entropy Eq mNH3 s3 s1 mAlu s3 s1 1Q3Tsource 1S3gen State 1 B21 v1 Vm 0001 001 01 m3kg v so 2phase P 8575 kPa x1 v vf vfg 01 0001638 014758 06665 u1 uf x1 ufg 27289 x1 10593 97891 kJkg s1 sf x1 sfg 10408 x1 40452 373693 kJkgK State 2 v2 14 v1 014 m3kg P 8575 kPa still 2phase so T2 20oC State 3 200oC v3 v2 P 1600 kPa u3 16765 kJkg s3 59734 kJkgK We get the work from the process equation see PV diagram 1W3 1W2 P1mv2 v1 8575 kPa 001 014 01 m3 0343 kJ The energy equation and the entropy equation give heat transfer and entropy generation 1Q3 mNH3 u3 u1 mAlu u3 u1 1W3 001 16765 97891 05 09 200 20 0343 8832 kJ 1S3gen mNH3s3 s1 mAlus3 s1 1Q3Tsource 001 59734 373693 05 09 ln47315 29315 8832 51315 002236 021543 01721 00657 kJK V P 1 2 3 NH P o cb 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6141 A cylinderpiston arrangement contains 01 kg R410A of quality x 02534 and at 20C Stops are mounted so Vstop 3V1 similar to Fig P6140 The system is now heated to the final temperature of 20C by a 50C source Find the total entropy generation CV The R410A mass out to source Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq S2 S1 dQT 1S2 gen 1Q2Tsource 1S2 gen Process P Constant if V Vstop V Vstop if P P1 State 1 u1 2792 x1 21807 8318 kJkg P1 Psat 3996 kPa v1 0000803 x1 0064 001702 m3kg s1 01154 x1 09625 02439 kJkgK State 1a vstop 3 v1 005106 m3kg vg at P 1 State 2 at 20C T1 vstop vg 001758 m3kg so superheated vapor Table B42 P2 600 kPa u2 27356 kJkg s2 11543 kJkgK 1W2 PdV P1 mv2 v1 3996 01 0051 0017 136 kJ 1Q2 mu2 u1 1W2 0127356 8318 136 20398 kJ 1S2gen m s2 s1 1Q2Tsource 01 11543 02439 20398 32315 00279 kJK See the work term from the process in the Pv diagram v P 1 1a 2 R410a P o cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6142 One kg of air at 300 K is mixed with 2 kg air at 400 K in a process at a constant 100 kPa and Q 0 Find the final T and the entropy generation in the process CV All the air Energy Eq U2 U1 0 W Entropy Eq S2 S1 0 1S 2 gen Process Eq P C W PV2 V1 Substitute W into energy Eq U2 U1 W U2 U1 PV2 V1 H2 H1 0 Due to the low T let us use constant specific heat H2 H1 mAh2 h1A mBh2 h1 B mACpT2 TA1 mBCpT2 TB1 0 T2 mA mB mATA1 mBTB1 1 3 TA1 2 3 TB1 36667 K Entropy change is from Eq 616 with no change in P 1S2 gen S2 S1 mACp ln T2 TA1 mBCp ln T2 TB1 1 1004 ln 36667 300 2 1004 ln 36667 400 020148 017470 00268 kJK Remark If you check the volume does not change and there is no work Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6143 Air in a rigid tank is at 900 K 500 kPa and it now cools to the ambient temperature of 300 K by heat loss to the ambient Find the entropy generation CV Air out to ambient No size given so do it per unit mass Energy Eq35 u2 u1 Cv T2 T1 1q2 1w2 Entropy Eq637 s2 s1 1q2Tamb 1s2 gen tot Process V constant v2 v1 also 1W2 0 Ideal gas P2 P1 T2 T1 500 300900 16667 kPa From Table A7 u1 67482 kJkg sT1 801581 kJkgK u2 21436 kJkg sT2 686926 kJkgK 1q2 u2 u1 21436 67482 46046 kJkg 1s2 gen tot s2 s1 1q2 Tamb sT2 sT1 R ln P2 P1 1q2 Tamb 686926 801581 0287 ln16667 500 46046 300 0661 kJkgK We could also have used constant specific heat being slightly less accurate Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6144 Two rigid insulated tanks are connected with a pipe and valve One tank has 05 kg air at 200 kPa 300 K and the other has 075 kg air at 100 kPa 400 K The valve is opened and the air comes to a single uniform state without external heat transfer Find the final T and P and the entropy generation Solution CV Total tank Control mass of constant volume Mass and volume m2 mA mB V VA VB Energy Eq U2 U1 m2 u2 mAuA1 mBuB1 1Q2 1W2 0 Entropy Eq S2 S1 m2 s2 mAsA1 mBsB1 1Q2T 1S2 gen Process Eq V constant 1W2 0 Insulated 1Q2 0 Ideal gas at A1 VA mARTA1PA1 05 0287 300 200 02153 m 3 Ideal gas at B1 VB mBRTB1 PB1 075 0287 400 100 0861 m3 State 2 m2 mA mB 125 kg V2 VA VB 10763 m3 Energy Eq u2 m2 mAuA1 mBuB1 and use constant specific heat T2 m2 mA TA1 mB m2 TB1 05 125 300 075 125 400 360 K P2 m2 RT2V 125 kg 0287 kJkgK 360 K 10763 m3 120 kPa S2 S1 mACP lnT2TA1 RlnP2PA1 mBCP lnT2TB1 RlnP2PB1 05 1004 ln360 300 0287 ln120 200 0751004 ln360 400 0287 ln120 100 05 13514 075 01581 05571 kJK 1S2 gen S2 S1 05571 kJK B A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6145 One kg of air at 100 kPa is mixed with 2 kg air at 200 kPa both at 300 K in a rigid insulated tank Find the final state P T and the entropy generation in the process CV All the air Energy Eq U2 U1 Q W Entropy Eq S2 S1 QT 1S 2 gen Process Eqs V C W 0 Q 0 States A1 B1 uA1 u B1 VA mART1PA1 VB mBRT1P B1 cb U2 U1 m2u2 mAuA1 mBuB1 0 u2 uA1 2uB13 u A1 State 2 T2 T1 300 K from u2 m2 mA mB 3 kg V2 m2RT1P2 VA VB mART1PA1 mBRT1P B1 Divide with mART1 and get 3P2 1PA1 2PB1 1 100 2 200 002 kPa1 P2 150 kPa Entropy change from Eq 616 with the same T so only P changes 1S2 gen S2 S1 mAR ln PA1 P2 mBR ln P2 PB1 0287 1 ln 150 100 2 ln 150 200 0287 04055 05754 0049 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6146 A rigid storage tank of 15 m3 contains 1 kg argon at 30C Heat is then transferred to the argon from a furnace operating at 1300C until the specific entropy of the argon has increased by 0343 kJkg K Find the total heat transfer and the entropy generated in the process Solution CV Argon out to 1300C Control mass m 1 kg Argon is an ideal gas with constant heat capacity Energy Eq35 m u2 u1 m Cv T2 T1 1Q2 1W2 Entropy Eq637 ms2 s1 1Q2Tres 1S2 gen tot Process V constant v2 v1 also 1W2 0 Properties Table A5 R 020813 Cv 0312 kJkg K State 1 T1 v1 Vm P1 mRT1V 42063 kPa State 2 s2 s1 0343 and change in s from Eq616 or Eq617 s2 s1 Cp ln T2 T1 R ln T2 T1 Cv ln T2 T1 T2 T1 exp s2 s1 Cv exp0343 0312 exp109936 30 Pv RT P2 P1 v2 v1 T2 T1 P2 P1 T2 30 T1 90945 K P2 30 P1 126189 kPa P v 1 2 2 1 T s v C P1 Heat transfer from energy equation 1Q2 1 0312 90945 30315 1892 kJ Entropy generation from entropy equation 1S2 gen tot ms2 s1 1Q2Tres 1 0343 1892 1300 273 0223 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6147 Argon in a light bulb is at 110 kPa 90oC The light is turned off so the argon cools to the ambient 20oC Disregard the glass and any other mass and find the specific entropy generation Solution CV Argon gas Neglect any heat transfer Energy Eq35 mu2 u1 1Q2 Entropy Eq637 s2 s1 dqT 1s2 gen 1q2Troom 1s2 gen Process v constant and ideal gas P2 P1 T2T 1 1q2 u2 u1 Cv T2 T1 0312 20 90 2184 kJkg Evaluate changes in s from Eq616 or 817 s2 s1 Cp ln T2T1 R ln P2 P1 Eq616 Cp ln T2T1 R ln T2 T1 Cv lnT2T1 Eq617 0312 ln 20 27390 273 006684 kJkg K 1s2 gen s2 s1 1q2Troom 006684 2184 29315 000766 kJkgK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6148 A rigid tank contains 2 kg of air at 200 kPa and ambient temperature 20C An electric current now passes through a resistor inside the tank After a total of 100 kJ of electrical work has crossed the boundary the air temperature inside is 80C Is this possible Solution CV Air in tank out to ambient Energy Eq35 mu2 u1 1Q2 1W2 1W2 100 kJ Entropy Eq637 ms2 s1 dQT 1S2 gen 1Q2Tamb 1S2 gen Process Constant volume and mass so v2 v 1 State 1 T1 20oC P1 200 kPa m1 2 kg State 2 T2 80oC v2 v 1 Ideal gas Table A5 R 0287 kJkgK Cv 0717 kJkgK Assume constant specific heat then energy equation gives 1Q2 mCvT2 T1 1W2 2 071780 20 100 140 kJ Change in s from Eq617 since second term drops out s2 s1 Cv ln T2T1 Rln v2 v1 v2 v1 ln v2 v1 0 s2 s1 Cvln T2T1 01336 kJkgK Now Eq637 1S2 gen ms2 s1 1Q2Tamb 2 01336 14 293 0315 kJK 0 Process is Possible Note P2 P1 T2 T1 in Eq616 s2 s1 Cp ln T2 T1 R ln P2 P1 results in the same answer as Eq617 P v 1 2 2 1 T s v C P 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6149 A pistoncylinder system contains 50 L of air at 300oC 100 kPa with the piston initially on a set of stops A total external constant force acts on the piston so a balancing pressure inside should be 200 kPa The cylinder is made of 2 kg of steel initially at 1300oC The system is insulated so that heat transfer occurs only between the steel cylinder and the air The system comes to equilibrium Find the final temperature and the entropy generation CV Steel water out to ambient T0 This is a control mass Energy Eq U2 U1 mairu2 u1 mstu2 u1 1Q2 1W2 Entropy Eq S2 S1 mairs2 s1 msts2 s1 dQT 1S2 gen 1S 2 gen Process 1Q2 0 and must be on PV diagram shown mair P1V1 RT1 100 005 0287 57315 00304 kg Since V1a V1 then T1a T1PfloatP1 57315 200100 11463 K Use constant Cv for air at 900 K Cv uT 0833 kJkgK from A7 To reach state 1a Uair mCvT 00304 0833 1146 573 145 kJ Ust mCvT 2 046 1146 1573 3928 kJ Conclusion from this is T2 is higher than T1a 1146 K piston lifts P2 P float Write the work as 1W2 P2 V2 V1 and use constant Cv in the energy Eq as mair Cv T2 T1 mst Cst T2 T1 P2mairv2 P2 V1 now P2 v2 RT2 for the air so isolate T2 terms as mair Cv R mCst T2 mair CvT1 air mCst T1 st P2V1 00304 112 2 046 T2 00304 0833 57315 2 046 157315 200 005 Solution gives T2 15427 K 1S2 gen S2 S1 mairs2 s1 msts2 s1 mair CP lnT2T1 air R lnP2P1 mst Cst lnT2T1 st 00304112 ln 15427 57315 0287 ln 200 100 2 046 ln 15427 157315 0027665 0017982 00097 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6150 A spring loaded piston cylinder contains 15 kg air at 27oC and 160 kPa It is now heated in a process where pressure is linear in volume P A BV to twice the initial volume where it reaches 900 K Find the work the heat transfer and the total entropy generation assuming a source at 900 K Solution CV Air out to the 900 K source Since air T is lower than the source temperature we know that this is an irreversible process Continuity Eq m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 dQT 1S2 gen 1Q2Tsource 1S2 gen Process P A BV State 1 T1 P1 Table A7 u1 21436 kJkg V1 mRT1 P1 15 0287 300 kJ 160 kPa 08072 m 3 State 2 T2 v2 2 v1 Table A7 u2 674824 kJkg P2 RT2 v2 RT22v1 T2 P1 2T1 P1 T22 T1 160 kPa 900 K 2 300 K 240 kPa From the process equation we can express the work as 1W2 PdV 05 P1 P2 V2 V1 05 P1 P2 V1 05 160 240 kPa 08072 m3 1614 kJ 1Q2 15 674824 21436 1614 8521 kJ Change in s from Eq619 and Table A7 values 1S2 gen mso T2 so T1 R ln P2 P1 1Q2TSOURCE 15 80158 68693 0287 ln 240 160 8521 900 1545 0947 0598 kJK v 2 1 T s P 1 P 1 2 900 300 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6151 A rigid container with volume 200 L is divided into two equal volumes by a partition shown in Fig P6151 Both sides contain nitrogen one side is at 2 MPa 200C and the other at 200 kPa 100C The partition ruptures and the nitrogen comes to a uniform state at 70C Assume the temperature of the surroundings is 20C determine the work done and the net entropy change for the process Solution CV A B no change in volume 1W2 0 mA1 PA1VA1RTA1 2000 0102968 4732 1424 kg mB1 PB1VB1RTB1 200 0102968 3732 01806 kg P2 mTOTRT2VTOT 16046 02968 343202 817 kPa From Eq616 S2 S1 mA1 s2 s1A1 mB1 s2 s1 B1 1424 kg 1042 ln 3432 4732 02968 ln 817 2000 kJkgK 01806 kg 1042 ln 3432 3732 02968 ln 817 200 kJkgK 01894 kJK 1Q2 U2 U1 1424 074570 200 01806 074570 100 14195 kJ From Eq637 1S2 gen S2 S1 1Q2T0 01894 kJK 14195 kJ 2932 K 01894 04841 02947 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6152 A constant pressure pistoncylinder contains 05 kg air at 300 K 400 kPa Assume the pistoncylinder has a total mass of 1 kg steel and is at the same temperature as the air at any time The system is now heated to 1600 K by heat transfer from a 1700 K source Find the entropy generation using constant specific heat for air CV Air and Steel Energy Eq U2 U1 mairu2 u1 mstu2 u1 1Q2 1W2 Entropy Eq637 mair s2 s1 msts2 s1 1Q2Tsource 1S2 gen Process P C 1W2 1 2 PdV P V2 V1 P mair v2 v1 1Q2 mairu2 u1air mstu2 u1st 1W2 mairh2 h1air mstu2 u1 st Use A3 u2 u1st C T2 T 046 kJkgK 1600 300 K 598 kJkg Use A5 h2 h1air CpT2 T1 1004 kJkgK 1600 300 K 13052 kJkg 1Q2 mairh2 h1air mstu2 u1st 05 kg 1305 kJkg 1 kg 598 kJkg 12506 kJ S2 S1 mair s2 s1 msts2 s1 05 kg 1004 kJkgK ln 1600 300 1 kg 046 kJkgK ln 1600 300 16104 kJK 1S2 gen S2 S1 1Q2Tsource 16104 125061700 0875 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6153 Do Problem 6152 using Table A7 CV Air and Steel Energy Eq U2 U1 mairu2 u1 mstu2 u1 1Q2 1W2 Entropy Eq637 mair s2 s1 msts2 s1 1Q2Tsource 1S2 gen Process P C 1W2 1 2 PdV P V2 V1 P mair v2 v1 1Q2 mairu2 u1air mstu2 u1st 1W2 mairh2 h1air mstu2 u1 st Use air tables A7 h2 h1air 175733 30047 145686 kJkg s2 s1air 869051 686926 0 182125 kJkgK No pressure correction as P2 P1 Use A3 u2 u1st CT2 T1 046 kJkgK 1600 300 K 598 kJkg 1Q2 mairh2 h1air mstu2 u1st 05 kg 145686 kJkg 1 kg 598 kJkg 132643 kJ S2 S1 mair s2 s1 msts2 s1 182125 1 kg 046 kJkgK ln 1600 300 259128 kJK 1S2 gen S2 S1 1Q2Tsource 259128 1326431700 1811 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6154 Nitrogen at 200oC 300 kPa is in a piston cylinder volume 5 L with the piston locked with a pin The forces on the piston require a pressure inside of 200 kPa to balance it without the pin The pin is removed and the piston quickly comes to its equilibrium position without any heat transfer Find the final P T and V and the entropy generation due to this partly unrestrained expansion Solution CV Nitrogen gas Energy Eq35 mu2 u1 1Q2 1W2 Peq dV P2 V2 V1 Entropy Eq637 ms2 s1 0 1S 2 gen Process 1Q2 0 already used P Peq after pin is out State 1 200 C 300 kPa State 2 P2 Peq 200 kPa m P1V1RT1 300 0005 02968 47315 001068 kg The energy equation becomes mu2 P2V2 mu1 P2V1 mh2 h2 u1 P2V1m u1 P2V1 RT1 P1V1 u1 P2P1 RT 1 Solve using constant Cp C v Cp T2 Cv T1 P2P1 RT1 T2 T1 Cv P2P1 R C p 47315 0745 200 300 02368 1042 42813 K V2 V1 T2 T1 P1P2 0005 42813 47315 300 200 000679 m3 1S2 gen ms2 s1 mCp ln T2T1 R ln P2 P1 P1V1 RT1 Cp ln T2T1 R ln P2 P1 001068 1042 ln 4281347315 02968 ln 200 300 0000173 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6155 The air in the tank of Problem 688 receives the heat transfer from a reservoir at 450 K Find the entropy generation due to the process from 1 to 3 CV Air out to reservoir Energy eq mu3 u1 1Q3 1W3 Entropy eq ms3 s1 1Q3 Tres 1S3 gen State 1 m P1V1 RT1 100 075 0287 300 kPa m3 kJkg 0871 kg Process 1 to 2 Constant volume heating dV 0 1W2 0 P2 P1 T2 T1 100 400 300 1333 kPa Process 2 to 3 Isothermal expansion dT 0 u3 u2 and P3 P2 V2 V3 1333 075 15 6667 kPa 2W3 2 P 3 dV P2V2 ln V2 V3 1333 075 ln2 693 kJ The overall process 1W3 1W2 2W3 2W3 693 kJ From the energy equation 1Q3 mu3 u1 1W3 m Cv T3 T1 1W3 0871 0717 400 300 693 1318 kJ 1S3 gen ms3 s1 1Q3 Tres m CP ln T1 T3 R ln P1 P3 1Q3 T res 0871 1004 ln 400 300 0287 ln 6667 100 1318 450 0060 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6156 One kg of carbon dioxide at 100 kPa 400 K is mixed with two kg carbon dioxide at 200 kPa 2000 K in a rigid insulated tank Find the final state P T and the entropy generation in the process using constant heat capacity from Table A5 CV All the carbon dioxide Continuity m2 mA mB 3 kg Energy Eq U2 U1 0 0 Entropy Eq S2 S1 0 1S 2 gen Process Eqs V C W 0 Q 0 VA mART1PA1 VB mBRT1P B1 cb U2 U1 m2u2 mAuA1 mBuB1 0 m2CvT2 mACvTA1 mBCvTB1 T2 mATA1 mBTB1 m2 1 3 400 2 3 2000 14667 K State 2 V2 m2RT2P2 VA VB mARTA1PA1 mBRTB1P B1 1 R 400100 2 R 2000200 24 R Substitute m2 T2 and solve for P2 P2 3 R T224 R 3 14667 24 1833 kPa Entropy change from Eq 616 s2 s1A Cp ln T2 TA1 R ln P2 PA1 0842 ln 14667 400 01889 ln 1833 100 097955 kJkgK s2 s1B Cp ln T2 TB1 R ln P2 PB1 0842 ln 14667 2000 01889 ln 1833 200 024466 kJkgK 1S2 gen S2 S1 mAs2 s1A mBs2 s1 B 1 097955 2 024466 049 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6157 One kg of carbon dioxide at 100 kPa 400 K is mixed with two kg carbon dioxide at 200 kPa 2000 K in a rigid insulated tank Find the final state P T and the entropy generation in the process using table A8 CV All the carbon dioxide Continuity m2 mA mB 3 kg Energy Eq U2 U1 0 0 Entropy Eq S2 S1 0 1S 2 gen Process Eqs V C W 0 Q 0 VA mART1PA1 VB mBRT1P B1 cb U2 U1 m2u2 mAuA1 mBuB1 0 u2 mAuA1 mBuB1 m2 1 3 22819 2 3 191267 135118 kJkg Interpolate in Table A8 T2 15172 K so T2 66542 kJkgK State 2 V2 m2RT2P2 VA VB mARTA1PA1 mBRTB1P B1 1 R 400100 2 R 2000200 24 R Substitute m2 T2 and solve for P2 P2 3 R T224 R 3 15172 24 18965 kPa Entropy change from Eq 619 s2 s1A so T2 so T1 R ln PA1 P2 66542 51196 01889 ln 18965 100 14137 kJkgK s2 s1B so T2 so T1 R ln PB1 P2 66542 70278 01889 ln 18965 200 036356 kJkgK 1S2 gen S2 S1 mAs2 s1A mBs2 s1 B 1 14137 2 036356 0687 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6158 Nitrogen at 600 kPa 127C is in a 05 m3 insulated tank connected to a pipe with a valve to a second insulated initially empty tank of volume 025 m3 shown in Fig P6158 The valve is opened and the nitrogen fills both tanks at a uniform state Find the final pressure and temperature and the entropy generation this process causes Why is the process irreversible Solution CV Both tanks pipe valve Insulated Q 0 Rigid W 0 Energy Eq35 mu2 u1 0 0 u2 u1 ua1 Entropy Eq637 ms2 s1 dQT 1S2 gen 1S2 gen dQ 0 1 P1 T1 Va m PVRT 600 025 02968 400 12635 kg 2 V2 Va Vb uniform state v2 V2 m u2 ua1 P v 1 2 2 1 T s 1 P P 2 Ideal gas u T u2 ua1 T2 Ta1 400 K P2 mR T2 V2 V1 V2 P1 ½ 600 300 kPa From entropy equation and Eq619 for entropy change Sgen ms2 s1 msT2 sT1 R lnP2 P1 m 0 R ln P2 P1 12635 02968 ln ½ 026 kJK Irreversible due to unrestrained expansion in valve P but no work out Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6159 A cylinderpiston contains carbon dioxide at 1 MPa 300C with a volume of 200 L The total external force acting on the piston is proportional to V3 This system is allowed to cool to room temperature 20C What is the total entropy generation for the process Solution CV Carbon dioxide gas of constant mass m2 m1 m out to ambient Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 dQT 1S2 gen 1Q2Tamb 1S2 gen Process P CV 3 or PV3 constant which is polytropic with n 3 State 1 T P m P1V1RT1 1000 02 018892 5732 1847 kg State 2 T state must be on process curve and ideal gas leads to Eq628 P2 P1T2T1 n n1 10002932573234 6048 kPa V2 V1T1T2 1 n1 016914 m 3 1W2 PdV P2V2 P1V11n 6048 016914 1000 02 1 3 244 kJ 1Q2 mu2 u1 1W 2 1847 0653 20 300 244 3621 kJ From Eq616 ms2 s1 18470842 ln 2932 5732 018892 ln 6048 1000 184704694 087 kJK SSURR 1Q2Tamb 3621 2932 1235 kJK From Eq637 or 839 1S2 gen ms2 s1 1Q2Tamb SNET SCO2 S SURR 087 1235 0365 kJK P v 1 2 T s 1 2 20 300 605 1000 Notice n 3 k 13 n k Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6160 The air in the engine cylinder of Problem 3156 looses the heat to the engine coolant at 100oC Find the entropy generation external to the air using constant specific heat Take CV as the air m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 dQT 1S2 gen 1Q2Tamb 1S2 gen Process Eq Pvn Constant polytropic From the ideal gas law and the process equation we can get State 2 P2 P1 v2 v1 n 4000 10 15 1265 kPa T2 T1 P2v2 P1v1 1527 273 1265 10 4000 5693 K From process eq 1W2 P dV m 1n P2v2 P1v1 mR 1n T2 T1 01 0287 1 15 5693 1800 7064 kJ From energy eq 1Q2 mu2 u1 1W2 mCvT2 T1 1W2 01 07175693 1800 7064 176 kJ ms2 s1 m CP lnT2T1 R lnP2P1 01 1004 ln 5693 1800 0287 ln 1265 4000 001645 kJK 1S2 gen ms2 s1 1Q2Tamb 0011645 17637315 00307 kJK P 1 2 T P C T 3 P v 1 2 T v 1 2 T T 2 1 T C v 05 P C v 15 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6161 A cylinderpiston contains 100 L of air at 110 kPa 25C The air is compressed in a reversible polytropic process to a final state of 800 kPa 500 K Assume the heat transfer is with the ambient at 25C and determine the polytropic exponent n and the final volume of the air Find the work done by the air the heat transfer and the total entropy generation for the process Solution CV Air of constant mass m2 m1 m out to ambient Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 dQT 1S2 gen 1Q2T0 1S2 gen Process Pv1 n P2v2 n Eq627 State 1 T1 P1 State 2 T2 P2 Thus the unknown is the exponent n m P1V1 RT1 110 010287 29815 01286 kg The relation from the process and ideal gas is in Eq628 T2T1 P2P1 n1 n 500 29815 800 110 n1 n n1 n 0260573 n 13524 V2 V1P1P2 1 n 01 110 800 073943 002306 m 3 The work is from Eq629 1W2 PdV 1 n P2V2 P1V1 800 002306 110 01 1 13524 21135 kJ Heat transfer from the energy equation 1Q2 mCvT2 T1 1W2 01286 0717 500 29815 21135 2523 kJ Entropy change from Eq616 s2 s1 CP0 lnT2T1 R lnP2P1 1004 ln 500 29815 0287 ln 800 110 00504 kJ kg K From the entropy equation also Eq637 1S2gen ms2 s1 1Q2T0 01286 00504 252329815 000198 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Rates or fluxes of entropy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6162 A room at 22oC is heated electrically with 1500 W to keep steady temperature The outside ambient is at 5oC Find the flux of S Q T into the room air into the ambient and the rate of entropy generation CV The room and walls out to the ambient T we assume steady state Energy Eq 0 W el in Q out Q out W el in 1500 W Entropy Eq 0 Q outT S gen tot Flux of S into room air at 22oC Q T 1500 29515 508 WK Flux of S into ambient air at 5oC Q T 1500 27815 5393 WK Entropy generation S gen tot Q outT 1500 27815 5393 WK Comment The flux of S into the outside air is what leaves the control volume and since the control volume did not receive any S it was all made in the process Notice most of the generation is done in the heater the room heat loss process generates very little S 5393 508 0313 WK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6163 A mass of 3 kg nitrogen gas at 2000 K V C cools with 500 W What is dSdt Assume that we do not generate any s in the nitrogen then Entropy Eq 642 S cv Q T 500 2000 W K 025 WK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6164 A heat pump see problem 549 should upgrade 5 MW of heat at 85oC to heat delivered at 150oC For a reversible heat pump what are the fluxes of entropy in and out of the heat pump CVTOT Assume reversible Carnot cycle Energy Eq Q L W Q H Entropy Eq 0 Q L TL Q H TH Q L TL Q H TH H Q W L Q HP 85 C o 150 C o The fluxes of entropy become the same as Q H TH Q L TL 5 27315 85 MW K 001396 MWK This is what constitutes a reversible process flux of S in flux out no generation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6165 Reconsider the heat pump in the previous problem and assume it has a COP of 25 What are the fluxes of entropy in and out of the heat pump and the rate of entropy generation inside it CV TOT Energy Eq Q L W Q H Entropy Eq 0 Q L TL Q H TH S gen tot H Q W L Q HP 85 C o 150 C o Definition of COP βHP W in Q H 25 βREF βHP 1 W in Q L 150 W in Q LβREF 5150 3333 MW Q H Q L W 5 MW 3333 MW 8333 MW Q L TL 5 27315 85 MW K 001396 MWK Q H TH 8333 27315 85 MW K 001969 MWK From the entropy equation S gen tot Q H TH Q L TL 001969 001396 MW K 573 kWK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6166 A radiant heating lamp powered by electricity has a surface temperature of 1000 K emitting 500 W The radiation is absorbed by surfaces at the ambient 18oC Find the total entropy generation and specify where it is made including how much CV1 Radiant element Energy Eq 0 W el in Q out Entropy Eq 0 Q outTsurf S gen CV1 Q out W el in 500 W S gen CV1 Q outTsurf 500 1000 05 WK CV2 Space between radiant element and 18oC surfaces this is the room air Energy Eq 0 Q in Q out Entropy Eq 0 Q in Tin Q out Tamb S gen CV2 Q in Q out 500 W S gen CV2 Q out Tamb Q in Tin 500 29115 500 1000 1217 WK The total entropy generation is the sum of the two contributions above which also matches with a total control volume that is the element and the room air CV Radiant element and space out to ambient 18oC Energy Eq 0 W el in Q out Entropy Eq 0 Q outTamb S gen CV tot Q out W el in 500 W S gen CV tot Q out Tamb 500 29115 1717 WK 500 W 1000 K 18 C CV1 CV2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6167 A heat pump with COP 4 uses 1 kW of power input to heat a 25oC room drawing energy from the outside at 15oC Assume the highlow temperatures in the heat pump are 45oC 0oC Find the total rates of entropy into and out of the heat pump the rate from the outside at 15oC and the rate to the room at 25oC Solution CVTOT Energy Eq Q L W Q H Entropy Eq 0 Q L TL Q H TH S gen CV tot H Q W L Q HP 15 C o 25 C o From definition of COP Q H COP W 4 1 kW 4 kW From energy equation Q L Q H W 4 1 kW 3 kW Flux into heat pump at 0oC TLHP Q L 3 27315 kW K 00110 kWK Flux out of heat pump at 45oC THHP Q H 4 31815 kW K 00126 kWK Flux out into room at TH 25oC Q H TH 4 29815 kW K 00134 kWK Flux from outside at 15oC TL Q L 3 28815 kW K 00104 kWK Comment Following the flow of energy notice how the flux from the outside at 15oC grows a little when it arrives at 0oC this is due to entropy generation in the low T heat exchanger The flux out of the heat pump at 45oC is larger than the flux in which is due to entropy generation in the heat pump cycle COP is smaller than Carnot COP and finally this flux increases due to entropy generated in the high T heat exchanger as the energy arrives at room T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6168 A window receives 500 W of heat transfer at the inside surface of 20oC and transmits the 500 W from its outside surface at 2oC continuing to ambient air at 5oC Find the flux of entropy at all three surfaces and the windows rate of entropy generation Flux of entropy S T Q S inside 500 29315 W K 17056 WK S win 500 27515 W K 18172 WK S amb 500 26815 W K 18646 WK Window only S gen win S win S inside 18172 17056 0112 WK If you want to include the generation in the outside air boundary layer where T changes from 2oC to the ambient 5oC then chose the control volume as CV tot and it becomes S gen tot S amb S inside 18646 17056 0159 WK Window Inside Outside 20 C 2 C 5 C o o o CV tot Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6169 An amount of power say 1000 kW comes from a furnace at 800C going into water vapor at 400C From the water the power goes to a solid metal at 200C and then into some air at 70C For each location calculate the flux of s through a surface as Q T What makes the flux larger and larger Solution T1 T2 T3 T4 furnace vapor metal air FURNACE AIR FLOW 1 2 4 3 Flux of s Fs Q T with T as absolute temperature Fs1 1000107315 0932 kWK Fs2 100067315 1486 kWK Fs3 100047315 211 kWK Fs4 100034315 291 kWK 1S2 gen for every change in T Q over T is an irreversible process T 800 400 200 70 C T amb 1073 673 476 646 K QT 0932 1486 2114 2915 kWK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6170 Room air at 23oC is heated by a 2000 W space heater with a surface filament temperature of 700 K shown in Fig P6170 The room at steady state looses the power to the outside which is at 7oC Find the rates of entropy generation and specify where it is made Solution For any CV at steady state the entropy equation as a rate form is Eq643 dScv dt 0 dQ T S gen CV Heater Element S gen dQ T 2000700 2857 WK CV Space between heater 700 K and room 23C S gen dQ T 2000 700 2000 23273 39 WK CV Wall between 23C inside and 7C outside S gen dQ T 2000 23273 2000 7 273 0389 WK Notice biggest S gen is for the largest change in 1T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6171 A car engine block receives 2 kW at its surface of 450 K from hot combustion gases at 1500 K Near the cooling channel the engine block transmits 2 kW out at its 400 K surface to the coolant flowing at 370 K Finally in the radiator the coolant at 350 K delivers the 2 kW to air which is at 25 C Find the rate of entropy generation inside the engine block inside the coolant and in the radiatorair combination For a CV at steady state we have the entropy equation as a rate form as Eq643 dScv dt 0 dQ T S gen CV1 Engine block receives 2 kW at 450 K and it leaves at 400 K S gen1 dQ T 2000 450 2000 400 0555 WK CV2 The coolant receives 2 kW at 370 K andf gives it out at 350 K S gen2 dQ T 2000 370 2000 350 0309 WK CV3 Radiator to air heat transfer S gen3 dQ T 2000 350 2000 29815 0994 WK Notice the biggest S gen is for the largest change 1T Gases Steel Glycol Air flow Radiator Remark The flux of S is Q T flowing across a surface Notice how this flux increases as the heat transfer flows towards lower and lower T T K 1500 450 370 29815 Q T WK 133 444 540 671 CV1 CV2 CV3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6172 A farmer runs a heat pump using 2 kW of power input It keeps a chicken hatchery at a constant 30oC while the room loses 10 kW to the colder outside ambient at 10oC Find the COP of the heat pump the rate of entropy generated in the heat pump and its heat exchangers and the rate of entropy generated in the heat loss process Solution CV Hatchery steady state To have steady state at 30oC for the hatchery Energy Eq 0 Q H Q Loss Q H Q Loss 10 kW COP COP Q H W 10 2 5 CV Heat pump steady state Energy eq 0 Q L W Q H Q L Q H W 8 kW Entropy Eq 0 Q L TL Q H TH S gen HP S gen HP Q H TH Q L TL 10 273 30 8 273 10 000473 kWK CV From hatchery at 30oC to the ambient 10oC This is typically the walls and the outer thin boundary layer of air Through this goes Q Loss Entropy Eq 0 Q Loss TH Q Loss Tamb S gen walls S gen walls Q Loss Tamb Q Loss TH 10 283 10 303 000233 kWK Q loss Q Q H L W 2 kW HP cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6173 An insulated cylinderpiston has an initial volume of 015 m3 and contains steam at 400 kPa 200oC The steam is expanded adiabaticly and the work output is measured very carefully to be 30 kJ It is claimed that the final state of the water is in the twophase liquid and vapor region What is your evaluation of the claim Solution CV Water Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq63 ms2 s1 dQT Process 1Q2 0 and reversible State 1 T P Table B13 v1 05342 u1 26468 s1 71706 kJkg K m V1 v1 015 05342 02808 kg With the assumed reversible process we have from entropy equation s2 s1 71706 kJkg K and from the energy equation u2 u1 1W2m 26468 30 02808 25400 kJkg State 2 given by u s check Table B11 sG at uG 2540 70259 s1 State 2 must be in superheated vapor region 1 T s P 1 130 o C 70259 u 2540 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6174 A piston cylinder has a water volume separated in VA 02 m3 and VB 03 m3 by a stiff membrane The initial state in A is 1000 kPa x 075 and in B it is 1600 kPa and 250C Now the membrane ruptures and the water comes to a uniform state at 200C with heat transfer from a 250C source Find the work the heat transfer and the total entropy generation in the process Take the water in A and B as CV Continuity m2 m1A m1B 0 Energy m2u2 m1Au1A m1Bu1B 1Q2 1W2 Entropy Eq m2 s2 mAsA1 mBsB1 1Q2Tres 1S2 gen Process P2 Peq constant P1A as piston floats and mp Po are constant State 1A Two phase Table B12 v1A 0001127 075 019332 0146117 m3kg u1A 76167 075 182197 212815 kJkg s1A 21386 075 44478 547445 kJkgK State 1B v1B 014184 m3kg u1B 269226 kJkg s1B 66732 kJkgK m1A V1Av1A 13688 kg m1B V1Bv1B 2115 kg State 2 1000 kPa 200oC sup vapor v2 020596 m3kg u2 26219 kJkg s2 66939 kJkgK m2 m1A m1B 34838 kg V2 m2v2 34838 020596 07175 m3 So now 1W2 P dV Peq V2 V1 1000 07175 05 2175 kJ 1Q2 m2u2 m1Au1A m1Bu1B 1W 2 34838 26219 13688 212815 2115 269226 2175 744 kJ 1S2 gen m2s2 mAsA1 mBsB1 1Q2Tres 34838 66939 13688 547445 2115 66732 744 52315 02908 kJK AH2O P o cb BH2O g p m Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6175 The water in the two tanks of Problem 3214 receives the heat transfer from a reservoir at 300oC Find the total entropy generation due to this process Two rigid tanks are filled with water Tank A is 02 m3 at 100 kPa 150oC and tank B is 03 m3 at saturated vapor 300 kPa The tanks are connected by a pipe with a closed valve We open the valve and let all the water come to a single uniform state while we transfer enough heat to have a final pressure of 300 kPa Give the two property values that determine the final state and heat transfer Take CV total A B out to reservoir neglect kinetic and potential energy Energy Eq m2 u2 mAuA1 mBuB1 1Q2 1W2 1Q2 Entropy Eq m2 s2 mAsA1 mBsB1 1Q2Tres 1S2 gen State A1 u 258275 kJkg v 193636 m3kg s 76133 kJkgK mA1 Vv 02193636 01033 kg State B1 u 254355 kJkg v 060582 m3kg s 69918 kJkgK mB1 Vv 03 060582 04952 kg The total volume and mass is the sum of volumes mass for tanks A and B m2 mA1 mB1 01033 04952 05985 kg V2 VA1 VB1 02 03 05 m3 v2 V2m2 05 05985 08354 m3kg State 2 P2 v2 300 kPa 08354 m3kg T2 27476C and u2 276732 kJkg s 760835 kJkgK From energy eq 1Q2 05985 276732 01033 258275 04952 254355 1299 kJ From entropy equation 1S2 gen m2 s2 mAsA1 mBsB1 1Q2Tres 05985 760835 01033 76133 04952 69918 1299 27315 300 00782 kJK B A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6176 A steel pistoncylinder of 1 kg contains 25 kg ammonia at 50 kPa 20oC Now it is heated to 50oC at constant pressure through the bottom of the cylinder from external hot gas at 200oC and we assume the steel has the same temperature as the ammonia Find the heat transfer from the hot gas and the total entropy generation Solution CV Ammonia plus space out to the hot gas Energy Eq35 mNH3u2 u1 msteelu2 u1 1Q2 1W2 Entropy Eq637 S2 S1 dQT 1S2gen 1Q2 Tgas 1S 2 gen S2 S1 mNH3s2 s1 msteels2 s1 Process P C 1W2 PmNH3v2 v1NH3 State 1 B22 v1 24463 m3kg h1 14346 kJkg s1 63187 kJkg K State 2 B22 v2 31435 m3kg h2 15835 kJkg s2 68379 kJkg K Substitute the work into the energy equation and solve for the heat transfer 1Q2 mNH3h2 h1 msteelu2 u1 25 15835 14346 1 046 50 20 40445 kJ 1S2 gen mNH3s2 s1 msteels2 s1 1Q2Tgas 25 68379 63187 1 046 ln32315 25315 40445 47315 0555 kJK P v 1 2 2 1 T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6177 Water in a pistoncylinder is at 1 MPa 500C There are two stops a lower one at which Vmin 1 m3 and an upper one at Vmax 3 m3 The piston is loaded with a mass and outside atmosphere such that it floats when the pressure is 500 kPa This setup is now cooled to 100C by rejecting heat to the surroundings at 20C Find the total entropy generated in the process CV Water Initial state Table B13 v1 035411 m3kg u1 31243 s1 77621 m Vv1 3035411 8472 kg P v 1 2 2 1 T s v C 500 1000 Final state 100C and on line in PV diagram Notice the following vg500 kPa 03749 v1 v1 vg154C Tsat500 kPa 152C T2 so now piston hits bottom stops State 2 v2 vbot Vbotm 0118 m3kg x2 0118 0001044167185 00699 u2 41891 00699208758 56498 kJkg s2 13068 006996048 173 kJkg K Now we can do the work and then the heat transfer from the energy equation 1W2 PdV 500V2 V1 1000 kJ 1w2 118 kJkg 1Q2 mu2 u1 1W2 226834 kJ 1q2 26775 kJkg Take CV total out to where we have 20C ms2 s1 1Q2T0 Sgen Sgen ms2 s1 1Q2T0 8472 173 77621 22683 29315 2627 kJK Swater Ssur Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6178 A pistoncylinder contains air at 300 K 100 kPa A reversible polytropic process with n 13 brings the air to 500 K Any heat transfer if it comes in is from a 325oC reservoir and if it goes out it is to the ambient at 300 K Sketch the process in a Pv and a Ts diagram Find the specific work and specific heat transfer in the process Find the specific entropy generation external to the air in the process Solution Process Pvn C P EA Ann1 AE A 500300 A1303E A 9148 2P1 T2 T1 A1E AwA2E A P dv P2v2P1v1 1n A R 1 nE A TA2E ATA1E A A 0287 1 13E A 500 300 1913 kJkg Energy equation A1E AqA2E A uA2E A uA1E A A1E AwA2E A CAvE A TA2E A TA1E A A1E AwA2E 0717 500 300 1913 4793 kJkg The A1E AqA2E A is negative and thus goes out Entropy is generated between the air and ambient sA2E A sA1E A A1E AqA2E ATAambE A A1E AsA2 genE A1E AsA2 genE A sA2E A sA1E A A1E AqA2E ATAambE A CApE A ln TA2E ATA1E A R ln PA2E APA1E A A1E AqA2E ATAambE A1E AsA2 genE A 1004 ln A500 300E A 0287 ln 9148 A 4793 300E A 051287 0635285 015977 003736 kJkg K P v 1 2 T s 1 2 300 500 100 915 Notice n 13 k 14 n k Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6179 Assume the heat transfer in problem 3213 came from a 200AoE AC reservoir What is the total entropy generation in the process CV Water in A and B Control mass goes through process 1 2 Continuity Eq mA2E A mAA1E A mAB1E A 0 mA2E A mAA1E A mAB1E A 05 05 1 kg Energy Eq UA2E A UA1E A A1E AQA2E A A1E AWA2E A Entropy Eq mA2E A sA2E A mAAE AsAA1E A mABE AsAB1E A A1E AQA2E ATAresE A A1E ASA2 genE A State A1 vAA1E A 0001067 xAA1E A 071765 0072832 VAA1E A mv 0036416 mA3E uAA1E A 53508 01 200214 73522 kJkg sAA1E A 16072 01 54455 215175 kJkgK State B1 vAB1E A 15493 mA3E Akg uAB1E A 296669 kJkg sAB1E A 82217 kJkgK VAB1E A mvAB1E A 077465 mA3E State 2 If VA2E A VAA1E A then PA2E A 200 kPa that is the piston floats For TA2E A PA2E A 150C 200 kPa superheated vapor uA2E A 257687 kJkg vA2E A 095964 mA3E Akg sA2E A 72795 kJkgK VA2E A mA2E AvA2E A 095964 mA3E A VAA1E A checks OK Process A1E AWA2E A PA2E A VA2E A VA1E A 200 095964 077465 0036416 29715 kJ From the energy and entropy equations A1E AQA2E A mA2E AuA2E A mAA1E AuAA1E A mAB1E AuAB1E A A1E AWA2E 1 257687 05 735222 05 296669 29715 75563 kJ A1E ASA2 genE A mA2E A sA2E A mAAE AsAA1E A mABE AsAB1E A A1E AQA2E ATAresE 1 72795 05 215175 05 82217 7556347315 0496 kJK The possible state 2 PV combinations are shown State a is 200 kPa vAaE A VAA1E AmA2E A 0036 and thus twophase TAaE A 120C less than TA2E V P 2 150 C a 467 200 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6180 A closed tank V 10 L containing 5 kg of water initially at 25C is heated to 150C by a heat pump that is receiving heat from the surroundings at 25C Assume that this process is reversible Find the heat transfer to the water and the work input to the heat pump CV Water from state 1 to state 2 Process constant volume reversible isometric State 1 v1 Vm 0002 x1 0002 000100343358 0000023 u1 10486 000002323049 10493 kJkg s1 03673 000002381905 036759 kJkg K Continuity eq same mass and V C fixes v2 2 T2 v2 v1 x2 0002 0001090039169 00023233 u2 63166 00023233192787 63614 kJkg s2 18417 0002323349960 18533 kJkg K Energy eq has W 0 thus provides heat transfer as 1Q2 mu2 u1 2656 kJ Entropy equation for the total tank plus heat pump control volume gives for a reversible process ms2 s1 QLT0 QL mT0s2 s1 5 29815 18533 036759 22148 kJ and then the energy equation for the heat pump gives WHP 1Q2 QL 2656 22148 4412 kJ Q W L Q T amb HP 1 2 HP Water P v 1 2 T 2 1 T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6181 A resistor in a heating element is a total of 05 kg with specific heat of 08 kJkgK It is now receiving 500 W of electric power so it heats from 20oC to 180oC Neglect external heat loss and find the time the process took and the entropy generation CV Heating element Energy Eq mu2 u1 1W2 in W electrical in t Entropy Eq ms2 s1 0 1S2 gen no heat transfer t mu2 u1 W electrical in m C T2 T1 W electrical in 05 kg 800 JkgK 180 20 K 500 Js 128 s 1S2 gen ms2 s1 m C ln T2 T1 05 kg 08 kJkgK ln 180 273 20 273 0174 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6182 Two tanks contain steam and they are both connected to a pistoncylinder as shown in Fig P6182 Initially the piston is at the bottom and the mass of the piston is such that a pressure of 14 MPa below it will be able to lift it Steam in A is 4 kg at 7 MPa 700C and B has 2 kg at 3 MPa 350C The two valves are opened and the water comes to a uniform state Find the final temperature and the total entropy generation assuming no heat transfer Solution Control mass All water mA mB Continuity Eq m2 mA mB 6 kg Energy Eq35 m2u2 mAuA1 mBuB1 1Q2 1W2 1W2 Entropy Eq637 m2s2 mAsA1 mBsB1 1S 2 gen B13 vA1 006283 uA1 34485 sA1 73476 VA 02513 m3 B13 vB1 009053 uB1 28437 sB1 67428 VB 01811 m3 The only possible P V combinations for state 2 are on the two lines Assume V2 VA VB P2 Plift 1W2 P2V2 VA VB Substitute into energy equation m2h2 mAuA1 mBuB1 P2VA VB 4 34485 2 28437 1400 04324 20 0868 kJ State 2 h2 m2h2m2 20 0868 kJ 6 kg 33478 kJkg P2 1400 kPa v2 02323 m3kg s2 7433 kJkgK T2 4419 C Check assumption V2 m2v2 1394 m3 VA VB OK 1S2 gen 6 7433 4 73476 2 67428 1722 kJK P V 2 2 A1 T s 1400 B1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6183 Assume the heat source in Problem 3217 is at 300C in a setup similar to Fig P6182 Find the heat transfer and the entropy generation CV A B C Only work in C total mass constant m2 m1 0 m2 mA1 m B1 U2 U1 1Q2 1W2 S2 S1 1Q2Tres 1S2 gen 1W2 PdV Plift V2 V1 1A v 063 02 m3kg xA1 02 00010608908 0223327 u 50348 x202576 95589 kJkg s 15275 x5602 27786 kJkgK 1B v 035202 m3kg u 263891 kJkg s 69665 kJkgK mB1 04035202 11363 kg and m2 3 11363 41363 kg V2 VA VB VC 1 m3 VC Locate state 2 Must be on PV lines shown State 1a 800 kPa v1a VAVB m 024176 m3kg 800 kPa v1a T 173C too low Assume 800 kPa 250C v 029314 m3kg v1a OK V2 m2v2 41363 kg 029314 m3kg 12125 m3 Final state is 800 kPa 250C u2 271546 kJkg s2 70384 kJkgK 1W2 Plift V2 V1 800 kPa 12125 1 m3 170 kJ 1Q2 m2u2 m1u1 1W2 m2u2 mA1uA1 mB1uB1 1W2 41363 271546 3 95589 11363 263891 170 11 232 28677 29986 170 5536 kJ 1S2 gen S2 S1 1Q2Tres m2s2 mA1sA1 mB1sB1 1Q2T res 41363 70384 3 27786 11363 69665 553657315 3202 kJK A B C V P 2 1a P2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6184 A cylinder fitted with a piston contains 05 kg of R134a at 60C with a quality of 50 percent The R134a now expands in an internally reversible polytropic process to ambient temperature 20C at which point the quality is 100 percent Any heat transfer is with a constanttemperature source which is at 60C Find the polytropic exponent n and show that this process satisfies the second law of thermodynamics Solution CV R134a Internally Reversible Polytropic Expansion PVn Const ContEq m2 m1 m Energy Eq mu2 u1 1Q2 1W2 Entropy Eq ms2 s1 dQT 1S 2 gen State 1 T1 60oC x1 05 Table B51 P1 Pg 16818 kPa v1 vf x1vfg 0000951 050010511 0006207 m3kg s1 sf x1sfg 12857 0504182 14948 kJkg K u1 uf x1ufg 28619 0512166 3471 kJkg State 2 T2 20oC x2 10 P2 Pg 5728 kPa Table B51 v2 vg 003606 m3kg s2 sg 17183 kJkgK u2 ug 38919 kJkg Process PVn Const P1 P2 v2 v1 n n ln P1 P2 ln v2 v1 06122 1W2 PdV P2V2 P1V1 1n 055728 003606 16818 00062071 06122 132 kJ 2nd Law for CV R134a plus wall out to source 1S2 gen ms2 s1 QH TH Check Snet 0 QH 1Q2 mu2 u1 1W2 342 kJ 1S2 gen 0517183 14948 34233315 00092 kJK 1S2 gen 0 Process Satisfies 2nd Law Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6185 A device brings 2 kg of ammonia from 150 kPa 20oC to 400 kPa 80oC in a polytropic process Find the polytropic exponent n the work and the heat transfer Find the total entropy generated assuming a source at 100oC Solution CV Ammonia of constant mass m2 m1 m out to source Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 dQT 1S2 gen 1Q2T 1S 2 gen Process P1v1 n P2v2 n Eq 827 State 1 Table B22 v1 079774 m3kg s1 57465 kJkg K u1 13033 kJkg State 2 Table B22 v2 04216 m3kg s2 59907 kJkg K u2 14680 kJkg ln P2P1 ln v1v2n n ln v1v2 ln 400 150 n ln 079774 04216 098083 n 063773 n 1538 The work term is integration of PdV as done in text leading to Eq629 1W2 m 1 n P2v2 P1v1 2 1 1538 400 04216 150 079774 18208 kJ Notice we did not use Pv RT as we used the ammonia tables 1Q2 mu2 u1 1W2 2 1468 13033 18208 1473 kJ From Eq637 1S2 gen ms2 s1 1Q2T 2 59907 57465 1473 37315 00936 kJK P v 1 2 T s 1 2 20 80 150 400 Notice n 154 k 13 n k Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6186 A rigid tank with 05 kg ammonia at 1600 kPa 160oC is cooled in a reversible process by giving heat to a reversible heat engine that has its cold side at ambient 20oC shown in Fig P6186 The ammonia eventually reaches 20oC and the process stops Find the heat transfer from the ammonia to the heat engine and the work output of the heat engine W Q H Q L HE CV total NH 3 Ambient CV Ammonia Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq63 ms2 s1 dQT T not constant Process v constant 1W2 0 State 1 T P Table B22 u1 15961 kJkg v1 012662 m3kg s1 57485 kJkgK State 2 T2 and v2 v1 Table B21 as v2 vg 2phase P2 Psat 8575 kPa x2 v2 vfvfg 012662 0001638014758 0846876 u2 uf x2 ufg 27289 0846876 10593 1170 kJkg s2 sf x2 sfg 10408 0846876 40452 44666 kJkgK From the energy equation QH 1Q2 mu2 u1 051170 15961 21305 kJ Take now CV total ammonia plus heat engine out to ambient Entropy Eq63 ms2 s1 QL Tamb QL mTamb s2 s1 05 29315 44666 57485 18789 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Now the CV heat engine can give the engine work from the energy equation Energy HE WHE QH QL 21305 18789 252 kJ Notice to get 1q2 T ds we must know the function Ts which we do not readily have for this process v P s T 2 1 1 2 v C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6187 A pistoncylinder with constant loading of piston contains 1 L water at 400 kPa quality 15 It has some stops mounted so the maximum possible volume is 11 L A reversible heat pump extracting heat from the ambient at 300 K 100 kPa heats the water to 300C Find the total work and heat transfer for the water and the work input to the heat pump Solution Take CV around the water and check possible PV combinations State 1 v1 0001084 015046138 007029 m3kg u1 60429 015 194926 89668 kJkg s1 17766 015 51193 25445 kJkg K m1 V1v1 0001007029 00142 kg Q W L Q T amb HP 1 2 HP water State a v 11 v1 077319 m3kg 400 kPa Sup vapor Ta 400oC T2 P V 1 2 2 1 T s a a v C State 2 Since T2 Ta then piston is not at stops but floating so P2 400 kPa T P v2 065484 m3kg V2 v2v1 V1 9316 L 1W2 P dV PV2 V1 400 kPa 9316 1 0001 m3 333 kJ 1Q2 mu2 u1 1W2 00142 28048 89668 333 3043 kJ Take CV as water plus the heat pump out to the ambient ms2 s1 QLTo QL mTo s2 s1 00142 kg 300 K 75661 25445 kJkgK 2139 kJ WHP 1Q2 QL 3043 2139 904 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6188 An uninsulated cylinder fitted with a piston contains air at 500 kPa 200C at which point the volume is 10 L The external force on the piston is now varied in such a manner that the air expands to 150 kPa 25 L volume It is claimed that in this process the air produces 70 of the work that would have resulted from a reversible adiabatic expansion from the same initial pressure and temperature to the same final pressure Room temperature is 20C a What is the amount of work claimed b Is this claim possible Solution CV Air R 0287 kJkgK Cp 1004 kJkg K Cv 0717 kJkg K State 1 T1 200oC P1 500 kPa V1 10 L 001 m3 m1 V1v1 P1V1RT1 00368 kg State 2 P2 150 kPa V2 25 L 0025 m3 ηs 70 Actual Work is 70 of Isentropic Work a Assume Reversible and Adiabatic Process s1 s2s T2s T1 P2 P1 k1 k 47315 150 500 3354 K Energy Eq 1Q2s mu2s u1 1W2s 1Q2s 0 Assume constant specific heat 1W2 s mCvT1 T2s 363 kJ 1W2 ac 071W2 s 254 kJ b Use Ideal Gas Law T2 ac T1P2V2 P1V1 3549 K Energy Eq 1Q2 ac mCvT2 ac T1 1W2 ac 058 kJ 2nd Law 1S2 gen ms2 s1 Qcv To QCV 1Q2 ac To 20oC s2 s1 Cp ln T2 T1 R ln P2 P1 00569 kJkgK 1S2 gen 000406 kJK 0 Process is Possible Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6189 A small halogen light bulb receives an electrical power of 50 W The small filament is at 1000 K and gives out 20 of the power as light and the rest as heat transfer to the gas which is at 500 K the glass is at 400 K All the power is absorbed by the room walls at 25oC Find the rate of generation of entropy in the filament in the total bulb including glass and the total room including bulb Solution W el 50 W Q RAD 10 W Q COND 40 W glass leads g a s Radiation Conduction We will assume steady state and no storage in the bulb air or room walls CV Filament steadystate Energy Eq331 dEcvdt 0 W el Q RAD Q COND Entropy Eq642 dScvdt 0 Q RAD TFILA Q COND TFILA S gen S gen Q RAD Q CONDTFILA W elTFILA 50 1000 005 WK CV Bulb including glass Q RAD leaves at 1000 K Q COND leaves at 400 K S gen dQ T 101000 40400 011 WK CV Total room All energy leaves at 25C Eq531 dEcvdt 0 W el Q RAD Q COND Eq642 dScvdt 0 Q TOT TWALL S gen S gen EA AQ ATOT A ETAWALL AE A 5025273 0168 WK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Solutions using the Pr and vr functions in Table A72 If you would like to see more of these please let me know clausumichedu and I can prepare more of the problem solutions using these functions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 692 uses Pr function A pistoncylinder shown in Fig P692 contains air at 1380 K 15 MPa with VA1E A 10 cmA3E A Acyl 5 cm2 The piston is released and just before the piston exits the end of the cylinder the pressure inside is 200 kPa If the cylinder is insulated what is its length How much work is done by the air inside Solution CV Air Cylinder is insulated so adiabatic Q 0 Continuity Eq mA2E A mA1E A m Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E A A1E AWA2E Entropy Eq637 msA2E A sA1E A dQT A1E ASA2 genE A 0 A1E ASA2 genE State 1 TA1E A PA1E A State 2 PA2E A So one piece of information is needed for the assume reversible process A1E ASA2 genE A 0 sA2E A sA1E A 0 State 1 Table A71 uA1E A 10952 kJkg Table A72 PAr1E A 34053 vAr1E A 27024 m PA1E AVA1E ARTA1E A A15000 10106 E0287 1380E A 0000379 kg State 2 PA2E A and from Entropy eq sA2E A sA1E A PAr2E A PAr1E APA2E APA1E A 3405320015000 45404 Interpolate in A72 to match the PAr2E A value TA2E A 447 K uA2E A 32085 kJkg vAr2E A 6567 VA2E A VA1E AvAr2E AvAr1E A 10 6567 27024 243 cmA3E A LA2E A VA2E A Acyl 2435 486 cm A1E AwA2E A uA1E A uA2E A 7744 kJkg A1E AWA2E A mA1E AwA2E A 02935 kJ We could also have done VA2E A VA1E A TA2E APA1E ATA1E APA2E A from ideal gas law and thus did not need the vr function for this problem Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6105 uses vr function A pistoncylinder contains air at 300 K 100 kPa It is now compressed in a reversible adiabatic process to a volume 7 times as small Use constant heat capacity and find the final pressure and temperature the specific work and specific heat transfer for the process Solution Here we use the vArE A function from Table A72 Expansion ratio vA2E A vA1E A 17 Process eq Rev adiabatic and ideal gas gives PvAnE A C with n k Since we know the v ratio and s is constant we use the vArE A function vAr1E A 17949 vAr2E A vAr1E A vA2E A vA1E A 179497 25641 Table A72 Interpolate TA2E A 6407 K PA2E A PA1E A TA2E A TA1E A vA1E AvA2E A 100 6407300 7 1495 kPa Adiabatic A1E AqA2E A 0 kJkg Polytropic process work term from the energy equation A1E AwA2E A uA2E A uA1E A 46637 21436 2520 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6 additional problem uses Pr function A mass of 1 kg of air contained in a cylinder at 15 MPa 1000 K expands in a reversible adiabatic process to 100 kPa Calculate the final temperature and the work done during the process using a Constant specific heat value from Table A5 b The ideal gas tables Table A7 Solution CV Air Continuity Eq mA2E A mA1E A m Energy Eq35 muA2E A uA1E A A1E AQA2E A A1E AWA2E A Entropy Eq637 msA2E A sA1E A dQT A1E ASA2 genE A Process A1E AQA2E A 0 A1E ASA2 genE A 0 sA2E A sA1E a Using constant Cp from Table A5 gives the power relation Eq623 TA2E A TA1E APA2E APA1E AA k1 k E A 1000A 01 15 E A 0286E A 4609 K A1E AWA2E A UA2E A UA1E A mCAVoE ATA1E A TA2E A 1 kg 0717 kjkgK 1000 4609 K 3865 kJ b Use the tabulated reduced pressure function that includes variable heat capacity from A72 so since sA2E A sA1E A we have PAr2E A PAr1E A PA2E APA1E A 9165 A01 15E A 611 Interpolation gives TA2E A 486 K and uA2E A 3494 kJkg A1E AWA2E A muA1E A uA2E A 1 kg 7592 3494 kJkg 4098 kJ ENGLISH UNIT PROBLEMS UPDATED JULY 2013 SOLUTION MANUAL CHAPTER 6 English units Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CHAPTER 6 SUBSECTION PROB NO ConceptStudy Guide Problems 190191 Entropy Clausius 192195 Reversible Processes 196206 Entropy of a Liquid or Solid 207212 Entropy of Ideal Gases 213219 Polytropic Processes 220221 Entropy Generation 222238 Rates or Fluxes of Entropy 239242 Review problems 243245 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Concept Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6190E Water at 20 psia 240 F receives 40 Btulbm in a reversible process by heat transfer Which process changes s the most constant T constant v or constant P ds dq T Look at the constant property lines in a Ts diagram Fig 65 The constant v line has a higher slope than the constant P line also at positive slope Thus both the constant P and v processes have an increase in T As T goes up the change in s is smaller for the same area heat transfer under the process curve in the Ts diagram as compared with the constant T process The constant T isothermal process therefore changes s the most In a reversible process the area below the process curve in the Ts diagram is the heat transfer 2 1 T s q Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6191E Saturated water vapor at 20 psia is compressed to 60 psia in a reversible adiabatic process Find the change in v and T Process adiabatic dq 0 Process reversible dsgen 0 Change in s ds dqT dsgen 0 0 0 thus s is constant Table F72 T1 22796 F v1 20091 ft3lbm s1 1732 Btulbm R Table F72 at 60 psia and s s1 1732 Btulbm R T 400 40 1732 17134 1736 17134 400 40 0823 4329 F v 8353 8775 8353 0823 8700 ft3lbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Entropy Clausius Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6192E Consider the steam power plant in Problem 5133E and show that this cycle satisfies the inequality of Clausius Solution Show Clausius dQ T 0 For this problem we have two heat transfer terms Boiler 1000 Btus at 1200 F 1660 R Condenser 580 Btus at 100 F 560 R dQ T QH TH QL TL 1000 1660 580 560 06024 10357 0433 Btus R 0 OK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6193E Find the missing properties of P v s and x for ammonia NH3 a T 190 F P 100 psia b T 80 F h 650 Btulbm c T 120 F v 16117 ft3lbm a Table F81 P Pg 708 psia superheated vapor so x undefined v 38804 401882 39496 ft3lbm s 13658 138342 13746 Btulbm R b Table F82 h hg 62962 Btulbm so superheated vapor x undefined found between 50 and 60 psia P 50 10 650 6514964957 65149 5776 psia v 65573 077604 54217 65573 5676 ft3lbm s 13588 077604 13348 13588 13402 Btulbm R c Table F81 v vg 10456 ft3lbm so superheated vapor Table F82 x undefined P 200 psia s 12052 Btulbm R v P s T a b c a b c Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6194E Find the missing properties and give the phase of the substance a H2O s 175 Btulbm R T 150 F h P x b H2O u 1350 Btulbm P 1500 lbfin2 T x s a Table F71 s sg 18683 BtulbmR so 2 phase P PsatT 3722 lbfin2 x s sfsfg 175 021516533 092845 h 11795 092845 10081 10539 Btulbm b Table F72 found between 1000 F and 1100 F x undefined T 1000 100 1350134043138716 134043 10205 F s 16001 02048 16398 16001 16083 Btulbm R b P v T s b a a T P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6195E Determine the missing property among P T s and x for R410A at a T 20 F v 31214 ft3lbm b T 60 F v 03121 ft3lbm c P 30 psia s 03425 BtulbmR a F91 v vg 14522 ft3lbm F92 superheated vapor so x is undefined very close to 20 psia s 02862 BtulbmR b F91 001451 vf v vg 03221 ft3lbm Twophase P Psat 18498 psia x v vfvfg 03121 001451 03076 096746 s sf x sfg 00744 096746 01662 02352 BtulbmR c Table F92 at 30 psia s sg so superheated vapor x is undefined and we find the state at T 160 F b P v T s b a a T P c c Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Reversible Processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6196E In a Carnot engine with water as the working fluid the high temperature is 500 F and as QL is received the water changes from saturated liquid to saturated vapor The water pressure at the low temperature is 147 lbfin2 Find TL cycle thermal efficiency heat added per poundmass and entropy s at the beginning of the heat rejection process T s 1 2 4 3 Constant T constant P from 1 to 2 Table F71 qH Tds T s2 s1 T s fg h2 h1 hfg 71476 Btulbm States 3 4 are twophase Table F71 TL T3 T4 212 F ηcycle 1 TLTH 1 212 45967 500 45967 0300 Table F81 s3 s2 sgTH 14335 Btulbm R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6197E Consider a Carnotcycle heat pump with R410A as the working fluid Heat is rejected from the R410A at 110 F during which process the R410A changes from saturated vapor to saturated liquid The heat is transferred to the R410A at 30 F a Show the cycle on a Ts diagram b Find the quality of the R410A at the beginning and end of the isothermal heat addition process at 30 F c Determine the coefficient of performance for the cycle a 1 2 3 4 110 30 T s Table F91 b State 3 is saturated liquid s4 s3 01115 Btulbm R 00526 x401955 x4 030128 State 2 is saturated vapor s1 s2 02261 Btulbm R 00526 x101955 x1 08875 c β qH wIN TH TH TL 56967 110 30 7121 depends only on the temperatures Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6198E Do Problem 6197E using refrigerant R134a instead of R410A a 1 2 3 4 110 30 T s b Table F101 State 3 is saturated liquid s4 s3 02882 Btulbm R 02375 x401749 x4 02899 State 2 is saturated vapor s1 s2 04087 Btulbm R 02375 x101749 x1 09788 c β qH wIN TH TH TL 56967 110 30 7121 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6199E R410A at 150 psia and 140 F is expanded in a piston cylinder to 75 psia 80 F in a reversible process Find the sign for both the work and the heat transfer for this process Solution 1w2 P dv so sign dv 1q2 T ds so sign ds F92 v1 05356 ft3lbm s1 02859 BtulbmR F92 v2 09911 ft3lbm s2 02857 BtulbmR dv 0 w is positive ds 0 0 q is negative nearly zero 2 1 P v T s 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6200E A pistoncylinder receives R410A at 75 psia and compresses it in a reversible adiabatic process to 300 psia 160 F Find the initial temperature CV R410A this is a control mass Energy Eq35 u2 u1 1q2 1w2 Entropy Eq63 s2 s1 dqT 1q2 T Process Adiabatic and reversible 1q2 0 so then s2 s 1 State 1 P1 s1 s2 02673 BtulbmR T1 20 40 20 02673 02595 02688 02595 3677 F 2 1 P v T s 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6201E A pistoncylinder contains 1 lbm of water at 40 psia 600 F and it now cools to 280 F in an isobaric process The heat goes into a heat engine which rejects heat to the ambient at 77 F shown in Fig P646 and the whole process is assumed reversible Find the heat transfer out of the water and the work given out by the heat engine CV H2O Energy Eq35 mu2 u1 1Q2 1W 2 Entropy Eq63 ms2 s1 dQT 0 Process P C W P dV PV2 V1 m P v2 v1 State 1 F72 s1 18621 BtulbmR h1 133343 Btulbm State 2 F73 s2 16857 BtulbmR h2 117659 Btulbm From the process equation and the energy equation 1Q2 mu2 u1 1W2 mh2 h1 1 lbm 117659 133343 Btulbm 15684 Btu CV Total Energy Eq35 mu2 u1 QL 1W2 W HE Entropy Eq63 ms2 s1 QLTamb 0 QL mTambs1 s2 1 lbm 5367 R 18621 16857 BtulbmR 9467 Btu Now the energy equation for the heat engine gives WHE 1Q2 QL 15684 9467 6217 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6202E A cylinder fitted with a piston contains ammonia at 120 F 20 quality with a volume of 60 in3 The ammonia expands slowly and during this process heat is transferred to maintain a constant temperature The process continues until all the liquid is gone Determine the work and heat transfer for this process CV Ammonia in the cylinder Energy Eq35 mu2 u1 1Q2 1W 2 Entropy Eq63 ms2 s1 dQT 0 Process T constant to x2 10 Table F81 P 2865 lbfin 2 W P dV PV2 V1 m P v2 v1 dQT 1Q2 T 1 2 T s 120 F NH 3 T1 120 F x1 020 V1 60 in3 v1 002836 02 10171 02318 ft3lbm s1 03571 02 07829 05137 Btulbm R m Vv 60 172802318 015 lbm State 2 Saturated vapor v2 1045 ft3lbm s2 1140 Btulbm R 1W2 2865144 778 015 1045 02318 647 Btu From the entropy equation 1Q2 T m s2 s1 5797 R 015 lbm 11400 05137 BtulbmR 5446 Btu or h1 17879 02 45384 26956 Btulbm h2 63263 Btulbm 1Q2 mh2 h1 01563263 26956 5446 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6203E One poundmass of water at 600 F expands against a piston in a cylinder until it reaches ambient pressure 147 lbfin2 at which point the water has a quality of 90 It may be assumed that the expansion is reversible and adiabatic a What was the initial pressure in the cylinder b How much work is done by the water Solution CV Water Process Rev Q 0 Energy Eq35 mu2 u1 1Q2 1W2 1W 2 Entropy Eq63 ms2 s1 dQT Process Adiabatic Q 0 and reversible s2 s 1 State 2 P2 147 lbfin2 x2 090 from Table F71 s2 03121 09 14446 16123 Btulbm R u2 1801 09 8975 9879 Btulbm State 1 Table F72 at T1 600 F s1 s2 P1 335 lbfin2 u1 12012 Btulbm From the energy equation 1W2 mu1 u2 1 lbm12012 9879 Btulbm 2133 Btu v P s T 2 1 1 2 T1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6204E A closed tank V 035 ft3 containing 10 lbm of water initially at 77 F is heated to 350 F by a heat pump that is receiving heat from the surroundings at 77 F Assume that this process is reversible Find the heat transfer to the water and the work input to the heat pump CV Water from state 1 to state 2 Process constant volume reversible isometric 1 v1 Vm 03510 0035 ft3lbm x1 2692105 u1 4511 Btulbm s1 008779 Btulbm R Continuity eq same mass and constant volume fixes v2 State 2 T2 v2 v1 x2 0035 001799 33279 000511 u2 32135 00051178845 32538 Btulbm s2 05033 0005111076 05088 Btulbm R Energy eq has zero work thus provides heat transfer as 1Q2 mu2 u1 1032538 4511 28027 Btu Entropy equation for the total control volume gives for a reversible process ms2 s1 QLT0 QL mT0s2 s1 HP Q WHP 1 2 Q L Tamb 10 lbm 77 45967R 05088 008779 BtulbmR 22594 Btu and the energy equation for the heat pump gives WHP 1Q2 QL 28027 22594 5433 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6205E A cylinder containing R134a at 60 F 30 lbfin2 has an initial volume of 1 ft3 A piston compresses the R134a in a reversible isothermal process until it reaches the saturated vapor state Calculate the required work and heat transfer to accomplish this process Solution CV R134a Continuity Eq m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq63 ms2 s1 dQT 1S 2 gen Process T constant reversible so 1S2 gen 0 State 1 T P Table F102 u1 16841 Btulbm s1 04321 Btulbm R m Vv1 117367 05758 lbm State 2 60 F sat vapor Table F101 u2 16628 Btulbm s2 04108 Btulbm R P v 1 2 T 2 1 T s As T is constant we can find Q by integration as 1Q2 Tds mTs2 s1 05758 lbm 5197 R 04108 04321 BtulbmR 6374 Btu The work is then from the energy equation 1W2 mu1 u2 1Q2 05758 lbm 16841 16628 Btulbm 6374 Btu 515 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6206E A rigid insulated vessel contains superheated vapor steam at 400 lbfin2 700 F A valve on the vessel is opened allowing steam to escape It may be assumed that the steam remaining inside the vessel goes through a reversible adiabatic expansion Determine the fraction of steam that has escaped when the final state inside is saturated vapor CV steam remaining inside tank Rev Adiabatic inside only ContEq m2 m1 m Energy Eq mu2 u1 1Q2 1W2 Entropy Eq ms2 s1 dQT 1S2 gen 0 0 P v 1 2 2 1 T s CV m2 State 1 Table F72 v1 16503 ft3lbm s1 16396 Btulbm R State 2 Table F71 s2 s1 16396 Btulbm R sg at P2 Interpolate in F71 P2 63861 lbfin2 v2 vg 6800 ft3lbm me m1 m1 m2 m1 1 m2 m1 1 v1 v2 1 16503 6800 0757 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Entropy of a Liquid or Solid Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6207E Two 5 lbm blocks of steel one at 500 F the other at 80 F come in thermal contact Find the final temperature and the change in the entropy of the steel CV Both blocks no external heat transfer C 011 BtulbmR from Table F2 Energy Eq U2 U1 mAu2 u1A mBu2 u1B 0 0 mACT2 TA1 mBCT2 TB1 T2 mA mB mATA1 mBTB1 1 2 TA1 1 2 TB1 290 F Entropy Eq637 S2 S1 mAs2 s1A mBs2 s1B 1S 2 gen Entropy changes from Eq611 S2 S1 mAC ln TA1 T2 mBC ln T2 TB1 5 011 ln 290 45967 500 45967 5 011 ln 290 45967 80 45967 013583 018077 00449 BtuR A B Heat transfer over a finite temperature difference is an irreversible process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6208E A foundry form box with 50 lbm of 400 F hot sand is dumped into a bucket with 2 ft3 water at 60 F Assuming no heat transfer with the surroundings and no boiling away of liquid water calculate the net entropy change for the masses CV Sand and water P const Energy Eq msandu2 u1sand mH2Ou2 u1H2O PV2 V1 msandhsand mH2OhH2O 0 mH2O 2 0016035 12473 lbm 50 lbm 019 BtulbmR T2 400 F 12473 lbm 10 BtulbmR T2 60 F 0 T2 84 F S2 S1 msand s2 s1 mH2O s2 s1 msand Csand lnT2T1 mH2O CH2O lnT2T1 50 019 ln 544 860 12473 10 ln 544 520 1293 BtuR Box holds the sand for form of the cast part Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6209E Heat transfer to a block of 3 lbm ice at 15 F melts it to liquid at 50 F in a kitchen Find the entropy change of the water Water changes state from nearly saturated solid to nearly saturated liquid The pressure is 1 atm but we approximate the state properties with saturated state at the same temperature State 1 Compressed saturated solid F74 s1 03093 BtulbmR State 2 Compressed saturated liquid F71 s2 00361 BtulbmR The entropy change is S ms2 s1 3 lbm 00361 03093 BtulbmR 10362 BtuR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6210E A rigid tank of 12 lbm steel contains 15 lbm of R134a at 100 F 80 psia The tank is placed in a refrigerator that brings it to 0 F Find the process heat transfer and the combined steel and R134a change in entropy CV The steel tank and the R134a The energy equation Eq 35 now becomes summing over the mass mst u2 u1st mR134a u2 u1R134a 1Q2 0 Process No change in volume so no work as used above Use specific heat from Table F2 for steel and Table F10 for R134a R134a v1 06617 ft3lbm u1 17406 Btulbm s1 04252 BtulbmR State 2 v2 v1 vg x2 v2 vfvfg 06617 001187 21340 0304513 u2 uf x2 ufg 7588 x2 8224 100923 Btulbm s2 sf x2 sfg 02178 x2 01972 027785 BtulbmR Now the heat transfer from the energy equation 1Q2 mR134au2 u1R134a mst Cst T2 T1 15 100923 17406 12 011 0 100 12291 Btu Steel msts2 s1st mstCst lnT2T1 12011 ln 4597 5597 002598 BtulbmR Entropy change for the total control volume steel and R134a S2 S1 mst s2 s1st mR134a s2 s1R134a 002598 15027785 04252 0247 BtuR 0 F Q 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6211E A 5lbm aluminum radiator holds 2 lbm of liquid R134a at 10 F The setup is brought indoors and heated with 220 Btu Find the final temperature and the change in entropy of all the mass Solution CV The aluminum radiator and the R134a Energy Eq35 m2u2 m1u1 1Q2 0 Process No change in volume so no work as used above The energy equation now becomes summing over the mass mal u2 u1al mR134a u2 u1R134a 1Q 2 Use specific heat from Table F2 and F3 malCal T2 T1 m R134aC R134a ln T2 T1 1Q 2 T2 T1 1Q2 malCal m R134aC R134a 220 5 0215 2 034 125 F T2 10 125 135 F Entropy change for solid F2 and liquid F3 from Eq611 S2 S1 mal s2 s1al mR134a s2 s1 R134a malCal ln T2T1 m R134aC R134a ln T2T1 5 0 215 2 034 ln 135 45967 10 45967 0414 BtuR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6212E Four pounds of liquid lead at 750 F are poured into a form It then cools at constant pressure down to room temperature at 68 F as heat is transferred to the room The melting point of lead is 620 F and the enthalpy change between the phases hif is 106 Btulbm The specific heats are in Table F2 and F3 Calculate the net entropy change for the lead Solution CV Lead constant pressure process mPbu2 u1Pb 1Q2 PV2 V1 We need to find changes in enthalpy u Pv for each phase separately and then add the enthalpy change for the phase change Cliq 0038 Btulbm R Csol 0031 Btulbm R Consider the process in several steps Cooling liquid to the melting temperature Solidification of the liquid to solid Cooling of the solid to the final temperature 1Q2 mPbh2 h1 mPbh2 h620sol hif h620f h750 4 0031 68 620 106 0038 620 750 6845 424 1976 13061 Btu S2 S1 mPb Cp sollnT210797 hif10797 CP liqln10797T1 4 0031 ln 5277 10797 106 10797 0038 ln 10796 12097 0145 BtuR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Entropy of Ideal Gases Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6213E Air inside a rigid tank is heated from 550 to 600 R Find the entropy increase s2 s1 What if it is from 2300 to 2350 R Process V C 1W2 Ø Entropy change from Eq617 a s2 s1 Cvo ln T1 T2 0171 ln 600 550 0015 BtulbmR b s2 s1 Cvo ln T1 T2 0171 ln 2350 2300 000368 BtulbmR From F5 case a Cv u T 686840 01717 BtulbmR see F4 0171 case b Cv u T 2153100 0215 BtulbmR 25 higher so result should have been 000463 BtulbmR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6214E R410A at 60 psia is brought from 60 F to 240 F in a constant pressure process Evaluate the change in specific entropy using Table F9 and using ideal gas with Cp 01935 BtulbmR Table F92 s1 02849 BtulbmR s2 03496 BtulbmR s2 s1 03496 02849 00647 BtulbmR Eq 616 s2 s1 Cpo ln T1 T2 01935 ln 240 460 60 460 00575 BtulbmR Two explanations for the difference are as the average temperature is higher than 77 F we could expect a higher value of the specific heat and secondly it is not an ideal gas if you calculate Z PvRT 094 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6215E Oxygen gas in a pistoncylinder at 500 R and 1 atm with a volume of 1 ft3 is compressed in a reversible adiabatic process to a final temperature of 1000 R Find the final pressure and volume using constant heat capacity from Table F4 and repeat with Table F6 Solution CV Air Assume a reversible adiabatic process Energy Eq35 u2 u1 0 1w2 Entropy Eq637 s2 s1 dqT 1s2 gen 0 Process Adiabatic 1q2 0 Reversible 1s2 gen 0 Properties Table F4 k 1393 With these two terms zero we have a zero for the entropy change So this is a constant s isentropic expansion process From Eq623 P2 P1 T2 T1 k k1 147 100050035445 1715 psia Using the ideal gas law to eliminate P from this equation leads to Eq624 V2 V1 T2 T1 1 1k 1 11393 1000 500 1 0171 ft 3 Using the ideal gas tables F6 we get s2 s1 so T2 so T1 R lnP2P1 or P2 P1 expso T2 so T1R P2 147 psia exp53475 484185198589 18755 psia V2 V1 T2 T1P1P2 1 ft3 1000 500 147 18755 0157 ft 3 P v 1 1 T s 1 P P 2 2 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6216E A handheld pump for a bicycle has a volume of 2 in3 when fully extended You now press the plunger piston in while holding your thumb over the exit hole so an air pressure of 45 lbfin2 is obtained The outside atmosphere is at Po To Consider two cases 1 it is done quickly 1 s and 2 it is done slowly 1 h a State assumptions about the process for each case b Find the final volume and temperature for both cases Solution CV Air in pump Assume that both cases result in a reversible process State 1 P0 T0 State 2 45 lbfin2 One piece of information must resolve the for a state 2 property Case I Quickly means no time for heat transfer Q 0 so a reversible adiabatic compression u2 u1 1w2 s2 s1 dqT 1s2 gen 0 With constant s and constant heat capacity we use Eq623 T2 T1 P2 P1 k1 k 5367 14 45 14696 04 7389 R Use ideal gas law PV mRT at both states so ratio gives V2 P1V1T2T1P2 0899 in3 Case II Slowly time for heat transfer so T constant T0 The process is then a reversible isothermal compression T2 T0 5367 R V2 V1P1P2 0653 in3 P v 1 2ii 2ii 1 T s 1 P P 2 2i 2i Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6217E A pistoncylinder contains air at 2500 R 2200 lbfin2 with V1 1 in3 Acyl 1 in2 as shown in Fig P692 The piston is released and just before the piston exits the end of the cylinder the pressure inside is 30 lbfin2 If the cylinder is insulated what is its length How much work is done by the air inside Solution CV Air Cylinder is insulated so adiabatic Q 0 Continuity Eq m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 1W 2 Entropy Eq614 ms2 s1 dQT 1S2 gen 0 1S 2 gen State 1 T1 P1 State 2 P2 So one piece of information is needed for the assume reversible process 1S2 gen 0 s2 s1 0 State 1 Table F5 u1 47433 Btulbm sT1 203391 Btulbm R m P1V1RT1 2200 10 5334 2500 12 1375 103 lbm State 2 P2 and from Entropy eq s2 s1 so from Eq619 sT2 sT1 R ln P2 P1 203391 5334 778 ln 30 2200 173944 Btulbm R Now interpolate in Table F5 to get T2 T2 840 40 173944 173463174653 173463 8162 R u2 137099 144114 137099 0404 13993 Btulbm V2 V1 T2 P1 T1P2 1 8162 2200 2500 30 2394 in3 L2 V2 Acyl 23941 2394 in 1W2 mu1 u2 1375 10347433 13993 046 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6218E A 25ft3 insulated rigid tank contains air at 110 lbfin2 75 F A valve on the tank is opened and the pressure inside quickly drops to 15 lbfin2 at which point the valve is closed Assuming that the air remaining inside has undergone a reversible adiabatic expansion calculate the mass withdrawn during the process CV Air remaining inside tank m2 ContEq m2 m Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 dQT 1S2 gen Process adiabatic 1Q2 0 and reversible 1S2 gen 0 P v 1 2 2 1 T s CV m 2 Entropy eq then gives s2 s1 and ideal gas gives the relation in Eq623 T2 T1P2P1 k1 k 535 R 15110 0286 3026 R m1 P1VRT1 110 psia 144 ft2 in2 25 ft35334 ftlbf lbmR 535 R 1388 lbm m2 P2VRT2 15 psia 144 ft2 in2 25 ft35334 ftlbf lbmR 3026 R 335 lbm me m1 m2 1053 lbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Polytropic Processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6219E A cylinderpiston contains 01 lbm methane gas at 15 psia 70 F The gas is compressed reversibly to a pressure of 120 psia Calculate the work required if the process is adiabatic Solution CV Methane gas of constant mass m2 m1 m and reversible process Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq63 37 ms2 s1 dQT 1S2 gen dQT Process 1Q2 0 s2 s 1 thus isentropic process s const and ideal gas gives relation in Eq623 T2 T1 P2P1 k1 k 5297 R 120 15 0230 8546 R 1W2 mCV0T2 T1 01 lbm 0415 BtulbmR 8546 5297 R 1348 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6220E Helium in a pistoncylinder at 70 F 20 psia is brought to 720 R in a reversible polytropic process with exponent n 125 You may assume helium is an ideal gas with constant specific heat Find the final pressure and both the specific heat transfer and specific work Solution CV Helium control mass Cv 0744 Btulbm R R 386 ft lbf lbm R Process Pvn C Pv RT Tvn1 C From the process equation and T1 70 460 530 R T2 720 R T1vn1 T2vn1 v2 v1 T1 T2 1n1 02936 P2 P1 v1 v2n 463 P2 694 lbfin2 The work is from Eq629 per unit mass 1w2 P dv C vn dv 1 1n P2 v2 P1 v1 R 1n T2 T1 386 778 025 BtulbmR 720 530 R 377 Btulbm The heat transfer follows from the energy equation 1q2 u2 u1 1w2 Cv T2 T1 1w2 0744720 530 377 2356 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6221E A cylinderpiston contains air at ambient conditions 147 lbfin2 and 70 F with a volume of 10 ft3 The air is compressed to 100 lbfin2 in a reversible polytropic process with exponent n 12 after which it is expanded back to 147 lbfin2 in a reversible adiabatic process a Show the two processes in Pv and Ts diagrams b Determine the final temperature and the net work a P T v s 1 2 3 1 2 3 P P 2 1 m P1V1RT1 147 144 10 5334 5297 07492 lbm b The process equation is expressed in Eq628 T2 T1P2P1 n1 n 5297 100 147 0167 7296 R The work is from Eq629 1w2 1 2 Pdv 1 n P2v2 P1v1 1 n RT2 T1 7781 120 53347296 5297 685 Btulbm Isentropic relation is from Eq623 T3 T2P3P2 k1 k 7297 R 147 100 0286 4216 R With zero heat transfer the energy equation gives the work 2w3 CV0T2 T3 01717296 4216 527 Btulbm wNET 07492685 527 118 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Entropy Generation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6222E Consider a heat transfer of 100 Btu from 2400 R hot gases to a steel container at 1200 R which has a heat transfer of the 100 Btu out to some air at 600 R Determine the entropy generation in each of the control volumes indicated in Fig P6115 There is no change in energy or entropy in the indicated control volumes so no storage effect There is a transfer of energy in and out of each CV and an associated transfer of entropy Take CV1 Take CV2 Energy Eq Energy Eq 0 Q Q 0 Q Q Entropy Eq Entropy Eq 0 Q TH Q TM Sgen CV1 0 Q TM Q TL S gen CV2 Sgen CV1 Q TM Q TH Sgen CV2 Q TL Q TM 100 1200 100 2400 00417 BtuR 100 600 100 1200 0083 BtuR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6223E A computer chip dissipates 2 Btu of electric work over time and rejects that as heat transfer from its 125 F surface to 70 F air How much entropy is generated in the chip How much if any is generated outside the chip CV1 Chip with surface at 125 F we assume chip state is constant Energy U2 U1 0 1Q2 1W2 Welectrical in Q out 1 Entropy S2 S1 0 Qout 1 Tsurf 1S 2 gen1 1S2 gen1 Qout 1 Tsurf Welectrical in Tsurf 2 Btu 125 4597 R 00034 BtuR CV2 From chip surface at 125 F to air at 70 F assume constant state Energy U2 U1 0 1Q2 1W2 Qout 1 Q out 2 Entropy S2 S1 0 Qout1 Tsurf Qout2 Tair 1S 2 gen2 1S2 gen2 Qout2 Tair Qout1 Tsurf 2 Btu 5297 R 2 Btu 5847 R 0000 36 BtuR 70 F air 125 F Q air flow cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6224E An insulated cylinderpiston contains R134a at 150 lbfin2 120 F with a volume of 35 ft3 The R134a expands moving the piston until the pressure in the cylinder has dropped to 15 lbfin2 It is claimed that the R134a does 180 Btu of work against the piston during the process Is that possible Solution CV R134a in cylinder Insulated so assume Q 0 State 1 Table F102 v1 03332 ft3lbm u1 17533 Btulbm s1 041586 Btulbm R m V1v1 3503332 10504 lbm Energy Eq35 mu2 u1 1Q2 1W2 0 180 u2 u1 1W2m 158194 Btulbm State 2 P2 u2 Table F102 T2 2 F s2 0422 Btulbm R Entropy Eq637 ms2 s1 dQT 1S2gen 1S2gen 1S2gen ms2 s1 10504 0422 041586 00645 BtuR This is possible since 1S2 gen 0 2 1 P v T s 1 2 s C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6225E Heat transfer from a 70 F kitchen to a block of 3 lbm ice at 15 F melts it to liquid at 50 F Find the entropy generation Water changes state from nearly saturated solid to nearly saturated liquid The pressure is 1 atm but we approximate the state properties with saturated state at the same temperature CV Ice out to the 20oC kitchen air Energy Eq35 mu2 u1 1Q2 1W2 1Q2 Entropy Eq637 ms2 s1 dQT 1S2 gen 1Q2T0 1S2 gen State 1 Compressed saturated solid F74 u1 15175 Btulbm s1 03093 BtulbmR State 2 Compressed saturated liquid F71 u2 3809 Btulbm s2 00746 BtulbmR Heat transfer from the energy Eq 1Q2 mu2 u1 3 3809 15175 56952 Btu From the entropy Eq 1S2 gen ms2 s1 1Q2T0 3 00746 03093 569525297 00765 BtuR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6226E A mass and atmosphere loaded pistoncylinder contains 4 lbm of water at 500 lbfin2 200 F Heat is added from a reservoir at 1200 F to the water until it reaches 1200 F Find the work heat transfer and total entropy production for the system and surroundings Solution CV Water out to surroundings at 1200 F This is a control mass Energy Eq35 U2 U1 1Q2 1W2 Entropy Eq637 ms2 s1 dQT 1S2 gen 1Q2Tres 1S 2 gen Process P constant so 1W2 PV2 V1 mPv2 v1 State 1 Table F73 v1 001661 ft3lbm h1 16918 Btulbm s1 02934 Btulbm R State 2 Table F72 v2 19518 ft3lbm h2 16298 Btulbm s2 18071 Btulbm R 2 1 P v T s 1 2 Work is found from the process area in PV diagram 1W2 mPv2 v1 4 50019518 001661 144 778 71637 Btu The heat transfer from the energy equation is 1Q2 U2 U1 1W2 mu2 u1 mPv2 v1 mh2 h1 1Q2 416298 16918 584248 Btu Entropy generation from entropy equation Eq637 1S2 gen ms2 s1 1Q2 Tres 418071 02934 584248 165967 2535 BtuR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6227E A 1 gallon jug of milk at 75 F is placed in your refrigerator where it is cooled down to the refrigerators inside temperature of 40 F Assume the milk has the properties of liquid water and find the entropy generated in the cooling process Solution CV Jug of milk Control mass at constant pressure Continuity Eq m2 m1 m Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 dQT 1S 2 gen State 1 Table F71 v1 vf 001606 ft3lbm h1 hf 43085 Btulbm sf 008395 Btulbm R State 2 Table F71 h2 hf 801 Btulbm s2 sf 00162 Btulbm R Process P constant 147 psia 1W2 mPv2 v1 V1 1 Gal 231 in3 m 231 001606 123 8324 lbm Substitute the work into the energy equation and solve for the heat transfer 1Q2 mh2 h1 8324 lbm 801 43085 Btulbm 292 Btu The entropy equation gives the generation as 1S2 gen ms2 s1 1Q2Trefrig 8324 00162 008395 292 500 0564 0584 002 BtuR MILK cb 40 F AIR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6228E A cylinderpiston contains water at 30 lbfin2 400 F with a volume of 1 ft3 The piston is moved slowly compressing the water to a pressure of 120 lbfin2 The loading on the piston is such that the product PV is a constant Assuming that the room temperature is 70 F show that this process does not violate the second law Solution CV Water cylinder out to room at 70 F Energy Eq35 mu2 u1 1Q2 1W 2 Entropy Eq637 ms2 s1 1Q2 Troom 1S2 gen Process PV constant Pmv v2 P1v1P 2 1w2 Pdv P1v1 lnv2v1 State 1 Table B13 v1 16891 ft3lbm u1 1144 Btulbm s1 17936 Btulbm R State 2 P2 v2 P1v1P2 30 16891120 4223 ft3lbm Table F73 T2 4254 F u2 11444 Btulbm s2 16445 BtulbmR 1w2 30 16891 144 778 ln 4223 16891 1300 Btu 1q2 u2 u1 1w2 11444 1144 130 1296 Btulbm 1s2gen s2 s1 1q2 Troom 16445 17936 1296 52967 00956 Btulbm R 0 satisfies entropy equation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6229E One pound mass of ammonia NH3 is contained in a linear springloaded pistoncylinder as saturated liquid at 0 F Heat is added from a reservoir at 225 F until a final condition of 200 lbfin2 160 F is reached Find the work heat transfer and entropy generation assuming the process is internally reversible Solution CV NH3 out to the reservoir Continuity Eq m2 m1 m Energy Eq35 E2 E1 mu2 u1 1Q2 1W 2 Entropy Eq637 S2 S1 dQT 1S2gen 1Q2Tres 1S 2gen Process P A BV linear in V 1W2 P dV 1 2 P1 P2V2 V1 1 2 P1 P2mv2 v1 State 1 Table F81 P1 304 psia v1 00242 ft3lbm u1 4246 Btulbm s1 00967 Btulbm R State 2 Table F82 sup vap P v 1 2 2 1 T s P2 v2 17807 ft3lbm u2 67736 Btulbm s2 12514 Btulbm R 1W2 1 2 304 200117807 00242144778 3745 Btu 1Q2 mu2 u1 1W2 167736 4246 3745 67235 Btu Sgen ms2 s1 1Q2Tres 112514 00967 67235 6847 0173 BtuR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6230E A hollow steel sphere with a 2ft inside diameter and a 01in thick wall contains water at 300 lbfin2 500 F The system steel plus water cools to the ambient temperature 90 F Calculate the net entropy change of the system and surroundings for this process CV Steel water out to ambient T0 This is a control mass Energy Eq U2 U1 1Q2 1W2 mH2Ou2 u1 msteelu2 u1 Entropy Eq S2 S1 dQT 1S2 gen 1Q2T0 1S2 gen Process V constant 1W2 0 Vsteel π 6200833 23 00526 ft3 msteel ρVsteel 490 00526 25763 lbm VH2O π6 23 4189 ft3 mH2O Vv 2372 lbm v2 v1 17662 0016099 x2 4677 x2 374103 u2 61745 Btulbm s2 01187 Btulbm R 1Q2 Usteel UH2O mCsteelT2T1 mH2Ou2 u1 25763 010790500 23726174 11595 1130 26039 3734 Btu S2 S1 msteels2 s1 mH2Os2 s1 25763 0107 ln550960 237201187 15701 4979 BtuR SSUR Q12T0 373454967 6793 BtuR 1S2 gen S2 S1 1Q2T0 SSYS SSUR 4979 6793 1814 BtuR Water Ambient Steel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6231E One lbm of air at 540 R is mixed with two lbm air at 720 R in a process at a constant 15 psia and Q 0 Find the final T and the entropy generation in the process CV All the air Energy Eq U2 U1 0 W Entropy Eq S2 S1 0 1S 2 gen Process Eq P C W PV2 V1 Substitute W into energy Eq U2 U1 W U2 U1 PV2 V1 H2 H1 0 Due to the low T let us use constant specific heat H2 H1 mAh2 h1A mBh2 h1 B mACpT2 TA1 mBCpT2 TB1 0 T2 mA mB mATA1 mBTB1 1 3 TA1 2 3 TB1 660 R Entropy change is from Eq 616 with no change in P and Table F4 for Cp 1S2 gen S2 S1 mACp ln T2 TA1 mBCp ln T2 TB1 1 024 ln 660 540 2 024 ln 660 720 00482 00418 00064 BtuR Remark If you check the volume does not change and there is no work Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6232E One lbm of air at 15 psia is mixed with two lbm air at 30 psia both at 540 R in a rigid insulated tank Find the final state P T and the entropy generation in the process CV All the air Energy Eq U2 U1 0 0 Entropy Eq S2 S1 0 1S 2 gen Process Eqs V C W 0 Q 0 States A1 B1 uA1 u B1 VA mART1PA1 VB mBRT1P B1 cb U2 U1 m2u2 mAuA1 mBuB1 0 u2 uA1 2uB13 u A1 State 2 T2 T1 540 R from u2 m2 mA mB 3 kg V2 m2RT1P2 VA VB mART1PA1 mBRT1P B1 Divide with mART1 and get 3P2 1PA1 2PB1 1 15 2 30 0133 psia1 P2 225 psia Entropy change from Eq 616 with the same T so only P changes 1S2 gen S2 S1 mAR ln PA1 P2 mBR ln P2 PB1 1 5334 ln 225 15 2 ln 225 30 5334 04055 05754 906 lbfftR 00116 BtuR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6233E A rigid container with volume 7 ft3 is divided into two equal volumes by a partition Both sides contain nitrogen one side is at 300 lbfin2 400 F and the other at 30 lbfin2 200 F The partition ruptures and the nitrogen comes to a uniform state at 160 F Assume the temperature of the surroundings is 68 F determine the work done and the net entropy change for the process Solution CV A B Control mass no change in volume 1W2 0 mA1 PA1VA1RTA1 300144 35 5515 8597 3189 lbm mB1 PB1VB1RTB1 30144 35 5515 6597 0416 lbm P2 mTOTRT2VTOT 3605 5515 6197144 7 1222 lbfin2 From Eq616 S2 S1 mA1 s2 s1A1 mB1 s2 s1 B1 3189 0249 ln 6197 8597 5515 778 ln 1222 300 0416 0249 ln 6197 6597 5515 778 ln 1222 30 00569 00479 01048 BtuR 1Q2 mA1u2 u1 mB1u2 u1 3189 0178160 400 0416 0178160 200 1392 Btu From Eq637 1S2 gen S2 S1 1Q2T0 01048 BtuR 1392 Btu 5277 R 0159 BtuR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6234E A constant pressure pistoncyl is 2 lbm of steel and it contains 1 lbm of air 540 R 60 psia The system is now heated to 2600 R by a 2800 R source and the steel has the same temperature as the air Find the entropy generation using constant specific heats CV Air and Steel Energy Eq U2 U1 mairu2 u1 mstu2 u1 1Q2 1W2 Entropy Eq637 mair s2 s1 msts2 s1 1Q2Tsource 1S2 gen Process P C 1W2 1 2 PdV P V2 V1 P mair v2 v1 1Q2 mairu2 u1air mstu2 u1st 1W2 mairh2 h1air mstu2 u1 st Use F2 u2 u1st C T2 T1 011 BtulbmR 2600 540 R 2266 Btulbm Use F4 h2 h1air CpT2 T1 024 BtulbmR 2600 540 R 4944 Btulbm 1Q2 mairh2 h1air mstu2 u1st 1 lbm 4944 Btulbm 2 lbm 2266 Btulbm 9476 Btu S2 S1 mair s2 s1 msts2 s1 1 lbm 024 BtulbmR ln 2600 540 2 lbm 011 BtulbmR ln 2600 540 0723 BtuR 1S2 gen S2 S1 1Q2Tsource 0723 94762800 0385 BtuR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6235E Do Problem 6234E using Table F5 CV Air and Steel Energy Eq U2 U1 mairu2 u1 mstu2 u1 1Q2 1W2 Entropy Eq637 mair s2 s1 msts2 s1 1Q2Tsource 1S2 gen Process P C 1W2 1 2 PdV P V2 V1 P mair v2 v1 1Q2 mairu2 u1air mstu2 u1st 1W2 mairh2 h1air mstu2 u1 st Use air tables F5 h2 h1air 674421 12918 54524 Btulbm s2 s1air 204517 163979 0 040538 BtulbmR No pressure correction as P2 P1 Use F2 u2 u1st CT2 T1 011 BtulbmR 2600 540 R 2266 Btulbm 1Q2 mairh2 h1air mstu2 u1st 1 lbm 54524 Btulbm 2 lbm 2266 Btulbm 99844 Btu S2 S1 mair s2 s1 msts2 s1 040538 2 lbm 011 kJkgK ln 2600 540 075115 Btulbm 1S2 gen S2 S1 1Q2Tsource 075115 998442800 0395 BtuR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6236E Nitrogen at 90 lbfin2 260 F is in a 20 ft3 insulated tank connected to a pipe with a valve to a second insulated initially empty tank of volume 10 ft3 The valve is opened and the nitrogen fills both tanks Find the final pressure and temperature and the entropy generation this process causes Why is the process irreversible CV Both tanks pipe valve Insulated Q 0 Rigid W 0 Energy Eq35 mu2 u1 0 0 u2 u1 ua1 Entropy Eq637 ms2 s1 dQT 1S2 gen 1S2 gen dQ 0 State 1 P1 T1 Va Ideal gas m PVRT 90 20 144 5515 720 6528 lbm State 2 V2 Va Vb uniform final state v2 V2 m u2 ua1 P v 1 2 2 1 T s 1 P P2 Ideal gas u T u2 ua1 T2 Ta1 720 R P2 mR T2 V2 V1 V2 P1 23 90 60 lbfin2 From entropy equation and Eq619 for entropy change Sgen ms2 s1 m sT2o sT1o R ln P2 P1 m 0 R ln P2 P1 6528 5515 1778 ln 23 0187 BtuR Irreversible due to unrestrained expansion in valve P but no work out Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful If not a uniform final state then flow until P2b P2a and valve is closed Assume no Q between A and B ma2 mb2 ma1 ma2 va2 mb2 vb2 ma1va1 ma2 sa2 mb2 sb2 ma1sa1 0 1S 2 gen Now we must assume ma2 went through rev adiabatic expansion 1 V2 ma2 va2 mb2 vb2 2 Pb2 Pa2 3 sa2 sa1 4 Energy eqs 4 Eqs 4 unknowns P2 Ta2 Tb2 x ma2 ma1 V2 ma1 x va2 1 x vb2 x R Ta2 P2 1 x R Tb2 P2 Energy Eq ma2 ua2 ua1 mb2 ub2 ua1 0 x Cv Ta2 Ta1 1 x Cv Tb2 Ta1 0 x Ta2 1 xTb2 Ta1 Vol constraint P2 V2 ma1 R x Ta2 1 x Tb2 Ta1 P2 ma1 R Ta1 V2 ma1 R Ta1 15Va1 23 Pa1 60 lbfin2 sa2 sa1 Ta2 Ta1 P2 Pa1k1 k 720 2302857 64124 R Now we have final state in A va2 R Ta2 P2 40931 ftlbm ma2 Va va2 4886 lbm x ma2 ma1 07485 mb2 ma1 ma2 1642 lbm Substitute into energy equation Tb2 Ta1 x Ta2 1 x 9544 R 1S2 gen mb2 sb2 sa1 mb2 Cp ln Tb2 Ta1 R ln P2 Pa1 1642 0249 ln 9544720 5515778 ln 23 01624 BtuR Comment This shows less entropy generation as the mixing part is out compared with the previous solution Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6237E A cylinderpiston contains carbon dioxide at 150 lbfin2 600 F with a volume of 7 ft3 The total external force acting on the piston is proportional to V3 This system is allowed to cool to room temperature 70 F What is the total entropy generation for the process CV Carbon dioxide gas of constant mass m2 m1 m out to ambient Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 dQT 1S2 gen 1Q2Tamb 1S2 gen Process P CV 3 or PV3 constant which is polytropic with n 3 State 1 P1 150 lbfin2 T1 600 F 1060 R V1 7 ft3 Ideal gas m P1V1 RT1 150 144 7 3510 1060 4064 lbm Process P CV3 or PV 3 const polytropic with n 3 P2 P1T2T1 n n1 150 530 1060 075 892 lbfin 2 V2 V1T1T2 1 n1 V1 P1 P2 T2 T1 7 150 892 530 1060 5886 1W2 PdV 1 n P2V2 P1V1 1 3 892 5886 150 7 144 778 243 Btu 1Q2 4064 0158 530 1060 243 3466 Btu ms2 s1 4064 0203 ln 530 1060 3510 778 ln 892 150 04765 BtuR SSURR 1Q2Tamb 3646 530 06879 BtuR From Eq637 or 639 1S2 gen ms2 s1 1Q2Tamb SNET SCO2 S SURR 04765 BtuR 06879 ΒtuR 02114 BtuR P v 1 2 T s 1 2 70 600 89 150 Notice n 3 k 13 n k Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6238E A cylinderpiston contains 4 ft3 of air at 16 lbfin2 77 F The air is compressed in a reversible polytropic process to a final state of 120 lbfin2 400 F Assume the heat transfer is with the ambient at 77 F and determine the polytropic exponent n and the final volume of the air Find the work done by the air the heat transfer and the total entropy generation for the process Solution CV Air of constant mass m2 m1 m out to ambient Energy Eq35 mu2 u1 1Q2 1W2 Entropy Eq637 ms2 s1 dQT 1S2 gen 1Q2T0 1S2 gen Process Pv1 n P2v2 n Eq627 State 1 T1 P1 State 2 T2 P2 Thus the unknown is the exponent n m P1V1RT1 16 4 1445334 537 0322 lbm The relation from the process and ideal gas is in Eq628 T2T1 P2P1 n1 n n1 n lnT2 T1 lnP2 P1 02337 n 1305 V2 V1P1P21n 4 162011305 0854 ft3 1W2 PdV 1 n P2V2 P1V1 120 0854 16 4 144 778 1 1305 2335 Btu lbm 1Q2 mu2 u1 1W2 mCvT2 T1 1W2 0322 0171 400 77 2335 556 Btu lbm s2 s1 Cp lnT2 T1 R lnP2 P1 024 ln 860537 5334778 ln 12016 00251 Btulbm R 1S2 gen ms2 s1 1Q2T0 0322 00251 556537 000226 BtuR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Rates or Fluxes of Entropy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6239E A room at 72 F is heated electrically with 1500 W to keep steady temperature The outside ambient is at 40 F Find the flux of S Q T into the room air into the ambient and the rate of entropy generation CV The room and walls out to the ambient T we assume steady state Energy Eq 0 W el in Q out Q out W el in 1500 W Entropy Eq 0 Q outT S gen tot Flux of S into room air at 22oC Q T 1500 5317 282 WR Flux of S into ambient air at 5oC Q T 1500 4997 30 WR Entropy generation S gen tot Q outT 1500 4997 30 WR Comment The flux of S into the outside air is what leaves the control volume and since the control volume did not receive any S it was all made in the process Notice most of the generation is done in the heater the room heat loss process generates very little S 30 282 018 WR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6240E A heat pump with COP 4 uses 1 kW of power input to heat a 78 F room drawing energy from the outside at 60 F Assume the highlow temperatures in the HP are 120 F32 F Find the total rates of entropy in and out of the heat pump the rate from the outside at 60 F and to the room at 78 F Solution CVTOT Energy Eq Q L W Q H Entropy Eq 0 Q L TL Q H TH S gen CV tot H Q W L Q HP 60 F 78 F From definition of COP Q H COP W 4 1 kW 4 kW From energy equation Q L Q H W 4 1 kW 3 kW Flux into heat pump at 32 F Q L TLHP 3 4917 kW R 00061 kWR Flux out of heat pump at 120 F Q H THHP 4 5797 kW R 00069 kWR Flux out into room at TH 78 F TH Q H 4 5377 kW R 000744 kWR Flux from outside at 60 F Q L TL 3 5197 kW R 000577 kWR Comment Following the flow of energy notice how the flux from the outside at 60 F grows a little when it arrives at 32 F this is due to entropy generation in the low T heat exchanger The flux out of the heat pump at 120 F is larger than the flux in which is due to entropy generation in the heat pump cycle COP is smaller than Carnot COP and finally this flux increases due to entropy generated in the high T heat exchanger as the energy arrives at room T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6241E A window receives 800 Btuh of heat transfer at the inside surface of 70 F and transmits the 800 Btuh from its outside surface at 36 F continuing to ambient air at 23 F Find the flux of entropy at all three surfaces and the windows rate of entropy generation Flux of entropy S T Q S inside 800 5297 Btu hR 151 BtuhR S win 800 4957 Btu hR 1614 BtuhR S amb 800 4827 Btu hR 1657 BtuhR Window Inside Outside 70 F 36 F 23 F Window only S gen win S win S inside 1614 151 0104 BtuhR If you want to include the generation in the outside air boundary layer where T changes from 36 F to the ambient 23 F then it is S gen tot S amb S inside 1657 151 0147 BtuhR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6242E A farmer runs a heat pump using 25 hp of power input It keeps a chicken hatchery at a constant 86 F while the room loses 20 Btus to the colder outside ambient at 50 F What is the rate of entropy generated in the heat pump What is the rate of entropy generated in the heat loss process Solution CV Hatchery steady state Power W 25 hp 25 25444 3600 1767 Btus To have steady state at 86 F for the hatchery Energy Eq 0 Q H Q Loss Q H Q Loss 20 Btus CV Heat pump steady state Energy eq 0 Q L W Q H Q L Q H W 18233 Btus Entropy Eq 0 Q L TL Q H TH S gen HP S gen HP Q H TH Q L TL 20 5457 18233 5097 0000 878 Btu s R CV From hatchery at 86 F to the ambient 50 F This is typically the walls and the outer thin boundary layer of air Through this goes Q Loss Entropy Eq 0 Q Loss TH Q Loss Tamb S gen walls S gen walls Q Loss Tamb Q Loss TH 20 5097 20 5457 000259 Btu s R Qleak Q QH L W 25 hp HP cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6243E Water in a pistoncylinder is at 150 lbfin2 900 F as shown in Fig P6177 There are two stops a lower one at which Vmin 35 ft3 and an upper one at Vmax 105 ft3 The piston is loaded with a mass and outside atmosphere such that it floats when the pressure is 75 lbfin2 This setup is now cooled to 210 F by rejecting heat to the surroundings at 70 F Find the total entropy generated in the process CV Water State 1 Table F72 v1 53529 ft3lbm u1 13302 btulbm s1 18381 Btulbm m Vv1 1055353 19615 lbm P v 1 2 2 1 T s v C 75 150 State 2 210 F and on line in Pv diagram Notice the following vgPfloat 5818 ft3lbm vbot Vminm 17843 TsatPfloat 3076 F T2 TsatPfloat V2 Vmin State 2 210 F v2 vbot x2 17843 0016727796 006359 u2 1781 0063598989 23526 btulbm s2 03091 00635914507 04014 btulbm R Now we can do the work and then the heat transfer from the energy equation 1W2 PdV PfloatV2 V1 7535 105 144 778 97172 Btu 1Q2 mu2 u1 1W2 1961523526 13302 97172 22449 Btu Take CV total out to where we have 70 F ms2 s1 1Q2T0 Sgen Sgen ms2 s1 1Q2T0 1961504014 18381 22449 52967 1420 BtuR Swater Ssur Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6244E A cylinderpiston contains 5 lbm of water at 80 lbfin2 1000 F The piston has crosssectional area of 1 ft2 and is restrained by a linear spring with spring constant 60 lbfin The setup is allowed to cool down to room temperature due to heat transfer to the room at 70 F Calculate the total water and surroundings change in entropy for the process State 1 Table F72 v1 10831 ft3lbm u1 13723 btulbm s1 19453 Btulbm R State 2 T2 on line in Pv diagram P P1 ksA2 cylV V1 Assume state 2 is twophase P2 PsatT2 03632 lbfin 2 1 2 P v v2 v1 P2 P1A2 cylmks 10831 03632 80112560 76455 ft3lbm vf x2vfg 001605 x2 867579 x2 0008793 u2 381 000879399564 4685 btulbm s2 00746 000879319896 00921 Btulbm R 1W2 1 2 P1 P2mv2 v1 5 280 0363276455 10831144 778 11846 Btu 1Q2 mu2 u1 1W2 54685 13723 11846 6746 Btu Stot Sgen tot ms2 s1 1Q2Troom 500921 19453 674652967 347 BtuR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 6245E A cylinder with a linear springloaded piston contains carbon dioxide gas at 300 lbfin2 with a volume of 2 ft3 The device is of aluminum and has a mass of 8 lbm Everything Al and gas is initially at 400 F By heat transfer the whole system cools to the ambient temperature of 77 F at which point the gas pressure is 220 lbfin2 Find the total entropy generation for the process Solution CO2 m P1V1RT1 300 2 1443510 860 2862 lbm V2 V1P1P2 T2 T1 2300220537860 1703 ft3 1W2 CO2 PdV 05P1 P2 V2 V1 300 2202 1703 2 144 778 1429 Btu 1Q2 CO2 mCV0T2T1 1W2 0156 286277 4001429 1585 Btu 1Q2 Al mC T2T1 8 02177 400 5426 Btu System CO2 Al Entropy Eq mCO2s2 s1CO2 mALs2 s1AL 1Q2T0 S gen tot 1Q2 5426 1585 70114 Btu mCO2s2 s1CO2 mALs2 s1AL 28620201 ln 537860 3510778 ln 220300 8 021 ln537860 023086 079117 1022 BtuR Sgen tot mCO2s2 s1CO2 mALs2 s1AL 1Q2T0 1022 70114 537 02837 BtuR Tamb Q CO2 Al Updated June 2013 SOLUTION MANUAL CHAPTER 7 Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CONTENT CHAPTER 7 SUBSECTION PROB NO InText Concept Questions a g Concept Problems 112 Steady State Reversible Processes Single Flow 1335 Multiple Devices and Cycles 3648 Transient Processes 4956 Reversible Shaft Work Bernoulli Equation 5781 Steady State Irreversible Processes 82114 Transient irreversible Processes 115127 Device efficiency 128157 Review Problems 158176 Problems resolved with Pr and vr from Table A72 17 31 34 55 80 174 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful InText Concept Questions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7a A reversible adiabatic flow of liquid water in a pump has increasing P How about T Solution Steady state single flow se si i e dq T sgen si 0 0 Adiabatic dq 0 means integral vanishes and reversible means sgen 0 so s is constant Properties for liquid incompressible gives Eq619 ds C T dT then constant s gives constant T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7b A reversible adiabatic flow of air in a compressor has increasing P How about T Solution Steady state single flow se si i e dq T sgen si 0 0 so s is constant Properties for an ideal gas gives Eq615 and for constant specific heat we get Eq616 A higher P means a higher T which is also the case for a variable specific heat recall Eq619 using the standard entropy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7c A compressor receives R134a at 10oC 200 kPa with an exit of 1200 kPa 50oC What can you say about the process Solution Properties for R134a are found in Table B5 Inlet state si 17328 kJkg K Exit state se 17237 kJkg K Steady state single flow se si i e dq T sgen Since s decreases slightly and the generation term can only be positive it must be that the heat transfer is negative out so the integral gives a contribution that is smaller than sgen Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7d A flow of water at some velocity out of a nozzle is used to wash a car The water then falls to the ground What happens to the water state in terms of V T and s let us follow the water flow It starts out with kinetic and potential energy of some magnitude at a compressed liquid state P T As the water splashes onto the car it looses its kinetic energy it turns in to internal energy so T goes up by a very small amount As it drops to the ground it then looses all the potential energy which goes into internal energy Both of theses processes are irreversible so s goes up If the water has a temperature different from the ambient then there will also be some heat transfer to or from the water which will affect both T and s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7e In a steady state single flow s is either constant or it increases Is that true Solution No Steady state single flow se si i e dq T sgen Entropy can only go up or stay constant due to sgen but it can go up or down due to the heat transfer which can be positive or negative So if the heat transfer is large enough it can overpower any entropy generation and drive s up or down Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7f If a flow device has the same inlet and exit pressure can shaft work be done The reversible work is given by Eq714 w v dP V2 i V2 e g Zi Ze For a constant pressure the first term drops out but the other two remains Kinetic energy changes can give work out windmill and potential energy changes can give work out a dam Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7g A polytropic flow process with n 0 might be which device As the polytropic process is Pvn C then n 0 is a constant pressure process This can be a pipe flow a heat exchanger flow heater or cooler or a boiler Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Concept Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 71 If we follow a mass element going through a reversible adiabatic flow process what can we say about the change of state Following a mass this is a control mass de dq dw 0 Pdv Pdv compressionexpansion changes e ds dqT dsgen 0 0 s constant isentropic process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 72 Which process will make the statement in concept question e on page 330 true Solution If the process is said to be adiabatic then Steady state adiabatic single flow se si sgen si Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 73 A reversible process in a steady flow with negligible kinetic and potential energy changes is shown in the diagrams Indicate the change he hi and transfers w and q as positive zero or negative i P v T s e i e dw v dP 0 P drops so work is positive out dq T ds 0 s is constant and process reversible so adiabatic he hi q w 0 w 0 so enthalpy drops Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 74 A reversible process in a steady flow of air with negligible kinetic and potential energy changes is shown in the diagrams Indicate the change he hi and transfers w and q as positive zero or negative i P v T s e i e dw v dP 0 P drops so work is positive out dq T ds 0 s is increasing and process reversible so q is positive he hi 0 as they are functions of T and thus the same Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 75 A reversible steady isobaric flow has 1 kW of heat added with negligible changes in KE and PE what is the work transfer P C Shaft work Eq 714 dw v dP ΔKE ΔPE T dsgen 0 0 0 0 0 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 76 An air compressor has a significant heat transfer out See example 94 for how high T becomes if there is no heat transfer Is that good or should it be insulated That depends on the use of the compressed air If there is no need for the high T say it is used for compressed air tools then the heat transfer will lower T and result in lower specific volume reducing the work For those applications the compressor may have fins mounted on its surface to promote the heat transfer In very high pressure compression it is done in stages between which is a heat exchanger called an intercooler This is a small compressor driven by an electric motor Used to charge air into car tires Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 77 Friction in a pipe flow causes a slight pressure decrease and a slight temperature increase How does that affect entropy Solution The friction converts flow work P drops into internal energy T up if single phase This is an irreversible process and s increases If liquid Eq 610 ds C T dT so s follows T If ideal gas Eq 614 ds Cp dT T R dP P both terms increase Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 78 To increase the work out of a turbine for a given inlet and exit pressure how should the inlet state be changed w v dP Eq714 For a given change in pressure boosting v will result in larger work term So for larger inlet T we get a larger v and thus larger work That is why we increase T by combustion in a gasturbine before the turbine section Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 79 An irreversible adiabatic flow of liquid water in a pump has higher P How about T Solution Steady state single flow se si i e dq T sgen si 0 sgen so s is increasing Properties for liquid incompressible gives Eq610 where an increase in s gives an increase in T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 710 The shaft work in a pump to increase the pressure is small compared to the shaft work in an air compressor for the same pressure increase Why The reversible work is given by Eq 714 or 715 if reversible and no kinetic or potential energy changes w v dP The liquid has a very small value for v compared to a large value for a gas Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 711 Liquid water is sprayed into the hot gases before they enter the turbine section of a large gasturbine power plant It is claimed that the larger mass flow rate produces more work Is that the reason No More mass through the turbine does give more work but the added mass is only a few percent As the liquid vaporizes the specific volume increases dramatically which gives a much larger volume flow throught the turbine and that gives more work output W m w m v dP m v dP V dP This should be seen relative to the small work required to bring the liquid water up to the higher turbine inlet pressure from the source of water presumably atmospheric pressure Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 712 A tank contains air at 400 kPa 300 K and a valve opens up for flow out to the outside which is at 100 kPa 300 K What happens to the air temperature inside As mass flows out of the tank the pressure will drop the air that remains basically goes through a simple adiabatic if process is fast enough expansion process so the temperature also drops If the flow rate out is very small and the process thus extremely slow enough heat transfer may take place to keep the temperature constant Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Steady state reversible processes single flow Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 713 A turbine receives steam at 6 MPa 600C with an exit pressure of 600 kPa Assume the stage is adiabatic and negelect kinetic energies Find the exit temperature and the specific work Solution CV Stage 1 of turbine The stage is adiabatic so q 0 and we will assume reversible so sgen 0 WT i e Energy Eq413 wT hi he Entropy Eq79 se si dqT sgen si 0 0 Inlet state B13 hi 365840 kJkg si 71676 kJkg K Exit state 600 kPa s si Table B13 T 2467C he 295019 kJkg wT 36584 295019 7082 kJkg v P s T i i e s e s 600 kPa 6 MPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 714 An evaporator has R410A at 20oC and quality 80 flowing in with the exit flow being saturated vapor at 20oC Consider the heating to be a reversible process and find the specific heat transfer from the entropy equation Entropy Eq79 se si dqT sgen si qT 0 q T se si T sg si Inlet si 01154 xi 09625 08854 kJkgK Exit sg 10779 kJkgK q 27315 20 K 10779 08854 kJkgK 4873 kJkg Remark It fits with he hi 1 xi hfg 02 24365 4873 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 715 Steam enters a turbine at 3 MPa 450C expands in a reversible adiabatic process and exhausts at 50 kPa Changes in kinetic and potential energies between the inlet and the exit of the turbine are small The power output of the turbine is 800 kW What is the mass flow rate of steam through the turbine Solution CV Turbine Steady single inlet and exit flows Adiabatic Q 0 Continuity Eq411 m i m e m Energy Eq412 m hi m he W T Entropy Eq 78 m si 0 m se Reversible S gen 0 Explanation for the work term is in Sect 73 Eq714 2 1 P v T s 1 2 Inlet state Table B13 hi 3344 kJkg si 70833 kJkg K Exit state Pe se si Table B12 saturated as se sg xe 70833 109165029 092148 he 34047 092148 230540 246485 kJkg m W TwT W Thi he 800 3344 246485 kW kJkg 091 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 716 The exit nozzle in a jet engine receives air at 1200 K 150 kPa with neglible kinetic energy The exit pressure is 80 kPa and the process is reversible and adiabatic Use constant heat capacity at 300 K to find the exit velocity Solution CV Nozzle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi he V2 e2 Zi Ze Entropy Eq79 se si dqT sgen si 0 0 Use constant specific heat from Table A5 CPo 1004 kJ kg K k 14 The isentropic process se si gives Eq623 Te Ti PePi k1 k 1200 K 80150 02857 10027 K The energy equation becomes V2 e2 hi he CP Ti Te Ve 2 CP Ti Te 21004120010027 1000 6294 ms P v T s e i i e Low V Hi P Low P Hi V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 717 Do the previous problem using the air tables in A7 The exit nozzle in a jet engine receives air at 1200 K 150 kPa with neglible kinetic energy The exit pressure is 80 kPa and the process is reversible and adiabatic Use constant heat capacity at 300 K to find the exit velocity Solution CV Nozzle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi he V2 e2 Zi Ze Entropy Eq79 se si dqT sgen si 0 0 Process q 0 sgen 0 as used above leads to se si Inlet state hi 12778 kJkg so Ti 83460 kJkg K The constant s is rewritten from Eq619 as so Te so Ti R lnPe Pi 83460 0287 ln 80150 81656 kJkg K Interpolate in A7 Te 1000 50 81656 81349 81908 81349 102746 K he 10462 11035 10463 81656 81349 81908 81349 10777 kJkg From the energy equation we have V2 e2 hi he so then Ve 2 hi he 212778 10777 kJkg 1000 JkJ 6326 ms P v T s e i i e Low V Hi P Low P Hi V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 718 A reversible adiabatic compressor receives 005 kgs saturated vapor R410A at 400 kPa and has an exit presure of 1800 kPa Neglect kinetic energies and find the exit temperature and the minimum power needed to drive the unit Solution CV Compressor Steady single inlet and exit flows Adiabatic Q 0 Continuity Eq411 m i m e m Energy Eq412 m hi m he W C Entropy Eq78 m si 0 m se Reversible S gen 0 Inlet state B 42 hi 27190 kJkg si 10779 kJkg K Exit state Pe se si Table B42 he 31433 kJkg Te 519C wc he hi 31433 27190 4243 kJkg W c Power In wcm 4243 kJkg 005kgs 212 kW Explanation for the work term is in Sect 73 Eq714 2 1 P v T s 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 719 In a heat pump that uses R134a as the working fluid the R134a enters the compressor at 150 kPa 10C and the R134a is compressed in an adiabatic process to 1 MPa using 4 kW of power input Find the mass flow rate it can provide assuming the process is reversible Solution CV Compressor Steady single inlet and exit flows Adiabatic Q 0 Continuity Eq411 m 1 m 2 m Energy Eq412 m h1 m h2 W C Entropy Eq78 m s1 0 m s2 Reversible S gen 0 Inlet state Table B52 h1 39384 kJkg s1 17606 kJkg K Exit state P2 1 MPa s2 s1 h2 4349 kJkg m W c wc W c h1 h2 4 kW 39384 4349 kJkg 0097 kgs Explanation for the work term is in Sect 73 Eq714 2 1 P v T s 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 720 Nitrogen gas flowing in a pipe at 500 kPa 200oC and at a velocity of 10 ms should be expanded in a nozzle to produce a velocity of 300 ms Determine the exit pressure and crosssectional area of the nozzle if the mass flow rate is 015 kgs and the expansion is reversible and adiabatic Solution CV Nozzle Steady flow no work out and no heat transfer Energy Eq413 hi V 2 i2 he V 2 e2 Entropy Eq79 si dqT sgen si 0 0 se Properties Ideal gas Table A5 CPo 1042 kJ kg K R 02968 kJ kg K k 140 he hi CPoTe Ti 1042Te 4732 102 300221000 Solving for exit T Te 430 K Process si se For ideal gas expressed in Eq623 Pe PiTeTi k k1 500 kPa 430 4732 35 3576 kPa ve RTePe 02968 kJkgK 430 K3576 kPa 035689 m3kg Ae mveVe 015 035689 300 ms kgs m3kg 178 104 m2 Inlet V Exit V cb i e i P v T s e i e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 721 A reversible isothermal expander a turbine with heat transfer has an inlet flow of carbon dioxide at 3 MPa 80oC and an exit flow at 1 MPa 80oC Find the specific heat transfer from the entropy equation and the specific work from the energy equation assuming ideal gas CV the expander control surface at 80oC Energy Eq413 0 hi he q w Entropy Eq79 0 si se dqT sgen si se qT 0 From entropy equation q T se si T CPo ln Ti Te R ln Pi Pe RT ln Pi Pe 01889 kJkgK 35315 K ln 1 3 7329 kJkg From energy equation w hi he q q 7329 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 722 Solve the previous Problem using Table B3 Energy Eq413 0 hi he q w Entropy Eq79 0 si se dqT sgen si se qT 0 Inlet state hi 42116 kJkg si 15385 kJkgK Exit state he 43523 kJkg se 1775 kJkgK From entropy equation q T se si 35315 1775 15385 8352 kJkg From energy equation w hi he q 42116 43523 8352 6945 kJkg Remark When it is not an ideal gas h is a fct of both T and P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 723 A compressor in a commercial refrigerator receives R410A at 25oC and unknown quality The exit is at 2000 kPa 60oC and the process assumed reversible and adiabatic Neglect kinetic energies and find the inlet quality and the specific work CV Compressor q 0 Energy Eq413 wC hi he Entropy Eq79 se si dqT sgen si 0 0 Exit state 2000 kPa 60oC se 10878 kJkgK si he 32062 kJkg Inlet state T s Table B41 xi 10878 0087110022 09985 hi 2108 xi 24869 2694 kJkg wC 2694 32062 512 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 724 A boiler section boils 3 kgs saturated liquid water at 2000 kPa to saturated vapor in a reversible constant pressure process Assume you do not know that there is no work Prove that there is no shaftwork using the first and second laws of thermodynamics Solution CV Boiler Steady single inlet and single exit flows Energy Eq413 hi q w he Entropy Eq79 si qT s e States Table B12 T Tsat 21242C 48557 K hi hf 90877 kJkg si 24473 kJkg K he hg 279951 kJkg se 63408 kJkg K From entropy equation q Tse si 48557 K 63408 24473 kJkgK 18906 kJkg From energy equation w hi q he 90877 18906 279951 01 kJkg It should be zero nonzero due to round off in values of s h and Tsat cb Often it is a long pipe and not a chamber Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 725 A compressor brings a hydrogen gas flow at 280 K 100 kPa up to a pressure of 1000 kPa in a reversible process How hot is the exit flow and what is the specific work input CV Compressor Assume q 0 Energy Eq413 wC hi he Cp Ti Te Entropy Eq79 se si dqT sgen si 0 0 So constant s gives the power relation in Eq 623 with k from A5 Te Ti PePi k1k 280 K 1000100 140911409 5463 K Now the work from the energy equation Cp from A5 wC 14209 kJkgK 280 5463 K 37839 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 726 Atmospheric air at 45C 60 kPa enters the front diffuser of a jet engine with a velocity of 900 kmh and frontal area of 1 m2 After the adiabatic diffuser the velocity is 20 ms Find the diffuser exit temperature and the maximum pressure possible Solution CV Diffuser Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi V 2 i2 he V 2 e2 and he hi CpTe Ti Entropy Eq79 si dqT sgen si 0 0 se Reversible adiabatic Specific heat and ratio of specific heats from Table A5 CPo 1004 kJ kg K k 14 the energy equation then gives 1004 Te 45 05900100036002 202 1000 3105 kJkg Te 1405 C 2591 K Constant s for an ideal gas is expressed in Eq623 we need the inverse relation here Pe Pi TeTi k k1 60 kPa 2591228135 936 kPa Fan 1 2 P v T s 2 1 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 727 A compressor is surrounded by cold R134a so it works as an isothermal compressor The inlet state is 0C 100 kPa and the exit state is saturated vapor Find the specific heat transfer and specific work Solution CV Compressor Steady single inlet and single exit flows Energy Eq413 hi q w he Entropy Eq79 si qT s e Inlet state Table B52 hi 4034 kJkg si 18281 kJkg K Exit state Table B51 he 39836 kJkg se 17262 kJkg K From entropy equation q Tse si 27315 K 17262 18281 kJkgK 2783 kJkg From energy equation w 4034 2783 39836 228 kJkg Explanation for the work term is in Sect 73 Eq 714 P v e i e i T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 728 A flow of 2 kgs saturated vapor R410A at 500 kPa is heated at constant pressure to 60oC The heat is supplied by a heat pump that receives heat from the ambient at 300 K and work input shown in Fig P728 Assume everything is reversible and find the rate of work input Solution CV Heat exchanger Continuity Eq m 1 m 2 Energy Eq m 1h1 Q H m 1h 2 Table B42 h1 27433 kJkg s1 10647 kJkg K h2 34232 kJkg s2 12959 kJkg K H Q W L Q T L HP 1 2 Notice we can find Q H but the temperature TH is not constant making it difficult to evaluate the COP of the heat pump CV Total setup and assume everything is reversible and steady state Energy Eq m 1h1 Q L W in m 1h 2 Entropy Eq m 1s1 Q LTL 0 m 1s2 TL is constant sgen 0 Q L m 1TL s2 s1 2 kgs 300 K 1 2959 10647 kJkgK 13872 kW W in m 1h2 h1 Q L 2 34232 27433 13872 274 kW Comment Net work is nearly zero due to the very low T the flow comes in at so the first heating of the flow actually generates work out and only the heating to above ambient T requires work input Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 729 A flow of 2 kgs hot exhaust air at 150oC 125 kPa supplies heat to a heat engine in a setup similar to the previous problem with the heat engine rejecting heat to the ambient at 290 K and the air leaves the heat exchanger at 50oC Find the maximum possible power out of the heat engine CV Total setup Assume everything is reversible Energy Eq m 1h1 Q L W HE m 1h 2 Entropy Eq m 1s1 0 Q LTL m 1s2 TL is constant sgen 0 Q L m 1 TL s1 s2 2 kgs 290 Κ 1004 kJkgK ln423323 157 kW From the energy equation W HE m 1h1 h2 Q L m 1 CPoT1 T2 Q L 2 kgs 1004 kJkgK 150 50 K 157 kW 438 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 730 A diffuser is a steadystate device in which a fluid flowing at high velocity is decelerated such that the pressure increases in the process Air at 120 kPa 30C enters a diffuser with velocity 200 ms and exits with a velocity of 20 ms Assuming the process is reversible and adiabatic what are the exit pressure and temperature of the air Solution CV Diffuser Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi V2 i 2 he V2 e2 he hi CPoTe Ti Entropy Eq79 si dqT sgen si 0 0 se Reversible adiabatic Use constant specific heat from Table A5 CPo 1004 kJ kg K k 14 Energy equation then gives CPoTe Ti 1004Te 3032 2002 20221000 Te 3229 K The isentropic process se si gives Eq623 Pe PiTeTi k k1 120 kPa 3229303235 1496 kPa P v T s e i i e Inlet Low V Exit Hi V Hi P A Low P A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 731 Air enters a turbine at 800 kPa 1200 K and expands in a reversible adiabatic process to 100 kPa Calculate the exit temperature and the specific work output using Table A7 and repeat using constant specific heat from Table A5 Solution i e Turbine air W CV Air turbine Adiabatic q 0 reversible sgen 0 Energy Eq413 wT hi he Entropy Eq79 se s i a Table A7 hi 12778 kJkg so Ti 834596 kJkg K The constant s process is written from Eq619 as so Te so Ti R ln Pe Pi 834596 0287 ln 100 800 77492 kJkg K Interpolate in A71 Te 706 K he 7199 kJkg w hi he 5579 kJkg b Table A5 CPo 1004 kJkg K R 0287 kJkg K k 14 then from Eq623 Te Ti PePi k1 k 1200 K 100 800 0286 6621 K w CPoTi Te 1004 kJkgK 1200 6621 K 5398 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 732 An expander receives 05 kgs air at 2000 kPa 300 K with an exit state of 400 kPa 300 K Assume the process is reversible and isothermal Find the rates of heat transfer and work neglecting kinetic and potential energy changes Solution CV Expander single steady flow Energy Eq m hi Q m he W Entropy Eq m si Q T m sgen m se Process T is constant and sgen 0 Ideal gas and isothermal gives a change in entropy by Eq 615 so we can solve for the heat transfer Q Tm se si m RT ln Pe Pi 05 kgs 300 K 0287 kJkgK ln 400 2000 693 kW From the energy equation we get W m hi he Q Q 693 kW P v T s e i i e Wexp i e Q Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 733 A highly cooled compressor brings a hydrogen gas flow at 300 K 100 kPa up to a pressure of 800 kPa in an isothermal process Find the specific work assuming a reversible process CV Compressor Isothermal Ti Te so that ideal gas gives hi he Energy Eq413 wC hi q he q Entropy Eq79 se si dqT sgen si qT 0 q Tse si T R lnPePi w q 41243 kJkgK 300 K ln8 2573 kJkg R is from Table A5 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 734 A compressor receives air at 290 K 95 kPa and a shaft work of 55 kW from a gasoline engine It should deliver a mass flow rate of 001 kgs air to a pipeline Find the maximum possible exit pressure of the compressor Solution CV Compressor Steady single inlet and exit flows Adiabatic Q 0 Continuity Eq411 m i m e m Energy Eq412 m hi m he W C Entropy Eq78 m si S gen m se Reversible S gen 0 W c m wc wc W m 55001 550 kJkg Use constant specific heat from Table A5 CPo 1004 kJkgK k 14 he hi 550 Te Ti 550 kJkg 1004 kJkgK Te 290 5501004 83781 K si se Pe Pi TeTi k k1 Eq623 Pe 95 kPa 8378129035 3893 kPa P v T s e i i e h 550 kJkg W C i e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 735 A reversible steady state device receives a flow of 1 kgs air at 400 K 450 kPa and the air leaves at 600 K 100 kPa Heat transfer of 900 kW is added from a 1000 K reservoir 50 kW rejected at 350 K and some heat transfer takes place at 500 K Find the heat transferred at 500 K and the rate of work produced Solution CV Device single inlet and exit flows Energy equation Eq412 m h1 Q 3 Q 4 Q 5 m h2 W Entropy equation with zero generation Eq78 m s1 Q 3T3 Q 4T4 Q 5 T5 m s2 1 2 T T 500 K 3 3 4 4 Q Q W Q 5 Solve for the unknown heat transfer using Table A71 and Eq 619 for change in s Q 5 T5 s2 s1 m T5 T4 Q 4 T5 T3 Q 3 500 1 75764 71593 0287 ln 100 450 500 350 50 500 1000 900 4244 714 450 458 kW Now the work from the energy equation is W 1 4013 6073 900 50 458 6898 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 736 A steam turbine in a powerplant receives 5 kgs steam at 3000 kPa 500oC 20 of the flow is extracted at 1000 kPa to a feedwater heater and the remainder flows out at 200 kPa Find the two exit temperatures and the turbine power output CV Turbine Steady flow and adiabatic q 0 Continuity Eq49 m 1 m 2 m 3 Energy Eq410 m 1h1 m 2h2 m 3h3 W Entropy Eq77 m 1s1 S gen m 2s2 m 3s3 W T 1 2 3 State 1 h1 3456 kJkg s1 7234 kJkgK We also assume turbine is reversible S gen 0 s1 s2 s3 State 2 Ps T2 3306oC h2 3116 kJkg State 3 Ps T3 1407oC h3 2750 kJkg W m 1h1 m 2h2 m 3h3 5 3456 1 3116 4 2750 3164 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 737 A reversible adiabatic compression of an air flow from 20oC 100 kPa to 200 kPa is followed by an expansion down to 100 kPa in an ideal nozzle What are the two processes How hot does the air get What is the exit velocity Solution W 1 2 13 T s 1 2 P P 1 2 Separate control volumes around compressor and nozzle For ideal compressor we have inlet 1 and exit 2 Adiabatic q 0 Reversible sgen 0 Energy Eq413 h1 0 wC h2 h2 h3 1 2V2 Entropy Eq79 s1 0T 0 s2 s2 0T s 3 So both processes are isentropic wC h2 h1 s2 s1 Properties Table A5 air CPo 1004 kJkg K R 0287 kJkg K k 14 Process gives constant s isentropic which with constant CPo gives Eq623 T2 T1 P2P1 k1 k 29315 200100 02857 3574 K wC CPoT2 T1 1004 3574 2932 64457 kJkg The ideal nozzle then expands back down to P1 constant s so state 3 equals state 1 The energy equation has no work but kinetic energy and gives 1 2V2 h2 h1 wC 64 457 Jkg remember conversion to J V3 2 64 457 359 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 738 A small turbine delivers 15 MW and is supplied with steam at 700C 2 MPa The exhaust passes through a heat exchanger where the pressure is 10 kPa and exits as saturated liquid The turbine is reversible and adiabatic Find the specific turbine work and the heat transfer in the heat exchanger Solution Continuity Eq411 Steady m 1 m 2 m 3 m WT 1 2 3 Q Turbine Energy Eq413 wT h1 h2 Entropy Eq79 s2 s1 sT gen Inlet state Table B13 h1 391745 kJkg s1 79487 kJkg K Ideal turbine sT gen 0 s2 s1 79487 sf2 x sfg2 State 3 P 10 kPa s2 sg saturated 2phase in Table B12 x2s s1 sf2sfg2 79487 064927501 09731 h2s hf2 x hfg2 1918 09731 23928 252035 kJkg wTs h1 h2s 139705 kJkg m W wTs 1500 1397 1074 kgs Heat exchanger Energy Eq413 q h3 h2 Entropy Eq79 s3 s2 dqT s He gen q h3 h2s 19183 252035 23285 kJkg Q m q 1074 kgs 23285 kJkg 2500 kW Explanation for the work term is in Sect 73 Eq714 2 1 P v T s 1 2 3 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 739 One technique for operating a steam turbine in partload power output is to throttle the steam to a lower pressure before it enters the turbine as shown in Fig P739 The steamline conditions are 2 MPa 400C and the turbine exhaust pressure is fixed at 10 kPa Assuming the expansion inside the turbine is reversible and adiabatic determine the specific turbine work for no throttling and the specific turbine work partload if it is throttled to 500 kPa Show both processes in a Ts diagram CV Turbine Full load reversible and adiabatic Entropy Eq79 reduces to constant s so from Table B13 and B12 s3 s1 71270 06492 x3a 75010 x3a 086359 h3a 19181 086359 239282 22582 kJkg Energy Eq413 for turbine 1w3a h1 h3a 32476 22582 9894 kJkg The energy equation for the part load operation gives the exit h Notice that we have constant h in the throttle process h2 h1 2b P h2b h1 32476 kJkg s2b 77563 kJkgK 3b P s s2b 77563 x3b 77563 064927501 09475 h3b 19181 x3b 239282 24590 kJkg wT 32476 24590 7886 kJkg 2b 1 2a T s 3a 3b h C WT 1 2 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 740 An adiabatic air turbine receives 1 kgs air at 1500 K 16 MPa and 2 kgs air at 400 kPa T2 in a setup similar to Fig P487 with an exit flow at 100 kPa What should the temperature T2 be so the whole process can be reversible The process is reversible if we do not generate any entropy Physically in this problem it means that state 2 must match the state inside the turbine so we do not mix fluid at two different temperatures we assume the pressure inside is exactly 400 kPa For this reason let us select the front end as CV and consider the flow from state 1 to the 400 kPa This is a single flow Entropy Eq79 s1 0T 0 s2 Property Eq619 s2 s1 0 so T2 so T1 R lnP2 P1 so T2 so T1 R lnP2 P1 861208 0287 ln 400 1600 82142 kJkgK From A71 T2 10718 K If we solve with constant specific heats we get from Eq623 and k 14 T2 T1 P2 P1k1k 1500 K 400160002857 10094 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 741 A turbo charger boosts the inlet air pressure to an automobile engine It consists of an exhaust gas driven turbine directly connected to an air compressor as shown in Fig P741 For a certain engine load the conditions are given in the figure Assume that both the turbine and the compressor are reversible and adiabatic having also the same mass flow rate Calculate the turbine exit temperature and power output Find also the compressor exit pressure and temperature Solution CV Turbine Steady single inlet and exit flows Process adiabatic q 0 reversible sgen 0 EnergyEq413 wT h3 h4 Entropy Eq78 s4 s 3 3 1 4 2 Engine W Compressor Turbine The property relation for ideal gas gives Eq623 k from Table A5 s4 s3 T4 T3P4P3 k1 k 9232 K 100 170 0286 7932 K The energy equation is evaluated with specific heat from Table A5 wT h3 h4 CP0T3 T4 10049232 7932 1305 kJkg WT mwT 1305 kW CV Compressor steady 1 inlet and 1 exit same flow rate as turbine Energy Eq413 wC h2 h1 Entropy Eq79 s2 s 1 Express the energy equation for the shaft and compressor having the turbine power as input with the same mass flow rate so we get wC wT 1305 kJkg CP0T2 T1 1004T2 3032 T2 4332 K The property relation for s2 s1 is Eq623 and inverted as P2 P1T2T1 k k1 100 kPa 4332 3032 35 3487 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 742 Two flows of air both at 200 kPa one has 3 kgs at 400 K and the other has 2 kgs at 290 K The two lines exchange energy through a number of ideal heat engines taking energy from the hot line and rejecting it to the colder line The two flows then leave at the same temperature Assume the whole setup is reversible and find the exit temperature and the total power out of the heat engines Solution HE H Q W L Q 1 2 3 4 HE H Q W L Q HE H Q W L Q CV Total setup Energy Eq410 m 1h1 m 2h2 m 1h3 m 2h4 W TOT Entropy Eq77 m 1s1 m 2s2 S gen dQ T m 1s3 m 2s4 Process Reversible S gen 0 Adiabatic Q 0 Assume the exit flow has the same pressure as the inlet flow then the pressure part of the entropy cancels out and we have Exit same T P h3 h4 he s3 s4 se m 1h1 m 2h2 m TOThe W TOT m 1s1 m 2s2 m TOTse so Te m 1 m TOT so T1 m 2 m TOT so T2 3 5 71593 2 5 68352 702966 kJkgK Table A7 Te 35198 K he 35278 kJkg W TOT m 1h1 he m 2 h2 he 34013 35278 229043 35278 2086 kW Note The solution using constant heat capacity writes the entropy equation using Eq616 the pressure terms cancel out so we get 3 5 Cp lnTeT1 2 5 Cp lnTeT2 0 lnTe 3lnT1 2 lnT25 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 743 A flow of 5 kgs water at 100 kPa 20oC should be delivered as steam at 1000 kPa 350oC to some application We have a heat source at constant 500oC If the process should be reversible how much heat transfer should we have CV Around unknown device out to the source surface Energy Eq m hi Q m he W Entropy Eq m si Q TS 0 m se TS is constant sgen 0 Inlet state si 02966 kJkgK Table B11 Exit state se 7301 kJkgK Table B13 Q m TS se s 5 kgs 77315 K 7301 02966 kJkgK 271 MW The theory does not say exactly how to do it As the pressure goes up we must have a pump or compressor and since the substance temperature is lower than the source temperature a reversible heat transfer must happen through some kind of heat engine receiving a Q from the source and delivering it to the flow extracting some work in the process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 744 A heatpowered portable air compressor consists of three components a an adiabatic compressor b a constant pressure heater heat supplied from an outside source and c an adiabatic turbine Ambient air enters the compressor at 100 kPa 300 K and is compressed to 600 kPa All of the power from the turbine goes into the compressor and the turbine exhaust is the supply of compressed air If this pressure is required to be 200 kPa what must the temperature be at the exit of the heater Solution P2 600 kPa P4 200 kPa Adiabatic and reversible compressor Process q 0 and sgen 0 Energy Eq413 h wc h2 Entropy Eq78 s2 s 1 For constant specific heat the isentropic relation becomes Eq823 T2 T1 P2 P1 k1 k 300 K 602857 5008 K wc CP0T2 T1 10045008 300 2015 kJkg Adiabatic and reversible turbine q 0 and sgen 0 Energy Eq413 h3 wT h4 Entropy Eq78 s4 s3 For constant specific heat the isentropic relation becomes Eq623 T4 T3P4P3 k1 k T3 20060002857 07304 T 3 Energy Eq for shaft wc wT CP0T3 T4 2015 kJkg 1004 kJkgΚ T31 07304 T3 7444 K 2 1 v T s 1 300 100 kPa 3 P 2 3 4 200 kPa 600 kPa 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 745 A twostage compressor having an interstage cooler takes in air 300 K 100 kPa and compresses it to 2 MPa as shown in Fig P746 The cooler then cools the air to 340 K after which it enters the second stage which has an exit pressure of 15 MPa Both stages are adiabatic and reversible Find q in the cooler total specific work and compare this to the work required with no intercooler Solution CV Stage 1 air Steady flow Process adibatic q 0 reversible sgen 0 Energy Eq413 wC1 h2 h1 Entropy Eq78 s2 s1 Assume constant CP0 1004 from A5 and isentropic leads to Eq623 T2 T1P2P1 k1 k 3002000100 0286 7067 K wC1 h1 h2 CP0T1 T2 1004 kJkgK 300 7067 K 4083 kJkg CV Intercooler no work and no changes in kinetic or potential energy q23 h3 h2 CP0T3 T2 1004340 7067 3682 kJkg CV Stage 2 Analysis the same as stage 1 So from Eq623 T4 T3P4P3 k1 k 340 K 152 0286 6046 K wC2 h3 h4 CP0T3 T4 1004340 6046 2657 kJkg Same flow rate through both stages so the total work is the sum of the two wcomp wC1 wC2 4083 2657 6828 kJkg For no intercooler P2 1574 MPa same analysis as stage 1 So Eq623 T2 300 K 15000100 0286 12556 K wcomp 1004 kJkgK 300 12556 K 9594 kJkg C1 C2 1 2 3 4 2 W Q 1 W cooler inter Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 746 A certain industrial process requires a steady supply of saturated vapor steam at 200 kPa at a rate of 05 kgs Also required is a steady supply of compressed air at 500 kPa at a rate of 01 kgs Both are to be supplied by the process shown in Fig P746 Steam is expanded in a turbine to supply the power needed to drive the air compressor and the exhaust steam exits the turbine at the desired state Air into the compressor is at the ambient conditions 100 kPa 20C Give the required steam inlet pressure and temperature assuming that both the turbine and the compressor are reversible and adiabatic Solution CV Each device Steady flow Both adiabatic q 0 and reversible sgen 0 3 1 4 2 Steam turbine Air compressor Compressor s4 s3 T4 T3P4P3 k1 k 2932 K 500 100 0286 4646 K WC m3h3 h4 01 kgs 1004 kJkgK 2932 4646 K 172 kW Turbine Energy WT 172 kW m1h1 h2 Entropy s2 s 1 Table B12 P2 200 kPa x2 1 h2 27066 kJkg s2 71271 kJkgK h1 27066 17205 27410 kJkg s1 s2 71271 kJkg K 300 kPa s s2 h 27830 kJkg Interpolate between the 200 and 300 kPa s T 300 kPa 200 kPa h 2741 2 1 P 200 300 200 2741 270663 27830 270663 245 kPa T 12023 16055 12023 2741 270663 27830 270663 1384C If you use the software you get At h1 s1 P1 242 kPaT1 1383C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 747 A certain industrial process requires a steady 075 kgs supply of compressed air at 500 kPa at a maximum temperature of 30C This air is to be supplied by installing a compressor and aftercooler Local ambient conditions are 100 kPa 20C Using an reversible compressor determine the power required to drive the compressor and the rate of heat rejection in the aftercooler Solution Air Table A5 R 0287 kJkgK Cp 1004 kJkg K k 14 State 1 T1 To 20oC P1 Po 100 kPa m 05 kgs State 2 P2 P3 500 kPa State 3 T3 30oC P3 500 kPa Compressor Assume Isentropic adiabatic q 0 and reversible sgen 0 From entropy equation Eq79 this gives constant s which is expressed for an ideal gas in Eq623 T2 T1 P2P1 k1 k 29315 K 50010002857 4646 K Energy Eq413 qc h1 h2 wc qc 0 assume constant specific heat from Table A5 wc CpT1 T2 1004 kJkgK 29315 4646 K 1720 kJkg W C m wC 129 kW Aftercooler Energy Eq413 q h2 h3 w w 0 assume constant specific heat Q m q m CpT3 T2 075 kgs 1004 kJkgK 30315 4646 K 1216 kW 1 3 2 Q cool Compressor W c Compressor section Aftercooler section Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 748 Consider a steam turbine power plant operating near critical pressure as shown in Fig P748 As a first approximation it may be assumed that the turbine and the pump processes are reversible and adiabatic Neglecting any changes in kinetic and potential energies calculate a The specific turbine work output and the turbine exit state b The pump work input and enthalpy at the pump exit state c The thermal efficiency of the cycle Solution WT QH QL WP in 1 2 3 4 P1 P4 20 MPa T1 700 C P2 P3 20 kPa T3 40 C a State 1 P T Table B13 h1 38091 kJkg s1 67993 kJkg K CV Turbine Entropy Eq79 s2 s1 67993 kJkg K Table B12 s2 08319 x2 70766 x2 08433 h2 2514 08433 235833 22401 Energy Eq413 wT h1 h2 1569 kJkg b State 3 P T Compressed liquid take sat liq Table B11 h3 1675 kJkg v3 0001008 m3kg Property relation v constant gives work from Eq715 as wP v3 P4 P3 000100820000 20 201 kJkg h4 h3 wP 1675 201 1876 kJkg c The heat transfer in the boiler is from energy Eq413 qboiler h1 h4 38091 1876 36215 kJkg wnet 1569 201 15489 kJkg ηTH wnetqboiler 15489 36215 0428 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Transient processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 749 A 10 m tall 2 m2 cross sectional area water tank is on a tower so the bottom is 5 m up from ground level and the top is open to the atmosphere It is initially empty and then filled by a pump taking water at ambient T 17oC 100 kPa from a small pond at ground level Assume the process is reversible and find the total pump work CV Pump pipe and water tank In this problem there clearly is a potential energy change but we will neglect any kinetic energy Pressure of the water in and pressure in the tank is the same Po which is the pressure where the volume water surface expands in the tank Continuity Eq415 m2 0 m in Energy Eq416 m2u2 gZ2 0 0 1W2 atm minhin gZin Wpump Entropy Eq713 m2s2 0 dQT 1S2 gen minsin minsin Process Adiabatic 1Q2 0 Process ideal 1S2 gen 0 s2 sin and Zin 0 From the entropy we conclude the T stays the same recall Eq611 and the work to the atmosphere is 1W2 atm Po V2 0 m2P2v2 So this will cancel the the incoming flow work included in minhin The energy equation becomes with Zin 0 m2u2 gZ2 m2P2v2 minuin Pinvin Wpump and then since state 2 Po Tin and state in Po Tin then u2 uin Left is m2gZ2 Wpump The final elevation is the average elevation for the mass Z2 5m 5m 10m Wpump m2g Z2 997 kgm3 10 2 m3 9807 ms2 10 m 1 955 516 J 1 9555 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 750 Air in a tank is at 300 kPa 400 K with a volume of 2 m3 A valve on the tank is opened to let some air escape to the ambient to a final pressure inside of 200 kPa Find the final temperature and mass assuming a reversible adiabatic process for the air remaining inside the tank Solution CV Total tank Continuity Eq415 m2 m1 mex Energy Eq416 m2u2 m1u1 mexhex 1Q2 1W 2 Entropy Eq712 m2s2 m1s1 mexsex dQT 1S 2 gen Process Adiabatic 1Q2 0 rigid tank 1W2 0 This has too many unknowns we do not know state 2 CV m2 the mass that remains in the tank This is a control mass Energy Eq35 m2u2 u1 1Q2 1W2 Entropy Eq614 m2s2 s1 dQT 1S2 gen Process Adiabatic 1Q2 0 Reversible 1S2 gen 0 s2 s1 Ideal gas and process Eq623 T2 T1 P2 P1 k1 k 400 K 20030002857 35625 K m2 RT2 P2V 200 2 0287 35625 kPa m3 kJkgK K 3912 kg Notice that the work term is not zero for mass m2 The work goes into pushing the mass mex out cb m 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 751 A 05 m3 tank contains carbon dioxide at 300 K 150 kPa is now filled from a supply of carbon dioxide at 300 K 150 kPa by a compressor to a final tank pressure of 450 kPa Assume the whole process is adiabatic and reversible Find the final mass and temperature in the tank and the required work to the compressor CV The tank and the compressor Continuity Eq415 m2 m1 min Energy Eq416 m2u2 m1u1 1Q2 1W2 minh in Entropy Eq713 m2s2 m1s1 dQT 1S2 gen minsin Process Adiabatic 1Q2 0 Process ideal 1S2 gen 0 s1 s in m2s2 m1s1 minsin m1 mins1 m2s1 s2 s1 Constant s so since the temperatures are modest use Eq623 T2 T1 P2 P1 k1 k 300 K 45015002242 38379 K m EA 150 kPa 05 mA3 A E01889 kJkgK 300 KE A 13235 kg 1 P1V1RT1 mA2E A PA2E AVA2E ARTA2E A 450 kPa 05 mA3E A01889 38379 kJkg 31035 kg mAinE A 178 kg A1E AWA2E A mAinE AhAinE A mA1E AuA1E A mA2E AuA2E A mAinE ARTAinE A mA1E AuA1E A uAinE A mA2E AuA2E A uAinE A 178 01889 300 mA1E A 0 31035 0653 38379 300 6893 kJ work must come in Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 752 A tank contains 1 kg of carbon dioxide at 6 MPa 60AoE AC and it is connected to a turbine with an exhaust at 1000 kPa The carbon dioxide flows out of the tank and through the turbine to a final state in the tank of saturated vapor is reached If the process is adiabatic and reversible find the final mass in the tank and the turbine work output CV The tank and turbine This is a transient problem Continuity Eq415 mA2E A mA1E A mAexE Energy Eq416 mA2E AuA2E A mA1E AuA1E A mAexE AhAexE A A1E AQA2E A A1E AWA2E Entropy Eq712 mA2E AsA2E A mA1E AsA1E A mAexE AsAexE A dQT A1E ASA2 genE Process Adiabatic A1E AQA2E A 0 reversible A1E ASA2 genE A 0 State 1 vA1E A 000801 mA3E Akg uA1E A 32251 kJkg sA1E A 12789 kJkgK State 2 Sat vapor 1 property missing CV mA2E A the mass that remains in the tank This is a control mass Process Adiabatic A1E AQA2E A 0 Reversible A1E ASA2 genE A 0 Entropy Eq614 mA2E AsA2E A sA1E A dQT A1E ASA2 genE A 0 0 sA2E A s1 this is the missing property State 2 TA2E A 1919AoE AC vA2E A 0018851 mA3E Akg uA2E A 28587 kJkg State exit sAexE A sA2E A s1 follows from entropy Eq for first CV with the use of the continuity equation Use 10045 kPa for 40AoE AC xAexE A 12789 013829 0924796 hAexE A 29817 kJkg Tank volume constant so V mA1E AvA1E A mA2E AvA2E mA2E A mA1E A vA1E A vA2E A 1 kg 000801 0018851 04249 kg From energy eq A1E AWA2E A mA1E AuA1E A mA2E AuA2E A mAexE AhAexE 1 32251 04249 28587 05751 29817 kg kJkg 2957 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 753 Air in a tank is at 300 kPa 400 K with a volume of 2 mA3E A A valve on the tank is opened to let some air escape to the ambient to a final pressure inside of 200 kPa At the same time the tank is heated so the air remaining has a constant temperature What is the mass average value of the s leaving assuming this is an internally reversible process Solution CV Tank emptying process with heat transfer Continuity Eq415 mA2E A m1 me Energy Eq416 mA2E AuA2E A m1u1 mehe A1E AQA2E Entropy Eq713 mA2E AsA2E A m1s1 mese A1E AQA2E AT 0 Process TA2E A T1 A1E AQA2E A in at 400 K Reversible A1E ASA2 genE A 0 State 1 Ideal gas m1 P1VRT1 300 20287 400 52265 kg State 2 200 kPa 400 K m2 P2VRT2 200 kPa 2 mA3E A0287 400 kJkg 34843 kg me 17422 kg From the energy equation A1E AQA2E A mA2E AuA2E A m1u1 mehe 34843 28649 52265 28649 17422 4013 17422 kg 4013 28649 kJkg 200 kJ mese m1s1 mA2E AsA2E A A1E AQA2E AT 52265715926 0287 ln 300100 34843715926 0287 ln 200100 200400 mese 35770 24252 05 12018 kJK se 1201817422 689817 68982 kJkg K Note that the exit state e in this process is for the air before it is throttled across the discharge valve The throttling process from the tank pressure to ambient pressure is a highly irreversible process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 754 An insulated 2 mA3E A tank is to be charged with R134a from a line flowing the refrigerant at 3 MPa The tank is initially evacuated and the valve is closed when the pressure inside the tank reaches 2 MPa The line is supplied by an insulated compressor that takes in R134a at 5C quality of 965 and compresses it to 3 MPa in a reversible process Calculate the total work input to the compressor to charge the tank Solution CV Compressor R134a Steady 1 inlet and 1 exit flow no heat transfer Energy Eq413 qc hA1E A hA1E A hA2E A wc Entropy Eq79 sA1E A dqT sAgenE A sA1E A 0 sA2E inlet TA1E A 5AoE AC xA1E A 0965 use Table B51 sA1E A sAfE A xA1E AsAfgE A 10243 096506995 16993 kJkg K hA1E A hAfE A xA1E AhAfgE A 20675 096519457 39451 kJkg exit PA2E A 3 MPa From the entropy eq sA2E A sA1E A 16993 kJkg K TA2E A 90AoE AC hA2E A 43619 kJkg wc hA1E A hA2E A 4168 kJkg CV Tank VATE A 2 m3 PATE A 2 MPa Energy Eq416 Q mAiE AhAiE A mA3E AuA3E A mA1E AuA1E A mehe W Process and states have Q 0 W 0 me 0 mA1E A 0 mA3E A mi Final state PA3E A 2 MPa uA3E A hAiE A hA2E A 43619 kJkg TA3E A 90AoE AC vA3E A 001137 m3kg sA3E A 1785 kJkgK mA3E A VATE AvA3E A 2001137 1759 kg The work term is from the specific compressor work and the total mass Wc mATE Awc 7 331 kJ Comment The filling process is not reversible note sA3E A sA2E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 755 An underground salt mine 100 000 mA3E A in volume contains air at 290 K 100 kPa The mine is used for energy storage so the local power plant pumps it up to 21 MPa using outside air at 290 K 100 kPa Assume the pump is ideal and the process is adiabatic Find the final mass and temperature of the air and the required pump work Solution CV The mine volume and the pump Continuity Eq415 mA2E A mA1E A mAinE Energy Eq416 mA2E AuA2E A mA1E AuA1E A A1E AQA2E A A1E AWA2E A mAinE AhAinE Entropy Eq713 mA2E AsA2E A mA1E AsA1E A AdQTEA A1E ASA2 genE A mAinE AsAinE Process Adiabatic A1E AQA2E A 0 Process ideal A1E ASA2 genE A 0 sA1E A sAinE mA2E AsA2E A mA1E AsA1E A mAinE AsAinE A mA1E A mAinE AsA1E A mA2E AsA1E A sA2E A sA1E Constant s Eq819 sAo T2E A sAo TiE A R lnPA2E A PAinE A sAo T2E A 683521 0287 ln 21 77090 kJkg K A7 TA2E A 680 K uA2E A 49694 kJkg mA1E A PA1E AVA1E ARTA1E A EA 100 kPa 10A5 A mA3 A E0287 kJkgK 290 KE A 120149 10A5E A kg mA2E A PA2E AVA2E ARTA2E A 100 kPa 2110A5E A mA3E A0287 680 kJkg 10760 10A5E A kg mAinE A 9558510A5E A kg A1E AWA2E A mAinE AhAinE A mA1E AuA1E A mA2E AuA2E mAinE A29043 mA1E A20719 mA2E A49694 2322 10A8E A kJ 2 1 i P v T s 1 i 2 T2 290 400 s C 100 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 756 R410A at 120AoE AC 4 MPa is in an insulated tank and flow is now allowed out to a turbine with a backup pressure of 800 kPa The flow continue to a final tank pressure of 800 kPa and the process stops If the initial mass was 1 kg how much mass is left in the tank and what is the turbine work assuming a reversible process Solution CV Total tank and turbine Continuity Eq415 mA2E A mA1E A mAexE Energy Eq416 mA2E AuA2E A mA1E AuA1E A mAexE AhAexE A A1E AQA2E A A1E AWA2E Entropy Eq712 mA2E AsA2E A mA1E AsA1E A mAexE AsAexE A dQT A1E ASA2 genE Process Adiabatic A1E AQA2E A 0 Reversible A1E ASA2 genE A 0 This has too many unknowns we do not know state 2 only PA2E A CV mA2E A the mass that remains in the tank This is a control mass Entropy Eq73 637 mA2E AsA2E A sA1E A dQT A1E ASA2 genE Process Adiabatic A1E AQA2E A 0 Reversible A1E ASA2 genE A 0 sA2E A s1 State 1 vA1E A 000897 mA3E Akg uA1E A 33139 kJkg sA1E A 11529 kJkgK State 2 Ps TA2E A 3323AoE AC vA2E A 037182 mA3E Akg uA2E A 28129 kJkg State exit sAexE A sA2E A s1 follows from entropy Eq for first CV using the continuity eq this is identical to state 2 hAexE A 31285 kJkg Tank volume constant so V mA1E AvA1E A mA2E AvA2E mA2E A mA1E A vA1E A vA2E A 1 kg 000897 037182 00241 kg From energy eq A1E AWA2E A mA1E AuA1E A mA2E AuA2E A mAexE AhAexE 1 33139 00241 28129 09759 31285 kg kJkg 193 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Reversible shaft work Bernoulli equation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 757 A river flowing at 05 ms across 1 m high and 10 m wide area has a dam that creates an elevation difference of 2 m How much energy can a turbine deliver per day if 80 of the potential energy can be extracted as work CV Dam from upstream to downstream 2 m lower elevation Continuity Eq AmE A constant AmE AeE A AmE AiE A AAeE AVAeE AvAeE A AAiE AVAiE AvAiE Find the mass flow rate AmE A AVAiE Av ρAVAiE A 997 kgmA3E A 1 10 mA2E A 05 ms 4985 kgs Energy Eq 0 AmE A hAiE A 05VAiE A2E A gZAiE A AmE A hAeE A 05VAeE A2E A gZAeE A AW E The velocity in and out is the same assuming same flow area and the two enthalpies are also the same since same T and P PA0E A This is consistent with Eq714 w gZAiE A ZAeE A loss AW E A 08 AmE A gZAiE A ZAeE A 08 4985 kgs 9807 msA2E A 2 m 78 221 Js 78 221 W 782 kW W AW E A t 782 kJs 24 60 60 s 676 GJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 758 How much liquid water at 15AoE AC can be pumped from 100 kPa to 300 kPa with a 3 kW motor Incompressible flow liquid water and we assume reversible Then the shaftwork is from Eq715 716 w v dP v P 0001 mA3E Akg 300 100 kPa 02 kJkg AmE A AW EwE A A 3 02E A A kW kJkgE A 15 kgs Remark The pump should also generate the kinetic energy so the AmE A will be smaller Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 759 A large storage tank contains saturated liquid nitrogen at ambient pressure 100 kPa it is to be pumped to 500 kPa and fed to a pipeline at the rate of 05 kgs How much power input is required for the pump assuming it to be reversible Solution CV Pump liquid is assumed to be incompressible Table B61 at PAiE A 1013 kPa vAFiE A 000124 mA3E Akg Eq715 wAPUMPE A wcv AvdPEA vAFiE APAeE A PAiE A 000124500 101 0494 kJkg i e liquid nitrogen A WEAPUMPE A A mEAwAPUMPE A 05 kgs 0494 kJkg 0247 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 760 Liquid water at 300 kPa 15C flows in a garden hose with a small ideal nozzle How high a velocity can be generated If the water jet is directed straight up how high will it go Solution Liquid water is incompressible and we will assume process is reversible Bernoullis Eq across the nozzle Eq716 vP A1 2E A VA2E A V A 2vPEA A 20001001 300101 1000EA 1996 ms Bernoullis Eq716 for the column A1 2E A VA2E A gZ Z A1 2E A VA2E Ag vPg 0001001 mA3E Akg 300 101 kPa 1000 PakPa 9807 msA2E 203 m Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 761 A wave comes rolling in to the beach at 2 ms horizontal velocity Neglect friction and find how high up elevation on the beach the wave will reach We will assume a steady reversible single flow at constant pressure and temperature for the incompressible liquid water The water will flow in and up the sloped beach until it stops V 0 so Bernoulli Eq717 leads to gzAinE A A1 2E AVA2E AinE A gzAexE A 0 zAexE A zAinE A A 1 2gE AVA2 inE A A 1 2 9807 ms2 E A 2A2E A msA2E A 0204 m Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 762 A small pump takes in water at 20C 100 kPa and pumps it to 25 MPa at a flow rate of 100 kgmin Find the required pump power input Solution CV Pump Assume reversible pump and incompressible flow With single steady state flow it leads to the work in Eq715 wApE A AvdPEA vAiE APAeE A PAiE A 0001002 mA3E Akg 2500 100 kPa 24 kJkg AWE ApE A AmE AwApE A A100 60E A Akgmin secminE A 24 kJkg 40 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 763 An irrigation pump takes water from a river at 10C 100 kPa and pumps it up to an open canal at a 50 m higher elevation The pipe diameter in and out of the pump is 01 m and the motor driving the pump is 5 hp Neglect kinetic energies and friction find the maximum possible mass flow rate CV the pump The flow is incompressible and steady flow The pump work is the difference between the flow work in and out and from Bernoullis eq for the pipe that is equal to the potential energy increase sincle pump inlet pressure and pipe outlet pressure are the same wApE A v P g Z 981 50 Jkg 049 kJkg The horsepower is converted from Table A1 W motor 5 hp 5 hp 0746 kWhp 373 kW Am E A W motor wApE A 373 kW 049 kJkg 76 kgs Comment Am E A AVv V Am v EAE A A 4m ρ π D2 E A A 4 76 997 π 012 E A 097 ms The power to generated the kinetic energy is Power Am E A 05 VA2E A 76 kgs 05 097A2E A msA2E A 357 W This is insignificant relative to the power needed for the potential energy increase Pump inlet and the pipe exit both have close to atmospheric pressure Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 764 A firefighter on a ladder 25 m above ground should be able to spray water an additional 10 m up with the hose nozzle of exit diameter 25 cm Assume a water pump on the ground and a reversible flow hose nozzle included and find the minimum required power Solution CV pump hose water column total height difference 35 m Continuity Eq43 611 AmE AinE A AmE AexE A ρAVAnozzleE Energy Eq412 AmE AwApE A AmE Ah VA2E A2 gzAinE A AmE Ah VA2E A2 gzAexE Process hAinE A hAexE A VAinE A VAexE A 0 zAexE A zAinE A 35 m ρ 1v 1vAfE wApE A gzAexE A zAinE A 98135 0 3432 Jkg The velocity in the exit nozzle is such that it can rise 10 m Make that column a CV for which Bernoulli Eq717 is gzAnozE A A1 2E AVA2E AnozE A gzAexE A 0 VAnozE A A 2gzex znozEA A 2 981 10EA 14 ms 10 m 35 m AmE A Aπ vf E A D 2 E A 2E AVAnozE A π4 0025A2E A 14 0001 6873 kgs AWE ApE A AmE AwApE A 6873 kgs 3432 Jkg 236 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 765 Saturated R410A at 10C is pumpedcompressed to a pressure of 20 MPa at the rate of 05 kgs in a reversible adiabatic process Calculate the power required and the exit temperature for the two cases of inlet state of the R410A a quality of 100 b quality of 0 Solution CV PumpCompressor AmE A 05 kgs R410A a State 1 Table B51 TA1E A 10AoE AC xA1E A 10 Saturated vapor PA1E A PAgE A 5731 kPa hA1E A hAgE A 27578 kJkg sA1E A sAgE A 10567 kJkgK Assume Compressor is isentropic sA2E A sA1E A 10567 kJkgK hA2E A 33418 kJkg TA2E A 52AoE AC Energy Eq413 qc hA1E A hA2E A wc qc 0 wcs hA1E A hA2E A 584 kJkg W C m wC 292 kW b State 1 TA1E A 10AoE AC x1 0 Saturated liquid This is a pump PA1E A 5731 kPa hA1E A hAfE A 4280 kJkg vA1E A vAfE A 0000827 m3kg Energy Eq413 qp hA1E A hA2E A wp qp 0 Assume Pump is isentropic and the liquid is incompressible Eq715 wps v dP vA1E APA2E A PA1E A 118 kJkg hA2E A hA1E A wp 4280 118 4398 kJkg PA2E A 2 MPa Assume State 2 is approximately a saturated liquid T2 92AoE AC W P m wP 059 kW 2a 1a P v T s 1a 2a 1b 2b 1b 2b Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 766 Liquid water at ambient conditions 100 kPa 25C enters a pump at the rate of 05 kgs Power input to the pump is 3 kW Assuming the pump process to be reversible determine the pump exit pressure and temperature Solution CV Pump Steady single inlet and exit flow with no heat transfer Energy Eq413 w hAiE A hAeE A A WEAA mEA 305 60 kJkg Using also incompressible media we can use Eq715 w AvdPEA vAiE APAeE A PAiE A 0001003PAeE A 100 from which we can solve for the exit pressure PAeE A 100 600001003 6082 kPa 6082 MPa W e i Pump A WEA 3 kW PAiE A 100 kPa TAiE A 25C A mEA 05 kgs Energy Eq hAeE A hAiE A w 10487 6 11087 kJkg Use Table B14 at 5 MPa TAeE A 253C Remark If we use the software we get A si 036736 se EAt se Pe E A TAeE A 251C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 767 The underwater bulb nose of a container ship has a velocity relative to the ocean water as 10 ms What is the pressure at the front stagnation point that is 2 m down from the water surface Solution CV A stream line of flow from the freestream to the stagnation point on the front of the bulb nose Eq717 vPAeE A PAiE A A1 2E A VA2 eE A VA2 iE A gZAeE A ZAiE A 0 Horizontal so ZAeE A ZAiE A and VAeE A 0 P A 1 2vE A VA2 iE A A 102 E0001001 2000E A 4995 kPa PAiE A PAoE A gHv 101 kPa 981 ms2 2 m0001001 m3kg 1000 JkJ 1206 kPa PAeE A PAiE A P 1206 4995 1706 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 768 A small water pump on ground level has an inlet pipe down into a well at a depth H with the water at 100 kPa 15C The pump delivers water at 400 kPa to a building The absolute pressure of the water must be at least twice the saturation pressure to avoid cavitation What is the maximum depth this setup will allow Solution CV Pipe in well no work no heat transfer From Table B11 P inlet pump 2 Psat 15C 21705 341 kPa Process Assume KE v constant Bernoulli Eq716 v P g H 0 i e H 1000 JkJ 0001001 m3kg 341 100 kPa 980665 ms2 H 0 H 986 m Since flow has some kinetic energy and there are losses in the pipe the height is overestimated Also the start transient would generate a very low inlet pressure due to the necessary dynamic forces it moves flow by suction Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 769 A pumpcompressor pumps a substance from 150 kPa 10C to 1 MPa in a reversible adiabatic process The exit pipe has a small crack so that a small amount leaks to the atmosphere at 100 kPa If the substance is a water b R 134a find the temperature after compression and the temperature of the leak flow as it enters the atmosphere neglecting kinetic energies Solution CV Compressor reversible adiabatic Eq413 hA1E A wAcE A hA2E A Eq78 sA1E A sA2E State 2 PA2E A sA2E A sA1E A CV Crack Steady throttling process Eq413 hA3E A hA2E A Eq78 sA3E A sA2E A sAgenE State 3 PA3E A hA3E A hA2E a Water 1 compressed liquid Table B11 wAcE A AvdPEA vAf1E APA2E A PA1E A 0001 m3kg 1000 150 kPa 085 kJkg hA2E A hA1E A wAcE A 4199 085 4284 kJkg TA2E A 102C PA3E A hA3E A hA2E A compressed liquid at 102C P v T s 1 2 1 2 3 3 States 1 and 3 are at 150 100 kPa and same v b R134a 1 superheated vapor Table B52 sA1E A 18220 kJkgK sA2E A sA1E A PA2E A TA2E A 831C hA2E A 45557 kJkg wAcE A hA2E A hA1E A 45557 41060 4497 kJkg PA3E A hA3E A hA2E A TA3E A 606C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 770 Atmospheric air at 100 kPa 17C blows at 60 kmh towards the side of a building Assume the air is nearly incompressible find the pressure and the temperature at the stagnation point zero velocity on the wall Solution CV A stream line of flow from the freestream to the wall Eq717 vPAeE APAiE A A1 2E A VA2 eE AVA2 iE A gZAeE A ZAiE A 0 V VAiE A 60 Akm hE A 1000 A m kmE A A 1 3600E A Ah sE A 16667 ms v A RTi EPi E A A0287 29015 100E A AkJkg kPaE A 08323 Am3 EkgE P A 1 2vE A VA2 iE A A 166672 E08323 2000E A A ms2 Em3kg JkJE A 017 kPa PAeE A PAiE A P 10017 kPa Then Eq623 for an isentropic process TAeE A TAiE A PAeE APAiE AA0286E A 29015 K 10005 2903 K Very small effect due to low velocity and air is light large specific volume v P s T 1 2 2 1 150 kPa 1 MPa 100 kPa h C 3 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 771 A small pump is driven by a 2 kW motor with liquid water at 150 kPa 10C entering Find the maximum water flow rate you can get with an exit pressure of 1 MPa and negligible kinetic energies The exit flow goes through a small hole in a spray nozzle out to the atmosphere at 100 kPa Find the spray velocity Solution CV Pump Liquid water is incompressible so work from Eq715 AW E A AmE Aw AmE AvPAeE A PAiE A AmE A AW E A vPAeE A PAiE A 20001003 mA3E Akg 1000 150 kPa 235 kgs CV Nozzle No work no heat transfer v constant Bernoulli Eq717 A1 2E AVA2 exE A vP 0001 mA3E Akg 1000 100 kPa 09 kJkg 900 Jkg VAexE A A 2 900 JkgEA 424 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 772 A speed boat has a small hole in the front of the drive with the propeller that sticks down into the water at a water depth of 025 m Assume we have a stagnation point at that hole when the boat is sailing with 40 kmh what is the total pressure there Solution CV A stream line of flow from the freestream to the wall Eq717 vPAeE APAiE A A1 2E A VA2 eE AVA2 iE A gZAeE A ZAiE A 0 VAiE A 40 Akm hE A 1000 A m kmE A A 1 3600E A Ah sE A 11111 ms P A 1 2vE A VA2 iE A A 111112 E0001001 2000E A 6166 kPa PAiE A PAoE A gHv 101 981 025 0001001 1000 10345 kPa PAeE A PAiE A P 10345 6166 165 kPa Remark This is fast for a boat Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 773 You drive on the highway with 120 kmh on a day with 17C 100 kPa atmosphere When you put your hand out of the window flat against the wind you feel the force from the air stagnating ie it comes to relative zero velocity on your skin Assume the air is nearly incompressible and find the air temperature and pressure right on your hand Solution Energy Eq413 A1 2E A VA2E A hAoE A hAstE TAstE A TAoE A A1 2E A VA2E ACApE A 17 A1 2E A 12010003600A2E A 11004 17 55551004 176C v RTAoE APAoE A 0287 290100 08323 mA3E Akg From Bernoulli Eq717 vP A1 2E A VA2E PAstE A PAoE A A1 2E A VA2E Av 100 555508323 1000 10067 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 774 A small dam has a pipe carrying liquid water at 150 kPa 20C with a flow rate of 2000 kgs in a 05 m diameter pipe The pipe runs to the bottom of the dam 15 m lower into a turbine with pipe diameter 035 m Assume no friction or heat transfer in the pipe and find the pressure of the turbine inlet If the turbine exhausts to 100 kPa with negligible kinetic energy what is the rate of work Solution CV Pipe Steady flow no work no heat transfer States compressed liquid B11 vA2E A vA1E A vAfE A 0001002 mA3E Akg Continuity Eq43 AmE A ρ AV AVv VA1E A AmE AvA1E A AA1E A 2000 kgs 0001002 mA3E Akg Aπ 4E A 05A2E A mA2E A 102 m sA1E VA2E A AmE AvA2E AAA2E A 2000 kgs 0001002 mA3E Akg Aπ 4E A 035A2E A mA2E A 2083 m sA1E From Bernoulli Eq716 for the pipe incompressible substance vPA2E A PA1E A A1 2E A VA2 2E A VA2 1E A g ZA2E A ZA1E A 0 PA2E A PA1E A A1 2E A VA2 1E A VA2 2E A g ZA1E A ZA2E Av 150 kPa A 1 2 1022 20832 980665 15 E1000 0001002E A A m2s2 EJkJ m3kgE 150 178 1322 kPa Note that the pressure at the bottom should be higher due to the elevation difference but lower due to the acceleration Now apply the energy equation Eq713 for the total control volume w v dP A1 2E A VA2 1E A VA2 3E A gZA1E A ZA3E A 0001002 100 150 A1 2E A102A2E A 980665 15 1000 025 kJkg AW E A AmE Aw 2000 kgs 025 kJkg 500 kW 1 2 Turbine 3 DAM Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 775 An air flow at 100 kPa 290 K 100 ms is directed towards a wall At the wall the flow stagnates comes to zero velocity without any heat transfer Find the stagnation pressure a assuming incompressible flow b assume an adiabatic compression Hint T comes from the energy equation Solution Ideal gas v RTAoE APAoE A 0287 290100 08323 mA3E Akg Kinetic energy A1 2E A VA2E A A1 2E A 100A2E A1000 5 kJkg a Reversible and incompressible gives Bernoulli Eq717 P A1 2E A VA2E Av 508323 6 kPa PAstE A PAoE A P 106 kPa b adiabatic compression Energy Eq413 A1 2E A VA2E A hAoE A hAstE 0 St cb hAstE A hAoE A A1 2E A VA2E A CApE AT T A1 2E A VA2E ACApE A 51004 5C TAstE A 290 5 295 K Entropy Eq79 assume also reversible process sAoE A sAgenE A A1T dqEA sAstE as dq 0 and sAgenE A 0 then it follows that s constant This relation gives Eq623 PAstE A PAoE A Tst ETo E A k k1 E A 100 295290A35E A 106 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 776 A flow of air at 100 kPa 300 K enters a device and goes through a polytropic process with n 13 before it exits at 800 K Find the exit pressure the specific work and heat transfer using constant specific heats Solution CV Steady state device single inlet and single exit flow Energy Eq413 hA1E A q hA2E A w Neglect kinetic potential energies Entropy Eq79 sA1E A dqT sAgenE A sA2E Te 800 K Ti 300 K Pi 100 kPa Process Eq628 Pe Pi Te TiA n n1 E A 100 800300 A 13 03 E A 7 012 kPa and the process leads to Eq718 for the work term w A n n1E A R Te Ti 1303 0287 kJkgK 800 300 K 6218 kJkg q he hi w CP Te Ti w 1004 kJkgK 800 300 K 6218 kJkg 1198 kJkg P v T s e i n 1 n 13 i e n 1 n k 14 n 13 Notice dP 0 so dw 0 ds 0 so dq 0 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 777 Solver the previous problem but use the air tables A7 Air at 100 kPa 300 K flows through a device at steady state with the exit at 1000 K during which it went through a polytropic process with n 13 Find the exit pressure the specific work and heat transfer Solution CV Steady state device single inlet and single exit flow Energy Eq413 hA1E A q hA2E A w Neglect kinetic potential energies Entropy Eq79 sA1E A dqT sAgenE A sA2E Te 800 K Ti 300 K Pi 100 kPa Process Eq628 Pe Pi Te TiA n n1 E A 100 800300 A 13 03 E A 7 012 kPa and the process leads to Eq718 for the work term w A n n1E A R Te Ti A13 03E A 1303 0287 kJkgK 800 300 K 6218 kJkg q he hi w 8222 3005 6218 1001 kJkg P v T s e i n 1 n 13 i e n 1 n k 14 n 13 Notice dP 0 so dw 0 ds 0 so dq 0 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 778 Helium gas enters a steadyflow expander at 800 kPa 300C and exits at 120 kPa The expansion process can be considered as a reversible polytropic process with exponent n 13 Calculate the mass flow rate for 150 kW power output from the expander Solution Wexp i e Q CV expander reversible polytropic process From Eq628 TAeE A TAiE A A Pe EPi E A A n1 n E A 5732 A 120 800 E A A 03 13 E A 370 K Work evaluated from Eq718 w AvdPEA AnR n1E A TAeE A TAiE A A13 207703 03E A kJkgK 370 5732 K 18289 kJkg A mEA A WEAw 150 kW 18289 kJkg 0082 kgs P v T s e i n 1 n 13 i e n 1 n k 1667 n 13 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 779 A flow of 4 kgs ammonia goes through a device in a polytropic process with an inlet state of 150 kPa 20C and an exit state of 400 kPa 60C Find the polytropic exponent n the specific work and heat transfer Solution CV Steady state device single inlet and single exit flow Energy Eq413 hA1E A q hA2E A w Neglect kinetic potential energies Entropy Eq79 sA1E A dqT sAgenE A sA2E Process Eq627 PA1E AvA1E AnE A PA2E AvA2E AnE A State 1 Table B22 vA1E A 079774 sA1E A 57465 kJkg K hA1E A 14229 kJkg State 2 Table B22 vA2E A 03955 sA2E A 58560 kJkg K hA2E A 15904 kJkg ln PA2E APA1E A n ln vA1E A vA2E A 098083 n 070163 n ln PA2E APA1E A ln vA1E A vA2E A 13979 From the process and the integration of v dP gives Eq718 wAshaftE A A n n1E A PA2E AvA2E A PA1E AvA1E A 35132 1582 11966 1354 kJkg q hA2E A w hA1E A 15904 14229 1354 321 kJkg P v T s 2 1 n 1 n 154 1 2 n 1 n k 13 n 154 Notice dP 0 so dw 0 ds 0 so dq 0 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 780 Calculate the air temperature and pressure at the stagnation point right in front of a meteorite entering the atmosphere 50 C 50 kPa with a velocity of 2000 ms Do this assuming air is incompressible at the given state and repeat for air being a compressible substance going through an adiabatic compression Solution Kinetic energy A1 2E A VA2E A A1 2E A 2000A2E A1000 2000 kJkg Ideal gas vAatmE A RTP 0287 kJkgK 223 K50 kPa 128 mA3E Akg a incompressible Energy Eq413 h A1 2E A VA2E A 2000 kJkg If A5 T hCApE A 1992 K unreasonable too high for that CApE Use A7 hAstE A hAoE A A1 2E A VA2E A 22322 2000 22233 kJkg TAstE A 1977 K Bernoulli incompressible Eq717 P PAstE A PAoE A A1 2E A VA2E Av 2000 ms2128 mA3E Akg 15625 kPa PAstE A 15625 50 16125 kPa b compressible TAstE A 1977 K the same energy equation From A71 sAo T stE A 89517 kJkg K sAo T oE A 65712 kJkg K Eq619 PAstE A PAoE A eEAsAo T st A sAo ET o ARE 50 exp A89517 65712 0287E A 200 075 kPa Notice that this is highly compressible v is not constant Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 781 An expansion in a gas turbine can be approximated with a polytropic process with exponent n 125 The inlet air is at 1200 K 800 kPa and the exit pressure is 125 kPa with a mass flow rate of 075 kgs Find the turbine heat transfer and power output Solution CV Steady state device single inlet and single exit flow Energy Eq413 hi q he w Neglect kinetic potential energies Entropy Eq79 si dqT sAgenE A se Process Eq628 Te Ti Pe PiA n1 n E A 1200 125800A 025 125 E A 82784 K so the exit enthalpy is from Table A71 he 8222 A2784 50E A8774 8222 85294 kJkg The process leads to Eq718 for the work term A WEA A mEAw A mEAnR n1E A Te Ti 075 A125 0287 025E A 82784 1200 4005 kW Energy equation gives A QEA A mEAq A mEAhe hi A WEA 07585294 127781 4005 31865 4005 819 kW P v T s e i n 1 n 125 i e n 1 n k 14 n 125 Notice dP 0 so dw 0 ds 0 so dq 0 Notice this process has some heat transfer in during expansion which is unusual The typical process would have n 15 with a heat loss Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Steady state irreversible processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 782 Consider a steam turbine with inlet 2 MPa 350AoE AC and an exhaust flow as saturated vapor 100 kPa There is a heat loss of 6 kJkg to the ambient Is the turbine possible Solution At the given states Table B13 si 69562 kJkg K se 73593 kJkg K Do the second law for the turbine Eq78 m ese m isi dAQ E AT S gen se si dqT sgen sgen se si dqT 73593 69562 6298 0383 kJkgK 0 Entropy goes up even if q goes out This is an irreversible process v P s T i i e ac e ac 100 kPa 2 MPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 783 A large condenser in a steam power plant dumps 15 MW by condensing saturated water vapor at 45AoE AC to saturated liquid What is the water flow rate and the entropy generation rate with an ambient at 25AoE AC Solution This process transfers heat over a finite temperature difference between the water inside the condenser and the outside ambient cooling water from the sea lake or river or atmospheric air CV The Condensing water flow Energy Eq 0 Am E A hg hf Q out Am E A Q out hfg A 15 000 239477E A A kW kJkgE A 6264 kgs CV The wall that separates the inside 45AoE AC water from the ambient at 25AoE AC Entropy Eq 71 for steady state operation Condensing water Sea water cb 45AoE AC 25AoE AC AdS dtE A 0 A Q E TE A S gen Q T45 Q T25 S gen S gen A 15 25 273E A AMW KE A A 15 45 273E A AMW KE A 317 AkW KE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 784 R410A at 5AoE AC 700 kPa is throttled so it becomes cold at 40AoE AC What is exit P and the specific entropy generation CV Throttle Steady state Process with q w 0 and VAiE A VAeE A ZAiE A ZAeE Energy Eq413 hAiE A hAeE A Inlet state Table B41 hAiE A 5022 kJkg sAiE A 01989 kJkgK slightly compressed liquid Exit state Table B41 since h hg 26283 kJkg it is twophase P Psat 175 kPa xAeE A A he hf e Ehfg e E A A5022 0 26283E A 019107 sAeE A 0 xAeE A 11273 02154 kJkgK sgen sAeE A sAiE A 02154 01989 00165 kJkgK 2 P v 1 h C T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 785 Ammonia is throttled from 15 MPa 35AoE AC to a pressure of 291 kPa in a refrigerator system Find the exit temperature and the specific entropy generation in this process The throttle process described in Example 65 is an irreversible process Find the entropy generation per kg ammonia in the throttling process Solution The process is adiabatic and irreversible The consideration with the energy given in the example resulted in a constant h and twophase exit flow Table B21 si 12792 kJkg K hi 3468 kJkg Table B21 x hi hfhfg 3468 1344112964 01638 se sf xe sfg 05408 01638 49265 134776 kJkg K We assumed no heat transfer so the entropy equation Eq79 gives sgen se si dqT 134776 12792 0 00686 kJkg K 1 2 e T s i 15 MPa 291 kPa h C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 786 A compressor in a commercial refrigerator receives R410A at 25AoE AC and x 1 The exit is at 1000 kPa and 20AoE AC Is this compressor possible Solution CV Compressor steady state single inlet and exit flow For this device we also assume no heat transfer and ZAiE A ZAeE W C i e cb From Table B41 hAiE A 26977 kJkg sAiE A 10893 kJkgK From Table B42 hAeE A 29549 kJkg sAeE A 1073 kJkgK Entropy gives sgen se si dqT 1073 10893 dqT negative The result is negative unless dq is negative it should go out but T T ambient so this compressor is impossible Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 787 R134a at 30AoE AC 800 kPa is throttled in a steady flow to a lower pressure so it comes out at 10AoE AC What is the specific entropy generation Solution The process is adiabatic and irreversible The consideration of the energy given in example 65 resulted in a constant h and twophase exit flow Table B41 hi 24179 kJkg si 1143 kJkgK compressed liquid State 2 10AoE AC he hi hg so twophase xe he hfhfg 0267 Table B41 se sf xe sfg 09507 0267 07812 116 kJkgK We assumed no heat transfer so the entropy equation Eq79 gives sgen se si dqT 116 1143 0 0017 kJkg K 1 2 e T s i 800 kPa 202 kPa h C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 788 Analyze the steam turbine described in Problem 484 Is it possible Solution CV Turbine Steady flow and adiabatic Continuity Eq49 m 1 m 2 m 3 Energy Eq410 m 1h1 m 2h2 m 3h3 AW E Entropy Eq77 m 1s1 S gen m 2s2 m 3s3 WT 1 2 3 States from Table B13 s1 72337 s2 73010 s3 75066 kJkgK S gen 1 7301 4 75066 5 72337 116 kWK 0 Since it is positive possible Notice the entropy is increasing through the turbine s1 s2 s3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 789 Two flowstreams of water one at 06 MPa saturated vapor and the other at 06 MPa 600C mix adiabatically in a steady flow process to produce a single flow out at 06 MPa 400C Find the total entropy generation for this process Solution Continuity Eq49 AmE A3E A AmE A1E A AmE A2E A Energy Eq410 AmE A3E AhA3E A AmE A1E AhA1E A AmE A2E AhA2E A 1 B12 hA1E A 27568 kJkg sA1E A 6760 kJkgK 2 B13 hA2E A 37009 kJkg sA2E A 82674 kJkgK 3 B13 hA3E A 32703 kJkg sA3E A 77078 kJkgK AmE A1E AAmE A3E A hA3E A hA2E A hA1E A hA2E A 0456 Entropy Eq77 AmE A3E AsA3E A AmE A1E AsA1E A AmE A2E AsA2E A ASE AgenE A ASE AgenE AAmE A3E A sA3E A AmE A1E AAmE A3E A sA1E A AmE A2E AAmE A3E A sA2E A 77078 04566760 054482674 0128 kJkg K 1 2 3 Mixing chamber 2 T s 3 1 600 kPa The mixing process generates entropy The two inlet flows could have exchanged energy they have different T through some heat engines and produced work the process failed to do that thus irreversible Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 790 A geothermal supply of hot water at 500 kPa 150C is fed to an insulated flash evaporator at the rate of 15 kgs A stream of saturated liquid at 200 kPa is drained from the bottom of the chamber and a stream of saturated vapor at 200 kPa is drawn from the top and fed to a turbine Find the rate of entropy generation in the flash evaporator Solution 1 2 3 Twophase out of the valve The liquid drops to the bottom Continuity Eq49 m 1 m 2 m 3 Energy Eq410 m 1h1 m 2h2 m 3h3 Entropy Eq77 m 1s1 S gen dAQ E AT m 2s2 m 3s3 Process AQ E A 0 irreversible throttle B11 h1 63218 kJkg s1 18417 kJkgK B12 h3 270663 kJkg s3 71271 kJkgK h2 50468 kJkg s2 153 kJkgK From the energy equation we solve for the flow rate m 3 m 1h1 h2h3 h2 15 00579 008685 kgs Continuity equation gives m 2 m 1 m 2 141315 kgs Entropy equation now leads to S gen m 2s2 m 3s3 m 1s1 141315 153 008685 7127 15 18417 001855 kWK v P s T 1 1 3 200 kPa 500 kPa 2 3 2 1a Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 791 A compressor in a commercial refrigerator receives R410A at 25AoE AC and x 1 The exit is at 2000 kPa and 80AoE AC Neglect kinetic energies and find the specific entropy generation Solution CV Compressor steady state single inlet and exit flow For this device we also assume no heat transfer q 0 and ZA1E A ZA2E WC i e cb Entropy Eq79 si dqT sAgenE A sAeE A si 0 sAgenE From Table B41 si 10893 kJkgK From Table B42 sAeE A 11537 kJkgK Entropy generation becomes sAgenE A sAeE A si 11537 10893 00644 kJkgK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 792 A steam turbine has an inlet of 2 kgs water at 1000 kPa and 400AoE AC with velocity of 15 ms The exit is at 100 kPa 150AoE AC and very low velocity Find the power produced and the rate of entropy generation Solution CV Turbine Steady flow and adiabatic Continuity Eq49 m 1 m 2 Energy Eq410 m 1h1 A1 2E AVA2E A m 2h2 AW E Entropy Eq77 m 1s1 S gen m 2s2 W T 1 2 States from Table B13 h1 326388 kJkg s1 74650 kJkgK h2 277638 kJkg s2 76133 kJkgK AW E A m 1h1 A1 2E AVA2E A h2 2 kgs 326388 A1 2E A A 152 E1000E A 277638 kJkg 975 kW S gen m 1s2 s1 2 kgs 76133 74650 kJkgK 03 kWK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 793 A factory generates compressed air from ambient 100 kPa 17AoE AC by compression to 1000 kPa 600 K after which it cools in a constant pressure cooler to 300 K by heat transfer to the ambient Find the specific entropy generation in the compressor and in the cooler Solution CV air compressor q 0 Continuity Eq Am E A2E A Am E A1E A Energy Eq413 0 hA1E A wAc inE A hA2E Entropy Eq 0 sA1E A sA2E A sAgen compE Table A7 State 1 hA1E A 29019 kJkg sAo T1E A 683521 kJkgK Table A7 State 2 hA2E A 60732 kJkg sAo T2E A 757638 kJkgK Table A7 State 3 hA3E A 30047 kJkg sAo T3E A 686926 kJkgK sAgen compE A sA2E A sA1E A sAo T2E A sAo T1E A R lnPA2E APA1E A 757638 683521 0287 ln1000100 0080 kJkgK CV cooler w A0E A Continuity Eq Am E A3E A Am E A1E A Energy Eq413 0 hA2E A qAoutE A hA3E Entropy Eq 0 sA2E A sA3E A qAoutE ATAambE A sAgen coolE qAoutE A hA2E A hA3E A 60732 30047 30685 kJkg sAgen coolE A sA3E A sA2E A qAoutE ATAambE A sAo T3E A sAo T2E A qAoutE ATAambE 686926 757638 A30685 290E A 0351 kJkg 1 3 2 Q cool Compressor W c Compressor section Cooler section Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 794 A mixing chamber receives 5 kgmin ammonia as saturated liquid at 20C from one line and ammonia at 40C 250 kPa from another line through a valve The chamber also receives 325 kJmin energy as heat transferred from a 40C reservoir This should produce saturated ammonia vapor at 20C in the exit line What is the mass flow rate in the second line and what is the total entropy generation in the process Solution CV Mixing chamber out to reservoir Continuity Eq49 AmE A1E A AmE A2E A AmE A3E Energy Eq410 AmE A1E AhA1E A AmE A2E AhA2E A AQE A AmE A3E AhA3E Entropy Eq77 AmE A1E AsA1E A AmE A2E AsA2E A AQE ATAresE A ASE AgenE A AmE A3E AsA3E 1 2 3 MIXING CHAMBER Q 2 P v 3 1 From Table B21 hA1E A 8905 kJkg sA1E A 03657 kJkgK From Table B22 hA2E A 15517 kJkg sA2E A 59599 kJkgK From Table B21 hA3E A 141805 kJkg sA3E A 56158 kJkgK From the energy equation AmE A2E A Am 1h1 h3 Q EE AhA3E A hA2E A 5 8905 141805 325 141805 15517 47288 kgmin AmE A3E A 52288 kgmin ASE AgenE A AmE A3E AsA3E A AmE A1E AsA1E A AmE A2E AsA2E A AQE ATAresE 52288 56158 5 03657 47288 59599 32531315 894 kJK min Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 795 Carbon dioxide at 300 K 200 kPa is brought through a steady device where it is heated to 600 K by a 700 K reservoir in a constant pressure process Find the specific work specific heat transfer and specific entropy generation Solution CV Heater and walls out to the source Steady single inlet and exit flows Since the pressure is constant and there are no changes in kinetic or potential energy between the inlet and exit flows the work is zero w 0 Continuity Eq411 Am E Ai Am E AeE A Am E Energy Eq413 hAiE A q hAeE Entropy Eq79 si dqT sAgenE A sAeE A si qTAsourceE A sAgenE Properties are from Table A8 so the energy equation gives q hAeE A hAiE A 50607 21438 2917 kJkg From the entropy equation sAgenE A sAeE A si qTAsourceE A 55279 48631 kJkgK 2917 kJkg 700 K 06648 04167 0248 kJkg K 1 2 Q 700 K 2 1 P v T s 1 2 T T 1 2 300 600 700 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 796 Methane at 1 MPa 300 K is throttled through a valve to 100 kPa Assume no change in the kinetic energy and ideal gas behavior What is the specific entropy generation Continuity Eq411 Am E Ai Am E AeE A Am E Energy Eq413 hAiE A 0 hAeE A Entropy Eq78 79 si dqT sAgenE A sAeE A si 0 sAgenE Properties are from Table B72 so the energy equation gives hAeE A hAiE A 61876 kJkg TAeE A 296 K sAeE A 115979 kJkgK sAgenE A sAeE A si 115979 104138 1184 kJkgK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 797 A heat exchanger that follows a compressor receives 01 kgs air at 1000 kPa 500 K and cools it in a constant pressure process to 320 K The heat is absorbed by ambient ait at 300 K Find the total rate of entropy generation Solution CV Heat exchanger to ambient steady constant pressure so no work Energy Eq412 AmE AhAiE A AmE AhAeE A AQE AoutE Entropy Eq77 AmE AsAiE A ASE AgenE A AmE AsAeE A AQE AoutE AT Using Table A5 and Eq825 for change in s AQE AoutE A AmE AhAiE A hAeE A AmE ACAPoE ATAiE A TAeE A 01 1004500 320 1807 kW ASE AgenE A AmE AsAeE A sAiE A AQE AoutE AT AmE ACAPoE A ln TAeE ATAiE A AQE AoutE AT 01 kgs 1004 kJkgK ln 320500 1807 kW300 K 00154 kWK Using Table A71 and Eq 619 for change in entropy h500 50336 kJkg h320 32058 kJkg sEATA500 AE A 738692 kJkg K sEATA320 AE A 693413 kJkg K AQE AoutE A AmE AhAiE A hAeE A 01 kgs 50336 32058 kJkg 1819 kW ASE AgenE A AmE AsAeE A sAiE A AQE AoutE AT 01 kgs 693413 738692 kJkgK 1819 kW300 K 00156 kWK i e Q 300 K out i e P v T s e i T T e i 320 500 300 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 798 A dual fluid heat exchanger has 5 kgs water enter at 40AoE AC 150 kPa and leaving at 10AoE AC 150 kPa The other fluid is glycol coming in at 10AoE AC 160 kPa and leaves at 10AoE AC 160 kPa Find the mass flow rate of glycol and the rate of entropy generation Continuity Eqs Each line has a constant flow rate through it Energy Eq410 Am E AH2OE A hA1E A Am E AglycolE A hA3E A Am E AH2OE A hA2E A Am E AglycolE A hA4E Entropy Eq77 0 Am E AH2OE A sA1E A Am E AglycolE A sA3E A Am E AH2OE A sA2E A Am E AglycolE A sA4E A S gen Process Each line has a constant pressure Table B1 hA1E A 16754 kJkg hA2E A 4199 kJkg sA1E A 05724 sA2E A 0151 kJkgK We could have used specific heat for the changes Table A4 CAP glyE A 242 kJkgK so hA4E A hA3E A CAP glyE A TA4E A TA3E A 242 10 10 484 kJkg sA4E A sA3E A CAP glyE A ln TA4E ATA3E A 242 ln2831526315 01773 kJkgK Am E AglycolE A Am E AH2OE A A h1 h2 Eh4 h3 E A 5 A16754 4199 484E A 1297 kgs S gen Am E AH2OE A sA2E A sA1E A Am E AglycolE A sA4E A sA3E A 5 kgs 0151 05724 kJkgK 1297 kgs 01773 kJkgK 0193 kWK CV Heat exchanger steady flow 1 inlet and 1 exit for glycol and water each The two flows exchange energy with no heat transfer tofrom the outside 3 glycol 1 water 4 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 799 Two flows of air both at 200 kPa one has 2 kgs at 400 K and the other has 1 kgs at 290 K The two flows are mixed together in an insulated box to produce a single exit flow at 200 kPa Find the exit temperature and the total rate of entropy generation Solution Continuity Eq49 m 1 m 2 m 3 2 1 3 kgs Energy Eq410 m 1h1 m 2h2 m 3h3 1 2 3 Entropy Eq77 m 1s1 m 2s2 S gen m 3s3 Using constant specific heats from A5 and Eq616 for s change Divide the energy equation with m 3CPo T3 m 1m 3T1 m 2m 3T2 A2 3E A 400 A1 3E A 290 36333 K S gen m 1s3 s1 m 2s3 s2 1 1004 ln A36333 400E A 2 1004 ln A36333 290E A kgs kJkgK 0356 kWK Using A71 and Eq619 for change in s h3 m 1m 3h1 m 2m 3h2 A2 3E A 4013 A1 3E A 29043 36434 kJkg From A71 T3 36444 K sEATA3 AE A 706216 kJkg K S gen 2 706216 715926 1 706216 683521 kgs kJkgK 003275 kWK The pressure correction part of the entropy terms cancel out as all three states have the same pressure Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7100 A condenser in a power plant receives 5 kgs steam at 15 kPa quality 90 and rejects the heat to cooling water with an average temperature of 17C Find the power given to the cooling water in this constant pressure process and the total rate of enropy generation when condenser exit is saturated liquid Solution CV Condenser Steady state with no shaft work term Energy Eq412 Am E A hAiE A AQE A Am E AhAeE A Entropy Eq78 Am E A sAiE A AQE AT ASE AgenE A Am E A sAeE Properties are from Table B12 hAiE A 22591 09 237314 236174 kJkg hAeE A 22591 kJkg sAiE A 07548 09 72536 7283 kJkg K sAeE A 07548 kJkgK AQE AoutE A AQE A Am E A hAiE A hAeE A 5236174 22591 10679 kW ASE AgenE A Am E A sAeE A sAiE A AQE AoutE AT 507548 7283 10679273 17 32641 36824 4183 kWK cb Often the cooling media flows inside a long pipe carrying the energy away Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7101 A large supply line has a steady flow of R410A at 1000 kPa 60AoE AC It is used in three different adiabatic devices shown in Fig P7101 a throttle flow an ideal nozzle and an ideal turbine All the exit flows are at 300 kPa Find the exit temperature and specific entropy generation for each device and the exit velocity of the nozzle Inlet state B42 hi 33575 kJkg si 12019 kJkgK CV Throttle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi hAeE A Zi ZAeE A and Vs are small Entropy Eq79 sAeE A si dqT sAgenE A si 0 sAgenE Exit state hAeE A hi 33575 kJkg TAeE A 479AoE AC sAeE A 1332 kJkgK sAgenE A sAeE A si 1332 12019 02 kJkg K CV Nozzle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi hAeE A VA2 eE A2 Zi ZAeE A Entropy Eq79 sAeE A si dqT sAgenE A si 0 0 The isentropic process sAeE A si gives from B42 TAeE A 46AoE AC sAgenE A 0 hAeE A 296775 kJkg The energy equation becomes VA2 eE A2 hi hAeE A 33575 296775 38975 kJkg VAeE A A 2 38975 1000EA 2792 ms Turbine Process Reversible and adiabatic same as for nozzle except w VAeE A 0 Energy Eq413 hi hAeE A w Zi ZAeE A TAeE A 46AoE AC sAgenE A 0 hAeE A 296775 kJkg P v T s e i i e Low V Hi P Low P Hi V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7102 A twostage compressor takes nitrogen in at 20AoE AC 150 kPa compresses it to 600 kPa 450 K then it flows through an intercooler where it cools to 320 K and the second stage compresses it to 3000 kPa 530 K Find the specific entropy generation in each of the two compressor stages The setup is like this CV Stage 1 nitrogen Steady flow Process adibatic q 0 sAgenE A 0 Energy Eq 413 wAC1E A hA2E A hA1E A Entropy Eq 710 0 sA1E A sA2E A sAgen C1E Assume constant CAP0E A 1042 from A5 and Eq616 for change in entropy sAgen C1E A sA2E A sA1E A CAP0E A lnTA2E ATA1E A R ln PA2E APA1E A 1042 ln45029315 02968 ln600150 00351 kJkgK sAgen C2E A sA4E A sA3E A CAP0E A lnTA4E ATA3E A R ln PA4E APA3E A 1042 ln530320 02968 ln3000600 00481 kJkgK C1 C2 1 2 3 4 2 W Q 1 W cooler inter Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7103 The intercooler in the previous problem uses cold liquid water to cool the nitrogen The nitrogen flow is 01 kgs and the liquid water inlet is 20AoE AC and setup to flow in the opposite direction as the nitrogen so the water leaves at 35AoE AC Find the flow rate of the water and the entropy generation in this intercooler Solution 1 3 2 4 H2O N2 A hydraulic motor driven compressor with intercooler in small pipe between the two stages CV Heat exchanger steady 2 flows in and two flows out Continuity eq AmE A1E A AmE A2E A AmE AH2OE A AmE A3E A AmE A4E A AmE AN2E Energy eq 0 AmE AN2E A hA3E A hA4E A AmE AH2OE A hA2E A hA1E A Entropy Eq77 0 Am E AN2E A sA3E A sA4E A Am E AH2OE A sA2E A sA1E A S gen Due to the lower range of temperature we will use constant specific heats from A5 and A4 so the energy equation is approximated as 0 AmE AH2OE ACApE AH2OE A TA2E A TA1E A AmE AN2E ACApE A TA3E A TA4E A Now solve for the water flow rate AmE AH2OE A AmE AN2E A CApE AN2E A TA3E A TA4E A CApE AH2OE A TA1E A TA2E A 01 kgs 1042 kJkgK 450 320 K 418 35 20 kJkg 0216 kgs AS E AgenE A Am E AN2E A sA4E A sA3E A Am E AH2OE A s1 sA2E A Am E AN2E ACAPE A ln T4T3 Am E AH2OE ACAPE A ln T1T2 01 kgs 1042 kJkgK ln A320 450E A 0216 kgs 418 kJkgK ln A308 293E 003552 004508 000956 kWK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7104 Air at 327C 400 kPa with a volume flow 1 mA3E As runs through an adiabatic turbine with exhaust pressure of 100 kPa Neglect kinetic energies and use constant specific heats Find the lowest and highest possible exit temperature For each case find also the rate of work and the rate of entropy generation Solution CV Turbine Steady single inlet and exit flows q 0 Inlet state T P vAiE A RTAiE A PAiE A 0287 600400 04305 mA3E Akg Am E A AV E AvAiE A 104305 2323 kgs The lowest exit T is for maximum work out ie reversible case Process Reversible and adiabatic constant s from Eq79 Eq823 TAeE A TAiE APAeE APAiE AA k1 k E A 600 100400 A02857E A 4038 K w hAiE A hAeE A CAPoE ATAiE A TAeE A 1004 600 4038 197 kJkg W T Am E Aw 2323 197 4576 kW and ASE AgenE A 0 Highest exit T occurs when there is no work out throttling q w hAiE A hAeE A 0 TAeE A TAiE A 600 K ASE AgenE A Am E A sAeE A sAiE A Am E AR ln A Pe EPi E A 2323 0287 ln A100 400E A 0924 kWK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7105 A counter flowing heat exchanger has one line with 2 kgs at 125 kPa 1000 K entering and the air is leaving at 100 kPa 400 K The other line has 05 kgs water coming in at 200 kPa 20C and leaving at 200 kPa What is the exit temperature of the water and the total rate of entropy generation Solution CV Heat exchanger steady flow 1 inlet and 1 exit for air and water each The two flows exchange energy with no heat transfer tofrom the outside 3 water 1 air 4 2 Energy Eq410 m AIRhAIR m H2OhH2O From A7 h1 h2 104622 4013 64492 kJkg From B12 h3 8394 kJkg s3 02966 kJkg K h4 h3 m AIRm H2Oh1 h2 20564492 257968 kJkg h4 h3 257968 266362 kJkg hg at 200 kPa T4 Tsat 12023C x4 266362 50468220196 09805 s4 153 x4 5597 701786 kJkg K From entropy Eq77 ASE AgenE A AmE AH2O sA4E A s3 m AIRsA2E A s1 05701786 02966 271593 81349 0287 ln 100125 33606 1823 154 kWK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7106 A large supply line has a steady air flow at 500 K 200 kPa It is used in three different adiabatic devices shown in Fig P985 a throttle flow an ideal nozzle and an ideal turbine All the exit flows are at 100 kPa Find the exit temperature and specific entropy generation for each device and the exit velocity of the nozzle CV Throttle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi hAeE A Zi ZAeE A and Vs are small Entropy Eq78 sAeE A si dqT sAgenE A si 0 sAgenE Since it is air we have h hT so same h means same TAeE A Ti 500 K sAgenE A sAeE A si sAo TeE A sAo TiE A R lnPAeE A PAiE A 0 0287 ln12 02 kJkg K CV Nozzle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi hAeE A VA2 eE A2 Zi ZAeE A Entropy Eq78 sAeE A si dqT sAgenE A si 0 0 Use constant specific heat from Table A5 CAPoE A 1004 A kJ kg KE A k 14 The isentropic process sAeE A si gives Eq823 TAeE A TAiE A PAeE APAiE AA k1 k E A 500 100200 A02857E A 410 K The energy equation becomes VA2 eE A2 hi hAeE A CAPE A TAiE A TAeE A VAeE A EA 2 CAP A TAi A TAe AEA A 21004 kJkgK 500410 K 1000 JkJEA 424 ms P v T s e i i e Low V Hi P Low P Hi V Turbine Process Reversible and adiabatic constant s from Eq78 Eq623 TAeE A TAiE APAeE APAiE AA k1 k E A 500 100200 A02857E A 410 K w hAiE A hAeE A CAPoE ATAiE A TAeE A 1004 500 410 90 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7107 In a heatdriven refrigerator with ammonia as the working fluid a turbine with inlet conditions of 20 MPa 70C is used to drive a compressor with inlet saturated vapor at 20C The exhausts both at 12 MPa are then mixed together The ratio of the mass flow rate to the turbine to the total exit flow was measured to be 062 Can this be true Solution Assume the compressor and the turbine are both adiabatic CV Total Continuity Eq411 AmE A5E A AmE A1E A AmE A3E Energy Eq410 AmE A5E AhA5E A AmE A1E AhA1E A AmE A3E AhA3E A Entropy AmE A5E AsA5E A AmE A1E AsA1E A AmE A3E AsA3E A ASE ACVgenE sA5E A ysA1E A 1ysA3E A ASE ACVgenE AAmE A5E Assume y AmE A1E AAmE A5E A 062 2 4 Turbine Compressor 5 1 3 State 1 Table B22 hA1E A 15427 kJkg sA1E A 4982 kJkg K State 3 Table B21 hA3E A 14181 kJkg sA3E A 5616 kJkg K Solve for exit state 5 in the energy equation hA5E A yhA1E A 1yhA3E A 062 15427 1 06214181 14954 kJkg State 5 hA5E A 14954 kJkg PA5E A 1200 kPa sA5E A 5056 kJkg K Now check the 2nd law entropy generation ASE ACVgenE AAmE A5E A sA5E A ysA1E A 1ysA3E A 01669 Impossible The problem could also have been solved assuming a reversible process and then find the needed flow rate ratio y Then y would have been found larger than 062 so the stated process can not be true Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7108 Repeat problem 7106 for the throttle and the nozzle when the inlet air temperature is 2500 K and use the air tables CV Throttle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi hAeE A Zi ZAeE A and Vs are small Entropy Eq79 sAeE A si dqT sAgenE A si 0 sAgenE Since it is air we have h hT so same h means same TAeE A Ti 2500 K sAgenE A sAeE A si sAo TeE A sAo TiE A R lnPAeE A PAiE A 0 0287 ln12 02 kJkg K CV Nozzle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi hAeE A VA2 eE A2 Zi ZAeE A Entropy Eq79 sAeE A si dqT sAgenE A si 0 0 The isentropic process sAeE A si gives Eq619 0 sAeE A si sAo TeE A sAo TiE A R lnPAeE A PAiE A sAo TeE A sAo TiE A R lnPAeE A PAiE A 924781 0287 ln 100200 904888 T 21366 K hAeE A 242286 kJkg The energy equation becomes VA2 eE A2 hi hAeE A 288306 242286 4602 kJkg VAeE A A 2 1000 4602EA 959 ms P v T s e i i e Low V Hi P Low P Hi V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7109 Carbon dioxide used as a natural refrigerant flows through a cooler at 10 MPa which is supercritical so no condensation occurs The inlet is at 200AoE AC and the exit is at 40AoE AC Assume the heat transfer is to the ambient at 20AoE AC and find the specific entropy generation CV Heat exchanger to ambient steady constant pressure so no work Energy Eq412 hAiE A hAeE A qAoutE Entropy Eq79 sAiE A sAgenE A sAeE A qAoutE AT Using Table B3 qAoutE A hAiE A hAeE A 51949 20014 31935 kJkg sAgenE A sAeE A sAiE A qAoutE AT 06906 15705 kJkgK A31935 kJkg 29315 KE A 02095 kJkgK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7110 Saturated liquid nitrogen at 600 kPa enters a boiler at a rate of 0005 kgs and exits as saturated vapor It then flows into a super heater also at 600 kPa where it exits at 600 kPa 280 K Assume the heat transfer comes from a 300 K source and find the rates of entropy generation in the boiler and the super heater Solution CV boiler steady single inlet and exit flow neglect KE PE energies in flow Continuity Eq m 1 m 2 m 3 Table B61 h1 8153 kJkg s1 3294 kJkgK h2 8685 kJkg s2 5041 kJkgK Table B62 h3 28905 kJkg s3 6238 kJkgK Energy Eq413 qboiler h2 h1 8685 8153 16838 kJkg AQ E Aboiler m 1qboiler 0005 16838 0842 kW Entropy Eq ASE AgenE A m 1 s2 s1 AQ E AboilerTsource 000550413294 0842300 00059 kWK CV Superheater same approximations as for boiler Energy Eq413 qsup heater h3 h2 28905 8685 2022 kJkg AQ E Asup heater m 2qsup heater 0005 kgs 2022 kJkg 101 kW Entropy Eq ASE AgenE A m 1 s3 s2 AQ E Asup heaterTsource 0005 kgs 62385041kJkgK 101 kW300 K 000262 kWK 1 2 3 Q Q boiler Super heater vapor cb 600 P 1 2 3 v T 1 2 3 v Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7111 A steam turbine in a power plant receives steam at 3000 kPa 500AoE AC The turbine has two exit flows one is 20 of the flow at 1000 kPa 350AoE AC to a feedwater heater and the remainder flows out at 200 kPa 200AoE AC Find the specific turbine work and the specific entropy generation both per kg flow in CV Steam turbine x 02 extraction fraction Energy Eq413 w hA1E A xhA2E A 1 xhA3E A Entropy Eq78 sA2E A sA1E A sAgen HPE A full flow rate Entropy Eq79 sA3E A sA2E A sAgen LPE A flow rate is fraction 1x Overall entropy gen sAgen HPE A sAgen HPE A 1 x sAgen LPE Inlet state Table B13 hA1E A 345648 kJkg sA1E A 72337 kJkg K Extraction state hA2E A 315765 kJkg sA2E A 73010 kJkg K Exit actual state Table B13 hA3E A 287046 sA3E A 75066 kJkg K Actual turbine energy equation w 345648 02 315765 08 287046 52858 kJkg sAgen totE A 7301 72337 08 75066 7301 0232 kJkgK v P s T 1 1 3s 3s 200 kPa 3 MPa 3 1 MPa 3 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7112 One type of feedwater heater for preheating the water before entering a boiler operates on the principle of mixing the water with steam that has been bled from the turbine For the states as shown in Fig P7112 calculate the rate of net entropy increase for the process assuming the process to be steady flow and adiabatic Solution CV Feedwater heater Steady flow no external heat transfer Continuity Eq49 A mEA1E A A mEA2E A A mEA3E Energy Eq410 A mEA1E AhA1E A A mEA3 E A A mEA1E AhA2E A A mEA3E AhA3E Properties All states are given by PT table B11 and B13 hA1E A 16842 hA2E A 2828 hA3E A 6758 all kJkg sA1E A 0572 sA2E A 6694 sA3E A 19422 all kJkg K 1 2 3 FEED WATER HEATER 2 T s 3 1 1 MPa Solve for the flow rate from the energy equation A mEA1E A A m3h3 h2 Eh1 h2E A A46758 2828 E16842 2828E A 3237 kgs A mEA2E A 4 3237 0763 kgs The second law for steady flow ASE ACVE A 0 and no heat transfer Eq77 ASE ACVgenE A ASE ASURRE A A mEA3E AsA3E A A mEA1E AsA1E A A mEA2E AsA2E 419422 32370572 07636694 08097 kJK s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7113 A coflowing same direction heat exchanger has one line with 05 kgs oxygen at 17C 200 kPa entering and the other line has 06 kgs nitrogen at 150 kPa 500 K entering The heat exchanger is very long so the two flows exit at the same temperature Use constant heat capacities and find the exit temperature and the total rate of entropy generation Solution CV Heat exchanger steady 2 flows in and two flows out 1 3 2 4 Energy Eq410 m O2h1 m N2h3 m O2h2 m N2h4 Same exit temperature so T4 T2 with values from Table A5 m O2CAP O2E AT1 m N2CAP N2E AT3 m O2CAP O2E A m N2CAP N2E AT2 T2 A05 0922 290 06 1042 500 05 0922 06 1042E A A44629 10862E 4109 K Entropy Eq77 gives for the generation ASE AgenE A AmE AO2sA2E A s1 m N2sA4E A s3 m O2CAPE A ln T2T1 m N2CAPE A ln T4T3 05 0922 ln 4109290 06 1042 ln 4109500 016064 01227 00379 kWK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7114 A supply of 5 kgs ammonia at 500 kPa 20C is needed Two sources are available one is saturated liquid at 20C and the other is at 500 kPa and 140C Flows from the two sources are fed through valves to an insulated mixing chamber which then produces the desired output state Find the two source mass flow rates and the total rate of entropy generation by this setup Solution CV mixing chamber valve Steady no heat transfer no work Continuity Eq49 AmE A1E A AmE A2E A AmE A3E A Energy Eq410 AmE A1E A hA1E A AmE A2E AhA2E A AmE A3E AhA3E Entropy Eq77 AmE A1E A sA1E A AmE A2E AsA2E A ASE AgenE A AmE A3E AsA3E 1 2 3 MIXING CHAMBER 2 T s 3 1 State 1 Table B21 hA1E A 2734 kJkg sA1E A 10408 kJkg K State 2 Table B22 hA2E A 17738 kJkg sA2E A 62422 kJkg K State 3 Table B22 hA3E A 14883 kJkg sA3E A 54244 kJkg K As all states are known the energy equation establishes the ratio of mass flow rates and the entropy equation provides the entropy generation AmE A1E A hA1E A AmE A3E A AmE A2E AhA2E A AmE A3E AhA3E A AmE A1E A AmE A3E A h3 h2 h1 h2 0952 kgs m 2 m 3 AmE A1E A 405 kgs AS E AgenE A 5 54244 095 10408 405 62422 kgs kJkgK 0852 kWK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Transient processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7115 Calculate the specific entropy generated in the filling process given in Example 410 Solution CV Cannister filling process where A1E AQA2E A 0 A1E AWA2E A 0 mA1E A 0 Continuity Eq415 mA2E A 0 mAinE A Energy Eq416 mA2E AuA2E A 0 mAinE AhAlineE A 0 0 uA2E A hAlineE Entropy Eq712 mA2E AsA2E A 0 mAinE AsAlineE A 0 A1E ASA2 genE A Inlet state 14 MPa 300C hi 30404 kJkg si 69533 kJkg K final state 14 MPa uA2E A hi 30404 kJkg TA2E A 452C sA2E A 745896 kJkg K A1E ASA2 genE A mA2E AsA2E A si A1E AsA2 genE A sA2E A si 745896 69533 0506 kJkg K 2 line T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7116 A 1mA3E A rigid tank contains 100 kg R410A at a temperature of 15C A valve on top of the tank is opened and saturated vapor is throttled to ambient pressure 100 kPa and flows to a collector system During the process the temperature inside the tank remains at 15C by heat transfer from the 20C ambient The valve is closed when no more liquid remains inside Calculate the heat transfer to the tank and total entropy generation in the process Solution CV Tank out to surroundings Rigid tank so no work term Continuity Eq415 mA2E A mA1E A mAeE A Energy Eq416 mA2E AuA2E A mA1E AuA1E A QACVE A mAeE AhAeE Entropy Eq712 mA2E AsA2E A mA1E AsA1E A QACVE ATASURE A mAeE AsAeE A Sgen State 1 Table B41 vA1E A VA1E AmA1E A 1100 0000904 xA1E A 001955 xA1E A 046527 uA1E A 8002 046527 1771 16242 kJkg sA1E A 03083 046527 06998 06339 hAeE A hAgE A 28279 kJkg State 2 vA2E A vAgE A 002045 uA2E A uAgE A 25712 sA2E A 10081 kJkg K Exit state hAeE A 28279 PAeE A 100 kPa TAeE A 1865C sAeE A 12917 kJkgK mA2E A 1002045 489 kg mAeE A 100 489 511 kg QACVE A mA2E AuA2E A mA1E AuA1E A mAeE AhAeE 489 25712 100 16242 511 28279 10 782 kJ Sgen mA2E AsA2E A mA1E AsA1E A mAeE AsAeE A QACVE ATASURE 489 10081 100 06339 511 12917 10 782 29315 1514 kJK e Q sat vap cv 789 P 1 2 e v T 1 2 e s h C P C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7117 A 02 mA3E A initially empty container is filled with water from a line at 500 kPa 200 C until there is no more flow Assume the process is adiabatic and find the final mass final temperature and the total entropy generation Solution CV The container volume and any valve out to line Continuity Eq415 mA2E A mA1E A mA2E A mAiE Energy Eq416 mA2E AuA2E A mA1E AuA1E A mA2E AuA2E A A1E AQA2E A A1E AWA2E A mAiE AhAiE A mAiE AhAiE Entropy Eq712 mA2E AsA2E A mA1E AsA1E A mA2E AsA2E A AdQTEA A1E ASA2 genE A mAiE AsAiE Process Adiabatic A1E AQA2E A 0 Rigid A1E AWA2E A 0 Flow stops PA2E A PAlineE State i hAiE A 285537 kJkg sAiE A 70592 kJkg K State 2 500 kPa uA2E A hAiE A 285537 kJkg Table B13 TA2E A 3329C sA2E A 75737 kJkg vA2E A 055387 m3kg mA2E A VvA2E A 02055387 0361 kg From the entropy equation A1E ASA2 genE A mA2E AsA2E A mA2E AsAiE 036175737 70592 0186 kJK 2 line T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7118 An initially empty 01 mA3E A cannister is filled with R410A from a line flowing saturated liquid at 5C This is done quickly such that the process is adiabatic Find the final mass liquid and vapor volumes if any in the cannister Is the process reversible Solution CV Cannister filling process where A1E AQA2E A A0E A A1E AWA2E A A0E A mA1E A A0E Continuity Eq415 mA2E A A0E A mAinE A Energy Eq416 mA2E AuA2E A A0E A mAinE AhAlineE A A0E A A0E A uA2E A hAlineE 1 Table B41 uAfE A 4965 uAfgE A 20175 hAfE A 5022 all kJkg 2 PA2E A PAlineE A uA2E A hAlineE A 2 phase uA2E A uAfE A uA2E A uAfE A xA2E AuAfgE xA2E A 5022 496520175 0002825 vA2E A vAfE A xA2E AvAfgE A 0000841 0002825003764 00009473 mA3E Akg mA2E A VvA2E A 10556 kg mAfE A 106262 kg mAgE A 0298 kg VAfE A mAfE A vAfE A 0089 mA3E A VAgE A mAgE AvAgE A 00115 mA3E Process is irreversible throttling sA2E A sAfE 2 line T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7119 A 1 L can of R134a is at room temperature 20AoE AC with a quality of 50 A leak in the top valve allows vapor to escape and heat transfer from the room takes place so we reach final state of 5AoE AC with a quality of 100 Find the mass that escaped the heat transfer and the entropy generation not including that made in the valve CV The can of R134a not including the nozzlevalve out to ambient 20AoE AC Continuity Eq mA2E A mA1E A mAeE A Energy Eq mA2E AuA2E A mA1E AuA1E A mAeE AhAeE A A1E AQA2E A A1E AWA2E Entropy Eq712 mA2E AsA2E A mA1E AsA1E A mAeE AsAeE A dQT A1E ASA2 genE Process Eq V constant A1E AWA2E A PdV 0 State 1 Tx vA1E A vAfE A xA1E AvAfgE A 0000817 05 003524 0018437 mA3E Akg uA1E A uAfE A xA1E AuAfgE A 22703 05 16216 30811 kJkg sA1E A sAfE A xA1E AsAfgE A 10963 05 0622 14073 kJkg mA1E A V vA1E A 0001 0018437 005424 kg State 2 Tx vA2E A vAgE A 005833 mA3E Akg uA2E A uAgE A 38085 kJkg sA2E A sAgE A 17239 kJkg K mA2E A VvA2E A 0001 005833 0017144 kg Exit state e Saturated vapor starting at 20AoE AC ending at 5AoE AC so we take an average hAeE A 05hAe1E A hAe2E A 05 40984 40132 40558 kJkg sAeE A 05sAe1E A sAe2E A 05 17183 17239 17211 kJkg K mAeE A mA1E A mA2E A 00371 kg The heat transfer from the energy equation becomes A1E AQA2E A mA2E AuA2E A mA1E AuA1E A mAeE AhAeE A 65293 167119 15047 4864 kJ A1E ASA2 genE A mA2E AsA2E A mA1E AsA1E A mAeE AsAeE A A1E AQA2E ATAambE 0029555 0076332 0063853 0016592 0000484 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7120 An empty can of 0002 mA3E A is filled with R134a from a line flowing saturated liquid R134a at 0C The filling is done quickly so it is adiabatic but after a while in storage the can warms up to room temperature 20C Find the final mass in the cannister and the total entropy generation Solution CV Cannister filling process where A1E AQA2E A A0E A A1E AWA2E A A0E A mA1E A A0E Continuity Eq415 mA2E A A0E A mAinE A Energy Eq416 mA2E AuA2E A A0E A mAinE AhAlineE A A0E A A0E A uA2E A hAlineE Inlet state Table B51 hAlineE A 200 kJkg sAlineE A 10 kJkg K State 2 PA2E A PAlineE A and uA2E A hAlineE A 200 kJkg uAfE xA2E A 200 19977 17824 000129 vA2E A 0000773 xA2E A 006842 0000861 mA3E Akg mA2E A V vA2E A 00020000861 2323 kg State 3 TA3E A 20C vA3E A vA2E A mA3E A mA2E A xA3E A 0000861 0000817003524 00012486 uA3E A 22703 x 16216 22723 kJkg sA3E A 10963 x 0622 10971 kJkgK Energy Eq416 mA2E AuA3E A mA2E AuA2E A A2E AQA3E A A0E A A2E AQA3E A mA2E AuA3E A uA2E A Entropy Eq712 mA2E AsA3E A A0E A mAinE AsAlineE A A2E AQA3E ATA3E A A1E ASA3 genE A A2E AQA3E A mA2E AuA3E A uA2E A 2323 22723 200 63255 kJ A1E ASA3 genE A mA2E AsA3E A sline A2E AQA3E ATA3E A 2323kg 10971 1 kJkgK 63255 kJ 29315 K 00098 kJK 2 line T s 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7121 A cook filled a pressure cooker with 3 kg water at 20C and a small amount of air and forgot about it The pressure cooker has a vent valve so if P 200 kPa steam escapes to maintain a pressure of 200 kPa How much entropy was generated in the throttling of the steam through the vent to 100 kPa when half the original mass has escaped Solution The pressure cooker goes through a transient process as it heats water up to the boiling temperature at 200 kPa then heats more as saturated vapor at 200 kPa escapes The throttling process is steady state as it flows from saturated vapor at 200 kPa to 100 kPa which we assume is a constant h process CV Pressure cooker no work Continuity Eq615 mA2E A mA1E A mAeE A Energy Eq616 mA2E A uA2E A mA1E AuA1E A mAeE A hAeE A A1E AQA2E A Entropy Eq712 mA2E AsA2E A mA1E AsA1E A mAeE A sAeE A dQT A1E ASA2 genE State 1 vA1E A vAfE A 0001002 mA3E Akg V mA1E AvA1E A 0003006 mA3E State 2 mA2E A mA1E A2 15 kg vA2E A VmA2E A 2vA1E A PA2E A 200 kPa Exit hAeE A hAgE A 270663 kJkg sAeE A sAgE A 71271 kJkg K So we can find the needed heat transfer and entropy generation if we know the CV surface temperature T If we assume T for water then A1E ASA2 genE A 0 which is an internally reversible externally irreversible process there is a T between the water and the source CV Valve steady flow from state e 200 kPa to state 3 at 100 kPa Energy Eq hA3E A hAeE A Entropy Eq sA3E A sAeE A AeE AsA3 genE A generation in valve throttle State 3 100 kPa hA3E A 270663 kJkg Table B13 TA3E A 9962 1509962 A270663 267546 277638 267546E A 1152C sA3E A 73593 76133 73593 030886 74378 kJkg K AeE ASA3 genE A mAeE AsA3E A sAeE A 15 74378 71271 0466 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7122 A 10 m tall 01 m diameter pipe is filled with liquid water at 20AoE AC It is open at the top to the atmosphere 100 kPa and a small nozzle is mounted in the bottom The water is now let out through the nozzle splashing out to the ground until the pipe is empty Find the water initial exit velocity the average kinetic energy in the exit flow and the total entropy generation for the process Total mass m ρAH ρ Aπ 4E A DA2E AH 998 Aπ 4E A 01A2E A 10 78383 kg Bernoulli A1 2E A VA2E A gH VA1E A 2gH1 A 2 9807 10EA 14 ms A1 2E A VA2 avgE A gHAavgE A g A1 2E A HA1E A 9807 5 49 mA2E AsA2E A Jkg All the energy average kinetic energy is dispersed in the ambient at 20AoE AC so SAgenE A AQ TE A A m 2TE A VA2 avgE A A78383 kg 49 Jkg 29315 KE A 131 JK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7123 A 200 L insulated tank contains nitrogen gas at 200 kPa 300 K A line with nitrogen at 500 K 500 kPa adds 40 more mass to the tank with a flow through a valve Use constant specific heats to find the final temperature and the entropy generation CV Tank no work and no heat transfer Continuity Eq615 mA2E A mA1E A mAiE A Energy Eq616 mA2E A uA2E A mA1E AuA1E A mAiE A hAiE A A1E AQA2E A Entropy Eq712 mA2E AsA2E A mA1E AsA1E A mAiE A sAiE A dQT A1E ASA2 genE Process Eq A1E AQA2E A 0 mA2E A 14 mA1E A mAiE A 04 mA1E Write hAiE A uAiE A RTAiE A then the energy equation contains only us so we can substitute the u differences with CAv0E A ΔT and divide by mA2E A CAv0E A to get TA2E A A m1 Em2 E A TA1E A A mi Em2 E ATAiE A A mi Em2 E A R Cv0 E A TAiE A 1 14E A 300 A04 14E A 500 A04 14E A A02968 0745E A 500 4140 K We need the pressure for the entropy PA2E A mA2E ARTA2E AV 14 PA1E A TA2E ATA1E A 14 200 414300 3864 kPa mA1E A A P1V1 ERT1 E A A 200 02 02968 300E A 04492 kg mA2E A 14 mA1E A 06289 kg The entropy generation with entropy difference from Eq616 becomes A1E ASA2 genE A mA2E AsA2E A mA1E AsA1E A mAiE A sAiE A mA2E AsA2E A sAiE A mA1E AsA1E A sAiE A mA2E A CAP0E AlnA T2 ETi E A R lnA P2 EPi E A mA1E A CAP0E A lnA T1 ETi E A R lnA P1 EPi E A 06289 1042 lnA414 500E A 02968 lnA3864 500E A 04492 1042 lnA300 500E A 02968 lnA200 500E A 00414 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7124 A 200 L insulated tank contains nitrogen gas at 200 kPa 300 K A line with nitrogen at 1500 K 1000 kPa adds 40 more mass to the tank with a flow through a valve Use table A8 to find the final temperature and the entropy generation CV Tank no work and no heat transfer Continuity Eq615 mA2E A mA1E A mAiE A Energy Eq616 mA2E A uA2E A mA1E AuA1E A mAiE A hAiE A A1E AQA2E A Entropy Eq712 mA2E AsA2E A mA1E AsA1E A mAiE A sAiE A dQT A1E ASA2 genE Process Eq A1E AQA2E A 0 mA2E A 14 mA1E A mAiE A 04 mA1E The energy equation can be solved for uA2E A to get uA2E A A m1 Em2 E A uA1E A A mi Em2 E A hAiE A A 1 14E A 22263 A04 14E A 12355 51202 kJkg A8 TA2E A 6798 K sAo T2E A 77088 kJkgK sAo T1E A 68463 sAo TiE A 86345 kJkgK We need the pressure for the entropy PA2E A mA2E ARTA2E AV 14 PA1E A TA2E ATA1E A 14 200 6798300 6345 kPa mA1E A A P1V1 ERT1 E A A 200 02 02968 300E A 04492 kg mA2E A 14 mA1E A 06289 kg The entropy generation with entropy difference from Eq619 becomes A1E ASA2 genE A mA2E AsA2E A mA1E AsA1E A mAiE A sAiE A mA2E AsA2E A sAiE A mA1E AsA1E A sAiE A mA2E A sAo T2E A sAo TiE A R lnA P2 EPi E A mA1E A sAo T1E A sAo TiE A R lnA P1 EPi E A 06289 77088 86345 02968 lnA6345 1000E A 04492 68463 86345 02968 lnA 200 1000E A 00914 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7125 Air from a line at 12 MPa 15C flows into a 500L rigid tank that initially contained air at ambient conditions 100 kPa 15C The process occurs rapidly and is essentially adiabatic The valve is closed when the pressure inside reaches some value P2 The tank eventually cools to room temperature at which time the pressure inside is 5 MPa What is the pressure P2 What is the net entropy change for the overall process Solution CV Tank Mass flows in so this is transient Find the mass first mA1E A PA1E AVRTA1E A A 100 05 0287 2882E A 0604 kg Fill to PA2E A then cool to TA3E A 15C PA3E A 5 MPa mA3E A mA2E A PA3E AVRTA3E A A 5000 05 0287 2882E A 30225 kg T s 100 kPa 5 MPa v C 2 1 12 MPa 3 line Mass mAiE A mA2E A mA1E A 30225 0604 29621 kg In the process 12 heat transfer 0 Energy Eq416 mAiE AhAiE A mA2E AuA2E A mA1E AuA1E A mAiE ACAP0E ATAiE A mA2E ACAv0E ATA2E A mA1E ACAv0E ATA1E TA2E A A296211004 060407172882 E30225 0717E A 4012 K PA2E A mA2E ARTA2E AV 30225 0287 401205 6960 MPa Consider now the total process from the start to the finish at state 3 Energy Eq416 QACVE A mAiE AhAiE A mA2E AuA3E A mA1E AuA1E A mA2E AhA3E A mA1E AhA1E A PA3E A PA1E AV But since TAiE A TA3E A TA1E A mAiE AhAiE A mA2E AhA3E A mA1E AhA1E QACVE A PA3E A PA1E AV 5000 10005 2450 kJ From Eq713 also Eqs724726 Sgen mA3E AsA3E A mA1E AsA1E A mAiE AsAiE A QACVE ATA0E A mA3E AsA3E A sAiE A mA1E AsA1E A sAiE A QACVE ATA0E 3022500287 ln A 5 12E A 060400287 ln A01 12E A 2450 2882 15265 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7126 An insulated pistoncylinder contains 01 mA3E A air at 250 kPa 300 K and it maintains constant pressure More air flows in through a valve from a line at 300 kPa 400 K so the volume increases 60 Use constant specific heats to solve for the final temperature and the total entropy generation CV Pistoncylinder volume no heat transfer Continuity Eq615 mA2E A mA1E A mAiE A Energy Eq616 mA2E A uA2E A mA1E AuA1E A mAiE A hAiE A A1E AWA2E A Entropy Eq712 mA2E AsA2E A mA1E AsA1E A mAiE A sAiE A dQT A1E ASA2 genE Process Eq A1E AQA2E A 0 VA2E A 16 VA1E A P C A1E AWA2E A PVA2E A VA1E A PA1E A16 1VA1E A 06 PA1E AVA1E A 06 mA1E ARTA1E mA1E A PA1E AVA1E ARTA1E A A 250 01 0287 300E A 02904 kg Write hAiE A uAiE A RTAiE A then the energy equation contains only us so we can substitute the u differences with CAv0E A ΔT and get mA2E ACAv0E ATA2E A mA1E ACAv0E ATA1E A mAiE ACAv0E ATAiE A mAiE ARTAiE A 06 mA1E ARTA1E mA2E ATA2E A mA1E ATA1E A mAiE ATAiE A 1 RCAv0E A 06 mA1E ATA1E ARCAv0E In this equation mA2E A TA2E A and mAiE A are unknows however we also have mA2E ATA2E A PA2E AVA2E A R 16 PA1E AVA1E AR 16 mA1E ATA1E so we can solve for the mass mAiE A mAiE ATAiE A 1 RCAv0E A 16 mA1E ATA1E A mA1E ATA1E A 06 mA1E ATA1E ARCAv0E mAiE ATAiE A 1 RCAv0E A 06 1 RCAv0E A mA1E ATA1E mAiE A 06 mA1E A TA1E A TAiE A 06 02904 300 400 013068 kg mA2E A mAiE A mA1E A 042108 kg TA2E A 16 mA1E A mA2E A TA1E A 331 K The entropy generation with entropy difference from Eq616 becomes A1E ASA2 genE A mA2E AsA2E A mA1E AsA1E A mAiE A sAiE A mA2E AsA2E A sAiE A mA1E AsA1E A sAiE A mA2E A CAP0E AlnA T2 ETi E A R lnA P2 EPi E A mA1E A CAP0E A lnA T1 ETi E A R lnA P1 EPi E A 042108 1004 lnA331 400E A 0287 lnA250 300E A 02904 1004 lnA300 400E A 0287 lnA250 300E A 00107 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7127 A balloon is filled with air from a line at 200 kPa 300 K to a final state of 110 kPa 300 K with a mass of 01 kg air Assume the pressure is proportional to the balloon volume as P 100 kPa CV Find the heat transfer tofrom the ambient at 300 K and the total entropy generation CV Balloon out to the ambient Assume mA1E A 0 Continuity Eq415 mA2E A 0 mAinE A Energy Eq416 mA2E AuA2E A 0 mAinE AhAinE A A1E AQA2E A A1E AWA2E A Entropy Eq712 mA2E AsA2E A 0 mAinE AsAinE A AdQ T EA A1E ASA2 genE A mAinE AsAinE A A1Q2 ETE A A1E ASA2 genE Process Eq P A C V A 100 kPa State 2 P T VA2E A mA2E ARTA2E A PA2E A A01 0287 300 110E A 0078273 mA3E PA2E A A CVA2E A C PA2E A 100 VA2E A 127758 kPamA3E Inlet state hAinE A hA2E A uA2E A PA2E AvA2E A sAinE A sA2E A R lnA Pin EP2 E A A1E AWA2E A AP dVEA AA CV dVEA A VA2E A 0 A1 2E AC VA2 2E A 0 100 0078273 A1 2E A 127758 0078273A2E 8219 kJ A1 2E APAoE A PA2E AVA2E A area in PV diagram A1E AQA2E A mA2E AuA2E A hAlineE A A1E AWA2E A PA2E AVA2E A A1E AWA2E 110 0078273 8219 0391 kJ A1E ASA2 genE A mA2E AsA2E A sAinE A A1Q2 ETE A mA2E A R ln A Pin EP2 E A A1Q2 ETE 01 0287 lnA200 110E A A0391 300E A 00185 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Device efficiency Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7128 A steam turbine inlet is at 1200 kPa 400AoE AC The exit is at 200 kPa What is the lowest possible exit temperature Which efficiency does that correspond to We would expect the lowest possible exit temperature when the maximum amount of work is taken out This happens in a reversible process so if we assume it is adiabatic this becomes an isentropic process Exit 200 kPa s sAinE A 73773 kJkg K T 1715AoE AC The efficiency from Eq727 measures the turbine relative to an isentropic turbine so the efficiency will be 100 v P s T i i e s e s 200 kPa 1200 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7129 A steam turbine inlet is at 1200 kPa 400AoE AC The exit is at 200 kPa What is the highest possible exit temperature Which efficiency does that correspond to The highest possible exit temperature would be if we did not get any work out ie the turbine broke down Now we have a throttle process with constant h assuming we do not have a significant exit velocity Exit 200 kPa h hAinE A 326066 kJkg T 392AoE AC Efficiency η w ws 0 v P s T h C i e i e Remark Since process is irreversible there is no area under curve in Ts diagram that correspond to a q nor is there any area in the Pv diagram corresponding to a shaft work Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7130 A steam turbine inlet is at 1200 kPa 400AoE AC The exit is at 200 kPa 200AoE AC What is the isentropic efficiency Inlet hAinE A 326066 kJkg sAinE A 73773 kJkg K Exit hex 287046 kJkg sex 75066 kJkg K Ideal Exit 200 kPa s sAinE A 73773 kJkg K hs 28126 kJkg wac hAinE A hex 326066 287046 3902 kJkg ws hAinE A hs 326066 28126 4481 kJkg η wac ws A3902 4481E A 0871 v P s T i e ac i e ac e s e s 200 kPa 1200 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7131 A compressor in a commercial refrigerator receives R410A at 25AoE AC and x 1 The exit is at 2000 kPa and 80AoE AC Neglect kinetic energies and find the isentropic compressor efficiency Solution CV Compressor steady state single inlet and exit flow For this device we also assume no heat transfer q 0 and ZA1E A ZA2E WC i e cb Energy Eq413 hi 0 wACE A hAeE A Entropy Eq79 si dqT sAgenE A sAeE A si 0 sAgenE From Table B41 hi 26977 kJkg si 10893 kJkgK From Table B42 hAeE A 34322 kJkg sAeE A 11537 kJkgK Energy equation gives wAC acE A hi hAeE A 26977 34322 7345 kJkgK The ideal compressor has an exit state es 2000 kPa 10893 kJkgK Table B42 TAe sE A 6045C hAe sE A 32113 kJkg wAC sE A 26977 32113 5136 kJkg The isentropic efficiency measures the actual compressor to the ideal one η wAC sE A wAC acE A 5136 7345 070 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7132 A steam turbine has an inlet of 2 kgs water at 1000 kPa and 400AoE AC with velocity of 15 ms The exit is at 100 kPa 150AoE AC and very low velocity Find the power produced and the rate of entropy generation Solution CV Turbine Steady flow and adiabatic Continuity Eq69 m 1 m 2 Energy Eq610 m 1h1 A1 2E AVA2E A m 2h2 AW E Entropy Eq97 m 1s1 S gen m 2s2 W T 1 2 States from Table B13 h1 326388 kJkg s1 74650 kJkgK h2 277638 kJkg s2 76133 kJkgK AW E A m 1 h1 A1 2E AVA2E A h2 2 kgs 326388 A1 2E A A 152 E1000E A 277638 kJkg 975 kW S gen m 1s2 s1 2 kgs 76133 7465 kJkgK 0297 kWK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7133 The exit velocity of a nozzle is 500 ms If ηAnozzleE A 088 what is the ideal exit velocity The nozzle efficiency is given by Eq 729 and since we have the actual exit velocity we get VA2 e sE A VA2 acE AηAnozzleE A Ve s VacA ηnozzle EA 500 A 088EA 533 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7134 An emergency drain pump should be able to pump 01 mA3E As liquid water at 15C 10 m vertically up delivering it with a velocity of 20 ms It is estimated that the pump pipe and nozzle have a combined isentropic efficiency expressed for the pump as 60 How much power is needed to drive the pump Solution CV Pump pipe and nozzle together Steady flow no heat transfer Consider the ideal case first it is the reference for the efficiency Energy Eq412 Am E Aihi VA2E Ai2 gZi W in Am E Aehe VA2E Ae2 gZe Solve for work and use reversible process Eq715 W ins Am E A he hi VA2E Ae VA2E Ai2 gZe Zi Am E A Pe Piv VA2E Ae2 gZ Am E A AV E Av 010001001 999 kgs W ins 9990 20A2E A2 11000 9807 101000 99902 009807 298 kW With the estimated efficiency the actual work Eq728 is W in actual W in sη 29806 497 kW 50 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7135 Find the isentropic efficiency of the R134a compressor in Example 48 Solution State 1 Table B52 hA1E A 3872 kJkg sA1E A 17665 kJkg K State 2ac hA2E A 4351 kJkg State 2s s 17665 kJkg K 800 kPa h 4318 kJkg T 468C wAc sE A hA2sE A h A1E A 4318 3872 446 kJkg wAacE A 501 50 kJkg η wAc sE A wAacE A 44650 089 v P s T i e ac e s e ac e s i Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7136 A gas turbine with air flowing in at 1200 kPa 1200 K has an exit pressure of 200 kPa and an isentropic efficiency of 87 Find the exit temperature Solution i e Turbine air W CV Ideal air turbine Adiabatic q 0 reversible sAgenE A 0 Energy Eq413 wATE A hAiE A hAeE A Entropy Eq79 sAeE A sAiE Table A7 hAiE A 12778 kJkg sAo TiE A 834596 kJkg K The constant s process is written from Eq619 as sAo TeE A sAo TiE A R ln A Pe EPi E A 834596 0287 lnA 200 1200 E A 783173 kJkg K Interpolate in A71 TAe sE A 7619 K hAe sE A 78052 kJkg wAT sE A hAiE A hAe sE A 127781 78052 4973 kJkg The actual turbine then has wAT acE A ηATE A wAT sE A 087 4973 43265 kJkg hAiE A hAe acE hAe acE A hAiE A wAT acE A 127781 43265 84516 kJkg Interpolate in A71 TAe acE A 8208 K If constant specific heats are used we get Table A5 CAPoE A 1004 kJkg K R 0287 kJkg K k 14 then from Eq623 TAe sE A TAiE A PAeE APAiE AA k1 k E A 1200 A 200 1200 E A 0286E A 7192 K wAT sE A CAPoE ATAiE A TAe sE A 10041200 7192 48272 kJkg The actual turbine then has wAT acE A ηATE A wAT sE A 087 48272 41997 kJkg CAPoE ATAiE A TAe acE A TAe acE A TAiE A wAT acE A CAPoE A 1200 419971004 7817 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7137 A gas turbine with air flowing in at 1200 kPa 1200 K has an exit pressure of 200 kPa Find the lowest possible exit temperature Which efficiency does that correspond to Solution Look at the Ts diagram for the possible processes We notice that the lowest exit T is for the isentropic process the ideal turbine i e Turbine air W Table A7 hAiE A 12778 kJkg sAo TiE A 834596 kJkg K The constant s process is written from Eq619 as sAo TeE A sAo TiE A R ln A Pe EPi E A 834596 0287 lnA 200 1200 E A 783173 kJkg K Interpolate in A71 TAe sE A 7619 K This is an efficiency of 100 1 2s T s 2a 1200 kPa 200 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7138 Repeat Problem 746 assuming the steam turbine and the air compressor each have an isentropic efficiency of 80 A certain industrial process requires a steady supply of saturated vapor steam at 200 kPa at a rate of 05 kgs Also required is a steady supply of compressed air at 500 kPa at a rate of 01 kgs Both are to be supplied by the process shown in Fig P941 Steam is expanded in a turbine to supply the power needed to drive the air compressor and the exhaust steam exits the turbine at the desired state Air into the compressor is at the ambient conditions 100 kPa 20C Give the required steam inlet pressure and temperature assuming that both the turbine and the compressor are reversible and adiabatic Solution CV Each device Steady flow Both adiabatic q 0 and actual devices sAgenE A 0 given by ηAsTE A and ηAscE A 3 1 4 2 Steam turbine Air compressor Air Eq623 TA4sE A TA3E APA4E APA3E AA k1 k E A 2932A 500 100 E A 0286E A 4646 K A WEACsE A A mEA3E AhA3E A hA4sE A 01 kgs 1004 kJkgK 2932 4646 K 1721 kW A WEACsE A A mEA3E AhA3E A hA4E A A WEACsE A ηAscE A 172080 215 kW Now the actual turbine must supply the actual compressor work The actual state 2 is given so we must work backwards to state 1 A WEATE A 215 kW A mEA1E AhA1E A hA2E A 05hA1E A 27066 hA1E A 27496 kJkg Also ηAsTE A 080 hA1E A hA2E AhA1E A hA2sE A 4327496 hA2sE A hA2sE A 26958 kJkg 26958 5047 xA2sE A27066 5047 xA2sE A 09951 sA2sE A 15301 0995171271 15301 70996 kJkg K sA1E A sA2sE A hA1E A PA1E A 269 kPa TA1E A 1435C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7139 Liquid water enters a pump at 15C 100 kPa and exits at a pressure of 5 MPa If the isentropic efficiency of the pump is 75 determine the enthalpy steam table reference of the water at the pump exit Solution CV pump A QEACVE A 0 KE 0 PE 0 2nd law reversible ideal process sAesE A sAiE A Eq715 for work term wAsE A A i es vdPEA vAiE APAeE A PAiE A 0001001 mA3E Akg 5000 100 kPa 4905 kJkg Real process Eq728 w wAsE AηAsE A 4905075 654 kJkg Energy Eq413 hAeE A hAiE A w 6299 654 6953 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7140 Ammonia is brought from saturated vapor at 300 kPa to 1400 kPa 140AoE AC in a steady flow adiabatic compressor Find the compressor specific work entropy generation and its isentropic efficiency CV Actual Compressor assume adiabatic and neglect kinetic energies Energy Eq413 wACE A hAiE A hAeE A Entropy Eq79 sAeE A sAiE A sAgenE States 1 B22 hAiE A 14317 kJkg sAiE A 54565 kJkgK 2 B22 hAeE A 17528 kJkg sAeE A 57023 kJkgK wACE A hAeE A hAiE A 17528 14317 3211 kJkg Ideal compressor We find the exit state from Ps State 2s PAeE A sAe sE A sAiE A 54565 kJkgK hAe sE A 165608 kJkg wAC sE A hA2sE A hAiE A 165608 14317 22438 kJkg ηACE A wAC sE A wACE A A22438 3211E A 0699 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7141 Find the isentropic efficiency of the nozzle in example 44 Solution CV adiabatic nozzle with known inlet state and velocity Inlet state B13 hAiE A 28501 kJkg sAiE A 69665 kJkg K Process ideal adiabatic and reversible Eq79 gives constant s ideal exit 150 kPa s xAesE A 69665 1433557897 09557 hAesE A hAfE A xAesE A hAfgE A 25949 kJkg VA 2 esE A2 hAiE A hAesE A VA 2 iE A2 28501 25949 50A2E A2000 25645 kJkg VAesE A 7162 ms From Eq730 ηAnozE A VA 2 eE A2 VA 2 esE A2 18025645 070 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7142 A centrifugal compressor takes in ambient air at 100 kPa 17C and discharges it at 450 kPa The compressor has an isentropic efficiency of 80 What is your best estimate for the discharge temperature Solution CV Compressor Assume adiabatic no kinetic energy is important Energy Eq413 w hA1E A hA2E A Entropy Eq79 sA2E A sA1E A sAgenE We have two different cases the ideal and the actual compressor We will solve using constant specific heat State 2 for the ideal sAgenE A 0 so sA2E A sA1E A and it becomes Eq823 TA2sE A TA1E A A P2 EP1 E A A k1 k E A 290 450 100A02857E A 4457 K wAsE A hA1E A hA2sE A CApE A TA1E A TA2sE A 1004 290 4457 1563 kJkg The actual work from definition Eq728 and then energy equation wAacE A wAsE Aη 1563 08 1954 kJkg hA1E A hA2E A CApE ATA1E A TA2E A TA2E A TA1E A wAacE A CApE 290 19541004 4846 K Solving using Table A71 instead will give State 1 Table A71 sAo T1E A 683521 kJkg K Now constant s for the ideal is done with Eq619 sAo T2sE A sAo T1E A R ln P2 P1 683521 0287 lnA450 100E A 726688 kJkg K From A71 TA2sE A 4421 K and hA2sE A 446795 kJkg wAsE A hA1E A hA2sE A 29043 446795 1564 kJkg The actual work from definition Eq728 and then energy equation wac wAsE Aη 1564 08 1955 kJkg hA2E A 1955 29043 48593 Table A71 TA2E A 483 K The answer is very close to the previous one due to the modest Ts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7143 A refrigerator uses carbon dioxide that is brought from 1 MPa 20AoE AC to 6 MPa using 2 kW power input to the compressor with a flow rate of 002 kgs Find the compressor exit temperature and its isentropic efficiency CV Actual Compressor assume adiabatic and neglect kinetic energies Energy Eq413 wACE A hA2E A hA1E A A W E mE A A 2 kW 002 kgsE A 100 kJkg Entropy Eq79 sA2E A sA1E A sAgenE States 1 B32 hA1E A 34231 kJkg sA1E A 14655 kJkgK 2 B32 hA2E A hA1E A wACE A 44231 kJkg TA2E A 1177AoE AC Ideal compressor We find the exit state from Ps State 2s PA2E A sA2sE A sA1E A 14655 kJkgK hA2sE A 43755 kJkg wAC sE A hA2sE A hA1E A 43755 34231 9524 kJkg ηACE A wAC sE A wACE A A9524 100E A 0952 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7144 The small turbine in Problem 733 was ideal Assume instead the isentropic turbine efficiency is 88 Find the actual specific turbine work and the entropy generated in the turbine Solution Continuity Eq411 Steady AmE A1E A AmE A2E A AmE A3E A AmE A Turbine Energy Eq413 wATE A hA1E A hA2E A W T 1 2 3 Q out Entropy Eq79 sA2E A sA1E A sAT genE Inlet state Table B13 hA1E A 391745 kJkg sA1E A 79487 kJkg K Ideal turbine sAT genE A 0 sA2E A sA1E A 79487 sAf2E A x sAfg2E State 2 P 10 kPa sA2E A sAgE A saturated 2phase in Table B12 xA2sE A sA1E A sAf2E AsAfg2E A 79487 064927501 09731 hA2sE A hAf2E A x hAfg2E A 1918 09731 23928 252035 kJkg wATsE A hA1E A hA2sE A 139705 kJkg Explanation for the reversible work term is in sect 73 Eq714 2s 1 P v T s 1 2s 3 3 2ac 2ac wATACE A η wATsE A 12299 kJkg hA1E A hA2ACE A hA2ACE A hA1E A wATACE A 26875 kJkg TA2ACE A 100C sA2ACE A 84479 kJkgK sAT genE A sA2ACE A sA1E A 04992 kJkg K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7145 Redo Problem 741 assuming the compressor and turbine in the turbocharger both have isentropic efficiency of 85 CV Turbine Steady single inlet and exit flows Process adiabatic q 0 reversible sAgenE A 0 EnergyEq413 wATE A hA3E A hA4E A Entropy Eq78 sA4E A sA3E 3 1 4 2 Engine W Compressor Turbine The property relation for ideal gas gives Eq623 k from Table A5 sA4E A sA3E A TA4E A TA3E APA4E APA3E AA k1 k E A 9232 K A 100 170 E A 0286E A 7932 K The energy equation is evaluated with specific heat from Table A5 wAT sE A hA3E A hA4E A CAP0E ATA3E A TA4E A 10049232 7932 1305 kJkg The actual turbine wAT acE A ηATE A wAT sE A 1305 085 1109 kJkg A WEATE A A mEAwATE A 1109 kJkg 01 kgs 1109 kW CV Compressor steady 1 inlet and 1 exit same flow rate as turbine Energy Eq413 wACE A hA2E A hA1E A Entropy Eq79 sA2E A sA1E Express the energy equation for the shaft and actual compressor having the actual turbine power as input with the same mass flow rate so we get wAC acE A wAC sE A ηACE A CAP0E ATA2 sE A TA1E AηACE A wAT acE A 1109 kJkg 1004TA2 acE A 3032 1004TA2 sE A 3032085 TA2 acE A 41366 K TA2 sE A 39709 K The property relation for sA2E A sA1E A only for ideal is Eq623 and inverted as PA2E A PA1E ATA2 sE ATA1E AA k k1 E A 100 kPa A 39709 3032 E A 35E A 257 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7146 A pump receives water at 100 kPa 15C and a power input of 15 kW The pump has an isentropic efficiency of 75 and it should flow 12 kgs delivered at 30 ms exit velocity How high an exit pressure can the pump produce Solution CV Pump We will assume the ideal and actual pumps have same exit pressure then we can analyse the ideal pump Specific work wac 1512 125 kJkg Ideal work Eq728 ws η wac 075 125 09375 kJkg As the water is incompressible liquid we get Energy Eq714 ws Pe Piv VA2E Ae2 Pe Pi0001001 30A2E A21000 Pe Pi0001001 045 Solve for the pressure difference Pe Pi ws 0450001001 487 kPa Pe 587 kPa Water pump from a car Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7147 A turbine receives air at 1500 K 1000 kPa and expands it to 100 kPa The turbine has an isentropic efficiency of 85 Find the actual turbine exit air temperature and the specific entropy increase in the actual turbine using Table A7 Solution CV Turbine steady single inlet and exit flow To analyze the actual turbine we must first do the ideal one the reference Energy Eq413 wATE A hA1E A hA2E A Entropy Eq79 sA2E A sA1E A sAgenE A sA1E Entropy change in Eq619 and Table A71 sAo T2E A sAo T1E A R lnPA2E APA1E A 861208 0287 ln1001000 795124 Interpolate in A7 TA2sE A 8492 hA2sE A 87656 wATE A 16358 87656 75924 kJkg Now we can consider the actual turbine from Eq727 and Eq413 wAT acE A ηATE A wATE A 085 75924 64535 hA1E A hA2acE A hA2acE A hA1E A wAT acE A 99045 TA2acE A 951 K The entropy balance equation is solved for the generation term sAgenE A sA2acE A sA1E A 8078 86121 0287 ln1001000 01268 kJkg K 1 2s P P T s 2 1 2ac Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7148 Air enters an insulated turbine at 50C and exits the turbine at 30C 100 kPa The isentropic turbine efficiency is 70 and the inlet volumetric flow rate is 20 Ls What is the turbine inlet pressure and the turbine power output Solution CV Turbine ηs 07 Insulated Air table A5 CApE A 1004 kJkg K R 0287 kJkg K k 14 Inlet Ti 50AoE AC V i 20 Ls 002 m3s Exit actual TAeE A 30AoE AC PAeE A 100 kPa 1st Law Steady state Eq413 qATE A hAiE A hAeE A wATE A qATE A 0 Assume Constant Specific Heat wATE A hAiE A hAeE A CApE ATAiE A TAeE A 803 kJkg wATsE A wη 1147 kJkg wATsE A CApE ATi TAesE A Solve for TAesE A 2089 K Isentropic Process Eq623 PAeE A PAiE A TAeE A TAiE AA k k1 E A PAiE A 461 kPa AmE A PAiE AV E ARTAiE A 461 kPa 002 m3s 0287 kJkgK 32315 K 0099 kgs W T AmE AwT 0099 kgs 803 kJkg 798 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7149 Carbon dioxide CO2 enters an adiabatic compressor at 100 kPa 300 K and exits at 1000 kPa 520 K Find the compressor efficiency and the entropy generation for the process Solution CV Ideal compressor We will assume constant heat capacity Energy Eq413 wAcE A hA1E A hA2E A Entropy Eq79 sA2E A sA1E A TA2sE A TA1E A P2 EP1 E A k1 k E A 300A 1000 100 E A 02242E A 5027 K wAcsE A CApE ATA1E A TA2sE A 08423005027 17067 kJkg CV Actual compressor wAcacE A CApE ATA1E A TA2acE A 0842300 520 1852 kJkg ηAcE A wAcsE AwAcacE A 170671852 092 Use Eq616 for the change in entropy sAgenE A sA2acE A sA1E A CApE A ln TA2acE ATA1E A R ln PA2E APA1E A 0842 ln520 300 01889 ln1000 100 0028 kJkg K P v T s e s i s C i e s e ac e ac P P e i Constant heat capacity is not the best approximation It would be more accurate to use Table A8 Entropy change in Eq619 and Table A8 sAo T2E A sAo T1E A R lnPA2E APA1E A 48631 01889 ln1000100 529806 Interpolate in A8 TA2sE A 481 K hA2sE A 382807 kJkg wAcsE A 382807 21438 16843 kJkg wAcacE A 42212 21438 20774 kJkg ηAcE A wAcsE AwAcacE A 1684320774 081 sAgenE A sA2acE A sA1E A 53767 48631 01889 ln10 00786 kJkgK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7150 A small air turbine with an isentropic efficiency of 80 should produce 270 kJkg of work The inlet temperature is 1000 K and it exhausts to the atmosphere Find the required inlet pressure and the exhaust temperature Solution CV Turbine actual energy Eq413 w hAiE A hAeacE A 270 kJkg Table A7 hAiE A 104622 hAeacE A 77622 kJkg TAeE A 7579 K CV Ideal turbine Eq727 and energy Eq413 wAsE A wηAsE A 27008 3375 hAiE A hAesE A hAesE A 70872 kJkg From Table A7 TAesE A 6955 K Entropy Eq79 sAiE A sAes E A adiabatic and reversible To relate the entropy to the pressure use Eq619 inverted and standard entropy from Table A71 PAeE APAiE A exp sAo TeE A sAo TiE A R exp7733 8134930287 02465 PAiE A PAeE A 02465 101302465 411 kPa P v T s e s i s C i e s e ac e ac P P e i If constant heat capacity were used TAeE A TAiE A wCApE A 1000 2701004 731 K CV Ideal turbine Eq727 and energy Eq413 wAsE A wηAsE A 27008 3375 kJkg hAiE A hAesE A CApE ATAiE A TAesE A TAesE A TAiE A wAsE ACApE A 1000 33751004 6638 K Eq79 adibatic and reversible gives constant s and relation is Eq623 PAeE APAiE A TAeE ATAiE Akk1 PAiE A 1013 10006638A35E A 425 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7151 A compressor in an industrial airconditioner compresses ammonia from a state of saturated vapor at 200 kPa to a pressure 800 kPa At the exit the temperature is measured to be 100AoE AC and the mass flow rate is 05 kgs What is the required motor size for this compressor and what is its isentropic efficiency CV Compressor Assume adiabatic and neglect kinetic energies Energy Eq413 w hA1E A hA2E A Entropy Eq79 sA2E A sA1E A sAgenE We have two different cases the ideal and the actual compressor States 1 B22 hA1E A 14196 kJkg vA1E A 05946 mA3E Akg sA1E A 55979 kJkg K 2ac B23 hA2ACE A 16706 kJkg vA2ACE A 021949 mA3E Akg 2s B23 P s sA1E A hA2sE A 16137 kJkg TA2sE A 766AoE AC ACTUAL wACACE A hA2ACE A hA1E A 16706 14196 251 kJkg W in Am E A wACACE A 05 kgs 251 kJkg 1255 kW IDEAL wAcsE A hA2sE A hA1E A 16137 14196 1941 kJkg Definition Eq728 ηAcE A wAcsE AwAcACE A 077 v P s T 1 2 ac 2 s 2 2 s 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7152 Repeat Problem 748 assuming the turbine and the pump each have an isentropic efficiency of 85 Solution W T Q H Q L W P in 1 2 3 4 PA1E A PA4E A 20 MPa TA1E A 700 C PA2E A PA3E A 20 kPa TA3E A 40 C ηAPE A ηATE A 85 a State 1 P T Table B13 hA1E A 38091 kJkg sA1E A 67993 kJkg K CV Turbine First we do the ideal then the actual Entropy Eq79 sA2E A sA1E A 67993 kJkg K Table B12 sA2E A 08319 xA2E A 70766 xA2E A 08433 hA2 sE A 2514 08433 235833 22401 kJkg Energy Eq413 wAT sE A hA1E A hA2 sE A 1569 kJkg wAT ACE A ηATE AwAT sE A 133365 hA1E A hA2 ACE hA2 ACE AhA1E A wAT ACE A 247545 kJkg xA2ACE A 247545 251423583 0943 TA2ACE A6006C b State 3 P T Compressed liquid take sat liq Table B11 hA3E A 16754 kJkg vA3E A 0001008 m3kg wAP sE A vA3E A PA4E A PA3E A 000100820000 20 201 kJkg wAPACE A wAPsE AηAρE A 201085 237 hA4ACE A hA3E hA4ACE A 1912 TA4ACE A 457C c The heat transfer in the boiler is from energy Eq413 qAboilerE A hA1E A hA4E A 38091 1912 36179 kJkg wAnetE A 133365 237 1310 kJkg ηATHE A wAnetE AqAboilerE A A 1310 36179E A 0362 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7153 Assume an actual compressor has the same exit pressure and specific heat transfer as the ideal isothermal compressor in Problem 727 with an isothermal efficiency of 80 Find the specific work and exit temperature for the actual compressor Solution CV Compressor Steady single inlet and single exit flows Energy Eq413 hAiE A q w hAeE A Entropy Eq79 sAiE A qT sAeE Inlet state Table B52 hAiE A 4034 kJkg sAiE A 18281 kJkg K Exit state Table B51 hAeE A 39836 kJkg sAeE A 17262 kJkg K q TsAeE A sAiE A 2731517262 18281 2783 kJkg w 4034 2783 39836 228 kJkg From Eq729 for a cooled compressor wAacE A wATE A η 22808 285 kJkg Now the energy equation gives hAeE A hi q wAacE A 4034 2783 285 40407 TAe acE A 6C PAeE A 294 kPa Explanation for the reversible work term is in Sect 73 Eqs 713 and 714 es i P v T s es i eac eac Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7154 A nozzle in a high pressure liquid water sprayer has an area of 05 cmA2E A It receives water at 350 kPa 20C and the exit pressure is 100 kPa Neglect the inlet kinetic energy and assume a nozzle isentropic efficiency of 85 Find the ideal nozzle exit velocity and the actual nozzle mass flow rate Solution CV Nozzle Liquid water is incompressible v constant no work no heat transfer Bernoulli Eq717 A1 2E AVA2 exE A 0 vPAiE A PAeE A 0001002 350 100 02505 kJkg VAexE A A 2 02505 1000 JkgEA 2238 m s 1 This was the ideal nozzle now we can do the actual nozzle Eq 730 A1 2E AVA2 ex acE A η A1 2E AVA2 exE A 085 02505 02129 kJkg VAex acE A A 2 02129 1000 JkgEA 2063 m s 1 AmE A ρAVAex acE A AVAex acE Av 05 10A4E A 2063 0001002 103 kgs These are examples of relatively low pressure spray systems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7155 Air flows into an insulated nozzle at 1 MPa 1200 K with 15 ms and mass flow rate of 2 kgs It expands to 650 kPa and exit temperature is 1100 K Find the exit velocity and the nozzle efficiency Solution CV Nozzle Steady 1 inlet and 1 exit flows no heat transfer no work Energy Eq413 hi 12VA2 iE A hAeE A 12VA2 eE A Entropy Eq79 si sAgenE A sAeE Ideal nozzle sAgenE A 0 and assume same exit pressure as actual nozzle Instead of using the standard entropy from Table A7 and Eq619 let us use a constant heat capacity at the average T and Eq623 First from A71 CAp 1150E A A127781 116118 1200 1100E A 1166 kJkg K CAvE A CAp 1150E A R 1166 0287 08793 k CAp 1150E ACAvE A 1326 Notice how they differ from Table A5 values TAe sE A Ti PAeE APiA k1 k E A 1200 A 650 1000 024585E A 10794 K A1 2E A VA2 e sE A A1 2E A VA2 iE A CTi TAe sE A A1 2E A 152 11661200 10794 1000 1125 1406196 140732 Jkg VAe sE A 5305 ms Actual nozzle with given exit temperature A1 2E AVA2 e acE A A1 2E AVA2 iE A hi hAe acE A 1125 11661200 1100 1000 1167125 Jkg VAe acE A 483 ms η noz A1 2E AVA2 e acE A A1 2E AVA2 iE A A1 2E AVA2 e sE A A1 2E AVA2 iE A hi he AChi he s A 116600 1406196E A 0829 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7156 A nozzle is required to produce a flow of air at 200 ms at 20C 100 kPa It is estimated that the nozzle has an isentropic efficiency of 92 What nozzle inlet pressure and temperature is required assuming the inlet kinetic energy is negligible Solution CV Air nozzle PAeE A TAeE Areal VAeE Areal ηAsE Areal For the real process hAiE A hAeE A VA 2 eE A2 or TAiE A TAeE A VA 2 eE A2CAP0E A 2932 200A2E A2 1000 1004 3131 K For the ideal process from Eq730 VA 2 esE A2 VA 2 eE A2ηAsE A 200A2E A2 1000 092 2174 kJkg and hAiE A hAesE A VA 2 esE A2 TAesE A TAiE A VA 2 esE A2CAP0E A 3131 21741004 2914 K The constant s relation in Eq623 gives PAiE A PAeE A TAiE ATAesE AA k k1 E A 100A 3131 2914 E A 350E A 1286 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7157 A watercooled air compressor takes air in at 20C 90 kPa and compresses it to 500 kPa The isothermal efficiency is 88 and the actual compressor has the same heat transfer as the ideal one Find the specific compressor work and the exit temperature Solution Ideal isothermal compressor exit 500 kPa 20C Reversible process dq T ds q TsAeE A sAiE A q TsAeE A sAiE A TsAo TeE A sAo T1E A R lnPAeE A PAiE A RT ln PAeE A PAiE A 0287 29315 ln 50090 1443 kJkg As same temperature for the ideal compressor hAeE A hAiE A w q 1443 kJkg wAacE A w η 16398 kJkg qAacE A q Now for the actual compressor energy equation becomes qAacE A hAiE A hAe acE A wAacE A hAe acE A hAiE A qAacE A wAacE A 1443 16398 197 kJkg CApE A TAe acE A TAiE A TAe acE A TAiE A 1971004 396C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7158 A flow of saturated liquid R410A at 200 kPa in an evaporator is brought to a state of superheated vapor at 200 kPa 20AoE AC Assume the process is reversible find the specific heat transfer and specific work CV Evaporator From the device we know that potential and kinetic energies are not important see chapter 6 Since the pressure is constant and the process is reversible from Eq714 w AvdPEA 0 0 0 0 From energy equation hi q w hAeE A hAeE A q hAeE A hi State i hi 418 kJkg State e hAeE A 31178 kJkg q hAeE A hi 31178 418 3076 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7159 A flow of R410A at 2000 kPa 40AoE AC in an isothermal expander is brought to a state of 1000 kPa in a reversible process Find the specific heat transfer and work CV Expander Steady reversible single inlet and exit flow Some q and w Energy Eq413 hi q w hAeE A Entropy Eq79 si dqT sAgenE A sAeE A Process T constant so dqT qT and reversible sAgenE A 0 State i hi 29549 kJkg si 10099 kJkgK State e hAeE A 31605 kJkg sAeE A 11409 kJkgK From entropy equation q T sAeE A si 31315 K 11409 10099 kJkgK 41023 kJkg From the energy equation w hi hAeE A q 29549 31605 41023 33046 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7160 A coflowing heat exchanger has one line with 2 kgs saturated water vapor at 100 kPa entering The other line is 1 kgs air at 200 kPa 1200 K The heat exchanger is very long so the two flows exit at the same temperature Find the exit temperature by trial and error Calculate the rate of entropy generation Solution CV Heat exchanger steady 2 flows in and two flows out No W no external Q 1 3 2 4 Flows m 1 m 2 AmE AH2OE A m 3 m 4 m air Energy AmE AH2OE A h2 h1 m air h3 h4 State 1 Table B12 h1 26755 kJkg State 2 100 kPa T2 State 3 Table A7 h3 12778 kJkg State 4 200 kPa T2 Only one unknown T2 and one equation the energy equation 2 h2 26755 112778 h4 1 kgs 2h2 h4 66288 kW At 500 K h2 29020 h4 50336 LHS 6307 too small At 700 K h2 33348 h4 71356 LHS 7383 too large Linear interpolation T2 560 K h2 30483 h4 56547 LHS 6662 Final states are with T2 5544 K 281 C H2O Table B13 h2 30368 kJkg s2 81473 s1 73593 kJkg K AIR Table A7 h4 55965 kJkg sT4 74936 sT3 83460 kJkg K The entropy balance equation Eq77 is solved for the generation term S gen AmE AH2OE A s2 s1 m air s4 s3 281473 73593 1 74936 83460 0724 kWK No pressure correction is needed as the air pressure for 4 and 3 is the same Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7161 Air at 100 kPa 17C is compressed to 400 kPa after which it is expanded through a nozzle back to the atmosphere The compressor and the nozzle both have an isentropic efficiency of 90 and are adiabatic The kinetic energy into and out of the compressor can be neglected Find the compressor work and its exit temperature and find the nozzle exit velocity W 1 3 5 CV Ideal compressor inlet 1 exit 2 Adiabatic q 0 Reversible sAgenE A 0 Energy Eq413 hA1E A 0 wACE A hA2E A Entropy Eq78 sA1E A 0T 0 sA2E wACsE A hA2E A hA1E A sA2E A sA1E A Properties use air Table A5 CAPoE A 1004 A kJ kg KE A R 0287 A kJ kg KE A k 14 Process gives constant s isentropic which with constant CAPoE A gives Eq623 TA2E A TA1E A PA2E APA1E AA k1 k E A 290 400100 A02857E A 4309 K wACsE A CAPoE ATA2E A TA1E A 1004 4309 290 14146 kJkg The ideal nozzle then expands back down to state 1 constant s The actual compressor discharges at state 3 however so we have wACE A wACsE AηACE A 15718 TA3E A TA1E A wACE ACApE A 4466 K Nozzle receives air at 3 and exhausts at 5 We must do the ideal exit at 4 first sA4E A sA3E A Eq623 TA4E A TA3E A PA4E APA3E AA k1 k E A 3005 K A1 2E A VAs E2 EA CApE ATA3E A TA4E A 14668 A1 2E A VAac E2 E A 132 kJkg VAacE A 5138 ms If we need it the actual nozzle exit 5 can be found TA5E A TA3E A VAac E2 E A2CApE A 315 K 1 2 3 4 5 P P P T s 2 3 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7162 A vortex tube has an air inlet flow at 20C 200 kPa and two exit flows of 100 kPa one at 0C and the other at 40C The tube has no external heat transfer and no work and all the flows are steady and have negligible kinetic energy Find the fraction of the inlet flow that comes out at 0C Is this setup possible Solution CV The vortex tube Steady single inlet and two exit flows No q or w Continuity Eq m 1 m 2 m 3 Energy m 1h1 m 2h2 m 3h3 Entropy m 1s1 S gen m 2s2 m 3s3 States all given by temperature and pressure Use constant heat capacity to evaluate changes in h and s Solve for x m 2m 1 from the energy equation m 3m 1 1 x h1 x h2 1x h3 x h1 h3h2 h3 T1 T3T2 T3 2040040 05 Evaluate the entropy generation assuming constant specific heat S genm 1 x s2 1xs3 s1 05s2 s1 05s3 s1 05 CApE A lnTA2E A TA1E A R lnPA2E A PA1E A 05CApE A lnT3 TA1E A R lnP3 PA1E A 05 1004 ln A27315 29315E A 0287 ln A100 200E A 05 1004 ln A31315 29315E A 0287 ln A100 200E A 01966 kJkg K 0 So this is possible Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7163 Air enters an insulated turbine at 50C and exits the turbine at 30C 100 kPa The isentropic turbine efficiency is 70 and the inlet volumetric flow rate is 20 Ls What is the turbine inlet pressure and the turbine power output CV Turbine ηs 07 Insulated Air Cp 1004 kJkgK R 0287 kJkgK k 14 Inlet Ti 50AoE AC V i 20 Ls 002 m3s Exit TAeE A 30AoE AC PAeE A 100 kPa a Energy Eq steady flow q hAiE A hAeE A wATE A q 0 Assume Constant Specific Heat wT hi hAeE A CApE ATi TAeE A 803 kJkg wATsE A wη 1147 kJkg wATsE A CApE ATAiE A TAesE A Solve for TAesE A 2089 K Isentropic Process PAeE A PAiE A TAeE A TAiE AA k k1 E A PAiE A 461 kPa b W T AmE AwT AmE A PAV E ART 0099 kgs W T 798 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7164 A stream of ammonia enters a steady flow device at 100 kPa 50C at the rate of 1 kgs Two streams exit the device at equal mass flow rates one is at 200 kPa 50C and the other as saturated liquid at 10C It is claimed that the device operates in a room at 25C on an electrical power input of 250 kW Is this possible Solution Control volume Steady device out to ambient 25C 1 2 3 W el Q Steady device cb Energy Eq410 AmE A1E AhA1E A AQE A AWE AelE A AmE A2E AhA2E A AmE A3E AhA3E Entropy Eq77 AmE A1E AsA1E A AQE ATAroomE A ASE AgenE A AmE A2E AsA2E A AmE A3E AsA3E State 1 Table B22 hA1E A 15812 kJkg sA1E A 64943 kJkg K State 2 Table B22 hA2E A 15766 kJkg sA2E A 61453 kJkg K State 3 Table B21 hA3E A 22697 kJkg sA3E A 08779 kJkg K From the energy equation AQE A 05 15766 05 22697 1 15812 250 9294 kW From the entropy equation ASE Agen 0561453 05 08779 1 64943 929429815 01345 kWK A0E A since ASE Agen A0E A this is possible Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7165 In a heatpowered refrigerator a turbine is used to drive the compressor using the same working fluid Consider the combination shown in Fig P9157 where the turbine produces just enough power to drive the compressor and the two exit flows are mixed together List any assumptions made and find the ratio of mass flow rates A mEA3E AA mEA1E A and TA5E A xA5E A if in twophase region if the turbine and the compressor are reversible and adiabatic Solution CV compressor sA2SE A sA1E A 10779 kJkg K hA2SE A 31743 kJkg wASCE A hA1E A hA2SE A 27189 31743 4554 kJkg CV turbine sA4SE A sA3E A 10850 kJkgK and PA4SE A hA4SE A 31972 kJkg wASTE A hA3E A hA4SE A 34129 31972 2157 kJkg As A wEATURBE A A wEACOMPE A A mEA3E AA mEA1E A A wSC EwST E A A4554 2157E A 2111 CV mixing portion A mEA1E AhA2SE A A mEA3E AhA4SE A A mEA1E A A mEA3E AhA5E 1 31743 2111 31972 3111 hA5E hA5E A 318984 kJkg TA5E A 587C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7166 A certain industrial process requires a steady 05 kgs supply of compressed air at 500 kPa at a maximum temperature of 30C This air is to be supplied by installing a compressor and aftercooler see Fig P946 Local ambient conditions are 100 kPa 20C Using an isentropic compressor efficiency of 80 determine the power required to drive the compressor and the rate of heat rejection in the aftercooler Air table A5 R 0287 kJkgK CApE A 1004 kJkgK k 14 State 1 TA1E A To 20AoE AC PA1E A Po 100 kPa AmE A 05 kgs State 2 PA2E A P3 500 kPa State 3 TA3E A 30AoE AC PA3E A 500 kPa We have ηAsE A 80 wACsE AwAC acE Compressor First do the ideal Isentropic TA2sE A TA1E A AP2P1E AA k1 k E A 29315 500100A 02857E A 4646 K Energy Eq qc hA1E A hA2E A wc qc 0 assume constant specific heat wcs CpT1 TA2sE A 1004 29315 4646 1720 kJkg ηAsE A wACsE AwAC acE A wAC acE A wACsE AηAsE A 215 W C m wC 1075 kW wAC acE A CApE A TA1E A TA2E A solve for TA2E A 5075 K Aftercooler Energy Eq q hA2E A hA3E A w w 0 assume constant specific heat q CApE A TA3E A TA2E A 100430315 5075 205 kJkg AQ E A AmE Aq 1025 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7167 Carbon dioxide flows through a device entering at 300 K 200 kPa and leaving at 500 K The process is steady state polytropic with n 38 and heat transfer comes from a 600 K source Find the specific work specific heat transfer and the specific entropy generation due to this process Solution CV Steady state device single inlet and single exit flow Energy Eq413 hi q he w Neglect kinetic potential energies Entropy Eq79 si dqT sAgenE A se Process Eq628 Pe Pi Te TiA n n1 E A 200500300A 38 28 E A 400 kPa and the process leads to Eq718 for the work term w A n n1E A R Te Ti A38 28E A 01889 500 300 513 kJkg Energy equation gives q he hi w 40152 21438 513 1358 kJkg Entropy equation gives CV out to source sAgenE A se si qTsource sAo TeE A sAo TiE A R lnPe Pi qTsource 53375 48631 01889 ln 400200 1358600 0117 kJkg K P v T s e i n 1 n 38 i e n 1 n k 129 n 38 Notice dP 0 so dw 0 ds 0 so dq 0 Notice process is externally irreversible T between source and CO2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7168 A flow of nitrogen 01 kgs comes out of a compressor stage at 500 kPa 500 K and is now cooled to 310 K in a counterflowing intercooler by liquid water at 125 kPa 15AoE AC which leaves at 22AoE AC Find the flowrate of water and the total rate of entropy generation Solution A hydraulic motor driven compressor with intercooler in small pipe between the two stages Continuity eq AmE A1E A AmE A2E A AmE AH2OE A AmE A3E A AmE A4E A AmE AN2E Energy eq 0 AmE AN2E A hA3E A hA4E A AmE AH2OE A hA2E A hA1E A Entropy Eq77 0 Am E AN2E A sA3E A sA4E A Am E AH2OE A sA2E A sA1E A S gen Due to the lower range of temperature we will use constant specific heats from A5 and A4 so the energy equation is approximated as 0 AmE AH2OE ACApE AH2OE A TA2E A TA1E A AmE AN2E ACApE A TA3E A TA4E A Now solve for the water flow rate AmE AH2OE A AmE AN2E A CApE AN2E A TA3E A TA4E A CApE AH2OE A TA1E A TA2E A 01 kgs 1042 kJkgK 500 310 K 418 22 15 kJkg 0677 kgs AS E AgenE A Am E AN2E A sA4E A sA3E A Am E AH2OE A s1 sA2E A Am E AN2E ACAPE A ln T4T3 Am E AH2OE ACAPE A ln T1T2 01 kgs 1042 kJkgK ln A310 500E A 0677 kgs 418 kJkgK ln A295 288E 004981 006796 00182 kWK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7169 An initially empty springloaded pistoncylinder requires 100 kPa to float the piston A compressor with a line and valve now charges the cylinder with water to a final pressure of 14 MPa at which point the volume is 06 mA3E A state 2 The inlet condition to the reversible adiabatic compressor is saturated vapor at 100 kPa After charging the valve is closed and the water eventually cools to room temperature 20C state 3 Find the final mass of water the piston work from 1 to 2 the required compressor work and the final pressure P3 Solution Wc in Process 12 transient adiabatic for CV compressor cylinder Assume process is reversible Continuity mA2E A 0 mAinE A Energy mA2E AuA2E A A0E A mAinE AhAinE A WAcE A A1E AWA2E Entropy Eq mA2E AsA2E A A0E A mAinE AsAinE A 0 sA2E A sAinE A Inlet state Table B12 hAinE A 26755 kJkg sAinE A 73594 kJkg K A1E AWA2E A APdVEA A1 2E A PAfloatE A PA2E AVA2E A A0E A A1 2E A 100140006 450 kJ State 2 PA2E A sA2E A sAinE A Table B13 vA2E A 02243 uA2E A 29844 kJkg mA2E A VA2E AvA2E A 0602243 2675 kg WAcE A mAinE AhAinE A mA2E AuA2E A A1E AWA2E A 2675 26755 29844 450 12763 kJ 1400 P V 100 2 3 0 06 State 3 must be on line 20C Assume 2phase PA3E A PAsatE A20C 2339 kPa less than PAfloatE A so compressed liquid Table B11 vA3E A vAfE A20C 0001002 VA3E A mA3E AvA3E A 000268 mA3E On line PA3E A 100 1400 100 00026806 1058 kPa P v 3 2 100 1400 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7170 Consider the scheme shown in Fig P7170 for producing fresh water from salt water The conditions are as shown in the figure Assume that the properties of salt water are the same as for pure water and that the pump is reversible and adiabatic a Determine the ratio A mEA7E AA mEA1E A the fraction of salt water purified b Determine the input quantities wAPE A and qAHE A c Make a second law analysis of the overall system CV Flash evaporator Steady flow no external q no work Energy Eq A mEA1E AhA4E A A mEA1E A A mEA7E AhA5E A A mEA7E AhA6E Table B11 or 6324 1 A mEA7E AA mEA1E A 41746 A mEA7E AA mEA1E A 26755 A mEA7E AA mEA1E A 00952 CV Pump steady flow incompressible liq wAPE A AvdPEA vA1E APA2E A PA1E A 0001001700 100 06 kJkg hA2E A hA1E A wAPE A 6299 06 636 kJkg CV Heat exchanger hA2E A A mEA7E AA mEA1E AhA6E A hA3E A A mEA7E AA mEA1E AhA7E 636 00952 26755 hA3E A 00952 14668 hA3E A 3043 kJkg CV Heater qAHE A hA4E A hA3E A 6324 3043 3281 kJkg CV entire unit entropy equation per unit mass flow rate at state 1 SACVgenE A qAHE ATAHE A 1 A mEA7E AA mEA1E AsA5E A A mEA7E AA mEA1E AsA7E A sA1E 328147315 09048 13026 00952 05053 02245 03088 kJKkgmA1E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7171 A rigid 10 mA3E A tank contains water initially at 120C with 50 liquid and 50 vapor by volume A pressurerelief valve on the top of the tank is set to 10 MPa the tank pressure cannot exceed 10 MPa water will be discharged instead Heat is now transferred to the tank from a 200C heat source until the tank contains saturated vapor at 10 MPa Calculate the heat transfer to the tank and show that this process does not violate the second law Solution CV Tank and walls out to the source Neglect storage in walls There is flow out and no boundary or shaft work Continuity Eq415 mA2E A mA1E A mAeE A Energy Eq416 mA2E A uA2E A mA1E AuA1E A mAeE AhAeE A A1E AQA2E A Entropy Eq713 mA2E AsA2E A mA1E AsA1E A mAeE AsAeE A dQT A1E ASA2 genE State 1 TA1E A 120AoE AC Table B11 vf 000106 m3kg mAliqE A 05VA1E AvAfE A 4717 kg vAgE A 08919 m3kg mAgE A 05V1vAgE A 056 kg mA1E A 47226 kg xA1E A mAgE AmA1E A 0001186 uA1E A uAfE A xA1E AuAfgE A 5035 000118620258 50588 kJkg sA1E A sAfE A xA1E AsAfgE A 15275 00011865602 15341 kJkgK State 2 PA2E A 10 MPa sat vap xA2E A 10 VA2E A 1m3 vA2E A vAgE A 019444 m3kg mA2E A VA2E AvA2E A 514 kg uA2E A uAgE A 25836 kJkg sA2E A sAgE A 65864 kJkgK Exit Pe 10 MPa sat vap xe 10 he hg 27781 kJkg se sAgE A 65864 kJkg me mA1E A mA2E A 46712 kg From the energy equation we get A1E AQA2E A mA2E A uA2E A mA1E AuA1E A mAeE AhAeE A 1 072 080 kJ From the entropy Eq713 or Eq724 with 725 and 726 we get A1E ASA2 genE A mA2E AsA2E A mA1E AsA1E A mAeE AsAeE A 1Q2 TH TH 200AoE AC 473 K A1E ASA2 genE A Snet 1204 kJ 0 Process Satisfies 2nd Law Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7172 A jetejector pump shown schematically in Fig P7172 is a device in which a lowpressure secondary fluid is compressed by entrainment in a highvelocity primary fluid stream The compression results from the deceleration in a diffuser For purposes of analysis this can be considered as equivalent to the turbinecompressor unit shown in Fig P7165 with the states 1 3 and 5 corresponding to those in Fig P7172 Consider a steam jetpump with state 1 as saturated vapor at 35 kPa state 3 is 300 kPa 150C and the discharge pressure PA5E A is 100 kPa a Calculate the ideal mass flow ratio A mEA1E AA mEA3E A b The efficiency of a jet pump is defined as η A mEA1E AA mEA3E AAactualE A A mEA1E AA mEA3E AAidealE A for the same inlet conditions and discharge pressure Determine the discharge temperature of the jet pump if its efficiency is 10 a ideal processes isen comp exp A expands 34s comp 12s E A then mix at const P sA4sE A sA3E A 70778 13026 xA4sE A 60568 xA4sE A 09535 hA4sE A 41746 09535 22580 25705 kJkg sA2sE A sA1E A 77193 TA2sE A 174C hA2sE A 28238 kJkg A mEA1E AhA2sE A hA1E A A mEA3E AhA3E A hA4sE A A mEA1E AA mEA3E AAIDEALE A A27610 25705 28238 26311E A 09886 b real processes with jet pump eff 010 A mEA1E AA mEA3E AAACTUALE A 010 09886 009886 Energy Eq A mEA1E AhA1E A A mEA3E AhA3E A A mEA1E A A mEA3E AhA5E 009886 26311 1 27610 109896 hA5E State 5 hA5E A 27493 kJkg PA5E A 100 kPa TA5E A 1365 AoE AC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7173 A horizontal insulated cylinder has a frictionless piston held against stops by an external force of 500 kN The piston crosssectional area is 05 m2 and the initial volume is 025 mA3E A Argon gas in the cylinder is at 200 kPa 100C A valve is now opened to a line flowing argon at 12 MPa 200C and gas flows in until the cylinder pressure just balances the external force at which point the valve is closed Use constant heat capacity to verify that the final temperature is 645 K and find the total entropy generation Solution The process has inlet flow no work volume constant and no heat transfer Continuity Eq415 mA2E A mA1E A mAiE A Energy Eq416 mA2E A uA2E A mA1E AuA1E A mAiE A hAiE A mA1E A PA1E AVA1E ARTA1E A 200 02502081 37315 0644 kg Force balance PA2E AA F PA2E A A500 05E A 1000 kPa For argon use constant heat capacities so the energy equation is mA2E A CAVoE A TA2E A mA1E A CAVoE A TA1E A mA2E A mA1E A CAPoE A T in We know PA2E A so only 1 unknown for state 2 Use ideal gas law to write mA2E ATA2E A PA2E AVA1E AR and mA1E A TA1E A PA1E AVA1E AR and divide the energy equation with CAVoE A to solve for the change in mass PA2E A VA1E A PA1E AVA1E AR mA2E A mA1E A CAPoE ACAVoE A T in mA2E A mA1E A PA2E A PA1E AVA1E AR k T in 1000 20002502081166747315 1219 kg mA2E A 1219 0644 1863 kg TA2E A PA2E AVA1E AmA2E AR 1000025186302081 645 K OK Entropy Eq712 mA2E AsA2E A mA1E AsA1E A mAiE AsAiE A 0 A1E ASA2 genE A1E ASA2 genE A mA1E AsA2E A sA1E A mA2E A mA1E AsA2E A sAiE A mA1E ACApE A lnA T2 ET1 E A R ln A P2 EP1 E A mA2E A mA1E ACApE A lnA T2 ETi E A R ln A P2 EPi E A 0644 052 ln A 645 37315E A 02081 ln A1000 200E A 1219 052 ln A 645 47315E A 02081 ln A1000 1200E A 003242 024265 021 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7174 Supercharging of an engine is used to increase the inlet air density so that more fuel can be added the result of which is an increased power output Assume that ambient air 100 kPa and 27C enters the supercharger at a rate of 250 Ls The supercharger compressor has an isentropic efficiency of 75 and uses 20 kW of power input Assume that the ideal and actual compressor have the same exit pressure Find the ideal specific work and verify that the exit pressure is 175 kPa Find the percent increase in air density entering the engine due to the supercharger and the entropy generation W c in ex CV Air in compressor steady flow Cont AmE AinE A AmE AexE A AmE A AVE AvAinE A 029 kgs Energy AmE AhAinE A AWE A AmE AhAexE A Assume AQE A 0 Entropy AmE AsAinE A ASE AgenE A AmE AsAexE vAinE A A RTin EPin E A 08614 mA3E Akg sAo TiE A 686975 kJkg K hAinE A 30062 kJkg ηAcE A wAC sE AwAC acE A A WEASE A A WEAACE A ηAcE A 15 kW wAC sE A A WEASE AAmE A 51724 kJkg wAC acE A 68966 kJkg Table A7 hAex sE A hAinE A wAC sE A 30062 51724 3523 kJkg TAex sE A 3515 K sAo TeE A 702830 kJkg K PAexE A PAinE A eEAsAo T ex A sAo ET in ARE A 100 exp A70283 686975 0287E A 17375 kPa The actual exit state is hAex acE A hAinE A wAC acE A 3696 kJkg TAex acE A 3686 K vAexE A RTAexE APAexE A 06088 mA3E Akg sAo Tex acE A 70764 ρAexE AρAinE A vAinE AvAexE A 0861406088 1415 or 415 increase sAgenE A sAexE A sAinE A 70764 686975 0287 lnA17375 100E A 00481 kJkg K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7175 A rigid steel bottle V 025 mA3E A contains air at 100 kPa 300 K The bottle is now charged with air from a line at 260 K 6 MPa to a bottle pressure of 5 MPa state 2 and the valve is closed Assume that the process is adiabatic and the charge always is uniform In storage the bottle slowly returns to room temperature at 300 K state 3 Find the final mass the temperature TA2E A the final pressure PA3E A the heat transfer A1E AQA3E A and the total entropy generation CV Bottle Flow in no work no heat transfer Continuity Eq415 mA2E A mA1E A mAinE A Energy Eq416 mA2E AuA2E A mA1E AuA1E A mAinE AhAinE State 1 and inlet Table A7 uA1E A 21436 kJkg hAinE A 26032 kJkg mA1E A PA1E AVRTA1E A 100 0250287 300 0290 kg mA2E A PA2E AVRTA2E A 5000 0250287 TA2E A 43554TA2E Substitute into energy equation uA2E A 000306 TA2E A 26032 kJkg Now trial and error on TA2E TA2E A 360 LHS 25863 low TA2E A 370 LHS 26588 high Interpolation TA2E A 3623 K LHS 2603 OK mA2E A 435543623 12022 kg PA3E A mA2E ARTA3E AV 4140 kPa Now use the energy equation from the beginning to the final state A1E AQA3E A mA2E AuA3E A mA1E AuA1E A mAinE AhAinE A 12022 029 21436 11732 26032 5392 kJ Entropy equation from state 1 to state 3 with change in s from Eq819 SAgenE A mA2E AsA3E A mA1E AsA1E A mAinE AsAinE A A1E AQA3E AT mA2E AsA3E A sAinE A mA1E AsA1E A sAinE A A1E AQA3E AT 1202268693 67256 R ln41406000 02968693 67256 R ln1006000 5392300 4423 kJK 2 1 v T s 1 260 300 v C 100 kPa 5 MPa 3 6 MPa line P 2 T2 3 line Problem could have been solved with constant specific heats from A5 in which case we would get the energy explicit in TA2E A no iterations Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7176 A certain industrial process requires a steady 05 kgs of air at 200 ms at the condition of 150 kPa 300 K This air is to be the exhaust from a specially designed turbine whose inlet pressure is 400 kPa The turbine process may be assumed to be reversible and polytropic with polytropic exponent n 120 a What is the turbine inlet temperature b What are the power output and heat transfer rate for the turbine c Calculate the rate of net entropy increase if the heat transfer comes from a source at a temperature 100C higher than the turbine inlet temperature Solution CV Turbine this has heat transfer PVn Constant n 12 Process polytropic Eq828 TAeE A TAiE A PAeE APAiE AA n1 n E A TAiE A 3533 K Energy Eq412 Am E Aih VA2E A2in AQ E A Am E Aexh VA2E A2ex W T Reversible shaft work in a polytropic process Eq714 and Eq718 wT v dP VA2 iE A VA2 eE A 2 A n n1E APeve Pivi VA2 iE A VA2 eE A 2 A n n1E AR Te Ti VA2 eE A 2 718 kJkg AW E AT Am E AwT 359 kW Assume constant specific heat in the energy equation AQ E A Am E ACAPE A Te Ti VA2 eE A 2 AW E AT 192 kW Entropy Eq77 or 923 with change in entropy from Eq816 dSnetdt S gen Am E Ase si AQ E AHE ATAHE A TH Ti 100 4533 K se si CAPE A lnTe Ti R lnPe Pi 01174 kJkg K dSnetdt 05 01174 1924533 00163 kWK P v T s e i n 1 n 12 i e n 1 n k 14 n 12 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Problems solved with Pr and vr functions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 717 uses Pr function Do the previous problem using the air tables in A7 The exit nozzle in a jet engine receives air at 1200 K 150 kPa with neglible kinetic energy The exit pressure is 80 kPa and the process is reversible and adiabatic Use constant heat capacity at 300 K to find the exit velocity Solution CV Nozzle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi hAeE A VA2 eE A2 Zi ZAeE A Entropy Eq79 sAeE A si dqT sAgenE A si 0 0 Process q 0 sAgenE A 0 as used above leads to sAeE A si Inlet state hi 12778 kJkg Pr i 19117 The constant s is done using the Pr function from A72 Pr e Pr i PAeE A PAiE A 19117 80150 101957 Interpolate in A7 TAeE A 1000 50 A101957 91651 11135 91651E A 102616 K hAeE A 10462 05232 11035 10462 10762 kJkg From the energy equation we have VA2 eE A2 hi hAeE A so then VAeE A E 2 hi hAe A A 212778 10762 kJkg 1000 JkJE A 635 ms P v T s e i i e Low V Hi P Low P Hi V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 731 uses Pr function Air enters a turbine at 800 kPa 1200 K and expands in a reversible adiabatic process to 100 kPa Calculate the exit temperature and the work output per kilogram of air using a The ideal gas tables Table A7 b Constant specific heat value at 300 K from table A5 Solution i e Turbine air W CV Air turbine Adiabatic q 0 reversible sAgenE A 0 Energy Eq413 wATE A hAiE A hAeE A Entropy Eq79 sAeE A sAiE a Table A7 hAiE A 12778 kJkg Pr i 19117 The constant s process is done using the Pr function from A72 Pr e Pr i PAeE A PAiE A 19117 A 100 800 E A 23896 Interpolate in A71 TAeE A 7057 K hAeE A 7197 kJkg w hAiE A hAeE A 12778 7197 5581 kJkg b Table A5 CAPoE A 1004 kJkg K R 0287 kJkg K k 14 then from Eq823 TAeE A TAiE A PAeE APAiE AA k1 k E A 1200 A 100 800 E A 0286E A 6621 K w CAPoE ATAiE A TAeE A 10041200 6621 5398 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 734 uses Pr function A compressor receives air at 290 K 95 kPa and a shaft work of 55 kW from a gasoline engine It should deliver a mass flow rate of 001 kgs air to a pipeline Find the maximum possible exit pressure of the compressor Solution CV Compressor Steady single inlet and exit flows Adiabatic AQ E A 0 Continuity Eq411 AmE AiE A AmE AeE A AmE A Energy Eq412 AmE AhAiE A AmE AhAeE A AWE ACE A Entropy Eq78 AmE AsAiE A S gen AmE AsAeE A Reversible S gen 0 AWE AcE A AmE Aw E AcE A Aw E AcE A AWE AAmE A 55001 550 kJkg Use Table A7 hi 29043 kJkg Pr i 09899 hAeE A hi Aw E AcE A 29043 550 84043 kJkg A7 TAeE A 8165 K Pr e 41717 PAeE A Pi Pr ePr i 95 4171709899 4003 kPa P v T s e i i e h 550 kJkg WC i e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 755 uses Pr function An underground saltmine 100 000 mA3E A in volume contains air at 290 K 100 kPa The mine is used for energy storage so the local power plant pumps it up to 21 MPa using outside air at 290 K 100 kPa Assume the pump is ideal and the process is adiabatic Find the final mass and temperature of the air and the required pump work Solution CV The mine volume and the pump Continuity Eq415 mA2E A mA1E A mAinE Energy Eq416 mA2E AuA2E A mA1E AuA1E A A1E AQA2E A A1E AWA2E A mAinE AhAinE Entropy Eq712 mA2E AsA2E A mA1E AsA1E A AdQTEA A1E ASA2 genE A mAinE AsAinE Process Adiabatic A1E AQA2E A 0 Process ideal A1E ASA2 genE A 0 sA1E A sAinE mA2E AsA2E A mA1E AsA1E A mAinE AsAinE A mA1E A mAinE AsA1E A mA2E AsA1E A sA2E A sA1E Constant s Pr2 Pr i PA2E A PAiE A 09899 A 2100 100 E A 207879 A72 TA2E A 680 K uA2E A 49694 kJkg mA1E A PA1E AVA1E ARTA1E A 10010A5E A0287 290 120149 10A5E A kg mA2E A PA2E AVA2E ARTA2E A 100 2110A5E A0287 680 10760 10A5E A kg mAinE A 9558510A5E A kg A1E AWA2E A mAinE AhAinE A mA1E AuA1E A mA2E AuA2E mAinE A29043 mA1E A20719 mA2E A49694 2322 10A8E A kJ 2 1 i P v T s 1 i 2 T2 290 400 s C 100 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 780 uses Pr function Calculate the air temperature and pressure at the stagnation point right in front of a meteorite entering the atmosphere 50 C 50 kPa with a velocity of 2000 ms Do this assuming air is incompressible at the given state and repeat for air being a compressible substance going through an adiabatic compression Solution Kinetic energy A1 2E A VA2E A A1 2E A 2000A2E A1000 2000 kJkg Ideal gas vAatmE A RTP 0287 22350 128 mA3E Akg a incompressible Energy Eq413 h A1 2E A VA2E A 2000 kJkg If A5 T hCApE A 1992 K unreasonable too high for that CApE Use A7 hAstE A hAoE A A1 2E A VA2E A 22322 2000 22233 kJkg TAstE A 1977 K Bernoulli incompressible Eq717 P PAstE A PAoE A A1 2E A VA2E Av 2000128 15625 kPa PAstE A 15625 50 16125 kPa b compressible TAstE A 1977 K the same energy equation From A72 Stagnation point PAr stE A 15803 Free PAr oE A 039809 PAstE A PAoE A A Pr st EPr o E A 50 A 15803 039809E A 198 485 kPa Notice that this is highly compressible v is not constant Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7174 uses Pr function Supercharging of an engine is used to increase the inlet air density so that more fuel can be added the result of which is an increased power output Assume that ambient air 100 kPa and 27C enters the supercharger at a rate of 250 Ls The supercharger compressor has an isentropic efficiency of 75 and uses 20 kW of power input Assume that the ideal and actual compressor have the same exit pressure Find the ideal specific work and verify that the exit pressure is 175 kPa Find the percent increase in air density entering the engine due to the supercharger and the entropy generation W c in ex CV Air in compressor steady flow Cont AmE AinE A AmE AexE A AmE A AVE AvAinE A 029 kgs Energy AmE AhAinE A AWE A AmE AhAexE A Assume AQE A 0 Entropy AmE AsAinE A ASE AgenE A AmE AsAexE Inlet state vAinE A RTAinE APAinE A 08614 mA3E Akg PAr inE A 11167 ηAcE A wAC sE AwAC acE A A WEASE A A WEAACE A ηAcE A 15 kW wAC sE A A WEASE AAmE A 51724 kJkg wAC acE A 68966 kJkg Table A7 hAex sE A hAinE A wAC sE A 30062 51724 3523 kJkg TAex sE A 3515 K PAr exE A 1949 PAexE A PAinE A PAr exE APAr inE A 100 1949 11167 1745 kPa The actual exit state is hAex acE A hAinE A wAC acE A 3696 kJkg TAex acE A 3686 K vAexE A RTAexE APAexE A 0606 mA3E Akg ρAexE AρAinE A vAinE AvAexE A 086140606 142 or 42 increase sAgenE A sAexE A sAinE A 70767 68693 0287 ln174100 00484 kJkg K UPDATED AUGUST 2013 SOLUTION MANUAL CHAPTER 7 English units Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CHAPTER 7 SUBSECTION PROB NO ConceptStudy Guide Problems 177 Steady Single Flow Devices 178189 Transient Processes 190192 Reversible Shaft Work Bernoulli 193202 Polytropic process 203204 Steady Irreversible Processes 205220 Transient Processes 221224 Device Efficiency 225234 Review Problems 235236 SOLUTION MANUAL CHAPTER 9 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Steady Single Flow Devices Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7177E A compressor receives R134a at 20 F 30 psia with an exit of 200 psia x 1 What can you say about the process Solution Properties for R134a are found in Table F10 Inlet state si 04157 Btulbm R Exit state se 04080 Btulbm R Steady state single flow se si i e dq T sgen Since s decreases slightly and the generation term can only be positive it must be that the heat transfer is negative out so the integral gives a contribution that is smaller than sgen Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7178E A condenser receives R410A at 0 F and quality 80 with the exit flow being saturated liquid at 0 F Consider the cooling to be a reversible process and find the specific heat transfer from the entropy equation Entropy Eq79 se si dqT sgen si qT 0 q T se si T sf si Inlet si 00306 xi 02257 021116 BtulbmR Exit sf 00306 BtulbmR q 45967 0 R 00306 021116 BtulbmR 830 Btulbm Remark It fits with he hi 1 xi hfg 08 10376 830 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7179E Steam enters a turbine at 450 lbfin2 900 F expands in a reversible adiabatic process and exhausts at 130 F Changes in kinetic and potential energies between the inlet and the exit of the turbine are small The power output of the turbine is 800 Btus What is the mass flow rate of steam through the turbine Solution CV Turbine Steady single inlet and exit flows Adiabatic Q 0 Continuity Eq411 m i m e m Energy Eq412 m hi m he W T Entropy Eq78 m si 0 m se Reversible S gen 0 Explanation for the work term is in Sect 73 Eq714 2 1 P v T s 1 2 Inlet state Table F72 hi 14683 btulbm si 17113 btulbm R Exit state se 17113 Btulbm R Te 130 F saturated xe 17113 0181717292 08846 he 9797 xe 101978 1000 Btulbm w hi he 14683 1000 4683 Btulbm m W w 800 Btus 4683 Btulbm 1708 lbms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7180E The exit nozzle in a jet engine receives air at 2100 R 20 psia with neglible kinetic energy The exit pressure is 10 psia and the process is reversible and adiabatic Use constant heat capacity at 77 F to find the exit velocity Solution CV Nozzle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi he V2 e2 Zi Ze Entropy Eq79 se si dqT sgen si 0 0 Use constant specific heat from Table F4 CPo 024 Btu lbm R k 14 The isentropic process se si gives Eq623 Te Ti PePi k1 k 2100 1020 02857 17227 R The energy equation becomes conversion 1 Btulbm 25 037 ft2s2 in A1 V2 e2 hi he CP Ti Te Ve 2 CP Ti Te 2024210017227 25 037 2129 fts P v T s e i i e Low V Hi P Low P Hi V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7181E Do the previous problem using Table F5 The exit nozzle in a jet engine receives air at 2100 R 20 psia with neglible kinetic energy The exit pressure is 10 psia and the process is reversible and adiabatic Use constant heat capacity at 77 F to find the exit velocity Solution CV Nozzle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi he V2 e2 Zi Ze Entropy Eq79 se si dqT sgen si 0 0 The constant s is rewritten from Eq619 as so Te so Ti R lnPe Pi 198461 5334 778 ln 1020 1937088 BtulbmR Interpolate in F5 Te 1750 50 1937088 193444 194209 193444 1750 50 034608 17673 R he 436205 449794 436205 034608 440908 Btulbm The energy equation becomes conversion 1 Btulbm 25 037 ft2s2 in A1 V2 e2 hi he 53257 440908 91662 Btulbm Ve 2 hi he 2 91662 25 037 2142 fts P v T s e i i e Low V Hi P Low P Hi V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7182E In a heat pump that uses R134a as the working fluid the R134a enters the compressor at 30 lbfin2 20 F In the compressor the R134a is compressed in an adiabatic process to 150 lbfin2 using 15 Btus of power Find the mass flow rate it can provide assuming the process is reversible Solution CV Compressor Steady single inlet and exit flows Adiabatic Q 0 Continuity Eq411 m 1 m 2 m Energy Eq412 m h1 m h2 W C Entropy Eq78 m s1 0 m s2 Reversible S gen 0 Inlet state Table F102 h1 16982 Btulbm s1 04157 Btulbm R Exit state P2 150 psia s2 h2 18446 Btulbm m W cwc W c h1 h2 15 16982 18446 0102 lbms Explanation for the work term is in Sect 73 Eq714 2 1 P v T s 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7183E A compressor in a commercial refrigerator receives R410A at 10 F and unknown quality The exit is at 300 psia 140 F and the process assumed reversible and adiabatic Neglect kinetic energies and find the inlet quality and the specific work CV Compressor q 0 Energy Eq413 wC hi he Entropy Eq79 se si dqT sgen si 0 0 Exit state 300 psia 140 F s si Table F92 he 13734 Btulbm se si 02582 BtulbmR Inlet state F91 T si se xi 02582 0023102362 099534 hi 1008 xi 1062 11578 Btulbm wC 11578 13734 2156 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7184E A compressor brings a hydrogen gas flow at 500 R 1 atm up to a pressure of 10 atm in a reversible process How hot is the exit flow and what is the specific work input CV Compressor Assume q 0 Energy Eq413 wC hi he Cp Ti Te Entropy Eq79 se si dqT sgen si 0 0 Properties from Table F4 and constant s from Eq623 Te Ti PePi k1k 500 R 101 140911409 9755 R Now the work from the energy equation Cp from F4 wC 3394 BtulbmR 500 9755 R 16138 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7185E A flow of 4 lbms saturated vapor R410A at 100 psia is heated at constant pressure to 140 F The heat is supplied by a heat pump that receives heat from the ambient at 540 R and work input shown in Fig P728 Assume everything is reversible and find the rate of work input Solution CV Heat exchanger Continuity Eq m 1 m 2 Energy Eq m 1h1 Q H m 1h 2 Table F92 h1 11938 Btulbm s1 02498 Btulbm R h2 14613 Btulbm s2 02994 Btulbm R H Q W L Q T L HP 1 2 Notice we can find Q H but the temperature TH is not constant making it difficult to evaluate the COP of the heat pump CV Total setup and assume everything is reversible and steady state Energy Eq m 1h1 Q L W m 1h2 Entropy Eq m 1s1 Q LTL 0 m 1s2 TL is constant sgen 0 Q L m 1TL s2 s1 4 540 02994 02498 10714 Btus W m 1h2 h1 Q L 4 14613 11938 10714 014 Btus Comment Net work is nearly zero due to the very low inlet T so the first heating of the flow actually generates work out and only the heating to above ambient T requires work input Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7186E A diffuser is a steadystate steadyflow device in which a fluid flowing at high velocity is decelerated such that the pressure increases in the process Air at 18 lbfin2 90 F enters a diffuser with velocity 600 fts and exits with a velocity of 60 fts Assuming the process is reversible and adiabatic what are the exit pressure and temperature of the air CV Diffuser Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi V2 i 2gc he V2 e2gc he hi CPoTe Ti Entropy Eq79 si dqT sgen si 0 0 se Reversible adiabatic Energy equation then gives conversion 1 Btulbm 25 037 ft2s2 from A1 CPoTe Ti 024Te 5497 6002 602 2 25 037 Te 5793 R The isentropic process se si gives Eq623 Pe PiTeTi k k1 18 5793 5497 35 216 lbfin 2 P v T s e i i e Inlet Low V Exit Hi V Hi P A Low P A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7187E An expander receives 1 lbms air at 300 psia 540 R with an exit state of 60 psia 540 R Assume the process is reversible and isothermal Find the rates of heat transfer and work neglecting kinetic and potential energy changes Solution CV Expander single steady flow Energy Eq m hi Q m he W Entropy Eq m si Q T m sgen m se Process T is constant and sgen 0 Ideal gas and isothermal gives a change in entropy by Eq 615 so we can solve for the heat transfer Q Tm se si m RT ln Pe Pi 1 lbms 540 R 5334 778 BtulbmR ln 60 300 596 Btus From the energy equation we get W m hi he Q Q 596 Btus P v T s e i i e Wexp i e Q Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7188E One technique for operating a steam turbine in partload power output is to throttle the steam to a lower pressure before it enters the turbine as shown in Fig P739 The steamline conditions are 200 lbfin2 600 F and the turbine exhaust pressure is fixed at 1 lbfin2 Assuming the expansion inside the turbine to be reversible and adiabatic determine the specific turbine work for no throttling and the specific turbine work partload if it is throttled to 60 psia Show both processes in a Ts diagram CV Turbine fullload reversible Entropy Eq79 reduces to constant s so from Table F72 and F71 s3a s1 16767 Btulbm R 0132 66 x3a 18453 x3a 08367 h3a 6974 08367 10360 9366 Btulbm 1w3a h1 h3a 132205 9366 38545 Btulbm The energy equation for the part load operation gives the exit h Notice that we have constant h in the throttle process h2 h1 2b P h2b h1 132205 btulbm s2b 180656 BtulbmR 3b P s s2b 180656 x3b 180656 0132 6618453 090711 h3b 6974 x3b 10360 10095 Btulbm wT 132205 10095 31254 Btulbm 2b 1 2a T s 3a 3b h C WT 1 2 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7189E An adiabatic air turbine receives 2 lbms air at 2700 R 240 psia and 4 lbms air at 60 psia T2 in a setup similar to Fig P487 with an exit flow at 15 psia What should the temperature T2 be so the whole process can be reversible The process is reversible if we do not generate any entropy Physically in this problem it means that state 2 must match the state inside the turbine so we do not mix fluid at two different temperatures we assume the pressure inside is exactly 60 psia For this reason let us select the front end as CV and consider the flow from state 1 to the 60 psia This is a single flow Entropy Eq79 s1 0T 0 s2 Property Eq619 s2 s1 0 so T2 so T1 R lnP2 P1 so T2 so T1 R lnP2 P1 205606 5334 778 ln 60 240 19610 BtulbmR From F5 T2 19288 R If we solve with constant specific heats we get from Eq623 and k 14 T2 T1 P2 P1k1k 2700 R 6024002857 1817 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7190E A 20 ft3 tank contains carbon dioxide at 540 R 20 psia is now filled from a supply of carbon dioxide at 540 R 20 psia by a compressor to a final tank pressure of 60 psia Assume the whole process is adiabatic and reversible Find the final mass and temperature in the tank and the required work to the compressor CV The tank and the compressor Continuity Eq415 m2 m1 min Energy Eq416 m2u2 m1u1 1Q2 1W2 minh in Entropy Eq713 m2s2 m1s1 dQT 1S2 gen minsin Process Adiabatic 1Q2 0 Process ideal 1S2 gen 0 s1 s in m2s2 m1s1 minsin m1 mins1 m2s1 s2 s1 Constant s so since the temperatures are modest use Eq623 T2 T1 P2 P1 k1 k 540 R 602002242 6908 R m1 P1V1RT1 20 20 144 351 540 3039 lbm m2 P2V2RT2 60 20 144351 6908 7127 lbm min 4088 lbm 1W2 minhin m1u1 m2u2 minRTin m1u1 uin m2u2 uin 4088 351 540778 m1 0 7127 0156 6908 540 6807 Btu work must come in Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7191E An underground saltmine 35 106 ft3 in volume contains air at 520 R 147 lbfin2 The mine is used for energy storage so the local power plant pumps it up to 310 lbfin2 using outside air at 520 R 147 lbfin2 Assume the pump is ideal and the process is adiabatic Find the final mass and temperature of the air and the required pump work Overnight the air in the mine cools down to 720 R Find the final pressure and heat transfer CV The mine volume and the pump Continuity Eq415 m2 m1 min Energy Eq416 m2u2 m1u1 1Q2 1W2 minh in Entropy Eq713 m2s2 m1s1 dQT 1S2 gen minsin Process Adiabatic 1Q2 0 Process ideal 1S2 gen 0 s1 s in m2s2 m1s1 minsin m1 mins1 m2s1 s2 s1 Constant s Eq619 so T2 so Ti R lnPe Pi Table F4 so T2 163074 5334 778 ln 310 147 183976 Btulbm R T2 1221 R u2 21313 Btulbm Now we have the states and can get the masses m1 P1V1RT1 147 35106 144 5334 520 2671105 lbm m2 P2V2RT2 310 35106 144 5334 1221 24 106 lbm min m2 m1 21319106 lbm 1W2 minhin m1u1 m2u2 21319106 12438 2671105 8873 24106 21313 2226 108 Btu pump work in Wpump 223 108 Btu 2W3 0 P3 P2T3T2 3107201221 1828 lbfin 2 2Q3 m2u3 u2 2410612317 21313 216 108 Btu 2 1 i P v T s 1 i 2 T 2 520 720 s C 147 psia 3 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7192E R410A at 240 F 600 psia is in an insulated tank and flow is now allowed out to a turbine with a backup pressure of 125 psia The flow continues to a final tank pressure of 125 psia and the process stops If the initial mass was 1 lbm how much mass is left in the tank and what is the turbine work assuming a reversible process Solution CV Total tank and turbine Continuity Eq415 m2 m1 mex Energy Eq416 m2u2 m1u1 mexhex 1Q2 1W 2 Entropy Eq712 m2s2 m1s1 mexsex dQT 1S 2 gen Process Adiabatic 1Q2 0 Reversible 1S2 gen 0 This has too many unknowns we do not know state 2 only P2 CV m2 the mass that remains in the tank This is a control mass Entropy Eq63 637 m2s2 s1 dQT 1S2 gen Process Adiabatic 1Q2 0 Reversible 1S2 gen 0 s2 s1 State 1 v1 01342 ft3lbm u1 13996 Btulbm s1 02703 BtulbmR State 2 Ps T2 8605 F v2 05722 ft3lbm u2 11945 Btulbm State exit sex s2 s1 follows from entropy Eq for first CV using the continuity eq this is identical to state 2 hex 132695 Btulbm Tank volume constant so V m1v1 m2v 2 m2 m1 v1 v2 1 01342 05722 02345 lbm From energy eq 1W2 m1u1 m2u2 mexh ex 1 13996 02345 11945 07655 132695 Btu 1037 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Reversible Shaft Work Bernoulli Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7193E A river flowing at 2 fts across 3 ft high and 30 ft wide area has a dam that creates an elevation difference of 7 ft How much energy can a turbine deliver per day if 80 of the potential energy can be extracted as work CV Dam from upstream to downstream 2 m lower elevation Continuity Eq m constant m e m i AeVeve AiVivi Find the mass flow rate m AViv ρAVi 622 lbmft3 3 30 ft2 2 fts 11 196 lbms Energy Eq 0 m hi 05Vi 2 gZi m he 05Ve 2 gZe W The velocity in and out is the same assuming same flow area and the two enthalpies are also the same since same T and P P0 This is consistent with Eq714 w gZi Ze loss W 08 m gZi Ze 08 11 196 lbms 32174 fts2 7 ft 2 017 232 25 037 Btus 8057 Btus Recall conversion 1 Btulbm 25 037 ft2s2 from A1 W W t 8057 Btus 24 60 60 s 696 MBtu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7194E How much liquid water at 60 F can be pumped from 147 psia to 35 psia with a 3 kW motor Incompressible flow liquid water and we assume reversible Then the shaftwork is from Eq715 716 w v dP v P 0016 ft3lbm 35 147 psia 4677 lbfftlbm 006 Btulbm W 3 kW 2844 Btus m w W 2844 006 474 lbms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7195E A wave comes rolling in to the beach at 6 fts horizontal velocity Neglect friction and find how high up elevation on the beach the wave will reach We will assume a steady reversible single flow at constant pressure and temperature for the incompressible liquid water The water will flow in and up the sloped beach until it stops V 0 so Bernoulli Eq717 leads to gzin 1 2V2 in gzex 0 zex zin 1 2gV2 in 1 2 32174 fts2 62 fts2 056 ft Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7196E A small pump takes in water at 70 F 147 lbfin2 and pumps it to 250 lbfin2 at a flow rate of 200 lbmmin Find the required pump power input Solution CV Pump Assume reversible pump and incompressible flow This leads to the work in Eq715 wp vdP viPe Pi 001605 ft3lbm 250 147 psi 144 in2ft2 778 lbfftBtulbm 07 Btulbm W p in m wp 200 60 lbms 07 Btulbm 233 Btus 33 hp Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7197E An irrigation pump takes water from a river at 50 F 1 atm and pumps it up to an open canal at a 150 ft higher elevation The pipe diameter in and out of the pump is 03 ft and the motor driving the pump is 5 hp Neglect kinetic energies and friction find the maximum possible mass flow rate CV the pump The flow is incompressible and steady flow The pump work is the difference between the flow work in and out and from Bernoullis eq for the pipe that is equal to the potential energy increase sincle pump inlet pressure and pipe outlet pressure are the same wp v P g Z 32174 fts2 150 ft 48261 fts2 48261 25037 Btulbm 019276 Btulbm The horsepower is converted from Table A1 W motor 5 hp 5 2544 Btuh 12 720 btuh 3533 Btus m W motor wp 3533 019276 1833 lbms Comment m AVv V A m v 4m ρ π D2 4 1833 622 π 032 417 fts The power to generated the kinetic energy is Power m 05 V2 1833 05 4172 25037 00064 Btus Recall conversion 1 Btulbm 25 037 ft2s2 from A1 This is insignificant relative to the power needed for the potential energy increase Pump inlet and the pipe exit both have close to atmospheric pressure Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7198E A fireman on a ladder 80 ft above ground should be able to spray water an additional 30 ft up with the hose nozzle of exit diameter 1 in Assume a water pump on the ground and a reversible flow hose nozzle included and find the minimum required power Solution CV pump hose water column total height difference 35 m Continuity Eq43 411 m in m ex ρAVnozzle Energy Eq412 m wp m h V22 gzin m h V22 gz ex Process hin hex Vin Vex 0 zex zin 110 ft ρ 1v 1vf wp gzex zin 32174 110 025 037 0141 Btulbm Recall the conversion 1 Btulbm 25 037 ft2s2 from Table A1 The velocity in the exit nozzle is such that it can rise 30 ft Make that column a CV for which Bernoulli Eq717 is gznoz 1 2V2 noz gzex 0 Vnoz 2gzex znoz 2 32174 30 4394 fts 30 ft 110 ft Assume v vF70F 001605 ft3lbm m π vf D 2 2 Vnoz π4 12144 4394 001605 1492 lbms Wpump mwp 1492 0141 36002544 3 hp Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7199E Saturated R410A at 10 F is pumpedcompressed to a pressure of 300 lbfin2 at the rate of 10 lbms in a reversible adiabatic steady flow process Calculate the power required and the exit temperature for the two cases of inlet state of the R410A a quality of 100 b quality of 0 Solution CV PumpCompressor m 1 lbms R410A a State 1 Table F101 x1 10 Saturated vapor P1 Pg 76926 psia h1 hg 11821 Btulbm s1 sg 02535 Btulbm R Assume Compressor is isentropic s2 s1 02535 Btulbm R h2 13454 Btulbm T2 1305 F Energy Eq413 qc h1 h2 wc qc 0 wcs h1 h2 11821 13454 1633 Btulbm W C m wC 163 Btus 231 hp b State 1 T1 10 F x1 0 Saturated liquid This is a pump P1 76926 psia h1 hf 170 Btulbm v1 vf 001316 ft3lbm Energy Eq413 qp h1 h2 wp qp 0 Assume Pump is isentropic and the liquid is incompressible Eq718 wps v dP v1P2 P1 001316 300 76926 144 42273 lbfftlbm 0543 Btulbm h2 h1 wp 170 0543 17543 Btulbm Assume State 2 is approximately a saturated liquid T2 115 F W P m wP 1 0543 0543 Btus 077 hp 2a 1a P v T s 1a 2a 1b 2b 1b 2b Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7200E Liquid water at ambient conditions 147 lbfin2 75 F enters a pump at the rate of 1 lbms Power input to the pump is 3 Btus Assuming the pump process to be reversible determine the pump exit pressure and temperature Solution CV Pump Steady single inlet and exit flow with no heat transfer Energy Eq413 w hi he W m 31 30 btulbm Using also incompressible media we can use Eq715 wP vdP viPe Pi 001606 ftlbmPe 147 psia from which we can solve for the exit pressure 3 001606Pe 147 144 778 Pe 10239 lbfin 2 W e i Pump W 3 Btus Pi 147 psia Ti 75 F m 1 lbms Energy Eq he hi wP 4309 3 4609 Btulbm Use Table F73 at 1000 psia Te 753 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7201E The underwater bulb nose of a container ship has a velocity relative to the ocean water as 30 fts What is the pressure at the front stagnation point that is 6 ft down from the water surface Solution CV A stream line of flow from the freestream to the wall Eq717 vPePi 1 2 V2 eV2 i gZe Zi 0 P 1 2v V2 i 302 001603 32174 144 2 606 psi Pi Po gHv 14695 6 001603 144 1729 psia Pe Pi P 1729 606 2335 psia This containership is under construction and not loaded The red line is the water line under normal load Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7202E A speed boat has a small hole in the front of the drive with the propeller that sticks down into the water at a water depth of 15 in Assume we have a stagnation point at that hole when the boat is sailing with 30 mih what is the total pressure there Solution CV A stream line of flow from the freestream to the wall Eq717 vPePi 1 2 V2 eV2 i gZe Zi 0 Vi 30 mi h 1467 fth smi 4401 fts P 1 2v V2 i 44012 001603 32174 144 2 1304 psi Pi Po gHv 14695 1512 001603 144 1524 psia Pe Pi P 1524 1304 2828 psia Remark This is fast for a boat Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7203E Helium gas enters a steadyflow expander at 120 lbfin2 500 F and exits at 18 lbfin2 The mass flow rate is 04 lbms and the expansion process can be considered as a reversible polytropic process with exponent n 13 Calculate the power output of the expander Solution Wexp i e Q CV expander reversible polytropic process From Eq628 Te Ti Pe Pi n1 n 960 18 120 03 13 6196 R Table F4 R 386 lbfftlbmR Work evaluated from Eq718 w vdP nR n 1 Te Ti 13 386 03 778 6196 960 7318 Btulbm W mw 04 7318 3600 2544 414 hp P v T s e i n 1 n 13 i e n 1 n k 1667 n 13 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7204E An expansion in a gas turbine can be approximated with a polytropic process with exponent n 125 The inlet air is at 2100 R 120 psia and the exit pressure is 18 psia with a mass flow rate of 2 lbms Find the turbine heat transfer and power output Solution CV Steady state device single inlet and single exit flow Energy Eq413 hi q he w Neglect kinetic potential energies Entropy Eq79 si dqT sgen se Process Eq628 Te Ti Pe Pi n1 n 2100 R 18120 025 125 14369 R so the exit enthalpy is from Table F5 hi 5326 Btulbm he 3430 369 40 3535 3430 3527 Btulbm The process leads to Eq718 for the work term W mw mnR n1 Te Ti 2 125 5334 025 778 14369 2100 4546 Btus Energy equation gives Q mq mhe hi W 23527 5326 4546 3598 4546 948 Btus P v T s e i n 1 n 125 i e n 1 n k 14 n 125 Notice dP 0 so dw 0 ds 0 so dq 0 Notice this process has some heat transfer in during expansion which is unusual The typical process would have n 15 with a heat loss Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Steady Irreversible Processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7205E Analyse the steam turbine described in Problem 4188E Is it possible CV Turbine Steady flow and adiabatic Continuity Eq49 m 1 m 2 m 3 Energy Eq410 m 1h1 m 2h2 m 3h3 W Entropy Eq77 m 1s1 S gen m 2s2 m 3s3 WT 1 2 3 States from Table F72 s1 16398 Btulbm R s2 16516 Btulbm R s3 sf x sfg 0283 095 15089 171 Btulbm R S gen 40 16516 160 1713 200 16398 122 Btus R Since it is positive possible Notice the entropy is increasing through turbine s1 s2 s3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7206E A large condenser in a steam power plant dumps 15 000 Btus at 115 F with an ambient at 77 F What is the entropy generation rate Solution This process transfers heat over a finite temperature difference between the water inside the condenser and the outside ambient cooling water from the sea lake or river or atmospheric air CV The wall that separates the inside 115 F water from the ambient at 77 F Entropy Eq 71 for steady state operation Condensing water Sea water cb 115 F 77 F dS dt 0 T Q S gen Q T115 Q T77 S gen S gen 15 000 5367 15 000 115 4597 Btu s R 185 Btu s R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7207E A compressor in a commercial refrigerator receives R410A at 10 F and x 1 The exit is at 150 psia and 60 F Is this compressor possible Solution CV Compressor steady state single inlet and exit flow For this device we also assume no heat transfer and Zi Z e W C i e cb From Table F91 hi 11629 Btulbm si 02592 BtulbmR From Table F92 he 12458 Btulbm se 02508 BtulbmR Entropy gives sgen se si dqT 02508 02592 dqT The result is negative unless dq is negative it should go out but T T ambient so this compressor is impossible Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7208E R134a at 90 F 125 psia is throttled in a steady flow to a lower pressure so it comes out at 10 F What is the specific entropy generation Solution The process is adiabatic and irreversible The consideration of the energy given in example 65 resulted in a constant h and twophase exit flow Table F101 hi 10534 Btulbm si 02757 BtulbmR compressed liquid State 2 10 F he hi hg so twophase xe he hfhfg 02956 Table F101 se sf xe sfg 02244 02956 01896 02804 BtulbmR We assumed no heat transfer so the entropy equation Eq79 gives sgen se si dqT 02804 02757 0 00047 BtulbmR 1 2 e T s i 125 psia 268 psia h C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7209E Two flowstreams of water one at 100 lbfin2 saturated vapor and the other at 100 lbfin2 1000 F mix adiabatically in a steady flow process to produce a single flow out at 100 lbfin2 600 F Find the total entropy generation for this process Solution Continuity Eq49 m 3 m 1 m 2 Energy Eq410 m 3h3 m 1h1 m 2h2 State properties from Table F72 h1 11878 h2 15321 h3 13293 all in Btulbm s1 16034 s2 19204 s3 17582 all in Btulbm R m 1m 3 h3 h2 h1 h2 0589 Entropy Eq77 m 3s3 m 1s1 m 2s2 S gen S genm 3 s3 m 1m 3 s1 m 2m 3 s2 17582 0589 16034 0411 19204 00245 Btu lbm R 1 2 3 Mixing chamber 2 T s 3 1 100 psia The mixing process generates entropy The two inlet flows could have exchanged energy they have different T through some heat engines and produced work the process failed to do that thus irreversible Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7210E A compressor in a commercial refrigerator receives R410A at 10 F and x 1 The exit is at 300 psia and 160 F Neglect kinetic energies and find the specific entropy generation Solution CV Compressor steady state single inlet and exit flow For this device we also assume no heat transfer q 0 and Z1 Z 2 WC i e cb Entropy Eq79 si dqT sgen se si 0 sgen From Table F91 si 02592 BtulbmR From Table F92 se 02673 BtulbmR Entropy generation becomes sgen se si 02673 02592 00081 BtulbmR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7211E A steam turbine has an inlet of 4 lbms water at 150 psia and 550 F with velocity of 50 fts The exit is at 1 atm 240 F and very low velocity Find the power produced and the rate of entropy generation Solution CV Turbine Steady flow and adiabatic Continuity Eq49 m 1 m 2 Energy Eq410 m 1h1 1 2V2 m 2h2 W Entropy Eq77 m 1s1 S gen m 2s2 W T 1 2 3 States from Table F72 h1 130005 Btulbm s1 16862 BtulbmR h2 116402 Btulbm s2 17764 BtulbmR W m 1h1 1 2V2 h2 4 130005 1 2 502 25037 116402 5461 Btus S gen m 1s2 s1 4 17764 16862 0361 BtusR Recall conversion 1 Btulbm 25 037 ft2s2 from A1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7212E A dual fluid heat exchanger has 10 lbms water enter at 104 F 20 psia and leaving at 50 F 20 psia The other fluid is glycol coming in at 14 F 22 psia and leaves at 50 F 22 psia Find the mass flow rate of glycol and the rate of entropy generation Continuity Eqs Each line has a constant flow rate through it Energy Eq410 m H2O h1 m glycol h3 m H2O h2 m glycol h 4 Entropy Eq77 0 m H2O s1 m glycol s3 m H2O s2 m glycol s4 S gen Process Each line has a constant pressure Table F7 h1 7203 h2 1805 Btulbm s1 01367 s2 00361 BtulbmR We could have used specific heat for the changes Table F3 CP gly 058 BtulbmR so h4 h3 CP gly T4 T3 058 50 14 2088 Btulbm s4 s3 CP gly ln T4T3 058 ln50974737 00425 BtulbmR m glycol m H2O h1 h2 h4 h3 10 7203 1805 2088 2585 lbms S gen m H2O s2 s1 m glycol s4 s3 10 lbms 00361 01367 BtulbmR 2585 lbms 00425 BtulbmR 0093 BtusR CV Heat exchanger steady flow 1 inlet and 1 exit for glycol and water each The two flows exchange energy with no heat transfer tofrom the outside 3 glycol 1 water 4 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7213E A factory generates compressed air from ambient 15 psia 62 F by compression to 150 psia 1080 R after which it cools in a constant pressure cooler to 540 R by heat transfer to the ambient Find the specific entropy generation in the compressor and in the cooler operation Solution CV air compressor q 0 Continuity Eq m 2 m 1 Energy Eq413 0 h1 wc in h2 Entropy Eq 0 s1 s2 sgen comp Table F5 State 1 h1 12479 Btulbm so T1 163151 BtulbmR Table F5 State 2 h2 261099 Btulbm so T2 180868 BtulbmR Table F5 State 3 h3 12918 Btulbm so T3 163979 BtulbmR sgen comp s2 s1 so T2 so T1 R lnP2P1 180868 163151 5334 778 ln1000100 0019 BtulbmR CV cooler w 0 Continuity Eq m 3 m 1 Energy Eq413 0 h2 qout h3 Entropy Eq 0 s2 s3 qoutTamb sgen cool qout h2 h3 261099 12918 131919 Btulbm sgen cool s3 s2 qoutTamb so T3 so T2 qoutTamb 163979 180868 131919 5217 0084 Btulbm 1 3 2 Q cool Compressor W c Compressor section Cooler section Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7214E A mixing chamber receives 10 lbmmin ammonia as saturated liquid at 0 F from one line and ammonia at 100 F 40 lbfin2 from another line through a valve The chamber also receives 340 Btumin energy as heat transferred from a 100F reservoir This should produce saturated ammonia vapor at 0 F in the exit line What is the mass flow rate at state 2 and what is the total entropy generation in the process Solution CV Mixing chamber out to reservoir Continuity Eq49 m 1 m 2 m 3 Energy Eq410 m 1h1 m 2h2 Q m 3h 3 Entropy Eq77 m 1s1 m 2s2 Q Tres S gen m 3s3 1 2 3 MIXING CHAMBER Q 2 P v 3 1 From Table F81 h1 426 Btulbm s1 00967 Btulbm R From Table F82 h2 66433 Btulbm s2 14074 Btulbm R From Table F81 h3 61092 Btulbm s3 13331 Btulbm R From the energy equation m 2 m 1h1 h3 Q h3 h2 10426 61092 340 61092 66433 1001 lbmmin m 3 1101 lbmmin S gen m 3s3 m 1s1 m 2s2 Q Tres 110113331 1000967 100114074 340 55967 437 Btu R min Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7215E A condenser in a power plant receives 10 lbms steam at 130 F quality 90 and rejects the heat to cooling water with an average temperature of 62 F Find the power given to the cooling water in this constant pressure process and the total rate of enropy generation when condenser exit is saturated liquid Solution CV Condenser Steady state with no shaft work term Energy Eq412 m hi Q m he Entropy Eq78 m si Q T S gen m se Properties are from Table F71 hi 980 09 10198 10158 Btulbm he 980 Btulbm si 01817 09 17292 17380 Btulbm R se 01817 Btulbm R Q out Q m hi he 1010158 980 9178 Btus S gen m se si Q outT 1001817 1738 91784597 62 15563 17592 203 BtusR cb Often the cooling media flows inside a long pipe carrying the energy away Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7216E A large supply line has a steady flow of R410A at 175 psia 140 F It is used in three different adiabatic devices shown in Fig P7101 a throttle flow an ideal nozzle and an ideal turbine All the exit flows are at 60 psia Find the exit temperature and specific entropy generation for each device and the exit velocity of the nozzle Inlet state F92 hi 14311 Btulbm si 02804 BtulbmR CV Throttle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi he Zi Ze and Vs are small Entropy Eq79 se si dqT sgen si 0 sgen Exit state he hi 14311 Btulbm Te 11887 F se 03076 kJkgK sgen se si 03076 02804 00272 BtulbmR CV Nozzle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi he V2 e2 Zi Ze Entropy Eq79 se si dqT sgen si 0 0 The isentropic process se si gives from F92 Te 4915 F sgen 0 he 12836 Btulbm The energy equation becomes V2 e2 hi he 14311 12836 1475 Btulbm Ve 2 1475 25037 859 fts Recall conversion 1 Btulbm 25 037 ft2s2 from A1 Turbine Process Reversible and adiabatic same as for nozzle except w Ve 0 Energy Eq413 hi he w Zi Ze Te 4915 F sgen 0 he 12836 Btulbm P v T s e i i e Low V Hi P Low P Hi V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7217E Air at 540 F 60 lbfin2 with a volume flow 40 ft3s runs through an adiabatic turbine with exhaust pressure of 15 lbfin2 Neglect kinetic energies and use constant specific heats Find the lowest and highest possible exit temperature For each case find also the rate of work and the rate of entropy generation Ti 540 F 1000 R vi RTi Pi 5334 100060 144 6174 ft3 lbm m V v i 40 ft3s 6174 ft3 lbm 6479 lbms a lowest exit T this must be reversible for maximum work out Process Reversible and adiabatic constant s from Eq79 Eq623 Te TiPePi k1 k 1000 R 15600286 673 R w 024 1000 673 7848 Btulbm W m w 5085 Btus S gen 0 b Highest exit T for no work out Te T i 1000 R W 0 S gen m se s i m R ln Pe P i 6479 5334 778 ln 1560 0616 BtusR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7218E A large supply line has a steady air flow at 900 R 2 atm It is used in three different adiabatic devices shown in Fig P7101 a throttle flow an ideal nozzle and an ideal turbine All the exit flows are at 1 atm Find the exit temperature and specific entropy generation for each device and the exit velocity of the nozzle CV Throttle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi he Zi Ze and Vs are small Entropy Eq78 se si dqT sgen si 0 sgen Since it is air we have h hT so same h means same Te Ti 900 R sgen se si so Te so Ti R lnPe Pi 0 5334 778 ln12 00475 BtulbmR CV Nozzle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi he V2 e2 Zi Ze Entropy Eq79 se si dqT sgen si 0 0 Use constant specific heat from Table F4 CPo 024 BtulbmR k 14 The isentropic process se si gives Eq623 Te Ti PePi k1 k 900 12 02857 7383 R The energy equation becomes V2 e2 hi he CP Ti Te Ve 2 CP Ti Te 20249007383 25037 1394 fts Recall conversion 1 Btulbm 25 037 ft2s2 from A1 P v T s e i i e Low V Hi P Low P Hi V Turbine Process Reversible and adiabatic constant s from Eq79 Eq623 Te TiPePi k1 k 900 12 02857 7383 R w hi he CPoTi Te 024 900 7383 388 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7219E Repeat the previous problem for the throttle and the nozzle when the inlet air temperature is 4000 R and use the air tables CV Throttle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi he Zi Ze and Vs are small Entropy Eq79 se si dqT sgen si 0 sgen Since it is air we have h hT so same h means same Te Ti 4000 R sgen se si so Te so Ti R lnPe Pi 0 5334 778 ln12 00475 BtulbmR CV Nozzle Steady single inlet and exit flow no work or heat transfer Energy Eq413 hi he V2 e2 Zi Ze Entropy Eq79 se si dqT sgen si 0 0 The isentropic process se si gives Eq619 0 se si so Te so Ti R lnPe Pi so Te so Ti R lnPe Pi 217221 5334778 ln 12 212469 T 3846 R he 9122 Btulbm The energy equation becomes V2 e2 hi he 1087988 9122 17579 Btulbm Ve 2 25 037 17579 2967 fts Recall conversion 1 Btulbm 25 037 ft2s2 from A1 P v T s e i i e Low V Hi P Low P Hi V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7220E A supply of 10 lbms ammonia at 80 lbfin2 80 F is needed Two sources are available one is saturated liquid at 80 F and the other is at 80 lbfin2 260 F Flows from the two sources are fed through valves to an insulated mixing chamber which then produces the desired output state Find the two source mass flow rates and the total rate of entropy generation by this setup Solution CV mixing chamber valve Steady no heat transfer no work Continuity Eq49 m 1 m 2 m 3 Energy Eq410 m 1 h1 m 2h2 m 3h3 Entropy Eq77 m 1 s1 m 2s2 S gen m 3s3 1 2 3 MIXING CHAMBER 2 T s 3 1 State 1 Table F81 h1 13168 Btulbm s1 02741 Btulbm R State 2 Table F82 h2 7485 Btulbm s2 14604 Btulbm R State 3 Table F82 h3 64563 Btulbm s3 12956 Btulbm R As all states are known the energy equation establishes the ratio of mass flow rates and the entropy equation provides the entropy generation m 1h1 m 3 m 1h2 m 3h3 m 1 m 3 h3 h2 h1 h2 10 10287 61682 1668 lbms m 2 m 3 m 1 8332 lbms S gen m 3s3 m 1s1 m 2s2 10 12956 1668 02741 8332 146 0331 BtusR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7221E An initially empty 5 ft3 tank is filled with air from 70 F 15 psia until it is full Assume no heat transfer and find the final mass and entropy generation Solution CV Tank valve out to line No boundaryshaft work m1 0 Q 0 and we recognize that this is a transient problem Continuity Eq415 m2 0 mi Energy Eq416 m2 u2 0 mi h i Entropy Eq712 m2s2 0 misi 0 1S 2 gen State 2 P2 Pi and u2 hi hline h2 RT2 ideal gas To reduce or eliminate guess use h2 hline CPoT2 Tline Energy Eq becomes CPoT2 Tline RT2 0 T2 Tline CPoCPo R Tline CPoCVo k Tline Use F4 CP 024 Btu lbm R k 14 T2 1470 460 742 R m2 P2VRT2 5334 ft lbflbm R 742R 15 psia 5 ft3144 inft2 02729 lbm 1S2 gen m2 s2 si m2 CP lnT2 Tline R lnP2 Pline 02729024 ln14 0 0351 BtuR Air 2 line P v T s 1 2 T T line 2 500 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7222E An empty cannister of volume 005 ft3 is filled with R134a from a line flowing saturated liquid R134a at 40 F The filling is done quickly so it is adiabatic How much mass of R134a is in the cannister How much entropy was generated Solution CV Cannister filling process where 1Q2 0 1W2 0 m1 0 Continuity Eq415 m2 0 min Energy Eq416 m2u2 0 minhline 0 0 u2 h line Entropy Eq713 m2s2 0 minsline 0 1S2 gen Inlet state Table F101 hline 8856 Btulbm sline 0244 Btulbm R State 2 P2 Pline and u2 hline 8856 Btulbm uf x2 u2 uf ufg 8856 8845 7516 0001464 v2 001253 x2 09395 0013905 ft3lbm m2 Vv2 0050013905 3596 lbm s2 0244 x2 01678 024425 Btulbm R 1S2 gen m2s2 sline 3596 024425 0244 00009 BtuR 2 line T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7223E A can of volume 8 ft3 is empty and filled with R410A from a line at 200 psia 100 F The process is adiabatic and stops at P 150 psia Use Table F9 to find the final temperature and the entropy generation Solution CV Cannister filling process where 1Q2 0 1W2 0 m1 0 Continuity Eq415 m2 0 min Energy Eq416 m2u2 0 minhline 0 0 u2 hline Entropy Eq712 m2s2 0 minsline 1S 2 gen 1 Table F92 hline 13184 Btulbm sline 02578 BtulbmR 2 P2 150 psia u2 hline T2 1528 F s2 029084 kJkgK v2 05515 ft3lbm m2 Vv2 14506 lbm 1S2 gen m2s2 minsline m2s2 sline 14506 029084 02578 0479 BtuR 2 line T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7224E Air from a line at 1800 lbfin2 60 F flows into a 20ft3 rigid tank that initially contained air at ambient conditions 147 lbfin2 60 F The process occurs rapidly and is essentially adiabatic The valve is closed when the pressure inside reaches some value P2 The tank eventually cools to room temperature at which time the pressure inside is 750 lbfin2 What is the pressure P2 What is the net entropy change for the overall process CV Tank Mass flows in so this is transient Find the mass first m1 P1VRT1 147 144 20 5334 520 1526 lbm Fill to P2 then cool to T3 60 F P3 750 psia m3 m2 P3VRT3 750 144 20 5334 520 77875 lbm T s 147 psia 750 psia v C 2 1 1800 psia 3 line Cont Eq mi m2 m1 77875 1526 76349 lbm The filling process from 1 to 2 T1 Ti 12 heat transfer 0 so Energy Eq mihi m2u2 m1u 1 miCP0Ti m2CV0T2 m1CV0T1 T2 76349 024 1526 0171 77875 0171 520 7257 R P2 m2RT2V 77875 5334 7257 144 20 1047 lbfin 2 Consider the overall process from 1 to 3 Energy Eq QCV mihi m2u3 m1u1 m2h3 m1h1 P3 P1V But since Ti T3 T1 mihi m2h3 m1h1 QCV P3 P1V 750 14720144778 2722 Btu From Eq713 also Eqs724726 1S2 gen m3s3 m1s1 misi QCVT0 m3s3 si m1s1 si QCVT 0 77875 0 5334 778 ln 750 1800 1526 0 5334 778 ln 147 1800 2722520 9406 BtuR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Device Efficiency Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7225E A steam turbine inlet is at 200 psia 800 F The exit is at 40 psia What is the lowest possible exit temperature Which efficiency does that correspond to We would expect the lowest possible exit temperature when the maximum amount of work is taken out This happens in a reversible process so if we assume it is adiabatic this becomes an isentropic process Exit 40 psia s sin 17659 Btulbm R T 4096 F The efficiency from Eq727 measures the turbine relative to an isentropic turbine so the efficiency will be 100 v P s T i i e s e s 40 psia 200 psia Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7226E A steam turbine inlet is at 200 psia 800 F The exit is at 40 psia What is the highest possible exit temperature Which efficiency does that correspond to The highest possible exit temperature would be if we did not get any work out ie the turbine broke down Now we have a throttle process with constant h assuming we do not have a significant exit velocity Exit 40 psia h hin 142531 Btulbm T 786 F Efficiency η w ws 0 v P s T h C i e i e Remark Since process is irreversible there is no area under curve in Ts diagram that correspond to a q nor is there any area in the Pv diagram corresponding to a shaft work Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7227E A steam turbine inlet is at 200 psia 800 F The exit is at 40 psia 600 F What is the isentropic efficiency from table F72 Inlet hin 142531 Btulbm sin 17659 Btulbm R Exit hex 133343 Btulbm sex 18621 Btulbm R Ideal Exit 40 psia s sin 17659 Btulbm R hs 124105 Btulbm wac hin hex 142531 133343 9188 Btulbm ws hin hs 142531 124105 1843 Btulbm η wac ws 9188 1843 0498 v P s T i e ac i e ac e s e s 40 psia 200 psia Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7228E The exit velocity of a nozzle is 1500 fts If ηnozzle 088 what is the ideal exit velocity The nozzle efficiency is given by Eq 730 and since we have the actual exit velocity we get V2 e s V2 acηnozzle Ve s Vac ηnozzle 1500 fts 088 1599 fts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7229E A small air turbine with an isentropic efficiency of 80 should produce 120 Btulbm of work The inlet temperature is 1800 R and it exhausts to the atmosphere Find the required inlet pressure and the exhaust temperature Solution CV Turbine actual energy Eq413 w hi heac 120 Table F5 hi 449794 Btulbm heac hi 120 329794 Btulbm Te 1349 R CV Ideal turbine Eq727 and energy Eq413 ws wηs 12008 150 hi hes hes 299794 Btulbm From Table F5 Tes 12327 R so Te 184217 Btulbm R Entropy Eq79 si ses adiabatic and reversible To relate the entropy to the pressure use Eq619 inverted and standard entropy from Table F5 PePi exp so Te so Ti R exp184217 194209 778 5334 02328 Pi Pe 02328 14702328 6314 psia If constant heat capacity was used Te Ti wCp 1800 120024 1300 R Tes Ti wsCp 1800 150024 1175 R Eq79 adibatic and reversible gives constant s and relation is Eq623 PePi TeTikk1 Pi 147 1800117535 654 psia P v T s e s i s C i e s e ac e ac P Pe i Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7230E Redo Problem 7198 if the water pump has an isentropic efficiency of 85 hose nozzle included Solution CV pump hose water column total height difference 110 ft Here V is velocity not volume Continuity Eq43 411 m in m ex ρAVnozzle Energy Eq412 m wp m h V22 gzin m h V22 gz ex Process hin hex Vin Vex 0 zex zin 110 ft ρ 1v 1vf wp gzex zin 32174 110 025 037 0141 Btulbm Recall the conversion 1 Btulbm 25 037 ft2s2 from Table A1 The velocity in the exit nozzle is such that it can rise 30 ft Make that column a CV for which Bernoulli Eq717 is gznoz 1 2V2 noz gzex 0 Vnoz 2gzex znoz 2 32174 30 4394 fts 30 ft 110 ft Assume v vF70F 001605 ft3lbm m π vf D 2 2 Vnoz π4 12144 4394 001605 1492 lbms Wpump mwpη 1492 0141 36002544085 35 hp Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7231E Air enters an insulated compressor at ambient conditions 147 lbfin2 70 F at the rate of 01 lbms and exits at 400 F The isentropic efficiency of the compressor is 70 What is the exit pressure How much power is required to drive the compressor Solution CV Compressor P1 T1 Tereal ηs COMP known assume constant CP0 Energy Eq413 for real w CP0Te Ti 024400 70 792 Btulbm Ideal ws w ηs 792 07 554 Btulbm Energy Eq413 for ideal 554 CP0Tes Ti 024Tes 530 Tes 761 R Constant entropy for ideal as in Eq623 Pe PiTesTi k k1 14776153035 521 lbfin2 WREAL mw 01 792 36002544 112 hp P v T s e s i s C i e s e ac e ac P P e i Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7232E A nozzle is required to produce a steady stream of R134a at 790 fts at ambient conditions 15 lbfin2 70 F The isentropic efficiency may be assumed to be 90 What pressure and temperature are required in the line upstream of the nozzle CV Nozzle steady flow and no heat transfer Actual nozzle energy Eq h1 h2 V 2 22 State 2 actual Table F102 h2 180975 Btulbm h1 h2 V 2 22 180975 2 25 037 7902 19344 Btulbm Recall 1 Btulbm 25 037 ft2s2 from Table A1 Ideal nozzle exit h2s h1 KEs 19344 2 25 037 7902 09 17959 Btulbm Recall conversion 1 Btulbm 25 037 ft2s2 from A1 State 2s P2 h2s T2s 6316 F s2s 04481 Btulbm R Entropy Eq ideal nozzle s1 s2s State 1 h1 s1 s2s Double interpolation or use software For 40 psia given h1 then s 04544 Btulbm R T 13447 F For 60 psia given h1 then s 04469 Btulbm R T 13813 F Now a linear interpolation to get P and T for proper s P1 40 20 04481 04544 04469 04544 568 psia T1 13447 13813 1344704481 04544 04469 04544 1375 F T s 1 2 2s s 1 h1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7233E A watercooled air compressor takes air in at 70 F 14 lbfin2 and compresses it to 80 lbfin2 The isothermal efficiency is 88 and the actual compressor has the same heat transfer as the ideal one Find the specific compressor work and the exit temperature Solution Ideal isothermal compressor exit 80 psia 70 F Reversible process dq T ds q Tse si q Tse si Tso Te so T1 R lnPe Pi RT ln Pe Pi 460 70 5334 778 ln 80 14 633 Btulbm As same temperature for the ideal compressor he hi w q 633 Btulbm wac w η 7193 Btulbm qac q Now for the actual compressor energy equation becomes qac hi he ac wac he ac hi qac wac 633 7193 863 Btulbm Cp Te ac Ti Te ac Ti 863024 1059 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7234E Repeat Problem 7199 for a pumpcompressor isentropic efficiency of 70 Solution CV PumpCompressor m 1 lbms R410A a State 1 Table F91 x1 10 Saturated vapor P1 Pg 76926 psia h1 hg 11821 Btulbm s1 sg 02535 Btulbm R Ideal Compressor is isentropic s2 s1 02535 Btulbm R h2 13454 Btulbm T2 1305 F Energy Eq413 qc h1 h2 wc qc 0 wcs h1 h2 11821 13454 1633 Btulbm Now the actual compressor wc AC wcsη 2333 h1 h2 AC h2 AC 13454 2333 15787 T2 217 F W C in m wC 233 Btus 33 hp b State 1 T1 10 F x1 0 Saturated liquid This is a pump P1 76926 psia h1 hf 170 Btulbm v1 vf 001316 ft3lbm Energy Eq413 qp h1 h2 wp qp 0 Ideal pump is isentropic and the liquid is incompressible Eq718 wps v dP v1P2 P1 001316 300 76926 144 42273 lbfftlbm 0543 Btulbm Now the actual pump wc AC wcsη 0776 h1 h2 AC h2 h1 wp 170 0776 17776 Btulbm Assume State 2 is approximately a saturated liquid T2 122 F W P in m wP 1 0776 0776 Btus 099 hp 2a 1a P v T s 1a 2a 1b 2b 1b 2b Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7235E Air at 1 atm 60 F is compressed to 4 atm after which it is expanded through a nozzle back to the atmosphere The compressor and the nozzle both have efficiency of 90 and kinetic energy inout of the compressor can be neglected Find the actual compressor work and its exit temperature and find the actual nozzle exit velocity W 1 3 5 1 2 3 4 5 T s Steady state separate control volumes around compressor and nozzle For ideal compressor we have inlet 1 and exit 2 Adiabatic q 0 Reversible sgen 0 Energy Eq h1 0 wC h2 Entropy Eq s1 0T 0 s 2 Ideal compressor wc h1 h2 s2 s1 The constant s from Eq 623 gives T2 T1 P2P1 k1 k 4597 60 R 4102857 772 R wC h2 h1 CPT2 T1 024 772 5197 6055 Btulbm Actual compressor wcAC wcsηc 673 Btulbm h1 h3 T3 T1 wcACCP 5197 673024 800 R Ideal nozzle s4 s3 so use Eq623 again T4 T3 P4P3 k1 k 800 1402857 5384 R Vs 22 h3 h4 CPT3 T4 024800 5384 6278 Btulbm VAC 2 2 Vs 2 ηNOZ2 6278 09 565 Btulbm VAC 2 565 25 037 1682 fts Remember conversion 1 Btulbm 25 037 ft2s2 from Table A1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7236E A rigid 35 ft3 tank contains water initially at 250 F with 50 liquid and 50 vapor by volume A pressurerelief valve on the top of the tank is set to 150 lbfin2 the tank pressure cannot exceed 150 lbfin2 water will be discharged instead Heat is now transferred to the tank from a 400 F heat source until the tank contains saturated vapor at 150 lbfin2 Calculate the heat transfer to the tank and show that this process does not violate the second law CV Tank and walls out to the source Neglect storage in walls There is flow out and no boundary or shaft work Continuity Eq415 m2 m1 me Energy Eq416 m2 u2 m1u1 mehe 1Q2 Entropy Eq712 m2s2 m1s1 mese dQT 1S 2 gen State 1 250 F Table F71 vf1 0017 vg1 138247 ft3lbm m LIQ V LIQ vf1 05 350017 10294 lbm m VAPV VAP vg1 05 35138247 1266 lbm m 1030 67 lbm x m VAP m LIQ m VAP 0001228 u uf1 x ufg1 21848 0001228 86941 21955 s sf1 x sfg1 03677 0001228 13324 036934 state 2 v2 vg 32214 ft3lbm u2 111031 h2 119377 Btulbm s2 1576 BtulbmR m2 Vv2 10865 lbm From the energy equation we get 1Q2 m2 u2 m1u1 mehe 10865 111031 10306721955 10198119377 1 003 187 Btu 1S2 gen m2 s2 m1s1 mese 1Q2 Tsource 10865 1576 103067 036934 10198 157 1003187860 772 BtuR Updated June 2013 SOLUTION MANUAL CHAPTER 8 Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Fundamentals of Thermodynamics 8th Edition Borgnakke and Sonntag CONTENT CHAPTER 8 SUBSECTION PROB NO InText Concept questions am ConceptStudy Guide Problems 115 Available energy reversible work 1634 Irreversibility 3552 Exergy 5380 Exergy Balance Equation 8198 Device Second Law Efficiency 99121 Review Problems 122139 Problems resolved using Pr and vr functions from table A72 43 69 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful InText Concept Questions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8a Can I have any energy transfer as heat transfer that is 100 available By definition the possible amount of work that can be obtained equals the exergy The maximum is limited to that out of a reversible heat engine if constant T then that is the Carnot heat engine W 1 To T Q So we get a maximum for an infinite high temperature T where we approach an efficiency of one In practice you do not have such a source the closest would be solar radiation and secondly no material could contain matter at very high T so a cycle process can proceed the closest would be a plasma suspended by a magnetic field as in a tokamak 8b Is electrical work 100 available Yes By definition work is 100 exergy 8c A nozzle does not have any actual work involved how should you then interpret the reversible work The purpose of the nozzle is not to make work but the reversible work can be used to generate kinetic energy in addition to what comes out of the actual nozzle The reversible work plus the actual kinetic energy would equal the possible total kinetic energy out of the reversible nozzle The other point is that the reversible nozzle has a heat transfer in if the actual nozzle is irreversible also unusual for a nozzle Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8d If an actual control volume process is reversible what can you say about the work term The work term is the maximum possible If the work is positive it is the maximum possible work out and if the work is negative it is the minimum work possible that must be supplied 8e Can entropy change in a control volume process that is reversible Yes it can A flow in or out at a state with a different entropy than the average entropy inside or a heat transfer will change the entropy All the terms that changes the entropy are transfer terms so there is no net increase generation of entropy 8f Energy can be stored as internal energy potential energy or kinetic energy Are those energy forms all 100 available The internal energy is only partly available a process like an expansion can give out work or if it cools by heat transfer out it is a Q out that is only partly available as work Potential energy like from gravitation mgH or a compressed spring or a charged battery are forms that are close to 100 available with only small losses present Kinetic energy like in a flywheel or motion of a mass can be transferred to work out with losses depending on the mechanical system Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8g We cannot create nor destroy energy can we create or destroy exergy Yes Every process that is irreversible to some degree destroys exergy This destruction is directly proportional to the entropy generation We cannot create exergy at most it can stay constant which is the case for a reversible process 8h In a turbine what is the source of exergy The flow into the turbine is the source it provides the conditions that allows the turbine to give work out If the turbine exit flow has useful exergy we can write the net source as the inlet flow exergy minus the exit flow exergy 8i In a pump what is the source of exergy The shaft work is the input that drives the pump which in turn pushes on the flow to generate a higher pressure exit flow The increase in the flow exergy is the desired output that is expressing the increase in P in terms of exergy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8j In a pump what gains exergy The higher pressure exit flow is the desired output When we express that in terms of exergy it becomes the increase in the flow exergy that is the output that is expressing the increase in P in terms of exergy 8k In a heat engine what is the source of exergy Generally it is the high temperature heat transfer However if the rejected heat transfer at the low temperature has any useful exergy we can also count the difference as the source 8l In a heat pump what is the source of exergy The work input is what drives the heat pump Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8m In Eq839 for the heat engine the source of exergy was written as a heat transfer How does the expression look like if the source is a flow of hot gas being cooled down as it gives energy to the heat engine Look at a heat exchanger that provides such a setup in Fig 83 CV Heat exchanger plus HE Energy Eq612 0 m 1h1 m 1h2 W Q L Entropy Eq72 0 m 1s1 m 1s2 Q LTo S gen Solve for Q L and sustitute into energy Eq Q L To m 1 s1 s2 ToS gen H Q W L Q Ambient HE 1 2 W m 1 h1 h2 To s1 s2 ToS gen m 1 ψ1 ψ2 ToS gen ηHE II m 1 ψ1 ψ2 ηHE II W rev So the HE work is a fraction of the flow exergy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful ConceptStudy Guide Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 81 Why does the reversible CV counterpart to the actual CV have the same storage and flow terms If you want to compare two devices they should be comparable in as many respects as possible to be useful Specifically when we want to find the possible work output all the aspects that can generate work are important and must be considered The flows in and out and the storage effect are closely tied to the possible work output and thus should be the same for the two control volumes Obviously one control volume that receives a high exergy flow or depletes its stored exergy can give out more work than a control volume that does not have such effects Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 82 Can one of the heat transfers in Eq856 be to or from the ambient Yes it can then the reversible heat transfer we find from the ambient to balance the entropy is in addition to the one already listed In many practical devices there is an actual heat transfer loss to the surroundings because of elevated temperatures inside the device Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 83 All the energy in the ocean is that available No Since the ocean is at the ambient T it is the ambient it is not possible to extract any work from it You can extract wave energy wind generated kinetic energy or run turbines from the tide flow of water moon generated kinetic energy However since the ocean temperature is not uniform there are a few locations where cold and warmer water flows close to each other like at different depths In that case a heat engine can operate due to the temperature difference Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 84 Does a reversible process change the exergy if there is no work involved Yes There can be heat transfer involved and that has an exergy associated with it which then equals the change of exergy of the substance A flow can enter and exit a device with different levels of exergy What is unchanged is the total CV plus surroundings exergy that remains constant but it may be redistributed Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 85 Is the reversible work between two states the same as ideal work for the device No It depends on the definition of ideal work The ideal device does not necessarily have the same exit state as the actual device An ideal turbine is approximated as a reversible adiabatic device so the ideal work is the isentropic work The reversible work is between the inlet state and the actual exit state that do not necessarily have the same entropy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 86 When is the reversible work the same as the isentropic work That happens when the inlet and exit states or beginning and end states have the same entropy For a reversible adiabatic process in a CV which then is isentropic the two work terms become the same Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 87 If I heat some cold liquid water to To do I increase its exergy No You decrease its exergy by bringing it closer to To where it has zero exergy if we neglect pressure effects Any substance at a T different from ambient higher or lower has a positive exergy since you can run a heat engine using the two temperatures as the hot and cold reservoir respectively For a T lower than the ambient it means that the ambient is the hot side of the heat engine Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 88 Are reversible work and exergy connected Yes They are very similar Reversible work is usually defined as the reversible work that can be obtained between two states inletexit or beginning to end Exergy is a property of a given state and defined as the reversible work that can be obtained by changing the state of the substance from the given state to the dead state ambient The only difference is the work term exchanged with the ambient if the control volume changes its volume W Po V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 89 Consider exergy associated with a flow The total exergy is based on the thermodynamic state the kinetic and potential energies Can they all be negative No By virtue of its definition kinetic energy can only be positive The potential energy is measured from a reference elevation standard sea level or a local elevation so it can be negative The thermodynamic state can only have a positive exergy the smallest it can be is zero if it is the ambient dead state Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 810 Verify that Eq829 reduces to Eq814 for a steady state process The expression for the reversible work is W rev Φ q m i ψ i m e ψe Φ CV Po V 1029 Definition of steady state is no storage effect so V 0 Φ CV 0 the last two terms in Eq829 drop out Eq814 is also written for a single flow so the summations are each a single term m i ψ i m e ψe m ψ i m ψe m ψ i ψe The heat transfer term is from Eq821 and with single flow Φ q 1 To Tj Q j m 1 To Tj qj 1021 Now we have W rev m 1 To Tj qj m ψ i ψe divide by the mass flow rate to get wrev 1 To Tj qj ψ i ψe The last difference in exergies are from Eq823 ψ i ψe htot i Tosi htot e Tose which when substituted gives Eq814 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 811 What is the second law efficiency of a Carnot Heat engine The Carnot Heat engine is by definition reversible and thus it delivers the maximum amount of work possible It has a second law eficiency of 100 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 812 What is the second law efficiency of a reversible heat engine Since the reversible heat engine has no entropy generation it produces the maximum work possible and the actual work is the reversible work so it has a second law efficiency of 100 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 813 For a nozzle what is the output and input source expressed in exergies For the nozzle a high pressure low velocity inlet flow generates a higher velocity in the exit flow at the expense of the pressure The desired output is the higher velocity expressed as kinetic energy which is part of exergy The source is the inlet high pressure expressed in flow exergy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 814 Is the exergy equation independent of the energy and entropy equations No The exergy equation is derived from the other balance equations by defining the exergy from the state properties and the reference dead state Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 815 Use the exergy balance equation to find the efficiency of a steady state Carnot heat engine operating between two fixed temperature reservoirs The exergy balance equation Eq838 for this case looks like 0 1 To TH Q H 1 To TL Q L W 0 0 0 0 Steady state LHS 0 and dVdt 0 no mass flow terms Carnot cycle so reversible and the destruction is then zero From the energy equation we have 0 Q H Q L W which we can subtract from the exergy balance equation to get 0 To TH Q H To TL Q L Solve for one heat transfer in terms of the other Q L TL TH Q H The work from the energy equation is W Q H Q L Q H 1 TL TH from which we can read the Carnot cycle efficiency as we found in Chapter 7 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Exergy Reversible work Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 816 A control mass gives out 10 kJ of energy in the form of a Electrical work from a battery b Mechanical work from a spring c Heat transfer at 500C Find the change in exergy of the control mass for each of the three cases Solution a Work is exergy Φ Wel 10 kJ b Work is exergy Φ Wspring 10 kJ c Give the heat transfer to a Carnot heat engine and W is exergy Φ 1 T0 TH Qout 1 29815 77315 10 614 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 817 A fraction of some power to a motor 1 2 kW is turned into heat transfer at 500 K 2 and then it dissipates in the ambient at 300 K 3 Give the rates of exergy along the process 123 Solution The exergy of an amount of heat transfer equals the possible work that can be extracted This is the work out of a Carnot heat engine with heat transfer to the ambient as the other reservoir The result is from Chapter 5 as also shown in Eq 81 and Eq 838 1 Φ W 2 kW 2 Φ W rev HE 1 To T Q 1 300 500 2 kW 08 kW 3 Φ W rev HE 1 To To Q 1 300 300 2 kW 0 kW As the energy transforms from wotrk to heat transfer it loses exergy and as the heat transfer goes to lower and lower T its exergy value drops ending as zero at the ambient T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 818 A refrigerator should remove 15 kW from the cold space at 10oC while it rejects heat to the kitchen at 25oC Find the reversible work The reversible work is related to the Carnot cycle work as the the two reservoirs are at constant temperatures EAW AE carnot E TL TH TL A 26315 25 10E A 752 W β β Q L Q L In general we have defined the reversible work with the standard sign definition AW revE A AQ AE L E β A 15 752E A kW 02 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 819 A heat engine receives 5 kW at 800 K and 10 kW at 1000 K rejecting energy by heat transfer at 600 K Assume it is reversible and find the power output How much power could be produced if it could reject energy at To 298 K Solution CV The heat engine this is in steady state Energy Eq 0 AQ AE 1 E AQ AE 2 E AQ AE L E AW E Entropy Eq 0 A Q AE 1 T1 E A Q AE 2 T2 E A Q AE L TL E 0 Q W L Q 1 HE Q 2 Now solve for AQ AE L E from the entropy equation AQ AE L E TL T1 AQ AE 1 E TL T2 AQ AE 2 E A600 800E A 5 A 600 1000E A 10 975 kW Substitute into the energy equation and solve for the work term AW E A AQ AE 1 E AQ AE 2 E AQ AE L E 5 10 975 525 kW For a low temperature of 298 K we can get AQ AE L2 E A298 600E A AQ AE L E 4843 kW AW E A AQ AE 1 E AQ AE 2 E AQ AE L E 5 10 4843 1016 kW Remark Notice the large increase in the power output Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 820 A household refrigerator has a freezer at TAFE A and a cold space at TACE A from which energy is removed and rejected to the ambient at TAAE A as shown in Fig P820 Assume that the rate of heat transfer from the cold space AQE ACE A is the same as from the freezer AQ E AFE A find an expression for the minimum power into the heat pump Evaluate this power when TAAE A 20C TACE A 5C TAFE A 10C and AQ E AFE A 3 kW Solution CV Refrigerator heat pump Steady no external flows except heat transfer Energy Eq AQ E AFE A AQ E AcE A W in AQ E AAE A amount rejected to ambient Q W A Q C REF Q F Reversible gives maximum work minimum work in as from Eq 81 or 811 on rate form standard sign notation for work AWE ArevE A AQ E AFE A A 1 TA ETF E A AQ E AcE A A 1 TA ETC E A AQ E AAE A A 1 TA ETA E 3 A 1 29315 26315 E A 3 A 1 29315 27815 E A 0 0504 kW negative so work goes in Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 821 An air compressor takes air in at the state of the surroundings 100 kPa 300 K The air exits at 400 kPa 200C using 100 kW of power Determine the minimum compressor work input CV Compressor Steady flow minimum work in is reversible work Energy Eq 0 hA1E A hA2E A wc wc hA2E A hA1E A 4758 30047 17533 kJkg AmE A AWE A wc 100 kW17533 057 kgs ψA1E A 0 at ambient conditions Get the properties from the air table A71 and correct standard entropy for the pressure sA0E A sA2E A sA ET0 EA sA ET2 EA R lnPA0E APA2E A 686926 73303 0287 ln100400 006317 kJkg K ψA2E A hA2E A hA0E A TA0E AsA0E A sA2E A 47579 300473 300 006317 156365 kJkg AWE AREVE A AmE AψA2E A ψA1E A 057 kgs 156365 0 kJkg 891 kW AWE Ac min Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 822 The compressor in a refrigerator takes refrigerant R134a in at 100 kPa 20C and compresses it to 1 MPa 60C With the room at 20C find the minimum compressor work Solution CV Compressor out to ambient Minimum work in is the reversible work Steady flow 1 inlet and 2 exit W C 1 2 Energy Eq wAcE A hA1E A hA2E A qArevE Entropy Eq sA2E A sA1E A AdqTEA sAgenE A sA1E A qArevE ATAoE A 0 qArevE A TAoE AsA2E A sA1E A wAc minE A hA1E A hA2E A TAoE AsA2E A sA1E A 38722 44189 29315 17818 17665 5467 44852 5018 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 823 Calculate the reversible work out of the twostage turbine shown in Problem 486 assuming the ambient is at 25C Compare this to the actual work which was found to be 1808 MW CV Turbine Steady flow 1 inlet and 2 exits Use Eq 811 for steady flow with q 0 for adiabatic turbine Supply state 1 20 kgs at 10 MPa 500AE AC Process steam 2 5 kgs 05 MPa 155AE AC Exit state 3 20 kPa x 09 Table B13 hA1E A 33737 hA2E A 27559 kJkg sA1E A 65966 sA2E A 68382 kJkg K W T 1 2 3 HP LP Table B12 hA3E A 2514 09 23583 23739 kJkg sA3E A 08319 09 70766 72009 kJkg K AWE ArevE A AmE A1E AhA1E A AmE A2E AhA2E A AmE A3E AhA3E A TA0E AAmE A1E AsA1E A AmE A2E AsA2E A AmE A3E AsA3E A 20 33737 5 27559 15 23739 29815 20 65966 5 68382 15 72009 2114 MW AWE AacE A AQE ArevE A 18084 kW 30627 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 824 Find the specific reversible work for a steam turbine with inlet 4 MPa 500C and an actual exit state of 100 kPa x 10 with a 25C ambient Solution Steam turbine TAoE A 25C 29815 K Inlet state Table B13 hAiE A 34452 kJkg sAiE A 7090 kJkg K Exit state Table B12 hAeE A 26755 kJkg sAeE A 73593 kJkg K From Eq814 asuming it to be adiabatic wArevE A hAiE A TAoE AsAiE A hAeE A TAoE AsAeE A hAiE A hAeE A TAoE AsAeE A sAiE A 34452 26755 298273593 70900 7697 803 8500 kJkg P v i i T s e e W T i e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 825 A steam turbine receives steam at 6 MPa 800C It has a heat loss of 497 kJkg and an isentropic efficiency of 90 For an exit pressure of 15 kPa and surroundings at 20C find the actual work and the reversible work between the inlet and the exit CV Reversible adiabatic turbine isentropic wATE A hAiE A hAesE A sAesE A sAiE A 76566 kJkg K hAiE A 41327 kJkg xAesE A 76566 0754872536 09515 hAesE A 22591 09515237314 24839 kJkg wATsE A 41327 24839 164879 kJkg CV Actual turbine wATacE A ηwATsE A 148391 kJkg hAiE A hAeacE A qAlossE A hAeacE A hAiE A qAlossE A wATacE A 41327 497 148391 25991 kJkg Actual exit state P h saturated vapor sAeacE A 80085 kJkg K CV Reversible process work from Eq814 qARE A TA0E AsAeacE A sAiE A 29315 80085 76566 10316 AkJ kgE wARE A hAiE A hAeacE A qARE A 41327 25991 10316 16368 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 826 A compressor in a refrigerator receives R410A at 150 kPa 40AoE AC and it brings it up to 600 kPa using an actual specific work of 5865 kJkg in an adiabatic compression Find the specific reversible work Energy Eq 0 hA1E A hA2E A w State 1 B42 hA1E A 26399 kJkg sA1E A 11489 kJkg K hA2E A hA1E A w 26399 5865 32264 kJkg State 2 B42 P hA2E A sA2E A 12152 kJkg K wArevE A ψ A1E A ψ A2E A hA1E A hA2E A TA0E A sA1E A sA2E A 5865 29815 11489 12152 5865 19767 389 kJkg P s T 1 2 s 2 ac 2 ac 2 s 1 40 40 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 827 Air flows through a constant pressure heating device shown in Fig P827 It is heated up in a reversible process with a work input of 200 kJkg air flowing The device exchanges heat with the ambient at 300 K The air enters at 300 K 400 kPa Assuming constant specific heat develop an expression for the exit temperature and solve for it by iterations CV Total out to T0 Energy Eq h1 qrev 0 wrev h2 Entropy Eq s1 qrev 0 T0 s2 qrev 0 T0s2 s1 h2 h1 T0s2 s1 wrev same as from Eq 814 Constant Cp gives CpT2 T1 T0Cp ln T2T1 200 The energy equation becomes T2 T0 ln T2 T1 T1 200 Cp T1 300 K Cp 1004 kJkg K T0 300 K T2 300 ln 300 T2 300 200 1004 4993 K Now trial and error on T2 At 600 K LHS 392 too low At 800 K LHS 50575 Linear interpolation gives T2 790 K LHS 4995 OK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 828 An adiabatic and reversible air compressor takes air in at 100 kPa 310 K The air exits at 600 kPa at the rate of 04 kgs Determine the minimum compressor work input and repeat for an inlet at 295 K instead Why is the work less for a lower inlet T Compressor Reversible adiabatic constant s Inlet at 310 K T2S T1 P2 P1 k1 k 310 600 100 02857 5172 K The reversible process ensures the minimum work input W m wS m CP0 T2S T1 04 kgs 1004 kJkgK 5172 310 K 832 kW Inlet at 295 K T2S T1 P2 P1 k1 k 295 600 100 02857 4922 K W m wS m CP0 T2S T1 04 1004 4922 295 04 kgs 1004 kJkgK 4922 295 K 792 kW The work term is reduced due to the reduced specific volume v1 recall Eq 715 for the indicated shaft work Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 829 An air flow of 5 kgmin at 1500 K 125 kPa goes through a constant pressure heat exchanger giving energy to a heat engine shown in Figure P829 The air exits at 500 K and the ambient is at 298 K 100 kPa Find the rate of heat transfer delivered to the engine and the power the engine can produce Solution CV Heat exchanger Continuity eq m 1 m 2 Energy Eq412 m 1h1 m 1h2 Q H Table A71 h1 16358 kJkg h2 50336 kJkg s1 861209 kJkg K s2 738692 kJkg K H Q W L Q Ambient HE 1 2 Q H m h1 h2 5 60 kg s 16358 50336 kJ kg 9437 kW Notice TH is not constant so we do not know the heat engine efficiency CV Total system for which we will write the second law Entropy Equation 78 m s1 S gen m s2 Q LTo Process Assume reversible S gen 0 and P C for air Q L To m s1 s2 298 K 5 60 kg s 861209 738692 kJ kg K 30425 kW Energy equation for the heat engine gives the work as W Q H Q L 9437 30425 639 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 830 Water at 800oC 15 MPa is flowing through a heat exchanger giving off energy to come out as saturated liquid water at 15 MPa in a steady flow process Find the specific heat transfer and the specific flowexergy the water has delivered Let us fix the states first from Table B12 and B13 State 1 in h 409243 kJkg s 72040 kJkg K State 2 ex h 161045 kJkg s 36847 T 34224oC 6154 K CV Heat exchanger no work Energy Eq qout hin hex 409243 161045 2482 kJkg Flow exergy ψ ψin ψex h Tosin h Tosex qout Tosin sex 2482 2981572040 36847 14327 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 831 A rock bed consists of 6000 kg granite and is at 70C A small house with lumped mass of 12000 kg wood and 1000 kg iron is at 15C They are now brought to a uniform final temperature with no external heat transfer by connecting the house and rock bed through some heat engines If the process is reversible find the final temperature and the work done in the process Solution Take CV Total rockbed and heat engine Energy Eq mrocku2 u1 mwoodu2 u1 mFeu2 u1 1W2 Entropy Eq mrocks2 s1 mwoods2 s1 mFes2 s1 0 mCrockln T1 T2 mCwoodln T2 T1 mCFeln T1 T2 0 6000 089 ln T234315 12000 126 ln T228815 1000 046 ln T228815 0 T2 3013 K Now from the energy equation 1W2 6000 kg 089 kJkgK 3013 34315 K 12000 126 1000 046 kJK 3013 28815 K 1W2 18 602 kJ W Q Q H L HE H O U S E cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 832 A constant pressure pistoncylinder has 1 kg of saturated liquid water at 100 kPa A rigid tank contains air at 1200 K 1000 kPa They are now thermally connected by a reversible heat engine cooling the air tank and boiling the water to saturated vapor Find the required amount of air and the work out of the heat engine CV Tank pistoncyl and the heat engine The minimum amount of air is when T2 of air equals water T 9962 C State 1 initial air 2 final air 3 initial water 4 final water For this CV only WHE and Wpist cross the control surface no heat transfer The entropy equation Eq63 and Eq637 for a reversible process becomes S2 S1tot mairs2 s1 mH2Os4 s3 0 0 mH2Os4 s3 mH2O sfg 1 kg 60568 kJkgK 60568 kJK s2 s1 Cv ln T1 T2 R lnv1 v2 Cv ln T2 T1 0717 ln 37277 1200 0838256 kJkgK Now solve for the air mass from entropy equation mair mH2Os4 s3 s2 s1 60568 0838256 72255 kg Let us find the two heat transfers Energy Eq water mH2Ou4 u3 QH2O WH2O WH2O Pmv4 v3 QH2O mH2Ou4 u3 WH2O mH2Oh4 h3 mH2O hfg 225802 kJ Energy Eq air mairu2 u1 Qair 0 mair Cv T2 T1 Qair 72255 kg 0717 1200 37277 42856 kJ Now the work out of the heat engine is WHE Qair QH2O 42856 225802 20276 kJ AIR water HE W HE Q air Q H2O Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 833 A pistoncylinder has forces on the piston so it keeps constant pressure It contains 2 kg of ammonia at 1 MPa 40C and is now heated to 100C by a reversible heat engine that receives heat from a 200C source Find the work out of the heat engine CV Ammonia plus heat engine Energy mamu2 u1 1Q2200 WHE 1W 2pist Entropy mams2 s1 1Q2Tres 0 1Q2 mams2 s1T res Process P const 1W2 Pv2 v1mam Substitute the piston work term and heat transfer into the energy equation WHE mams2 s1Tres mamh2 h1 H Q W L Q HE 200 C o NH 3 cb Table B22 h1 15085 kJkg s1 51778 kJkg K h2 16643 kJkg s2 56342 kJkg K WHE 2 56342 5177847315 16643 15085 1203 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 834 A basement is flooded with 16 m3 of water at 15oC It is pumped out with a small pump driven by a 075 kW electric motor The hose can reach 8 m vertically up and to ensure the water can flow over the edge of a dike it should have a velocity of 20 ms at that point generated by a nozzle see Fig P834 Find the maximum flow rate you can get and how fast the basement can be emptied CV Pump plus hose and nozzle single steady state flow For maximum flow rate assume reversible process so from Eq814 wrev To s2 s1 h2 tot h1 tot 0 Since we have no heat transfer and reversible s is constant and with a liquid flow T is also constant so h2 h1 We could also have used Bernoulli equation wrev 1 2V 2 2 gH 0 1 2 202 981 8 2785 Jkg m W wrev 075 kW 02785 kJkg 2693 kgs m Vv 16 m3 0001 m3kg 16 000 kg Δt m m 16 000 kg 2693 kgs 5941 sec 99 min Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Irreversibility Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 835 A 20oC room is heated with a 2000 W electric baseboard heater What is the rate of irreversibility Let us not consider any storage in the room or any heat loss but just the effect of the heating process Electric power comes in to the heater and leaves as heat transfer to the room air CV heater Entropy Eq 0 Q Tair S gen S gen Q Tair Irreversibility I To S gen 29315 K 2000 W 29315 K 2000 W Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 836 A refrigerator removes 15 kW from the cold space at 10oC using 750 W of power input while it rejects heat to the kitchen at 25oC Find the rate of irreversibility CV Refrigerator Energy Eq 0 Q L W Q H Q H Q L W Entropy Eq 0 Q L TL Q H TH S gen S gen Q H TH Q L TL Q H Q L W 15 kW 075 kW 225 kW S gen 2250 W 29815 K 1500 W 26315 K 18464 WK Irreversibility I To S gen 29815 K 18464 WK 550 W Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 837 Calculate the irreversibility for the condenser in Problem 7100 assuming an ambient temperature at 17C Solution CV Condenser Steady state with no shaft work term Energy Equation 412 m hi Q m he Entropy Equation 78 m si Q T S gen m se Properties are from Table B12 hi 22591 09 237314 236174 kJkg he 22591 kJkg si 07548 09 72536 7283 kJkg K se 07548 kJkg K From the energy equation Q out Q m hi he 5236174 22591 10679 kW From the entropy equation S gen m se si Q outT 507548 7283 1067927315 17 32641 36805 4164 kWK From Eq813 I To S gen 29015 4164 1208 kW cb Often the cooling media flows inside a long pipe carrying the energy away Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 838 A throttle process is an irreversible process Assume an air flow at 1000 kPa 400 K runs through a valve out to ambient 100 kPa Find the reversible work and irreversibility assuming an ambient temperature at 25C Solution CV Throttle Steady state adiabatic q 0 and no shaft work w 0 Energy Eq413 he hi so temperature is constant se si so Te so Ti R lnPePi 0287 ln 1000 100 06608 kJkg K The reversible work is the difference in exergy expressed in Eq611 814 and 829 per unit mass flow wrev ψi ψe hi Tosi he Tose hi he Tose si 0 29815 06608 197 kJkg i wrev w 197 0 197 kJkg e P v i h C T e T s i h C P Process is in the ideal gas region Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 839 A compressor in a refrigerator receives R410A at 150 kPa 40oC and it brings it up to 600 kPa 40oC in an adiabatic compression Find the specific work reversible work entropy generation and irreversibility States 1 B42 h1 26399 kJkg s1 11489 kJkg K 2 B42 h2 32264 kJkg s2 12152 kJkg K CV Compressor steady state q 0 no KE PE energies Energy Eq w h1 h2 26399 32264 5865 kJkg Entropy Eq 0 s1 s2 sgen sgen s2 s1 12152 11489 00663 kJkgK wrev ψ 1 ψ 2 h1 h2 T0 s1 s2 5865 29815 11489 12152 5865 19767 389 kJkg Irreversibility i To sgen 29815 K 00663 kJkgK 1977 kJkg P s 1 2 s 2 ac 2 ac 2 s 1 40 40 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 840 A constant pressure pistoncylinder contains 2 kg of water at 5 MPa and 100oC Heat is added from a reservoir at 600oC to the water until it reaches 600oC We want to find the total irreversibility in the process Solution CV Piston cylinder out to the reservoir incl the walls Energy Eq mu2 u1 1Q2 1W 2 Entropy Eq ms2 s1 1Q2Tres 1S2 gen State 1 h1 42271 kJkg s1 1303 kJkg K State 2 h2 366647 kJkg s2 72588 kJkg K Process P C 1W2 PV2 V1 1 Q 2 H O 2 600 C o From the energy equation we get 1Q2 mu2 u1 1W2 mh2 h1 2366647 42271 64875 kJ From the entropy equation we get 1S2 gen ms2 s1 1Q2 Tres 272588 1303 64875 273 600 44816 kJ K Now the irreversibility is from Eq 818 1I2 m 1i2 To 1S2 gen 29815 K 44816 kJ K 1336 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 841 A constant flow of steel parts 2 kgs at 20oC goes into a furnace where they are heat treated to 900oC by a source at an average 1250 K Find the reversible work and the irreversibility in this process Energy Eq 0 m hi he Q Entropy Eq 0 m si se Q Ts S gen Q m he hi m CP Te Ti 2 kgs 046 kJkgK 880 K 8096 kW S gen m se si Q Ts m CP lnTeTi Q Ts 2 kgs 046 kJkgK ln1173 293 8096 kW 1250 K 06285 kWK I T0S gen 293 K 06285 kWK 18415 kW The reversible work is equal to the irreversibility plus the actual work Eq812 W rev I W ac I 18415 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 842 Calculate the reversible work and irreversibility for the process described in Problem 3146 assuming that the heat transfer is with the surroundings at 20C Solution P v 2 1 Linear spring gives 1W2 PdV 1 2P1 P2V2 V1 1Q2 mu2 u1 1W 2 Equation of state PV mRT State 1 V1 mRT1P1 2 01889 67315 500 05087 m 3 State 2 V2 mRT2P2 2 01889 31315 300 03944 m 3 1W2 1 2500 30003944 05087 4572 kJ From Figure 326 CpTavg 525 R 099 Cv 0803 Cp R For comparison the value from Table A5 at 300 K is Cv 0653 kJkg K 1Q2 mCvT2 T1 1W2 2 080340 400 4572 6239 kJ 1Wrev 2 ToS2 S1 U2 U1 1Q2 1 ToTH Toms2 s1 1Wac 2 1Q2 ToTo TomCP lnT2 T1 R lnP2 P1 1Wac 2 1Q 2 29315 2 099 ln313673 01889 ln300500 4572 6239 3878 4572 6239 1904 kJ 1I2 1Wrev 2 1Wac 2 1904 4572 2361 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 843 An air compressor receives atmospheric air at T0 17C 100 kPa and compresses it up to 1400 kPa The compressor has an isentropic efficiency of 88 and it loses energy by heat transfer to the atmosphere as 10 of the isentropic work Find the actual exit temperature and the reversible work CV Compressor Isentropic wcins hes hi ses si From table A71 and entropy equation we get s o Te s s o Ti R ln PePi 683521 0287 ln14 759262 kJkgK Back interpolate in Table A7 hes 61723 kJkg wcins 61723 29043 3268 kJkg Actual wcinac wcinsηc 37136 kJkg qloss 3268 kJkg wcinac hi heac q loss heac 29043 37136 3268 6291 kJkg Teac 621 K Reversible wrev hi heac T0seac si 29043 6291 29015 76120 68357 33867 22538 1133 kJkg Since qloss is also to the atmosphere it is the net q exchanged with the ambient that explains the change in s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 844 Two flows of air both at 200 kPa of equal flow rates mix in an insulated mixing chamber One flow is 1 kgs at 1500 K and the other is 2 kgs at 300 K Find the irreversibility in the process per kilogram of air flowing out CV Mixing chamber Continuity Eq m 1 m 2 m 3 Energy Eq m 1h1 m 2h2 m 3h3 Entropy Eq m 1s1 m 2s2 S gen m 3s3 Properties from Table A7 h3 h1 2h23 16358 2 300473 74558 kJkg s T3 77848 kJkg K linear interpolation From the entropy equation same P so no correction for that S genm 3 s3 s1 2s23 77848 861208 2 6869263 03346 kJkg K i Im 3 T0 S genm 3 29815 03346 9976 kJkg 1 2 3 MIXING CHAMBER 2 T s 3 1 200 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 845 Fresh water can be produced from saltwater by evaporation and subsequent condensation An example is shown in Fig P845 where 150kgs saltwater state 1 comes from the condenser in a large power plant The water is throttled to the saturated pressure in the flash evaporator and the vapor state 2 is then condensed by cooling with sea water As the evaporation takes place below atmospheric pressure pumps must bring the liquid water flows back up to P0 Assume that the saltwater has the same properties as pure water the ambient is at 20C and that there are no external heat transfers With the states as shown in the table below find the irreversibility in the throttling valve and in the condenser State 1 2 5 7 8 T C 30 25 23 17 20 h kJkg 12577 25472 965 7137 8396 s kJkg K 04369 8558 03392 02535 02966 CV Valve P2 PsatT2 T3 3169 kPa Continuity Eq m 1 mex m 2 m 3 Energy Eq h1 he Entropy Eq s1 sgen se he h1 xe 12577 1048724423 0008558 se 03673 0008558 81905 04374 kJkg K m 2 1 xem 1 148716 kgs sgen se s1 04374 04369 0000494 kJkg K I m T0sgen 150 29315 0000494 2172 kW CV Condenser Energy Eq m 2h2 m 7h7 m 2h5 m 7h8 m 7 m 2 h2 h5h8 h7 148716 25472 965 8396 7137 28 948 kg s Entropy Eq m 2s2 m 7s7 S gen m 2s5 m 7s 8 I T0S gen T0 7s8 s7 m 2s5 s2 m 2931514871603392 8558 2894802966 02535 29315 25392 7444 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 846 A rock bed consists of 6000 kg granite and is at 70C A small house with lumped mass of 12000 kg wood and 1000 kg iron is at 15C They are now brought to a uniform final temperature by circulating water between the rock bed and the house Find the final temperature and the irreversibility of the process assuming an ambient at 15C CV Total Rockbed and house No work no Q irreversible process Energy Eq mCrockT2 70 mCwood mCFeT2 15 0 T2 290C 3022 K Entropy Eq S2 S1 mis2 s1i 0 S gen Sgen mis2 s1i 5340 ln 3022 34315 15580 ln 3022 28815 6313 kJK 1I2 T01S2gen 28815 6313 18 191 kJ Q H O U S E cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 847 A computer CPU chip consists of 50 g silicon 20 g copper and 50g polyvinyl chloride plastic It now heats from ambient 25oC to 70oC in an adiabatic process as the computer is turned on Find the amount of irreversibility CV CPU chip The process has electrical work input and no heat transfer Entropy Eq S2 S1 mis2 s1i dQT 1S2 gen 1S2 gen Irreversibility 1I2 T0 1S2 gen For the solid masses we will use the specific heats Table A3 and they all have the same temperature so mis2 s1i miCi lnT2 T1i ln T2T1 miCi miCi 005 07 002 042 005 096 00914 kJK 1S2 gen S2 S1 00914 ln 34315 28815 0016 kJK 1I2 T0 1S2 gen 29815 K 0016 kJK 477 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 848 A car airconditioning unit has a 05kg aluminum storage cylinder that is sealed with a valve and it contains 2 L of refrigerant R134a at 500 kPa and both are at room temperature 20C It is now installed in a car sitting outside where the whole system cools down to ambient temperature at 10C What is the irreversibility of this process CV Aluminum and R134a Energy Eq mAlu2 u1Al mRu2 u1R 1Q2 1W2 1W2 0 Entropy Eq mALs2 s1Al mRs2 s1R 1Q2T0 1S2 gen Table A3 for specific heat CvAl CpAl 09 kJkg K u2 u1Al CvAlT2 T1 0910 20 27 kJkg s2 s1Al CpAl lnT2T1 09 ln2631529315 009716 kJkg K Table B52 v1 004226 m3kg u1 3905 kJkg s1 17342 kJkg K mR134a Vv1 00473 kg v2 v1 004226 T2 x2 004226 0000755009845 04216 u2 18657 042161857 2649 kJkg s2 09507 0421607812 12801 kJkg K 1Q2 05 27 004732649 3905 1944 kJ 1S2 gen 05 009716 0047312801 17342 1944 26315 0003815 kJK 1I2 T0 1S2 gen 26315 0003815 10 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 849 R134a is filled into an insulated 02 m3 initially empty container from a line at 500 kPa saturated vapor until the flow stops by itself Find the final mass and temperature in the container and the total irreversibility in the process Solution CV Cannister filling process where 1Q2 0 1W2 0 m1 0 Continuity Eq420 m2 0 min Energy Eq421 m2u2 0 minhline 0 0 u2 hline Entropy Eq713 m2s2 0 minsline 0 1S2 gen Inlet state Table B51 hline 40745 kJkg sline 17198 kJkg K State 2 P2 Pline and u2 hline 40745 kJkg ug T2 40C v2 004656 m3kg s2 17971 kJkg K m2 V v2 02004656 4296 kg 1S2 gen m2s2 sline 4296 17971 17198 0332 kJK CO2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 850 The water cooler in Problem 524 operates steady state Find the rate of exergy destruction irreversibility Energy equation for heat exchanger Q L m h1 h2 m CP T1 T2 04 kgs 418 kJkgK 35 15 K 3344 kW Q W L T H H Q REF 1 2 cb β COP Q L W W Q L β 3344 3 1115 kW Energy equation for the refrigerator Q H Q L W 3344 kW 1115 kW 4459 kW CV Total Entropy Eq 0 m s1 s2 Q HTH S gen S gen Q HTH m s1 s2 Q HTH m Cp lnT1T2 4459 kW 29815 K 04 kgs 418 kJkgK ln 30815 28815 00857 kWK I T0 S gen 29815 K 00857 kWK 2555 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 851 Air enters the turbocharger compressor see Fig P851 of an automotive engine at 100 kPa 30C and exits at 200 kPa The air is cooled by 50C in an intercooler before entering the engine The isentropic efficiency of the compressor is 75 Determine the temperature of the air entering the engine and the irreversibility of the compressioncooling process Solution a Compressor First ideal which is reversible adiabatic constant s T2S T1 P2 P1 k1 k 30315 200 100 02857 3695 K wS CP0T1 T2S 100430315 3695 66565 kJkg Now the actual compressor w wSηS 66565075 8875 kJkg CPT1 T2 T2 39155 K T3to engine T2 TINTERCOOLER 39155 50 34155 K 684C b Irreversibility from Eq815 with rev work from Eq814 q 0 at TH s3 s1 1004 ln 34155 30315 0287 ln 200 100 00792 kJ kg K i Ts3 s1 h3 h1 w wrev h1 Ts1 h3 Ts3 Ts3 s1 CPT3 T1 CPT1 T2 3031500792 100450 262 kJkg 3 1 Exhaust 2 Engine W Compressor Cooler C Q C cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 852 A constant pressure pistoncylinder has 1 kg of saturated liquid water at 100 kPa A rigid tank contains air at 1000 K 1000 kPa They are now thermally connected by conduction through the walls cooling the air tank and bringing the water to saturated vapor Find the required amount of air and the irreversibility of the process assuming no external heat transfer State 1 initial air 2 final air 3 initial water 4 final water we assume T4 T 2 Take CV as each of the masses This gives Q from the two energy equations Q mH2Ou4 u3 WH2O mH2Oh4 h3 mH2O hfg 225802 kJ Q mairu2 u1 mair Cv T2 T1 mair mH2O hfg Cv T1 T2 225802 0717 1000 37277 50209 kg The entropy equation Eq837 for an irreversible process becomes S2 S1tot mairs2 s1 mH2Os4 s3 0 1S2 gen mH2Os4 s3 mH2O sfg 1 kg 60568 kJkgK 60568 kJK s2 s1 Cv ln T1 T2 R lnv1 v2 Cv ln T2 T1 0717 ln 37277 1000 070753 kJ kgK Now solve for the entropy generation 1S2 gen mairs2 s1 mH2Os4 s3 50209 070753 60568 25044 kJK 1I2 T0 1S2 gen 29815 25044 7466 kJ AIR water Q Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Exergy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 853 The automatic transmission in a car receives 25 kW shaft work and gives out 23 kW to the drive shaft The balance is dissipated in the hydraulic fluid and metal casing all at 45oC which in turn transmits it to the outer atmosphere at 20oC Find all the exergy transfer rates Solution CV Total unit Steady state and surface at 45oC Exergy transfer 25 kW 2 kW 23 kW Φ H 1 To TH Q H Φ W W The two shaft works are 100 exergy Φ W W 25 kW coming in and Φ W W 23 kW leaving The remaingning balance of energy leaves as heat transfer from the 45oC surface Φ out with Q 1 To TSurface Q out 1 29315 27315 45 2 kW 0157 kW Here we used 29315 for the reference T as it was the ambient Once the heat transfer is redistributed at the ambient T the exergy transfer is zero Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 854 A heat engine receives 1 kW heat transfer at 1000 K and gives out 400 W as work with the rest as heat transfer to the ambient What are the fluxes of exergy in and out Exergy flux in Φ H 1 To TH Q H 1 29815 1000 1 kW 0702 kW Exergy flux out Φ L 1 To TL Q L 0 TL To The other exergy flux out is the power Φ out W 04 kW H Q 1 kW W 400 W L Q T 1000 K amb HE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 855 In a refrigerator 1 kW is removed from the 10oC cold space and 13 kW is moved into the 30oC warm space Find the exergy fluxes including direction associated with the two heat transfers Φ H 1 To TH Q H 1 29815 30315 13 kW 0021 kW Φ L 1 To TL Q L 1 29815 26315 1 kW 0133 kW Ie the flux goes into the cold space Why As you cool it T To and you increase its exergy it is further away from the ambient H Q 13 kW W 300 W L Q 1 kW T 30 C L REF The refrigerator gives 0133 kW of exergy to the cold space and 0021 kW to the hot space in terms of exergy driven by a work input of W 13 1 03 kW that is a second law efficiency of about 50 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 856 A steady stream of R410A at ambient temperature 20oC and 800 kPa enters a solar collector and it exits at 80oC 600 kPa Calculate the change in exergy of the R410a Solution inlet exit SOLAR COLLECTOR Inlet TP Table B42 hi 30002 kJkg si 11105 kJkg K Exit TP Table B42 he 35967 kJkg se 13265 kJkg K From Eq823 or flow terms in Eq838 ψie ψe ψi he hi T0se si 35967 30002 2931513265 11105 367 kJkg Remark it is negative due to the pressure loss Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 857 A heat pump has a coefficient of performance of 2 using a power input of 3 kW Its low temperature is To and the high temperature is 80oC with an ambient at To Find the fluxes of exergy associated with the energy fluxes in and out First let us do the energies in and out COP β Q H W Q H β W 2 3 kW 6 kW Energy Eq Q L Q H W 6 3 3 kW Exergy flux in Φ L 1 To TL Q L 0 TL To Exergy flux in Φ W W 3 kW Exergy flux out Φ H 1 To TH Q H 1 29815 35315 6 kW 0935 kW Remark The process then destroys 3 0935 kW of exergy H Q W 3 kW L Q T o 80 C HP o Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 858 A flow of air at 1000 kPa 300 K is throttled to 500 kPa What is the irreversibility What is the drop in flow exergy A throttle process is constant enthalpy if we neglect kinetic energies Process he hi so ideal gas Te T i Entropy Eq se si sgen so Te so Ti R ln Pi Pe 0 R ln Pi Pe sgen 0287 ln 500 1000 02 kJkg K Eq815 i To sgen 298 02 596 kJkg The drop in exergy is exergy destruction which is the irreversibility ψ i 596 kJkg P P high low i e cb 1000 500 P v i e T s P P i e i e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 859 A power plant has an overall thermal efficiency of 40 receiving 100 MW of heat transfer from hot gases at an average of 1300 K and rejects heat transfer at 50oC from the condenser to a river at ambient temperature 20oC Find the rate of both energy and exergy a from the hot gases and b from the condenser CV High T heat exchanger Energy flux in Q H 100 MW Exergy flux in Φ H 1 To TH Q H 1 29315 1300 100 MW 7745 MW CV condenser Overall power plant W ηQ H so from energy equation Energy flux out Q L Q H W 1 η Q H 06 100 MW 60 MW Exergy flux out Φ L 1 To TL Q L 1 29315 32315 60 MW 557 MW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 860 Find the change in exergy from inlet to exit of the condenser in Problem 748 Solution Condenser of Prob 748 has inlet equal to turbine exit State 2 P2 20 kPa s2 s1 67993 kJkg K x2 67993 0831970766 08433 h2 22401 kJkg State 3 P2 P3 T3 40C Compressed liquid assume satliq same T Table B11 h3 1675 kJkg s3 05724 kJkg K From Eq823 or from flow terms in Eq838 ψ3 ψ2 h3 Tos3 h2 Tos2 h3 h2 Tos3 s2 1675 22401 298205724 67993 20726 18569 2157 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 861 Calculate the change in exergy kW of the two flows in Problem 7105 Solution The two flows in the heat exchanger exchanges energy and thus also exergy exergy Fist find state 4 Air A7 h1 104622 h2 4013 kJkg so T1 81349 so T2 71593 kJkg K 3 water 1 air 4 2 Water B11 h3 8394 kJkg s3 02966 kJkg K Energy Eq610 m AIRhAIR m H2OhH2O h4 h3 m AIRm H2Oh1 h2 20564492 257968 kJkg h4 h3 257968 266362 hg at 200 kPa T4 Tsat 12023C x4 266362 50468220196 09805 s4 153 x4 5597 701786 kJkg K We consider each flow separately and for each flow exergy is Eq823 include mass flow rate as in Eq838 use To 20 C For the air flow m 1ψ1 ψ2 m 1 h1 h2 To s1 s2 2 104622 4013 293281349 71593 0287 ln 125 100 2 64492 26722 7554 kW For the water flow m 3ψ4 ψ3 m 3 h4 h3 To s4 s3 05 266362 8394 2932701786 02966 05 257968 19707 3047 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 862 A steadyflow device receives R410A at 40oC 800 kPa and it exits at 40oC 100 kPa Assume a reversible isothermal process Find the change in specific exergy CV The steady flow device assume T0 25oC Inlet TP Table B42 hi 31942 kJkg si 11746 kJkg K Exit TP Table B42 he 33012 kJkg se 14380 kJkg K From Eq822 or 1023 ψie ψe ψi he hi T0se si 33012 31942 298214380 11746 6785 kJkg The device is like an expander P drops so work is out and s increases so heat transfer comes in Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 863 Consider the springtime melting of ice in the mountains which gives cold water running in a river at 2C while the air temperature is 20C What is the exergy of the water relative to the temperature of the ambient Solution ψ h1 h0 T0s1 s0 flow exergy from Eq822 Approximate both states as saturated liquid from Table B11 ψ 8392 8396 29315003044 02966 2457 kJkg Why is it positive As the water is brought to 20C it can be heated with qL from a heat engine using qH from atmosphere TH T0 thus giving out work Clipart from Microsoft Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 864 Nitrogen flows in a pipe with velocity 300 ms at 500 kPa 300C What is its exergy with respect to an ambient at 100 kPa 20C Solution From the exergy or exergy in Eq822 ψ h1 h0 12V 2 1 T0s1 s0 CpT1 T0 12V 2 1 T0Cp ln T1 T0 R ln P1 P0 1042300 20 3002 2000 29315 1042 ln 57315 29315 02968 ln 500 100 272 kJkg Notice that the high velocity does give a significant contribution Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 865 Compressed air for machines and tools in a plant is generated by a central compressor receiving air at 100 kPa 300 K 1 kgs delivering it at 600 kPa to a buffer tank and a distribution pipe After flowing through the tank and pipe the air is at the ambient 300 K at its point of use Assume a reversible adiabatic compressor and find the compressor exit temperature and the increase in air exergy through the compressor CV Compressor Energy 0 h1 h2 wC Entropy 0 s1 s2 0 s2 s1 s 0 T2 s 0 T1 R lnP2P1 find s 0 T2 and then T 2 or with constant specific heat T2 T1 P2 P1 k1 k 300 600 100 02857 5005 K Exergy increase through the compressor matches with the compressor work ψ2 ψ1 h2 h1 T0s2 s1 h2 h1 CP T2 T1 wC 1004 kJkgK 5005 300 K 2013 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 866 For the air system in the previous problem find the increase in the air exergy from the inlet to the point of use How much exergy was lost in the flow after the compressor exit From inlet state 1 to final point of use state 3 ψ3 ψ1 h3 h1 T0s3 s1 0 T00 R lnP3P1 T0 R lnP3P1 29815 K 0287 kJkgK ln 600100 1533 kJkg Compressor exit reversible adiabatic constant s process s2 s 1 T2 T1 P2 P1 k1 k 300 600 100 02857 5005 K So then ψlost ψ2 ψ3 ψ3 ψ1 ψ2 ψ1 ψ3 ψ1 h2 h1 ψ3 ψ1 CP0T2 T1 1533 kJkg 1004 kJkgK 5005 300 K 48 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 867 Calculate the exergy of the water at the initial and final states of Problem 6130 and the irreversibility of the process State properties 1 u1 8394 kJkg s1 02966 kJkg K v1 0001 m3kg 2 u2 31243 kJkg s2 77621 kJkg K v2 0354 m3kg 0 uo 10486 kJkg so 03673 kJkg K vo 0001003 m3kg Process transfers 1Wac 2 203 kJ 1Qac 2 32434 kJ TH 87315 K φ u Tos uo Toso Po v vo φ1 8394 2981502966 10486 2981503673 100 0001002 0001003 0159 kJkg φ2 31243 2981577621 10486 2981503673 100 035411 0001003 850 kJkg 1I2 mφ1 φ2 1 ToTH1Qac 2 1Wac 2 Po V2 V1 84984 1 29815 87315 32434 203 100 03541 0001 84984 21359 203 3531 1118 kJ Sgen 375 kJK ToSgen 1118 kJ so OK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 868 A geothermal source provides 10 kgs of hot water at 500 kPa 145C flowing into a flash evaporator that separates vapor and liquid at 200 kPa Find the three fluxes of exergy inlet and two outlets and the irreversibility rate CV Flash evaporator chamber Steady flow with no work or heat transfer Cont Eq m 1 m 2 m 3 Energy Eq m 1h1 m 2h2 m 3h3 Entropy Eq m 1s1 S gen m 2s2 m 3s3 1 2 3 Vap Liq B11 ho 10487 so 03673 B14 h1 6107 s1 17902 B12 h2 270663 s2 71271 h3 50468 s3 1530 h1 xh2 1 x h3 x m 2m 1 h1 h3 h2 h3 004815 m 2 xm 1 04815 kgs m 3 1xm 1 95185 kgs S gen 04815 71271 95185 153 10 17902 0093 kWK Flow exergy Eq822 ψ h Tos ho Toso h ho Tos so ψ1 6107 10487 29815 17902 03673 8159 kJkg ψ2 270663 10487 29815 71271 03673 58633 kJkg ψ3 50468 10487 29815 153 03673 5315 kJkg m 1 ψ1 81592 kW m 2ψ2 2823 kW m 3ψ3 5059 kW Balance of flow terms in Eq838 I m 1 ψ1 m 2 ψ2 m 3ψ3 277 kW To S gen Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 869 An air compressor is used to charge an initially empty 200L tank with air up to 5 MPa The air inlet to the compressor is at 100 kPa 17C and the compressor isentropic efficiency is 80 Find the total compressor work and the change in exergy of the air CV Tank compressor Transient process with constant inlet conditions no heat transfer Continuity m2 m1 min m1 0 Energy m2u2 minhin 1W2 Entropy m2s2 minsin 1S 2 gen Reversible compressor 1S2 GEN 0 s2 sin State 1 v1 RT1P1 08323 m3kg the ambient state State inlet Table A71 hin 29043 kJkg s o Tin 683521 kJkg K Eq619 s o T2 s o Tin R ln P2 Pin 683521 0287 ln 5000 100 795796 Table A71 T2s 8546 K u2s 63725 kJkg 1w2s hin u2s 29043 63725 34682 kJkg Actual compressor 1w2AC 1w2sηc 43353 kJkg u2AC hin 1w2AC 29043 43353 72396 kJkg T2AC 9585 K s o T2 ac 808655 kJkg K State 2 u P v2 RT2P2 005502 m3kg so m2 V2v2 3635 kg 1W2 m2 1w2AC 15759 kJ m2φ2 φ1 m2u2 u1 P0v2 v1 T0s2 s1 3635 72396 20719 100005502 08323 290808655 683521 0287 ln5000100 14604 kJ W cb 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 870 Find the exergy at all 4 states in the power plant of Problem 748 with an ambient at 298 K Solution Flow exergy from Eq822 neglecting kinetic and potential energy is ψ h h0 T0s s0 so we need hs for all four states W T Q H W P in Q L 3 2 1 4 P1 P4 20 MPa T1 700 C h1 38091 kJkg s1 67993 kJkg K P2 P3 20 kPa T3 40 C State 3 P T Comp liquid take sat liquid Table B11 h3 1675 kJkg v3 0001008 m3kg CV Turbine Entropy Eq78 s2 s1 67993 kJkg K Table B12 s2 08319 x2 70766 x2 08433 h2 2514 08433 235833 22401 kJkg wT h1 h2 38091 22401 1569 kJkg CV Pump property relation v constant gives work from Eq715 as wP v3 P4 P3 000100820000 20 201 kJkg h4 h3 wP 1675 201 1876 kJkg Flow exergy from Eq823 and notice that since turbine work and pump work are reversible they represent also change in avalability ψ1 h1 h0 T0s1 s0 38091 10487 298 67993 03673 17875 kJkg ψ2 h2 h0 T0s2 s0 ψ1 wT 17875 1569 2185 kJkg ψ3 h3 h0 T0s3 s0 1675 10487 29805724 03673 151 kJkg ψ4 h4 h0 T0s4 s0 ψ3 wP 151 201 2161 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 871 An electric stove has one heating element at 300oC getting 750 W of electric power It transfers 90 of the power to 1 kg water in a kettle initially at 20oC 100 kPa the rest 10 leaks to the room air The water at a uniform T is brought to the boiling point At the start of the process what is the rate of exergy transfer by a electrical input b from heating element and c into the water at Twater a Work is exergy Φ W 750 W b Heat transfer at 300oC is only partly exergy Φ 1 To TH Q 1 29315 27315 300 750 366 W c Water receives heat transfer at 20oC as 90 of 750 W 675 W Φ 1 To Twater Q 1 29315 27315 20 675 0 W 750 W at 300oC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 872 Air flows at 1500 K 100 kPa through a constant pressure heat exchanger giving energy to a heat engine and comes out at 500 K What is the constant temperature the same heat transfer should be delivered at to provide the same exergy Solution CV Heat exchanger Continuity eq m 1 m 2 Energy Eq412 m 1h1 m 1h2 Q H Table A71 h1 16358 kJkg h2 50336 kJkg s1 861209 kJkg K s2 738692 kJkg K H Q W L Q Ambient HE 1 2 qout h1 h2 16358 50336 11324 kJkg Exergy from heat transfer at T ψ 1 To TH qout ψ1 ψ2 Eq823 and the flow terms per unit mass flow in Eq838 ψ1 ψ2 h1 h2 To s1 s2 11324 29815 86121 738692 11324 3563 7671 kJkg 1 To TH ψ1 ψ2 qout 7671 11324 06774 To TH 03226 TH 924 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 873 A wooden bucket 2 kg with 10 kg hot liquid water both at 85C is lowered 400 m down into a mineshaft What is the exergy of the bucket and water with respect to the surface ambient at 20C CV Bucket and water Both thermal exergy and potential energy terms v1 v0 for both wood and water so work to atm is zero Use constant heat capacity table A3 for wood and table B11 sat liq for water From Eq828 φ1 φ0 mwoodu1 u0 T0s1 s0 mH2Ou1 u0 T0s1 s0 mtotgz1 z0 2 kg 126 kJ kg K 85 20 K 29315 K 126 kJ kg K ln27315 85 29315 10 kg 35582 8394 29311342 02966 kJkg 12 kg 9807 ms2 400 m 1000 JkJ 1585 26338 4707 2322 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 874 A flow of 01 kgs hot water at 70oC is mixed with a flow of 02 kgs cold water at 20oC in a shower fixture What is the rate of exergy destruction irreversibility for this process Solution Continuity Eq49 m 1 m 2 m 3 01 02 03 kgs Energy Eq410 m 1h1 m 2h2 m 3h3 Entropy Eq77 m 1s1 m 2s2 S gen m 3s3 Use specific heat from A4 and Eq611 for s change or Table B11 saturated liquid values Divide the energy equation with m 3CP T3 m 1m 3T1 m 2m 3T2 1 3 70 2 3 20 3667oC S gen m 1s3 s1 m 2s3 s2 01 418 ln 3098234315 02 418 ln 3098229315 000353 kWK Φ Destruction I To S gen 29815 000353 105 kW You could also have decided to use 20oC for the ambient T in this problem 1 2 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 875 A 1 kg block of copper at 350C is quenched in a 10 kg oil bath initially at ambient temperature of 20C Calculate the final uniform temperature no heat transfer tofrom ambient and the change of exergy of the system copper and oil Solution CV Copper and oil Cco 042 kJkg K Coil 18 kJkg K m2u2 m1u1 1Q2 1W2 0 mcoCcoT2 T1co mCoilT2 T1oil 1 042 T2 350 10 18 T2 20 0 1842 T2 507 T 275C 30065 K For each mass copper and oil we neglect work term v C so Eq828 is φ2 φ1 u2 u1 Tos2 s1 mC T2 T1 Toln T2 T1 mcvφ2 φ1cv moil φ2 φ1oil 042 3225 29315 ln 30065 62315 10 18 75 29315 ln 30065 29315 45713 1698 440 kJ Oil Cu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 876 A water kettle has 1 kg of saturated liquid water at Po It is on an electric stove that heats it from a hot surface at 500 K Water vapor escapes from the kettle and when the last liquid drop disappears the stove is turned off Find the destruction of exergy two places a between the hot surface and the water and b between the electrical wire input and the hot surface CV The water this is a control mass We bring it from saturated liquid to saturated vapor same pressure Energy eq mu2 u1 1Q2 1W2 1Q2 m h2 h1 mhfg 2257 kJ CV Kettle bottom consider the hot surface to water heat transfer ψin 1 To TH surface 1Q2 1 29815 500 2257 kJ 911 kJ ψout 1 To Twater 1Q2 1 29815 37315 2257 kJ 4536 kJ The destruction is the drop in exergy ψdestruction ψin ψout 911 4536 4574 kJ CV Wire and hot plate ψin Welectrical 1Q2 2257 kJ ψout 1 To TH surface 1Q2 1 29815 500 2257 kJ 911 kJ ψdestruction ψin ψout 2257 911 1346 kJ Comment We could also have found the destruction of exergy by finding the entropy generation terms and then ψdestruction To Sgen Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 877 A 200 L insulated tank contains nitrogen gas at 200 kPa 300 K A line with nitrogen at 500 K 500 kPa adds 40 more mass to the tank with a flow through a valve Use constant specific heats to find the final temperature and the exergy destruction CV Tank no work and no heat transfer Continuity Eq420 m2 m1 mi Energy Eq421 m2 u2 m1u1 mi hi 1Q2 Entropy Eq712 m2s2 m1s1 mi si dQT 1S 2 gen Process Eq 1Q2 0 m2 14 m1 mi 04 m 1 Write hi ui RTi then the energy equation contains only us so we can substitute the u differences with Cv0 ΔT and divide by m2 Cv0 to get T2 m2 m1 T1 mi m2Ti m2 mi R Cv0 Ti 1 14 300 04 14 500 04 14 02968 0745 500 4140 K We need the pressure for the entropy P2 m2RT2V 14 P1 T2T1 14 200 414300 3864 kPa m1 RT1 P1V1 200 02 02968 300 04492 kg m2 14 m1 06289 kg The entropy generation with entropy difference from Eq616 becomes 1S2 gen m2s2 m1s1 mi si m2s2 si m1s1 si m2 CP0ln T2 Ti R ln P2 Pi m1 CP0 ln T1 Ti R ln P1 Pi 06289 1042 ln414 500 02968 ln3864 500 04492 1042 ln300 500 02968 ln200 500 00414 kJK Φdestruction To 1S2 gen 29815 00414 kJ 1234 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 878 A 10kg iron disk brake on a car is initially at 10C Suddenly the brake pad hangs up increasing the brake temperature by friction to 110C while the car maintains constant speed Find the change in exergy of the disk and the energy depletion of the cars gas tank due to this process alone Assume that the engine has a thermal efficiency of 35 Solution All the friction work is turned into internal energy of the disk brake Energy eq mu2 u1 1Q2 1W2 1Q2 mFeCFeT2 T1 1Q2 10 045 110 10 450 kJ Neglect the work to the surroundings at P0 so change in exergy is from Eq827 φ mu2 u1 T0ms2 s1 Change in s for a solid Eq611 ms2s1 mC lnT2T1 10 045 ln 38315 28315 1361 kJK φ 450 28315 1361 6463 kJ Wengine ηthQgas 1Q2 Friction work Qgas 1Q2ηth 450035 12857 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 879 Water as saturated liquid at 200 kPa goes through a constant pressure heat exchanger as shown in Fig P879 The heat input is supplied from a reversible heat pump extracting heat from the surroundings at 17C The water flow rate is 2 kgmin and the whole process is reversible that is there is no overall net entropy change If the heat pump receives 40 kW of work find the water exit state and the increase in exergy of the water CV Heat exchanger heat pump m 1 m 2 2 kgmin m 1h1 Q 0 W in m 1h2 m 1s1 Q 0T0 m 1s2 Substitute Q 0 into energy equation and divide by m 1 h1 T0s1 win h2 T0s2 LHS 5047 29015 15301 40602 12607 kJkg State 2 P2 h2 T0s2 12607 kJkg At sat vap hg T0sg 6388 so state 2 is superheated vapor at 200 kPa At 600oC h2 T0s2 370396 29015 87769 115734 kJkg At 700oC h2 T0s2 392766 29015 90194 131068 kJkg Linear interpolation T2 667C ψ h2 T0s2 h1 T0s1 win 1200 kJkg 12607 5047 29015 15301 1200 kJkg 2 1 T s T o w q o 1 Q W 0 Q Ambient HP 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 880 Ammonia 2 kg at 400 kPa 40oC is in a pistoncylinder together with an unknown mass of saturated liquid ammonia at 400 kPa The piston is loaded so it keeps constant pressure and the two masses are allowed to mix to a final uniform state of saturated vapor without external heat transfer Find the total exergy destruction in the process CV All the ammonia Continuity Eq m2 mA mB 0 Energy Eq m2u2 mAuA mBuB Q W Entropy Eq m2s2 mAsA mBsB dQT 1S2 gen Process Q 0 P C so W PV2 VA VB State A hA 15436 kJkg sA 57111 kJkgK mA 2 kg State B hB 17165 kJkg sB 06793 kJkgK mB kg State 2 h2 14402 kJkg s2 53559 kJkgK m2 mA mB Insert the process equation into the energy equation m2h2 mAhA mBhB 0 Now solve the energy equation for the unknown mass mB mB mA h2 hA hB h2 2 kg 14402 15436 17165 14402 0163 kg Solve for entropy generation 1S2 gen m2s2 mAsA mBs B 2163 53559 2 57111 0163 06793 0051886 kJK Exergy destruction ψdestruction To Sgen 29815 0051886 1547 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Exergy Balance Equation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 881 Apply the exergy equation to solve Problem 818 A refrigerator should remove 15 kW from the cold space at 10oC while it rejects heat to the kitchen at 25oC Find the reversible work The exergy balance equation is Eq838 dΦ dt 1 To T Q cv W cv Po dV dt m iψi m eψe ToS gen 0 1 To TL Q L 1 To TH Q H W cv 0 0 0 0 The two ddt terms are zero due to steady state no flow and reversible give zero for the last three terms so we get the equation we can solve for the work W cv 1 To TL Q L 1 To TH Q H To solve we need the heat transfer Q H as we do have Q L 15 kW We need to use the entropy equation to get Q H THTL Q L 29815 26315 15 kW 16995 kW W cv 1 To TL Q L 1 To TH Q H Q L Q H To TH Q H To TL Q L Q L Q H 15 16995 01995 kW 02 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 882 Apply the exergy equation to solve Problem 835 with To 20oC A 20oC room is heated with a 2000 W electric baseboard heater What is the rate of irreversibility The exergy balance equation is Eq838 dΦ dt 1 To T Q cv W cv Po dV dt m iψi m eψe ToS gen For the heater we have steady state ddt terms are zero one Q cv out at To electrical work in and no flow terms leaving us with 0 1 To To Q cv W electrical ToS gen I To S gen W electrical 2000 W This assumes that To 20oC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 883 Estimate some reasonable temperatures to use and find all the fluxes of exergy in the refrigerator given in Example 52 We will assume the following temperatures Ambient T 20oC usually it is the kitchen air Low T T 5oC refrigerator T 10oC freezer Φ W W 150 W Φ H 1 To TH Q H 1 Tamb Tamb Q H 0 Φ L 1 To TL Q L 1 293 278 250 135 W Ie the flux goes into the cold space Why As you cool it T To and you increase its exergy it is further away from the ambient Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 884 Find the specific flow exergy in and out of the steam turbine in Example 71 assuming an ambient at 293 K Use the exergy balance equation to find the reversible specific work Does this calculation of specific work depend on To Solution The specific flow exergy is from Eq 822 ψi hi 1 2 V2 i Tosi ho Toso Reference state ho 8394 kJkg so 02966 kJkg K ho Toso 29638 kJkg The properties are listed in Example 71 so the specific flow exergies are ψi 30512 125 293 71228 29638 96843 kJkg ψe 26550 20 293 71228 29638 59098 kJkg The reversible work is from Eq829 with q 0 and steady state so w ψi ψe 96843 59098 37745 kJkg The offset To terms drop out as we take the difference and also si se ψi ψe hi he Tosi se hi he Notice since the turbine is reversible we get the same as in Example 71 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 885 Apply the exergy equation to solve Problem 836 A refrigerator removes 15 kW from the cold space at 10oC using 750 W of power input while it rejects heat to the kitchen at 25oC Find the rate of irreversibility The exergy balance equation is Eq838 dΦ dt 1 To T Q cv W cv Po dV dt m iψi m eψe ToS gen 0 1 To TL Q L 1 To TH Q H W cv 0 0 0 ToS gen The two ddt terms are zero due to steady state and no flow Now solve for the irreversibility notice the Q H term drops out since TH To I To S gen 1 To TL Q L W cv 1 29815 26315 15 kW 075 kW 02 kW 075 kW 055 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 886 Evaluate the steady state exergy fluxes due to a heat transfer of 50 W through a wall with 600 K on one side and 400 K on the other side What is the exergy destruction in the wall Solution Exergy flux due to a Q term Eq836 Φ Q 1 To T Q Φ 1 1 To T1 Q 1 298 600 50 252 W Φ 2 1 To T2 Q 1 298 400 50 1275 W 250 W 1 600 K 2 400 K Steady state state so no storage and Eq838 is 0 Φ 1 Φ 2 Φ destr Φ destr Φ 1 Φ 2 252 1275 1245 W Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 887 A counterflowing heat exchanger cools air at 600 K 400 kPa to 320 K using a supply of water at 20C 200 kPa The water flow rate is 01 kgs and the air flow rate is 1 kgs Assume this can be done in a reversible process by the use of heat engines and neglect kinetic energy changes Find the water exit temperature and the power out of the heat engines HE H Q W L Q 1 3 2 4 HE H Q W L Q HE H Q W L Q water air CV Total Energy eq m ah1 m H2Oh3 m ah2 m H2Oh4 W Entropy Eq m as1 m H2Os3 m as2 m H2Os4 sgen 0 Table A7 h1 607316 kJkg sT1 757638 kJkg K Table A7 h2 320576 kJkg sT2 693413 kJkg K Table B11 h3 8396 kJkg s3 02966 kJkg K From the entropy equation we first find state 4 s4 m am H2Os1 s2 s3 101757638 693413 02966 67191 4 P4 P3 s4 Table B12 x4 6719115305597 09271 h4 50468 09271 220196 25461 kJkg T4 12020C From the energy equation W m ah1 h2 m H2Oh3 h4 160732 32058 018396 25461 4053 kW The work could be found from the exergy balance equation requires state 4 as W m aψ1 ψ2 m H2Oψ3 ψ4 Since the entropies including m s balance out it is the same as the energy Eq Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 888 Consider the condenser in Problem 748 Find the specific energy and exergy that are given out assuming an ambient at 20oC Find also the specific exergy destruction in the process Solution WT QH QL WP in 1 2 3 4 Condenser from state 2 to state 3 P2 P3 20 kPa T3 40 C State 1 P T Table B13 h1 38091 kJkg s1 67993 kJkg K CV Turbine Entropy Eq78 s2 s1 67993 kJkg K Table B12 s2 08319 x2 70766 x2 08433 h2 2514 08433 235833 22401 kJkg State 3 P T Compressed liquid take sat liq Table B11 h3 16754 kJkg s3 05724 kJkg K CV Condenser Energy Eq qL h2 h3 22401 16754 207256 kJkg Exergy Eq ψ ψ2 ψ3 h2 h3 Tos2 s3 207256 2931567993 05724 2471 kJkg going out Since all the exergy that goes out ends up at the ambient where it has zero exergy the destruction equals the outgoing exergy ψdestr ψ 2471 kJkg Notice the condenser gives out a large amount of energy but little exergy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 889 Apply the exergy equation to find the exergy destruction for Problem 854 Exergy flux in Φ H 1 To TH Q H 1 29815 1000 1 kW 0702 kW Exergy flux out Φ L 1 To TL Q L 0 TL To The other exergy flux out is the power Φ out W 06 kW H Q 1 kW W 600 W L Q T 1000 K amb HE The exergy balance equation Eq838 for the HE becomes steady state 0 Φ H Φ L W Φ destruction Φ destruction Φ H Φ L W 702 W 0 600 W 102 W Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 890 A flow of 1 kgs of air at 300 K is mixed with a flow of 2 kgs air at 1500 K in an insulated pipe junction at a pressure of 100 kPa Find the exit temperature and the exergy destruction Solution Continuity Eq49 m 1 m 2 m 3 1 2 3 kgs Energy Eq410 m 1h1 m 2h2 m 3h3 1 2 3 Entropy Eq77 m 1s1 m 2s2 S gen m 3s3 Due to high T at state 2 use A7 and Eq619 for s change h3 m 1m 3h1 m 2m 3h2 1 3 30047 2 3 16358 119069 kJkg From A71 T3 11254 K sEATA3 AE A 827073 kJkg K S gen 2827073 861208 1827073 686926 071877 kWK AΦ E Adestruction To S gen 29815 071877 2143 kW The pressure correction part of the entropy terms cancel out as all three states have the same pressure Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 891 The condenser in a power plant cools 10 kgs water at 10 kPa quality 90 so it comes out as saturated liquid at 10 kPa The cooling is done by oceanwater coming in at ambient 15AoE AC and returned to the ocean at 20AoE AC Find the transfer out of the water and the transfer into the oceanwater of both energy and exergy 4 terms Solution CV Water line No work but heat transfer out Energy Eq AQE Aout Am E A h1 h2 10234535 19181 21 535 kW CV Ocean water line No work but heat transfer in equals water heattransfer out Energy Eq q hA4E A hA3E A 8394 6298 2096 kJkg Am E Aocean AQE Aout q 21 535 2096 10274 kgs Exergy out of the water follows from a flow term in Eq838 Φ out Am E Aψ1 ψ2 Am E A h1 h2 To s1 s2 10 234535 19181 2881574001 06492 20823 kW Exergy into the ocean water Φ ocean Am E AoceanψA4E A ψA3E A Am E Aocean hA4E A hA3E A TAoE AsA4E A sA3E A 10274 2096 2881502966 02245 1894 kW Notice there is a large amount of energy exchanged but very little exergy cb 1 2 3 4 Often the cooling media flows inside a long pipe carrying the energy away Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 892 Consider the car engine in Example 51 and assume the fuel energy is delivered at a constant 1500 K The 70 of the energy that is lost is 40 exhaust flow at 900 K and the remainder 30 heat transfer to the walls at 450 K goes on to the coolant fluid at 370 K finally ending up in atmospheric air at ambient 20AoE AC Find all the energy and exergy flows for this heat engine Find also the exergy destruction and where that is done From the example in the text we get Q L 07 Q H 233 kW This is separated into two fluxes Q L1 04 Q H 133 kW 900 K Q L2 03 Q H 100 kW 450 K Q L3 100 kW 370 K Q L4 100 kW 293 K Gases Steel Glycol Air flow 1500 K 450 K 370 K 293 K Radiator Assume all the fuel energy is delivered at 1500 K then that has an exergy of Φ QH 1 To TH Q H 1 A 293 1500E A 333 2679 kW Work is exergy Φ W AWE A 100 kW The exhaust flow exergy becomes Φ QL1 1 To TL1 Q H 1 A293 900E A 133 897 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful The heat transfer through the coolant keeps dropping in temperature and are Φ QL2 1 To TL2 Q QL2 1 A293 450E A 100 349 kW Φ QL3 1 To TL3 Q QL3 1 A293 370E A 100 208 kW Φ QL4 1 To TL4 Q QL4 1 A293 293E A 100 0 kW The destruction terms are the drops in exergy fluxes For the inside of the cylinder Φ Destruction inside Φ QH AWE A Φ QL1 Φ QL2 433 kW Outside the cylinder the wall that separates the gas from the coolant gives Φ Destruction wall Φ QL2 Φ QL3 141 kW The wall in the radiator separating the coolant from the air gives Φ Destruction radiator Φ QL3 Φ QL4 208 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 893 A pistoncylinder has forces on the piston so it keeps constant pressure It contains 2 kg of ammonia at 1 MPa 40C and is now heated to 100C by a reversible heat engine that receives heat from a 200C source Find the work out of the heat engine using the exergy balance equation Solution To evaluate it we need the change in exergy Eq828 ΦA2E A Φ A1E A mAamE AuA2E A uA1E A PoVA2E A VA1E A mAamE ATosA2E A sA1E A The work W WAHEE A A1E AWA2pistE A is from the integrated exergy Eq838 for a reversible process with no flow terms and use of Eq828 W PoVA2E A VA1E A 1 To THA1E AQA2E A Φ A2E A ΦA1E A 0 1 To THA1E AQA2E A mAamE AuA2E A uA1E A mAamE ATosA2E A sA1E A Now we must evaluate the three terms on the RHS and the work A1E AWA2pistE A State 1 uA1E A 13698 kJkg vA1E A 013868 mA3E Akg sA1E A 51778 kJkg K State 2 uA2E A 14905 kJkg vA2E A 017389 mA3E Akg sA2E A 56342 kJkg K A1E AWA2pistE A mAamE APvA2E A vA1E A 2 1000 017389 013868 7042 kJ CV Heat engine and ammonia otherwise we involve another Q Entropy mAamE AsA2E A sA1E A A1E AQA2E ATAHE A 0 A1E AQA2E A TAHE A mAamE AsA2E A sA1E A 47315 2 56342 51778 43189 kJ Substitute this heat transfer into the work term H Q W L Q HE 200 C o NH3 cb W 1 A29815 47315E A 43189 21490513698 2298155634251778 15974 2414 27215 19049 kJ WAHEE A W A1E AWA2pistE A 19049 7042 1200 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 894 An airconditioner on a hot summer day removes 8 kW of energy from a house at 21AoE AC and pushes energy to the outside which is at at 31AoE AC The house has 15 000 kg mass with an average specific heat of 095 kJkgK In order to do this the cold side of the airconditioner is at 5AoE AC and the hot side is at 40AoE AC The air conditioner refrigerator has a COP that is 60 of a corresponding Carnot refrigerator Find the actual airconditioner COP the power required to run the airconditioner the rate of exergy destruction inside the air conditioner and the total rate of exergy destruction due to the airconditioner and house Any heat transfer must go from a higher to a lower T domain H Q W L Q REF 5 C o 31 C amb 40 C 21 C o Q leak from atm 31 C House A steady state refrigerator definition of COP COP βREF Q L AWE A Q L Q H Q L 06 βCarnot Carnot βCarnot A TL ETH TL E A A5 27315 40 5E A 795 βREF 06 795 477 AWE A Q L βREF 8 kW 477 168 kW Q H Q L AWE A 968 kW CV Refrigerator AC unit Energy Eq E CV 0 AQ E ALE A AW E A AQ E AHE Entropy Eq S CV 0 A Q L ETL E A A Q H ETH E A S gen ac Φ destr To S gen ac 29815 A 968 40 27315E A A 8 5 27315E A 064 kW CV Total house AC unit AQ E ALeakE A and AQ E AHE A comes from and leaves to atm Energy Eq E CV 0 AQ E AleakE A AW E A AQ E AHE Entropy Eq S CV 0 A Q leak ETatm E A A Q H ETatm E A S gen tot S gen tot A W ETatm E Φ destr To S gen tot To A W ETatm E A 29815 K A168 kW 30415 KE A 165 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 895 If the airconditioner in the previous problem is turned off how fast does the house heat up in Ksec Once the AC unit is turned off we do not cool the house so heat leaks in from the atm at a rate of 8 kW that is what we had to remove to keep steady state Energy Eq E CV AQ E AleakE A 8 kW mAhouseE A CAPE A AdT dtE AdT dtE A AQ E AleakE A mAhouseE A CAPE A A 8 kW 15 000 095 kJKE A 056 10A3E A Ks Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 896 A disk brake of 2 kg steel and 1 kg brake pads is at 20AoE AC The brakes are now stopping a car so they dissipate energy by friction and heats up to T2 200AoE AC Assume the brake pads have specific heat of 06 kJkgK After this process the disk and pads now slowly cools to the ambient 20AoE AC T3 Find the exergy destruction in the braking process 1 to 2 and that in the cooling process 2 to 3 Control volume around the disk brake assembly Energy equation UA2E A UA1E A A1E AQA2E A A1E AWA2E A A1E AWA2E A mAstE ACAstE A TA2E A TA1E A mApadE ACApadE A TA2E A TA1E A 2 046 1 06 kJK 200 20 K 2736 kJ Entropy equation SA2E A SA1E A 0 A1E ASA2 genE A A1E ASA2 genE A mAstE ACAstE A lnTA2E A TA1E A mApadE ACApadE A lnTA2E A TA1E A 2 046 1 06 ln A473 293E A 07277 kJK A1E AΦA2 destructionE A TAoE A A1E ASA2 genE A 29815 K 07277 kJK 217 kJ The cooling process 2 to 3 extend CV out to ambient TAoE A UA3E A UA2E A A2E AQA3E A 2736 kJ UA2E A UA1E A A2E ASA3 genE A SA3E A SA2E A A2E AQA3E A TAoE A 07277 A2736 293E A 02056 kJK A2E AΦA3 destructionE A TAoE A A2E ASA3 genE A 29815 K 02056 kJK 613 kJ Brake pads Disk Hydraulic line fluid acts on piston that pushes the brake pad Clipart from Microsoft Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 897 A small house kept at 20oC inside loses 12 kW to the outside ambient at 0oC A heat pump is used to help heat the house together with possible electrical heat The heat pump is driven by a motor of 25 kW and it has a COP that is one quarter of a Carnot heat pump unit Find the actual heat pump COP and the exergy destruction in the whole process CV House Energy 0 AQ E AHE A AW E AelE A AQ E ALossE Definition of COP β COPAHPE A A Q H EW E A A1 4E A A TH ETH TL E A A1 4E A A29315 20 0E A 3664 AQ E AHE A COPAHPE A AW E AHPE A 3664 25 kW 916 kW AW E AelE A AQ E ALossE A AQ E AHE A 12 916 284 kW CV Total Energy 0 AQ E ALE A AW E AHPE A AW E AelE A AQ E ALossE A AQ E ALossE A AQ LE A AW E AHPE A AW E AelE A 25 284 534 kW Entropy 0 AQ LE ATALE A AQ lossE ATALE A S gen Φ destr To S gen To AQ lossE A AQ LE A TALE A 27315 A 534 27315E A 534 kW Comment We must use To 27315 K here as the AQ LE A and AQ lossE A were exchanged with the ambient Otherwise the destruction becomes larger than the supplied work terms because the exergy exchanged with the ambient is nonzero Q loss Q Q H L W 25 kW HP cb W el Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 898 A farmer runs a heat pump using 2 kW of power input It keeps a chicken hatchery at a constant 30AoE AC while the room loses 10 kW to the colder outside ambient at 10AoE AC Find the COP of the heat pump the rate of exergy destruction in the heat pump and its heat exchangers and the rate of exergy destruction in the heat loss process CV Hatchery steady state at 30AoE AC for the hatchery Energy Eq 0 AQ E AHE A AQ E ALossE A AQ E AHE A AQ E ALossE A 10 kW COP AQ E AHE A AW E A 10kW 2 kW 5 CV Heat pump steady state Energy eq 0 AQ E ALE A AW E A AQ E AHE A AQ E ALE A AQ E AHE A AW E A 8 kW Entropy Eq 0 A Q L ETL E A A Q H ETH E A S gen HP S gen HP A Q H ETH E A A Q L ETL E A A 10 273 30E A A 8 273 10E A 000473 kWK Φ destr HP To S gen HP 28315 K 000473 kWK 134 kW CV From hatchery at 30AoE AC to the ambient 10AoE AC This is typically the walls and the outer thin boundary layer of air Through this goes AQ E ALossE A Entropy Eq 0 A Q Loss ETH E A A Q Loss ETamb E A S gen walls S gen walls A Q Loss ETamb E A A Q Loss ETH E A A 10 283E A A 10 303E A 000233 kWK Φ destr walls To S gen HP 28315 K 000233 kWK 066 kW Comment Notice the two exergy destruction terms exactly equals work input It was necessary to use To as ambient T otherwise the flux of exergy associated with the heat transfers AQ E ALE A and AQ E ALossE A would be nonzero and a storage effect present Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Q loss Q Q H L W 2 kW HP cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Device SecondLaw Efficiency Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 899 A heat engine receives 1 kW heat transfer at 1000 K and gives out 400 W as work with the rest as heat transfer to the ambient Find its first and second law efficiencies First law efficiency is based on the energies ηAIE A AW E AAQ E AH A04 1E A 04 The second law efficiency is based on work out versus exergy in Exergy flux in AΦ E AH 1 To TH AQ E AH A 1 29815 1000 E A 1 kW 0702 kW ηAIIE A A W AE Φ H E A 04 0702E A 057 Notice the exergy flux in is equal to the Carnot heat engine power output given 1 kW at 1000 K and rejecting energy to the ambient H Q 1 kW W 600 W L Q T 1000 K amb HE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8100 A heat exchanger increases the exergy of 3 kgs water by 1650 kJkg using 10 kgs air coming in at 1400 K and leaving with 600 kJkg less exergy What are the irreversibility and the second law efficiency CV Heat exchanger steady flow 1 inlet and 1 exit for air and water each The two flows exchange energy with no heat transfer tofrom the outside The irreversibility is the destruction of exergy so AI E A AΦ E Adestruction AΦ E Ain AΦ E Aout 10 600 3 1650 1050 kW The second law efficiency Eq832 ηAIIE A AΦ E Aout AΦ E Ain A3 1650 10 600E A 0825 3 1 4 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8101 Find the second law efficiency of the heat pump in problem 857 The second law efficiency is a ratio of exergies namely what we want out divided by what we have to put in Exergy from first term on RHS Eq 838 Φ H 1 To TH Q H Q H β W 2 2 kW 4 kW η E A 1 29815 35315 E A A4 2E A 031 II 1 To TH Φ H W Q H W H Q W 2 kW L Q T o 80 C HP o Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8102 A steam turbine inlet is at 1200 kPa 500AoE AC The actual exit is at 300 kPa with an actual work of 407 kJkg What is its second law efficiency The second law efficiency is the actual work out measured relative to the reversible work out Eq 829 Steam turbine TAoE A 25C 29815 K Inlet state Table B13 hAiE A 347628 kJkg sAiE A 76758 kJkg K Actual turbine energy Eq hAeE A hAiE A wAacE A 347628 407 306928 kJkg Actual exit state Table B13 TAeE A 300AoE AC sAeE A 77022 kJkg K From Eq829 for steady state and adiabatic process wArevE A hAiE A TAoE AsAiE A hAeE A TAoE AsAeE A hAiE A hAeE A TAoE AsAeE A sAiE A 347628 306928 2981577022 76758 407 787 4149 kJkg ηAIIE A wAacE AwArevE A 407 4149 098 1200 300 P v i e T s P i P e i e 300 500 8103 Find the second law efficiency for the compressor in problem 821 An air compressor takes air in at the state of the surroundings 100 kPa 300 K The air exits at 400 kPa 200C using 100 kW of power Determine the minimum compressor work input CV Compressor Steady flow minimum work in is reversible work Energy Eq 0 hA1E A hA2E A wc wc hA2E A hA1E A 4758 30047 17533 kJkg ψA1E A 0 at ambient conditions Get the properties from the air table A71 and correct standard entropy for the pressure sA0E A sA2E A sA ET0 EA sA ET2 EA R lnPA0E APA2E A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 686926 73303 0287 ln100400 006317 kJkg K ψA2E A hA2E A hA0E A TA0E AsA0E A sA2E A 47579 300473 300 006317 156365 kJkg wAREVE A ψA2E A ψA1E A 156365 0 kJkg 156365 kJkg 2nd law efficiency ηAIIE A A wrev Ewac E A A156365 17533E A 0892 A wac i Ewac E A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8104 Find the isentropic efficiency and the second law efficiency for the compressor in Problem 826 A compressor in a refrigerator receives R410A at 150 kPa 40AoE AC and it brings it up to 600 kPa using an actual specific shaft work of 5865 kJkg in an adiabatic compression Find the specific reversible work CV Compressor Assume adiabatic and neglect kinetic energies Energy Eq 0 hA1E A hA2E A w Entropy Eq78 sA2E A sA1E A sAgenE State 1 B42 hA1E A 26399 kJkg sA1E A 11489 kJkg K hA2ACE A hA1E A w 26399 5865 32264 kJkg State 2ac B42 P hA2E A sA2E A 12152 kJkg K State 2s B42 P s sA1E A hA2sE A 30267 kJkg We have two different compressors IDEAL wAcsE A hA2sE A hA1E A 3868 kJkg ACTUAL wACACE A hA2ACE A hA1E A 5865 kJkg Definition Eq727 ηAcE A wAcsE AwAcACE A 0659 Rev work Eq829 wArevE A ψA1E A ψA2ACE A hA1E A hA2ACE A TA0E AsA1E A sA2ACE A 5865 29815 11489 12152 5865 19767 389 kJkg 2nd law efficiency ηAIIE A A wrev Ewac E A A 389 5865E A 0663 A wac i Ewac E A P s 1 2 s 2 ac 2 ac 2 s 1 40 40 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8105 A steam turbine has inlet at 4 MPa 500C and actual exit of 100 kPa x 10 Find its first law isentropic and its second law efficiencies Solution CV Steam turbine Energy Eq413 w hi he Entropy Eq79 se si sgen Inlet state Table B13 hi 34452 kJkg si 70900 kJkg K Exit actual state Table B12 he 26755 se 73593 kJkg K Actual turbine energy equation w hi he 7697 kJkg Ideal turbine reversible process so sgen 0 giving ses si 70900 13025 xes 60568 xes 09555 hes 4174 09555 22580 25750 kJkg The energy equation for the ideal gives ws hi hes 8702 kJkg The first law efficiency is the ratio of the two work terms ηs wws 0885 The reversible work for the actual turbine states is Eq829 wrev hi he Tose si 7697 298273593 70900 7697 803 8500 kJkg Second law efficiency Eq831 η2nd Law wwrev 76978500 0906 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8106 Find the second law efficiency for the compressed air system in Problem 865 Consider the total system from the inlet to the final point of use In the total system the exergy increase in the air from inlet to final exit point is the gain and the source is the compressor work input Compressor Reversible adiabatic constant s s2 s1 T2 T1 P2 P1 k1 k 300 600 100 02857 5005 K ψ2 ψ1 h2 h1 T0s2 s1 h2 h1 CP0T2 T1 1004 5005 300 2013 kJkg wC From inlet state 1 to final point of use state 3 ψ3 ψ1 h3 h1 T0s3 s1 0 T00 R lnP3P1 T0 R lnP3P1 29815 0287 ln 600100 1533 kJkg ηII ψ2 ψ1 ψ3 ψ1 1533 2013 076 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8107 A turbine receives steam at 3000 kPa 500oC and has two exit flows one at 1000 kPa 350oC with 20 of the flow and the remainder at 200 kPa 200oC Find the isentropic and the second law efficiencies CV Steam turbine x 02 extraction fraction Energy Eq413 w h1 xh2 1 xh3 Entropy Eq79 se s1 sgen Inlet state Table B13 h1 345648 kJkg s1 72337 kJkg K Extraction state h2 315765 kJkg s2 73010 kJkg K Exit actual state Table B13 h3 287046 s3 75066 kJkg K Actual turbine energy equation w 345648 02 315765 08 287046 52858 kJkg Ideal turbine reversible process so sgen 0 giving s2s s1 72337 h2s 311743 kJkg s3s s1 72337 h3s 275012 kJkg The energy equation for the ideal turbine gives ws h1 xh2s 1 xh3s 345648 02 311743 08 275012 6329 kJkg The first law efficiency ηs wws 0835 The reversible work for the actual turbine states is Eq829 wrev h1 xh2 1 xh3 To s1 xs2 1 xs3 52858 298272337 02 7301 08 75066 52858 691 59768 kJkg Second law efficiency Eq831 ηII wwrev 5285859768 0884 v P s T 1 1 3s 3s 200 kPa 3 MPa 3 1 MPa 3 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8108 Steam enters a turbine at 25 MPa 550C and exits at 5 MPa 325C at a flow rate of 70 kgs Determine the total power output of the turbine its isentropic efficiency and the second law efficiency Solution hi 33356 kJkg si 61765 kJkg K he 29965 kJkg se 63289 kJkg K Actual turbine wTac hi he 3391 kJkg W m wTac 70 3391 23 740 kW Isentropic turbine has a different exit state ses si hes 29066 kJkg wTs hi hes 429 kJkg Rev turbine has the same exit state as the actual turbine wrev wTac T0se si 3391 4544 38454 kJkg Eq727 ηT wTacwTs 3391429 079 Eq831 ηII wTacwrev 339138454 088 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8109 A heat engine operating with an evironment at 298 K produces 5 kW of power output with a first law efficiency of 50 It has a second law efficiency of 80 and TL 310 K Find all the energy and exergy transfers in and out Solution From the definition of the first law efficiency conversion efficiency Q H W ηI 5 05 10 kW Energy Eq Q L Q H W 10 5 5 kW Exergy output Φ W W 5 kW From the definition of the second law efficiency ηII W Φ H this requires that we assume the exergy rejected at 310 K is lost and not counted otherwise the efficiency should be ηII W Φ H Φ L Exergy from source Φ H 1 To TH Q H W ηII 5 08 625 kW Exergy rejected Φ L 1 To TL Q L 1 298 310 5 0194 kW Notice from the Φ H form we could find the single characteristic TH as 1 To TH 625 kW Q H 0625 TH 795 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8110 A flow of nitrogen 01 kgs comes out of a compressor stage at 500 kPa 500 K and is now cooled to 310 K in a counterflowing intercooler by liquid water at 125 kPa 15oC which leaves at 22oC Find the flowrate of water and the second law efficiency for the heat exchanger Solution Continuity eq m 1 m 2 m H2O m 3 m 4 m N2 Energy eq 0 m N2 h3 h4 m H2O h2 h1 Entropy Eq77 0 m N2 s3 s4 m H2O s2 s1 S gen Due to the lower range of temperature we will use constant specific heats from A5 and A4 so the energy equation is approximated as 0 m H2OCpH2O T2 T1 m N2Cp T3 T4 Now solve for the water flow rate m H2O m N2 CpN2 T3 T4 CpH2O T1 T2 01 kgs 1042 kJkgK 500 310 K 418 22 15 kJkg 0677 kgs The second law efficiency is the exergy gain over the exergy source so we need exergy change in both flows m N2 ψ4 ψ3 m N2CP T4 T3 T0 ln T4T3 01 kgs 1042 kJkgK 310 500 29815 ln 310 500 K 495 kW m H2O ψ1 ψ2 m H2OCP T1 T2 T0 ln T1T2 0677 kgs 418 kJkgK 22 15 29815 ln295 288 K 045 kW The nitrogen loses exergy as it cools but the water also loses why The water is brought closer to T0 as it is being heated up which is closer to zero exergy Both exergies drops and equals total exergy destruction T0 Sgen 54 kW ηII 0 A hydraulic motor driven compressor intercooler in pipe between the two stages Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8111 Air flows into a heat engine at ambient conditions 100 kPa 300 K as shown in Fig P8111 Energy is supplied as 1200 kJ per kg air from a 1500 K source and in some part of the process a heat transfer loss of 300 kJkg air happens at 750 K The air leaves the engine at 100 kPa 800 K Find the first and the second law efficiencies CV Engine out to reservoirs hi q1500 q750 he w wac hi q1500 q750 he 30047 1200 300 82220 37827 kJkg ηTH wq1500 03152 For second law efficiency also a q tofrom ambient si q1500TH q0T0 q750Tm s e q0 T0se si T0Tmq750 T0THq 1500 300 788514 686925 0287 ln100 100 300 750 300 3001500 1200 184764 kJkg wrev hi he q1500 q750 q0 wac q0 56303 kJkg ηII wacwrev 3782756303 0672 Comment You could also have defined the second law efficiency as Exergy in ϕH 1T0TH q1500 080 1200 960 kJkg ηII wac ϕH 37827 960 0394 This measure places zero value of the loss at 750 K zero exergy value for the two flow terms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8112 Air enters a compressor at ambient conditions 100 kPa 300 K and exits at 800 kPa If the isentropic compressor efficiency is 85 what is the secondlaw efficiency of the compressor process Solution s 300 K 1 2s 2 100 kPa 800 kPa T Ideal isentropic Eq623 T2s 30080286 5438 K ws 10045438 300 2446 kJkg w ws ηs 2446 085 2878 kJkg Actual exit temperature T2 T1 w CP0 300 2878 1004 5868 K Eq616 s2 s1 1004 ln5868300 0287 ln 8 007645 kJkgK Exergy Eq823 ψ2 ψ1 h2 h1 T0s2 s1 2878 300007645 2649 kJkg 2nd law efficiency Eq832 or 834 but for a compressor ηII w ψ2 ψ1 2649 2878 092 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8113 Consider that the air in the previous problem after the compressor flows in a pipe to an air tool and at that point the temperature has dropped to ambient 300 K and an air pressure of 750 kPa What is the second law efficiency for the total system CV Compressor Energy 0 h1 h2 wC Entropy 0 s1 s2 0 T2 T1 P2 P1 k1 k 300 800 100 02857 5434 K Exergy increase through the compressor matches with the ideal compressor work wC s h2 h1 CP T2 T1 1004 5434 300 kJkg 2444 kJkg The actual compressor work is wC ac wC s ηC s 2444085 2875 kJkg From inlet state 1 to final point of use state 3 ψ3 ψ1 h3 h1 T0s3 s1 0 T00 R lnP3P1 T0 R lnP3P1 29815 K 0287 kJkgK ln 750100 1724 kJkg So then the second law efficiency is the gain ψ3 ψ1 over the source wC ac as ηII wC ac ψ3 ψ1 1724 2875 06 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8114 A compressor is used to bring saturated water vapor at 1 MPa up to 15 MPa where the actual exit temperature is 650C Find the irreversibility and the secondlaw efficiency Solution Inlet state Table B12 hi 277808 kJkg si 65864 kJkg K Actual compressor Table B13 heac 371232 kJkg seac 68223 kJkg K Energy Eq Actual compressor wcac heac hi 9342 kJkg From Eq813 i T0seac si 29815 68223 65864 7033 kJkg From Eq815 wrev i wcac 7033 9342 86387 kJkg ηII wrevwcac 863879342 0925 v P s T 15 MPa i e ac i e ac 1 MPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8115 Use the exergy equation to analyze the compressor in Example 48 to find its second law efficiency assuming an ambient at 20oC CV The R134a compressor Steady flow We need to find the reversible work and compare that to the actual work Notice the heat loss goes out to ambient T and has thus 0 exergy Exergy eq 838 0 m ψ1 ψ2 Wcomp rev 0 W rev comp m h2 h1 To s2 s1 01 kgs 4351 3872 29315 K 17768 17665 kJ kg K 45 kW ηII W rev comp W ac comp 45 5 090 For a real device this is a little high Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8116 Calculate the second law efficiency of the coflowing heat exchanger in Problem 7113 with an ambient at 17C Solution CV Heat exchanger steady 2 flows in and two flows out 1 3 2 4 First solve for the exit temperature in Problem 7113 CV Heat exchanger steady 2 flows in and two flows out Energy Eq410 m O2h1 m N2h3 m O2h2 m N2h4 Same exit tempearture so T4 T2 with values from Table A5 m O2CP O2T1 m N2CP N2T3 m O2CP O2 m N2CP N2T2 T2 05 0922 290 06 1042 500 05 0922 06 1042 44629 10862 4109 K The second law efficiency for a heat exchanger is the ratio of the exergy gain by one fluid divided by the exergy drop in the other fluid We thus have to find the change of exergy in both flows For each flow exergy in Eq823 include mass flow rate as in Eq838 For the oxygen flow m O2ψ2 ψ1 m O2 h2 h1 To s2 s1 m O2 CPT2 T1 To CP lnT2 T1 R lnP2 P1 m O2CP T2 T1 TolnT2 T1 05 0922 4109 290 290 ln4109290 9148 kW For the nitrogen flow m N2ψ3 ψ4 m N2CP T3 T4 TolnT3 T4 06 1042 500 4109 290 ln5004109 20122 kW From Eq832 η E A 0513 2nd Law m O2ψ1 ψ2 m N2ψ3 ψ4 9148 20122 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8117 An aircompressor receives air at 290 K 100 kPa and brings it up to a higher pressure in an adiabatic process The actual specific work is 210 kJkg and the isentropic efficiency is 82 Find the exit pressure and the second law efficiency CV Compressor Assume adiabatic and neglect kinetic energies Energy Eq413 w hA1E A hA2E A Entropy Eq79 sA2E A sA1E A sAgenE We have two different cases the ideal and the actual compressor wACACE A hA2ACE A hA1E A 210 kJkg wAcsE A ηAcE TA2ACE A TA1E A wACACE A CAPE A 290 210 1004 4992 K For the isentropic compressor we get wAcsE A ηAcE A wACACE A 082 210 1722 kJkg hA2sE A hA1E A TA2sE A TA1E A wAcsE A CAPE A 290 17221004 4615 K PA2E A PA1E A TA2sE A TA1E AA kk1E A 100 4615290A35E A 5084 kPa Now the wArevE A ψA1E A ψA2ACE A hA1E A hA2ACE A TAoE AsA2ACE A sA1E A wACACE A TAoE A CAPE A lnTA2acE A TA1E A R ln PA2E A PA1E A 210 29815 1004 lnA4992 290E A 0287 lnA5084 100E A 210 234 1866 kJkg ηII wArevE A wACACE A A1866 210E A 088 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8118 A flow of 2 kgs water at 1000 kPa 80AoE AC goes into a constant pressure boiler where the water is heated to 400AoE AC Assume the hot gases that heats the water is air coming in at 1200 K and leaving at 900 K as in a counter flowing heat exchanger Find the total rate of irreversibility in the process and the second law efficiency of the boiler setup Solution CV Heat exchanger steady flow 1 inlet and 1 exit for air and water each The two flows exchange energy with no heat transfer tofrom the outside We need to find the air mass flow rate 1 water 3 air 4 2 Energy Eq m H2OhA2E A hA1E A m airhA3E A hA4E A m air m H2O A h2 h1 Eh3 h4 E A 2 A326388 33564 127781 93315E A 16992 kgs Exergy increase of the water flow m H2OψA2E A ψA1E A m H2OhA2E A hA1E A TAoE AsA2E A sA1E A 2 326388 33564 298157465 10746 2 292824 19053 20459 kW Exergy decrease of the air flow m airψA3E A ψA4E A m airhA3E A hA4E A TAoE AsA3E A sA4E A 16992 127781 93315 29815834596 801581 41839 kW The irreversibility is the destruction of exergy AI E A AΦ E Adestruction AΦ E Ain AΦ E Aout 41839 kW 20459 kW 2138 kW ηA2nd LawE A A m AE H2Oψ2 ψ1 m airψ3 ψ4 E A20459 41839E A 0489 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8119 A heat exchanger brings 10 kgs water from 100AoE AC to 500AoE AC at 2000 kPa using air coming in at 1400 K and leaving at 460 K What is the second law efficiency Solution CV Heat exchanger steady flow 1 inlet and 1 exit for air and water each The two flows exchange energy with no heat transfer tofrom the outside We need to find the air mass flow rate 1 water 3 air 4 2 Energy Eq m H2OhA2E A hA1E A m airhA3E A hA4E A m air m H2O A h2 h1 Eh3 h4 E A 10 A346755 42045 151527 46234E A 28939 kgs Exergy increase of the water flow m H2OψA2E A ψA1E A m H2OhA2E A hA1E A TAoE AsA2E A sA1E A 10 346755 42045 2981574316 13053 10 30471 182656 12 205 kW Exergy decrease of the air flow m airψA3E A ψA4E A m airhA3E A hA4E A TAoE AsA3E A sA4E A 28939 151527 46234 29815852891 730142 19 880 kW ηA2nd LawE A A m AE H2Oψ2 ψ1 m airψ3 ψ4 E A12 205 19 880E A 0614 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8120 Calculate the second law efficiency of the counter flowing heat exchanger in Problem 7105 with an ambient at 20C Solution CV Heat exchanger steady flow 1 inlet and 1 exit for air and water each The two flows exchange energy with no heat transfer tofrom the outside 3 water 1 air 4 2 Heat exchanger Prob 7105 with TAoE A 20C solve first for state 4 Energy Eq410 m AIRhAIR m H2OhH2O From A7 h1 h2 104622 4013 64492 kJkg From B12 h3 8394 kJkg s3 02966 kJkg K h4 h3 m AIRm H2Oh1 h2 20564492 257968 kJkg h4 h3 257968 266362 hg at 200 kPa T4 Tsat 12023C x4 266362 50468220196 09805 s4 153 x4 5597 701786 kJkg K We need the change in exergy for each flow from Eq823 ψA1E A ψA2E A hA1E A hA2E A TAoE AsA2E A sA1E A 10462 4013 293271593 81349 0287 ln100125 6449 2932091156 3776 kJkg ψA4E A ψA3E A hA4E A hA3E A TAoE AsA4E A sA3E A 26636 839 293270179 02966 25799 19707 6090 Efficiency from Eq832 ηA2nd LawE A Am E AwψA4E A ψA3E AAm E AAψA1E A ψA2E A 05 60902 3776 0403 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8121 A steam turbine receives 5 kgs steam at 400AoE AC 10 MPa One flow of 08 kgs is extracted at 3 MPa as saturated vapor and the remainder runs out at 1500 kPa with a quality of 0975 Find the second law efficiency of the turbine CV Turbine Steady flow and adiabatic q 0 Continuity Eq49 m 1 m 2 m 3 Energy Eq410 m 1h1 m 2h2 m 3h3 AW E Entropy Eq77 m 1s1 S gen m 2s2 m 3s3 W T 1 2 3 State 1 h1 309646 kJkg s1 62119 kJkgK State 2 h2 280414 kJkg s2 61869 kJkgK State 3 h3 84487 0975 194728 27435 kJkg s3 2315 0975 41298 634156 kJkgK AW E A m 1h1 m 2h2 m 3h3 5 309646 08 280414 42 27435 1716 kW This work is now compared to the reversible work possible AW revE A m 1ψ1 m 2ψ2 m 3ψ3 m 1h1 m 2h2 m 3h3 TAoE A m 1s1 m 2s2 m 3s3 1716 kW 29815 5 62119 08 61869 42 634156 kW 1716 kW 1564 kW 1872 kW ηA2nd LawE A AW E A AW revE A A1716 1872E A 0917 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8122 Calculate the irreversibility for the process described in Problem 4154 assuming that heat transfer is with the surroundings at 17C Solution CV Cylinder volume out to To 17 oC Continuity Eq415 mA2E A mA1E A mAinE A Energy Eq416 mA2E AuA2E A mA1E AuA1E A mAinE AhAlineE A A1E AQA2E A A1E AWA2E Entropy Eq712 mA2E A sA2E A mA1E AsA1E A misi A1E AQA2E A To A1E ASA2 genE A Process PA1E A is constant to stops then constant V to state 2 at PA2E State 1 PA1E A TA1E A mA1E A A P1V ERT1 E A A 300 025 0287 2902E A 090 kg State 2 Open to PA2E A 400 kPa TA2E A 350 K mA2E A A 400 1 0287 350E A 3982 kg mAiE A 3982 090 3082 kg Only work while constant P A1E AWA2E A PA1E AVA2E A VA1E A 3001 025 225 kJ Energy eq A1E AQA2E A mA2E AuA2E A mA1E AuA1E A A1E AWA2E A mAiE AhAiE 3982 0717 350 090 0717 2902 225 3082 1004 600 8192 kJ Entropy eq gives To A1E ASA2 genE A I To mA1E A sA2E A sA1E A mi sA2E A si A1E AQA2E 2901509Cp ln A350 290E A R ln A400 300E A 3082Cpln A350 600E A R ln A400 500E A 8192 kJ 29015 00956 14705 8192 4203 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8123 The hightemperature heat source for a cyclic heat engine is a steady flow heat exchanger where R134a enters at 80C saturated vapor and exits at 80C saturated liquid at a flow rate of 5 kgs Heat is rejected from the heat engine to a steady flow heat exchanger where air enters at 150 kPa and ambient temperature 20C and exits at 125 kPa 70C The rate of irreversibility for the overall process is 175 kW Calculate the mass flow rate of the air and the thermal efficiency of the heat engine CV R134a Heat Exchanger m R134a 5 kgs Table B51 Inlet T1 80oC sat vapor x1 10 h1 hg 429189 kJkg s1 sg 16862 kJkgK Exit TA2E A 80oC sat liquid xA2E A 00 hA2E A hAfE A 322794 kJkg sA2E A sAfE A 13849 kJkgK H Q W L Q HE 1 2 3 4 CV Air Heat Exchanger CApE A 1004 kJkgK R 0287 kJkgK Inlet T3 20oC P3 150 kPa Exit T4 70oC P4 125 kPa s4 s3 Cp ln T4 T3 R lnP4 P3 02103 kJkgK Entropy Eq for the total system as control volume since we know AI E A AI E A To S net m R134a s2 s1 m airs4 s3 m air I To m R134as2 s1 s4 s3 A17529315 5 13849 16862 E02103E A A kWK kJkgKE 100 kgs Energy Eq for each line AQ E A AmE AhAinE A AmE AhAexE A AW E A AW E A 0 R134a 1 AQ E A2 AQ E AH m R134ah2 h1 532 kW Air AQ E AL 3 AQ E A4 m airh4 h3 m air CApE AT4 T3 5018 kW Control volume heat engine AW E Anet AQ E AH AQ E AL 532 5018 302 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful ηth AW E Anet AQ E AH 0057 or 57 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8124 A cylinder with a linear springloaded piston contains carbon dioxide gas at 2 MPa with a volume of 50 L The device is of aluminum and has a mass of 4 kg Everything Al and gas is initially at 200C By heat transfer the whole system cools to the ambient temperature of 25C at which point the gas pressure is 15 MPa Find the exergy at the initial and final states and the destruction of exergy in the process State 1 TA1E A 200 oC vA1E A V1 m 005 11186 00447 mA3E Akg State 2 vA2E A vA1E A 2 15 29815 47315 003756 mA3E Akg The metal does not change volume so the combined is using Eq827 as φA1E A mgasφgas mAlφAl mgasuA1E AuoTosA1E A socv mgasPovA1E Avo mAluA1E Auo TosA1E AsoAl mgasCv TA1E A To mgasTo Cp ln A T1 ToE A R ln A P1 PoE A mgasPo vA1E A vo mAl C TA1E A To ToC ln TA1E ATo Al φA1E A 11186 065320025 29815 0842 ln A47315 29815E A 018892 ln A2000 100E A 100 00447 05633 4 090 200 25 29815 ln A47315 29815E A 12888 1343 2632 kJ φA2E A 11186 065325 25 29815 0842 ln A29815 29815E A 018892 ln A1500 100E A 100 003756 05633 4 09 25 25 29815 ln A29815 29815E A 11182 0 11182 kJ The irreversibility is as in Eq838 integrated over time no flow terms A1E AIA2E A φA1E A φA2E A 1 TA0E ATAHE A A1E AQA2E A A1E AWA2E AAC Pom VA2E A VA1E A 2632 11182 0 14 100 11186 003756 00447 16458 kJ A1E ASAs genE A 0552 kJK To A1E ASAs genE A 16458 kJ so OK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful T amb Q CO 2 Al Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8125 A twostage compressor takes nitrogen in at 20AoE AC 150 kPa compresses it to 600 kPa 450 K then it flows through an intercooler where it cools to 320 K and the second stage compresses it to 3000 kPa 530 K Find the specific exergy increase and the specific exergy destruction in each of the two compressor stages The setup is like this CV Stage 1 nitrogen Steady flow Process adibatic q 0 sAgenE A 0 Energy Eq 413 wAC1E A hA2E A hA1E A Entropy Eq 710 0 sA1E A sA2E A sAgen C1E Assume constant CAP0E A 1042 from A5 and Eq616 for change in entropy sAgen C1E A sA2E A sA1E A CAP0E A lnTA2E ATA1E A R ln PA2E APA1E A 1042 ln45029315 02968 ln600150 00351 kJkgK ψdestruction AC1E A To sAgen C1E A 29815 K 00351 kJkgK 1046 kJkg ψA2E A ψA1E A hA2E A hA1E A To sA2E A sA1E A CAP0E ATA2E A TA1E A To sA2E A sA1E A 1042450 29315 29815 00351 15298 kJkg sAgen C2E A sA4E A sA3E A CAP0E A lnTA4E ATA3E A R ln PA4E APA3E A 1042 ln530320 02968 ln3000600 00481 kJkgK ψdestruction AC2E A To sAgen C2E A 29815 K 00481 kJkgK 1434 kJkg ψA4E A ψA3E A hA4E A hA3E A To sA4E A sA3E A CAP0E ATA4E A TA3E A To sA4E A sA3E A 1042530 320 29815 00481 20448 kJkg Comment Notice how the exergy increases are the enthalpy increase minus the exergy destruction C1 C2 1 2 3 4 2 W Q 1 W cooler inter Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8126 The intercooler in the previous problem uses cold liquid water to cool the nitrogen The nitrogen flow is 01 kgs and the liquid water inlet is 20AoE AC and setup to flow in the opposite direction as the nitrogen so the water leaves at 35AoE AC Find the flow rate of the water and the exergy destruction in this intercooler CV Heat exchanger steady 2 flows in and two flows out Continuity eq AmE A1E A AmE A2E A AmE AH2OE A AmE A3E A AmE A4E A AmE AN2E Energy eq 0 AmE AN2E A hA3E A hA4E A AmE AH2OE A hA2E A hA1E A Entropy Eq77 0 Am E AN2E A sA3E A sA4E A Am E AH2OE A sA2E A sA1E A S gen Due to the lower range of temperature we will use constant specific heats from A5 and A4 so the energy equation is approximated as 0 AmE AH2OE ACApE AH2OE A TA2E A TA1E A AmE AN2E ACApE A TA3E A TA4E A Now solve for the water flow rate AmE AH2OE A AmE AN2E A CApE AN2E A TA3E A TA4E A CApE AH2OE A TA1E A TA2E A 01 kgs 1042 kJkgK 450 320 K 418 35 20 kJkg 0216 kgs AS E AgenE A Am E AN2E A sA4E A sA3E A Am E AH2OE A s1 sA2E A Am E AN2E ACAPE A ln T4T3 Am E AH2OE ACAPE A ln T1T2 01 kgs 1042 kJkgK ln A320 450E A 0216 kgs 418 kJkgK ln A308 293E 003552 004508 000956 kWK AΦ E Adestruction AI E A To S gen 29815 K 000956 kWK 285 kW 1 3 2 4 H2O N2 A hydraulic motor driven compressor with intercooler in small pipe between the two stages Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8127 Find the irreversibility in the cooling process of the glass plate in Problem 4144 In a glass factory a 2 m wide sheet of glass at 1500 K comes out of the final rollers that fix the thickness at 5 mm with a speed of 05 ms Cooling air in the amount of 20 kgs comes in at 17AoE AC from a slot 2 m wide and flows parallel with the glass Suppose this setup is very long so the glass and air comes to nearly the same temperature a coflowing heat exchanger what is the exit temperature Energy Eq m glasshAglass 1E A m airhair 2 m glasshAglass 3E A m airhair 4 AmE AglassE A ρAV E A ρAV 2500 2 0005 05 125 kgs AmE AglassE ACAglassE A T3 T1 m air CAPaE A T4 T2 T4 T3 CAglassE A 080 kJkg K CAPaE A 1004 kJkg K T3 m glassCglass T1 m airCPa T2 m glassCglass m airCPa A1250801500 201004290 125080 201004E 6923 K To find the irreversibility we need the entropy generation term Entropy Eq m glasssAglass 1E A m airsair 2 S gen m glasssAglass 3E A m airsair 4 S gen m glass sA3E A sA1E Aglass m airs4 s2air m glass CAglassE A lnT3 T1 m air CAPaE A lnT4 T2 125 08 ln6923 1500 20 1004 ln6923290 7732 kWK 17472 kWK 974 kWK AI E A AΦ E Adestruction To S gen 29815 K 974 kWK 2904 kW 1 2 3 4 Air Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8128 Air in a pistoncylinder arrangement is at 110 kPa 25C with a volume of 50 L It goes through a reversible polytropic process to a final state of 700 kPa 500 K and exchanges heat with the ambient at 25C through a reversible device Find the total work including the external device and the heat transfer from the ambient CV Total out to ambient Energy mAaE AuA2E A uA1E A A1E AQA2E A A1E AWA2totE A Entropy mAaE AsA2E A sA1E A A1E AQA2E ATA0E A 0 mAaE A 110 0050287 29815 00643 kg A1E AQA2E A TA0E AmAaE AsA2E A sA1E A 29815 0064373869 68631 0287 ln 700110 014 kJ A1E AWA2totE A A1E AQA2E A mAaE AuA2E A uA1E A 014 00643 359844 213037 958 kJ This is identical to the reversible work in Eq817 TAjE A TA0E A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8129 A rigid container with volume 200 L is divided into two equal volumes by a partition Both sides contains nitrogen one side is at 2 MPa 300C and the other at 1 MPa 50C The partition ruptures and the nitrogen comes to a uniform state at 100C Assuming the surroundings are at 25C find the actual heat transfer and the irreversibility in the process Solution CV Total container Continuity Eq mA2E A mAAE A mABE A 0 Energy Eq mAAE AuA2E A uA1E AAAE A mABE AuA2E A uA1E AABE A A1E AQA2E A A1E AWA2E Entropy Eq mAAE AsA2E A sA1E AAAE A mABE AsA2E A sA1E AABE A A1E AQA2E ATAsurE A A1E ASAs genE Process V C A1E AWA2E A 0 From the initial state we get the mass as mA2E A mAAE A mABE A A PA1VA ERTA1 E A A PB1VB ERTB1 E A A 200001 0296857315E A A 100001 0296832315E A 1176 1043 2219 kg PA2E A mA2E ARTA2E AVAtotE A 2219 02968 3731502 12288 kPa From the energy equation we get the heat transfer as the change in U A1E AQA2E A mAAE ACAvE ATA2E A TA1E AAAE A mABE ACAvE ATA2E A TA1E AABE 1176 0745 100 300 1043 0745 100 50 1364 kJ The entropy changes are found from Eq816 sA2E A sA1E AAAE A 1042 lnA37315 57315E A 02968 lnA12288 2000E A 009356 kJkg K sA2E A sA1E AABE A 1042 lnA37315 32315E A 02968 lnA12288 1000E A 00887 kJkg K The entropy generation follows from the entropy equation A1E ASA2genE A 1176 009356 1043 00887 136429815 04396 kJK Now the irreversibility comes from Eq 1013 A1E AIA2E A TA0E A A1E ASA2genE A 13108 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8130 Consider the irreversible process in Problem 6182 Assume that the process could be done reversibly by adding heat enginespumps between tanks A and B and the cylinder The total system is insulated so there is no heat transfer to or from the ambient Find the final state the work given out to the piston and the total work to or from the heat enginespumps CV Water mAAE A mABE A heat engines No Qexternal only A1E AWA2cylE A WAHEE Continuity Eq mA2E A mAA1E A mAB1E A 6 kg Energy Eq mA2E AuA2E A mAA1E AuAA1E A mAB1E AuAB1E A A1E AWA2cylE A WAHEE Entropy Eq mA2E AsA2E A mAA1E AsAA1E A mAB1E AsAB1E A A0E A A0E vAA1E A 006283 mA3E Akg uAA1E A 34485 kJkg sAA1E A 73476 kJkgK vAB1E A 009053 mA3E Akg uAB1E A 28437 kJkg sAB1E A 67428 kJkgK VAAE A mAA1E A vAA1E A 02513 mA3E A VABE A mAB1E AvAB1E A 01811 mA3E mA2E AsA2E A 473476 267428 42876 kJK sA2E A 7146 kJkg K If PA2E A PAliftE A 14 MPa then VA2E A VAAE A VABE A 04324 mA3E A vA2E A 007207 mA3E Akg PAliftE A sA2E A vA2E A 020135 VA2E A 1208 mA3E A VA2E A OK PA2E A PAliftE A 14 MPa uA2E A 28742 kJkg A1E AWA2cylE A PAliftE AVA2E A VAAE A VABE A 14001208 04324 108584 kJ WAHEE A mAA1E AuAA1E A mAB1E AuAB1E A mA2E AuA2E A A1E AWA2cylE 4 34478 2 28437 6 28742 108584 11476 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8131 Consider the heat engine in Problem 8111 The exit temperature was given as 800 K but what are the theoretical limits for this temperature Find the lowest and the highest assuming the heat transfers are as given For each case give the first and second law efficiency The lowest exhaust temperature will occur when the maximum amount of work is delivered which is a reversible process Assume no other heat transfers then Energy Eq 0 hAiE A qA1500E A qA750E A hAeE A w Entropy Eq 0 sAiE A qAHE ATAHE A A0E A sAeE A qAmE ATAmE sAeE A sAiE A qAHE ATAHE A qAmE ATAmE A sATe E E A sATi E E A R lnPAeE APAiE A sATe E E A sATi E E A R lnPAeE APAiE A qAHE ATAHE A qAmE ATAmE 686926 0287 ln100100 12001500 300750 726926 kJkg K Table A71 TAemin E A 446 K hAeE A 4479 kJkg wArevE A hAiE A qA1500E A qA750E A hAeE A 30047 1200 300 4479 75257 kJkg ηI ηATHE A A w rev Eq1500 E A A75257 1200E A 0627 The second law efficiency measures the work relative to the source of exergy and not qA1500E A So ηII wrev 1 ToTHq1500 A 75257 1 30015001200E A A75257 960E A 0784 Comment This measure places zero value of the loss at 750 K zero exergy value for the two flow terms The maximum exhaust temperature occurs with no work out hAeE A hAiE A qAHE A qAmE A hAeE A 300473 1200 300 12005 kJkg Table A71 TAemaxE A 1134 K Now wac 0 so ηI ηII 0 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8132 A small air gun has 1 cmA3E A air at 250 kPa 27AoE AC The piston is a bullet of mass 20 g What is the potential highest velocity with which the bullet can leave Solution The exergy of the air can give the bullet kinetic energy expressed in the exergy change Eq828 no heat transfer and reversible ΦA2E A Φ A1E A muA2E A uA1E A PoVA2E A VA1E A mTosA2E A sA1E A A1E AWA2E A PoVA2E A VA1E A Ideal gas so m PVRT A250 1 106 E0287 300E A 29 10A6E A kg The second state with the lowest exergy to give maximum velocity is the dead state and we take To 20AoE AC Now solve for the exergy change ΦA2E A Φ A1E A muA2E A uA1E A mTosA2E A sA1E A mCAvE ATA1E A TA2E A mTo CApE A lnA T2 ET1 E A R lnA P2 EP1 E A 29 10A6E A 071727 20 29315 1004 lnA293 300E A 0287 lnA100 250E A 00002180 kJ 0218 J A1 2E A mAbulletE AVA2 exE VAexE A A 2 02180020EA 467 ms Comment Notice that an isentropic expansion from 250 kPa to 100 kPa will give the final air temperature as 2309 K but less work out The above process is not adiabatic but Q is transferred from ambient at To Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8133 Consider the nozzle in Problem 7154 What is the second law efficiency for the nozzle A nozzle in a high pressure liquid water sprayer has an area of 05 cmA2E A It receives water at 350 kPa 20C and the exit pressure is 100 kPa Neglect the inlet kinetic energy and assume a nozzle isentropic efficiency of 85 Find the ideal nozzle exit velocity and the actual nozzle mass flow rate Solution CV Nozzle Liquid water is incompressible v constant no work no heat transfer Bernoulli Eq717 A1 2E AVA2 exE A 0 vPAiE A PAeE A 0001002 350 100 02505 kJkg VAexE A A 2 02505 1000 JkgEA 2238 m s1 This was the ideal nozzle now we can do the actual nozzle Eq 730 A1 2E AVA2 ex acE A η A1 2E AVA2 exE A 085 02505 02129 kJkg The second law efficiency is the actual nozzle compared to a reversible process between the inlet and actual exit state However no work so the reversible exit state with PAeE A TAex acE A then must have the reversible possible kinetic energy Energy actual nozzle hAiE A 0 hAeE A A1 2E AVA2 ex acE A same Z no q and no w The reversible process has zero change in exergies from Eq838 as ψAiE A ψAeE A hAiE A 0 TAoE A sAiE A hAeE A A1 2E AVA2 ex revE A TAoE AsAeE A1 2E AVA2 ex revE A hAiE A hAeE A TAoE A sAeE A sAiE A A1 2E AVA2 ex acE A TAoE A sAgenE A We can not get properties for these states accurately enough by interpolation so use constant specific heat and incompressibility hAex acE A hAiE A A1 2E AVA2 ex acE A uAex acE A PAeE Av uAiE A PAiE Av A1 2E AVA2 ex acE uAex acE A uAiE A PAiE Av PAeE Av A1 2E AVA2 ex acE A CTAex acE A TAiE A TAex acE A TAiE A PAiE Av PAeE Av A1 2E AVA2 ex acE AC 20 350 100 0001002 02129 418 2001C A1 2E AVA2 ex revE A A1 2E AVA2 ex acE A TAoE A sAgenE A 02129 29315 418 ln A29316 29315E A 02547 kJkg ηII 0212902547 084 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8134 Consider the light bulb in Problem 6189 What are the fluxes of exergy at the various locations mentioned What are the exergy destruction in the filament the entire bulb including the glass and the entire room including the bulb The light does not affect the gas or the glass in the bulb but it gets absorbed on the room walls A small halogen light bulb receives an electrical power of 50 W The small filament is at 1000 K and gives out 20 of the power as light and the rest as heat transfer to the gas which is at 500 K the glass is at 400 K All the power is absorbed by the room walls at 25AoE AC Find the rate of generation of entropy in the filament in the total bulb including glass and the total room including bulb Solution AW E AelE A 50 W AQ E ARADE A 10 W AQ E ACONDE A 40 W glass leads g as We will assume steady state and no storage in the bulb air or room walls CV Filament steadystate Energy Eq33 dEAcvE Adt 0 AW E AelE A AQ E ARADE A AQ E ACONDE Entropy Eq642 dSAcvE Adt 0 EA AQ ARAD A ETAFILA AE A EA AQ ACOND A ETAFILA AE A S gen AI E A TAoE AS gen AQ E ARADE A AQ E ACONDE A To TFILA TAoE A W el ETFILA E A 298 A 50 1000E A 149 W CV Bulb including glass AQ E ARADE A leaves at 1000 K AQ E ACONDE A leaves at 400 K AI E A TAoE AS gen TAoE A dAQ E AT 298101000 40400 328 W CV Total room All energy leaves at 25C Eq33 dEAcvE Adt 0 AW E AelE A AQ E ARADE A AQ E ACONDE Eq642 dSAcvE Adt 0 EA AQ ATOT A ETAWALL AE A S gen Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful AI E A TAoE AS gen TAoE A EA AQ ATOT A ETAWALL AE A 298 A 50 25273E A 50 W Since the room is at TAoE A then all the incoming exergy is destroyed Flux into filament Flux AW E AelE A 50 W Flux leaving the filament Flux AQ E ARADE A AQ E ACONDE A1 To TFILA 50 W 149 W 351 W Flux leaving the glass Flux AQ E ARADE A1 To TFILA AQ E ACONDE A1 To Tglass 10 1 A 298 1000E A W 40 1 A298 400E A W 702 W 102 W 1722 W Flux at walls Flux 0 wall is at TAoE A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8135 Air in a pistoncylinder arrangement shown in Fig P8135 is at 200 kPa 300 K with a volume of 05 mA3E A If the piston is at the stops the volume is 1 mA3E A and a pressure of 400 kPa is required The air is then heated from the initial state to 1500 K by a 1700 K reservoir Find the total irreversibility in the process assuming surroundings are at 20C Solution Energy Eq mu2 u1 1Q2 1W2 Entropy Eq ms2 s1 A dQTEA 1S2 gen Process P P0 αVV0 if V Vstop Information Pstop P0 αVstopV0 Eq of state Tstop T1PstopVstopP1V1 1200 T2 So the piston will hit the stops V2 Vstop P2 T2Tstop Pstop 15001200 400 500 kPa 25 P1 State 1 m2 m1 P1V1 RT1 A 200 05 0287 300E 1161 kg 2 P v 1a 1 v vstop 1 Air Q Tres 1W2 A1 2E AP1 PstopVstop V1 A1 2E A200 4001 05 150 kJ 1Q2 mu2 u1 1W2 1161120525 21436 150 1301 kJ s2 s1 sA o T2E A sA o T1E A R lnP2P1 86121 68693 0287 ln 25 148 kJkg K Take control volume as total out to reservoir at TRES 1S2 gen tot ms2 s2 1Q2TRES 1161 148 13011700 0953 kJK A1E AIA2E A TA0E A 1S2 gen E A 29315 0953 279 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8136 Air enters a steadyflow turbine at 1600 K and exhausts to the atmosphere at 1000 K The second law efficiency is 85 What is the turbine inlet pressure CV Turbine exits to atmosphere so assume Pe 100 kPa Inlet TAiE A 1600 K Table A7 hAiE A 17573 kJkg sAo iE A 8 69051 kJkg K Exit TAeE A 1000 K hAeE A 10462 kJkg sAo eE A 8 13493 kJkg K Energy Eq q hAiE A hAeE A w q 0 w hAiE A hAeE A 7111 kJkg Second law efficiency expresses work as a fraction of exergy change ψAiE A ψAeE A wηII 71111085 8366 kJkg hAiE A hAeE A TosAiE A sAeE A hAiE A hAeE A w 71111 kJkg assume To 25oC si sAeE A 71111 8366 29815 0420896 kJkgK sAiE A sAeE A sAo iE A sAo eE A R lnPAiE APAeE A 04209 kJkg K R ln PAiE APAeE A 869051 813493 0420896 0976476 kJkgK PAiE APAeE A 30035 PAiE A 3003 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8137 A 1 kg rigid steel tank contains 12 kg of R134a at 20oC and 500 kPa Now the setup is placed in a freezer that brings it to 20oC The freezer operates in a 20oC kitchen and has a coefficient of performance that is half that of a Carnot refrigerator Find the heat transfer out of the R134a the extra work input to the refrigerator due to this process and the total irreversibility including the refrigerator CV The R134a and the steel Energy Eq mRu2 u1R mstu2 u1st 1Q2 1W2 Entropy Eq mRs2 s1R msts2 s1st A dQTEA 1S2 gen Process V C so 1W2 0 State 1 v1 004226 mA3E Akg u1 39052 kJkg s1 17342 kJkgK State 2 v2 v1 20oC 2 phase state x2 v2 vfvfg 004226 0000738014576 028486 u2 17365 x2 19285 22859 kJkg s2 09007 x2 08388 113964 kJkgK 1Q2 mRu2 u1R mstu2 u1st 12 22859 39052 1 04620 20 2127 kJ Refrigerator β 05 βCarnot 05 TL TH TL 05 A 25315 20 20E A 3164 Wref QLβ 2127 kJ 3164 6723 kJ CV Total incl refrigerator Storage effect in R134a and steel otherwise transfer terms Energy Eq mRu2 u1R mstu2 u1st Wref QH Entropy Eq mRs2 s1R msts2 s1st QHT0 1S2 gen QH Wref mu2 u1 6723 2127 27993 kJ 1S2 gen mRs2 s1R msts2 s1st QH T0 1211396417342 1 046 lnA25315 29315E A A27993 29315E 0713472 006748 0954904 017395 kJK A1E AIA2E A TA0E A A 1S2 gen E A 29315 017395 5099 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8138 A pistoncylinder arrangement has a load on the piston so it maintains constant pressure It contains 1 kg of steam at 500 kPa 50 quality Heat from a reservoir at 700C brings the steam to 600C Find the secondlaw efficiency for this process Note that no formula is given for this particular case so determine a reasonable expression for it Solution 1 Table B12 P1 x1 v1 0001093 0503738 0188 mA3E Akg h1 64021 05210847 16945 kJkg s1 18606 0549606 4341 kJkg K 2 P2 P1T2 v2 08041 h2 37017 kJkg s2 83521 kJkg K Energy Eq mu2 u1 1Q2 1W2 1Q2 PV2 V1 1Q2 mu2 u1 Pmv2 v1 mh2 h1 20072 kJ 1W2 Pmv2 v1 30805 kJ 1W2 to atm P0mv2 v1 6161 kJ Useful work out 1W2 1W2 to atm 24644 kJ φreservoir 1 T0Tres1Q2 A 1 29815 97315 E A 20072 13922 kJ ηII Wnetφ 0177 Remark You could argue that the stored exergy exergy should be accounted for in the second law efficiency but it is not available from this device alone Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8139 A jet of air at 200 ms flows at 25AoE AC 100 kPa towards a wall where the jet flow stagnates and leaves at very low velocity Consider the process to be adiabatic and reversible Use the exergy equation and the second law to find the stagnation temperature and pressure Solution CV From free flow to stagnation point Reversible adiabatic steady flow Exergy Eq838 0 Am E Aψi Am E Aψe Φ destr Entropy Eq 0 Am E Asi Am E Ase Am E AdqT Am E Asgen Am E Asi Am E Ase 0 0 Process Reversible Φ destr 0 sgen 0 adiabatic q 0 From exergy Eq ψe ψi 0 he Tose hi Tosi A1 2E AVA2 iE From entropy Eq se si so entropy terms drop out Exergy eq now leads to he hi A1 2E AVA2 iE A Te Ti A1 2E AVA2 iE A Cp Te 25 A1 2E A A 2002 Jkg E1004 Jkg KE A 4492AoE AC Eq823 Pe Pi TeTe A k k1 E A 100 A 273 4492 273 25 14 04E A 1254 kPa State i is the free stream state State e is the stagnation state i e cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Problems Solved Using Pr and vr Functions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 843 Solved using the isentropic Pr function in A72 An air compressor receives atmospheric air at T0 17C 100 kPa and compresses it up to 1400 kPa The compressor has an isentropic efficiency of 88 and it loses energy by heat transfer to the atmosphere as 10 of the isentropic work Find the actual exit temperature and the reversible work CV Compressor Isentropic wAcinsE A hAesE A hAiE A sAesE A sAiE Table A7 PAresE A PAriE A PAeE APAiE A 09917 14 13884 hAesE A 61751 kJkg wAcinsE A 61751 29058 32693 kJkg Actual wAcinacE A wAcinsE AηAcE A 37151 qAlossE A 32693 kJkg wAcinacE A hAiE A hAeacE A qAlossE hAeacE A 29058 37151 32693 6294 kJkg TAeacE A 621 K Reversible wArevE A hAiE A hAeacE A TA0E AsAeacE A sAiE A 29058 6294 29015 76121 68357 33882 22542 1134 kJkg Since qAlossE A is also to the atmosphere the above term is the net heat transfer and the reversible work requires the heat transfer to be reversible Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 869 Solved using the isentropic Pr function in A72 An air compressor is used to charge an initially empty 200L tank with air up to 5 MPa The air inlet to the compressor is at 100 kPa 17C and the compressor isentropic efficiency is 80 Find the total compressor work and the change in exergy of the air Solution CV Tank compressor Transient process with constant inlet conditions no heat transfer Continuity mA2E A Am1E A mAinE A Am1E A 0 Energy mA2E AuA2E A mAinE AhAinE A A1E AWA2E A Entropy mA2E AsA2E A mAinE AsAinE A A1E ASA2 genE Reversible compressor A1E ASA2 GENE A 0 sA2E A sAinE A State 1 vA1E A RTA1E APA1E A 08323 mA3E Akg State inlet Table A71 hAinE A 29043 kJkg sA o TinE A 68352 kJkg K Table A72 PArinE A 09899 used for constant s process Table A72 PAr2E A PArinE APA2E APAinE A 09899 5000100 49495 TA2sE A 855 K uA2sE A 6372 kJkg A1E AwA2sE A hAinE A uA2sE A 29043 6372 34677 kJkg Actual compressor A1E AwA2ACE A A1E AwA2sE AηAcE A 43346 kJkg uA2ACE A hAinE A A1E AwA2ACE A 29043 43346 72389 kJkg Backinterpolate in Table A71 TA2ACE A 958 K sAo T2ACE A 80867 kJkg K vA2E A RTA2E APA2E A 0055 mA3E Akg State 2 A u PE A mA2E A VA2E AvA2E A 3636 kg A1E AWA2E A mA2E AA1E AwA2ACE A 1576 kJ mA2E AφA2E A φA1E A mA2E AuA2E A uA1E A PA0E AvA2E A vA1E A TA0E AsA2E A sA1E A 3636 72389 20719 1000055 08323 29080867 68352 0287 ln5000100 14603 kJ Here we used Eq619 for the change in entropy UPDATED AUGUST 2013 SOLUTION MANUAL CHAPTER 8 English Units Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CHAPTER 8 SUBSECTION PROB NO Available Energy Reversible Work 126133 Irreversibility 134139 Exergy exergy 140149 Exergy Balance Equation 150151 Device 2nd Law Efficiency 152160 Review Problems 161165 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Exergy Reversible work Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8140E A control mass gives out 1000 Btu of energy in the form of a Electrical work from a battery b Mechanical work from a spring c Heat transfer at 700 F Find the change in exergy of the control mass for each of the three cases Solution a Work is exergy Φ Wel 1000 Btu b Work is exergy Φ Wspring 1000 Btu c Give the heat transfer to a Carnot heat engine and W is exergy Φ 1 T0 TH Qout 1 537 1160 1000 Btu 537 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8141E A fraction of some power to a motor 1 2 kW is turned into heat transfer at 800 R 2 and then it dissipates in the ambient at 540 R 3 Give the rates of exergy along the process 123 Solution The exergy of an amount of heat transfer equals the possible work that can be extracted This is the work out of a Carnot heat engine with heat transfer to the ambient as the other reservoir The result is from Chapter 5 as also shown in Eq 81 and Eq 838 1 Φ W 2 kW 1896 Btus 2 Φ W rev HE 1 To T Q 1 540 800 2 kW 065 kW 0616 Btus 3 Φ W rev HE 1 To To Q 1 540 540 2 kW 0 kW 0 Btus As the energy transforms from wotrk to heat transfer it loses exergy and as the heat transfer goes to lower and lower T its exergy value drops ending as zero at the ambient T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8142E A refrigerator should remove 15 Btus from the cold space at 15 F while it rejects heat to the kitchen at 77 F Find the reversible work The reversible work is related to the Carnot cycle work as the the two reservoirs are at constant temperatures EAW AE carnot E TL TH TL A15 4597 77 15E A 765 W β β Q L Q L In general we have defined the reversible work with the standard sign definition AW revE A AQ AE L E β A 15 765E A Btus 02 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8143E A heat engine receives 15 000 btuh at 1400 R and 30 000 btuh at 1800 R rejecting energy by heat transfer at 900 R Assume it is reversible and find the power output How much power could be produced if it could reject energy at To 540 R Solution CV The heat engine this is in steady state Energy Eq 0 AQ AE 1 E AQ AE 2 E AQ AE L E AW E Entropy Eq 0 A Q AE 1 T1 E A Q AE 2 T2 E A Q AE L TL E 0 Q W L Q 1 HE Q 2 Now solve for AQ AE L E from the entropy equation AQ AE L E TL T1 AQ AE 1 E TL T2 AQ AE 2 E A 900 1400E A 15 000 A 900 1800E A 30 000 24 643 Btuh Substitue into the enrgy equation and solve for the work term AW E A AQ AE 1 E AQ AE 2 E AQ AE L E 15 000 30 000 24 643 20 357 Btuh For a low temperature of 540 R we can get AQ AE L2 E A540 900E A AQ AE L E 14 786 btuh AW E A AQ AE 1 E AQ AE 2 E AQ AE L E 15 000 30 000 14 786 30 214 Btuh Remark Notice the large increase in the power output Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8144E The compressor in a refrigerator takes refrigerant R134a in at 15 lbfinA2E A 0 F and compresses it to 125 lbfinA2E A 100 F With the room at 70 F find the reversible heat transfer and the minimum compressor work Solution CV Compressor out to ambient Minimum work in is the reversible work Steady flow 1 inlet and 2 exit WC 1 2 Energy Eq wAcE A hA1E A hA2E A qArevE Entropy Eq sA2E A sA1E A AdqTEA sAgenE A sA1E A qArevE ATAoE A 0 qArevE A TAoE AsA2E A sA1E A qArevE A 52967 R 041262 042288 BtulbmR 543 Btulbm wAc minE A hA1E A hA2E A TAoE AsA2E A sA1E A 167193 181059 543 193 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8145E A compressor in a refrigerator receives R410A at 20 psia 40 F and it brings it up to 100 psia using an actual specific work of 2348 Btulbm in an adiabatic compression Find the specific reversible work Energy Eq 0 hA1E A hA2E A w State 1 F92 hA1E A 11373 Btulbm sA1E A 02772 BtulbmR hA2E A hA1E A w 11373 2348 13721 kJkg State 2 F92 P hA2E A 13721 Btulbm TA2E A 100 F sA2E A 02840 BtulbmR wArevE A ψ A1E A ψ A2E A hA1E A hA2E A TA0E A sA1E A sA2E A 11373 13721 5367 02772 02840 2348 365 1983 Btulbm P s T 1 2 s 2 ac 2 ac 2 s 1 40 100 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8146E Air flows through a constant pressure heating device as shown in Fig P827 It is heated up in a reversible process with a work input of 85 Btulbm air flowing The device exchanges heat with the ambient at 540 R The air enters at 540 R 60 lbfinA2E A Assuming constant specific heat develop an expression for the exit temperature and solve for it CV Total out to TA0E A Energy Eq hA1E A qA0 Erev EA wArevE A hA2E Entropy Eq sA1E A qA0 Erev EATA0E A sA2E A qA0 Erev EA TA0E AsA2E A sA1E A hA2E A hA1E A TA0E AsA2E A sA1E A wArevE A same as from Eq 814 Constant CApE A gives CApE ATA2E A TA1E A TA0E ACApE A ln TA2E ATA1E A 85 The energy equation becomes TA2E A TA0E A lnTA2E ATA1E A TA1E A 85CApE TA1E A 540 R CApE A 024 Btulbm R TA0E A 540 R TA2E A 540 ln TA2E A540 540 85024 89417 R Now trial and error on TA2E At 1400 R LHS 88556 Rtoo low At 1420 R LHS 8979 R Interpolate to get TA2E A 1414 R LHS 89419 R OK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8147E An adiabatic and reversible air compressor takes air in at 15 psia 560 R The air exits at 90 psia at the rate of 08 lbms Determine the minimum compressor work input and repeat for an inlet at 530 R instead Why is the work less for a lower inlet T Compressor Reversible adiabatic constant s Inlet at 560 R TA2SE A TA1E AA P2 EP1 E AA k1 k E A 560 R A90 15E AA 02857E A 9343 R The reversible process ensures the minimum work input AW E A Am E AwASE A Am E ACAP0E A TA2SE A TA1E A 08 024 9343 560 7186 Btus Inlet at 530 R TA2SE A TA1E AA P2 EP1 E AA k1 k E A 530 R A90 15E AA 02857E A 8843 R AWE A Am E AwASE A Am E ACAP0E A TA2SE A TA1E A 08 024 8843 530 680 Btus The work term is reduced due to the reduced specific volume vA1E A recall Eq 715 for the indicated shaft work Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8148E A rock bed consists of 12 000 lbm granite and is at 160 F A small house with lumped mass of 24 000 lbm wood and 2000 lbm iron is at 60 F They are now brought to a uniform final temperature with no external heat transfer by connecting the house and rock bed through some heat engines If the process is reversible find the final temperature and the work done during the process Take CV Total rockbed and heat engine Energy Eq mArockE AuA2E A uA1E A mAwoodE AuA2E A uA1E A mAFeE AuA2E A uA1E A A1E AWA2E Entropy Eq mArockE AsA2E A sA1E A mAwoodE AsA2E A sA1E A mAFeE AsA2E A sA1E A 0 mCArockE Aln A T2 ET1 E A mCAwoodE Aln A T2 ET1 E A mCAFeE Aln A T2 ET1 E A 0 12000 0212 ln TA2E A61967 24000 033 ln TA2E A51967 2000 011 ln TA2E A51967 0 TA2E A 5419 R Now from the energy equation A1E AWA2E A 12 000 02125419 61967 24 000 033 2000 0115419 51967 A1E AWA2E A 16 895 Btu W Q Q H L HE H O U S E cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8149E A basement is flooded with 250 ftA3E A of water at 60 F It is pumped out with a small pump driven by a 075 kW electric motor The hose can reach 25 ft vertically up and to ensure the water can flow over the edge of a dike it should have a velocity of 45 fts at that point generated by a nozzle see Fig P834 Find the maximum flow rate you can get and how fast the basement can be emptied CV Pump plus hose and nozzle single steady state flow For maximum flow rate assume reversible process so from Eq814 wArevE A TAoE A sA2E A sA1E A hA2 totE A hA1 totE A 0 Since we have no heat transfer and reversible s is constant and with a liquid flow T is also constant so hA2E A hA1E A We could also have used Bernoulli equation wArevE A A1 2E AVA 2 2E A gH 0 A1 2E A 45A2E A25 037 1 25778 004044 0032134 007257 Btulbm Recall 1 Btulbm 25 037 ftA2E AsA2E A from conversion in Table A1 Am E A AW E A wArevE A A0751055 Btus E007257 BtulbmE A 9796 lbms m Vv 250 ftA3E A 001603 ftA3E Albm 15 596 lbm Δt m Am E A A15 596 lbm 9796 lbmsE A 1592 sec 265 min Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8150E A heat engine receives 3500 Btuh heat transfer at 1800 R and gives out 2000 Btuh as work with the rest as heat transfer to the ambient What are the fluxes of exergy in and out Exergy flux in AΦ E AH 1 To TH AQ E AH A 1 5367 1800 E A 3500 Btuh 2456 btuh Exergy flux out AΦ E AL 1 To TL AQ E AL 0 TL To The other exergy flux out is the power AΦ E Aout AW E A 2000 Btuh H Q 3500 Btuh W 2000 Btuh L Q T 1800 R amb HE ΦH 2456 Btuh W 2000 Btuh ΦL 0 T 1800 R amb HE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8151E A flow of air at 150 psia 540 R is throttled to 75 psia What is the irreversibility What is the drop in flow exergy assuming an ambient at 77 F A throttle process is constant enthalpy if we neglect kinetic energies Process hAeE A hAiE A so ideal gas TAeE A TAiE Entropy Eq sAeE A sAiE A sAgenE A sAo TeE A sAo TiE A R ln A Pe EPi E A 0 R ln A Pe EPi E sAgenE A A5334 778E A ln A 75 150E A 00475 Btulbm R Eq815 i TAoE A sAgenE A 5367 00475 2549 Btulbm The drop in exergy is exergy destruction which is the irreversibility ψ i 2549 Btulbm P P high low i e cb 150 75 P v i e T s P P i e i e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Irreversibility Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8152E A compressor in a refrigerator receives R410A at 20 psia 40 F and it brings it up to 100 psia 100 F in an adiabatic compression Find the specific work entropy generation and irreversibility States 1 F92 hA1E A 11373 Btulbm sA1E A 02772 BtulbmR 2 F92 hA2E A 13721 Btulbm sA2E A 02840 BtulbmR CV Compressor steady state q 0 no KE PE energies Energy Eq w hA1E A hA2E A 11373 13721 2348 Btulbm Entropy Eq 0 sA1E A sA2E A sAgenE A sAgenE A sA2E A sA1E A 02840 02772 00068 BtulbmR Irreversibility i TAoE A sAgenE A 5367 R 00068 BtulbmR 365 Btulbm v P s T 1 2 s 2 ac 2 ac 2 s 1 40 100 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8153E A constant pressure pistoncylinder contains 4 lbm of water at 1000 psia and 200 F Heat is added from a reservoir at 1200 F to the water until it reaches 1200 F We want to find the total irreversibility in the process Solution CV Piston cylinder out to the reservoir incl the walls Energy Eq muA2E A uA1E A A1E AQA2E A A1E AWA2E Entropy Eq msA2E A sA1E A A1E AQA2E ATAresE A A1E ASA2 genE State 1 hA1E A 16807 Btulbm sA1E A 0294 Btulbm R State 2 hA2E A 161967 Btulbm sA2E A 1726 Btulbm R Process P C A1E AWA2E A PVA2E A VA1E A 1 Q 2 H O 2 1200 F From the energy equation we get A1E AQA2E A muA2E A uA1E A A1E AWA2E A mhA2E A hA1E A 4161967 16807 58064 Btu From the entropy equation we get A1E ASA2 genE A msA2E A sA1E A A1Q2 ETres E A 41726 0294 A 58064 4597 1200E A 22295 BtuR Now the irreversibility is from Eq 818 A1E AIA2E A m A1E AiA2E A TAoE A A1E ASA2 genE A 5367 R 22295 BtuR 1197 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8154E A constant flow of steel parts 4 lbms at 77 F goes into a furnace where they are heat treated to 1600 F by a source at an average 2100 R Find the reversible work and the irreversibility in this process Energy Eq 0 AmE AhAiE A hAeE A AQE A Entropy Eq 0 AmE AsAiE A sAeE A AQE A TAsE A ASE AgenE AQE A AmE AhAeE A hAiE A AmE ACAPE A TAeE A TAiE A 4 lbms 011 BtulbmR 1523 R 670 Btus ASE AgenE A AmE AsAeE A sAiE A AQE A TAsE A AmE ACAPE A lnTAeE ATAiE A AQE A TAsE 4 lbms 011 BtulbmR ln2060537 670 2100 BtusR 02725 BtusR AIE A TA0E ASE AgenE A 537 R 02725 BtusR 1463 Btus The reversible work is equal to the irreversibility plus the actual work Eq812 AW revE A AIE A W ac AIE A 1463 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8155E A cylinder with a piston restrained by a linear spring contains 4 lbm of carbon dioxide at 70 psia 750 F It is cooled to 75 F at which point the pressure is 45 psia Find the reversible work and the irreversibility assuming the heat transfer is with surroundings at 68 F P v 2 1 Linear spring gives A1E AWA2E A APdVEA A1 2E APA1E A PA2E AVA2E A VA1E A A1E AQA2E A muA2E A uA1E A A1E AWA2E Equation of state PV mRT State 1 VA1E A mRTA1E APA1E A 4 351 12097 70 144 16849 ftA3E State 2 VA2E A mRTA2E APA2E A 4 351 5347 45 144 11585 ftA3E A1E AWA2E A A1 2E A70 45 psi 11585 16849 ftA3E A A 144 in2ft2 778 lbfftBtuE 5602 Btu Table F4 at 77 F is CAvE A 0156 BtulbmR A1E AQA2E A mCAvE ATA2E A TA1E A A1E AWA2E A 4 0156 75 750 5602 4772 Btu A1E AWArev 2E A TAoE ASA2E A SA1E A UA2E A UA1E A A1E AQA2E A 1 TAoE ATAHE A TAoE AmsA2E A sA1E A A1E AWAac 2E A A1E AQA2E A TAoE ATAoE TAoE AmCAPE A lnTA2E A TA1E A R lnPA2E A PA1E A A1E AWAac 2E A A1E AQA2E 5277 4 0201 ln A 5347 12097E A A351 778E A ln A45 70E A 5602 4772 3043 5602 4772 1169 Btu A1E AIA2E A A1E AWArev 2E A A1E AWAac 2E A 1169 5602 1729 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8156E Fresh water can be produced from saltwater by evaporation and subsequent condensation An example is shown in Fig P845 where 300lbms saltwater state 1 comes from the condenser in a large power plant The water is throttled to the saturated pressure in the flash evaporator and the vapor state 2 is then condensed by cooling with sea water As the evaporation takes place below atmospheric pressure pumps must bring the liquid water flows back up to P0 Assume that the saltwater has the same properties as pure water the ambient is at 68 F and that there are no external heat transfers With the states as shown in the table below find the irreversibility in the throttling valve and in the condenser State 1 2 5 7 8 T F 86 77 74 63 68 h Btulbm 5408 10951 4209 3108 3609 s Btulbm R 01043 2044 00821 00613 00708 CV Valve AmE A1E A mAexE A AmE A2E A AmE A3E A Energy Eq hA1E A hAeE A Entropy Eq sAiE A sAgenE A sAeE PA2E A PAsatE ATA2E A TA3E A 04641 psia hAeE A hA1E A xAeE A 5408 450810500 0008571 se 008769 0008571 19565 01045 Btulbm R AmE A2E A 1 xAeE A AmE AiE A 1 0008571 300 29744 lbms sgen se si 01045 01043 00002 Btulbm R AIE A AmE ATA0E AsAgenE A 300 lbms 528 R 00002 BtulbmR 3168 Btus CV Condenser Energy Eq AmE A2E AhA2E A AmE A7E AhA7E A AmE A2E AhA5E A AmE A7E AhA8E A AmE A7E A AmE A2E A hA2E A hA5E AhA8E A hA7E A 29744 A10951 4209 3609 3108E A 62516 Albm sE Entropy Eq AmE A2E AsA2E A AmE A7E AsA7E A ASE AgenE A AmE A2E AsA5E A AmE A7E AsA8E AIE A TA0E ASE AgenE A TA0E AAm 2s5 s2 m E 7s8 s7 EA 5282974400821 2044 6251600708 00613 528 10354 5467 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8157E A rock bed consists of 12 000 lbm granite and is at 160 F A small house with lumped mass of 24 000 lbm wood and 2000 lbm iron is at 60 F They are now brought to a uniform final temperature by circulating water between the rock bed and the house Find the final temperature and the irreversibility in the process assuming an ambient at 60 F CV Total Rockbed and house No work no Q irreversible process Energy eq mArockE AuA2E A uA1E A mAwoodE AuA2E A uA1E A mAFeE AuA2E A uA1E A 0 Entropy Eq mArockE AsA2E A sA1E A mAwoodE AsA2E A sA1E A mAFeE AsA2E A sA1E A SAgenE mCArockE Aln A T2 ET1 E A mCAwoodE Aln A T2 ET1 E A mCAFeE Aln A T2 ET1 E A SAgenE Energy eq mCArockE ATA2E A 160 mCAwoodE A mCAFeE ATA2E A 60 0 12 000 0212 TA2E A 160 24 000 033 2000 011TA2E A 60 0 2544 TA2E A 160 7920 220TA2E A 60 0 TA2E A 838 F 5435 R SAgenE A Amis2 s1i EA mCArockE Aln A T2 ET1 E A mCAwoodE Aln A T2 ET1 E A mCAFeE Aln A T2 ET1 E 2544 ln A 5435 61967E A 7920 220 ln A 5435 51967E A 3526 BtuR A1E AIA2E A TA0E A SAgenE A 51967 3526 18 324 Btu Q H O U S E cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8158E Air enters the turbocharger compressor of an automotive engine at 147 lbfinA2E A 90 F and exits at 25 lbfinA2E A as shown in Fig P851 The air is cooled by 90 F in an intercooler before entering the engine The isentropic efficiency of the compressor is 75 Determine the temperature of the air entering the engine and the irreversibility of the compressioncooling process Solution a Compressor First ideal which is reversible adiabatic constant s TA2sE A TA1E AA P2 EP1 E AA k1 k E A 550 R A 25 147E AA 0286E A 6402 R wAsE A CAP0E ATA2sE A TA1E A 0246402 550 2165 Btulbm Now the actual compressor w wAsE AηAsE A 2165075 2887 CAP0E ATA2E A TA1E A 024TA2E A 550 TA2E A 6703 R Cool down 90 F TA3E A 6703 90 5803 R b Irreversibility from Eq815 with rev work from Eq814 q 0 at TH sA3E A sA1E A 024 ln A5803 550E A A5334 778E A ln A 25 147E A 00235 Btulbm R i TsA3E A sA1E A hA3E A hA1E A w TsA3E A sA1E A CAP E ATA3E A TA1E A CAP E ATA1E A TA2E A 55000235 02490 87 Btulbm 3 1 Exhaust 2 Engine W Compressor Cooler C Q C cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Exergy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8159E A heat pump has a coefficient of performance of 2 using a power input of 15000 Btuh Its low temperature is To and the high temperature is 180 F with ambient at To Find the fluxes of exergy associated with the energy fluxes in and out First let us do the energies in and out COP β A Q AE H W E AQ E AH β AW E A 2 15 000 Btuh 30 000 Btuh Energy Eq AQ E AL AQ E AH AW E A 30 000 15 000 15 000 Btuh Exergy flux in AΦ E AL 1 To TL AQ E AL 0 TL To Exergy flux in AΦ E AW AW E A 15 000 Btuh Exergy flux out AΦ E AH 1 To TH AQ E AH A 1 5367 6397 E A 30 000 4830 Btuh Remark It destroys 15 000 4830 10 170 Btuh of exergy H Q W 15 000 Btuh L Q T o 180 F HP cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8160E A steady flow device receives R410A at 125 psia 100 F and it exits at 15 psia 100 F Assume a reversible isothermal process Find the change in specific exergy Solution inlet exit SOLAR COLLECTOR cb Inlet Table F92 hAiE A 13596 Btulbm sAiE A 02763 Btulbm R Exit Table F92 hAeE A 14110 Btulbm sAeE A 03411 Btulbm R From Eq823 or flow terms in Eq838 ψAieE A ψAeE A ψAiE A hAeE A hAiE A TA0E AsAeE A sAiE A 14110 13596 5367 03411 02763 296 Btulbm Remark it is negative due to the pressure loss Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8161E Consider the springtime melting of ice in the mountains which gives cold water running in a river at 34 F while the air temperature is 68 F What is the exergy of the water relative to the temperature of the ambient ψ hA1E A hA0E A TA0E AsA1E A sA0E A flow exergy from Eq822 Approximate both states as saturated liquid ψ 19973 36088 52767 000405 007081 1136 Btulbm Why is it positive As the water is brought to 68 F it can be heated with qALE A from a heat engine using qAHE A from atmosphere TAHE A TA0E A thus giving out work Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8162E Compressed air for machines and tools in a plant is generated by a central compressor receiving air at 15 psia 540 R 1 lbms delivering it at 90 psia to a buffer tank and a distribution pipe After flowing through the tank and pipe the air is at the ambient 540 R at its point of use Assume a reversible adiabatic compressor and find the compressor exit temperature and the increase in air exergy through the compressor CV Compressor Energy 0 hA1E A hA2E A wACE A Entropy 0 s A1E A s A2E A 0 s A2E A s A1E A s A 0 T2E A s A 0 T1E A R lnP A2E AP A1E A find s A 0 T2E A and then TA2E or with constant specific heat TA2E A TA1E AA P2 EP1 E AA k1 k E A 540 R A90 15E AA 02857E A 901 R Exergy increase through the compressor matches with the compressor work ψA2E A ψA1E A hA2E A hA1E A TA0E As A2E A s A1E A hA2E A hA1E A CAP E A TA2E A TA1E A wACE 024 BtulbmR 901 540 R 866 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8163E For the air system in the previous problem find the increase in the air exergy from the inlet to the point of use How much exergy was lost in the flow after the compressor exit From inlet state 1 to final point of use state 3 ψA3E A ψA1E A hA3E A hA1E A TA0E AsA3E A sA1E A 0 TA0E A0 R lnPA3E APA1E A TA0E A R lnPA3E APA1E A 537 R 5334778 BtulbmR ln 9015 6597 Btulbm Compressor exit reversible adiabatic constant s process sA2E A sA1E TA2E A TA1E AA P2 EP1 E AA k1 k E A 540 R A90 15E AA 02857E A 901 R So then ψAlostE A ψA2E A ψA3E A ψA3E A ψA1E A ψA2E A ψA1E A ψA3E A ψA1E A hA2E A hA1E ψA3E A ψA1E A CAP0E ATA2E A TA1E A 6597 kJkg 024 BtulbmR 901 540 R 2067 btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8164E A geothermal source provides 20 lbms of hot water at 80 lbfinA2E A 280 F flowing into a flash evaporator that separates vapor and liquid at 30 lbfinA2E A Find the three fluxes of exergy inlet and two outlets and the irreversibility rate CV Flash evaporator chamber Steady flow with no work or heat transfer Cont Eq m 1 m 2 m 3 Energy Eq m 1h1 m 2h2 m 3h3 Entropy Eq m 1s1 S gen m 2s2 m 3s3 1 2 3 Vap Liq Properties from Table F71 ho 4508 h1 24917 h2 11643 h3 2189 Btulbm so 008769 s1 04098 s2 16997 s3 036815 Btulbm R h1 xh2 1 x h3 x m 2m 1 h1 h3 h2 h3 003202 m 2 xm 1 064 lbms m 3 1xm 1 1936 lbms S gen 064 16997 1936 036815 20 04098 00192 BtusR Flow exergy Eq822 ψ h Tos ho Toso h ho Tos so ψ1 24917 4508 537 04098 008769 31117 Btulbm ψ2 11643 4508 537 16997 008769 25357 Btulbm ψ3 2189 4508 537 036815 008769 2321 Btulbm m 1 ψ1 6223 Btus m 2ψ2 1623 Btus m 3ψ3 4493 Btus Balance of flow terms in Eq838 AI E A m 1 ψ1 m 2 ψ2 m 3ψ3 107 Btus To S gen Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8165E An air compressor is used to charge an initially empty 7ftA3E A tank with air up to 750 lbfinA2E A The air inlet to the compressor is at 147 lbfinA2E A 60 F and the compressor isentropic efficiency is 80 Find the total compressor work and the change in energy of the air CV Tank compressor constant inlet conditions Continuity mA2E A 0 mAinE A Energy mA2E AuA2E A mAinE AhAinE A A1E AWA2E Entropy mA2E AsA2E A mAinE AsAinE A A1E ASA2 GENE To use isentropic efficiency we must calc ideal device State 1 vA1E A RTA1E APA1E A 13103 ftA3E Albm the ambient state Reversible compressor A1E ASA2 GENE A 0 sA2E A sAinE sA o T2E A sA o TinE A R lnA P2 EPin E A 16307 A5334 778E A ln A750 147E A 19003 A Btu lbm RE TA2sE A 1541 R uA2sE A 27449 Btulbm A1E AwA2sE A hAinE A uA2sE A 12438 27449 15011 Btulbm Actual compressor A1E AwA2ACE A A1E AwA2sE AηAcE A 18764 Btulbm uA2ACE A hAinE A A1E AwA2ACE A 312 TA2ACE A 1729 R Final state 2 A u PE A vA2E A RTA2E APA2E A 0854 ft3lbm mA2E A VA2E AvA2E A 82 lbm A1E AWA2E A mA2E AA1E AWA2ACE A 1539 Btu mA2E AφA2E A φA1E A mA2E A uA2E A uA1E A PA0E AvA2E A vA1E A TA0E AsA2E A sA1E A 82 312 88733 1470854 13103A144 778E A 52019311 163074 A5334 778E A lnA750 147E A 82 17394 14263 Btu W cb 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8166E An electric stove has one heating element at 600 F getting 750 W of electric power It transfers 90 of the power to 2 lbm water in a kettle initially at 70 F 1 atm the rest 10 leaks to the room air The water at a uniform T is brought to the boiling point At the start of the process what is the rate of exergy transfer by a electrical input b from heating element and c into the water at Twater We take here the reference T to be the room 70 F 52967 R a Work is exergy AΦ E A AW E A 750 W b Heat transfer at 600 F is only partly exergy AΦ E A 1 To TH AQ E A A 1 52967 45967 600 E A 750 375 W c Water receives heat transfer at 70 F as 90 of 750 W 675 W AΦ E A 1 To Twater AQ E A A 1 52967 45967 70 E A 675 0 W cb 750 W at 600 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8167E A wood bucket 4 lbm with 20 lbm hot liquid water both at 180 F is lowered 1300 ft down into a mineshaft What is the exergy of the bucket and water with respect to the surface ambient at 70 F CV Bucket and water Both thermal exergy and potential energy terms vA1E A vA0E A for both wood and water so work to atm is zero Use constant heat capacity table F2 for wood and table F71 sat liquid for water From Eq828 φA1E A φA0E A mAwoodE AuA1E A uA0E A TA0E AsA1E A sA0E A mH2OuA1E A uA0E A TA0E AsA1E A sA0E A mAtotE AgzA1E A zA0E A 403180 70 03 530 ln A640 530E A 2014776 3809 5300263 0074 24 32174 1300 25 037 1205 1993 401 17125 Btu Recall 1 Btulbm 25 037 ftA2E AsA2E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8168E A 20lbm iron disk brake on a car is at 50 F Suddenly the brake pad hangs up increasing the brake temperature by friction to 230 F while the car maintains constant speed Find the change in exergy of the disk and the energy depletion of the cars gas tank due to this process alone Assume that the engine has a thermal efficiency of 35 All the friction work is turned into internal energy of the disk brake muA2E A uA1E A A1E AQA2E A A1E AWA2E A A1E AQA2E A mAFeE ACAFeE ATA2E A TA1E A A1E AQA2E A 20 0107 230 50 3852 Btu Change in s for a solid Eq811 msA2E A sA1E A mC lnTA2E A TA1E A 20 0107 ln A 690 510 E A 06469 BtuR No change in kinetic or potential energy no volume change and neglect the work to the surroundings at PA0E A so change in exergy is from Eq827 φ muA2E A uA1E A TA0E AmsA2E A sA1E A 3852 510 06469 5528 Btu WAengineE A ηAthE AQAgasE A A1E AQA2E A Friction work QAgasE A A1E AQA2E AηAthE A 3852035 1100 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8169E Apply the exergy equation to find the exergy destruction in Problem 8150E Exergy flux in AΦ E AH 1 To TH AQ E AH A 1 5367 1800 E A 3500 Btuh 2456 Btuh Exergy flux out AΦ E AL 1 To TL AQ E AL 0 TL To The other exergy flux out is the power AΦ E Aout AW E A 2000 Btuh H Q 3500 Btuh W 2000 Btuh L Q T 1800 R amb HE cb ΦH 2456 Btuh W 2000 Btuh ΦL 0 T 1800 R amb HE The exergy balance equation Eq838 for the HE becomes steady state 0 AΦ E AH AΦ E AL AW E A AΦ E Adestruction AΦ E Adestruction AΦ E AH AΦ E AL AW E A 2456 Btuh 0 2000 Btuh 456 Btuh Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8170E The condenser in a power plant cools 20 lbms water at 120 F quality 90 so it comes out as saturated liquid at 120 F The cooling is done by oceanwater coming in at 60 F and returned to the ocean at 68 F Find the transfer out of the water and the transfer into the oceanwater of both energy and exergy 4 terms Solution CV Water line No work but heat transfer out Energy Eq AQE Aout Am E A h1 h2 20101099 8799 18 460 Btus CV Ocean water line No work but heat transfer in equals water heattransfer out Energy Eq q hA4E A hA3E A 3609 2808 80 Btulbm Am E Aocean AQE Aout q 18 460 80 2308 kgs Exergy out of the water follows from flow term Eq838 we will use To 60 F Φ out Am E Aψ1 Am E Aψ2 Am E A h1 h2 To s1 s2 20 101099 8799 519717567 01646 1912 Btus Exergy into the ocean water Φ ocean Am E AoceanψA4E A ψA3E A Am E Aocean hA4E A hA3E A TAoE AsA4E A sA3E A 2308 80 519700708 00555 112 Btus Notice there is a large amount of energy exchanged but very little exergy cb 1 2 3 4 Often the cooling media flows inside a long pipe carrying the energy away Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Device 2nd Law Efficiency Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8171E A heat engine receives 3500 Btuh heat transfer at 1800 R and gives out 1400 Btuh as work with the rest as heat transfer to the ambient Find its first and second law efficiencies First law efficiency is based on the energies ηAIE A AW E AAQ E AH A1400 3500E A 04 The second law efficiency is based on work out versus exergy in Exergy flux in AΦ E AH 1 To TH AQ E AH A 1 5367 1800 E A 3500 Btuh 2456 btuh ηAIIE A A W AE Φ H E A1400 2456E A 057 Notice the exergy flux in is equal to the Carnot heat engine power output given 3500 Btuh at 1800 R and rejecting energy to the ambient H Q 3500 Btuh W 1400 Btuh L Q T 1800 R amb HE ΦH 2456 Btuh W 1400 Btuh ΦL 0 T 1800 R amb HE Energy terms Exergy terms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8172E A heat exchanger increases the exergy of 6 lbms water by 800 btulbm using 20 lbms air coming in at 2500 R and leaving with 250 Btulbm less exergy What are the irreversibility and the second law efficiency CV Heat exchanger steady flow 1 inlet and 1 exit for air and water each The two flows exchange energy with no heat transfer tofrom the outside 3 water 1 air 4 2 The irreversibility is the destruction of exergy exergy so AI E A AΦ E Adestruction AΦ E Ain AΦ E Aout 20 250 6 800 200 Btus The second law efficiency Eq832 ηAIIE A AΦ E Aout AΦ E Ain A 6 800 20 250E A 096 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8173E Find the second law efficiency of the heat pump in problem 8159E The second law efficiency is a ratio of exergies namely what we want out divided by what we have to put in Exergy from first term on RHS Eq 838 AΦ E AH 1 To TH AQ E AH AQ E AH β AW E A 2 15 000 Btuh 30 000 Btuh ηAIIE A A Φ AE H W E 1 To TH A Q AE H W E A 1 5367 6397 E A A30 000 15 000E A 032 H Q W 15 000 Btuh L Q T o 180 F HP cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8174E Find the isentropic efficiency and the second law efficiency for the compressor in Problem 8145 CV Compressor Assume adiabatic and neglect kinetic energies Energy Eq413 w hA1E A hA2E A Entropy Eq78 sA2E A sA1E A sAgenE We have two different cases the ideal and the actual compressor State 1 F92 hA1E A 11373 Btulbm sA1E A 02772 BtulbmR hA2E A hA1E A w 11373 2348 13721 kJkg State 2 F92 P hA2E A 13721 Btulbm TA2E A 100 F sA2E A 02840 BtulbmR State 2s B42 P s sA1E A hA2sE A 13344 Btulbm Actual compressor wACACE A hA2ACE A hA1E A 13721 11373 2348 Btulbm Ideal work wAcsE A hA2sE A hA1E A 13344 11373 1971 Btulbm Definition Eq727 ηAcE A wAcsE AwAcACE A 0839 Rev work Eq829 wArevE A ψ A1E A ψ A2E A hA1E A hA2E A TA0E A sA1E A sA2E A 11373 13721 5367 02772 02840 2348 365 1983 Btulbm 2nd law efficiency ηAIIE A A wrev Ewac E A A1983 2348E A 08445 A wac i Ewac E A P s T 1 2 s 2 ac 2 ac 2 s 1 40 100 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8175E A steam turbine has an inlet at 600 psia and 900 F and actual exit of 1 atm with x 10 Find its first law isentropic and the secondlaw efficiencies CV Steam turbine Energy Eq413 w hAiE A hAeE A Entropy Eq79 sAeE A sAiE A sAgenE Inlet state Table F72 hAiE A 146292 Btulbm sAiE A 16766 Btulbm R Exit actual state F72 hAeE A 115049 Btulbm sAeE A 17567 Btulbm R Actual turbine energy equation w hAiE A hAeE A 31243 Btulbm Ideal turbine reversible process so sAgenE A 0 giving sAesE A sAiE A 16766 03121 xAesE A 14446 xAesE A 094455 hAesE A 18013 094455 97035 109667 The energy equation for the ideal gives wAsE A hAiE A hAesE A 36625 Btulbm The first law efficiency is the ratio of the two work terms ηAsE A wwAsE A 3124336625 0853 The reversible work for the actual turbine states is Eq829 wArevE A hAiE A hAeE A TAoE AsAeE A sAiE A 31243 536717567 16766 31243 4299 3554 Btulbm Second law efficiency Eq831 ηA2nd LawE A wwArevE A 312433554 0879 v P s T i i e s e s 147 psia 600 psia e ac Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8176E A heat engine operating with an evironment at 540 R produces 17 000 Btuh of power output with a first law efficiency of 50 It has a second law efficiency of 80 and TL 560 R Find all the energy and exergy transfers in and out Solution From the definition of the first law efficiency Q H AWE A η A17 000 05E A 34 000 Btuh Energy Eq Q L Q H AWE A 34 000 17 000 17 000 Btuh Φ W AWE A 17 000 Btuh From the definition of the second law efficiency ηII AWE AΦ H this requires that we assume the exergy rejected at 560 R is lost and not counted otherwise the efficiency should be ηII AWE AΦ H Φ L Exergy from source Φ H 1 To TH Q H A17 000 08E A 21 250 Btuh Exergy rejected Φ L 1 To TL Q L 1 A540 560E A 17 000 607 Btuh Notice from the Φ H form we could find the single characteristic TH as 1 To TH 21 250 Btuh Q H 0625 TH 1440 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8177E Air flows into a heat engine at ambient conditions 147 lbfinA2E A 540 R as shown in Fig P8111 Energy is supplied as 540 Btu per lbm air from a 2700 R source and in some part of the process a heat transfer loss of 135 Btu per lbm air happens at 1350 R The air leaves the engine at 147 lbfinA2E A 1440 R Find the first and the secondlaw efficiencies CV Engine out to reservoirs hAiE A qAHE A qALE A hAeE A w Table F5 hAiE A 12918 Btulbm sA o TiE A 163979 Btulbm R hAeE A 353483 Btulbm sA o TeE A 188243 Btulbm R wAacE A 12918 540 135 353483 1807 Btulbm ηATHE A wqAHE A 1807540 0335 For second law efficiency also a q tofrom ambient sAiE A qAHE ATAHE A qA0E ATA0E A qAlossE ATAmE A sAeE qA0E A TA0E A se si qlossTm qHTHE 540A 188243 163979 135 1350 540 E2700 E A 7702 Btulbm wArevE A hAiE A hAeE A qAHE A qAlossE A qA0E A wAacE A qA0E A 2577 Btulbm ηAIIE A wAacE AwArevE A 18072577 070 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8178E Air enters a compressor at ambient conditions 15 psia 540 R and exits at 120 psia If the isentropic compressor efficiency is 85 what is the secondlaw efficiency of the compressor process s 540 R 1 2s 2 15 psia 120 psia T Ideal isentropic Eq623 TA2sE A 540 8A 02857E A 9781 R wAsE A 024 9781 540 10514 Btulbm w A ws E ηs E A A10514 085E A 1237 Btulbm Actual exit temperature TA2E A TA1E A A w CP0 E A 540 A1237 024E A 1055 R Eq616 sA2E A sA1E A 024 ln1055540 5334778 ln 8 001817 BtulbmR Exergy Eq823 ψA2E A ψA1E A hA2E A hA1E A TA0E AsA2E A sA1E A 1237 537 001817 11394 Btulbm 2nd law efficiency Eq832 or 834 but for a compressor ηAIIE A A ψ2 ψ1 EwE A A11394 1237E A 092 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8179E Consider that the air in the previous problem after the compressor flows in a pipe to an air tool and at that point the temperature has dropped to ambient 540 R and an air pressure of 110 psia What is the second law efficiency for the total system CV Compressor Energy 0 hA1E A hA2E A wACE A Entropy 0 s A1E A s A2E A 0 TA2E A TA1E AA P2 EP1 E AA k1 k E A 540 A120 15E AA 02857E A 9781 R Exergy increase through the compressor matches with the ideal compressor work wAC s E A hA2E A hA1E A CAP E ATA2E A TA1E A 0249781 540 Btulbm 10514 Btulbm The actual compressor work is wAC acE A wAC s E A ηAC s E A 10514085 1237 Btulbm From inlet state 1 to final point of use state 3 ψA3E A ψA1E A hA3E A hA1E A TA0E AsA3E A sA1E A 0 TA0E A0 R lnPA3E APA1E A TA0E A R lnPA3E APA1E A 537 R 5334778 BtulbmR ln 11015 7336 Btulbm So then the second law efficiency is the gain ψA3E A ψA1E A over the source wAC acE A as ηAIIE A A ψ3 ψ1 EwC ac E A A7336 1237E A 059 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8180E A compressor is used to bring saturated water vapor at 103 lbfinA2E A up to 2000 lbfinA2E A where the actual exit temperature is 1200 F Find the irreversibility and the second law efficiency Inlet state Table F71 hAiE A 11884 Btulbm sAiE A 1601 Btulbm R Actual compressor F72 hAeE A 15986 Btulbm sAeE A 16398 Btulbm R Energy Eq actual compressor wAcacE A hAeE A hAiE A 4102 Btulbm Eq813 i TA0E AsAeE A sAiE A 53667 16398 1601 2082 Btulbm Eq815 wArevE A i wAcacE A 2082 4102 3894 Btulbm ηAIIE A wArevE AwAcacE A 3894 4102 0949 v P s T i i 103 psia 2000 psia e ac e ac Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8181E A coflowing same direction heat exchanger has one line with 05 lbms oxygen at 68 F and 30 psia entering and the other line has 12 lbms nitrogen at 20 psia and 900 R entering The heat exchanger is long enough so that the two flows exit at the same temperature Use constant heat capacities and find the exit temperature and the second law efficiency for the heat exchanger assuming ambient at 68 F Solution CV Heat exchanger steady 2 flows in and two flows out 1 3 2 4 Energy Eq410 m O2h1 m N2h3 m O2h2 m N2h4 Same exit tempearture so T4 T2 with values from Table F4 m O2CAP O2E AT1 m N2CAP N2E AT3 m O2CAP O2E A m N2CAP N2E AT2 T2 A05 022 5277 12 0249 900 05 022 12 0249E A A32697 04088E A 800 R The second law efficiency for a heat exchanger is the ratio of the exergy gain by one fluid divided by the exergy drop in the other fluid For each flow exergy is Eq823 include mass flow rate as in Eq838 For the oxygen flow m O2ψ2 ψ1 m O2 h2 h1 To s2 s1 m O2 CAPE AT2 T1 To CAPE A lnTA2E A TA1E A R lnPA2E A PA1E A m O2CAPE A T2 T1 TolnTA2E A TA1E A 05 022 800 5277 5367 ln8005277 5389 Btus For the nitrogen flow m N2ψ3 ψ4 m N2CAPE A T3 T4 TolnT3 T4 12 0249 900 800 5367 ln900800 10992 Btus From Eq832 ηA2nd LawE A A m AE O2ψ1 ψ2 m N2ψ3 ψ4 E A 5389 10992E A 049 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8182E Calculate the irreversibility for the process described in Problem 4211E assuming that the heat transfer is with the surroundings at 61 F CV Cylinder volume out to To 61 F Continuity Eq415 mA2E A mA1E A mAinE A Energy Eq416 mA2E AuA2E A mA1E AuA1E A mAinE AhAlineE A A1E AQA2E A A1E AWA2E Entropy Eq712 mA2E A sA2E A mA1E AsA1E A misi A1E AQA2E A To A1E ASA2 genE A Process PA1E A is constant to stops then constant V to state 2 at PA2E State 1 PA1E A TA1E A mA1E A A P1V ERT1 E A A 45 9 144 5334 5197E A 2104 lbm State 2 Open to 60 lbfinA2E A TA2E A 630 R Table F5 hi 26613 btulbm u1 8868 Btulbm u2 10762 Btulbm AIR Only work as piston moves V changes while P P1 until V Vstop 1W2 APdVEA P1Vstop V1 45 36 9A144 778E A 2249 Btu m2 P2V2RT2 A60 36 144 5334 630E A 9256 lbm mi 7152 lbm A1E AQA2E A m2u2 m1u1 mi hi 1W2 9256 10762 2104 8868 7152 26613 2249 8689 Btu Use from table F4 Cp 024 R 5334 778 006856 Btulbm R I To A1E ASA2 genE A To mA1E A sA2E A sA1E A mi sA2E A si A1E AQA2E 52072104Cp ln A 630 5197E A R ln A60 45E A 7152Cpln A 630 1100E A R ln A60 75E A 8689 5207 005569 08473 8689 4567 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8183E Calculate the exergy of the system aluminum plus gas at the initial and final states of Problem 6245E and also the irreversibility State 1 TA1E A 400 F vA1E A 22862 06988 PA1E A 300 psi Ideal gas vA2E A vA1E A300 220537 860 0595 vo 8904 RTo Po The metal does not change volume so the terms as Eq822 are added φA1E A mgasφgas mAlφAl A1E mgasCv TA1E A To mgasTo Cp ln A T1 ToE A R ln A P1 PoE A mgasPo vA1E A vo mAl C TA1E A To ToC ln TA1E A To Al φA1E A 2862015640077 5370201 ln A860 537E A A351 778E A ln A300 147E A 147 06988 8904 A144 778E A 8 021 400 77 537 ln A860 537E A 14396 11778 26174 Btu φA2E A 2862 015677 77 537 0201 ln A537 537E A A351 778E A ln A220 147E A 1470595 8904 A144 778E A 8 021 77 77 537 ln A537 537E A 12291 0 12291 Btu A1E AWA2 CO2 E A A PdVEA 05PA1E A PA2E AVA2E A VA1E A 300 2202 1703 2 A144 778E A 1429 Btu A1E AIA2E A φA1E A φA2E A 1 ToTH A1E AQA2E A A1E AWA2E AAC Pom VA2E A VA1E A 26174 12291 0 1429 147 2862 A144 778E A 0595 06988 1523 Btu Tamb Q CO2 Al Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8184E Air in a pistoncylinder arrangement shown in Fig P8135 is at 30 lbfinA2E A 540 R with a volume of 20 ft3 If the piston is at the stops the volume is 40 ft3 and a pressure of 60 lbfinA2E A is required The air is then heated from the initial state to 2700 R by a 3400 R reservoir Find the total irreversibility in the process assuming surroundings are at 70 F Solution Energy Eq mu2 u1 1Q2 1W2 Entropy Eq ms2 s1 A dQTEA 1S2 gen Process P P0 αVV0 if V Vstop Information Pstop P0 αVstopV0 Eq of state Tstop T1PstopVstopP1V1 2160 T2 So the piston will hit the stops V2 Vstop P2 T2Tstop Pstop 27002160 60 75 psia 25 P1 State 1 m2 m1 P1V1 RT1 A30 20 144 5334 540E 30 lbm 2 P v 1a 1 v vstop 1 Air Q Tres 1W2 A1 2E AP1 PstopVstop V1 A1 2E A30 6040 20 1666 Btu 1Q2 mu2 u1 1W2 3518165 9216 1666 14446 Btu s2 s1 sA o T2E A sA o T1E A R lnP2P1 20561 16398 5334778 ln 25 03535 Btulbm R Take control volume as total out to reservoir at TRES 1S2 gen tot ms2 s2 1Q2TRES 06356 BtuR A1E AIA2E A TA0E A 1S2 gen E A 530 06356 337 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8185E A pistoncylinder arrangement has a load on the piston so it maintains constant pressure It contains 1 lbm of steam at 80 lbfinA2E A 50 quality Heat from a reservoir at 1300 F brings the steam to 1000 F Find the secondlaw efficiency for this process Note that no formula is given for this particular case so determine a reasonable expression for it 1 P1 x1 v1 27458 ftA3E Albm h1 732905 Btulbm s1 10374 Btulbm R 2 P2 P1T2 v2 10831 ftA3E Albm h2 15326 Btulbm s2 19453 Btulbm R mu2 u1 1Q2 1W2 1Q2 PV2 V1 1Q2 mu2 u1 Pmv2 v1 mh2 h1 7997 Btu 1W2 Pmv2 v1 11972 Btu 1W2 to atm P0mv2 v1 22 Btu Useful work out 1W2 1W2 to atm 11972 22 9772 Btu φreservoir 1 T0 Tres 1Q2 A 1 53667 175967 E A 7997 556 Btu nII Wnetφ A9772 556E A 0176 Remark You could argue that the stored exergy exergy should be accounted for in the second law efficiency but it is not available from this device alone Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 8186E The exit nozzle in a jet engine receives air at 2100 R 20 psia with negligible kinetic energy The exit pressure is 10 psia and the actual exit temperature is 1780 R What is the actual exit velocity and the second law efficiency Solution CV Nozzle with air has no work no heat transfer Energy eq hAiE A hAeE A A1 2E AVA2 exE Entropy Eq sAiE A sAgenE A sAeE A A1 2E AVA2 exE A hAiE A hAeE A 53257 44436 8821 Btulbm VAexE A A 2 8821 25 037EA 2102 ft s 1 Recall 1 Btulbm 25 037 ftA2E AsA2E A This was the actual nozzle Now we can do the reversible nozzle which then must have a q Energy eq hAiE A q hAeE A A1 2E AVA2 ex revE Entropy Eq sAiE A qTAoE A sAeE A q TAoE A sAeE A sAiE A q TAoE A C ln A Te ETi E A R ln A Pe EPi E A 5367 024 ln A1780 2100E A A 5334 77817E A ln A10 20E A 4198 Btulbm A1 2E AVA2 ex revE A hAiE A q hAeE A 8821 4198 92408 Btulbm ηAIIE A A1 2E AVA2 exE A A1 2E AVA2 ex revE A 8821 92408 095 Notice the reversible nozzle is not isentropic there is a heat transfer Updated June 2013 SOLUTION MANUAL CHAPTER 9 Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CONTENT CHAPTER 9 SUBSECTION PROB NO Intext concept questions af ConceptStudy guide problems 112 Rankine cycles power plants Simple cycles 1332 Reheat cycles 3339 Open feedwater heaters 4049 Closed feedwater heaters 5060 Nonideal cycles 6175 Combined Heat and Power 7682 Refrigeration cycles 83105 Extended refrigeration cycles 106110 Ammonia absorption cycles 111115 Exergy Concepts 116128 refrigeration cycles 129132 Combined cycles 133137 Review Problems 138146 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful InText Concept Questions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9a Consider a Rankine cycle without superheat How many single properties are needed to determine the cycle Repeat the answer for a cycle with superheat a No superheat Two single properties High pressure or temperature and low pressure or temperature This assumes the condenser output is saturated liquid and the boiler output is saturated vapor Physically the high pressure is determined by the pump and the low temperature is determined by the cooling medium and the heat echanger b Superheat Three single properties High pressure and temperature and low pressure or temperature This assumes the condenser output is saturated liquid Physically the high pressure is determined by the pump and the high temperature by the heat transfer from the hot source The low temperature is determined by the cooling medium and the heat echanger 9b Which component determines the high pressure in a Rankine cycle What determines the low pressure The high pressure in the Rankine cycle is determined by the pump The low pressure is determined as the saturation pressure for the temperature you can cool to in the condenser Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9c What is the difference between an open and a closed feedwater heater The open feedwater heater mixes the two flows at the extraction pressure and thus requires two feedwater pumps The closed feedwater heater does not mix the flows but let them exchange energy it is a two fluid heat exchanger The flows do not have to be at the same pressure The condensing source flow is dumped into the next lower pressure feedwater heater or the condenser or it is pumped up to line pressure by a drip pump and added to the feedwater line 9d In a cogenerating power plant what is cogenerated The electricity is cogenerated The main product is a steam supply Such an arrangement is also known as Combined Heat and Power CHP Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9e A refrigerator in my 20oC kitchen uses R134a and I want to make ice cubes at 5oC What is the minimum high P and the maximum low P it can use Since the R134a must give heat transfer out to the kitchen air at 20oC it must at least be that hot at state 3 From Table B51 P3 P2 Psat 573 kPa is minimum high P Since the R134a must absorb heat transfer at the freezers 5oC it must at least be that cold at state 4 From Table B51 P1 P4 Psat 245 kPa is maximum low P 9f How many parameters are needed to completely determine a standard vapor compression refrigeration cycle Two parameters The high pressure and the low pressure This assumes the exit of the condenser is saturated liquid and the exit of the evaporator is saturated vapor Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful ConceptStudy Guide Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 91 Is a steam power plant running in a Carnot cycle Name the four processes No It runs in a Rankine cycle 12 An isentropic compression constant s Pump 23 An isobaric heating constant P Boiler 34 An isentropic expansion constant s Turbine 41 An isobaric cooling heat rejection constant P Condenser Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 92 Raising the boiler pressure in a Rankine cycle for fixed superheat and condenser temperatures in what direction do these change turbine work pump work and turbine exit T or x Turbine work about the same P up but v down Turbine exit T same if it was twophase down if sup vapor Turbine exit x down Pump work up T s 1 2 3 4 3 4 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 93 For other properties fixed in a Rankine cycle raising the condenser temperature causes changes in which work and heat transfer terms This results in less turbine work out An increase in heat rejection A small reduction in pump work A small reduction in boiler heat addition P v 1 2 3 4 1 2 4 T s 1 2 3 4 2 1 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 94 Mention two benefits of a reheat cycle The reheat raises the average temperature at which you add heat The reheat process brings the states at the lower pressure further out in the superheated vapor region and thus raises the quality if twophase in the last turbine section Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 95 What is the benefit of the moisture separator in the powerplant of Problem 4121 You avoid larger droplets in the turbine and raise the quality for the later stages 96 Instead of the moisture separator in Problem 4121 what could have been done to remove any liquid in the flow A reheat could be done to reboil the liquid and even superheat it Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 97 Can the energy removed in a power plant condenser be useful Yes In some applications it can be used for heating buildings locally or as district heating Other uses could be to heat greenhouses or as general process steam in a food process or paper mill These applications are all based on economics and scale The condenser then has to operate at a higher temperature than it otherwise would Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 98 If the district heating see Fig11 should supply hot water at 90oC what is the lowest possible condenser pressure with water as the working substance The condenser temperature must be higher than 90oC for which the saturation pressure is 7014 kPa P 7014 kPa Pipes in the power plant Pipes in the ground for distribution Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 99 What is the mass flow rate through the condensate pump in Fig 914 We need to check the continuity equation around several CVs Do control volume around HP turbine Number in 1000 kgh 0 320 28 28 12 out to LP turbine out to LP turbine 252 000 kgh which matches with Fig 227 000 in condenser 25 000 from trap Condensate pump main has 252 000 kgh Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 910 A heat pump for a 20oC house uses R410A and the outside is at 5oC What is the minimum high P and the maximum low P it can use As the heat pump must be able to heat at 20oC that becomes the smallest possible condensing temperature and thus P Psat 1444 kPa It must absorb heat from 5oC and thus must be colder in the evaporation process so P Psat 679 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 911 A heat pump uses carbon dioxide and it is required that it condenses at a minimum of 22oC and receives energy from the outside on a winter day at 10oC What restrictions does that place on the operating pressures The high pressure P Psat 6003 kPa close to critical P 7377 kPa The low pressure P Psat 2649 kPa Notice for carbon dioxide that the low pressure is fairly high Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 912 Since any heat transfer is driven by a temperature difference how does that affect all the real cycles relative to the ideal cycles Heat transfers are given as Q CA T so to have a reasonable rate the area and the temperature difference must be large The working substance then must have a different temperature than the ambient it exchanges energy with This gives a smaller temperature difference for a heat engine with a lower efficiency as a result The refrigerator or heat pump must have the working substance with a higher temperature difference than the reservoirs and thus a lower coefficient of performance COP The smaller CA is the larger T must be for a certain magnitude of the heat transfer rate This can be a design problem think about the front end air intake grill for a modern car which is very small compared to a car 20 years ago Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Simple Rankine cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 913 A steam power plant as shown in Fig 93 operating in a Rankine cycle has saturated vapor at 30 MPa leaving the boiler The turbine exhausts to the condenser operating at 10 kPa Find the specific work and heat transfer in each of the ideal components and the cycle efficiency Solution Cycle is determined by P3 x3 P1 P4 assume x1 0 CV Pump Reversible and adiabatic Energy wp h2 h1 Entropy s2 s1 since incompressible it is easier to find work positive in as wp v dP v1 P2 P1 000101 3000 10 302 kJkg h2 h1 wp 19181 302 19483 kJkg CV Boiler qH h3 h2 280414 19483 26093 kJkg CV Turbine wT h3 h4 s4 s3 s4 s3 61869 06492 x4 7501 x4 07383 h4 19181 07383 239282 195834 kJkg wT 280414 195834 8458 kJkg CV Condenser qL h4 h1 195834 19181 17665 kJkg ηcycle wnet qH wT wp qH 8458 30 26093 0323 Q W T 3 2 4 1 Condenser Boiler Turbine W P Q B T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 914 Consider a solarenergypowered ideal Rankine cycle that uses water as the working fluid Saturated vapor leaves the solar collector at 1000 kPa and the condenser pressure is 10 kPa Determine the thermal efficiency of this cycle Solution CV H2O ideal Rankine cycle Cycle is determined by P3 x3 P1 P4 assume x1 0 State 3 1000 kPa sat vap h3 277808 kJkg s3 65864 kJkg K CV Turbine adiabatic and reversible so second law gives s4 s3 65864 06492 x4 75010 x4 07915 h4 19181 07915 239282 208573 kJkg The energy equation gives wT h3 h4 277808 208573 69235 kJkg CV pump and incompressible liquid gives work into pump wP v1P2 P1 0001011000 10 10 kJkg h2 h1 wP 19181 10 19281 kJkg CV boiler gives the heat transfer from the energy equation as qH h3 h2 277808 19281 25853 kJkg The cycle net work and efficiency are found as wNET wT wP 69235 10 69135 kJkg ηTH wNETqH 6913525853 0267 Q W T 3 2 4 1 Condenser Solar Turbine W P Q RAD collector T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 915 The power plant in the previous problem is augmented with a natural gas burner to superheat the water to 300oC before entering the turbine Find the cycle efficiency with this configuration and the specific heat transfer added by the natural gas burner Solution CV H2O ideal Rankine cycle Cycle is determined by P3 T3 P1 P4 assume x1 0 State 3 1000 kPa 300oC h3 305115 kJkg s3 71228 kJkg K CV Turbine adiabatic and reversible so second law gives s4 s3 71228 06492 x4 75010 x4 086303 h4 19181 086303 239282 225688 kJkg The energy equation gives wT h3 h4 305115 225688 79427 kJkg CV pump and incompressible liquid gives work into pump wP v1P2 P1 0001011000 10 10 kJkg h2 h1 wP 19181 10 19281 kJkg CV boiler gives the heat transfer from the energy equation as qH h3 h2 305115 19281 28583 kJkg qH gas h3 g 1000 kPa 305115 277808 27307 kJkg The cycle net work and efficiency are found as wNET wT wP 79427 10 79327 kJkg ηTH wNETqH 7932728583 0277 Q W T 3 2 4 1 Condenser Solar Turbine W P Q RAD collector Q burner T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 916 A utility runs a Rankine cycle with a water boiler at 30 MPa and the cycle has the highest and lowest temperatures of 450C and 60C respectively Find the plant efficiency and the efficiency of a Carnot cycle with the same temperatures Solution The states properties from Tables B11 and B13 1 60oC x 0 h1 25111 v1 0001017 m3kg Psat 1994 kPa 3 30 MPa 450oC h3 3344 kJkg s3 70833 kJkg K CV Pump Reversible and adiabatic Energy wp h2 h1 Entropy s2 s1 since incompressible it is easier to find work positive in as wp v dP v1 P2 P1 0001017 3000 1994 303 kJkg h2 h1 wp 25111 303 25414 kJkg CV Boiler qH h3 h2 3344 25414 308986 kJkg CV Turbine wT h3 h4 s4 s3 s4 s3 70833 08311 x4 70784 x4 08833 h4 25111 08833 235848 233436 kJkg wT 3344 233436 10096 kJkg CV Condenser qL h4 h1 233436 25111 208325 kJkg ηcycle wnet qH wT wp qH 10096 303 308986 0326 ηcarnot 1 TL TH 1 27315 60 27315 450 054 Q W T 3 2 4 1 Condenser Boiler Turbine W P Q B T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 917 The power plant in the previous problem has a too low quality in the low pressure turbine section so the plant wants to increase the superheat What should the superheat be so the quality of the water in the turbine stays above 92 Cycle is determined by P3 T1 x4 assume x1 0 Consider the turbine and its exhaust at condenser 60C 92 quality CV Turbine wT h3 h4 s4 s3 x4 092 s4 s3 08311 x4 70784 734323 kJkgK Look in B13 3 MPa match s T3 500 100 734323 72337 75084 72337 5399C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 918 A power plant for a polar expedition uses ammonia which is heated to 80oC at 1000 kPa in the boiler and the condenser is maintained at 15oC Find the cycle efficiency Solution Standard Rankine cycle with superheat From the listed information we get from Table B22 State 1 h1 11166 kJkg v1 0001519 m3kg P1 2363 kPa s 04538 kJkgK State 3 h3 16146 kJkg s3 54971 kJkgK CV Tubine Energy wTs h3 h4 Entropy s4 s3 54971 kJkg K x4 sfg s4 sf 54971 04538 50859 09916 h4 11166 09916 13129 141356 kJkg wTs 16146 141356 20104 kJkg CV Pump wP v dP v1P2 P1 00015191000 2363 116 kJkg h2 h1 wP 11166 116 1128 kJkg CV Boiler qH h3 h2 16146 1128 15018 kJkg ηCYCLE wNETqH 20104 116 15018 0133 P v 1 2 3 4 T s 1 2 3 4 Comment The cycle efficiency is low due to the low high temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 919 A Rankine cycle with R410A has the boiler at 3 MPa superheating to 180oC and the condenser operates at 800 kPa Find all four energy transfers and the cycle efficiency Standard Rankine cycle given by P2 P3 T3 P1 assume x1 0 From Table B4 State 1 P x v1 0000855 m3kg h1 5776 kJkg at 0oC State 3 P T h3 44509 kJkg s3 13661 kJkgK State 4 800 kPa s s3 h4 38597 kJkg interpolated superheated vapor CV Pump wP v dP v1P2 P1 0000855 3000 800 1881 kJkg h2 h1 wP 5776 1881 5964 kJkg CV Boiler qH h3 h2 44509 5964 38545 kJkg CV Tubine Energy wTs h3 h4 44509 38597 5912 kJkg CV Condenser qL h4 h1 38597 5776 32821 kJkg ηCYCLE wNETqH 5912 1881 38545 0148 P v 1 2 3 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 920 A steam power plant has a high pressure of 3 MPa and it maintains 60oC in the condenser A condensing turbine is used but the quality should not be lower than 90 at any state in the turbine For a turbine power output of 8 MW find the work and heat transfer in all components and the cycle efficiency Solution Based on the standard Rankine cycle and Table B1 State 1 Sat liquid P1 1994 kPa h1 25111 kJkg v1 0001017 m3kg Consider CV pump Energy h2 h1 wp v1 P2 P1 0001017 3000 1994 303 kJkg State 2 P2 3000 kPa h2 h1 wp 25111 303 2541 kJkg State 4 P4 P1 1994 kPa x 09 s4 sf x4 sfg 08311 09 70784 720166 kJkgK h4 hf x4 hfg 25111 09 235848 237374 kJkg Consider the turbine for which s4 s3 State 3 Table B22 3000 kPa s3 720166 kJkg K h3 34325 kJkg Turbine wT h3 h4 34325 237374 10588 kJkg Boiler qH h3 h2 34325 2541 31784 kJkg Condenser qL h4 h 237374 2511 21226 kJkg Efficiency ηTH wNETqH wT wPqH 10588 303 31784 0332 Scaling m W TwT 8 000 kW 10588 kJkg 7555 kgs W p m wp 229 kW Q H m qH 24 MW Q L m qL 16 MW Q W T 3 2 4 1 Condenser Boiler Turbine W P Q B P v 1 2 3 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 921 A supply of geothermal hot water is to be used as the energy source in an ideal Rankine cycle with R134a as the cycle working fluid Saturated vapor R134a leaves the boiler at a temperature of 85C and the condenser temperature is 40C Calculate the thermal efficiency of this cycle Solution CV Pump use R134a Table B5 wP h2 h1 1 2 vdP v1P2P1 000087329262 10170 167 kJkg h2 h1 wP 25654 167 25821 kJkg CV Boiler qH h3 h2 42810 25821 16989 kJkg CV Turbine s4 s3 16782 11909 x4 05214 x4 09346 h4 25654 09346 16328 40914 kJkg Energy Eq wT h3 h4 4281 40914 1896 kJkg wNET wT wP 1896 167 1729 kJkg ηTH wNETqH 172916989 0102 W T Q H W P in Q L 3 2 1 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 922 Do Problem 921 with R410A as the working fluid and boiler exit at 4000 kPa 70C A supply of geothermal hot water is to be used as the energy source in an ideal Rankine cycle with R134a as the cycle working fluid Saturated vapor R134a leaves the boiler at a temperature of 85C and the condenser temperature is 40C Calculate the thermal efficiency of this cycle Solution CV Pump use R410A Table B4 wP h2 h1 1 2 vdP v1P2P1 00010254000 24207 1619 kJkg h2 h1 wP 12409 1619 12571 kJkg CV Boiler qH h3 h2 28788 12571 16217 kJkg CV Turbine s4 s3 093396 04473 x4 05079 x4 09582 h4 12409 09582 15904 27648 kJkg wT h3 h4 28788 27648 114 kJkg ηTH wNETqH 114 16216217 0060 W T Q H W P in Q L 3 2 1 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 923 A low temperature power plant operates with R410A maintaining 20oC in the condenser a high pressure of 3 MPa with superheat Find the temperature out of the boilersuperheater so the turbine exit temperature is 60oC and find the overall cycle efficiency State 1 P1 3996 kPa v1 0000803 m3kg h1 2824 kJkg State 4 P4 P1 400 kPa h4 34358 kJkg s4 13242 kJkgK State 3 3 MPa s s4 h3 42656 kJkg T3 1636oC Pump wp v1 P2 P1 0000803 m3kg 3000 3996 kPa 209 kJkg Boiler qH h3 h2 42656 2824 209 39623 kJkg Turbine wT h3 h4 42656 34358 8298 kJkg Efficiency ηTH wNETqH wT wPqH 8298 209 39623 0204 P v 1 2 3 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 924 A low temperature power plant should produce a turbine work of 25 MW It operates with carbon dioxide maintaining 10oC in the condenser a high pressure of 6 MPa and it superheats to 100oC Find the turbine exit temperature the mass flow rate required and the overall cycle efficiency State 1 v1 0001017 m3kg h1 6362 kJkg P1 26487 kPa State 3 h3 42169 kJkg s3 14241 kJkgK State 4 26487 kPa s s3 Do a double linear interpolation in B32 At 2000 kPa 14775oC h 36284 At 3000 kPa 44096oC h 38298 Then at 26487 kPa T4 3379oC h4 3759 kJkg use CATT3 entry P T for accuracy then T 345oC h 3763 kJkg CV Pump wP v dP v1P2 P1 0001017 6000 26487 3408 kJkg h2 h1 wP 6362 3408 6703 kJkg CV Boiler qH h3 h2 42169 6703 35466 kJkg CV Tubine Energy wTs h3 h4 42169 3759 4579 kJkg Scaling m W TwT 2 500 kW 4579 kJkg 546 kgs ηCYCLE wNETqH 4579 3408 35466 0129 Q W T 3 2 4 1 Condenser Boiler Turbine W P Q B T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Here are the CATT3 entries I used for locating state 4 Start out at P 26487 kPa 26487 MPa T 34oC the s is then too low For some reason the entry Ps does not work it iterates out of bounce to be fixed in next edition of the CATT program Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 925 Do Problem 921 with ammonia as the working fluid A supply of geothermal hot water is to be used as the energy source in an ideal Rankine cycle with R134a as the cycle working fluid Saturated vapor R134a leaves the boiler at a temperature of 85C and the condenser temperature is 40C Calculate the thermal efficiency of this cycle Solution CV Pump use Ammonia Table B2 wP h2 h1 1 2 vdP v1P2 P1 000172546086 15549 527 kJkg h2 h1 wP 37143 527 3767 kJkg CV Boiler qH h3 h2 14478 3767 10711 kJkg CV Turbine s4 s3 43901 13574 x4 35088 x4 08643 h4 37143 08643 10988 132113 kJkg Energy Eq wT h3 h4 14478 132113 12667 kJkg wNET wT wP 12667 527 1214 kJkg ηTH wNETqH 121410711 0113 W T Q H W P in Q L 3 2 1 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 926 Geothermal water can be used directly as a source to a steam turbine Consider 10 kgs water at 500 kPa 150oC brought to a flash chamber where it is throttled to 200 kPa as shown in Fig P926 From the chamber saturated vapor at 200 kPa flows to the turbine with an exit at 10 kPa From state 4 it is cooled in a condenser and pumped back into the ground Determine quality at the turbine exit and the power that can be obtained from the turbine Solution 1 2 3 Twophase out of the valve The liquid drops to the bottom Continuity Eq49 m 1 m 2 m 3 Energy Eq410 m 1h1 m 2h2 m 3h3 Entropy Eq77 m 1s1 S gen dQ T m 2s2 m 3s3 Process Q 0 irreversible throttle B11 h1 63218 kJkg s1 18417 kJkg K B12 h3 270663 kJkg s3 71271 kJkg K h2 50468 kJkg From the energy equation we solve for the flow rate out to turbine m 3 m 1h1 h2h3 h2 10 00579 0579 kgs The flow through the turbine is rev and adiabatic thus isentropic State 4 s4 s3 71271 06492 x4 75010 x4 08636 h4 19181 x4 239282 225825 kJkg wT h3 h4 270663 225825 44838 kJkg W T m 3 wT 0579 kgs 44838 kJkg 2596 kW v P s T 1 1 3 200 kPa 500 kPa 2 3 2 4 10 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 927 Some geothermal location have higher pressure and temperature so assume the geothermal power plant as in problem 926 but with a supply as saturated liquid at 250oC For this case the flash chamber operates at 1000 kPa For a supply of 10 kgs determine the mass flow that goes through the turbine and the power output Solution 1 2 3 Twophase out of the valve The liquid drops to the bottom Continuity Eq49 m 1 m 2 m 3 Energy Eq410 m 1h1 m 2h2 m 3h3 Entropy Eq77 m 1s1 S gen dQ T m 2s2 m 3s3 Process Q 0 irreversible throttle B11 h1 108534 kJkg B12 h3 277808 kJkg hfg 201529 kJkg h2 76279 kJkg From the energy equation we solve for the flow rate out to turbine m 3 m 1 h1 h2 h3 h2 10 108534 76279 201529 10 016 16 kgs The flow through the turbine is rev and adiabatic thus isentropic State 4 s4 s3 65864 06492 x4 75010 x4 079152 h4 19181 x4 239282 2085775 kJkg wT h3 h4 277808 2085775 6923 kJkg W T m 3 wT 16 kgs 6923 kJkg 11077 kW v P s T 1 1 3 1000 kPa 3973 kPa 2 3 2 4 10 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 928 With a higher supply pressure and temperature of the geothermal source it is possible to have two flash evaporators as shown in Fig P928 Assume the supply is saturated water at 250oC and the first chamber flashes to 2000 kPa and saturated liquid at state 2 is flashed to 500 kPa with the saturated vapor out added to the turbine which has an exit state of 10 kPa with quality 78 For a supply of 10 kgs determine the mass flow that goes through the turbine and the power output The enthalpies in kJkg for the states in the flow diagram are h1 108534 h2 90877 h3 279951 hfg 2000 kPa 189074 h6 64021 h5 274867 hfg 500 kPa 210847 h4 19181 078 239282 205821 To determine the mass flow rates we need the split in the two chambers The fractions of the vapor output flows equals the quality in the flow right after the valve Flash Chamber 1 m 1 h1 m 3 h3 m 1 m 3 h2 x1a m 3m 1 h1 h2hfg 2000 kPa 108534 90877 189074 0093387 m 3 x1a m 1 0093387 10 kgs 093387 kgs Flash Chamber 2 m 2 h2 m 5 h5 m 2 m 5 h6 x2a m 5m 2 h2 h6hfg 500 kPa 90877 64021 210847 0127372 m 5 x2a m 2 0127372 10 093387 kgs 115477 kgs CV Turbine Energy Eq gives power out put W T m 3 h3 m 5 h5 m 3 m 5 h4 093387 279951 115477 274867 208864 205821 14896 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 929 A smaller power plant produces 25 kgs steam at 3 MPa 600oC in the boiler It cools the condenser with ocean water coming in at 15oC and returned at 18oC so the condenser exit is at 50oC Find the net power output and the required mass flow rate of ocean water Solution The states properties from Tables B11 and B13 1 45oC x 0 h1 18842 kJkg v1 000101 m3kg Psat 959 kPa 3 30 MPa 600oC h3 368234 kJkg s3 75084 kJkg K CV Pump Reversible and adiabatic Energy wp h2 h1 Entropy s2 s1 since incompressible it is easier to find work positive in as wp v dP v1 P2 P1 000101 m3kg 3000 96 kPa 302 kJkg CV Turbine wT h3 h4 s4 s3 s4 s3 75084 06386 x4 75261 x4 09128 h4 18842 09128 239477 23744 kJkg wT 368234 23744 130794 kJkg W NET m wT wp 25 130794 302 326 MW CV Condenser qL h4 h1 23744 18842 2186 kJkg Q L m qL 25 kgs 2186 kJkg 5465 MW m ocean Cp T m ocean Q L Cp T 54 650 418 18 15 4358 kgs Q W T 3 2 4 1 Condenser Boiler Turbine W P Q B T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 930 Consider an ideal Rankine cycle using water with a highpressure side of the cycle at a supercritical pressure Such a cycle has a potential advantage of minimizing local temperature differences between the fluids in the steam generator such as the instance in which the hightemperature energy source is the hot exhaust gas from a gasturbine engine Calculate the thermal efficiency of the cycle if the state entering the turbine is 30 MPa 550C and the condenser pressure is 10 kPa What is the steam quality at the turbine exit Solution For the efficiency we need the net work and steam generator heat transfer CV Pump For this high exit pressure we use Table B14 compressed liquid State 1 s1 06492 kJkg K h1 19181 kJkg Entropy Eq s2 s1 h2 2225 kJkg wp h2 h1 3069 kJkg CV Turbine Assume reversible and adiabatic Entropy Eq s4 s3 60342 06492 x4 7501 x4 07179 Very low for a turbine exhaust h4 19181 x4 239282 190963 h3 327536 kJkg wT h3 h4 13657 kJkg Steam generator qH h3 h2 30529 kJkg wNET wT wp 13657 3069 1335 kJkg η wNETqH 1335 30529 0437 P v 1 2 3 4 T s 1 2 3 4 10 kPa 30 MPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 931 Consider the ammonia Rankinecycle power plant shown in Fig P931 The plant was designed to operate in a location where the ocean water temperature is 25C near the surface and 5C at some greater depth The mass flow rate of the working fluid is 1000 kgs a Determine the turbine power output and the pump power input for the cycle b Determine the mass flow rate of water through each heat exchanger c What is the thermal efficiency of this power plant Solution a CV Turbine Assume reversible and adiabatic s2 s1 50863 08779 x2 43269 x2 09726 h2 22708 09726 122509 14186 kJkg wT h1 h2 146029 14186 4169 kJkg W T m wT 1000 4169 41 690 kW Pump wP v3P4 P3 00016857 615 0387 kJkg W P m wP 1000 0387 387 kW b Consider condenser heat transfer to the low T water Q to low T H2O 100014186 22708 11915106 kW m low T H2O 11915106 2938 2098 141 850 kgs h4 h3 wP 22708 039 22747 kJkg Now consider the boiler heat transfer from the high T water Q from high T H2O 1000146029 22747 12328106 kW m high T H2O 12328106 10487 9650 147 290 kgs c ηTH W NETQ H 41 690 387 12328106 0033 W T Q H W P in Q L 3 2 1 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 932 Do problem 931 with carbon dioxide as the working fluid Consider the ammonia Rankinecycle power plant shown in Fig P931 The plant was designed to operate in a location where the ocean water temperature is 25C near the surface and 5C at some greater depth The mass flow rate of the working fluid is 1000 kgs a Determine the turbine power output and the pump power input for the cycle b Determine the mass flow rate of water through each heat exchanger c What is the thermal efficiency of this power plant a CV Turbine Assume reversible and adiabatic s2 s1 10406 04228 x2 06963 x2 088726 h2 11283 088726 19715 28775 kJkg wT h1 h2 29496 28775 7206 kJkg W T m wT 1000 7206 7 206 kW Pump wP v3P4 P3 00011615729 4502 1425 kJkg W P m wP 1000 1425 1425 kW b Consider condenser heat transfer to the low T water Q to low T H2O 100028775 11283 02749 106 kW m low T H2O 02749 106 2938 2098 32 728 kgs h4 h3 wP 11283 1425 11426 kJkg Now consider the boiler heat transfer from the high T water Q from high T H2O 100029496 11426 01807 106 kW m high T H2O 01807 106 10487 9650 21 589 kgs c ηTH W NETQ H 7 206 1425 01807 106 0032 W T Q H W P in Q L 3 2 1 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Reheat Cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 933 The power plant in problem 916 has a too low quality in the low pressure turbine section so the plant wants to apply reheat at 800 kPa What should the superheat for the reheat be so the turbine quality stays above 92 The reheat cycle has expansion in the turbine from state 3 to 800 kPa then reheat to state 5 and then expansion in the turbine to the new exhaust state 4 We can then relate states 4 and 5 as follows CV Turbine wT2 h5 h4 s4 s5 and x4 092 s4 s5 08311 x4 70784 734323 kJkgK State 5 800 kPa s 734323 kJkgK From Table B13 at 800 kPa T5 300 50 734323 72327 74088 72327 3314oC T s 1 2 3 4 4 5 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 934 Consider the supercritical cycle in problem 930 and assume the turbine first expands to 3 MPa then a reheat to 500oC with a further expansion in the low pressure turbine to 10 kPa Find the combined specific turbine work and the total specific heat transfer in the boiler For the efficiency we need the net work and steam generator heat transfer CV Pump For this high exit pressure we use Table B14 State 1 s1 06492 kJkg K h1 19181 kJkg Entropy Eq s2 s1 h2 2225 kJkg State 3 h3 327536 kJkg s3 60342 kJkgK CV Turbine section 1 Assume reversible and adiabatic Entropy Eq s4 s3 60342 26456 x4 35412 x4 0956907 h4 100841 x4 179573 272676 kJkg State 5 h5 345648 kJkg s5 72337 kJkgK CV Turbine section 2 Assume reversible and adiabatic Entropy Eq s6 s5 72337 06492 x6 7501 x6 087782 h6 19181 x6 239282 229227 kJkg Steam generator qH h3 h2 h5 h4 327536 2225 345648 272676 305286 72972 37826 kJkg Turbine wT h3 h4 h5 h6 327536 272676 345648 229227 5486 116421 17128 kJkg P v 1 2 3 4 5 6 T s 1 2 3 4 10 kPa 30 MPa 5 6 3 MPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 935 A smaller power plant produces steam at 3 MPa 600oC in the boiler It keeps the condenser at 45oC by transfer of 10 MW out as heat transfer The first turbine section expands to 500 kPa and then flow is reheated followed by the expansion in the low pressure turbine Find the reheat temperature so the turbine output is saturated vapor For this reheat find the total turbine power output and the boiler heat transfer Q W T 3 2 4 1 Condenser Boiler Turbine W P Q H 5 6 L cb s 3 MPa 959 kPa 1 4 6 2 T 3 5 500 kPa The states properties from Tables B11 and B13 1 45oC x 0 h1 18842 kJkg v1 000101 m3kg Psat 959 kPa 3 30 MPa 600oC h3 368234 kJkg s3 75084 kJkg K 6 45oC x 1 h6 258319 kJkg s6 81647 kJkg K CV Pump Reversible and adiabatic Energy wp h2 h1 Entropy s2 s1 since incompressible it is easier to find work positive in as wp v dP v1 P2 P1 000101 3000 959 302 kJkg h2 h1 wp 18842 302 19144 kJkg CV HP Turbine section Entropy Eq s4 s3 h4 309326 kJkg T4 314oC CV LP Turbine section Entropy Eq s6 s5 81647 kJkg K state 5 State 5 500 kPa s5 h5 354755 kJkg T5 529oC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CV Condenser Energy Eq qL h6 h1 hfg 239477 kJkg m Q L qL 10 000 239477 4176 kgs Both turbine sections W Ttot m wTtot m h3 h4 h5 h6 4176 368234 309326 354755 258319 6487 kW Both boiler sections Q H m h3 h2 h5 h4 4176 368234 19144 354755 309326 16 475 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 936 A smaller power plant produces 25 kgs steam at 3 MPa 600oC in the boiler It cools the condenser with ocean water so the condenser exit is at 50oC There is a reheat done at 500 kPa up to 400oC and then expansion in the low pressure turbine Find the net power output and the total heat transfer in the boiler Solution The states properties from Tables B11 and B13 1 50oC x 0 h1 20931 kJkg v1 0001012 m3kg Psat 1235 kPa 3 30 MPa 600oC h3 368234 kJkg s3 75084 kJkg K 5 500 kPa 400oC h5 327183 kJkg s5 77937 kJkg K CV Pump Reversible and adiabatic Incompressible flow so Energy wp h2h1 v1P2 P1 0001012 3000 1235 302 kJkg CV LP Turbine section Entropy Eq s6 s5 77937 kJkg K twophase state x6 s6 sfsfg 77937 07037 73725 09617 h6 20931 09617 238275 25008 kJkg Both turbine sections wTtot h3 h4 h5 h6 368234 309326 327183 25008 136011 kJkg W net W T W p m wTtot wp 25 136011 302 33 927 kW Both boiler sections Q H m h3 h2 h5 h4 25 368234 19144 327183 309326 91 737 kW Q W T 3 2 4 1 Condenser Boiler Turbine W P Q H 5 6 L cb s 3 MPa 124 kPa 1 4 6 2 T 3 5 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 937 Consider an ideal steam reheat cycle where steam enters the highpressure turbine at 30 MPa 400C and then expands to 08 MPa It is then reheated to 400C and expands to 10 kPa in the lowpressure turbine Calculate the cycle thermal efficiency and the moisture content of the steam leaving the lowpressure turbine Solution CV Pump reversible adiabatic and assume incompressible flow wP v1P2 P1 000101 m3kg 3000 10 kPa 302 kJkg h2 h1 wP 19181 302 19483 kJkg Q W T 3 2 4 1 Condenser Boiler Turbine W P Q H 5 6 L cb s 3 MPa 10 kPa 1 4 6 2 T 3 5 CV HP Turbine section P3 3 MPa T3 400oC h3 323082 kJkg s3 69211 kJkg K s4 s3 h4 28916 kJkg CV LP Turbine section State 5 400oC 08 MPa h5 32671 kJkg s5 75715 kJkg K Entropy Eq s6 s5 75715 kJkg K twophase state x6 sfg s6 sf 75715 06492 7501 092285 0923 h6 19181 092285 239282 2400 kJkg wTtot h3 h4 h5 h6 323082 2891632671 2400 12378 kJkg qH1 h3 h2 323082 19483 3036 kJkg qH qH1 h5 h4 3036 32671 28916 34115 kJkg ηCYCLE wTtot wPqH 12378 30234115 0362 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 938 The reheat pressure affects the operating variables and thus turbine performance Repeat Problem 935 twice using 06 and 10 MPa for the reheat pressure Solution Q W T 3 2 4 1 Condenser Boiler Turbine W P Q H 5 6 L cb s 3 MPa 96 kPa 1 4 6 2 T 3 5 CV Pump reversible adiabatic and assume incompressible flow wP v1P2 P1 000101 m3kg 3000 959 kPa 302 kJkg h2 h1 wP 18842 302 19144 kJkg State 3 3 MPa 400oC h3 323082 kJkg s3 69211 kJkg K Low T boiler section qH1 h3 h2 323082 19144 303938 kJkg State 4 P4 s4 s3 For P4 1 MPa h4 294085 kJkg state 4 is sup vapor State 5 400oC P5 P4 h5 32639 kJkg s5 7465 kJkg K For P4 06 MPa h4 27932 kJkg state 4 is sup vapor State 5 400oC P5 P4 h5 32703 kJkg s5 77078 kJkg K State 6 45oC s6 s5 x6 s6 sfsfg Total turbine work wTtot h3 h4 h5 h6 Total boiler HTr qH qH1 h5 h 4 Cycle efficiency ηCYCLE wTtot wPq H P4P5 x6 h6 wT qH ηCYCLE 1 0907 23605 11934 33624 03540 06 09393 24378 12701 35165 03603 Notice the small change in efficiency Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 939 The effect of a number of reheat stages on the ideal steam reheat cycle is to be studied Repeat Problem 935 using two reheat stages one stage at 12 MPa and the second at 02 MPa instead of the single reheat stage at 08 MPa CV Pump reversible adiabatic and assume incompressible flow work in wP v1P2 P1 000101 m3kg 3000 9593 kPa 302 kJkg h2 h1 wP 18842 302 19144 kJkg P4 P5 12 MPa P6 P7 02 MPa 3 h3 323082 kJkg s3 69211 kJkg K 4 P4 s4 s3 sup vap h4 29853 5 h5 32607 kJkg s5 73773 kJkg K 6 P6 s6 s5 sup vapor h6 28112 kJkg s 3 5 7 o 3 MPa 400 C 959 kPa 1 8 4 6 2 T 7 h7 32765 kJkg s7 82217 kJkg K 8 P8 s8 s7 sup vapor h8 2602 kJkg used CATT3 Total turbine work same flow rate through all sections wT h3 h4 h5 h6 h7 h8 2455 4495 6745 13695 kJkg Total heat transfer in boiler same flow rate through all sections qH h3 h2 h5 h4 h7 h6 3039 3198 4653 38245 kJkg Cycle efficiency ηTH qH wT wP 13695 302 38245 0357 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Open Feedwater Heaters Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 940 An open FWH receives steam at 1 MPa 200oC from the turbine and 1 MPa 100o C water from the feed water line Find the required fraction of the extraction flow in the turbine The setup follows Fig912 State enthalpies h2 41902 h6 282786 all kJkg h3 76279 x 0 at 1 MPa Analyze the FWH which leads to Eq95 y m 6 m 5 the extraction fraction y h3 h2 h6 h2 76279 41902 282786 41902 01427 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 941 A power plant for a polar expedition uses ammonia and the boiler exit is 80oC 1000 kPa and the condenser operates at 15oC A single open feed water heater operates at 400 kPa with an exit state of saturated liquid Find the mass fraction extracted in the turbine CV Feedwater heater States given and fixed from knowing Fig 912 5 h5 16146 kJkg s5 54971 kJkgK 3 h3 171226 kJkg 1 h1 11166 kJkg v1 0001519 m3kg Analyze the pump h2 h1 wP1 h1 v1 P2 P1 11166 0001519 m3kg 400 2363 kPa 11191 kJkg Analyze the turbine 6 400 kPa s6 s1 h6 14796 kJkg Analyze the FWH leads to Eq95 y h3 h2 h6 h2 171226 11191 14796 11191 00401 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 942 Find the cycle efficiency for the cycle in Problem 941 CV Feedwater heater States given and fixed from knowing Fig 912 5 h5 16146 kJkg s5 54971 kJkgK 3 h3 171226 kJkg 1 h1 11166 kJkg v1 0001519 m3kg Analyze the pump wP1 v1 P2 P1 0001519 m3kg 400 2363 kPa 0249 kJkg h2 h1 wP1 11166 0249 111909 kJkg Analyze the turbine 6 400 kPa s6 s5 h6 14796 kJkg 7 15 C s7 s5 h7 141356 kJkg Analyze the FWH leads to Eq95 y h3 h2 h6 h2 171226 111909 14796 11909 00434 wT h5 y h6 1 y h7 16146 00434 14796 1 00434 141356 19815 kJkg Pump 2 gives wP2 v3 P4 P3 000156 m3kg 1000 400 kPa 0936 kJkg Net work wnet wT wP2 1 y wP1 19815 0936 100434 0249 19698 kJkg Boiler qH h5 h4 h5 h3 wP2 16146 17165 0936 1442 kJkg Cycle efficiency η wnet qH 19698 1442 01366 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 943 A steam power plant has high and low pressures of 20 MPa and 10 kPa and one open feedwater heater operating at 1 MPa with the exit as saturated liquid The maximum temperature is 800C and the turbine has a total power output of 5 MW Find the fraction of the flow for extraction to the feedwater and the total condenser heat transfer rate The physical components and the Ts diagram is as shown in Fig 912 in the main text for one open feedwater heater The same state numbering is used From the Steam Tables State 5 P T h5 40698 kJkg s5 70544 kJkg K State 1 P x 0 h1 19181 kJkg v1 000101 m3kg State 3 P x 0 h3 7628 kJkg v3 0001127 m3kg Pump P1 wP1 v1P2 P1 000101 990 1 kJkg h2 h1 wP1 19281 kJkg Turbine 56 s6 s5 h6 30137 kJkg wT56 h5 h6 40698 30137 10561 kJkg Feedwater Heater m TOT m 5 y m 5 h6 1 y m 5h2 m 5h3 y h3 h2 h6 h2 7628 19281 30137 19281 02021 To get state 7 into condenser consider turbine s7 s6 s5 x7 70544 0649375009 085391 h7 19181 085391 239282 22351 kJkg Find specific turbine work to get total flow rate W T m TOTh5 ym TOTh6 1 ym TOTh7 m TOT h5 yh6 1 y h7 m TOT 16773 m TOT 500016773 298 kgs Q L m TOT 1y h7h1 298 0797922351 19181 4858 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 944 A low temperature power plant operates with R410A maintaining 20oC in the condenser a high pressure of 3 MPa with superheat to 80oC There is one open feed water heater operating at 800 kPa with an exit as saturated liquid at 0oC Find the extraction fraction of the flow out of the turbine and the turbine work per unit mass flowing through the boiler 1 TURBINE COND FWH P2 4 2 3 5 6 7 P1 STEAM GEN HP LP cb s 1 2 3 4 5 6 7 400 kPa 3 MPa 800 kPa T State 1 x1 0 h1 2824 kJkg v1 0000803 m3kg State 3 x3 0 h3 5776 kJkg v3 0000855 m3kg State 5 h5 3291 kJkg s5 10762 kJkg K State 6 s6 s5 T6 102oC h6 2903 kJkg State 7 s7 s5 x7 s7 sfsfg 09982 h7 2715 kJkg CV Pump P1 wP1 h2 h1 v1P2 P1 0000803 800 400 032 kJkg h2 h1 wP1 2824 032 2856 kJkg CV Feedwater heater Call m 6 m tot y the extraction fraction Energy Eq 1 y h2 y h6 1 h3 y h3 h2 h6 h2 5776 2856 2903 2856 01116 CV Turbine W T m TOT h5 y m TOT h6 1 y m TOT h7 wT h5 y h6 1 y h7 3291 01116 2903 1 01116 2715 5552 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 945 A Rankine cycle operating with ammonia is heated by some low temperature source so the highest T is 120oC at a pressure of 5000 kPa Its low pressure is 1003 kPa and it operates with one open feedwater heater at 2033 kPa The total flow rate is 5 kgs Find the extraction flow rate to the feedwater heater assuming its outlet state is saturated liquid at 2033 kPa Find the total power to the two pumps 1 TURBINE COND FWH P2 4 2 3 5 6 7 P1 STEAM GEN HP LP cb s 1 2 3 4 5 6 7 1 MPa 5 MPa 203 MPa T State 1 x1 0 h1 29825 kJkg v1 0001658 m3kg State 3 x3 0 h3 42148 kJkg v3 0001777 m3kg State 5 h5 42148 kJkg s5 47306 kJkg K State 6 s6 s5 x6 s6 sfsfg 099052 h6 146153 kJkg CV Pump P1 wP1 h2 h1 v1P2 P1 00016582033 1003 1708 kJkg h2 h1 wP1 29825 1708 29996 kJkg CV Feedwater heater Call m 6 m tot y the extraction fraction Energy Eq 1 y h2 y h6 1 h3 y h3 h2 h6 h2 42148 29996 146153 29996 01046 m extr y m tot 01046 5 0523 kgs m 1 1 y m tot 1 01046 5 4477 kgs CV Pump P2 wP2 h4 h3 v3P4 P3 0001777 5000 2033 5272 kJkg Total pump work W p m 1wP1 m tot wP2 4477 1708 5 5272 34 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 946 A power plant with one open feedwater heater has a condenser temperature of 45C a maximum pressure of 5 MPa and boiler exit temperature of 900C Extraction steam at 1 MPa to the feedwater heater is mixed with the feedwater line so the exit is saturated liquid into the second pump Find the fraction of extraction steam flow the two specific pump work inputs and the turbine work per unit mass flow in the boiler Solution The complete diagram is as in Figure 912 in the main text From turbine 6 1 2 3 From condenser Pump 1 Pump 2 4 To boiler FWH State out of boiler 5 h5 437882 kJkg s5 79593 kJkg K CV Turbine reversible adiabatic s7 s6 s5 State 6 P6 s6 h6 36406 kJkg T6 574oC State 7 s7 s5 x7 s7 sfsfg 097271 h7 251784 kJkg CV Pump P1 wP1 h2 h1 v1P2 P1 0001011000 96 10 kJkg h2 h1 wP1 18842 10 18942 kJkg CV Feedwater heater Call m 6 m tot y the extraction fraction Energy Eq 1 y h2 y h6 1 h3 y h3 h2 h6 h2 76279 18942 36406 18942 01661 CV Pump P2 wP2 h4 h3 v3P4 P3 00011275000 1000 45 kJkg CV Turbine W T m TOT h5 y m TOT h6 1 y m TOT h7 wT h5 y h6 1 y h7 437882 01661 36406 1 01661 251784 16745 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 947 A steam power plant operates with a boiler output of 20 kgs steam at 2 MPa 600C The condenser operates at 50C dumping energy to a river that has an average temperature of 20C There is one open feedwater heater with extraction from the turbine at 600 kPa and its exit is saturated liquid Find the mass flow rate of the extraction flow If the river water should not be heated more than 5C how much water should be pumped from the river to the heat exchanger condenser Solution The setup is as shown in Fig 912 1 50oC sat liq v1 0001012 m3kg h1 20931 kJkg 2 600 kPa s2 s 1 3 600 kPa sat liq h3 hf 67054 kJkg 5 P T h5 36901 kJkg s5 77023 kJkg K CONDENSER 7 1 From river To river To pump 1 Ex turbine 6 600 kPa s6 s5 h6 32700 kJkg CV P1 wP1 v1P2 P1 0001012 m3kg 600 1235 kPa 0595 kJkg h2 h1 wP1 2099 kJkg CV FWH y h6 1 y h2 h3 y EA hA3 A hA2 A EhA6 A hA2 AE A A67054 2099 32700 2099E A 01505 Am E A6E A y Am E A5E A 01505 20 3 kgs CV Turbine sA7E A sA6E A sA5E A xA7E A 09493 hA7E A 247117 kJkg CV Condenser qALE A hA7E A hA1E A 247117 20931 226186 kJkg The heat transfer out of the water from 7 to 1 goes into the river water AQ E ALE A 1 y Am E AqALE A 085 20 226186 38 429 kW Am E AH2OE A hAH2OE A Am E AH2OE A hAf25E A hAf20E A Am E A 2093 Am E A 38 429 2093 1836 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 948 In one type of nuclear power plant heat is transferred in the nuclear reactor to liquid sodium The liquid sodium is then pumped through a heat exchanger where heat is transferred to boiling water Saturated vapor steam at 5 MPa exits this heat exchanger and is then superheated to 600C in an external gasfired superheater The steam enters the turbine which has one opentype feedwater extraction at 04 MPa The condenser pressure is 10 kPa Determine the heat transfer in the reactor and in the superheater to produce a net power output of 5 MW Solution The complete cycle diagram is similar to Figure 912 except the boiler is separated into a section heated by the reactor and a super heater section 1 TURBINE COND FWH P2 4 2 3 5 6 SUPER HEATER REACTOR Q 7 8 P1 s 1 2 3 4 5 6 8 7 10 kPa 5 MPa 04 MPa T CV Pump P1 wAP1E A 0001008400 75 04 kJkg hA2E A hA1E A wAP1E A 1688 04 1692 kJkg CV Pump P2 wAP2E A 00010845000 400 50 kJkg hA4E A hA3E A wAP2E A 6047 50 6097 kJkg CV Turbine to get exit state properties sA7E A sA6E A 72589 PA7E A 04 MPa TA7E A 2212AoE AC hA7E A 29045 kJkg sA8E A sA6E A 72589 06492 xA8E A 7501 xA8E A 088118 hA8E A 19181 088118 239282 23003 kJkg CV Feedwater heater FWH to get the extraction fraction xA7E A Divide the equations with the total mass flow rate Am E A3E A Am E A4E A Am E A5E A Am E A6E Continuity xA2E A xA7E A xA3E A 10 Energy Eq xA2E AhA2E A xA7E AhA7E A hA3E xA7E A 60471692290451692 01592 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CV Turbine to get the total specific work Full flow from 6 to 7 and the fraction 1 xA7E A from 7 to 8 wATE A hA6E A hA7E A 1 xA7E AhA7E A hA8E A 3666529045 084082904523003 12700 kJkg CV Pumps P1 has xA1E A 1 xA7E A P2 has the full flow xA3E A 1 wAPE A xA1E AwAP1E A xA3E AwAP2E A 08408 04 1 50 53 kJkg wANETE A 12700 53 12647 AmE A 500012647 3954 kgs CV Reactor this has the full flow AQ E AREACTE A AmE AhA5E A hA4E A 3954 27943 6097 8638 kW CV Superheater this has the full flow AQ E ASUPE A AmE AhA6E A hA5E A 3954 36665 27943 3449 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 949 Consider an ideal steam regenerative cycle in which steam enters the turbine at 30 MPa 400C and exhausts to the condenser at 10 kPa Steam is extracted from the turbine at 08 MPa for an open feedwater heater The feedwater leaves the heater as saturated liquid The appropriate pumps are used for the water leaving the condenser and the feedwater heater Calculate the thermal efficiency of the cycle and the net work per kilogram of steam Solution This is a standard Rankine cycle with an open FWH as shown in Fig912 CV Pump P1 wAP1E A hA2E A hA1E A vA1E APA2E A PA1E A 000101800 10 0798 kJkg hA2E A hA1E A wAP1E A 19181 0798 19261 kJkg CV FWH Call Am E A6E A Am E AtotE A y the extraction fraction 1 y hA2E A y hA6E A 1 hA3E A y EA hA3 A hA2 A EhA6 A hA2 AE A A 7211 19261 28916 19261E A 01958 CV Pump P2 wAP2E A hA4E A hA3E A vA3E APA4E A PA3E A 00011153000 800 245 kJkg hA4E A hA3E A wAP2E A 7211 245 72355 kJkg CV Boiler qAHE A hA5E A hA4E A 323082 72355 25073 kJkg CV Turbine Entropy Eq sA7E A sA6E A sA5E A 69211 kJkg K PA6E A sA6E A hA6E A 28916 kJkg superheated vapor sA7E A sA6E A sA5E A 69211 xA7E A A69211 06492 7501E A 083614 hA7E A 19181 xA7E A 239282 219255 kJkg Turbine has full flow in HP section and fraction 1y in LP section AW E ATE A Am E A5E A hA5E A hA6E A 1 y hA6E A hA7E A wATE A 323082 28916 1 01988 28916 219255 8993 kJkg P2 has the full flow and P1 has the fraction 1y of the flow wAnetE A wATE A 1 y wAP1E A wAP2E A 8993 1 019880798 245 8962 kJkg ηAcycleE A wAnetE A qAHE A 8962 25073 0357 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Closed Feedwater Heaters Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 950 Write the analysis continuity and energy equations for the closed feed water heater with a drip pump as shown in Fig913 Take the control volume to have state 4 out so it includes the drip pump Find the equation for the extraction fraction CV Feedwater heater plus drip pump Continuity Eq Am E A6E A Am E A2E A Am E A4E Energy Eq Am E A6E AhA6E A Am E A2E AhA2E A Am E A6E A wAP dripE A Am E A4E AhA4E CV drip pump wAP dripE A vA6E A PA4E A PA6E A hA6bE A hA6aE A hA6aE A hAf at P6E Divide the energy equation with the full flow rate Am E A4E A to get Energy Eq y hA6E A 1 y hA2E A y wAP dripE A hA4E Now solve for the fraction y Am E A6E A Am E A4E y A h4 h2 Eh6 h2 wP drip E A So to use this expression we assume we know states 2 4 and 6 and have analyzed the drip pump Of course if you assume state 3 is at the temperature of the condensing steam at state 6a then take CV to exclude the drip pump and the expression for y is different Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 951 A closed FWH in a regenerative steam power cycle heats 20 kgs of water from 100AoE AC 20 MPa to 200AoE AC 20 MPa The extraction steam from the turbine enters the heater at 4 MPa 275AoE AC and leaves as saturated liquid What is the required mass flow rate of the extraction steam Solution The schematic from Figure 913 has the feedwater from the pump coming at state 2 being heated by the extraction flow coming from the turbine state 6 so the feedwater leaves as saturated liquid state 4 and the extraction flow leaves as condensate state 6a 2 4 6 6a From table B1 h kJkg B14 100C 20 MPa hA2E A 43404 B14 200C 20 MPa hA4E A 86047 B13 4 MPa 275C hA6E A 288039 B12 4 MPa sat liq hA6aE A 108729 CV Feedwater Heater Energy Eq AmE A2hA2E A AmE A6hA6E A AmE A2hA4E A AmE A6hA6aE Since all four states are known we can solve for the extraction flow rate AmE A6 AmE A2 A h2 h4 Eh6a h6 E A 20 A 43404 86047 108729 288039E A kgs 4756 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 952 Find the specific turbine work from the extraction flow for the cycle in Problem 951 and the specific heat transfer in the boiler operating at 20 MPa assuming the extraction flow goes through a drip pump and added to the feedwater line Solution The schematic from Figure 913 has the feedwater from the pump coming at state 2 20 MPa 100C being heated by the extraction flow coming from the turbine state 6 4 MPa 275C so the feedwater leaves at state 3 20 MPa 200C and the extraction flow leaves as condensate state 6a 4 MPa x 0 goes through the drip pump and added to the feedwater line to produce a state 4 20 MPa T going to the boiler From B1 h in kJkg hA2E A 43404 hA3E A 86047 hA6E A 288039 hA6aE A 108729 vA6aE A vAfE A 0001252 mA3E Akg sA6E A sA5E A 62145 kJkgK From turbine 6 1 2 6a From condenser Pump 1 Pump 2 3 4 6b CV Feedwater Heater Energy Eq AmE A2hA2E A AmE A6hA6E A AmE A2hA3E A AmE A6hA6aE Since all four states are known we can solve for the extraction flow rate AmE A6 AmE A2 A h2 h3 Eh6a h6 E A 20 A 43404 86047 108729 288039E A kgs 4756 kgs CV drip pump wAP2E A vA6aE A PA6bE A PA6aE A 0001252 20 000 4000 2003 kJkg hA6bE A hA6aE A wAP2E A 108729 2003 110732 kJkg CV Junction 6b 3 4 hA4E A m 2h3 m 6h6b m 2 m 6 A20 86047 4756 110732 20 4756E A 90789 kJkg We must assume state 5 is 20 MPa s sA6E A so hA5E A 329759 kJkg then wAT ext flowE A h5 h6 329759 288039 4172 kJkg qAH ext flowE A h5 h4 329759 90789 23897 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 953 Assume the cycle in problem 951 has a condenser operating at 20 kPa what would the quality at the turbine exit be Find the reheat temperature if reheat is at the 4 MPa extraction pressure and the turbine should have a minimum quality of 90 Assume rev adiabatic turbine so 4 MPa 275C sA5E A sA6E A sA7E A 62145 kJkgK xA7E A A62145 08319 70766E A 07606 If we should have the new state 7 as xA7E A 09 then the entropy is sA7E A 08319 09 70766 720084 kJkgK this is then also entropy after reheat at 4 MPa sA8E A sA7E A so then TA8E A 500 100 A720084 709 73688 709E A 5398C 1 TURBINE COND FWH 2 34 5 6 7 P1 STEAM GEN HP LP cb 6a s 1 2 6a 5 6 7 20 kPa 20 MPa 4 MPa T 8 34 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 954 A power plant with one closed FWH has a condenser temperature of 45C a maximum pressure of 5 MPa and boiler exit temperature of 900C Extraction steam at 1 MPa to the feedwater heater condenses and is pumped up to the 5 MPa feedwater line where all the water goes to the boiler at 200C Find the fraction of extraction steam flow and the two specific pump work inputs Solution sA1E A 06387 kJkg K hA1E A 18842 kJkg vA1E A 000101 mA3E Akg sA4E A 21386 kJkg K hA4E A 76279 kJkg vA4E A 0001127 mA3E Akg T6 hA6E A 85385 kJkg From turbine 3 1 2 4 From condenser Pump 1 Pump 2 5 6 7 CV Turbine Reversible adiabatic so constant s from inlet to extraction point sA3E A sAINE A 79593 kJkg K TA3E A 5738 hA3E A 36406 kJkg CV P1 wAP1E A vA1E APA2E A PA1E A 000101 mA3E Akg 5000 96 kPa 504 kJkg hA2E A hA1E A wAP1E A 18842 504 19346 kJkg CV P2 wAP2E A vA4E APA7E A PA4E A 0001127 mA3E Akg 5 1 MPa 4508 kJkg hA7E A hA4E A wAP2E A 76279 451 76730 kJkg CV FWH and pump 2 The extraction fraction is y AmE A3AmE A6 Continuity Eq AmE A6 AmE A2E A AmE A3 1 1 y y Energy 1 yhA2E A y hA3E A y wAP2E A hA6E y h6 h2 h3 wP2 h2 A 85385 19346 36406 4508 19346E A 01913 AmE A3AmE A6 y 01913 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 955 Do Problem 943 with a closed feedwater heater instead of an open and a drip pump to add the extraction flow to the feed water line at 20 MPa Assume the temperature is 175C after the drip pump flow is added to the line One main pump brings the water to 20 MPa from the condenser Solution vA1E A 000101 mA3E Akg hA1E A 19181 kJkg TA4E A 175AoE AC hA4E A 75166 kJkg hA6aE A hAf 1MPaE A 76279 kJkg vA6aE A 0001127 mA3E Akg From turbine 3 1 2 4 From condenser Pump 1 Pump 2 6 6b 6a Turbine section 1 sA6E A sA5E A 70544 kJkg K PA6E A 1 MPa hA6E A 30137 kJkg CV Pump 1 wAP1E A hA2E A hA1E A vA1E APA2E A PA1E A 00010120 000 10 2019 kJkg hA2E A hA1E A wAP1E A 19181 2019 2120 kJkg CV Pump 2 wAP2E A hA6bE A hA6aE A vA6aE APA6bE A PA6aE A 000112720 000 1000 2141 kJkg CV FWH P2 select the extraction fraction to be y Am E A6E A Am E A4E y hA6E A 1 y hA2E A y wAP2E A hA4E A y EA hA4 A hA2 A E hA6 A hA2 A wAP2 AE A A 75166 2120 30137 2120 2141E A 0191 Turbine sA7E A sA6E A sA5E A PA7E A 10 kPa xA7E A A70544 06493 75009E A 085391 h7 19181 085391 239282 22351 kJkg wATE A hA5E A hA6E A 1 y hA6E A hA7E A 40698 30137 0809 30137 22351 1686 kJkg AW E ATE A 5000 kW Am E A5E A wATE A Am E A5E A 1686 kJkg Am E A5E A 2966 kgs AQ E ALE A Am E A5E A1 y hA7E A hA1E A 2966 0809 22351 19181 4903 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 956 Repeat Problem 949 but assume a closed instead of an open feedwater heater A single pump is used to pump the water leaving the condenser up to the boiler pressure of 30 MPa Condensate from the feedwater heater is going through a drip pump and added to the feedwater line so state 4 is at TA6E A Solution CV Turbine 2nd law sA5E A sA6E A sA7E A 69211 kJkg K hA5E A 323082 hA6E A 28916 xA7E A 69211 064927501 083614 hA7E A 19181 xA7E A 239282 219255 kJkg Assume feedwater heater exit state 4 at the T of the condensing steam CV Pump 1 wAP1E A hA2E A hA1E A vA1E APA2E A PA1E A 0001013000 10 302 kJkg hA2E A hA1E A wAP1E A 19181 302 19483 kJkg TA4E A TAsatE A PA6E A 17043C hA4E A hAfE A hA6aE A 7211 kJkg CV Pump 2 the drip pump wAP2E A hA6bE A hA6aE A vA6aE APA6bE A PA6aE A 00011153000 800 245 kJkg CV FWH P2 select the extraction fraction to be y Am E A6E A Am E A4E y hA6E A 1 y hA2E A y wAP2E A hA4E A y EA hA4 A hA2 A E hA6 A hA2 A wAP2 AE A A 7211 19483 28916 19483 245E A 0195 Turbine work with full flow from 5 to 6 fraction 1 y flows from 6 to 7 wATE A hA5E A hA6E A 1 y hA6E A hA7E A 323082 28916 0805 28916 219255 90195 kJkg wAnetE A wATE A 1ywAP1E A xwAP2E A 90195 0805 302 0195 245 8990 kJkg qAHE A hA5E A hA4E A 323082 7211 25097 kJkg ηAcycleE A wAnetE A qAHE A 8990 25097 0358 P1 1 2 4 5 6 COND 3 TURBINE BOILER FW HTR P2 6b Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 957 Repeat Problem 949 but assume a closed instead of an open feedwater heater A single pump is used to pump the water leaving the condenser up to the boiler pressure of 30 MPa Condensate from the feedwater heater is drained through a trap to the condenser Solution CV Turbine 2nd law sA4E A sA5E A sA6E A 69211 kJkg K hA4E A 323082 hA5E A 28916 xA6E A 69211 064927501 083614 hA6E A 19181 xA6E A 239282 219255 kJkg P 1 2 4 5 6 COND 3 TURBINE BOILER FW HTR Trap 7 Assume feedwater heater exit at the T of the condensing steam CV Pump wAPE A hA2E A hA1E A vA1E APA2E A PA1E A 0001013000 10 302 kJkg hA2E A hA1E A wAPE A 19181 302 19483 kJkg TA3E A TAsatE A PA5E A 17043C hA3E A hAfE A hA7E A 7211 kJkg CV FWH Am E A5E A Am E A3E A y Energy Eq hA2E A y hA5E A hA3E A hA7E A y y EA hA3 A hA2 A E hA5 A hAf 800 AE A A7211 19483 28916 7211E A 02425 Turbine work with full flow from 4 to 5 fraction 1y flows from 5 to 6 wATE A hA4E A hA5E A 1 yhA5E A hA6E A 323082 28916 07575 28916 219255 86875 kJkg wAnetE A wATE A wAPE A 86875 302 8657 kJkg qAHE A hA4E A hA3E A 323082 7211 25097 kJkg ηAcycleE A wAnetE A qAHE A 8657 25097 0345 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 958 A Rankine cycle feeds 5 kgs ammonia at 2 MPa 140AoE AC to the turbine which has an extraction point at 800 kPa The condenser is at 20AoE AC and a closed feed water heater has an exit state 3 at the temperature of the condensing extraction flow and it has a drip pump The source for the boiler is at constant 180AoE AC Find the extraction flow rate and state 4 into the boiler PA1E A 1902 kPa hA1E A 8905 kJkg vA1E A 0001504 mA3E Akg sA5E A 55022 kJkg K hA5E A 17382 kJkg TA6aE A TAsat 800 kPaE A 1785AoE AC hA6aE A 26418 kJkg CV Turbine Reversible adiabatic so constant s from inlet to extraction point sA6E A sAINE A 55022 kJkg K TA6E A 634AoE AC hA6E A 158089 kJkg CV P1 wAP1E A vA1E APA2E A PA1E A 00015042000 1902 2722 kJkg hA2E A hA1E A wAP1E A 91772 kJkg CV P2 wAP2E A vA6aE A PA4E A PA6E A 00016108 2000 800 1933 kJkg hA6bE A hA6aE A wAP2E A 26611 kJkg CV Total FWH and pump notice h3 hA6aE A as we do not have table for this state The extraction fraction is y AmE A6AmE A4E A Energy 1 yhA2E A yh6 1 yh3 yhA6aE y h3 h2 h3 h2 h6 h6a A 26418 91772 26418 91772 158089 26418E A 01158 AmE A6 y AmE A4E A 01158 5 05789 kgs CV The junction after FWH and pump 2 hA4E A 1yh3 y hA6bE A 1 01158 26418 01158 26611 2644 kJkg Interpolate in Table B21 TA4E A 1792AoE AC From turbine 6 1 2 6a From condenser Pump 1 Pump 2 3 4 6b Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 959 Assume the powerplant in Problem 945 has one closed feedwater heater instead of the open FWH The extraction flow out of the FWH is saturated liquid at 2033 kPa being dumped into the condenser and the feedwater is heated to 50AoE AC Find the extraction flow rate and the total turbine power output State 1 xA1E A 0 hA1E A 29825 kJkg vA1E A 0001658 mA3E Akg State 3 hA3E A hAfE A PA3E APAsatE AvAfE A 42148 500020330001777 42675 kJkg State 5 hA5E A 42148 kJkg sA5E A 47306 kJkg K State 6 sA6E A sA5E A xA6E A sA6E A sAfE AsAfgE A 099052 hA6E A 146153 kJkg State 6a xA6aE A 0 hA6aE A 42148 kJkg State 7 sA7E A sA5E A xA7E A sA7E A sAfE AsAfgE A 09236 hA7E A 137443 kJkg CV Pump P1 wAP1E A hA2E A hA1E A vA1E APA2E A PA1E A 00016585000 1003 6627 kJkg hA2E A hA1E A wAP1E A 29825 6627 30488 kJkg CV Feedwater heater Call Am E A6E A Am E AtotE A y the extraction fraction Energy Eq hA2E A y hA6E A 1 hA3E A y hA6aE A y A h3 h2 Eh6 h6a E A A 42675 30488 146153 42148E A 01172 Am E AextrE A y Am E AtotE A 01172 5 0586 kgs Total turbine work AW E ATE A Am E AtotE AhA5E A hA6E A 1 yAm E AtotE A hA6E A hA7E A 515863 146153 5 0586146153 137443 1008 kW 1 TURBINE COND FWH 2 34 5 6 7 P1 STEAM GEN HP LP cb 6a s 1 2 6a 34 5 6 7 1 MPa 5 MPa 203 MPa T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 960 Assume a variation of the cycle in Problem 949 with a closed FWH at 08 MPa and one open FWH at 100 kPa A pump is used to bring the water leaving the condenser up to 100 kPa for an open FWH and a second pump brings the feedwater up to 30 MPa as shown in Fig P960 Condensate from the closed FWH is drained through a trap to the open FWH Calculate the thermal efficiency of the cycle and the specific net work Solution This is a Rankine cycle with an open FWH and a closed FWH shown in Fig P960 Here state 6 is out of boiler state 8 is extraction at 100 kPa state 9 is out of turbine States 1 10 kPa x 0 hA1E A 19181 kJkg vA1E A 0001010 mA3E Akg 3 100 kPa x 0 hA3E A 41744 kJkg vA3E A 0001043 mA3E Akg 5 3 MPa T TA7aE A 17043AoE AC hA5E A 7211 kJkg hA7aE A 6 3 MPa 400AoE AC hA6E A 323082 kJkg sA6E A 69211 kJkgK 7 800 kPa s sA6E A hA7E A 289162 kJkg sup vapor 7a 800 kPa x 0 hA7aE A hAfE A 7211 kJkg 8 100 kPa s sA6E A xA8E A 69211 1302560568 092765 hA8E A 41744 xA8E A 225802 251209 kJkg 9 10 kPa s sA6E A xA9E A 69211 064927501 083614 hA9E A 19181 xA9E A 239282 219255 kJkg CV Pump P1 wAP1E A hA2E A hA1E A vA1E APA2E A PA1E A 000101100 10 00909 kJkg hA2E A hA1E A wAP1E A 19181 00909 1919 kJkg CV Pump P2 wAP2E A hA4E A hA3E A vA3E APA4E A PA3E A 00010433000 100 3025 kJkg hA4E A hA3E A wAP2E A 41744 3025 420465 kJkg Since the open FWH has a flow in from the trap we must do that first CV Closed FWH Call Am E A7E A Am E AtotE A yA1E A the extraction fraction Energy Eq 0 hA4E A hA5E A yA1E AhA7E A hA7aE A yA1E A EA hA5 A hA4 A EhA7 A hA7a AE A A7211 420465 289162 7211E A 01385 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CV Open FWH 3 flows in 1 flow out Energy Eq 0 yA2E A hA8E A yA1E A hA7aE A 1 yA1E A yA2E A hA2E A hA3E yA2E A A h3 y1 h7a 1y1 h2 Eh8 h2 E A A 41744 y17211 1 y1 1919 E251209 1919E A 00656 CV Boiler qAHE A hA6E A hA5E A 323082 7211 25097 kJkg CV Turbine wATE A hA6E A yA1E A hA7E A yA2E A hA8E A 1 yA1E A yA2E A hA9E 323082 01385 289162 00656 251209 07959 219255 920487 kJkg P2 has the full flow and P1 has the fraction 1 yA1E A yA2E A of the flow wAnetE A wATE A 1 yA1E A yA2E A wAP1E A wAP2E A 920487 07959 00909 3025 91739 kJkg ηAcycleE A wAnetE A qAHE A 91739 25097 0366 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Nonideal Cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 961 A steam power cycle has a high pressure of 30 MPa and a condenser exit temperature of 45C The turbine efficiency is 85 and other cycle components are ideal If the boiler superheats to 800C find the cycle thermal efficiency Solution Basic Rankine cycle as shown in Figure 93 in the main text CV Turbine wT h3 h4 s4 s3 sTGEN Ideal Table B13 s4 s3 79862 kJkg K xA4sE A 79862 0638675261 09763 hA4sE A hAfE A x hAfgE A 18842 09763 239477 25264 kJkg wATsE A hA3E A hA4sE A 4146 25264 16196 kJkg Actual wTAC η wTS 085 16196 137666 kJkg CV Pump wP v dP v1P2 P1 000101 3000 96 302 kJkg hA2E A hA1E A wAPE A 18842 302 19144 kJkg CV Boiler qH h3 h2 4146 19144 39546 kJkg η wTAC wPqH 137666 30239546 0347 P v 1 2 3 4s 4ac T s 1 2 3 4s 4ac Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 962 A Rankine cycle with water superheats to 500AoE AC at 3 MPa in the boiler and the condenser operates at 100AoE AC All components are ideal except the turbine which has an exit state measured to be saturated vapor at 100AoE AC Find the cycle efficiency with a an ideal turbine and b the actual turbine Standard Rankine cycle 1234s for ideal turbine Modified Rankine cycle 1234ac for actual turbine Table B11 vA1E A 0001044 mA3E Akg hA1E A 41902 kJkg hA3E A 345648 kJkg sA3E A 72337 kJkgK State 4s sA4sE A sA3E A 72337 sAfE A xA4sE A sAfgE A 13068 xA4sE A 60480 xA4sE A 097998 hA4sE A hAfE A xA4sE AhAfgE A 41902 xA4sE A 225703 26309 kJkg State 4ac hA4acE A hAgE A 267605 kJkg CV Pump Assume adiabatic reversible and incompressible flow wApsE A v dP vA1E APA2E A PA1E A 303 kJkg hA2E A hA1E A wApE A 41902 303 42205 kJkg CV Boiler qABE A hA3E A hA2E A 345648 42205 30344 kJkg CV Turbine wATsE A hA3E A hA4sE A 345648 26309 82558 kJkg Efficiency ηAthE A wAnetE A qABE A 82558 30330344 0271 Actual turbine wATacE A hA3E A hA4acE A 345648 267605 78043 kJkg Efficiency ηAthE A wAnetE A qABE A 78043 30330344 0256 P v 1 2 3 4s 4ac T s 1 2 3 4s 4ac Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 963 For the steam power plant described in Problem 913 assume the isentropic efficiencies of the turbine and pump are 85 and 80 respectively Find the component specific work and heat transfers and the cycle efficiency Solution This is a standard Rankine cycle with actual nonideal turbine and pump CV Pump Rev Adiabatic wAPsE A hA2sE A hA1E A vA1E APA2E A PA1E A 0001013000 10 302 kJkg sA2sE A sA1E wAPacE A wAPsE A ηAPE A 30208 3775 kJkg hA2aE A hA1E hA2aE A wAPacE A hA1E A 3775 19181 19558 kJkg CV Boiler qAHE A hA3E A hA2aE A 280414 19558 260856 kJkg CV Turbine wATE A hA3E A hA4E A sA4E A sA3E sA4E A sA3E A 61869 06492 xA4E A 7501 xA4E A 07383 hA4E A 19181 07383 239282 195834 kJkg wATsE A 280414 195834 8458 kJkg wATacE A wATsE A ηATE A 7189 hA3E A hA4aE hA4aE A hA3E A wATacE A 280414 7189 208524 kJkg CV Condenser qALE A hA4aE A hA1E A 208524 19181 18934 kJkg ηAcycleE A A wTac wPac Eq H E A A7189 378 260856E A 0274 This compares to 032 for the ideal case Q W T 3 2 4 1 Condenser Boiler Turbine W P Q B T s 1 2 3 4s 4ac state 2s and 2ac nearly the same Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 964 Steam enters the turbine of a power plant at 5 MPa and 400C and exhausts to the condenser at 10 kPa The turbine produces a power output of 20 000 kW with an isentropic efficiency of 85 What is the mass flow rate of steam around the cycle and the rate of heat rejection in the condenser Find the thermal efficiency of the power plant and how does this compare with a Carnot cycle Solution AW E ATE A 20 000 kW and ηATsE A 85 State 3 hA3E A 31956 kJkg sA3E A 66458 kJkgK State 1 PA1E A PA4E A 10 kPa sat liq xA1E A 0 TA1E A 458AoE AC hA1E A hAfE A 1918 kJkg vA1E A vAfE A 000101 mA3E Akg CV Turbine energy Eq qATE A hA3E A hA4E A wATE A qATE A 0 wATE A hA3E A hA4E A Assume Turbine is isentropic sA4sE A sA3E A 66458 kJkgK sA4sE A sAfE A xA4sE A sAfgE A solve for xA4sE A 07994 hA4sE A hAfE A xA4sE AhAfgE A 19181 xA4sE A 239282 21046 kJkg wATsE A hA3E A hA4sE A 1091 kJkg wATE A ηATsE AwATsE A 9273 kJkg Am E A EA AW AT A EwAT AE A 21568 kgs hA4E A hA3E A wATE A 22683 kJkg CV Condenser energy Eq hA4E A hA1E A qAcE A wAcE A wAcE A 0 qAcE A hA4E A hA1E A 20765 kJkg AQ E AcE A Am E A qAcE A 44 786 kW CV Pump Assume adiabatic reversible and incompressible flow wApsE A v dP vA1E APA2E A PA1E A 504 kJkg energy Eq hA2E A hA1E A wApE A 1968 kJkg CV Boiler energy Eq qABE A hA2E A hA3E A wABE A wABE A 0 qABE A hA3E A hA2E A 29988 kJkg wAnetE A wATE A wAPE A 9223 kJkg ηAthE A wAnetE A qABE A 0307 Carnot cycle TAHE A TA3E A 400AoE AC TALE A TA1E A 458AoE AC ηAthE A EA TAH A TAL A ETAH AE A 0526 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 965 Consider the boiler in Problem 921 where the geothermal hot water brings the R 134a to saturated vapor Assume a counter flowing heat exchanger arrangement The geothermal water temperature should be equal to or greater than the R134a temperature at any location inside the heat exchanger The point with the smallest temperature difference between the source and the working fluid is called the pinch point If 2 kgs of geothermal water is available at 95C what is the maximum power output of this cycle for R134a as the working fluid hint split the heat exchanger CV into two so the pinch point with T 0 T 85C appears 2 kgs of water is available at 95 AoE AC for the boiler The restrictive factor is the boiling temperature of 85 C Therefore break the process up from 23 into two parts as shown in the diagram sat liq at 85 C o D Q BC 2 3 Q AB B liq H2O at 85 C o sat vap R134a 85 C o 95 C o C R134a A LIQUID HEATER BOILER liquid H2O out liquid liquid H2O CV Pump wAPE A vA1E APA2E APA1E A 000087329262 10170 167 kJkg h2 hA1E A wAPE A 25654 167 25821 kJkg Write the energy equation for the first section AB and D3 AQ E AABE A AmE AH2OE AhAAE A hABE A 239794 35588 8412 kW AmE AR134AE A4281 33265 AmE AR134AE A 08813 kgs To be sure that the boiling temp is the restrictive factor calculate TACE A from the energy equation for the remaining section AQ E AACE A 0881333265 25821 6560 kW 235588 hACE A hACE A 3231 kJkg TACE A 772C TA2E A 40C OK CV Turbine sA4E A sA3E A 16782 11909 xA4E A 05214 xA4E A 09346 hA4E A 25654 09346 16328 40914 kJkg Energy Eq wATE A hA3E A hA4E A 4281 40914 1896 kJkg Cycle wANETE A wATE A wAPE A 1896 167 1729 kJkg AW E ANETE A AmE AR134AE AwNET 08813 1729 1524 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 966 Do the previous problem with ammonia as the working fluid A flow with 2 kgs of water is available at 95AoE AC for the boiler The restrictive factor is the boiling temperature of 85AoE AC Therefore break the process up from 2 3 into two parts as shown in the diagram sat liq at 85 C o D Q BC 2 3 Q AB B liq H2O at 85 C o sat vap 85 C o 95 C o C A LIQUID HEATER BOILER liquid H2O out liquid liquid H2O Amn Amn State 1 40AoE AC 15549 kPa vA1E A 0001725 mA3E Akg CV Pump wAPE A vA1E APA2E A PA1E A 000172546086 15549 527 kJkg h2 hA1E A wAPE A 37143 527 3767 kJkg AQ E AABE A AmE AH2OE AhAAE A hABE A 239794 35588 8412 kW AmE AamnE A14478 60921 AmE AamnE A 0100 kgs To verify that TD T3 is the restrictive factor find TC AQ E AAC 010060921 3767 2325 2035588 hC hACE A 34425 kJkg TACE A 825AoE AC T2 40AoE AC OK CV Turbine sA4E A sA3E A 43901 13574 xA4E A 35088 xA4E A 08643 hA4E A 37143 08643 10988 132113 kJkg Energy Eq wATE A hA3E A hA4E A 14478 132113 12667 kJkg wANETE A wATE A wAPE A 12667 527 1214 kJkg AW E ANETE A AmE AamnE AwNET 01 1214 1214 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 967 A concentrated solar power plant receives the energy from molten salt coming in at 560AoE AC and leaving at 300AoE AC in a counterflow heat exchanger where the water comes in at 3 MPa 60AoE AC and leaves at 450AoE AC 3 MPa The molten salt has 5 kgs flow with Cp 15 kJkgK What is the possible water flow rate the rate of energy transfer and rate of entropy generation Continuity Eqs Each line has a constant flow rate through it Energy Eq410 0 Am E AH2OE A hA1E A hA2E A Am E AsaltE A hA3E A hA4E A Entropy Eq77 0 Am E AH2OE A sA1E A sA2E A Am E AsaltE A sA3E A sA4E A S gen Process Each line has a constant pressure Table B1 hA1E A 25361 kJkg hA2E A 3344 kJkg sA1E A 08295 sA2E A 70833 kJkgK From the energy leaving the salt we get AQ E A Am E AsaltE A hA3E A hA4E A Am E AsaltE A CAP saltE A TA3E A TA4E A 5 15 560 300 1950 kW Am E AH2OE A Am E AsaltE A A h3 h4 Eh2 h1 E A A Q Eh2 h1 E A A 1950 kW 3344 25361 kJkgE A 0631 kgs sA4E A sA3E A CAP saltE A ln TA4E ATA3E A 15 ln5731583315 0561 kJkgK S gen Am E AH2OE A sA2E A sA1E A Am E AsaltE A sA4E A sA3E A 0631 kgs 70833 08295 kJkgK 5 kgs 0561 kJkgK 114 kWK CV Heat exchanger steady flow 1 inlet and 1 exit for salt and water each The two flows exchange energy with no heat transfer tofrom the outside 3 salt 1 water 4 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 968 Look at the pinch point for the heat exchanger in Problem 967 the same way it is explained in Problem 965 and determine the maximum possible water flow rate for a pinch point ΔT 10 K CV Part of the heat exchanger from the salt inlet to state 3a for the salt and then from state 1a for the water to its outlet at state 2 Pinch point TA3aE A TA1aE A 10 2439AoE AC 3 salt 1 water 4 2 3a salt 1a H2O Energy Eq 0 Am E AH2OE A hA1aE A hA2E A Am E AsaltE A hA3E A hA3aE A AQ E A Am E AsaltE A hA3E A hA3aE A Am E AsaltE A CAP saltE A TA3E A TA3aE A 5 15 560 2439 237075 kW Am E AH2OE A A Q Eh2 h1 E A A 237075 kW 3344 100841 kJkgE A 1015 kgs Check also for the salt outlet condition take the missing CV Energy Eq Am E AH2OE A hA1aE A hA1E A Am E AsaltE A CAP saltE A TA3aE A TA4E A 1015 100841 25361 5 15 2439 TA4E A Solve for TA4E A TA4E A 2439 102 1417AoE AC This temperature is too low the salt would solidify so the pinch point ΔT should be much higher Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 969 A steam power plant operates with a high pressure of 4 MPa and has a boiler exit of 600AoE AC receiving heat from a 750AoE AC source The ambient at 20AoE AC provides cooling to maintain the condenser at 60AoE AC all components are ideal except for the turbine which has an isentropic efficiency of 92 Find the ideal and the actual turbine exit qualities Find the actual specific work and specific heat transfer in all four components Solution A standard Rankine cycle with an actual nonideal turbine Boiler exit hA3E A 367444 kJkg sA3E A 73688 kJkg K Condenser exit hA1E A 25111 kJkg Ideal Turbine 4s 50C s sA3E A xA4sE A 73688 0831170784 09236 hA4sE A 25111 09236 235848 242943 kJkg wATsE A hA3E A hA4sE A 124501 kJkg Actual turbine wATacE A ηAT sE AwATsE A 092 124501 11454 kJkg hA3E A hA4acE A hA4acE A hA3E A wATacE A 367444 11454 252904 kJkg xA4acE A 252904 25111235848 096585 Pump wAPE A vA1E A PA2E A PA1E A 00010174000 1994 405 kJkg hA2E A hA1E A wAPE A 25111 405 25516 kJkg qAHE A hA3E A hA2E A 367444 25516 34193 kJkg qALE A hA4acE A hA1E A 252904 25111 22779 kJkg P v 1 2 3 4s 4ac T s 1 2 3 4s 4ac Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 970 For the previous Problem find also the specific entropy generation in the boiler heat source setup CV Boiler out to the source Entropy Eq sA2E A A qH ETsource E A sAgenE A sA3E A State 1 Sat liquid hA1E A 25111 kJkg sA1E A 08311 kJkgK Pump wAPE A vA1E A PA2E A PA1E A 00010174000 1994 405 kJkg State 2 hA2E A hA1E A wAPE A 25111 405 25516 kJkg sA2E A sA1E State 3 hA3E A 367444 kJkg sA3E A 73688 kJkgK Boiler qAHE A hA3E A hA2E A 367444 25516 34193 kJkg sAgenE A sA3E A sA2E A A qH ETsource E A 73688 08311 A 34193 750 273E A 3195 kJkgK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 971 Consider the power plant in Problem 941 Assume the high temperature source is a flow of liquid water at 120AoE AC into a heat exchanger at constant pressure 300 kPa and that the water leaves at 90AoE AC Assume the condenser rejects heat to the ambient which is at 20AoE AC List all the places that have entropy generation and find the entropy generated in the boiler heat exchanger per kg ammonia flowing Solution a The hot waterammonia boiler b The condensing ammonia 15AoE AC to the ambient 20AoE AC heat transfer c The feedwater heater has mixing of a flow at state 6 with a flow at state 2 State 3 xA3E A 0 hA3E A 17165 kJkg vA3E A 000156 mA3E Akg sA3E A 06793 kJkgK State 5 hA5E A 16146 kJkg sA5E A 54971 kJkg K CV Pump P2 rev and adiabatic so sA4E A sA3E A wAP2E A hA4E A hA3E A vA3E APA4E A PA3E A 0001561000 400 0936 kJkg hA5E A hA4E A hA5E A hA3E A wAP2E A 16146 17165 0936 1442 kJkg CV Boiler Energy Eq 0 Am E AammE A hA5E A hA4E A Am E AH2OE A hAinE A hAexE A Am E AH2OE A Am E AammE A A h5 h4 Ehin hex E A A 1442 50369 3769E A 11373 Entropy Eq 0 Am E AammE A sA4E A sA5E A Am E AH2OE A sAinE A sAexE A S gen sAammE A sA5E A sA4E A Am E AH2OE A Am E AammE AsAinE A sAexE A 54971 06793 1137315275 11924 1007 kJkgK 1 TURBINE COND FWH P2 4 2 3 5 6 7 P1 BOILER HP LP cb H2O in H2O ex s 1 2 3 4 5 6 7 236 kPa 1000 kPa 400 kPa T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 972 Repeat Problem 943 assuming the turbine has an isentropic efficiency of 85 The physical components and the Ts diagram is as shown in Fig 912 in the main text for one open feedwater heater The same state numbering is used From the Steam Tables State 5 P T h5 40698 kJkg s5 70544 kJkg K State 1 P x 0 h1 19181 kJkg v1 000101 mA3E Akg State 3 P x 0 h3 7628 kJkg v3 0001127 mA3E Akg Pump P1 wP1 v1P2 P1 000101 990 1 kJkg h2 h1 wP1 19281 kJkg Turbine 56 s6 s5 h6 30137 kJkg wT56s h5 h6 40698 30137 10561 kJkg wT56AC 10561 085 89769 kJkg wT56AC h5 h6AC h6AC h5 wT56AC 40698 89769 317211 kJkg Feedwater Heater AmE ATOT AmE A5 xAmE A5h6AC 1 xAmE A5h2 AmE A5h3 x h3 h2 h6 h2 A 7628 19281 317211 19281E A 01913 To get the turbine work apply the efficiency to the whole turbine ie the first section should be slightly different s7s s6s s5 x7s 70544 0649375009 085391 h7s 19181 085391 239282 22351 kJkg wT57s h5 h7s 40698 22351 18347 kJkg wT57AC wT57s T 15595 h5 h7AC h7AC 25103 kJkg Find specific turbine work to get total flow rate AmE ATOT W T xwT56 1xwT57 A 5000 01913 89769 08087 15595E A 3489 kgs AQE AL AmE ATOT1 xh7 h1 3489 0808725103 19181 6542 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 973 Steam leaves a power plant steam generator at 35 MPa 400AoE AC hA1E A 32223 kJkg sA1E A 68405 kJkg K and enters the turbine at 34 MPa 375AoE AC hA2E A 31657 kJkg sA2E A 67675 kJkg K The isentropic turbine efficiency is 88 and the turbine exhaust pressure is 10 kPa Condensate leaves the condenser and enters the pump at 35AoE AC 10 kPa The isentropic pump efficiency is 80 and the discharge pressure is 37 MPa The feedwater enters the steam generator at 36 MPa 30AoE AC h 1290 kJkg Calculate the thermal efficiency of the cycle and the entropy generation for the process in the line between the steam generator exit and the turbine inlet assuming an ambient temperature of 25AoE AC ST GEN 3 6 COND TURBINE 1 2 P 4 5 η 088 T s 400 C 375 C 2 5s o o 6 4 3s 3 5 35 MPa 34 MPa 10 kPa 1 1 hA1E A 32223 kJkg sA1E A 68405 kJkg K 2 hA2E A 31657 kJkg sA2E A 67675 kJkg K 3s sA3SE A sA2E A xA3SE A 08157 hA3SE A 21436 kJkg wATsE A hA2E A hA3sE A 31657 21436 10221 kJkg wATacE A ηATsE A wATsE A 8994 kJkg 3ac hA3E A hA2E A wATacE A 22663 kJkg wAPsE A vfP5 P4 00010063700 10 37 kJkg wAPacE A wAPsE AηAPsE A 46 kJkg qAHE A hA1E A hA6E A 32223 1290 30933 kJkg η wANETE AqAHE A 8994 4630933 0289 CV Line from 1 to 2 w A0EA Energy Eq q hA2E A hA1E A 31657 32223 566 kJkg Entropy Eq sA1E A sAgenE A qTA0E A sA2E A sAgenE A sA2E A sA1E A qTA0E A 67675 68405 56629815 0117 kJkg K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 974 Find the entropy generation per unit mass leaving the open FWH in Problem 940 The setup follows Fig912 States hA2E A 41902 hA6E A 282786 hA3E A 76279 all kJkg sA2E A 13068 sA6E A 66939 sA3E A 21386 x 0 at 1 MPa all kJkgK Analyze the FWH which leads to Eq95 y AmE A6 AmE A5 the extraction fraction y A h3 h2 Eh6 h2 E A A 76279 41902 282786 41902E A 01427 Entropy eq scaled to flow at 3 0 y sA6E A 1 y sA2E A sA3E A sAgenE sAgenE A sA3E A y sA6E A 1 y sA2E A 21386 01427 66939 1 01427 13068 0063 kJkgK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 975 Find the rate of entropy generation in the closed FWH in Problem 951 The schematic from Figure 913 has the feedwater from the pump coming at state 2 being heated by the extraction flow coming from the turbine state 6 so the feedwater leaves as saturated liquid state 4 and the extraction flow leaves as condensate state 6a 2 4 6 6a From table B1 h kJkg s kJkgK B14 100C 20 MPa hA2E A 43404 12917 B14 200C 20 MPa hA4E A 86047 23031 B13 4 MPa 275C hA6E A 288039 62145 B12 4 MPa sat liq hA6aE A 108729 27963 CV Feedwater Heater Energy Eq AmE A2hA2E A AmE A6hA6E A AmE A2hA4E A AmE A6hA6aE Entropy Eq 0 AmE A2sA2E A AmE A6sA6E A AmE A2sA4E A AmE A6sA6aE A ASE AgenE Since all four states are known we can solve for the extraction flow rate AmE A6 AmE A2 A h2 h4 Eh6a h6 E A 20 A 43404 86047 108729 288039E A kgs 4756 kgs ASE AgenE A AmE A2sA4E A sA2E A AmE A6 sA6aE A sA6E A 20 23031 12917 4756 27963 62145 20228 16257 3971 kWK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Cogeneration Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 976 A cogenerating steam power plant as in Fig 919 operates with a boiler output of 25 kgs steam at 7 MPa 500C The condenser operates at 75 kPa and the process heat is extracted as 5 kgs from the turbine at 500 kPa state 6 and after use is returned as saturated liquid at 100 kPa state 8 Assume all components are ideal and find the temperature after pump 1 the total turbine output and the total process heat transfer Solution Pump 1 Inlet state is saturated liquid hA1E A 16879 kJkg vA1E A 0001008 mA3E Akg wAP1E A v dP vA1E A PA2E A PA1E A 0001008 100 75 0093 kJkg wAP1E A hA2E A hA1E A hA2E A hA1E A wAP1E A 16888 kJkg TA2E A 403C Turbine hA5E A 34103 kJkg sA5E A 67974 kJkg K PA6E A sA6E A sA5E A xA6E A 09952 hA6E A 27386 kJkg PA7E A sA7E A sA5E A xA7E A 08106 hA7E A 21190 kJkg From the continuity equation we have the full flow from 5 to 6 and the remainder after the extraction flow is taken out flows from 6 to 7 AW E ATE A AmE A5E A hA5E A hA6E A 080AmE A5E A hA6E A hA7E A 25 34103 27386 20 27386 2119 16 7925 12 392 29185 MW AQ E AprocE A AmE A6E AhA6E A hA8E A 527386 41746 11606 MW 3 2 4 Steam generator Q H 1 W P2 W T Turbine Q Condenser Mixer Thermal process 5 6 7 8 P2 P1 L W P1 T s 1 2 3 7 6 5 8 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 977 A steam power plant has 4 MPa 500C into the turbine and to have the condenser itself deliver the process heat it is run at 101 kPa How much net power as work is produced for a process heat of 13 MW Solution From the Rankine cycle we have the states 1 101 kPa x A0E A vA1E A 0001043 mA3E Akg hA1E A 4186 kJkg 3 4 MPa 500C hA3E A 34452 kJkg sA3E A 7090 kJkg K CV Turbine sA4E A sA3E A xA4E A 7090 130686048 09562 hA4E A 41902 09562 225703 25772 kJkg wATE A hA3E A hA4E A 34452 25772 868 kJkg CV Pump wAPE A vA1E APA2E A PA1E A 00010434000 101 407 kJkg wAPE A hA2E A hA1E A hA2E A 41902 407 42309 kJkg CV Condenser qALoutE A hA4E A hA1E A 25772 41902 21582 kJkg AmE A AQ E AprocE A qALoutE A 13 000 kW 21582 kJkg 6023 kgs AW E ATE A AmE A wATE A wAPE A 6023 868 407 5203 kW Q W T 3 2 4 1 Condenser Boiler Turbine W P Q B T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 978 Steam must be generated at 150AoE AC for process heat in a food production facility This is done in a combined heat and power system as extraction steam from the turbine Assume the standard cycle has a turbine inlet of 3 MPa 450C and 60C in the condenser What pressure should be used for the extraction so there is a maximum of process heat at 150C available and the least amount of turbine work lost Most energy will be available if the extraction flow can condense at 150C Since the flow through the turbine follows from 5 to 6 to 7 the extraction point is selected as the saturation pressure for the 150C PA6E A PAsat 150CE A 476 kPa As sA5E A 70833 sAg 150CE A 68378 kJkgK state 6 is superheated vapor 3 2 4 Steam generator Q H 1 W P2 W T Turbine Q Condenser Mixer Thermal process 5 6 7 8 P2 P1 L W P1 T s 1 2 3 7 6 5 8 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 979 A 15 kgs steady supply of saturatedvapor steam at 500 kPa is required for drying a wood pulp slurry in a paper mill It is decided to supply this steam by cogeneration that is the steam supply will be the exhaust from a steam turbine Water at 20C 100 kPa is pumped to a pressure of 5 MPa and then fed to a steam generator with an exit at 400C What is the additional heat transfer rate to the steam generator beyond what would have been required to produce only the desired steam supply What is the difference in net power Solution Desired exit State 4 PA4E A 500 kPa sat vap xA4E A 10 TA4E A 1519C hA4E A hAgE A 27487 kJkg sA4E A sAgE A 68212 kJkgK Inlet State 20C 100 kPa hA1E A hAfE A 8394 kJkg vA1E A vAfE A 0001002 mA3E Akg Without Cogeneration The water is pumped up to 500 kPa and then heated in the steam generator to the desired exit T CV Pump wAPwoE A vA1E A PA4E A PA1E A 04 kJkg hA2E A hA1E A wAPwoE A 843 kJkg CV Steam Generator qAwoE A hA4E A hA2E A 26644 kJkg With Cogeneration The water is pumped to 5 MPa heated in the steam generator to 400C and then flows through the turbine with desired exit state CV Pump wAPwE A vdP vA1E A PA2E A PA1E A 491 kJkg hA2E A hA1E A wAPwE A 8885 kJkg CV Steam Generator Exit 400C 5 MPa hA3E A 319564 kJkg qAwE A hA3E A hA2E A 319564 8885 31068 kJkg CV Turbine Inlet and exit states given wAtE A hA3E A hA4E A 319564 27487 44694 kJkg Comparison Additional Heat Transfer qAwE A qAwoE A 31068 26644 4424 kJkg AQ E AextraE A AmE AqAwE A qAwoE A 6636 kW Difference in Net Power wAdiffE A wAtE A wAPwE A wAPwoE A wAdiffE A 44694 491 04 4424 kJkg AW E AdiffE A AmE AwAdiffE A 6636 kW By adding the extra heat transfer at the higher pressure and a turbine all the extra heat transfer can come out as work it appears as a 100 efficiency Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 980 In a cogenerating steam power plant the turbine receives steam from a high pressure steam drum h 34459 kJkg s 69108 kJkgK and a lowpressure steam drum h 28554 kJkg s 70592 kJkgK as shown in Fig P980 For the turbine calculation assume a mass weighted average entropy and neglect entropy generation by mixing The condenser is made as two closed heat exchangers used to heat water running in a separate loop for district heating The hightemperature heater adds 30 MW and the lowtemperature heater adds 31 MW to the district heating water flow Find the power cogenerated by the turbine and the temperature in the return line to the deaerator Solution Inlet states from Table B13 hA1E A 34459 kJkg sA1E A 69108 kJkg K hA2E A 28554 kJkg sA2E A 70592 kJkg K AmE ATOTE A AmE A1E A AmE A2E A 27 kgs Assume a reversible turbine and the two flows can mix without s generation 3 4 1 2 T W Turbine Energy Eq410 AmE A1E AhA1E A AmE A2E AhA2E A AmE A3E AhA3E A AmE A4E AhA4E A AW E ATE Entropy Eq77 AmE A1E AsA1E A AmE A2E AsA2E A AmE ATOTE AsAmixE A sAMIXE A 69383 kJkg K State 3 sA3E A sAMIXE A hA3E A 26324 kJkg xA3E A 0966 State 4 sA4E A sAMIXE A hA4E A 24135 kJkg xA4E A 0899 AW E ATE A 22 34459 5 28554 13 26324 14 24135 22 077 kW 22 MW District heating line AQ E ATOTE A AmE AhA95E A hA60E A 60 935 kW OK this matches close enough CV Both heaters Am E A3E AhA3E A AmE A4E AhA4E A AQ E ATOTE A AmE ATOTE AhAEXE 13 26324 14 24135 60 935 70752 27 hAEXE hAEXE A 262 hAfE A TAEXE A 625C Remark We could have computed the expansion from state 1 to PA2E A followed by a mixing process to find a proper state 2a from which we expand down to PA3E A and PA4E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 981 A boiler delivers steam at 10 MPa 550C to a twostage turbine as shown in Fig 919 After the first stage 25 of the steam is extracted at 14 MPa for a process application and returned at 1 MPa 90C to the feedwater line The remainder of the steam continues through the lowpressure turbine stage which exhausts to the condenser at 10 kPa One pump brings the feedwater to 1 MPa and a second pump brings it to 10 MPa Assume all components are ideal If the process application requires 5 MW of power how much power can then be cogenerated by the turbine Solution 5 hA5E A 35009 sA5E A 67561 kJkg K First ideal turbine T1 6 sA6E A sA5E A hA6E A 29324 kJkg wAT1E A hA5E A hA6E A 5685 kJkg Ideal turbine T2 State 7 sA7E A sA6E A sA5E A x A67561 06492 7501E A 08141 hA7E A 21399 kJkg P2 P1 C 1 2 8 4 3 5 6 7 T1 T2 Boiler Process heat 5 MW wAT2E A hA6E A hA7E A 29324 21399 7925 kJkg Now do the process heat requirement 8 hA8E A 3773 kJkg approx from the compressed liq Table at 500 kPa qAPROCE A hA6E A hA8E A 29324 3773 25551 kJkg AmE A6E A AQ E AqAPROCE A 5000 25551 19569 kgs 025 AmE ATOTE A AmE ATOTE A AmE A5E A 78275 kgs AmE A7E A AmE A5E A AmE A6E A 58706 kgs AW E ATE A AmE A5E AhA5E A AmE A6E AhA6E A AmE A7E AhA7E A 78275 35009 19569 29324 58706 21399 9102 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 982 A smaller power plant produces 25 kgs steam at 3 MPa 600C in the boiler It cools the condenser to an exit of 45C and the cycle is shown in Fig P982 An extraction is done at 500 kPa to an open feedwater heater in addition a steam supply of 5 kgs is taken out and not returned The missing 5 kgs water is added to the feedwater heater from a 20C 500 kPa source Find the needed extraction flow rate to cover both the feedwater heater and the steam supply Find the total turbine power output Solution The states properties from Tables B11 and B13 1 45oC x 0 hA1E A 18842 kJkg vA1E A 000101 mA3E Akg Psat 959 kPa 5 30 MPa 600oC hA5E A 368234 kJkg sA5E A 75084 kJkg K 3 500 kPa x 0 hA3E A 64021 kJkg 8 hA8E A 8441 kJkg 6 500 kPa sA6E A sA5E A from HP turbine hA6E A 309326 kJkg CV Pump 1 Reversible and adiabatic Incompressible so v constant Energy wAp1E A hA2E A hA1E A v dP vA1E APA2E A PA1E A 000101 500 96 0495 kJkg hA2E A hA1E A wAp1E A 18842 0495 188915 kJkg CV Turbine sections Entropy Eq sA7E A sA5E A 75084 kJkg K twophase state sA7E A 75084 06386 xA7E A 75261 xA7E A 09128 hA7E A 18842 09128 239477 23744 kJkg CV Feedwater heater including the makeup water flow y AmE A6E AAmE A5E A Energy eq AmE A8E AhA8E A AmE A5E A AmE A6E AhA2E A AmE A6E A AmE A8E AhA6E A AmE A5E AhA3E Divide by AmE A5E A and solve for y y A h3 h2 h6 h8 m E 8 m 5 h6 h2 E A A64021 188915 309326 8441525 E309326 188915E 03626 AmE A6E A y AmE A5E A 03626 25 9065 kgs CV Turbine energy equation AW E ATE A AmE A5E AhA5E A AmE A6E AhA6E A AmE A7E AhA7E A 25 368234 9065 309326 16935 23744 26 182 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Refrigeration cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 983 A refrigeration cycle as in Fig 923 can be used for cooling or for heating purposes using one of the two heat exchangers Suppose a refrigerator should cool meat at 10C in a 30C hot kitchen what are the minimum high and maximum low pressures in the cycle if the working substance is a R134a or b R410A Ideal refrigeration cycle TAcondE A 30C TAevapE A 10AoE AC TA1E Properties from Tbl B4 B5 T s 1 2 3 4 30 10 The condensing temperature TA3E A must be minimum 30C to move a heat transfer to the kitchen air The low temperature in the evaporator must at most 10C to be able to extract energy from the cold space R134a B5 PAhiE A PA3E A PAsatE A30C 771 kPa PAlowE A PA1E A PAsatE A10C 2017 kPa R410A B4 PAhiE A PA3E A PAsatE A30C 1885 kPa PAlowE A PA1E A PAsatE A10C 573 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 984 A refrigerator with R134a as the working fluid has a minimum temperature of 10C and a maximum pressure of 1 MPa Assume an ideal refrigeration cycle as in Fig 923 Find the specific heat transfer from the cold space and that to the hot space and the coefficient of performance Solution Exit evaporator sat vapor 10C from B51 hA1E A 39228 sA1E A 17319 kJkgK Exit condenser sat liquid 1 MPa from B51 hA3E A 25560 kJkg Compressor sA2E A sA1E A PA2E A from B52 hA2E A 42568 kJkg Evaporator qALE A hA1E A hA4E A hA1E A hA3E A 39228 25560 1367 kJkg Condenser qAHE A hA2E A hA3E A 42568 25560 1701 kJkg COP β qALE AwAcE A qALE AqAHE A qALE A 409 Ideal refrigeration cycle PAcondE A PA3E A PA2E A 1 MPa TAevapE A 10AoE AC TA1E Properties from Table B5 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 985 Repeat the previous Problem with R410A as the working fluid Will that work in an ordinary kitchen Solution Exit evaporator sat vapor 10C from B41 hA1E A 27578 sA1E A 10567 kJkgK Exit condenser sat liquid 1 MPa from B41 hA3E A 6892 kJkg Compressor sA2E A sA1E A PA2E A from B42 hA2E A 29081 kJkg Evaporator qALE A hA1E A hA4E A hA1E A hA3E A 27578 6892 2069 kJkg Condenser qAHE A hA2E A hA3E A 29081 6892 2219 kJkg COP β qALE AwAcE A qALE AqAHE A qALE A 138 Ideal refrigeration cycle PAcondE A PA3E A PA2E A 1 MPa TAevapE A 10AoE AC TA1E Properties from Table B4 T s 1 2 3 4 The 1 MPa is too small the condensing temperature is 725C and the qAHE A in the condenser can not be rejected to a kitchen normally at 20C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 986 The natural refrigerant carbon dioxide has a fairly low critical temperature Find the high temperature the condensing temperature and the COP if it is used in a standard cycle with high and low pressures of 6 MPa and 3 MPa Exit evaporator x 1 and 3 MPa from B32 hA1E A 32071 kJkg sA1E A 12098 kJkgK Exit condenser saturated liquid 6 MPa from B31 TA3E A 22C hA3E A 150 kJkg Exit compressor 6 MPa s sA1E A so interpolate in B32 TA2E A 459C hA2E A 34824 kJkg COP β A qL Ewc E A A h1 h3 Eh2 h1 E A 62 Remark The condensing T is too low for a standard refrigerator it can not push the heat transfer to the room which may be hotter than that Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 987 Consider an ideal refrigeration cycle that has a condenser temperature of 45C and an evaporator temperature of 15C Determine the coefficient of performance of this refrigerator for the working fluids R134a and R410A Solution Ideal refrigeration cycle TAcondE A 45AoE AC TA3E TAevapE A 15AoE AC TA1E T s 1 2 3 4 Property for R134a B5 R410A B4 Compressor hA1E A kJkg 3892 2739 sA2E A sA1E A kJkg K 17354 10671 PA2E A MPa 116 27283 TA2E A AoE AC 518 715 hA2E A kJkg 4299 32272 wACE A hA2E A hA1E 407 4882 Exp valve hA3E A hA4E A kJkg 26411 13361 Evaporator qALE A hA1E A hA4E 1251 14029 β qALE AwACE 307 287 For state 2 an interpolation between 1 and 12 MPa is needed for 116 MPa At 1 MPa s 17354 T 459 C and h 4268 kJkg At 12 MPa s 17354 T 533 C and h 4307 kJkg For state 2 an interpolation between 2 and 3 MPa is needed for 2728 MPa At 2 MPa s 10671 T 5468 C and h 313942 kJkg At 3 MPa s 10671 T 7780 C and h 3260 kJkg It would make more sense to use the CATT3 program Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 988 Do problem 987 with ammonia as the working fluid Solution Ideal refrigeration cycle TAcondE A 45AoE AC TA3E TAevapE A 15AoE AC TA1E T s 1 2 3 4 State 1 hA1E A 14246 kJkg sA1E A 55397 kJkgK State 2 s sA1E A PA2E A 1782 kPa TA2E A 1351AoE AC hA2E A 17313 kJkg For state 2 an interpolation between 16 and 2 MPa is needed for 1782 MPa At 16 MPa s 55397 T 1260 C and h 17122 kJkg At 20 MPa s 55397 T 1461C and h 17541 kJkg It would make more sense to use the CATT3 program wACE A hA2E A hA1E A 17313 14246 3067 kJkg State 34 hA4E A hA3E A 3963 kJkg qALE A hA1E A hA4E A 10283 kJkg β qALE AwACE A 3353 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 989 A refrigerator receives 500 W of electrical power to the compressor driving the cycle flow of R134a The refrigerator operates with a condensing temperature of 40AoE AC and a low temperature of 5AoE AC Find the COP for the cycle and its cooling capacity Solution Ideal refrigeration cycle TAcondE A 40AoE AC TA3E TAevapE A 5AoE AC TA1E Properties from Table B5 T s 1 2 3 4 Exit evaporator sat vapor 5C from B51 hA1E A 3953 kJkg sA1E A 17288 kJkgK Compressor sA2E A sA1E A PA2E A from B52 TA2E A 45AoE AC hA2E A 425 kJkg wACE A hA2E A hA1E A 425 3953 297 kJkg Exit condenser sat liquid PA2E A PA3E A 1017 kPa from B51 hA3E A 25650 kJkg Evaporator qALE A hA1E A hA4E A hA1E A hA3E A 3953 25650 1388 kJkg Condenser qAHE A hA2E A hA3E A 425 2565 1685 kJkg COP β qALE AwAcE A qALE AqAHE A qALE A 1388 297 467 AQ E ALE A β AW E A 467 500 W 2335 W 23 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 990 A new airconditioner using R410A is used in heat pump mode The high pressure is 2000 kPa and the low pressure is 400 kPa It warms a house at 20C driven by an electric power input of 2 kW in an ambient at 5C Find the COP the heating rate and the rate of entropy generation for the heat pump Exit evaporator sat vapor 400 kPa from B42 hA1E A 2719 sA1E A 10779 kJkgK Exit condenser sat liquid 2 MPa 3231AoE AC from B41 hA3E A 11021 kJkg Compressor sA2E A sA1E A PA2E A from B42 hA2E A 31743 kJkg Evaporator qALE A hA1E A hA4E A hA1E A hA3E A 2719 11021 16169 kJkg Condenser qAHE A hA2E A hA3E A 31743 11021 20722 kJkg Compressor wAcE A hA2E A hA1E A 31743 2719 4553 kJkg COP β qAHE AwAcE A 207224553 455 Heating rate AQ E AHE A β AW E A 455 2 91 kW CV The AC unit including the two heat exchangers Energy 0 qALE A qAHE A wAcE Entropy 0 qALE ATAambE A qAHE ATAhouseE A sAgenE AS genE A AQ E AHE ATAhouseE A AQ E ALE ATAambE A A 91 kW 29315 KE A A91 2 kW E26815 KE A 000456 kWK Ideal refrigeration cycle PAcondE A PA3E A PA2E A 2 MPa TAevapE A 1998AoE AC TA1E Properties from Table B4 T s 1 2 3 4 Comment The entropy generation does not include the heat loss from the house to the ambient this is why we heat a house if that was included then qALE A comes in at TAambE A and qAHE A leaves at TAambE A so the total entropy generation becomes wTAambE A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 991 Consider the heat pump in the previous problem Repeat the questions for a system with an actual compressor with an exit state of 2000 kPa 65C Exit evaporator sat vapor 400 kPa from B42 hA1E A 2719 sA1E A 10779 kJkgK Ideal Compressor sA2E A sA1E A PA2E A from B42 hA2E A 31743 kJkg wAcE A hA2E A hA1E A 31743 2719 4553 kJkg Actual compressor exit hA2 acE A 32627 kJkg sA2E A 11043 kJkgK Actual Compressor wAc acE A hA2 acE A hA1E A 32627 2719 5437 kJkg Exit condenser sat liquid 2 MPa 3231AoE AC from B41 hA3E A 11021 kJkg Condenser qAHE A hA2 acE A hA3E A 32627 11021 21606 kJkg Evaporator qALE A hA1E A hA4E A hA1E A hA3E A 2719 11021 16169 kJkg COP β qAHE AwAc acE A 216065437 3974 Heating rate AQ E A β AW E A 3974 2 kW 795 kW CV The AC unit including the two heat exchangers Energy 0 qALE A qAHE A wAcE Entropy 0 qALE ATAambE A qAHE ATAhouseE A sAgenE AS genE A AQ E AHE ATAhouseE A AQ E ALE ATAambE A A795 kW 29315 KE A A795 2 kW E26815 KE A 000493 kWK Actual refrigeration cycle PAcondE A PA3E A PA2E A 2 MPa TAevapE A 1998AoE AC TA1E Properties from Table B4 T s 1 2 s 3 4 2 ac Comment The entropy generation does not include the heat loss from the house to the ambient this is why we heat a house if that was included then qALE A comes in at TAambE A and qAHE A leaves at TAambE A so the total entropy generation becomes wTAambE A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 992 A heat pump for heat upgrade uses ammonia with a low temperature of 25AoE AC and a high pressure of 5000 kPa If it receives 1 MW of shaft work what is the rate of heat transfer at the high temperature State 1 Table B21 hA1E A 14635 kJkg sA1E A 50293 kJkgK State 3 Table B21 hA3E A hAfE A 6319 kJkg Entropy compressor sA2E A sA1E A TA2E A 156AoE AC hA2E A 17091 kJkg Energy eq compressor wACE A hA2E A hA1E A 2456 kJkg Energy condenser qAHE A hA2E A hA3E A 10772 kJkg Scaling to power input AQ E AHE A qAHE A A W IN EwC E A 10772 kJkg A1000 2456E A kgs 4386 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 993 Reconsider the heat pump in the previous problem Assume the compressor is split into two first compress to 2000 kPa then take heat transfer out at constant P to reach saturated vapor then compress to the 5000 kPa Find the two rates of heat transfer at 2000 kPa and at 5000 kPa for a total of 1 MW shaft work input Ideal heat pump TAevapE A 25AoE AC TA1E T s 1 2c 3 4 2a 2b State 1 Table B21 hA1E A 14635 kJkg sA1E A 50293 kJkgK State 3 Table B21 hA3E A hAfE A 6319 kJkg Entropy compressor 1 sA2aE A sA1E A TA2aE A 753AoE AC hA2aE A 15591 kJkg Energy eq compressor 1 wAC1E A hA2aE A hA1E A 15591 14635 956 kJkg Exit heat exchanger 1 hA2bE A 14715 kJkg sA2bE A 4768 kJkgK Entropy compressor 2 sA2cE A sA2bE A TA2cE A 1243AoE AC hA2cE A 160137 kJkg Energy eq compressor 2 wAC2E A hA2cE A hA2bE A 160137 14715 12987 kJkg Total power input AW E AINE A Am E A wAC1E A wAC2E A Am E A A W IN EwC1 wC2 E A A 1000 956 12987E A 44352 kgs Heat exchanger 1 AQ E AH1E A Am E AhA2aE A hA2bE A 443521559114715 3885 kW Heat exchanger 2 AQ E AH2E A Am E AhA2cE A hA3E A 443521601376319 42998 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 994 A refrigerator with R134a as the working fluid has a minimum temperature of 10C and a maximum pressure of 1 MPa The actual adiabatic compressor exit temperature is 50C Assume no pressure loss in the heat exchangers Find the specific heat transfer from the cold space and that to the hot space the coefficient of performance and the isentropic efficiency of the compressor Solution State 1 Inlet to compressor sat vapor 10C hA1E A 39228 sA1E A 17319 kJkgK State 2 Actual compressor exit hA2ACE A 43124 kJkg State 3 Exit condenser sat liquid 1 MPa hA3E A 25560 kJkg State 4 Exit valve hA4E A hA3E A CV Evaporator qALE A hA1E A hA4E A hA1E A hA3E A 39228 25560 1367 kJkg CV Ideal Compressor wACSE A hA2SE A hA1E A sA2SE A sA1E State 2s 1 MPa s 17319 kJkg K TA2SE A 449C hA2SE A 4257 kJ kg wACSE A hA2SE A hA1E A 3342 kJkg CV Actual Compressor wACE A hA2ACE A hA1E A 3896 kJkg β A qL EwC E A 351 ηACE A wACSE AwACE A 0858 CV Condenser qAHE A hA2ACE A hA3E A 1863 kJkg Ideal refrigeration cycle with actual compressor PAcondE A PA3E A PA2E A 1 MPa TA2ACE A 50AoE AC TAevapE A 10AoE AC TA1E Properties from Table B5 T s 1 2s 3 4 2ac Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 995 An airconditioner in the airport of Timbuktu runs a cooling system using R410A with a high pressure of 1800 kPa and a low pressure of 200 kPa It should cool the desert air at 45oC down to 15oC Find the cycle COP Will the system work Solution Ideal refrigeration cycle Pcond P2 P3 1800 kPa Pevap P 1 T s 1 2 3 4 State 1 T1 370oC h1 26427 kJkg s1 11192 kJkgK State 2 P2 1800 kPa s2 s1 T2 636oC h2 3279 kJkg wC h2 h1 3279 26427 6363 kJkg State 34 P4 P3 x3 0 T3 2822oC h4 h3 103096 kJkg qL h1 h4 26427 103096 16117 kJkg β qLwC 161176363 253 The heat rejection from 23 to ambient at 45oC has T3 2822oC not hot enough so the system will not work The high pressure must be so also T3 45oC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 996 A heat pump uses R410a with a high pressure of 3000 kPa and an evaporator operating at 10oC so it can absorb energy from underground water layers at 4oC Find the COP and the temperature it can deliver energy at Solution Ideal refrigeration cycle R410A Pcond P2 P3 3000 kPa Tevap 10oC T1 T s 1 2 3 4 State 1 h1 27578 kJkg s1 10567 kJkgK State 2 P2 s2 s1 T2 753oC h2 32243 kJkg wC h2 h1 32243 27578 4665 kJkg State 34 h3 h4 x3 0 T3 4907oC h4 h3 14178 kJkg qH h2 h3 32243 14178 18065 kJkg β qHwC 18065 4665 387 It can deliver heat at about 49oC T3 minus a ΔT for the heat transfer rate a smaller amount of heat is delivered between T2 and T3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 997 Consider an ideal heat pump that has a condenser temperature of 50C and an evaporator temperature of 0C Determine the coefficient of performance of this heat pump for the working fluids R134a and ammonia Solution Ideal heat pump Tcond 50oC T3 Tevap 0oC T1 T s 1 2 3 4 CV Property for From Table R134a B5 NH3 B2 h1 kJkg 39836 144232 Compressor s2 s1 kJkgK 17262 53313 P2 MPa 13181 20333 T2 oC 551 1156 h2 kJkg 42955 167284 wC h2 h1 3119 23052 Exp valve h3 h4 kJkg 27183 42158 Condenser qH h2 h3 15772 125126 β qHw C 506 5428 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 998 A refrigerator in a meat warehouse must keep a low temperature of 15C and the outside temperature is 20C It uses ammonia as the refrigerant which must remove 5 kW from the cold space Find the flow rate of the ammonia needed assuming a standard vapor compression refrigeration cycle with a condenser at 20C Solution Basic refrigeration cycle T1 T4 15C T3 20C Table B3 h4 h3 2743 kJkg h1 hg 14246 kJkg Q L m amm qL m amm h1 h4 qL 14246 2743 11503 kJkg m amm 50 kW 11503 kJkg 000435 kgs Ideal refrigeration cycle Tcond 20oC Tevap 15oC T1 Properties from Table B2 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 999 A refrigerator has a steady flow of R410A as saturated vapor at 20C into the adiabatic compressor that brings it to 1400 kPa After the compressor the temperature is measured to be 60C Find the actual compressor work and the actual cycle coefficient of performance Solution Table B41 h1 27189 kJkg s1 10779 kJkg K P2 P3 1400 kPa T3 1888C h4 h3 hf 8745 kJkg h2 ac 33007 kJkg CV Compressor actual Energy Eq wC ac h2 ac h1 33007 27189 5818 kJkg CV Evaporator Energy Eq qL h1 h4 h1 h3 27189 8745 18444 kJkg β qL wC ac 18444 5818 317 Ideal refrigeration cycle with actual compressor Tcond 1888oC Tsat 1400 kPa T2 60oC Tevap 20oC T1 Properties from Table B4 T s 1 2s 3 4 2ac Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9100 The air conditioner in a car uses R134a and the compressor power input is 15 kW bringing the R134a from 2017 kPa to 1200 kPa by compression The cold space is a heat exchanger that cools atmospheric air from the outside 30C down to 10C and blows it into the car What is the mass flow rate of the R134a and what is the low temperature heat transfer rate How much is the mass flow rate of air at 10C Standard Refrigeration Cycle Table B5 h1 39228 kJkg s1 17319 kJkg K h4 h3 266 kJkg CV Compressor assume ideal m 1 m 2 wC h2 h1 s2 s1 s gen P2 s s1 h2 4295 kJkg wC 372 kJkg m wC W C m 15 372 00403 kgs CV Evaporator Q L m h1 h4 0040539228 266 521 kW CV Air Cooler m airhair Q L m airCpT m air Q L CpT 521 100420 026 kg s Ideal refrigeration cycle Pcond 1200 kPa P 3 Pevap 2017 kPa P 1 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9101 A refrigerator in a laboratory uses R134a as the working substance The high pressure is 1200 kPa the low pressure is 165 kPa and the compressor is reversible It should remove 500 W from a specimen currently at 10C not equal to TL in the cycle that is inside the refrigerated space Find the cycle COP and the electrical power required Solution State 1 165 kPa x 1 Table B51 h1 3892 kJkg s1 17354 kJkg K State 3 1200 kPa 4631C x 0 Table B51 h3 26613 kJkg CV Compressor Energy Eq wC h2 h1 Entropy Eq s2 s1 sgen s1 State 2 12 MPa s2 s1 17354 kJkg T2 534oC h2 43067 kJkg wC h2 h1 43067 3892 4147 kJkg Energy Eq evaporator qL h1 h4 h1 h3 3892 26613 12307 kJkg COP Refrigerator β qL wC 12307 4147 297 Power W IN Q L β 500 W 297 1683 W Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9102 Consider the previous problem and find the two rates of entropy generation in the process and where they occur Solution State 1 165 kPa x 1 Table B51 h1 3892 kJkg s1 17354 kJkg K State 3 1200 kPa x 0 Table B51 h3 26613 kJkg s3 12204 kJkg K Energy Eq evaporator qL h1 h4 h1 h3 3892 26613 12307 kJkg Mass flow rate m Q L qL 05 12307 000406 kgs CV Valve Energy Eq h4 h3 26613 kJkg x4 h4 hfh fg x4 26613 18019 2090 04112 s4 sf x4 sfg 09258 x4 08096 12587 kJkg K Entropy Eq sgen s4 s3 12587 12204 00383 kJkg K S gen valve m sgen 000406 00383 1000 0155 WK There is also entropy generation in the heat transfer process from the specimen at 10C to the refrigerant T1 15C Tsat 165 kPa S gen inside Q L 1 Tspecimen 1 TL 500 1 25815 1 26315 00368 WK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9103 A refrigerator using R134a is located in a 20C room Consider the cycle to be ideal except that the compressor is neither adiabatic nor reversible Saturated vapor at 20C enters the compressor and the R134a exits the compressor at 50C The condenser temperature is 40C The mass flow rate of refrigerant around the cycle is 02 kgs and the coefficient of performance is measured and found to be 23 Find the power input to the compressor and the rate of entropy generation in the compressor process Solution Table B5 P2 P3 Psat 40C 1017 kPa h4 h3 25654 kJkg s2 17472 kJkg K h2 43087 kJkg s1 17395 kJkg K h1 38608 kJkg β qL wC wC qL β h1 h4 β 38608 25654 23 5632 W C m wC 1126 kW CV Compressor h1 wC q h2 qin h2 h1 wC 43087 38608 5632 1153 kJkg ie a heat loss s1 dQT sgen s2 sgen s2 s1 q To 17472 17395 1153 29315 0047 kJkg K S gen m sgen 02 0047 00094 kW K Ideal refrigeration cycle with actual compressor Tcond 40oC T2 50oC Tevap 20oC T1 Properties from Table B5 T s 1 2s 3 4 2ac Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9104 A small heat pump unit is used to heat water for a hotwater supply Assume that the unit uses ammonia and operates on the ideal refrigeration cycle The evaporator temperature is 15C and the condenser temperature is 60C If the amount of hot water needed is 01 kgs determine the amount of energy saved by using the heat pump instead of directly heating the water from 15 to 60C Solution Ideal ammonia heat pump T1 15oC T3 60oC From Table B21 h1 14563 kJkg s2 s1 51444 kJkg K P2 P3 2614 MPa h3 4728 kJkg T s 1 2 3 4 Entropy compressor s2 s1 T2 1116oC h2 1643 kJkg Energy eq compressor wC h2 h1 1867 kJkg Energy condenser qH h2 h3 11702 kJkg To heat 01 kgs of water from 15oC to 60oC Q H2O m h 0125111 6298 1881 kW Using the heat pump W IN Q H2OwCqH 18811867 11702 30 kW a saving of 158 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9105 In an actual refrigeration cycle using R134a as the working fluid the refrigerant flow rate is 005 kgs Vapor enters the compressor at 150 kPa 10oC h1 3942 kJkg s1 1739 kJkgK and leaves at 12 MPa 75oC h2 4542 kJkg s2 1805 kJkgK The power input to the nonadiabatic compressor is measured and found be 24 kW The refrigerant enters the expansion valve at 115 MPa 40oC h 2564 kJkg and leaves the evaporator at 160 kPa 15oC h 3898 kJkg Determine the entropy generation in the compression process the refrigeration capacity and the coefficient of performance for this cycle Solution Actual refrigeration cycle 1 compressor inlet T1 10oC P1 150 kPa 2 compressor exit T2 75oC P2 12 MPa 3 Expansion valve inlet T3 40oC P3 115 MPa 5 evaporator exit T5 15oC P5 160 kPa T 1 3 5 2 s 4 Table B5 CATT3 h1 3942 kJkg s1 1739 kJkgK h2 4542 kJkg s2 1805 kJkgK CV Compressor h1 qCOMP wCOMP h2 s1 dqT sgen s2 wCOMP W COMPm 24005 480 kJkg qCOMP h2 wCOMP h1 4542 480 3942 12 kJkg sgen s2 s1 q To 1805 1739 1229815 00258 kJkgK CV Evaporator qL h5 h4 3898 2564 1334 kJkg Q L m qL 005 1334 667 kW COP β q qL wCOMP 1334 480 278 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Extended refrigeration cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9106 One means of improving the performance of a refrigeration system that operates over a wide temperature range is to use a twostage compressor Consider an ideal refrigeration system of this type that uses R410A as the working fluid as shown in Fig 925 Saturated liquid leaves the condenser at 40oC and is throttled to 20o C The liquid and vapor at this temperature are separated and the liquid is throttled to the evaporator temperature 50oC Vapor leaving the evaporator is compressed to the saturation pressure corresponding to 20oC after which it is mixed with the vapor leaving the flash chamber It may be assumed that both the flash chamber and the mixing chamber are well insulated to prevent heat transfer from the ambient Vapor leaving the mixing chamber is compressed in the second stage of the compressor to the saturation pressure corresponding to the condenser temperature 40oC Determine the COP of the system Compare to the COP of a simple ideal refrigeration cycle operating over the same condenser and evaporator ranges as those of the twostage compressor unit studied in this problem T2 907 oC h2 35035 kJkg ROOM COND EVAP Flash Chamber L Q H Q COMP ST2 COMP ST1 MIXCHAM COLD SPACE satliq 40 C o SATVAP 50 C o 20 C o satvapor 20 C o 1 2 3 4 5 6 7 8 9 satliq R410A refrigerator with 2stage compression CV expansion valve upper loop h2 h1 12409 2824 x2 24365 x2 03934 m3 x2m2 x2m1 03934 kg for m11 kg m6 m1 m3 06066 kg CV expansion valve lower loop 1 2 T 3 4 s 5 6 7 8 9 40 50 20 C o Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful h7 h6 2824 138 x7 2716 x7 015478 QL m6 h8 h7 06066 25780 2824 13925 kJ qL QL m1 13925 kJkgm 1 CV 1st stage compressor s8 s9 11568 kJkgK P9 PSAT 20 oC 03996 MPa T9 27 oC h9 29282 kJkg CV mixing chamber assume constant pressure mixing Energy Eq m6h9 m3h3 m1h 4 or h4 06066 29282 03934 27189 2846 kJkg h4 P4 400 kPa T4 63 oC s4 11261 kJkg K CV 2nd stage compressor P4 400 kPa P9 P3 P5 Psat 40oC 24207 MPa s5 s4 T5 813oC h5 33942 kJkg CV condenser Energy Eq qH h5 h1 33942 12409 21533 kJkg β2 stage qLqH qL 1392521533 13925 183 b 1 stage compression h3 h4 12409 kJkg h1 25780 kJkg qL h1 h4 1337 kJkg s1 s2 11568 P2 24207 MPa T2 907 oC h2 35035 1 2 T 3 4 s 40 C o 50 C o qH h2 h3 35035 12409 22626 kJkg β1 stage qLqH qL 133722626 1337 144 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9107 A cascade system with one refrigeration cycle operating with R410A has an evaporator at 40oC and a high pressure of 1200 kPa The high temperature cycle uses R134a with an evaporator at 0oC and a high pressure of 1200 kPa Find the ratio of the two cycles mass flow rates and the overall COP R134a C 1 COND 3 4 0 C o P 1200 kPa 2 sat vapor R410a C EVAP 2 1 T 40 C o 1 P 1200 kPa 3 sat vapor 3 sat liquid sat liquid 4 R134a cycle R410A cycle ToC P h s ToC P h s 1 0 294 39836 17262 1 40 175 26283 11273 2 4699 1200 423296 17262 2 4505 1200 31765 11273 3 4631 1200 26613 3 1343 1200 7866 4 0 26613 4 40 175 7866 m m h1 h4 h2 h3 39836 26613 31765 7866 05533 qL h1 h4 26283 7866 18417 kJkg W TOTm h2 h1 m m h2 h1 31765 26283 1 05533 423296 39836 9989 kJ kg β Q LW TOT 184179989 1844 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9108 A cascade system is composed of two ideal refrigeration cycles as shown in Fig 928 The hightemperature cycle uses R410A Saturated liquid leaves the condenser at 40C and saturated vapor leaves the heat exchanger at 20C The lowtemperature cycle uses a different refrigerant R23 Saturated vapor leaves the evaporator at 80C h 330 kJkg and saturated liquid leaves the heat exchanger at 10C h 185 kJkg R23 out of the compressor has h 405 kJkg Calculate the ratio of the mass flow rates through the two cycles and the COP of the system R410a C 1 COND 3 4 20 C o T 40 C o 3 2 sat vapor R23 C EVAP 2 1 T 80 C o 1 T 10 C o 3 sat vapor 3 sat liquid sat liquid 4 T s 1 2 3 4 T s 1 2 3 4 R410A ToC P h s ToC P h s 1 20 0400 27189 10779 1 80 012 330 176 2 71 2421 32261 10779 2 50 190 405 176 3 40 2421 12409 3 10 190 185 4 20 12409 4 80 012 185 m m h1 h4 h2 h3 27189 12409 405 185 06718 qL h1 h4 330 185 145 kJkg W TOTm h2 h1 m m h2 h1 405330 1 06718 3226127189 1505 kJ kg β Q LW TOT 1451505 096 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9109 A split evaporator is used to provide cooling of the refrigerator section and separate cooling of the freezer section as shown in Fig P9109 Assume constant pressure in the two evaporators How does the COP QL1 QL2W compare to a refrigerator with a single evaporator at the lowest temperature Throttle processes h4 h3 h5 h6 Refrigerator qL R h5 h4 freezer qL F h1 h6 Add the two heat transfers qL R qL F h5 h4 h1 h6 h1 h 4 which is the same as for the standard cycle expanding to the lowest pressure COPsplit COPstd h1 h4 h2 h1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9110 A refrigerator using R410A is powered by a small natural gas fired heat engine with a thermal efficiency of 25 as shown in Fig P9110 The R410A condenses at 40C and it evaporates at 20C and the cycle is standard Find the two specific heat transfers in the refrigeration cycle What is the overall coefficient of performance as QLQ1 Solution Evaporator Inlet State is saturated liqvap with h4 h3 12409 kJkg The exit state is saturated vapor with h1 27189 kJkg qL h1 h4 h1 h3 1478 kJkg Compressor Inlet State 1 and Exit State 2 about 242 MPa wC h2 h1 s2 s1 10779 kJkgK 2 T2 70C h2 3226 kJkg wC h2 h1 5071 kJkg Condenser Brings it to saturated liquid at state 3 qH h2 h3 3226 12409 1985 kJkg Overall Refrigerator β qL wC 1478 5071 2915 Heat Engine W HE ηHEQ 1 W C Q L β Q L Q 1 ηβ 025 2915 0729 Ideal refrigeration cycle Tcond 40oC T3 Tevap 20oC T1 Properties from Table B4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Ammonia absorption cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9111 Notice the configuration in Fig929 has the left hand side column of devices substitute for a compressor in the standard cycle What is an expression for the equivalent work output from the left hand side devices assuming they are reversible and the high and low temperatures are constant as a function of the pump work W and the two temperatures The left hand side devices works like a combination of a heat engine with some additional shaft work input We can analyze this with a control volume around all the devices that substitute for the compressor in the standard cycle CV Pump absorber heat exchanger and generator This CV has an inlet flow at state 1 and exit flow at state 2 with numbers as in the standard cycle Energy Eq 0 h1 w qH qL h2 all per mass flow at 1 and 2 Entropy Eq 0 s1 TH qH TL qL s2 0 Now solve for qL from the entropy equation and substitute into the energy equation qL TL s1 s2 TL TH qH w 1 TL TH qH h2 h1 TL s2 s1 The high T heat transfer acts as if it was delivered to a Carnot heat engine and the Carnot heat engine work output was added to the shaft work w That sum gives the increase in exergy from state 1 to state 2 Notice in the standard cycle s2 s1 and the last term is zero Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9112 As explained in the previous Problem the ammonia absorption cycle is very similar to the setup sketched in Problem 9110 Assume the heat engine has an efficiency of 30 and the COP of the refrigeration cycle is 30 what is then the ratio of the cooling to the heating heat transfer QLQ1 Heat Engine W ηQ1 03 Q1 Refrigerator β QL W QL β W β ηQ1 So now QLQ1 β η 3 03 09 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9113 Give an estimate for the overall COP of an ammonia absorption cycle used as a chiller to cool water to 5oC in an 25oC ambient when the small pump work is neglected A heat source is available at 100oC Find also the efficiency of the heat engine part and the COP of the refrigeration cycle part The heat engine operates between the 100C source and the 25C ambient so ηHE 1 ToTs 1 29815 37315 020 To estimate the COP of the basic ammonia cycle consider the highlow temperatures to be as T2 25C and T4 5C From B21 we get P2 Psat 25C 1003 kPa h3 h2 29825 kJkg h4 14473 kJkg Assume T1 100C P1 P2 h1 16643 kJkg COP qL wC h4 h2h1 h4 14473 29825 16643 14473 53 Neglecting the small pump work we get COP COP ηHE 53 02 106 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9114 Consider a small ammonia absorption refrigeration cycle that is powered by solar energy and is to be used as an air conditioner Saturated vapor ammonia leaves the generator at 50C and saturated vapor leaves the evaporator at 10C If 3000 kJ of heat is required in the generator solar collector per kilogram of ammonia vapor generated determine the overall performance of this system Solution NH3 absorption cycle sat vapor at 50oC exits the generator sat vapor at 10oC exits the evaporator qH qGEN 3000 kJkg NH3 out of gen T s 1 2 Exit generator Evaporator exit CV Evaporator qL h2 h1 hg 10oC hf 50oC 14522 4216 10306 kJkg COP qLqH 103063000 034 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9115 The performance of an ammonia absorption cycle refrigerator is to be compared with that of a similar vaporcompression system Consider an absorption system having an evaporator temperature of 10C and a condenser temperature of 50C The generator temperature in this system is 150C In this cycle 042 kJ is transferred to the ammonia in the evaporator for each kilojoule transferred from the hightemperature source to the ammonia solution in the generator To make the comparison assume that a reservoir is available at 150C and that heat is transferred from this reservoir to a reversible engine that rejects heat to the surroundings at 25C This work is then used to drive an ideal vaporcompression system with ammonia as the refrigerant Compare the amount of refrigeration that can be achieved per kilojoule from the hightemperature source with the 042 kJ that can be achieved in the absorption system Solution T s 1 2 3 4 T1 10 oC h1 14308 kJkg s1 54673 kJkgK h4 h3 42148 kJkg For the rev heat engine ηTH 1 T LT H 1 2982 4232 0295 WC ηTH Q H 0295 kJ For the NH3 refrig cycle P2 P3 2033 kPa Use 2000 kPa Table s2 s1 54673 kJkgK T2 135C h2 1724 kJkg wC h2 h1 1724 14308 2932 kJkg qL h1 h4 14308 42148 10093 kJkg β qLwC 10093 2932 344 QL βwC 344 0295 1015 kJ This is based on the assumption of an ideal heat engine refrigeration cycle Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Exergy Concepts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9116 If we neglect the external irreversibilities due to the heat transfers over finite temperature differences in a power plant how would you define its second law efficiency The first law efficiency is a conversion efficiency as ηI wnet qH wnet h3 h2 The second law efficiency is the same ratio but expressed in exergy ηII output source wnet φH wnet φ3 φ2 or wnet φH φL The last expression must be used if the heat rejection at the low T is assigned any exergy value normally not Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9117 Find the exergy of the water at all four states in the Rankine cycle described in Problem 916 Assume that the hightemperature source is 500C and the low temperature reservoir is at 25C Determine the flow of exergy in or out of the reservoirs per kilogram of steam flowing in the cycle What is the overall cycle second law efficiency Solution States and properties are from solution to problem 916 Reference State 100 kPa 25C so 03673 kJkg K ho 10487 kJkg ψ1 h1 ho Tos1 so 25111 10487 2981508311 03673 7958 kJkg ψ2 25414 10487 2981508311 03673 ψ1 303 1099 kJkg ψ3 3344 10487 2981570833 03673 123675 kJkg ψ4 ψ3 wTs 123675 10096 2271 kJkg ψH 1 ToTHqH 06144 308986 18984 kJkg ψL 1 ToToqC 0 kJkg ηII wNETψH 10096 30318984 053 Notice TH T3 TL T4 T1 so cycle is externally irreversible Both qH and qC are over finite T Energy Transfers kJkg Exergy Transfers kJkg HE cb T T q 3090 H o q 2083 H L w 1010 w 3 HE cb T T φ 1898 H o φ 219 H L w 1010 w 3 φ 0 φ 1226 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9118 A condenser is maintained at 60oC by cooling it with atmospheric air coming in at 20oC and leaving at 35 oC The condenser must reject 25 MW from the water to the air Find the flow rate of air and the second law efficiency of the heat exchanger CV Condenser a dual steady flow 1 inlet and 1 exit for air and water each The two flows exchange energy with no heat transfer tofrom the outside 4 water a air 1 b Energy Eq 0 m H2O h4 h1 m air ha hb m H2O h4 h1 Q m air hb ha m air Cp Tb Ta m air Q Cp Tb Ta 25 000 kW 1004 15 kJkg 1660 kgs Since the water is at constant T the change in entropy equals Q T and the exergy terms for the water and air becomes ΔΦ water Q 1 T0 TH2O 25 MW 1 29315 33315 30016 kW sb sa Cp ln TbTa 1004 ln 3081529315 00501 kJkgK ΔΦ air m air hb ha T0 sb sa Q m airT0 sb sa 25 000 1660 29315 00501 61989 kW ηII ΔΦ air ΔΦ water 61989 30016 02065 Comment We used T0 20oC 29315 K since the air comes in at that T in order not to straddle the T0 with the air temperatures Recall exergy is positive for Ts higher and lower than T0 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9119 Find the flows and fluxes of exergy in the condenser of Problem 929 Use those to determine the second law efficiency To 15C For this case we select To 15C 288 K the ocean water temperature The states properties from Tables B11 and B13 1 45oC x 0 h1 18842 kJkg 3 30 MPa 600oC s3 75084 kJkg K 5 6 4 1 cb CV Turbine wT h3 h4 s4 s3 s4 s3 75084 06386 x4 75261 x4 09128 h4 18842 09128 239477 23744 kJkg CV Condenser qL h4 h1 23744 18842 2186 kJkg Q L m qL 25 2186 5465 MW m ocean Cp T m ocean Q L Cp T 54 650 418 3 4358 kgs The net drop in exergy of the water is Φ water m water h4 h1 Tos4 s1 25 23744 1884 285 75084 06386 54 650 48 947 5703 kW The net gain in exergy of the ocean water is Φ ocean m oceanh6 h5 Tos6 s5 m oceanCpT6 T5 ToCp ln T6 T5 4358 41815 12 285 418 ln 273 15 273 12 54 650 54 364 286 kW The second law efficiency is ηII Φ ocean Φ water 286 5703 005 In reality all the exergy in the ocean water is destroyed as the 15C water mixes with the ocean water at 12C after it flows back out into the ocean and the efficiency does not have any significance Notice the small rate of exergy relative to the large rates of energy being transferred Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9120 Find the second law efficiency for the open FWH in Problem 940 The setup follows Fig912 State enthalpies kJkg h2 41902 h6 282786 h3 76279 x 0 at 1 MPa State entropies kJkgK s2 13068 s6 66939 s3 21386 Analyze the FWH which leads to Eq95 y m 6 m 5 the extraction fraction m 5 m 3 y h3 h2 h6 h2 76279 41902 282786 41902 01427 The second law efficiency for this FWH can be done similar to a coflowing heat exchanger The extraction flow fraction y of total flow provides the source of exergy as it goes from state 6 to state 3 The feedwater flow with fraction 1 y increase its exergy from state 2 to state 3 ψ3 ψ2 h3 h2 Tos3 s2 76279 41902 2981521386 13068 95769 kJkg ψ6 ψ3 h6 h3 Tos6 s3 282786 76279 2981566939 21386 706907 kJkg ηII 1 yψ3 ψ2 yψ6 ψ3 08573 95769 01427 706907 0814 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9121 Find the second law efficiency for the closed FWH in Problem 951 2 4 6 6a From table B1 h kJkg s kJkgK B14 100C 20 MPa h2 43404 12917 B14 200C 20 MPa h4 86047 23031 B13 4 MPa 275C h6 288039 62145 B12 4 MPa sat liq h6a 108729 27963 CV Feedwater Heater Energy Eq m 2h2 m 6h6 m 2h4 m 6h 6a Since all four states are known we can solve for the extraction flow rate m 6 m 2 h2 h4 h6a h6 20 43404 86047 108729 288039 kgs 4756 kgs The feedwater from state 2 to state 4 gains exergy from the extraction flow which provides the source as it goes from state 6 to state 6a ψ4 ψ2 h4 h2 Tos4 s2 86047 43404 2981523031 12917 124881 kJkg ψ6 ψ6a h6 h6a Tos6 s6a 288039 108729 2981562145 27963 773964 kJkg ηII m 2 ψ4 ψ2 m 6 ψ6 ψ6a 20 124881 4756 773964 0679 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9122 For problem 958 consider the boilersuperheater Find the exergy destruction in this setup and the second law efficiency for the boilersource setup A Rankine cycle feeds 5 kgs ammonia at 2 MPa 140oC to the turbine which has an extraction point at 800 kPa The condenser is at 20oC and a closed feed water heater has an exit state 3 at the temperature of the condensing extraction flow and it has a drip pump The source for the boiler is at constant 180oC Find the extraction flow rate and state 4 into the boiler The boiler has flow in at state 4 and out at state 5 with the source providing a q at 180oC Assume state 4 is saturated liquid at T4 so h4 hf 4 T4 1792oC State 4 h4 2644 kJkg hf 4 s4 sf 4 100705 kJkgK State 5 h5 17382 kJkg s5 55022 kJkg K Energy Eq qH h5 h4 17382 2644 14738 kJkg Entropy Eq s4 qHTH sgen s5 sgen s5 s4 qHT H S gen m sgen 5 55022 100705 14738 180 27315 5 12428 621 kWK The flow increase in exergy is ψ5 ψ4 h5 h4 To s5 s4 14738 29815 55022 100705 13357 kJkg The exergy provided by the source is φH 1 To TH qH 1 29815 180 27315 14738 5041 kJkg Second law efficiency is ηII gain in exergy source exergy input 13357 5041 0265 1 TosgenφH 1 29815 12428 5041 180 C 4 5 CV T s 5 4 6 7 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9123 The power plant using ammonia in Problem 971 has a flow of liquid water at 120oC 300 kPa as a heat source the water leaves the heat exchanger at 90oC Find the second law efficiency of this heat exchanger CV The liquid water source ammonia boiler heat exchanger hliq 50369 3769 12679 kJkg sat liquid at 120oC and 90oC sliq 15275 11924 03351 kJkgK The energy equation establishes the ratio of the mass flow rates Q H m water hliq m NH3 qH m water m NH3 qH hliq 144212679 11373 Now the second law efficiency is the ratio of exergy pickup over exergy source ηII m water ψ m NH3 ψ m water hliq Tosliq m NH3 qH Tos5 s3 From the power plant cycle we have state 3 State 3 x3 0 h3 17165 kJkg s3 06793 kJkgK v3 000156 m3kg State 5 h5 16146 kJkg s5 54971 kJkg K CV Pump P2 wP2 h4 h3 v3P4 P3 0001561000 400 0936 kJkg qH h5 h4 16146 17165 0936 1442 kJkg s5 s3 54971 06793 481775 kJkgK ηII 1 11373 1442 25315 481775 12679 25315 03351 0466 cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9124 A concentrated solar power plant receives the energy from molten salt coming in at 560oC and leaving at 300oC in counterflow heat exchanger where the water comes in at 3 MPa 60oC and leaves at 450oC 3 MPa The molten salt has 5 kgs flow with CP salt 15 kJkgK What is the possible water flow rate and the rate of energy transfer Find the second law efficiency of this heat exchanger Continuity Eqs Each line has a constant flow rate through it Energy Eq410 0 m H2O h1 h2 m salt h3 h4 Entropy Eq77 0 m H2O s1 s2 m salt s3 s4 S gen Process Each line has a constant pressure Table B1 h1 25361 kJkg h2 3344 kJkg s1 08295 s2 70833 kJkgK ψ2 ψ1 h2 h1 T0s2 s1 3344 25361 29815 70833 08295 122582 kJkg For the salt we get h3 h4 CP salt T3 T4 15 560 300 390 kJkg s3 s4 CP salt ln T3T4 15 ln8331557315 0561 kJkgK ψ3 ψ4 h3 h4 T0s3 s4 390 29815 0561 22274 kJkgK m H2O m salt h3 h4 h2 h1 390 3344 25361 0631 kgs ηII m saltψ3 ψ4 m H2Oψ2 ψ1 0631 122582 5 22274 069 CV Heat exchanger steady flow 1 inlet and 1 exit for salt and water each The two flows exchange energy with no heat transfer tofrom the outside 3 salt 1 water 4 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9125 What is the second law efficiency of the heat pump in Problem 992 A heat pump for heat upgrade uses ammonia with a low temperature of 25oC and a high pressure of 5000 kPa If it receives 1 MW of shaft work what is the rate of heat transfer at the high temperature State 1 h1 14635 kJkg s1 50293 kJkgK State 3 h3 hf 6319 kJkg s3 21100 kJkgK Entropy compressor s2 s1 T2 156oC h2 17091 kJkg Energy eq compressor wC h2 h1 2456 kJkg Energy condenser qH h2 h3 10772 kJkg Exergy output ψH ψ2 ψ3 h2 h3 T0s2 s3 10772 298 50293 21100 20725 kJkg ηII output source ψH wC ac 20725 2456 0844 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9126 Steam is supplied in a line at 3 MPa 700C A turbine with an isentropic efficiency of 85 is connected to the line by a valve and it exhausts to the atmosphere at 100 kPa If the steam is throttled down to 2 MPa before entering the turbine find the actual turbine specific work Find the change in exergy through the valve and the second law efficiency of the turbine Take CV as valve and a CV as the turbine Valve h2 h1 39117 kJkg s2 s1 77571 kJkg K h2 P2 s2 79425 kJkg K ψ1 ψ2 h1h2 T0s1s2 0 298157757179425 553 kJkg So some potential work is lost in the throttling process Ideal turbine s3 s2 h3s 292913 kJkg wTs 98257 kJkg wTac h2 h3ac ηwTs 8352 kJkg h3ac 39117 8352 30765 s3ac 8219 kJkg K wrev ψ2 ψ3ac h2 h3ac T0s2 s3ac 8352 2981579425 8219 91763 kJkg ηII wTac wrev 835291763 091 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9127 The condenser in a refrigerator receives R134a at 700 kPa 50C and it exits as saturated liquid at 25C The flowrate is 01 kgs and the condenser has air flowing in at ambient 15C and leaving at 35C Find the minimum flow rate of air and the heat exchanger secondlaw efficiency CV Total heat exchanger Energy Eq610 m 1h1 m ah3 m 1h2 m ah4 m a m 1 h1 h2 h4 h3 01 43689 23459 100435 15 1007 kgs 1 2 Air in 3 Air out 4 Availability from Eq824 ψ1 ψ2 h1 h2 T0s1 s2 43689 23459 2881517919 11201 87208 kJkg ψ4 ψ3 h4 h3 T0s4 s3 100435 15 28815 1004 ln 30815 28815 0666 kJkg Efficiency from Eq830 ηII m a ψ4 ψ3 m 1 ψ1 ψ2 1007 0666 01 87208 077 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9128 A flow of steam at 10 MPa 550C goes through a twostage turbine The pressure between the stages is 2 MPa and the second stage has an exit at 50 kPa Assume both stages have an isentropic efficiency of 85 Find the second law efficiencies for both stages of the turbine CV T1 h1 35009 kJkg s1 67561 kJkg K Isentropic s2s s1 h2s 30179 kJkg wT1s h1 h2s 483 kJkg Actual T1 wT1ac ηT1 wT1s 41055 h1 h2ac h2ac h1 wT1ac 309035 kJkg s2ac 68782 kJkg K CV T2 s3s s2ac 68782 x3s 68782109165029 08899 h3s 34047 08899 23054 23922 kJkg wT2s h2ac h3s 69815 wT2ac ηT2 wT2s 5934 kJkg h3ac 24969 x3ac 24969 3404723054 09354 s3ac 1091 09354 65029 71736 kJkg K Actual T1 iT1ac T0s2acs1 2981568782 67561 364 kJkg w R T1 wT1ac i 447 kJkg ηII wT1acw R T1 0918 Actual T2 iT2ac T0s3acs2ac 2981571736 68782 8807 kJkg w R T2 wT2ac iT2ac 6815 ηII wT2acw R T2 0871 T1 T2 2 1 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9129 A new airconditioner using R410A is used in heat pump mode The high pressure is 2000 kPa and the low pressure is 400 kPa It warms a house at 20oC driven by an electric power input of 2 kW in an ambient at 5oC Find the destruction of exergy 4 places 1 inside the heat pump 2 in the high T heat exchanger 3 in the low T heat exchanger and 4 in the house wallswindows etc that separates the inside from the outside of the house Exit evaporator x 1 400 kPa from B42 h1 2719 s1 10779 kJkgK Exit condenser x 0 2 MPa from B41 h3 11021 kJkg s3 040378 kJkgK Exit valve h3 h4 2824 x4 24366 x4 033638 s4 043917 Compressor s2 s1 P2 from B42 h2 31743 kJkg Evaporator qL h1 h4 h1 h3 2719 11021 16169 kJkg Condenser qH h2 h3 31743 11021 20722 kJkg Compressor wc h2 h1 31743 2719 4553 kJkg m W wc 2 kW4553 kJkg 004393 kgs COP β qHwc 207224553 455 Heat transfers Q H β W 455 2 91 kW Q L 91 2 71 kW CV The AC unit entropy only made in valve Φ destr HP T0 S gen m T0 s4 s3 004393 kgs 29815 K 043917 040378 kJkgK 046 kW CV High T heat exchanger with Q from T2 T3 32oC to room 20oC Entropy 0 m s2 s3 Q HThouse S gen S gen Q HThouse m s2 s3 Φ destr Hi T T0 S gen 29815 91 29315 004393 10779 040378 0426 kW CV Low T heat exchanger here T1 1998oC is constant as heat is absorbed the term Q LT1 is also equal to m s1 s4 Φ destr Low T T0 S gen T0Q LT1 Q LTamb 29815 71 25317 71 26815 0467 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CV House walls that leaks heat out from 20oC to ambient 5oC Φ destr walls T0 S gen T0 Q HTamb Q HThouse 29815 91 26815 91 29315 0863 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9130 An airconditioner using R410A is used in cooling mode The high pressure is 3000 kPa and the low pressure is 800 kPa It cools a house at 20oC with a rate of 12 kW and the outside ambient is at 35oC Find the destruction of exergy 4 places 1 inside the refrigerator 2 in the high T heat exchanger 3 in the low T heat exchanger and 4 in the house wallswindows etc that separates the inside from the outside of the house The exergy destruction can be found from the exergy balance or from the entropy generation for each control volume Q leak Q Q H L W REF cb CV2 CV1 CV3 CV4 Ideal refrigeration cycle T s 1 2 3 4 The control volumes 14 correspond to the exergy destruction locations To answer the first one and to find the COP and the high T heat transfer we must analyze the refrigeration cycle State 1 800 kPa x 1 h1 27914 kJkg s1 10367 kJkgK T1 005oC State 2 3000 kPa s s1 h2 31558 kJkg T2 705 oC State 3 3000 kPa x 0 h3 14178 kJkg s3 05011 kJkgK T3 4907 oC State 4 800 kPa h h3 x4 14178 578365221304 037931 s4 022667 x4 081003 053392 kJkgK Evaporator qL h1 h4 h1 h3 27914 14178 13736 kJkg Condenser qH h2 h3 31558 14178 1738 kJkg Compressor wc h2 h1 31558 27914 3644 kJkg m Q L qL 12 kW13736 kJkg 008736 kgs COP β qLwc 137363644 37695 W Q L β 12 kW37695 3183 kW Q H W Q L 15183 kW CV1 The only entropy generation is in the valve Φ destr ref T0 S gen m T0 s4 s3 008736 kgs 29815 K 053392 05011 kJkgK 0855 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CV2 The heat transfer goes to the ambient from the 23 condenser process Entropy Eq 0 m s2 s3 Q HTamb S gen Φ destr Hi T T0 S gen T0 Q HTamb m s2 s3 29815 1518330815 00873610367 05011 074 kW CV3 Low T heat exchanger here T1 005oC is constant as heat is absorbed the term Q LT1 is also equal to m s1 s4 Φ destr Low T T0 S gen T0Q LT1 Q LThouse 29815 12 2732 12 29315 0891 kW CV House walls that leaks heat from ambient 35oC to house 20oC Φ destr walls T0 S gen T0 Q LThouse Q LTamb 29815 12 29315 12 30815 0594 kW The 4 destruction terms do not quite add up to the work input 3183 kW as the reference T 29815 K 25C had we used 30815 K instead they would add to the work term exactly Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9131 Assume the house in the previous problem has a combined 12000 kg hard wood 2500 kg gypsum plates Cp 1 kJkgK and 750 kg steel all of which is at 20oC If the air conditioner is turned off how fast does the house heat up oCs Since the AC cools the house at a rate of 12 kW it means there must be a leak into the house of 12 kW coming from the warmer ambient As the AC is truned off it then heats with a rate of 12 kW As the house T goes up that rate drops exponentially in time until the house reaches the ambient 35C Energy Eq house dE dt dU dt m CdT dt Q leak The thermal capacity is m C mCwood mCgypsum mCsteel 12 000 126 2500 1 750 046 17 965 kJK Then dT dt Q leak m C 12 kW 17 965 kJK 0000668 Ks Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9132 Assume the house in problem 9129 has a combined 12000 kg hard wood 2500 kg gypsum plates Cp 1 kJkgK and 750 kg steel all of which is at 20oC If the heat pump is turned off how fast does the house cool down oCs Since the heat pump heats the house at a rate of 91 kW it means there must be a leak out of the house of 91 kW going to the colder ambient As the heat pump is truned off the house then cools with a rate of 91 kW As the house T drops the rate drops exponentially in time until the house reaches the ambient 5C Energy Eq house dE dt dU dt m CdT dt Q leak The thermal capacity is m C mCwood mCgypsum mCsteel 12 000 126 2500 1 750 046 17 965 kJK Then dT dt Q leak m C 91 kW 17 965 kJK 0000506 Ks Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Combined Cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9133 A binary system power plant uses mercury for the hightemperature cycle and water for the lowtemperature cycle as shown in Fig 1220 The temperatures and pressures are shown in the corresponding Ts diagram The maximum temperature in the steam cycle is where the steam leaves the superheater at point 4 where it is 500C Determine the ratio of the mass flow rate of mercury to the mass flow rate of water in the heat exchanger that condenses mercury and boils the water and the thermal efficiency of this ideal cycle The following saturation properties for mercury are known P MPa Tg C hf kJkg hg kJkg sf kJkgK sg kJkgK 004 309 4221 33564 01034 06073 160 562 7537 36404 01498 04954 Solution For the mercury cycle sd sc 04954 01034 xd 05039 xd 07779 hb ha wP HG ha since vF is very small qH hc ha 36404 4221 32183 kJkg qL hd ha 27048 4221 22827 kJkg For the steam cycle s5 s4 70097 06493 x5 75009 x5 08480 h5 19183 0848 23928 22208 wP v1P2 P1 0001014688 10 47 kJkg h2 h1 wP 1918 47 1965 qH from Hg h3 h2 27699 1965 26004 qH ext source h4 h3 34374 27969 6405 CV Hg condenser H2O boiler 1st law mHghd ha mH2Oh3 h2 mHgmH2O 27969 1965 27048 4221 11392 qH TOTAL mHgmH2Ohc hb h4 h3 for 1 kg H2O 11392 32183 6405 43068 kJ All qL is from the H2O condenser qL h5 h1 22208 1918 20290 kJ wNET qH qL 43068 20290 22778 kJ ηTH wNETqH 2277843068 0529 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9134 A Rankine steam power plant should operate with a high pressure of 3 MPa a low pressure of 10 kPa and the boiler exit temperature should be 500C The available hightemperature source is the exhaust of 175 kgs air at 600C from a gas turbine If the boiler operates as a counter flowing heat exchanger where the temperature difference at the pinch point is 20C find the maximum water mass flow rate possible and the air exit temperature Solution CV Pump wP h2 h1 v1P2 P1 0001013000 10 302 kJkg h2 h1 wP 19183 302 19485 kJkg Heat exchanger water states State 2a T2a TSAT 2339 C h2a 100842 kJkg State 3 h3 34565 kJkg Heat exchanger air states inlet hairin 90316 kJkg State 2a hairT2a 20 53128 kJkg HEAT EXCH i e a 2a 3 2 Air temperature should be 2539C at the point where the water is at state 2a CV Section 2a3 ia m H2Oh3 h2a m airhi ha m H2O 175 90316 53128 34565 100842 26584 kgs Take CV Total m H2Oh3 h2 m airhi he he hi m H2Oh3 h2m air 9036 2658434565 19485175 40813 kJkg Te 4067 K 1336 C Te T2 465 C OK 1 T 3 2 s 2a Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9135 Consider an ideal dualloop heatpowered refrigeration cycle using R134a as the working fluid as shown in Fig P9135 Saturated vapor at 90C leaves the boiler and expands in the turbine to the condenser pressure Saturated vapor at 15C leaves the evaporator and is compressed to the condenser pressure The ratio of the flows through the two loops is such that the turbine produces just enough power to drive the compressor The two exiting streams mix together and enter the condenser Saturated liquid leaving the condenser at 45C is then separated into two streams in the necessary proportions Determine the ratio of mass flow rate through the power loop to that through the refrigeration loop Find also the performance of the cycle in terms of the ratio Q L Q H Solution T1 15 oC sat vap Table B51 T6 90oC sat vapor P5 P6 32445 MPa Table B51 T3 45oC sat liquid P2 P3 P7 11602 MPa h1 3892 h3 h4 26411 h6 4257 kJkg CV Turbine s7 s6 16671 12145 x7 04962 x7 09121 h7 26411 09121 15785 40808 kJkg CV Compressor computer tables are used for this due to value of P or you do a double linear interpolation between 1000 and 1200 kPa to get 11602 kPa s2 s1 17354 P2 T2 5227oC h2 4299 kJkg CV turbine compressor Continuity Eq m 1 m 2 m 6 m 7 Energy Eq m 1h1 m 6h6 m 1h2 m 6h 7 B O I L C O N D E V A P T U R B C O M P 1 2 7 6 3 4 5 P Q L T 3 4 s 6 7 2 1 5 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful m 6m 1 h2 h1 h6 h7 4299 3892 4257 40808 231 CV pump wP v3P5 P3 00008932445 11602 1855 kJkg h5 h3 wP 26411 1855 26596 kJkg CV evaporator Q L m 1 h1 h4 CV boiler Q H m 6 h6 h5 β Q L Q H m 1h1 h4 m 6h6 h5 3892 26411 231 4257 26596 034 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9136 For a cryogenic experiment heat should be removed from a space at 75 K to a reservoir at 180 K A heat pump is designed to use nitrogen and methane in a cascade arrangement see Fig 928 where the high temperature of the nitrogen condensation is at 10 K higher than the lowtemperature evaporation of the methane The two other phase changes take place at the listed reservoir temperatures Find the saturation temperatures in the heat exchanger between the two cycles that gives the best coefficient of performance for the overall system The nitrogen cycle is the bottom cycle and the methane cycle is the top cycle Both are standard refrigeration cycles THm 180 K T3m TLN 75 K T4N T1N TLm T4m T1m T3N 10 Trial and error on T3N or TLm For each cycle we have wC h2 h1 s2 s1 qH h2 h3 qL h1 h4 h1 h3 Nitrogen T4 T1 75 K h1 74867 kJkg s1 54609 kJkg K N2 T3 h3 P2 h2 wc qH qL a 120 17605 25125 20296 1281 22057 9247 b 115 34308 19388 18835 1135 22266 10918 c 110 48446 14672 17388 990 22233 12331 Methane T3 180 K h3 05 kJkg P2 328655 MPa CH4 T4 h1 s1 h2 wc qH qL a 110 221 9548 5403 3193 5408 2215 b 105 2122 9691 5811 3689 5816 2127 c 100 2029 9851 6297 4268 6302 2034 The heat exchanger that connects the cycles transfers a Q Q Hn qHn m n Q Lm qLm m m m mm n qHnqLm The overall unit then has Q L 75 K m n qLn W tot in m nwcn m mwcm β Q L 75 KW tot in qLnwcn m mm nwcm Case m mm n wcnm mm nwcm β a 0996 44606 0207 b 1047 49965 0219 c 1093 56549 0218 A maximum coefficient of performance is between case b and c Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9137 For Problem 9134 determine the change of exergy of the water flow and that of the air flow Use these to determine a second law efficiency for the boiler heat exchanger From solution to 9134 m H2O 26584 kgs h2 19485 kJkg s2 06587 kJkg K h3 34565 kJkg s3 72338 sTi 79820 sTe 71762 kJkg K hi 90316 kJkg he 40813 kJkg ψ3 ψ2 h3 h2 T0s3 s2 130128 kJkg ψi ψe hi he T0sTi sTe 25478 kJkg ηII ψ3 ψ2 m H2O ψi ψe m air 130128 26584 25478 175 0776 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9138 Do Problem 931 with R134a as the working fluid in the Rankine cycle Consider the ammonia Rankinecycle power plant shown in Fig P931 a plant that was designed to operate in a location where the ocean water temperature is 25C near the surface and 5C at some greater depth The mass flow rate of the working fluid is 1000 kgs a Determine the turbine power output and the pump power input for the cycle b Determine the mass flow rate of water through each heat exchanger c What is the thermal efficiency of this power plant Solution a Turbine s2 s1 17183 10485 x2 06733 x2 09948 h2 21358 09948 19065 40324 kJkg wT h1 h2 40984 40324 66 kJkg W T m wT 6600 kW Pump wP v3P4 P3 00007945728 4158 0125 kJkg wP wP ηS 0125 W P m wP 125 kW b Consider the condenser heat transfer to the low T water Q to low T H2O 100040324 21358 189 660 kW m low T H2O 189660 2938 2098 22 579 kgs h4 h3 wP 21358 0125 21371 kJkg Now consider the boiler heat transfer from the high T water Q from high T H2O 100040984 21371 196 130 kW m high T H2O 196130 10487 9650 23 432 kgs c ηTH W NETQ H 6600 125 196130 0033 W T Q H W P in Q L 3 2 1 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9139 A simple steam power plant is said to have the four states as listed 1 20oC 100 kPa 2 25oC 1 MPa 3 1000oC 1 MPa 4 250oC 100 kPa with an energy source at 1100oC and it rejects energy to a 0oC ambient Is this cycle possible Are any of the devices impossible Solution The cycle should be like Figure 93 for an ideal or Fig915 for an actual pump and turbine in the cycle We look the properties up in Table B1 State 1 h1 8394 s1 02966 State 2 h2 10487 s2 03673 State 3 h3 46376 s3 89119 State 4 h4 29743 s4 80332 We may check the overall cycle performance Boiler qH h3 h2 46376 10487 45327 kJkg Condenser qL h4 h1 29743 8394 28904 kJkg ηcycle qnet qH qH qL qH 16423 45327 0362 ηcarnot 1 TL TH 1 27315 27315 1100 080 ηcycle OK Check the second law for the individual devices CV Boiler plus wall to reservoir sgen s3 s2 Tres qH 89119 03673 45327 1373 524 kJkg K 0 OK CV Condenser plus wall to reservoir sgen s1 s4 Tres qL 02966 80332 28904 273 2845 kJkg K 0 OK CV Pump wp h2 h1 2093 kJkg sgen s2 s1 03673 02966 00707 kJkg K 0 OK CV Turbine wT h3 h4 46376 29743 16633 kJkg sgen s4 s3 80332 89119 08787 kJkg K sgen 0 NOT POSSIBLE W T Q H W P in Q L 3 2 1 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9140 A supercritical power plant has a high pressure of 30 MPa and the boiler heats the water to 500oC with 45oC in the condenser To avoid a quality in the turbine of less than 92 determine the pressures at which to make reheat look only at the Ps listed in Table B1 do not interpolate between them if the reheat takes it up to 400oC Let us work backwards from the lowest towards the highest pressures What state should the steam come from if the final exit state is at 45oC and x 092 State 45oC and x 092 s 06386 092 75261 75626 kJkgK The reheat should then produce 400oC s 75626 kJkgK B12 At P 800 kPa 400oC s 75715 kJkgK At 800 kPa x 092 s 20461 092 46166 62934 kJkgK The next reheat should then produce 400oC s 62934 kJkgK B12 At P 8000 kPa 400oC s 63633 kJkgK At 8000 kPa x 092 s 32067 092 25365 55403 kJkgK At 30 MPa 500oC s 57904 kJkgK so this is fine Comment In reality you do not want any twophase states in the high pressure sections of the turbine so if that is a condition you expand from the inlet to a state the closest to sat vapor 8000 kPa reheat to 400 C then expand again to near sat vapor giving about 2000 kPa reheat to 400 C and expand again for a final reheat at 500 kPa The processes for the different steps are shown generated by CATT3 using the plot process from the Options menu drop down select plot proces Starting from the 45oC and x 092 a process is plotted with constant s to 400oC Then starting with 400oC 800 kPa a constant pressure process is plotted to x 092 followed by another constant s process up to 400oC Again a constant pressure process from 400oC 8000 kPa to x 092 followed by a constant s process from the original 30 MPa 500oC down to 8000 kPa to verify the final step The curves do not match precisely as the table entries were chosen to match the hand calculations The curves are shown on the following page Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Notice how close to the saturated vapor curve the x 092 states are Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9141 A dairy farmer needs to heat a 01 kgs flow of milk from room temperature 20oC to 60oC to pasteurize it and then send the flow through a cooler bringing it to 10o C for storage He wants to by a heat pump to do this and finds one using R134a with a high pressure of 3 MPa and a low pressure of 300 kPa The farmer is very clever and uses the heat pump to heat the milk in a heat exchanger after which the milk flows through the evaporator heat exchanger to cool it Find the power required to run the heat pump so it can do both the heating and the cooling assuming milk has the properties of water Is there any excess heating or cooling capacity What is the total rate of exergy destruction by running this system Do the refrigeration cycle State 1 300 kPa x 1 h1 39869 kJkg s1 17259 kJkgK T1 056oC State 2 3000 kPa s s1 h2 44590 kJkg T2 955 oC State 3 3000 kPa x 0 h3 335225 kJkg s3 14186 kJkgK T3 862oC Evaporator qL h1 h4 h1 h3 39869 335225 63465 kJkg Condenser qH h2 h3 44590 335225 110675 kJkg Compressor wc h2 h1 44590 39869 4721 kJkg COP β qLwc 63465 4721 1344 Since we need to cool more than we need to heat the heat pump needs to be sized to the cooling From the definition of COP and the milk heat transfer required QL mCpT2 T3 01 418 60 10 209 kW β W QH milk mCpT2 T1 01 418 60 20 1672 kW W Q L β 209 kW1344 1555 kW Q H W Q L 3645 kW QH extra Q H QH milk 3645 1672 1973 kW For exergy we have work in and flow in with zero To heat dumped into the ambient also assumed zero and then the cold flow out with a positive exergy Φflow out m h3 h1 Tos3 s1 mCpT3 T1 To lnT3T1 01 418 10 293 ln283293 0073 kW Φdestr W Φflow out 1555 0073 1548 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful If we take control volume total as shown then the s generation can be found 0 m s1 s3 QH QH milkTo Sgen and then Φdestr To Sgen giving the same result Comment Clearly the extra heat out at 86oC could be useful and the cycle can be run with a slightly lower high pressure as it only need to heat the milk to 60oC HP o T QL QH W QH extra Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9142 Consider an ideal combined reheat and regenerative cycle in which steam enters the highpressure turbine at 30 MPa 400C and is extracted to an open feedwater heater at 08 MPa with exit as saturated liquid The remainder of the steam is reheated to 400C at this pressure 08 MPa and is fed to the low pressure turbine The condenser pressure is 10 kPa Calculate the thermal efficiency of the cycle and the net work per kilogram of steam Solution In this setup the flow is separated into fractions x and 1x after coming out of T1 The two flows are recombined in the FWH CV T1 s6 s5 69211 kJkg K h6 28916 kJkg wT1 h5 h6 323082 28916 33922 kJkg CV Pump 1 wP1 h2 h1 v1P2 P1 000101800 10 0798 kJkg h2 h1 wP1 19181 0798 19261 kJkg s 1 2 3 5 6 7 8 400 C o 4 10 kPa T CV FWH h3 hf 7211 P1 P2 1 2 4 5 6 7 8 COND FWH 3 T1 T2 x 1x 1x Energy equation per unit mass flow exit at 3 x h6 1 x h2 h3 x h3 h2 h6 h2 7211 19261 28916 19261 01958 CV Pump 2 wP2 h4 h3 v3P4 P3 00011153000 800 245 kJkg h4 h3 wP2 7211 245 72355 kJkg CV Boilersteam generator including reheater Total flow from 4 to 5 only fraction 1x from 6 to 7 qH h5 h4 1 xh7 h6 25073 30195 28093 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CV Turbine 2 s8 s7 75715 kJkg K x8 75715 064927501 092285 h8 hf x8 hfg 19181 092285 239282 24000 kJkg wT2 h7 h8 326707 240002 86705 kJkg Sum the work terms to get net work Total flow through T1 only fraction 1x through T2 and P1 and after FWH we have the total flow through P2 wnet wT1 1 x wT2 1 x wP1 wP2 3392 6973 064 245 103341 kJkg ηcycle wnet qH 103341 28093 0368 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9143 An ideal steam power plant is designed to operate on the combined reheat and regenerative cycle and to produce a net power output of 10 MW Steam enters the highpressure turbine at 8 MPa 550C and is expanded to 06 MPa at which pressure some of the steam is fed to an open feedwater heater and the remainder is reheated to 550C The reheated steam is then expanded in the lowpressure turbine to 10 kPa Determine the steam flow rate to the highpressure turbine and the power required to drive each of the pumps a s 1 2 3 5 6 7 8 550 C o 4 10 kPa T P P 1 2 4 5 6 7 8 COND HTR 3 6a T1 T2 HI P LOW P b wP12 000101600 10 06 kJkg h2 h1 wP12 1918 06 1924 kJkg wP34 0001018000 600 81 kJkg h4 h3 wP34 6706 81 6787 h5 35210 kJkg s6 s5 68778 T6 18232 oC h6 28100 kJkg h7 35919 s8 s7 81348 06493 x8 75009 x8 09979 h8 19183 09979 23928 25797 kJkg CV heater Cont m6a m2 m3 1 kg Energy Eq m6ah6 m2h2 m3h3 m6a 6706 1924 28100 1924 01827 m2 m7 1 m6a 08173 CV turbine wT h5 h6 1 m6ah7 h8 3521 2810 0817335919 25797 15382 kJkg CV pumps wP m2wP12 m4wP34 0821406 181 86 kJkg wNet 15382 86 15296 kJkg m5 m 5 W NetwNet 1000015296 653 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9144 An industrial application has the following steam requirement one 10kgs stream at a pressure of 05 MPa and one 5kgs stream at 14 MPa both saturated or slightly superheated vapor It is obtained by cogeneration whereby a high pressure boiler supplies steam at 10 MPa 500C to a reversible turbine The required amount is withdrawn at 14 MPa and the remainder is expanded in the lowpressure end of the turbine to 05 MPa providing the second required steam flow a Determine the power output of the turbine and the heat transfer rate in the boiler b Compute the rates needed were the steam generated in a lowpressure boiler without cogeneration Assume that for each 20C liquid water is pumped to the required pressure and fed to a boiler Solution 1 H O IN 2 20 C o W P BOILER Q H HP TURB LP TURB HPT W W LPT 10 MPa 500 C o 14 MPa 5 kgs STEAM 05 MPa 10 kgs STEAM 2 3 4 5 P a With cogeneration highpressure turbine s4 s3 65966 kJkg K T4 2199 oC h4 28526 kJkg wS HPT h3 h4 33737 28526 5211 kJkg lowpressure turbine s5 s4 65966 18607 x5 49606 x5 09547 h5 64023 09547 21085 26532 kJkg wS LPT h4 h5 28526 26532 1994 kJkg W TURB 15 5211 10 1994 9810 kW W P 15 0001002 10 000 23 1503 kW h2 h1 wP 8396 1002 940 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Q H m 1h3 h2 1533737 940 49 196 kW b Without cogeneration This is to be compared to the amount of heat required to supply 5 kgs of 14 MPa sat vap plus 10 kgs of 05 MPa sat vap from 20oC water 1 2 3 W P1 Q 2 3 4 5 6 W P2 Q 5 6 5 kgs 10 kgs Sat vapor 14 MPa Sat vapor 05 MPa 20 C o 20 C o Pump 1 and boiler 1 wP 00010021400 23 140 kJkg h2 h1 wP 8396 140 854 kJkg 2Q 3 m 1h3 h2 527900 854 13 523 kW W P1 5 140 7 kW Pump 2 and boiler 2 h5 h4 wP2 8396 0001002500 23 845 kJkg 5Q 6 m 4h6 h5 1027487 845 26 642 kW W P2 10 05 5 kW Total Q H 13523 26642 40 165 kW Notice here that the extra heat transfer is about 9000 kW to run the turbines but that provides 9800 kW of work for electricity a 100 conversion of the extra Q to W Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9145 The effect of a number of open feedwater heaters on the thermal efficiency of an ideal cycle is to be studied Steam leaves the steam generator at 20 MPa 600C and the cycle has a condenser pressure of 10 kPa Determine the thermal efficiency for each of the following cases A No feedwater heater B One feedwater heater operating at 1 MPa C Two feedwater heaters one operating at 3 MPa and the other at 02 MPa a no feed water heater wP 1 2 vdP 00010120000 10 202 kJkg h2 h1 wP 1918 202 2120 s4 s3 65048 06493 x4 75009 x4 078064 ST GEN P 1 2 TURBINE COND 4 3 h4 19183 0780 64 23928 20597 wT h3 h4 35376 20597 14779 kJkg wN wT wP 14779 202 14577 qH h3 h2 35376 2120 33256 T s 1 2 3 10 kPa 600 C o 4 20 MPa ηTH qH wN 14577 33256 0438 b one feedwater heater wP12 0001011000 10 10 kJkg h2 h1 wP12 1918 10 1928 wP34 0001127 20000 1000 214 kJkg h4 h3 wP34 7628 214 7842 s6 s5 65048 21387 x6 44478 S T G E N P 1 T U R B I N E C O N D H T R P 4 2 3 5 6 7 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful x6 09816 h6 7628 09816 20153 27411 CV heater const m3 m6 m2 10 kg 1st law m6h6 m2h2 m3h 3 m6 7628 1928 27411 1928 02237 m2 07763 h7 20597 h4 of part a CV turbine wT h5 h6 m2h6 h7 35376 27411 0776327411 20597 13255 kJkg CV pumps wP m1wP12 m3wP34 0776310 1214 222 kJkg wN 13255 222 13033 kJkg CV steam generator qH h5 h4 35376 7842 27534 kJkg ηTH wNqH 1303327534 0473 c two feedwater heaters wP12 000101 200 10 02 kJkg h2 wP12 h1 1918 02 1920 wP34 0001061 3000 200 30 kJkg h4 h3 wP34 5047 30 5077 T s 1 2 3 10 kPa 600 C o 4 20 MPa 5 6 7 1 MPa S T G E N P P P H P H T R L P H T R 10 1 3 2 4 5 6 7 8 9 C O N D T U R B I N E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful wP56 000121720000 3000 207 kJkg h6 h5 wP56 10084 207 10291 s8 s7 65048 at P8 3 MPa T8 2932 oC h8 29748 s9 s8 65048 15301 x9 55970 T s 1 2 3 4 5 6 7 8 9 10 600 C o 10 kPa 02 MPa 3 MPa 80 MPa x9 08888 h9 5047 0888 22019 24618 kJkg CV high pressure heater cont m5 m4 m8 10 kg 1st law m5h5 m4h4 m8h8 m8 10084 5077 29748 5077 02030 m4 07970 CV low pressure heater cont m9 m2 m3 m4 1st law m9h9 m2h2 m3h 3 m9 24618 1920 079705047 1920 01098 m2 07970 01098 06872 CV turbine wT h7 h8 1 m8h8 h9 1 m8 m9h9 h10 35376 29748 079729748 24618 0687224618 20597 12480 kJkg CV pumps wP m1wP12 m3wP34 m5wP56 0687202 079730 1207 232 kJkg wN 12480 232 12248 kJkg CV steam generator qH h7 h6 35376 10291 25085 kJkg ηTH wNqH 1224825085 0488 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9146 A jet ejector a device with no moving parts functions as the equivalent of a coupled turbinecompressor unit see Problems 7165 and 7172 Thus the turbinecompressor in the dualloop cycle of Fig P9135 could be replaced by a jet ejector The primary stream of the jet ejector enters from the boiler the secondary stream enters from the evaporator and the discharge flows to the condenser Alternatively a jet ejector may be used with water as the working fluid The purpose of the device is to chill water usually for an airconditioning system In this application the physical setup is as shown in Fig P9146 Using the data given on the diagram evaluate the performance of this cycle in terms of the ratio QLQH a Assume an ideal cycle b Assume an ejector efficiency of 20 see Problem 7172 T1 T7 10 oC T2 150 oC T4 30 oC T9 20 oC Assume T5 T10 from mixing streams 4 9 P3 P4 P5 P8 P9 P10 PG 30 oC 4246 kPa P11 P2 PG 150oC 4758 kPa P1 P6 P7 PG 10oC 12276 kPa Cont m 1 m 9 m 5 m 10 m 5 m 6 m 7 m 1 m 7 m 8 m 9 m 10 m 11 m 2 m 3 m 4 a m 1 m 2 m 3 ideal jet ejector s1 s1 s2 s2 1 2 at P3 P4 then m 1h1 h1 m 2h2 h2 JET EJECT BOIL HP P LP P COND CHILL FLASH CH 2 11 1 3 4 10 9 7 8 Q L Q H LIQ 10 C o VAP 10 C o 30 C o VAP 150 C o 20 C o 6 5 T 3 4 s 510 6 8 7 9 11 2 1 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful From s2 s2 04369 x2 80164 x2 07985 h2 12579 07985 24305 20665 kJkg From s1 s1 89008 T1 112 C h1 27104 kJkg m 1m 2 27465 20665 27104 25198 35677 Also h4 12579 kJkg h7 4201 kJkg h9 8396 kJkg Mixing of streams 4 9 5 10 m 1 m 2h4 m 7h9 m 7 m 1 m 2h5 10 Flash chamber since h6 h5 m 7m 1h5 10 m 1h1 m 7h1 using the primary stream m 2 1 kgs 45677 12579 m 7 8396 m 7 45677h5 m 7 35677h5 35677 25198 m 7 4201 Solving m 7 202627 h5 8488 kJkg LP pump wLP P 000104246 12276 0003 kJkg h8 h7 wLP P 4201 0003 4201 kJkg Chiller Q L m 7h9h8 2026278396 4201 8500 kW for m 2 1 HP pump wHP P 00010024758 4246 047 kJkg h11 h10 wHP P 8488 047 8535 kJkg Boiler Q 11 m 11h2 h11 127465 8535 26611 kW Q LQ H 850026611 3194 b Jet eject eff m 1m 2ACTm 1m 2IDEAL 020 m 1m 2ACT 02 35677 07135 using m 2 1 kgs 17135 12579 m 7 8396 m 7 17135h5 m 7 07135h5 07135 25198 m 7 4201 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Solving m 7 39762 h5 h10 8569 kJkg Then Q L 397628396 4201 1668 kW h11 8569 047 8616 kJkg Q H 127465 8616 26603 kW Q LQ H 166826603 0627 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Computer Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9192 a A refrigerator with R12 as the working fluid has a minimum temperature of 10C and a maximum pressure of 1 MPa Assume an ideal refrigeration cycle as in Fig 923 Find the specific heat transfer from the cold space and that to the hot space and the coefficient of performance Solution Exit evaporator sat vapor 10C from CATT3 h1 18319 s1 07019 kJkgK Exit condenser sat liquid 1 MPa from CATT3 h3 7622 kJkg Compressor s2 s1 P2 from CATT3 h2 2101 kJkg Evaporator qL h1 h4 h1 h3 18319 7622 107 kJkg Condenser qH h2 h3 2101 7622 1339 kJkg COP β qLwc qLqH qL 398 Ideal refrigeration cycle Pcond P3 P2 1 MPa Tevap 10oC T1 Properties from CATT3 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9192 b Consider an ideal refrigeration cycle that has a condenser temperature of 45C and an evaporator temperature of 15C Determine the coefficient of performance of this refrigerator for the working fluid R12 Solution Ideal refrigeration cycle Tcond 45oC T3 Tevap 15oC T1 T s 1 2 3 4 Property for R12 Compressor h1 kJkg 18097 s2 s1 kJkg K 07051 P2 MPa 10843 T2 oC 547 h2 kJkg 21263 wC h2 h1 3166 Exp valve h3 h4 kJkg 7971 Evaporator qL h1 h4 10126 β qLw C 3198 The properties are from the computer program CATT3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9192 c A refrigerator in a meat warehouse must keep a low temperature of 15C and the outside temperature is 20C It uses R12 as the refrigerant which must remove 5 kW from the cold space Find the flow rate of the R12 needed assuming a standard vapor compression refrigeration cycle with a condenser at 20C Solution Basic refrigeration cycle T1 T4 15C T3 20C Computer Tables CATT3 h4 h3 5487 kJkg h1 hg 18097 kJkg Q L m R12 qL m R12h1 h4 qL 18097 5487 1261 kJkg m R12 50 1261 003965 kgs Ideal refrigeration cycle Tcond 20oC Tevap 15oC T1 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9192 d In an actual refrigeration cycle using R12 as the working fluid the refrigerant flow rate is 005 kgs Vapor enters the compressor at 150 kPa 10C and leaves at 12 MPa 75C The power input to the compressor is measured and found be 24 kW The refrigerant enters the expansion valve at 115 MPa 40C and leaves the evaporator at 175 kPa 15C Determine the entropy generation in the compression process the refrigeration capacity and the coefficient of performance for this cycle Solution Actual refrigeration cycle 1 compressor inlet T1 10oC P1 150 kPa 2 compressor exit T2 75oC P2 12 MPa 3 Expansion valve inlet T3 40oC P3 115 MPa 5 evaporator exit T5 15oC P5 175 kPa T 1 3 5 2 s 4 CATT3 h1 1848 s1 07324 h2 2267 s2 0741 CV Compressor h1 qCOMP wCOMP h2 s1 dqT sgen s2 wCOMP W COMPm 24005 480 kJkg qCOMP h2 wCOMP h1 2267 480 1848 61 kJkg sgen s2 s1 q To 0741 07324 6129815 0029 kJ kg K CV Evaporator qL h5 h4 1815 7459 1069 kJkg Q L m qL 005 1069 5346 kW COP β qLwCOMP 1069480 223 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9193 a Do Problem 921 with R22 as the working fluid A supply of geothermal hot water is to be used as the energy source in an ideal Rankine cycle with R134a as the cycle working fluid Saturated vapor R134a leaves the boiler at a temperature of 85C and the condenser temperature is 40C Calculate the thermal efficiency of this cycle Solution CV Pump use R22 Computer Table in CATT3 wP h2 h1 1 2 vdP v1P2P1 00008844037 1534 221 kJkg h2 h1 wP 9427 221 9648 kJkg CV Boiler qH h3 h2 25369 9648 15721 kJkg CV Turbine s4 s3 07918 03417 x4 05329 x4 08446 h4 9427 08446 16688 23522 kJkg wT h3 h4 25369 23522 1847 kJkg ηTH wNETqH 1847 22115721 01034 W T Q H W P in Q L 3 2 1 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9193 b Do problem 965 with R22 as the working fluid A flow with 2 kgs of water is available at 95oC for the boiler The restrictive factor is the boiling temperature of 85oC Therefore break the process up from 2 3 into two parts as shown in the diagram sat liq at 85 C o D Q BC 2 3 Q AB B liq H2O at 85 C o sat vap 85 C o 95 C o C A LIQUID HEATER BOILER liquid H2O out liquid liquid H2O R22 R22 Properties from CATT3 Q AB m H2OhA hB 239794 35588 8412 kW m R2225369 16509 m R22 0949 kgs To verify that TD T3 is the restrictive factor find TC Q AC 094916509 9648 6511 2035588 hC hC 32332 kJkg TC 772oC OK State 1 40oC 15335 kPa v1 0000884 m3kg CV Pump wP v1P2 P1 000088440368 15335 221 kJkg CV Turbine s4 s3 07918 03417 x4 05329 x4 08446 h4 9427 08446 16688 23522 kJkg Energy Eq wT h3 h4 25369 23522 1847 kJkg Cycle wNET wT wP 1847 221 1626 kJkg W NET m R22wNET 0949 1626 1543 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9193 c Consider an ideal refrigeration cycle that has a condenser temperature of 45C and an evaporator temperature of 15C Determine the coefficient of performance of this refrigerator for the working fluid R22 Solution Ideal refrigeration cycle Tcond 45oC T3 Tevap 15oC T1 T s 1 2 3 4 Property for R22 Compressor h1 kJkg 24413 s2 s1 kJkg K 09505 P2 MPa 1729 T2 oC 744 h2 kJkg 28926 wC h2 h1 4513 Exp valve h3 h4 kJkg 10098 Evaporator qL h1 h4 14315 β qLw C 3172 The properties are from CATT3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9193 SI version of d The refrigerant R22 is used as the working fluid in a conventional heat pump cycle Saturated vapor enters the compressor of this unit at 10C its exit temperature from the compressor is measured and found to be 85C If the compressor exit is at 2 MPa what is the compressor isentropic efficiency and the cycle COP Solution R22 heat pump Computer Table State 1 TEVAP 10oC x 1 h1 25342 kJkg s1 09129 kJkg K State 2 T2 P2 h2 29517 kJkg T s 1 2s 3 4 2 CV Compressor Energy Eq wC ac h2 h1 29517 25342 4175 kJkg State 2s 2 MPa s2S s1 09129 kJkg T2S 69oC h2S 2802 kJkg Efficiency η wC s wC ac h2S h1 h2 h1 2802 25342 29517 25342 06414 CV Condenser Energy Eq qH h2 h3 29517 1096 18557 kJkg COP Heat pump β qH wC ac 18557 4175 444 ENGLISH UNIT PROBLEMS UPDATED JULY 2013 SOLUTION MANUAL CHAPTER 9 English Units Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CHAPTER 9 SUBSECTION PROB NO Rankine Cycles 147157 reheat feedwater heaters real 158167 Refrigeration Cycles 168173 Availability and Combined Cycles 174180 Review Problems 181195 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Rankine cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9147E A steam power plant as shown in Fig 93 operating in a Rankine cycle has saturated vapor at 600 lbfin2 leaving the boiler The turbine exhausts to the condenser operating at 223 lbfin2 Find the specific work and heat transfer in each of the ideal components and the cycle efficiency Solution For the cycle as given 1 h1 9797 Btulbm v1 001625 ft3lbm 3 h3 hg 120406 Btulbm s3 sg 14464 Btulbm R CV Pump Reversible and adiabatic Energy wp h2 h1 Entropy s2 s1 since incompressible it is easier to find work positive in as wP v dP v1P2 P1 001625 ft3lbm 600 22 psi 144 in2ft2 139885 lbfft 18 Btulbm h2 h1 wP 9797 18 9977 Btulbm CV Boiler qH h3 h2 120406 9977 11043 Btulbm CV Tubine wT h3 h4 s4 s3 s4 s3 14464 01817 x4 17292 x4 07314 h4 9797 07314 101978 84384 Btulbm wT 120406 84384 36022 Btulbm ηCYCLE wT wPqH 36022 1811043 0325 CV Condenser qL h4 h1 84384 9797 7459 Btulbm Q WT 3 2 4 1 Condenser Boiler Turbine WP QB T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9148E Consider a solarenergypowered ideal Rankine cycle that uses water as the working fluid Saturated vapor leaves the solar collector at 150 psia and the condenser pressure is 095 lbfin2 Determine the thermal efficiency of this cycle H2O ideal Rankine cycle CV turbine State 3 Table F71 h3 119491 Btulbm s3 15704 Btulbm R CV Turbine adiabatic and reversible so second law gives s4 s3 15704 01296 x4 18526 x4 077772 h4 6804 077772 103698 87452 Btulbm The energy equation gives wT h3 h4 119491 87452 32039 Btulbm CV pump and incompressible liquid gives work into pump wP vdP v1P2 P1 001613 ft3lbm 150 095 psi 144 in2ft2 3462 lbfft 044 Btulbm h2 h1 wP 6804 044 6848 Btulbm CV boiler gives the heat transfer from the energy equation as qH h3 h2 119491 6848 112643 Btulbm The cycle net work and efficiency are found as wNET wT wP 32039 044 31995 Btulbm ηTH wNETqH 31995112643 0284 Q WT 3 2 4 1 Condenser Solar Turbine WP QRAD collector T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9149E The power plant in the previous problem is augmented with a natural gas burner to superheat the water to 600 F before entering the turbine Find the cycle efficiency with this configuration and the specific heat transfer added by the natural gas burner Solution CV H2O ideal Rankine cycle State 3 150 psia 600 F h3 132569 Btulbm s3 17110 Btulbm R CV Turbine adiabatic and reversible so second law gives s4 s3 17110 01296 x4 18526 x4 085361 h4 6804 085361 103698 95322 Btulbm The energy equation gives wT h3 h4 132569 95322 37247 Btulbm CV pump and incompressible liquid gives work into pump wP vdP v1P2 P1 001613 ft3lbm 150 095 psi 144 in2ft2 3462 lbfft 044 Btulbm h2 h1 wP 6804 044 6848 Btulbm CV boiler gives the heat transfer from the energy equation as qH h3 h2 132569 6848 125721 Btulbm The cycle net work and efficiency are found as wNET wT wP 37247 044 37203 Btulbm ηTH wNETqH 37203125721 0296 Q W T 3 2 4 1 Condenser Solar Turbine W P Q RAD collector T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9150E A Rankine cycle with R410A has the boiler at 600 psia superheating to 340 F and the condenser operates at 100 psia Find all four energy transfers and the cycle efficiency State 1 v1 001349 ft3lbm h1 219 Btulbm at 2384 F State 3 h3 1839 Btulbm s3 03091 BtulbmR State 4 100 psia s s3 h4 1520 Btulbm interpolated sup vap CV Pump wP v dP v1P2 P1 001349 ft3lbm 600 100 psi 144 in2ft2 9713 lbfft 125 Btulbm h2 h1 wP 219 125 2315 Btulbm CV Boiler qH h3 h2 1839 2315 16075 Btulbm CV Tubine energy wTs h3 h4 1839 1520 319 Btulbm CV Condenser qL h4 h1 1520 219 1301 Btulbm ηCYCLE wNETqH 319 125 16075 0191 P v 1 2 3 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9151E A supply of geothermal hot water is to be used as the energy source in an ideal Rankine cycle with R134a as the cycle working fluid Saturated vapor R134a leaves the boiler at a temperature of 180 F and the condenser temperature is 100 F Calculate the thermal efficiency of this cycle Solution CV Pump use R134a Table F10 P1 13893 psia P2 P3 4004 psia h3 18436 Btulbm s3 0402 Btulbm R h1 10886 Btulbm v1 001387 ft3lbm wP h2 h1 1 2 vdP v1P2P1 0013874004 13893 144 778 0671 Btulbm h2 h1 wP 10886 0671 10953 Btulbm CV Boiler qH h3 h2 18436 10953 7483 Btulbm CV Turbine s4 s3 0402 x4 0402 0281901272 09442 h4 17608 Btulbm Energy Eq wT h3 h4 8276 Btulbm wNET wT wP 8276 0671 7605 Btulbm ηTH wNET qH 76057483 0102 WT QH WP in QL 3 2 1 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9152E Do Problem 9151E with R410A as the working fluid boiler exit at 160 F 600 psia Standard Rankine cycle with properties from the R410A tables h1 5158 Btulbm v1 001619 ft3lbm P1 33254 psia P2 P3 600 psia h3 12506 Btulbm s3 02248 Btulbm R CV Pump wP v1P2P1 001619 600 33254 144 778 080 Btulbm h2 h1 wP 5158 080 5238 Btulbm CV Turbine s4 s3 x4 02248 0103801256 09634 h4 11932 Btulbm wT h3 h4 574 Btulbm CV Boiler qH h3 h2 12506 5238 7268 Btulbm ηTH wT wPqH 574 0807268 0068 WT QH WP in QL 3 2 1 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9153E A low temperature power plant operates with R410A maintaining 60 psia in the condenser a high pressure of 400 psia with superheat Find the temperature out of the boilersuperheater so the turbine exit temperature is 20 F and find the overall cycle efficiency State 1 P1 60 psia v1 00129 ft3lbm h1 127 Btulbm State 4 P4 P1 60 psia h4 34358 Btulbm s4 13242 BtulbmR State 3 400 psia s s4 h3 1472 Btulbm T3 1903 F Pump wp v1 P2 P1 00129 400 60 144 778 209 Btulbm Boiler qH h3 h2 42656 127 08 1337 Btulbm Turbine wT h3 h4 1472 1221 251 Btulbm Efficiency ηTH wNETqH wT wPqH 251 08 1337 0182 P v 1 2 3 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9154E A smaller power plant produces 50 lbms steam at 400 psia 1100 F in the boiler It cools the condenser with ocean water coming in at 60 F and returned at 65 F so that the condenser exit is at 110 F Find the net power output and the required mass flow rate of the ocean water Solution The states properties from Tables F71 and F72 1 110 F x 0 h1 7801 Btulbm v1 001617 ft3lbm Psat 128 psia 3 400 psia 1100 F h3 157744 Btulbm s3 17989 Btulbm R CV Pump Reversible and adiabatic Energy wp h2 h1 Entropy s2 s1 since incompressible it is easier to find work positive in as wp v dP v1 P2 P1 001617 400 13144 778 119 Btulbm CV Turbine wT h3 h4 s4 s3 s4 s3 17989 01473 x4 18101 x4 09124 h4 7801 09124 103128 101895 Btulbm wT 157744 101895 5585 Btulbm W NET m wT wp 50 5585 119 27 866 Btus CV Condenser qL h4 h1 101895 7801 94094 Btulbm Q L m qL 50 94094 47 047 Btus m ocean Cp T m ocean Q L Cp T 47 047 10 5 9409 lbms Q WT 3 2 4 1 Condenser Boiler Turbine WP QB T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9155E Consider a simple ideal Rankine cycle using water at a supercritical pressure Such a cycle has a potential advantage of minimizing local temperature differences between the fluids in the steam generator such as the instance in which the hightemperature energy source is the hot exhaust gas from a gas turbine engine Calculate the thermal efficiency of the cycle if the state entering the turbine is 8000 lbfin2 1300 F and the condenser pressure is 095 lbfin2 What is the steam quality at the turbine exit Solution For the efficiency we need the net work and steam generator heat transfer State 1 s1 01296 Btulbm R h1 6804 Btulbm State 3 h3 15475 Btulbm s3 14718 Btulbm R CV Pump For this high exit pressure we use Table F73 to get state 2 Entropy Eq s2 s1 h2 9169 Btulbm wp h2 h1 9169 6804 2365 Btulbm CV Turbine Assume reversible and adiabatic Entropy Eq s4 s3 14718 01296 x418526 x4 07245 Very low for a turbine exhaust h4 6804 x4 103698 75129 Btulbm wT h3 h4 7962 Btulbm Steam generator qH h3 h2 15475 9169 14558 Btulbm wNET wT wp 7962 2365 7726 Btulbm η wNETqH 7726 14558 053 P v 1 2 3 4 T s 1 2 3 4 095 psia 8000 psia Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9156E A Rankine cycle uses ammonia as the working substance and powered by solar energy It heats the ammonia to 320 F at 800 psia in the boilersuperheater The condenser is water cooled and the exit is kept at 70 F Find T P and x if applicable for all four states in the cycle NH3 ideal Rankine cycle State 1 Table F81 T 70 F x 0 P1 12885 psia h1 12021 Btulbm v1 02631 ft3lbm CV Pump wP h2 h1 v1P2 P1 002631 ft3lbm 800 12885 psi 144 in2ft2 254275 lbfft 327 Btulbm h2 h1 wP 12021 327 12348 Btulbm hf P2 Psatvf T2 T1 70 F with the CATT3 we do P2 s2 s1 State 3 320 F 800 psia superheated vapor s3 11915 Btulbm CV turbine s4 s3 11915 02529 x4 09589 x4 09788 P4 P1 12885 psia T4 T1 70 F Q WT 3 2 4 1 Condenser Solar Turbine WP QRAD collector T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9157E Assume the powerplant in Problem 9156E should deliver 1000 Btus What is the mass flow rate of ammonia State 3 320 F 800 psia h3 7347 Btulbm s3 11915 Btulbm CV turbine Energy wTs h3 h4 Entropy 0 s3 s4 0 s4 s3 11915 02529 x4 09589 x4 09788 h4 12021 09788 50789 6173 Btulbm wTs h3 h4 7347 6173 1174 Btulbm W T m wTs m W T wTs 1000 Btus 1174 Btulbm 8518 lbms Q WT 3 2 4 1 Condenser Solar Turbine WP QRAD collector T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9158E Consider an ideal steam reheat cycle in which the steam enters the highpressure turbine at 600 lbfin2 700 F and then expands to 150 lbfin2 It is then reheated to 700 F and expands to 2225 lbfin2 in the lowpressure turbine Calculate the thermal efficiency of the cycle and the moisture content of the steam leaving the lowpressure turbine Solution Basic Rankine cycle with a reheat section For this cycle from Table F7 State 3 Superheated vapor h3 135062 Btulbm s3 15871 Btulbm R State 1 Saturated liquid h1 9797 Btulbm v1 001625 ft3lbm CV Pump Adiabatic and reversible Use incompressible fluid so wP v dP v1P2 P1 001625600 22144 778 18 Btulbm h2 h1 wP 9797 18 9977 Btulbm CV HP Tubine section wT1 h3 h4 s4 s3 h4 120893 Btulbm wT1 135062 120893 14169 Btulbm CV LP Tubine section wT2 h5 h6 s6 s5 State 5 h5 137655 Btulbm s5 17568 Btulbm R State 6 s6 s5 17568 01817 x6 17292 x6 09109 h6 9797 09109 101978 102689 Btulbm wT2 137655 102689 34966 Btulbm CV Boiler qH1 h3 h2 135062 9977 125085 Btulbm Total cycle qH qH1 h5 h4 125085 137655 120893 14185 Btulbm wTtot wT1 wT2 14169 34966 49135 Btulbm Overall efficiency ηCYCLE wTtot wPqH 49135 1814203 0345 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Q W T 3 2 4 1 Condenser Boiler Turbine W P Q H 5 6 L cb T s 1 2 3 4 5 6 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9159E An open FWH receives steam at 150 psia 400 F from the turbine and 150 psia 200 F water from the feed water line Find the required fraction of the extraction flow in the turbine The setup follows Fig912 State enthalpies h2 16807 h6 121951 all Btulbm h3 33074 x 0 at 150 psia 35847 F Analyze the FWH which leads to Eq95 y m 6 m 5 the extraction fraction y h3 h2 h6 h2 33074 16807 121951 16807 01547 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9160E Consider an ideal steam regenerative cycle in which steam enters the turbine at 600 lbfin2 700 F and exhausts to the condenser at 2225 lbfin2 Steam is extracted from the turbine at 150 lbfin2 for an open feedwater heater The feedwater leaves the heater as saturated liquid The appropriate pumps are used for the water leaving the condenser and the feedwater heater Calculate the thermal efficiency of the cycle and the net work per poundmass of steam From Table F72 h5 135062 Btulbm s5 15871 Btulbm R h1 9797 Btulbm v1 001625 ft3lbm Interpolate to get h3 33067 Btulbm v3 001809 ft3lbm ST GEN P1 P2 FW HTR COND TURBINE 4 3 5 6 7 1 2 CV Pump1 wP12 001625150 22144 778 044 Btulbm h2 h 1 h2 h1 wP12 9841 Btulbm CV Pump2 22 psi 150 psi 600 psi 1 2 4 3 7 6 5 T s wP34 001809600 150144778 1507 Btulbm h4 h3 wP34 33218 Btulbm CV Turbine high pressure section Entropy Eq s6 s5 15871 Btulbm R h6 120893 Btulbm CV feedwater heater call the extraction fraction y m 6m 3 Continuity Eq m 3 m 6 m 2 Energy Eq m 6h6 m 2h2 m 3h3 yh6 1 yh2 h3 y6 h3 h2h6 h2 y 33067 9841120893 9841 02091 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CV Turbine from 5 to 7 s7 s5 x7 15871 0181717292 08127 h7 9797 08127 101978 92675 Btulbm The HP turbine section has the full flow the LP section has fraction 1y wT h5 h6 1 yh6 h7 135062 120893 07909120893 92675 36487 Btulbm CV pumps P2 has the full flow and P1 has the fraction 1 y of the flow wP 1 ywP12 wP34 07909 044 1 1507 1855 Btulbm wNET wT wP 36487 1855 3630 Btulbm CV steam generator qH h5 h4 135062 33218 101844 Btulbm ηTH wNETqH 363101844 0356 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9161E A closed feedwater heater in a regenerative steam power cycle heats 40 lbms of water from 200 F 2000 lbfin2 to 450 F 2000 lbfin2 The extraction steam from the turbine enters the heater at 600 lbfin2 550 F and leaves as saturated liquid What is the required mass flow rate of the extraction steam Solution The schematic from Figure 913 has the feedwater from the pump coming at state 2 being heated by the extraction flow coming from the turbine state 6 so the feedwater leaves as saturated liquid state 4 and the extraction flow leaves as condensate state 6a 2 4 6 6a From the steam tables F7 F73 h2 1726 Btulbm F73 h4 43113 Btulbm F72 h6 125536 Btulbm Interpolate for this state F71 h6a 47156 Btulbm CV Feedwater Heater Energy Eq m 2h2 m 6h6 m 2h4 m 6h 6a Since all four state are known we can solve for the extraction flow rate m 6 m 2 h2 h4 h6a h6 40 lbms 1726 43113 47156 125536 132 lbm s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9162E A Rankine cycle feeds 10 lbms ammonia at 300 psia 280 F to the turbine which has an extraction point at 125 psia The condenser is at 0 F and a closed feed water heater has an exit state 3 at the temperature of the condensing extraction flow and it has a drip pump The source for the boiler is at constant 350 F Find the extraction flow rate and state 4 into the boiler P1 30415 psia h1 426 Btulbm v1 00242 ft3lbm s5 13062 BtulbmR h5 74407 Btulbm T6a Tsat 125 psia 6828 F h6a 11825 Btulbm From turbine 6 1 2 6a From condenser Pump 1 Pump 2 3 4 6b CV Turbine Reversible adiabatic so constant s from inlet to extraction point s6 s5 13062 BtulbmR T6 1481 F h6 6798 Btulbm CV P1 wP1 v1P2 P1 00242300 30415 144778 121 Btulbm h2 h1 wP1 4381 Btulbm CV P2 wP2 v6a P4 P6 002625 300 125144778 085 Btulbm h6b h6a wP2 1191 Btulbm CV Total FWH and pump notice h3 h6a Pv h6b no table for this state State 3 h3 hf P3Psatvf 11825 002625 300125 144778 1191 Btulbm The extraction fraction is y m 6m 4 Energy 1 yh2 yh6 1 yh3 yh 6a y h3 h2 h3 h2 h6 h6a 1191 4381 1191 4381 6798 11825 01182 m 6 y m 4 01182 10 1182 lbms CV The junction after FWH and pump 2 h4 1yh3 y h6b 1 01182 1191 01182 1191 1191 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9163E Do the previous problem with a closed FWH that has a trap and drain to the condenser for the extraction flow State 1 x1 0 h1 426 Btulbm v1 00242 ft3lbm State 3 h3 hf P3Psatvf 11825 002625 300125 144778 1191 Btulbm State 5 h5 74407 Btulbm s5 13062 BtulbmR State 6 s6 s5 h6 67983 Btulbm State 6a x6a 0 h6a 11825 Btulbm CV P1 wP1 v1P2 P1 00242300 30415 144778 121 Btulbm h2 h1 wP1 4381 Btulbm CV Feedwater heater Call m 6 m tot y the extraction fraction Energy Eq h2 y h6 1 h3 y h6a y h3 h2 h6 h6a 1191 4381 67983 11825 01341 m extr y m tot 01341 10 1341 lbms h4 h3 1191 Btulbm 1 TURBINE COND FWH 2 34 5 6 7 P1 STEAM GEN HP LP cb 6a s 1 2 6a 34 5 6 7 304 psia 300 psia 125 psia T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9164E The steam power cycle in Problem 9147E has an isentropic efficiency of the turbine of 85 and that for the pump it is 80 Find the cycle efficiency and the specific work and heat transfer in the components States numbered as in fig 93 of text CV Pump wPS v1P2 P1 001625600 22144778 18 Btulbm wPAC 1808 2245 Btulbm h2 h1 wPAC 9797 2245 1002 Btulbm CV Turbine wTS h3 h4s s4 s3 14464 Btulbm R s4 s3 14464 01817 x4 17292 x4 07314 h4 9797 07314 101978 84384 Btulbm wTS 120406 84384 36022 Btulbm wTAC h3 h4AC 36022 085 3062 h4AC 89786 Btulbm still twophase CV Boiler qH h3 h2 120406 1002 11039 Btulbm qL h4AC h1 89786 9797 7999 Btulbm ηCYCLE wT wPqH 3062 224511039 0275 Compared to 360221811043 0325 in the ideal case Q WT 3 2 4 1 Condenser Boiler Turbine WP QB T s 1 2 3 4s 4ac state 2s and 2ac nearly the same Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9165E A concentrated solar power plant receives the energy from molten salt coming in at 1000 F and leaving at 600 F in a counterflow heat exchanger where the water comes in at 400 psia 140 F and leaves at 800 F 400 psia The molten salt has 10 lbms flow with Cp 036 BtulbmR What is the possible water flow rate the rate of energy transfer and rate of entropy generation Continuity Eqs Each line has a constant flow rate through it Energy Eq410 0 m H2O h1 h2 m salt h3 h4 Entropy Eq77 0 m H2O s1 s2 m salt s3 s4 S gen Process Each line has a constant pressure Table F7 h2 141659 Btulbm s1 01985 s2 16844 BtulbmR h1 hf P3Psatvf 10796 40029 001629144778 10915 Btulbm From the energy leaving the salt we get Q m salt h3 h4 m salt CP salt T3 T4 10 036 1000 600 1440 Btus m H2O m salt h3 h4 h2 h1 Q h2 h1 1440 Btus 141659 10915 Btulbm 11 lbms s4 s3 CP salt ln T4T3 036 ln1059714597 0115 BtulbmR S gen m H2O s2 s1 m salt s4 s3 11 lbms 16844 01985 Btu lbmR 10 lbms 0115 Btu lbmR 048 BtusR CV Heat exchanger steady flow 1 inlet and 1 exit for salt and water each The two flows exchange energy with no heat transfer tofrom the outside 3 salt 1 water 4 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9166E Steam leaves a power plant steam generator at 500 lbfin2 650 F and enters the turbine at 490 lbfin2 625 F h 1314 Btulbm s 15752 BtulbmR The isentropic turbine efficiency is 88 and the turbine exhaust pressure is 17 lbfin2 Condensate leaves the condenser and enters the pump at 110 F 17 lbfin2 The isentropic pump efficiency is 80 and the discharge pressure is 520 lbfin2 The feedwater enters the steam generator at 510 lbfin2 100 F h 681 Btulbm Calculate the thermal efficiency of the cycle and the entropy generation of the flow in the line between the steam generator exit and the turbine inlet assuming an ambient temperature of 77 F ST GEN 3 6 COND TURBINE 1 2 P 4 5 η 088 T s 650 F 625 F 2 5s 6 4 3s 3 5 500 psia 490 psia 17 psia 1 h1 1328 Btulbm s1 1562 BtulbmR s3S s2 15752 016483 x3S 17686 x3S 079745 h3S 881 0797 45 10254 9058 Btulbm wTs h2 h3S 13140 9058 4082 Btulbm wTac ηTs wTs 088 4082 3592 Btulbm h3 h2 wTac 13140 3592 9548 Btulbm wPs 0016166 520 17 144778 155 Btulbm wPac wPsηPs 155080 194 Btulbm qH h1 h6 13280 681 12599 Btulbm ηTH wNETqH 3592 19412599 0284 CV Line from 1 to 2 w 0 Energy Eq q h2 h1 1314 1328 14 Btulbm Entropy Eq s1 sgen qT0 s2 sgen s2 s1 qT0 15752 1586 145367 00153 Btulbm R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9167E A boiler delivers steam at 1500 lbfin2 1000 F to a reversible twostage turbine as shown in Fig 919 After the first stage 25 of the steam is extracted at 200 lbfin2 for a process application and returned at 150 lbfin2 190 F to the feedwater line The remainder of the steam continues through the lowpressure turbine stage which exhausts to the condenser at 2225 lbfin2 One pump brings the feedwater to 150 lbfin2 and a second pump brings it to 1500 lbfin2 If the process application requires 5000 Btus of power how much power can then be cogenerated by the turbine CV Turbine T1 5 h5 149032 s5 16001 BtulbmR 6 Rev and adiabatic s6 s5 Table F72 Sup vapor h6 12466 Btulbm wT1 h5 h6 2437 Btulbm State 7 s7 s6 s5 x7 16001 01817 17292 082026 P2 P1 C 1 2 8 4 3 5 6 7 T1 T2 Boiler Process heat 5000 Btus h7 9797 082026 101978 93445 Btulbm wT2 h6 h7 12466 93445 3121 Btulbm 8 Compressed liquid use sat liq same T h8 15802 Btulbm CV process unit Assume no work only heat out qPROC h6 h8 12466 15802 10886 Btulbm m 6 Q qPROC 5000 Btus 10886 Btulbm 4593 lbms 025 m TOT m TOT m 5 18372 lbms m 7 m 5 m 6 13779 lbms CV Total turbine W T m 5h5 m 6h6 m 7h7 18372 149032 4593 12466 13779 93445 lbm s Btu lbm 8779 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Refrigeration Cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9168E A car airconditioner refrigerator in 70 F ambient uses R134a and I want to have cold air at 20 F produced What is the minimum high P and the maximum low P it can use Since the R134a must give heat transfer out to the ambient at 70 F it must at least be that hot at state 3 Ideal Ref Cycle Tcond 70 F T 3 Tevap 20 F Use Table F10 for R134a 1 2 T 3 4 s From Table F101 P3 P2 Psat 8595 psia is minimum high P Since the R134a must absorb heat transfer at the cold air 20 F it must at least be that cold at state 4 From Table F101 P1 P4 Psat 3329 psia is maximum low P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9169E Consider an ideal refrigeration cycle that has a condenser temperature of 110 F and an evaporator temperature of 5 F Determine the coefficient of performance of this refrigerator for the working fluids R134a and R410A Ideal Ref Cycle Tcond 110 F T 3 Tevap 5 F Use Table F10 for R134a Use Table F9 for R410A 1 2 T 3 4 s Property for R134a R410A h1 Btulbm 1675 1178 s2 s1 BtulbmR 04149 02549 P2 lbfin 2 1611 3804 T2 F 1237 1573 h2 Btulbm 1847 1384 h3 h4 Btulbm 1124 561 wC h2 h1 172 206 qL h1 h4 551 617 β qLwC 320 30 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9170E Find the high temperature the condensing temperature and the COP if ammonia is used in a standard refrigeration cycle with high and low pressures of 800 psia and 300 psia respectively Exit evaporator x 1 and 300 psia from F82 T1 1232 F h1 63263 Btulbm s1 11356 BtulbmR Exit condenser saturated liquid 800 psia from F81 T3 20065 F h3 28417 Btulbm Exit compressor 800 psia s s1 so interpolate in F82 T2 269 F h2 69265 Btulbm wC h2 h1 6002 Btulbm qL h1 h4 h1 h3 34846 Btulbm COP β qL wc h1 h3 h2 h1 34846 6002 62 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9171E A refrigerator receives 500 W of electrical power to the compressor driving the cycle flow of R134a The refrigerator operates with a condensing temperature of 100 F and a low temperature of 10 F Find the COP for the cycle and its cooling capacity Solution Ideal refrigeration cycle Tcond 100 F T 3 Tevap 10 F T 1 Use Table F10 T s 1 2 3 4 State 1 h1 16509 Btulbm s1 04162 BtulbmR State 2 s s1 P2 P3 138926 psia h2 18402 Btulbm wC h2 h1 18402 16509 18934 Btulbm State 34 h4 h3 10886 Btulbm qL h1 h4 16509 10886 5623 Btulbm COP β qLwC 297 The cooling capacity is the rate at which it can cool Q L COP W 297 500 W 1485 W 1407 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9172E Consider an ideal heat pump that has a condenser temperature of 120 F and an evaporator temperature of 30 F Determine the coefficient of performance of this heat pump for the working fluids R410A and ammonia Ideal Heat Pump Tcond 120 F Tevap 30 F Use Table F8 for NH3 Use Table F9 for R410A 1 2 T 3 4 s Property R410A NH3 h1 Btulbm 1199 6195 s2 s 1 02482 12768 P2 lbfin 2 4333 2865 T2 F 1582 2393 h2 Btulbm 1359 7194 h3h4 Btulbm 608 1788 wC h2h1 160 999 qH h2h3 751 5406 β qHw C 469 541 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9173E The refrigerant R134a is used as the working fluid in a conventional heat pump cycle Saturated vapor enters the compressor of this unit at 50 F its exit temperature from the compressor is measured and found to be 185 F If the compressor exit is 300 psia what is the isentropic efficiency of the compressor and the coefficient of performance of the heat pump R134a heat pump T2 185 F TEVAP 50 F State 1 Table F91 h1 1738 Btulbm s1 04113 Btulbm R 1 2 T 3 4 s 2S State 2 h2 1938 Btulbm Compressor work wC h2 h1 1938 1738 200 Btulbm Isentropic compressor s2S s1 04113 Btulbm R State 2s P2 s T2S 1669 F h2S 1880 Btulbm Ideal compressor work wC s h2S h1 1880 1738 142 Btulbm The efficiency is the ratio of the two work terms ηS COMP wC wC s 142 200 071 The condenser has heat transfer as h3 hf at 300 psia qH h2 h3 1938 1301 637 Btulbm and a coefficient of performance of β qHwC 319 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Exergy and Combined Cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9174E A Rankine cycle maintains 130 F in the condenser which is cooled by a 70 F reservoir The steam out of the boiler is at 600 psia 700 F being heated from a 900 F source Determine the flux of exergy in or out of the reservoirs per unit mass flowing in the cycle Find the overall cycle second law efficiency Solution Cycle from problem 9181E State 3 Superheated vapor h3 135062 Btulbm s3 15871 Btulbm R State 1 Sat liq h1 9797 Btulbm v1 001625 ft3lbm s 01817 Btulbm R CV Pump Adiabatic and reversible Use incompressible fluid so wP v dP v1P2 P1 001625 ft3lbm 600 22 psi 144 in2ft2 139885 lbfft 18 Btulbm h2 h1 wP 9797 18 9977 Btulbm CV Boiler qH h3 h2 135062 9977 125085 Btulbm CV Tubine s4 s3 15871 Btulbm R 01817 x4 17292 x4 08127 h4 9797 08127 101978 92675 Btulbm wT h3 h4 135062 92675 42387 Btulbm CV Condenser qL h4 h1 92675 9797 8288 Btulbm We can do the fluxes of exergy evaluarted at the reservoir Ts ψH 1 T0TH qH 1 53667 135967 125085 75713 Btulbm ψL 1 T0TL qL 1 53667 52967 8288 1095 Btulbm as TL T0 you decrease ψ The overall second law efficiency is based on the exergy delivered from the high T source ηII ψH wnet ψH wT wP 42387 18 75713 0557 Notice TH T3 TL T4 T1 so cycle is externally irreversible Both qH and qL are over finite T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Energy Transfers Btulbm Exergy Transfers Btulbm Comment The exergy into the cycle from the high temperature source must be found from the flow exergies in the water as the temperature is varying HE cb T T q 1251 H q 829 H L w 424 w 18 L HE cb T T φ 757 H φ 745 H L w 424 w 18 φ 11 φ 497 L Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9175E Find the flows and fluxes of exergy in the condenser of Problem 9154E Use those to determine the 2nd law efficiency A smaller power plant produces 50 lbms steam at 400 psia 1100 F in the boiler It cools the condenser with ocean water coming in at 60 F and returned at 65 F so that the condenser exit is at 110 F Find the net power output and the required mass flow rate of the ocean water Solution The states properties from Tables F71 5 6 4 1 cb State 1 110 F x 0 h1 7801 Btulbm s1 01473 Btulbm R State 3 400 psia 1100 F h3 157744 Btulbm s3 17989 Btulbm R CV Turbine wT h3 h4 s4 s3 s4 s3 17989 01473 x4 18101 x4 09124 h4 7801 09124 103128 101895 Btulbm CV Condenser qL h4 h1 101895 7801 94094 Btulbm Q L m qL 50 94094 47 047 Btus m ocean Cp T m ocean Q L Cp T 47 047 10 5 9409 lbms The specific flow exergy for the two states are from Eq1024 neglecting kinetic and potential energy ψ4 h4 h0 T0s4 s0 ψ1 h1 h0 T0s1 s0 The net drop in exergy of the water is Φ water m water h4 h1 Tos4 s1 50 101895 7801 5147 17989 01473 47 047 42 504 4543 Btus Notice that the reference values drops out Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful The net gain in exergy of the ocean water is Φ ocean m oceanh6 h5 Tos6 s5 m oceanCpT6 T5 ToCp ln T6 T5 9409 10 65 60 5147 10 ln 4597 65 4597 60 47 047 46 370 677 Btus The second law efficiency is ηII Φ ocean Φ water 677 4543 015 In reality all the exergy in the ocean water is destroyed as the 65 F water mixes with the ocean water at 60 F after it flows back out into the ocean and the efficiency does not have any significance Notice the small rate of exergy relative to the large rates of energy being transferred Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9176E Find the flows of exergy into and out of the feedwater heater in Problem 9160E State 1 x1 0 h1 9797 Btulbm v1 001625 ft3lbm s 01817 State 3 x3 0 h3 3307 Btulbm s3 05142 Btulbm R State 5 h5 135052 Btulbm s5 15871 Btulbm R State 6 s6 s5 15871 Btulbm R h6 120893 Btulbm CV Pump P1 wP1 h2 h1 v1P2 P1 001625150 2225144 778 044 Btulbm h2 h1 wP1 9797 044 9841 Btulbm s2 s1 01817 Btulbm R CV Feedwater heater Call m 6 m tot x the extraction fraction Energy Eq 1 x h2 x h6 1 h3 x h3 h2 h6 h2 3307 9841 120893 9841 02092 2 6 3 x 1x FWH Ref State 147 psia 77 F so 008774 Btulbm R ho 4508 Btulbm ψ2 h2 ho Tos2 so 9841 4508 5366701817 008774 290 Btulbm ψ6 120893 4508 5366715871 008774 3592 Btulbm ψ3 3307 4508 5366705142 008774 5675 Btulbm The rate of exergy flow scaled with maximum flow rate is then Φ 2m 3 1 x ψ2 07908 290 2297 Btulbm Φ 6m 3 xψ6 02092 3592 75144 Btulbm Φ 3m 3 ψ3 5675 Btulbm The mixing is destroying 2297 75144 5675 207 Btulbm of exergy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9177E For Problem 9162E consider the boilersuperheater Find the exergy destruction and the second law efficiency for the boilersource setup The boiler has flow in at state 4 and out at state 5 with the source providing a q at 350 F Assume state 4 is saturated liquid at T4 so h4 hf 4 T4 684 F State 4 h4 1184 Btulbm hf 4 s4 sf 4 02494 BtulbmR State 5 h5 7441 Btulbm s5 13062 BtulbmR Energy Eq qH h5 h4 7441 1184 6257 Btulbm Entropy Eq s4 qHTH sgen s5 sgen s5 s4 qHT H S gen m sgen 10 13062 02494 6257 350 4597 10 02841 2841 BtusR The flow increase in exergy is ψ5 ψ4 h5 h4 To s5 s4 6257 5367 13062 02494 5862 Btulbm The exergy provided by the source is φH 1 To TH qH 1 5367 350 4597 6257 21096 Btulbm Second law efficiency is ηII gain in exergy source exergy input 5862 21096 0265 1 TosgenφH 1 5367 02841 21096 T s 5 4 6 7 1 350 F 4 5 CV Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9178E Steam is supplied in a line at 400 lbfin2 1200 F A turbine with an isentropic efficiency of 85 is connected to the line by a valve and it exhausts to the atmosphere at 147 lbfin2 If the steam is throttled down to 300 lbfin2 before entering the turbine find the actual turbine specific work Find the change in exergy through the valve and the second law efficiency of the turbine CV Valve Energy Eq h2 h1 163179 Btulbm Entropy Eq s2 s1 18327 Btulbm R State 2 h2 P2 s2 186407 Btulbm R Ideal turbine s3 s2 h3s 121228 Btulbm wTs h2 h3s 41951 Btulbm Actual turbine wTac ηTwTs 35658 Btulbm h3ac h2 wTac 127521 Btulbm s3ac 19132 Btulbm R ψ2 ψ1 h2 h1 T0s2 s1 0 53667186407 18327 16835 Btulbm wrev ψ2 ψ3 163179 127521 53667186407 19132 38295 Btulbm ηII wacwrev 3565838295 0931 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9179E Find the two heat transfer rates the total cycle exergy destruction and the second law efficiency for the refrigerator in problem 9171E A refrigerator receives 500 W of electrical power to the compressor driving the cycle flow of R134a The refrigerator operates with a condensing temperature of 100 F and a low temperature of 10 F Find the COP for the cycle State 1 h1 16509 Btulbm s1 04162 BtulbmR State 2 s2 s1 P2 P3 1389 psia h2 18402 Btulbm wC h2 h1 18402 16509 1893 Btulbm State 3 h3 10886 Btulbm State 4 h4 h3 10886 Btulbm qH h2 h3 18402 10886 7516 Btulbm qL h1 h4 5623 Btulbm m W wC 05 3412 1893 9012 lbmhr Q H 9012 7516 6773 Btuhr Q L 9012 5623 5067 Btuhr Entropy Eq for total CV 0 qL TL qH TH sgen i To sgen To TH qH TL qL 5367 7516 5597 5623 4497 496 Btulbm I m i 9012 lbmhr 496 Btulbm 447 Btuhr The cold space gain in availability ΔψL 1 TL To qL 1 5367 4497 5623 1088 Btulbm ηII output source ψL wC 1088 1893 0575 Remark The high temperature reservoir also gains availability so ΔψH 1 TH To qL 1 5367 5597 7516 309 Btulbm the total balance is wC ΔψH ΔψL i 309 1088 496 1893 OK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9180E Consider an ideal dualloop heatpowered refrigeration cycle using R134a as the working fluid as shown in Fig P9135 Saturated vapor at 200 F leaves the boiler and expands in the turbine to the condenser pressure Saturated vapor at 0 F leaves the evaporator and is compressed to the condenser pressure The ratio of the flows through the two loops is such that the turbine produces just enough power to drive the compressor The two exiting streams mix together and enter the condenser Saturated liquid leaving the condenser at 110 F is then separated into two streams in the necessary proportions Determine the ratio of mass flow rate through the power loop to that through the refrigeration loop Find also the performance of the cycle in terms of the ratio QLQH BOIL COND E V A P TURB COMP 1 2 7 6 3 4 5 P Q L T 3 4 s 6 7 2 1 5 T P h s Computer tables for F lbfin2 Btulbm Btulbm R properties 1 0 222 1668 04154 P2P3PSAT at 110 F 2 1611 04154 P5P6PSAT at 200 F 3 110 1611 1124 0067 45 s2s104154 4 0 212 1124 h2 1850 5 5036 Pump work 6 200 5036 1817 03955 wP h5h3 7 110 1611 03955 v3P5P3 wP 0014155036 1611144 778 0897 Btulbm h5 1124 0897 1133 Btulbm s7 s6 03955 02881 x701203 x7 08928 h7 1124 08928 685 1736 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CV turbine compressor Continuity Eq m 1 m 2 m 6 m 7 Energy Eq m 1h1 m 6h6 m 2h2 m 7h 7 m 6m 1 h2 h1 h6 h7 1850 1668 1817 1736 2247 CV evaporator Q L m 1h1h4 CV boiler Q H m 6h6h5 β Q LQ H m 1h1h4 m 6h6h5 1668 1124 2247 1817 1133 0354 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9181E The power plant in Problem 9147E is modified to have a superheater section following the boiler so the steam leaves the super heater at 600 lbfin2 700 F Find the specific work and heat transfer in each of the ideal components and the cycle efficiency Solution For this cycle from Table F7 State 3 Superheated vapor h3 135062 Btulbm s3 15871 Btulbm R State 1 Saturated liquid h1 9797 Btulbm v1 001625 ft3lbm CV Pump Adiabatic and reversible Use incompressible fluid so wP v dP v1P2 P1 001625600 22144 778 18 Btulbm h2 h1 wP 9797 18 9977 Btulbm CV Boiler qH h3 h2 135062 9977 125085 Btulbm CV Tubine wT h3 h4 s4 s3 s4 s3 15871 Btulbm R 01817 x4 17292 x4 08127 h4 9797 08127 101978 92675 Btulbm wT 135062 92675 42387 Btulbm ηCYCLE wT wPqH 42387 18125085 0337 CV Condenser qL h4 h1 92675 9797 8288 Btulbm P v 1 2 3 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9182E Consider a small ammonia absorption refrigeration cycle that is powered by solar energy and is to be used as an air conditioner Saturated vapor ammonia leaves the generator at 120 F and saturated vapor leaves the evaporator at 50 F If 3000 Btu of heat is required in the generator solar collector per poundmass of ammonia vapor generated determine the overall performance of this system NH3 absorption cycle sat vapor at 120 F exits the generator Sat vapor at 50 F exits the evaporator qH qGEN 3000 Btulbm NH3 out of generator 1 2 T s GEN EXIT EVAP EXIT 120F 50 F qL h2 h1 hG 50 F hF 120 F 62428 17879 44549 Btulbm qLqH 445493000 01485 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9183E Consider an ideal combined reheat and regenerative cycle in which steam enters the highpressure turbine at 500 lbfin2 700 F and is extracted to an open feedwater heater at 120 lbfin2 with exit as saturated liquid The remainder of the steam is reheated to 700 F at this pressure 120 lbfin2 and is fed to the lowpressure turbine The condenser pressure is 2225 lbfin2 Calculate the thermal efficiency of the cycle and the net work per poundmass of steam 5 h5 135666 s5 16112 7 h7 137817 s7 17825 3 h3 hf 31259 v3 001788 CV T1 s5 s6 h6 120976 wT1 h5 h6 135666 120976 1469 Btulbm CV Pump 1 wP1 h2 h1 v1P2 P1 001623120 2 0354 P P 1 2 4 5 6 7 8 COND HTR 3 T1 T2 x 1x 1x h2 h1 wP1 9373 0354 9408 Btulbm CV FWH x h6 1 x h2 h3 x EA hA3 A hA2 A E hA6 A hA2 AE A A 31259 9408 120976 9408E A 01958 CV Pump 2 s 1 2 3 5 6 7 8 700 F 4 2 psi T wAP2E A hA4E A hA3E A vA3E APA4E A PA3E A 001788500 120144778 126 Btulbm hA4E A hA3E A wAP2E A 31259 126 31385 Btulbm qAHE A hA5E A hA4E A 1 xhA7E A hA6E A 104281 13543 11782 Btulbm CV Turbine 2 sA7E A sA8E A xA8E A 17825 0181717292 09257 hA8E A hAfE A xA8E A hAfgE A 9797 09257 101978 104198 wAT2E A hA7E A hA8E A 137817 104198 33619 wAnetE A wAT1E A 1 x wAT2E A 1 x wAP1E A wAP2E 1469 27036 0285 126 41572 kJkg ηAcycleE A wAnetE A qAHE A 41572 11782 0353 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9184E In one type of nuclear power plant heat is transferred in the nuclear reactor to liquid sodium The liquid sodium is then pumped through a heat exchanger where heat is transferred to boiling water Saturated vapor steam at 700 lbfin2 exits this heat exchanger and is then superheated to 1100 F in an external gasfired superheater The steam enters the turbine which has one opentype feedwater extraction at 60 lbfin2 The isentropic turbine efficiency is 87 and the condenser pressure is 095 lbfin2 Determine the heat transfer in the reactor and in the superheater to produce a net power output of 1000 Btus P 1 TURBINE COND HTR P 4 2 3 5 6 SUP HT REACT Q 7 8 T s 1 2 3 1100 F 4 5 6 7 8 7s 8s 700 lbfin 2 60 lbfin 2 1 lbfin 2 A W E ANETE A 1000 Btus ηASTE A 087 CV Pump 1 wAP12E A 001613660 1144778 018 Btulbm hA2E A hA1E A wAP12E A 6973 018 6991 Btulbm CV Pump 2 wAP34E A 0017378700 60144778 206 Btulbm hA4E A hA3E A wAP34E A 26224 206 2643 Btulbm CV HP Turbine section sA7SE A sA6E A 17682 PA7E A TA7SE A 5008 F hA7SE A 12834 Btulbm hA7E A hA6E A ηASTE AhA6E A hA7SE A 16258 08716258 12834 13279 CV LP Turbine section sA8SE A sA6E A 17682 01296 xA8SE A 18526 xA8SE A 08845 hA8SE A 6804 08845 103698 98525 Btulbm hA8E A hA6E A ηASTE AhA6E A hA8SE A 16258 08716258 98525 10685 Btulbm CV Feedwater heater Continuity Eq mA2E A mA7E A mA3E A 10 lbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Energy Eq mA2E AhA2E A mA7E AhA7E A mA3E AhA3E mA7E A 262246991 132796991 01529 CV turbine wATE A hA6E A hA7E A 1 mA7E AhA7E A hA8E A 1625813279 084711327910685 5176 Btulbm CV pumps wAP E A mA1E AwAP12E A mA3E AwAP34E A 08471018 1206 22 Btulbm wANETE A 5176 22 5154 Btulbm AmE A 10005154 1940 lbms CV reactor AQ E AREACTE A AmE AhA5E A hA4E A 194 1202 2643 1819 Btus CV superheater AQ E ASUP E A AmE AhA6E A hA5E A 194 16258 1202 822 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9192eE Consider an ideal heat pump that has a condenser temperature of 120 F and an evaporator temperature of 30 F Determine the coefficient of performance of this heat pump for the working fluid R12 Ideal Heat Pump TAcondE A 120 F TAevapE A 30 F Use computer table for R12 1 2 T 3 4 s Propertyparameter R12 hA1E A Btulbm 8042 sA2E A sA1E A BtulbmR 01665 PA2E A lbfinA2E 1723 TA2E A F 1322 hA2E A Btulbm 910 hA3E A hA4E A Btulbm 36011 wACE A hA2E A hA1E 1058 qAHE A hA2E A hA3E 54995 β qAHE AwACE A 5198 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9193dE Do Problem 9151 with R22 as the working fluid Standard Rankine cycle with properties from the computer R22 tables hA1E A 39267 Btulbm vA1E A 001404 ftA3E Albm PA1E A 2106 psia PA2E A PA3E A 5548 psia hA3E A 11007 Btulbm sA3E A 01913 Btulbm R CV Pump wAP E A vA1E APA2E APA1E A 001404 55482106A144 778E A 0894 Btulbm hA2E A hA1E A wAP E A 39267 0894 4016 Btulbm CV Turbine sA4E A sA3E A xA4E A 01913 007942013014 09442 hA4E A 101885 Btulbm wATE A hA3E A hA4E A 8185 Btulbm CV Boiler qAHE A hA3E A hA2E A 11007 4016 6991 Btulbm ηATHE A wATE A wAP E AqAHE A 8185 089415721 0104 WT QH WP in QL 3 2 1 4 T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 9195E The simple steam power plant in Problem 4180E shown in Fig P4121 has a turbine with given inlet and exit states Find the exergy at the turbine exit state 6 Find the second law efficiency for the turbine neglecting kinetic energy at state 5 Use CATT3 to solve Properties from Problem 4180E and s values from F72 hA6E A 1029 hA5E A 14556 ho 4508 all in Btulbm sA5E A 16408 sA6E A 18053 so 008769 all in Btulbm R Kinetic energy at state 6 KEA6E A 05VA2 6E A 600A2E A 2 25 037 719 Btulbm Recall 1 Btulbm 25 037 ftA2E AsA2E ψA6E A hA6E A KEA6E A ho To sA6E A so 6126 Btulbm wrev ψ5 ψA6E A hA5E A hA6E A To sA5E A sA6E A 5152 Btulbm wAC hA5E A hA6E A 4269 Btulbm ηII wac wrev 4269 5152 0829 v P s T 5 5 6 6 Updated June 2013 SOLUTION MANUAL CHAPTER 10 Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CONTENT CHAPTER 10 SUBSECTION PROB NO InText Concept Questions ag Concept problems 118 Brayton cycles gas turbines 1929 Regenerators Intercoolers nonideal cycles 3043 Ericsson cycle 4445 Jet engine cycles 4656 Air standard refrigeration cycles 5762 Otto cycles 6388 Diesel cycles 89101 Stirling and Carnot cycles 102108 Atkinson and Miller cycles 109116 Combined cycles 117121 Exergy Concepts 122130 Review 131140 Problems resolved with the Pr vr functions from A72 37 42 80 see also 87 88 101 108 121 and 140 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful InText Concept Questions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10a The Brayton cycle has the same 4 processes as the Rankine cycle but the Ts and Pv diagrams look very different why is that The Brayton cycle have all processes in the superheated vapor close to ideal gas region The Rankine cycle crosses in over the twophase region 10b Is it always possible to add a regenerator to the Brayton cycle What happens when the pressure ratio is increased No When the pressure ratio is high the temperature after compression is higher than the temperature after expansion The exhaust flow can then not heat the flow into the combustor 10c Why would you use an intercooler between compressor stages The cooler provides two effects It reduces the specific volume and thus reduces the work in the following compressor stage It also reduces the temperature into the combustor and thus lowers the peak temperature This makes the control of the combustion process easier no autoignition or uncontrollable flame spread it reduces the formation of NOx that takes place at high temperatures and lowers the cooling requirements for the chamber walls Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10d The jet engine does not produce shaft work how is power produced The turbine produces just enough shaft work to drive the compressor and it makes a little electric power for the aircraft The power is produced as thrust of the engine In order to exhaust the gases at high speed they must be accelerated so the high pressure in the turbine exit provides that force high P relative to ambient The high P into the turbine is made by the compressor which pushes the flow backwards and thus has a net resulting force forwards on the blades transmitted to the shaft and the aircraft The outer housing also has a higher pressure inside that gives a net component in the forward direction Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10e How is the compression in the Otto cycle different from the Brayton cycle The compression in an Otto cycle is a volume reduction dictated by the piston motion The physical handles are the volumes V1 and V2 The compression in a Brayton cycle is the compressor pushing on the flow so it determines the pressure The physical control is the pressure P2 determined by how much torque you drive the shaft with Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10f How many parameters do you need to know to completely describe the Otto cycle How about the Diesel cycle Otto cycle State 1 2 parameters and the compression ratio CR and the energy release per unit mass in the combustion a total of 4 parameters With that information you can draw the diagrams in Figure 102 Another way of looking at it is four states 8 properties minus the four process equations s2 s1 v3 v2 s4 s3 and v4 v1 gives 4 unknowns Diesel cycle Same as for the Otto cycle namely 4 parameters The only difference is that one constant v process is changed to a constant P process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10g The exhaust and inlet flow processes are not included in the Otto or Diesel cycles How do these necessary processes affect the cycle performance Due to the pressure loss in the intake system and the dynamic flow process we will not have as much mass in the cylinder or as high a P as in a reversible process The exhaust flow requires a slightly higher pressure to push the flow out through the catalytic converter and the muffler higher back pressure and the pressure loss in the valve so again there is a loss relative to a reversible process Both of these processes subtract a pumping work from the net work out of the engine and a lower charge mass gives less power not necessarily lower efficiency than otherwise could be obtained Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful ConceptStudy Guide Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 101 Is a Brayton cycle the same as a Carnot cycle Name the four processes No The Brayton cycle approximates a gas turbine 12 An isentropic compression constant s Compressor 23 An isobaric heating constant P Combustor 34 An isentropic expansion constant s Turbine 41 An isobaric cooling heat rejection constant P Heat exchanger Comment This cycle is the same as the Rankine cycle 4 processes but it takes place in the ideal gas region of states The last process does not exist in the real gas turbine which is an open cycle Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 102 Why is the back work ratio in the Brayton cycle much higher than in the Rankine cycle Recall the expression for shaft work in a steady flow device w v dP The specific volume in the compressor is not so much smaller than the specific volume in the turbine of the Brayton cycle as it is in the pump liquid compared to turbine superheated vapor in the Rankine cycle Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 103 For a given Brayton cycle the cold air approximation gave a formula for the efficiency If we use the specific heats at the average temperature for each change in enthalpy will that give a higher or lower efficiency The specific heats are increasing functions of temperature As the expression for the efficiency is from p464 η 1 h4 h1 h3 h2 1 CP 41 CP 32 T4 T1 T3 T2 The average T from 4 to 1 is lower than the average T from 2 to 3 and therefore the ratio of the two specific heats is lower than one yielding a higher efficiency than the cold air approximation gives for a given set of temperatures However the temperature ratios are different from the cold air calculation for the same compression ratio so the final result is a little more complex Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 104 Does the efficiency of a jet engine change with altitude since the density varies No just like the standard Brayton cycle the simple model performance depends only on the compression ratio 105 Why are the two turbines in Figures 1089 not connected to the same shaft Such a configuration gives a little more flexibility in the control of the cycle under varying loads The two turbines would then not have to run at the same speed for various power output levels Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 106 Why is an air refrigeration cycle not common for a household refrigerator The capacity of the air cycle per mass flowing through the system is very small compared with the vapor compression cycle The cycle also includes the expander which is one more piece of equipment that will add cost and maintenance requirements to the system Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 107 Does the inlet state P1 T1 have any influence on the Otto cycle efficiency How about the power produced by a real car engine Very little The efficiency for the ideal cycle only depends on compression ratio when we assume cold air properties The us are slightly nonlinear in T so there will be a small effect In a real engine there are several effects The inlet state determines the density and thus the total mass in the chamber The more mass the more energy is released when the fuel burns the peak P and T will also change which affects the heat transfer loss to the walls and the formation of NOx sensitive to T The combustion process may become uncontrollable if T is too high knocking Some increase in P1 like that done by a turbocharger or supercharger increases the power output and if high it must be followed by an intercooler to reduce T1 If P1 is too high the losses starts to be more than the gain so there is an optimum level Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 108 For a given compression ratio does an Otto cycle have higher or lower efficiency than a diesel cycle This is actually not clear from the formulas so we need to refer to the Ts diagram where we see that the average T at which the heat is added is higher for the Otto cycle than the diesel cycle for the same compression ratio However since the diesel cycle runs with much higher compressions ratios than the Otto cycle most typical diesel cycles have higher efficiency than typical Otto cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 109 How many parameters do you need to know to completely describe the Atkinson cycle How about the Miller cycle Four parameters for the Atkinson cycle A total of 8 properties minus four known process equations give 4 unknowns The Miller cycle has one additional process so that requires one more parameter for a total of five That is 10 properties with 5 known process equations leaving 5 unknowns Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1010 Why would one consider a combined cycle system for a power plant For a heat pump or refrigerator Dual cycle or combined cycle systems have the advantage of a smaller difference between the high and low ranges for P and T The heat can be added at several different temperatures reducing the difference between the energy source T and the working substance T The working substance vapor pressure at the desired T can be reduced from a high value by adding a topping cycle with a different substance or have a higher low pressure at very low temperatures Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1011 Can the exhaust flow from a gas turbine be useful Usually the temperature in the exhaust flow is fairly high compared to ambient conditions so we could use the hot exhaust flow energy It could be used directly for heating purposes in process applications or heating of buildings As a topping cycle it can be used to heatboil water for use in a Rankine cycle 1012 Where may a heat engine driven refrigerator be useful Any remote location where electricity is not available Since a rotating shaft is available in a car engine the car AC unit is driven by a belt engaged with a magnetic clutch so you do not have to have an electric motor Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1013 Since any heat transfer is driven by a temperature difference how does that affect all the real cycles relative to the ideal cycles Heat transfers are given as Q CA T so to have a reasonable rate the area and the temperature difference must be large The working substance then must have a different temperature than the ambient it exchanges energy with This gives a smaller temperature difference for a heat engine with a lower efficiency as a result The refrigerator or heat pump must have the working substance with a higher temperature difference than the reservoirs and thus a lower coefficient of performance COP The smaller CA is the larger T must be for a certain magnitude of the heat transfer rate This can be a design problem think about the front end air intake grill for a modern car which is very small compared to a car 20 years ago Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1014 In an Otto cycle the cranking mechanism dictates the volume given the crank position Can you say something similar for the Brayton cycle Yes in the Brayton cycle the compressor determines the high pressure the more compressor work input there is the harder it pushes on the gas leading to a higher exit pressure So in the Brayton cycle the pressure ratio is one of the design parameters just like the volume compression ratio is in a piston cylinder type of cycle Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1015 For all the gas cycles it is assumed the ideal compression and expansions are isentropic This is approximated with a polytropic process having n k The expansion after combustion will have some heat loss due to high temperature so what does that imply for the value of n In the expansion of the hot gases we can expect a heat loss to the cooler walls so a q out means a decrease in entropy The polytropic processes as plotted in Fig 613 shows what happens In an expansion volume goes up so P and T drops then if entropy must decrease it means n k Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1016 For all the gas cycles it is assumed the ideal compression and expansions are isentropic This is approximated with a polytropic process having n k The compression in a diesel engine leads to high temperatures and thus will have some heat loss so what does that imply for the value of n In the compression of air P and T go up see Fig 613 The start of the process may have n k as the temperature of the air is not too different from the wall temperature As the air gets hotter it will lose heat to the walls and thus entropy will decrease Moving up and to the left in the Ts diagram corresponds to a process with n k Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1017 If we compute the efficiency of an Otto or Diesel cycle we get something like 60 for a compression ratio of 101 Is a real engine close to this Not at all A gasoline engine has an efficiency of about 35 pushing it to 40 for a hybrid configuration Diesel engines can be about 4045 with bigger two stroke engines like for ships having the higher efficiencies Scale helps in that heat loss and flow losses scale with area whereas power scales with volume This also means that very small engines have the smallest efficiency as the losses become significant Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1018 A hybrid powertrain couples a batterymotor with an internal combustion engine Mention a few factors that make this combination a little more efficient 1 The combustion engine can be operated in a narrow range of RPM and load throttle conditions This means it can be optimized for that narrow range 2 The combustion engine does not have to accelerate and go through a transient period with extra fuel or decelerate and idle with no useful power output 3 The combustion engine can be smaller and operate at full throttle cutting pumping losses and fully use all components that do not need to be designed for a larger capacity In a standard car all systems must be designed for the maximum expected load which is used 2 of the time if ever thus adding mass increasing warm up time etc 4 With electric motors driving the wheels regenerative braking can be used This means the motors act as electrical generators and charge the battery during braking Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Brayton Cycles Gas Turbines Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1019 In a Brayton cycle the inlet is at 300 K 100 kPa and the combustion adds 800 kJkg The maximum temperature is 1400 K due to material considerations Find the maximum permissible compression ratio and for that the cycle efficiency using cold air properties Solution Combustion h3 h2 qH 2w3 0 and Tmax T3 1400 K T2 T3 qHCP 1400 K 800 kJkg 1004 kJkgK 6032 K Reversible adiabatic compression leads to constant s from Eq623 P2 P1 T2T1 k k1 603230035 1153 Reversible adiabatic expansion leads to constant s from Eq623 T4 T3 P4P3 k1 k T3 T1T2 1400 300 6032 6963 K For net work we get wT CP T3 T4 10041400 6963 7065 kJkg wC CP T2 T1 10046032 300 3044 kJkg wnet wT wC 7065 3044 4021 kJkg η wnet qH 4021 800 0503 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1020 A Brayton cycle has compression ratio of 151 with a high temperature of 1600 K and the inlet at 290 K 100 kPa Use cold air properties and find the specific heat addition and specific net work output Brayton cycle so this means Minimum T T1 290 K Maximum T T3 1600 K Pressure ratio P2P1 15 Compression in compressor s2 s 1 Implemented in Eq623 T2 T1P2P1 k1 k 290 K 150286 62865 K Energy input is from the combustor qH CP0T3 T2 1004 1600 62865 9752 kJkg Do the overall cycle efficiency and the net work η W net Q H wnet qH 1 rk1k p 1 150414 05387 wNET η qH 05387 9752 52534 kJkg 1 2 3 4 P P 100 kPa T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1021 A large stationary Brayton cycle gasturbine power plant delivers a power output of 100 MW to an electric generator The minimum temperature in the cycle is 300 K and the exhaust temperature is 750 K The minimum pressure in the cycle is 100 kPa and the compressor pressure ratio is 14 to 1 Calculate the power output of the turbine What fraction of the turbine output is required to drive the compressor What is the thermal efficiency of the cycle Solution Brayton cycle so this means Minimum T T1 300 K Exhaust T T4 750 K Pressure ratio P2P1 14 Solve using constant CP0 1 2 3 4 P P 100 kPa T s Compression in compressor s2 s1 Implemented in Eq623 T2 T1P2P1 k1 k 300 K 140286 6381 K wC h2 h1 CP0T2 T1 1004 6381 300 3395 kJkg Expansion in turbine s4 s3 Implemented in Eq623 T3 T4 P3P4 k1 k 750 K 140286 1594 K wT h3 h4 CP0T3 T4 1004 1594 750 8474 kJkg wNET 8474 3395 5079 kJkg Do the overall net and cycle efficiency m W NETwNET 100 000 kW5079 kJkg 1969 kgs W T m wT 1969 kgs 8474 kJkg 16685 MW wCwT 33958474 040 Energy input is from the combustor qH CP0T3 T2 1004 1594 6381 9597 kJkg ηTH wNETqH 50799597 0529 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1022 Consider an ideal airstandard Brayton cycle in which the air into the compressor is at 100 kPa 20C and the pressure ratio across the compressor is 121 The maximum temperature in the cycle is 1100C and the air flow rate is 10 kgs Assume constant specific heat for the air value from Table A5 Determine the compressor work the turbine work and the thermal efficiency of the cycle Solution 1 2 3 4 P v s s 1 2 3 4 P P 100 kPa T s Compression ratio P2 P1 12 Max temperature T3 1100oC m 10 kgs The compression is reversible and adiabatic so constant s From Eq623 T2 T1 P2 P1 k1 k 2932 K 120286 5968 K Energy equation with compressor work in wC 1w2 CP0T2 T1 10045968 2932 3048 kJkg The expansion is reversible and adiabatic so constant s From Eq623 T4 T3 P4 P3 k1 k 13732 K 1 12 0286 6747 K Energy equation with turbine work out wT CP0T3 T4 100413732 6747 7013 kJkg Scale the work with the mass flow rate W C m wC 3048 kW W T m wT 7013 kW Energy added by the combustion process qH CP0T3 T2 100413732 5968 7795 kJkg ηTH wNETqH 7013 30487795 0509 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1023 A Brayton cycle has air into the compressor at 95 kPa 290 K and has an efficiency of 50 The exhaust temperature is 675 K Find the pressure ratio and the specific heat addition by the combustion for this cycle The efficiency is given by the compression ratio so we get η 1 rk1k p 05 rp P2P1 105 k k1 2 35 11314 The compression is reversible and adiabatic so constant s From Eq623 T2 T1 rk1k p 290 K 11314 0414 580 K The expansion process gives T3 T4 rk1k p 675 K 11314 0414 1350 K Energy added by the combustion process qH CP0T3 T2 1004 kJkgK 1350 580 K 773 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1024 A Brayton cycle has inlet at 290 K 90 kPa and the combustion adds 1000 kJkg How high can the compression ratio be so the highest temperature is below 1700 K Use cold air properties to solve 1 2 3 4 P v s s 1 2 3 4 P P 90 kPa T s Compression ratio rp P2 P1 Max temperature T3 1700 K Combustion adds 2q3 h3 h2 Let us work back from state 3 to 2 as 2q3 h3 h2 CP0T3 T2 T2 T3 2q3 CP0 1700 K 1000 kJkg 1004 kJkgK 704 K The compression is reversible and adiabatic so constant s From Eq623 T2 T1 rk1k p rp T1 T2 k k1 704 290 35 2229 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1025 Assume a state of 1400 kPa 2100 K into the turbine section of a Brayton cycle with an adiabatic expansion to 100 kPa and the compressor inlet temperature is 300 K Find the missing temperatures in the cycle using Table A7 and then give the average value of k ratio of specific heats for the compression and expansion processes Compression process so T2 so T1 R lnP2P1 686926 0287 ln14 762667 kJkgK T2 6294 K and h2 6383 kJkg Expansion process so T4 so T3 R lnP4P3 902721 0287 ln114 826980 kJkgK T4 11245 K and h4 11896 kJkg In the compression process CP h2 h1 T2 T1 6383 30047 6294 300 10256 kJkgK k CP CP R 10256 10256 0287 13886 In the expansion process CP h3 h4 T3 T4 237682 11896 2100 11245 1217 kJkgK k CP CP R 1217 1217 0287 1308 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1026 Repeat Problem 1024 using Table A7 1 2 3 4 P v s s 1 2 3 4 P P 90 kPa T s Compression ratio P2 P1 Max temperature T3 1700 K Combustion adds 2q3 h3 h2 Let us work back from state 3 to 2 as A7 h3 187976 kJkg 2q3 h3 h2 h2 h3 2q3 187976 1000 87976 kJkg From A7 so T2 795477 kJkgK The compression is reversible and adiabatic so constant s From Eq619 so T2 so T1 R lnP2P1 so P2P1 exp so T2 so T1R exp 795477 683521 0287 494 This is too high for a real compressor Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1027 A Brayton cycle produces net 50 MW with an inlet state of 17oC 100 kPa and the pressure ratio is 141 The highest cycle temperature is 1600 K Find the thermal efficiency of the cycle and the mass flow rate of air using cold air properties Inlet state is state 1 P2P1 14 and T3 1600 K Compression Reversible and adiabatic so constant s from Eq623 T2 T1P2P1k1k 290 K 1402857 61638 K ηCYCLE wNETqH 1 T1T2 1 290 61638 05295 Combustion constant pressure qH h3 h2 CP0 T3 T2 1004 kJkgK 1600 61638 K 98756 kJkg wNET ηCYCLE qH 05295 98755 kJkg 5229 kJkg m W wNET 50 000 kW 5229 kJkg 9562 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1028 A Brayton cycle produces 14 MW with an inlet state of 17oC 100 kPa and a compression ratio of 161 The heat added in the combustion is 1160 kJkg What are the highest temperature and the mass flow rate of air assuming cold air properties Solution Temperature after compression is T2 T1 rk1k p 290 K 160414 64035 K The highest temperature is after combustion T3 T2 qHCp 64035 K 1160 kJkg 1004 kJkgK 17957 K Efficiency is from Eq101 η W net Q H wnet qH 1 rk1k p 1 160414 0547 From the required power we can find the needed heat transfer Q H W net η 14 000 0547 kW 25 594 kW m Q H qH 25 594 kW 1160 kJkg 2206 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1029 Do Problem 1023 using Table A7 this becomes trial and error Solution We have states 1 and 4 which mean we can find qL and since we have efficiency η 1 qLqH we can find qH From A71 h1 29043 kJkg so T1 683521 kJkg K h4 686785 kJkg so T4 7701065 kJkgK The heat rejection becomes qL h4 h1 686785 29043 396355 kJkg qH qL1 η 2 qL 79271 kJkg h3 h2 With a known compression ratio we can get states 2 and 4 from states 1 and 3 The compression is reversible and adiabatic so constant s From Eq619 s2 s1 so T2 so T1 RlnP2P1 683521 0287 ln rp We get a similar relation for the expansion process s4 s3 so T3 so T4 RlnP3P4 7701065 0287ln rp From the constant specific heat η 1 rk1k p rp 113 so guess rp 12 so T2 683521 0287 ln 12 754838 h2 5908 kJkg so T3 7701065 0287 ln 12 841423 h3 136244 kJkg h3 h2 7716 kJkg which is too low should be 79271 kJkg Guess rp 14 and repeat the calculations so T2 683521 0287 ln 14 759262 h2 61728 kJkg so T3 7701065 0287 ln 14 8458474 h3 141976 kJkg h3 h2 80248 kJkg which is too high should be 79271 kJkg Linear interpolation rp 12 14 12 79271 7716 80248 7716 1337 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Regenerators Intercoolers and Nonideal Cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1030 Would it be better to add an ideal regenerator to the Brayton cycle in problem 1028 Inlet state is state 1 P2P1 16 and qH 1160 kJkg Compression Reversible and adiabatic so constant s from Eq623 T2 T1P2P1k1k 290 Κ 1602857 64034 K Get the high temperature T3 T2 qH CP0 64034 K 1160 kJkg 1004 kJkgK 17957 K Expansion isentropic T4 T3 P3P4k1k 17957 K 1602857 8132 K T2 Since T4 T2 some regeneration can be done We could find not needed see p471 ηCYCLE 1 T3 T1 P2P1k1k 1 T2 T3 0643 This is better than the regular cycle ηCYCLE 1 rp k1k 0547 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1031 A Brayton cycle with an ideal regenerator has inlet at 290 K 90 kPa with the highest P T as 1170 kPa 1700 K Find the specific heat transfer and the cycle efficiency using cold air properties 1 2 3 4 P v s s 1 2 3 4 P 90 kPa T s x y Compression ratio P2 P1 Max temperature T3 1700 K Combustion adds qH h3 hx Let us work from state 1 to 2 as constant s Eq623 T2 T1 P2 P1 k1 k 290 K 11709002857 6035 K Energy equation with compressor work in wC h2 h1 CP0T2 T1 10046035 290 3135 kJkg qL The expansion is reversible and adiabatic so constant s From Eq623 T4 T3 P4 P3 k1 k 1700 K 90 1170 02857 8169 K T 2 Since the exhaust T4 T2 a regenerator can be used Now the heat transfer added is qH h3 hx CP0T3 Tx 10041700 8169 8866 kJkg wT ηTH wNETqH 8866 3135 8866 0646 We could also have found efficiency see p471 as ηTH 1 T3 T1 P1 P2 k1 k 1 T2 T3 1 6035 1700 0645 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1032 An ideal regenerator is incorporated into the ideal airstandard Brayton cycle of Problem 1022 Find the thermal efficiency of the cycle with this modification Consider an ideal airstandard Brayton cycle in which the air into the compressor is at 100 kPa 20C and the pressure ratio across the compressor is 121 The maximum temperature in the cycle is 1100C and the air flow rate is 10 kgs Assume constant specific heat for the air value from Table A5 Determine the compressor work the turbine work and the thermal efficiency of the cycle Solution 1 2 3 4 P v s s 1 2 3 4 P 100 kPa T s x y Compression ratio P2 P1 12 Max temperature T3 1100oC m 10 kgs The compression is reversible and adiabatic so constant s From Eq623 T2 T1 P2 P1 k1 k 2932 K 120286 5968 K Energy equation with compressor work in wC h2 h1 CP0T2 T1 10045968 2932 3048 kJkg The expansion is reversible and adiabatic so constant s From Eq623 T4 T3 P4 P3 k1 k 13732 K 1 12 0286 6747 K Energy equation with turbine work out wT CP0T3 T4 100413732 6747 7013 kJkg Ideal regenerator TX T4 6747 K qH h3 hX 100413732 6747 7013 kJkg w T ηTH wNETqH 7013 30487013 0565 We could also have found efficiency see p471 as ηTH 1 T3 T1 P1 P2 k1 k 1 T2 T3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1033 Consider an ideal gasturbine cycle with a pressure ratio across the compressor of 12 to 1 The compressor inlet is at 300 K and 100 kPa and the cycle has a maximum temperature of 1600 K An ideal regenerator is also incorporated into the cycle Find the thermal efficiency of the cycle using cold air 298 K properties Solution 1 2 3 4 P v s s 1 2 3 4 P 100 kPa T s x y Compression ratio P2 P1 12 Max temperature T3 1600 K The efficiency with an ideal regenerator T4 Tx is from page 471 ηTH 1 T3 T1 P1 P2 k1 k 1 300 1600 12 02857 0619 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1034 Assume the compressor in Problem 1028 has an intercooler that cools the air to 330 K operating at 500 kPa followed by a second stage of compression to 1600 kPa Find the specific heat transfer in the intercooler and the total combined work required Solution CV Stage 1 1 2 Reversible and adiabatic gives constant s which from Eq623 gives T2 T1 P2P1k1k 290 K 500100 02857 4593 K wc1in CP T2 T1 10044593 290 1870 kJkg CV Intercooler 2 3 Constant pressure cooling qout h2 h3 CP T2 T3 1004 4593 330 1298 kJkg CV Stage 2 3 4 Reversible and adiabatic gives constant s which from Eq623 gives T4 T3 P4P3k1k 330 K 1600500 02857 4601 K wc2in CP T4 T3 10044601 330 1306 kJkg wtot wc1 wc2 187 1306 318 kJkg The intercooler reduces the work for stage 2 as T is lower and so is specific volume The reduction in work due to the intercooler is shaded in the Pv diagram s T P v 1 2 3 4 1 2 3 4 5 100 kPa 500 kPa 1600 kPa 5 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1035 A twostage air compressor has an intercooler between the two stages as shown in Fig P1035 The inlet state is 100 kPa 290 K and the final exit pressure is 16 MPa Assume that the constant pressure intercooler cools the air to the inlet temperature T3 T1 It can be shown that the optimal pressure P2 P1P412 for minimum total compressor work Find the specific compressor works and the intercooler heat transfer for the optimal P2 Solution Optimal intercooler pressure P2 100 1600 400 kPa 1 h1 29043 kJkg so T1 683521 kJkg K CV C1 wC1 h2 h1 s2 s1 leading to Eq619 so T2 so T1 R lnP2P1 683521 0287 ln 4 72331 kJkg K T2 4303 K h2 43205 kJkg wC1 43205 29043 1416 kJkg CV Cooler T3 T1 h3 h 1 qOUT h2 h3 h2 h1 wC1 1416 kJkg CV C2 T3 T1 s4 s3 and since so T3 so T1 P4P3 P2P 1 so T4 so T3 R lnP4P3 so T2 so we have T4 T2 Thus we get wC2 wC1 1416 kJkg s T P v 1 2 3 4 1 2 3 4 100 kPa 400 kPa 1600 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1036 The gasturbine cycle shown in Fig P1036 is used as an automotive engine In the first turbine the gas expands to pressure P5 just low enough for this turbine to drive the compressor The gas is then expanded through the second turbine connected to the drive wheels The data for the engine are shown in the figure and assume that all processes are ideal Determine the intermediate pressure P5 the net specific work output of the engine and the mass flow rate through the engine Find also the air temperature entering the burner T3 and the thermal efficiency of the engine a Consider the compressor s2 s1 T2 T1 P2 P1 k1 k 300 K 60286 5008 K wC w12 CP0T2 T1 10045008 300 2016 kJkg Consider then the first turbine work wT1 wC 2016 kJkg CP0T4 T5 1004 kJkgK 1600 K T5 T5 13992 K s5 s4 P5 P4 T5 T4 k k1 600 kPa 13992 1600 35 375 kPa b s6 s5 T6 T5 P6 P5 k1 k 13992 K 100 375 0286 9588 K The second turbine gives the net work out wT2 CP0T5 T6 100413992 9588 4422 kJkg m W NETwT2 1504422 0339 kgs c Ideal regenerator T3 T6 9588 K qH CP0T4 T3 1004 kJkgK 1600 9588 K 6438 kJkg ηTH wNETqH 44226438 0687 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1037 Repeat Problem 1035 when the intercooler brings the air to T3 320 K The corrected formula for the optimal pressure is P2 P1P4 T3T1nn112 see Problem 7245 where n is the exponent in the assumed polytropic process Solution The polytropic process has n k isentropic so nn 1 1404 35 P2 P1P4 T3T1nn112 100 1600 kPa 32029035 4752 kPa CV C1 s2 s1 T2 T1 P2P1 k1 k 290 K 475210002857 45267 K wC1 h2 h1 CpT2 T1 1004 kJkgK 45267 290 K 1633 kJkg CV Cooler qOUT h2 h3 1004 kJkgK 45267 320 K 1332 kJkg CV C2 s4 s3 T4 T3 P4P3 k1 k 320 K 1600475202857 45267 K wC2 h4 h3 CpT2 T1 1004 kJkgK 45267 320 K 1332 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1038 Repeat Problem 1021 but include a regenerator with 75 efficiency in the cycle A large stationary Brayton cycle gasturbine power plant delivers a power output of 100 MW to an electric generator The minimum temperature in the cycle is 300 K and the exhaust temperature is 750 K The minimum pressure in the cycle is 100 kPa and the compressor pressure ratio is 14 to 1 Calculate the power output of the turbine What fraction of the turbine output is required to drive the compressor What is the thermal efficiency of the cycle Solution Both compressor and turbine are reversible and adiabatic so constant s Eq623 relates then T to P assuming constant heat capacity Compressor T2 T1P2P1 k1 k 300 K 140286 6381 K wC h2 h1 CP0T2 T1 1004 6381 300 3395 kJkg Turbine s4 s3 T3 T4 P3P4 k1 k 750 K 140286 1594 K wT h3 h4 CP0T3 T4 1004 1594 750 8474 kJkg wNET 8474 3395 5079 kJkg m W NETwNET 100 000 kW 5079 kJkg 1969 kgs W T m wT 1969 kgs 8474 kJkg 16685 MW wCwT 33958474 040 1 2 3 4 P 100 kPa T s x x For the regenerator ηREG 075 hX h2 hX h2 T4 T2 TX T2 750 6381 TX 6381 TX 7220 K Turbine and compressor work not affected by regenerator Combustor needs to add less energy with the regenerator as qH CP0T3 TX 10041594 722 8755 kJkg ηTH wNETqH 50798755 058 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1039 An air compressor has inlet of 100 kPa 290 K and brings it to 500 kPa after which the air is cooled in an intercooler to 340 K by heat transfer to the ambient 290 K Assume this first compressor stage has an isentropic efficiency of 85 and it is adiabatic Using constant specific heat and find the compressor exit temperature and the specific entropy generation in the process CV Stage 1 air Steady flow Isentropic compressor is done first Process adiabatic q 0 reversible sgen 0 Energy Eq613 wC1 h2 h1 Entropy Eq98 s2 s1 Assume constant CP0 1004 from A5 and isentropic leads to Eq623 T2s T1P2P1 k1 k 290 K 500100 0286 4593 K wC1s h1 h2s CPoT1 T2 1004290 4593 170 kJkg Now the actual compressor work becomes wC1 ac wC1sηC1s 170085 200 kJkg h1 h2ac CPoT1 T2ac T2ac T1 wC1 acCPo 290 2001004 4892 K sgen s2ac s1 CPo ln T2ac T1 R ln P2 P1 1004 ln4892290 0287500100 00631 kJkgK C1 1 2 3 Q 1 W cooler inter Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1040 A twostage compressor in a gas turbine brings atmospheric air at 100 kPa 17oC to 500 kPa then cools it in an intercooler to 27oC at constant P The second stage brings the air to 2500 kPa Assume both stages are adiabatic and reversible Find the combined specific work to the compressor stages Compare that to the specific work for the case of no intercooler ie one compressor from 100 to 1000 kPa Solution CV Stage 1 1 2 Reversible and adiabatic gives constant s which from Eq623 gives T2 T1 P2P1k1k 290 K 500100 02857 4593 K wc1in CP T2 T1 10044593 290 1700 kJkg CV Stage 2 3 4 Reversible and adiabatic gives constant s which from Eq623 gives T4 T3 P4P3k1k 300 K 2500500 02857 4751 K wc2in CP T4 T3 1004 kJkgK 4751 300 K 1758 kJkg wtot wc1 wc2 170 1758 3458 kJkg The intercooler reduces the work for stage 2 as T is lower and so is specific volume CV One compressor 1 5 Reversible and adiabatic gives constant s which from Eq623 gives T5 T1 P5P1k1k 290 K 2500100 02857 7274 K win CP T5 T1 1004 kJkgK 7274 290 K 439 kJkg The reduction in work due to the intercooler is shaded in the Pv diagram s T P v 1 2 3 4 1 2 3 4 5 100 kPa 500 kPa 2500 kPa 5 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1041 Repeat Problem 1021 but assume that the compressor has an isentropic efficiency of 85 and the turbine an isentropic efficiency of 88 Solution The original cycle was reversible Brayton cycle so this means Minimum T T1 300 K Exhaust T T4s 750 K Pressure ratio P2P1 14 Solve using constant CP0 1 2s 3 4s P P 100 kPa T s 4 2 Ideal compressor s2 s1 Implemented in Eq623 T2s T1P2P1 k1 k 300 K 140286 6381 K wCs h2 h1 CP0T2 T1 1004 6381 300 3395 kJkg Actual compressor wC wSCηSC 3395085 3994 kJkg CP0T2T1 T2 T1 wcCP0 300 39941004 6978 K Ideal turbine s4 s3 Implemented in Eq623 T3 T4s P3P4s k1 k 750 K 1410286 1594 K wTs h3 h4 CP0T3 T4 1004 1594 750 8474 kJkg Actual turbine wT ηST wST 088 8474 7457 kJkg CP0T3T4 T4 T3 wTCP0 1594 74571004 8513 K Do the overall net and cycle efficiency wNET wT wC 7457 3994 3463 kJkg m W NETwNET 100000 kW 3463 kJkg 2888 kgs W T m wT 2888 kgs 7457 kJkg 21536 MW wCwT 39947457 0536 Energy input is from the combustor qH CP0T3 T2 1004 kJkgK 1594 6978 K 8998 kJkg ηTH wNETqH 34638998 0385 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1042 A gas turbine with air as the working fluid has two ideal turbine sections as shown in Fig P1042 the first of which drives the ideal compressor with the second producing the power output The compressor input is at 290 K 100 kPa and the exit is at 450 kPa A fraction of flow x bypasses the burner and the rest 1 x goes through the burner where 1200 kJkg is added by combustion The two flows then mix before entering the first turbine and continue through the second turbine with exhaust at 100 kPa If the mixing should result in a temperature of 1000 K into the first turbine find the fraction x Find the required pressure and temperature into the second turbine and its specific power output CVComp wC h2 h1 s2 s1 Reversible and adiabatic gives constant s which from Eq623 gives T2 T1 P2P1k1k 290 K 450100 02857 4457 K h2 44775 kJkg wC 44775 29043 1573 kJkg CVBurner h3 h2 qH 44775 1200 164775 kJkg T3 1510 K CVMixing chamber 1 xh3 xh2 hMIX 104622 kJkg x h3 hMIX h3 h2 164775 104622 164775 44775 05013 W T1 W Cin wT1 wC 1573 h3 h4 h4 104622 1573 8889 kJkg T4 860 K P4 PMIXT4TMIXkk1 450 kPa 860100035 265 kPa s4 s5 T5 T4 P5P4k1k 860 K 10026502857 651 K h5 6612 kJkg wT2 h4 h5 8889 6612 2277 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1043 A gas turbine cycle has two stages of compression with an intercooler between the stages Air enters the first stage at 100 kPa 300 K The pressure ratio across each compressor stage is 4 to 1 and each stage has an isentropic efficiency of 82 Air exits the intercooler at 330 K Calculate the temperature at the exit of each compressor stage and the total specific work required Solution State 1 P1 100 kPa T1 300 K State 3 T3 330 K P2 4 P1 400 kPa P4 4 P3 1600 kPa Energy Eq wc1 h1 h2 wc1 h2 h1 CPT2 T1 Ideal C1 constant s Eq623 T2s T1 P2P1k1k 4458 K wc1 s h2s h1 CPT2s T1 1464 kJkg Actual Eq728 wc1 wc1 sη 1464082 1785 kJkg T2 T1 wc1CP 4778 K Ideal C2 constant s Eq623 T4s T3 P4P3k1k 4904 K wc2 s h4s h3 CPT4s T3 161 kJkg Actual Eq728 wc2 wc2 sη 1964 kJkg T4 T3 wc2 CP 5256 K Total work in w wc1 wc2 1785 1964 3749 kJkg 1 2s 2ac 3 4s 4ac s T P v 1 2ac 2s 3 4s 4ac Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Ericsson Cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1044 Consider an ideal airstandard Ericsson cycle that has an ideal regenerator as shown in Fig P1044 The high pressure is 15 MPa and the cycle efficiency is 60 Heat is rejected in the cycle at a temperature of 350 K and the cycle pressure at the beginning of the isothermal compression process is 150 kPa Determine the high temperature the compressor work and the turbine work per kilogram of air P2 P3 15 MPa T1 T2 350 K P1 150 kPa 2q3 4q1 ideal reg qH 3q4 wT qH rp P2P1 10 ηTH ηCARNOT TH 1 TLTH 06 T3 T4 TH 875 K qL wC v dP RT1 ln P2 P1 0287 350 ln 1500 150 2313 kJkg wT qH v dP RT3 lnP4P3 5782 kJkg P v 1 2 3 4 T T P P 1 2 3 4 T T P P s T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1045 An airstandard Ericsson cycle has an ideal regenerator Heat is supplied at 1000C and heat is rejected at 80C Pressure at the beginning of the isothermal compression process is 70 kPa The heat added is 700 kJkg Find the compressor work the turbine work and the cycle efficiency Solution Identify the states Heat supplied at high temperature T3 T4 1000C 127315 K Heat rejected at low temperature T1 T2 80C 35315 K Beginning of the compression P1 70 kPa Ideal regenerator 2q3 4q1 qH 3q4 700 kJkg wT qH 700 kJkg ηTH ηCARNOT 1 35315 127315 07226 wNET ηTH qH 07226 700 5058 kJkg wC qL qH wNET 700 5058 1942 kJkg P v 1 2 3 4 T T P P 1 2 3 4 T T P P s T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Jet Engine Cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1046 The Brayton cycle in Problem 1021 is changed to be a jet engine Find the exit velocity using cold air properties A large stationary Brayton cycle gasturbine power plant delivers a power output of 100 MW to an electric generator The minimum temperature in the cycle is 300 K and the exhaust temperature is 750 K The minimum pressure in the cycle is 100 kPa and the compressor pressure ratio is 14 to 1 Calculate the power output of the turbine What fraction of the turbine output is required to drive the compressor What is the thermal efficiency of the cycle Solution Brayton cycle so this means Minimum T T1 300 K Exhaust T T5 750 K Pressure ratio P2P1 14 Solve using constant CP0 1 2 3 4 P P 100 kPa T s 5 Compression in compressor s2 s1 Implemented in Eq623 T2 T1P2P1 k1 k 300 K 140286 6381 K wC h2 h1 CP0T2 T1 1004 kJkgK 6381 300 K 3395 kJkg Turbine wT wC and nozzle s5 s4 s3 Implemented in Eq623 T3 T5 P3P5 k1 k 750 K 1410286 1594 K h3 h5 CP0T3 T5 1004 kJkgK 1594 750 K 8474 kJkg 12V5 2 h3 h5 wC 8474 3395 5079 kJkg V5 2 1000 JkJ 5079 kJkg 1008 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1047 Consider an ideal airstandard cycle for a gasturbine jet propulsion unit such as that shown in Fig 109 The pressure and temperature entering the compressor are 90 kPa 290 K The pressure ratio across the compressor is 14 to 1 and the turbine inlet temperature is 1500 K When the air leaves the turbine it enters the nozzle and expands to 90 kPa Determine the velocity of the air leaving the nozzle Solution COMPR TURBINE BURNER NOZ 1 2 3 4 5 1 2 3 4 P P 90 kPa T s 5 CV Compressor Reversible and adiabatic s2 s1 From Eq620 623 T2 T1 P2 P1 k1 k 290 K 1402857 6164 K wC h2 h1 CP0 T2 T1 1004 kJkgK 6164 290 K 3277 kJkg CV Turbine wT h3 h4 wC and s4 s3 T4 T3 wCCP0 1500 32771004 11736 K CV Nozzle s5 s4 s3 so from Eq623 T5 T3 P5 P3 k1 k 1500 K 90 1260 02857 7057 K Now the energy equation 12V5 2 h4 h5 CP0 T4 T5 1004 11736 7057 46977 kJkg V5 2 1000 JkJ 46977 kJkg 969 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1048 The turbine section in a jet engine receives gas assume air at 1200 K 800 kPa with an ambient atmosphere at 80 kPa The turbine is followed by a nozzle open to the atmosphere and all the turbine work drives a compressor Find the turbine exit pressure so the nozzle has an exit velocity of 800 ms Hint take the CV around both turbine and nozzle Solution CV Reversible and adiabatic turbine and nozzle This gives constant s from Eq623 we can relate the Ts and Ps State 3 1200 K 800 kPa State 5 80 kPa s5 s 3 Eq623 T5 T3 P5P3k1k 1200 K 80800 02857 62156 K Energy h3 0 h5 12V5 2 wT h4 wT w E ATA3E A TA5E A 12VA3 2 AE T h3 h5 12V C 5 2 P 1004 kJkgK 1200 62156 K 12 800A2E AJkg 1000 JkJ 58075 320 26075 kJkg CV Nozzle alone to establish state 4 same s as state 5 and 3 hA4E A hA5E A 12VA5 2 AEE hA3E A wATE TA4E A TA5E A 12VA5 2 AECAPE A 62156 3201004 94029 K PA4E A PA3E A TA4E ATA3E AAkk1E A 800 kPa 940291200A35E A 3407 kPa TURBINE NOZZLE 3 4 w T 5 T s 5 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1049 Given the conditions in the previous problem what pressure could an ideal compressor generate not the 800 kPa but higher CV Reversible and adiabatic turbine and nozzle This gives constant s from Eq623 we can relate the Ts and Ps State 3 1200 K 800 kPa State 5 80 kPa sA5E A sA3E Eq623 TA5E A TA3E A PA5E APA3E AAk1kE A 1200 K 80800 A02857E A 62156 K Energy hA3E A 0 hA5E A 12VA5 2 AE wATE A hA4E A wATE wATE A hA3E A hA5E A 12VA5 2 AEE CAPE ATA3E A TA5E A 12VA3 2 AE 10041200 62156 12 800A2E A1000 58075 320 26075 kJkg CV Compressor wAcE A hA2E A hA1E A wATE A 26075 kJkg TA2E A TA1E A wAcE A CAPE A 270 K 26075 kJkg 1004 kJkgK 52971 K Reversible adiabatic compressor constant s gives relation in Eq623 PA2E A PA1E A TA2E ATA1E AAkk1E A 85 kPa 52971270A35E A 899 kPa TURBINE NOZZLE 3 4 w T 5 T s 5 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1050 Consider a turboprop engine where the turbine powers the compressor and a propeller Assume the same cycle as in Problem 1046 with a turbine exit temperature of 900 K Find the specific work to the propeller and the exit velocity COMPR TURBINE BURNER NOZ 1 2 3 4 5 1 2 3 4 P P 100 kPa T s 5 Compression in compressor sA2E A sA1E A Implemented in Eq623 TA2E A TA1E AAP2P1E AA k1 k E A 300 K 14A0286E A 6381 K wACE A hA2E A hA1E A CAP0E ATA2E A TA1E A 1004 kJkgK 6381 300 K 3395 kJkg CV Turbine wATE A hA3E A hA4E A wACE A wApropE A and sA4E A sA3E A sA5E A TA3E A TA5E A PA3E APA5E AA k1 k E A 750 K 141A0286E A 1594 K wApropE A CAP0E A TA3E A TA4E A wACE A 10041594 900 3395 3573 kJkg CV Nozzle sA5E A sA4E A sA3E A so from Eq623 Now the energy equation 12VA5 2 AE hA4E A hA5E A CAP0E A TA4E A TA5E A 1004 900 750 1506 kJkg V5 A 2 1000 1506EA 549 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1051 Consider an air standard jet engine cycle operating in a 280 K 100 kPa environment The compressor requires a shaft power input of 4000 kW Air enters the turbine state 3 at 1600 K 2 MPa at the rate of 9 kgs and the isentropic efficiency of the turbine is 85 Determine the pressure and temperature entering the nozzle at state 4 If the nozzle efficiency is 95 determine the temperature and velocity exiting the nozzle at state 5 Solution CV Shaft AW E ATE A AmE AhA3E A hA4E A AW E ACE CV Turbine hA3E A hA4E A AW E ACE A AmE A 4000 kW 9 kgs 4444 kJkg TA4E A TA3E A wATaE A CApE A 1600 44441004 11574 K Work back to the ideal turbine conditions Eq727 wATsE A wATaE A η 4444085 52282 hA3E A hA4sE A CApE A TA3E A TA4sE A TA4sE A 1600 522821004 10793 K PA4E A PA3E A TA4sE ATA3E AAkk1E A 2000 kPa 107931600A35E A 5042 kPa CV Nozzle First the ideal reversible and adiabatic ie constant s Eq623 TA5sE A TA4E A PA5E APA4E AAk1kE A 11574 K 1005042Ak1kE A 729 K Energy Eq 12VA2 5sE A hA4E A hA5sE A CApE ATA4E A TA5sE A 1004 kJkgK 11574 729 K 4301 kJkg Now consider the actual nozzle Eq730 05VA2 5aE A η05VA2 5sE A 4086 kJkg VA5aE A A 2 1000 JkJ 4086 kJkgE A 904 ms TA5aE A TA4E A 05VA2 5aE ACApE A 11574 4086 1004 750 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1052 Solve the previous problem using the air tables Solution done with the air tables A7 CV Shaft AW E ATE A AmE AhA3E A hA4E A AW E ACE CV Turbine hA3E A hA4E A AW E ACE A AmE A 40009 4444 kJkg hA4E A 17573 4444 13129 kJkg Work back to the ideal turbine conditions Eq727 wATaE A wACE A 4444 wATsE A wATaE A η 52282 hA3E A hA4sE hA4sE A 12345 TA4sE A 1163 K sAo T4sE A 83091 kJkg K sA4sE A sA3E A 0 sAo T4sE A sAo T3E A R lnPA4E APA3E A 0 83091 86905 0287 lnPA4E A2000 PA4E A 530 kPa State 4 from A71 hA4E A 13129 TA4E A 12298 K sAo T4E A 83746 kJkg K First consider the reversible adiabatic isentropic nozzle so from Eq619 sA5sE A sA4E A 0 sAo T5sE A sAo T4E A R lnPA5E APA4E A sAo T5sE A 83746 0287 ln100530 78960 kJkg K Table A71 TA5sE A 8081 K hA5sE A 8310 kJkg 05VA2 5sE A hA4E A hA5sE A 13129 8310 4819 kJkg Now consider the actual nozzle Eq730 05VA2 5aE A η05VA2 5sE A 45781 kJkg VA5aE A 957 ms hA5aE A hA4E A 05VA2 5aE A 13129 45781 8551 kJkg TA5aE A 830 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1053 A jet aircraft is flying at an altitude of 4900 m where the ambient pressure is approximately 50 kPa and the ambient temperature is 20C The velocity of the aircraft is 280 ms the pressure ratio across the compressor is 141 and the cycle maximum temperature is 1450 K Assume the inlet flow goes through a diffuser to zero relative velocity at state 1 Find the temperature and pressure at state 1 Solution 1 2 3 4 P P 50 kPa T s 5 x Ambient TAXE A 20AoE AC 25315 K PAXE A 50 kPa PA5E also VAXE A 280 ms Assume that the air at this state is reversibly decelerated to zero velocity and then enters the compressor at 1 PA2E APA1E A 14 TA3E A 1450 K CV Diffuser section state 1 is the stagnation state Energy Eq TA1E A TAXE A V 2 X 2 1000 Cp 25315 A 2802 E2 1000 1004E A 2922 K Eq623 PA1E A PAXE A T1 ETX E A k k1 E A 50 kPa A 2922 25315 E A 35E A 826 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1054 The turbine in a jet engine receives air at 1250 K 15 MPa It exhausts to a nozzle at 250 kPa which in turn exhausts to the atmosphere at 100 kPa The isentropic efficiency of the turbine is 85 and the nozzle efficiency is 95 Find the nozzle inlet temperature and the nozzle exit velocity Assume negligible kinetic energy out of the turbine CV Turbine First the ideal reversible and adiabatic ie constant s Eq623 TAesE A TAiE A PAeE APAiE A 1250 K 2501500Ak1kE A 7492 K Energy Eq wATsE A hAiE A hAesE A CApE A TAiE A TAesE A 1004 1250 7492 5028 kJkg Eq727 wATACE A wATsE A ηATE A 4274 kJkg hAiE A hAeACE A CApE A TAiE A TAeACE A TAeACE A 1250 4274 1004 8243 K CV Nozzle First the ideal reversible and adiabatic ie constant s Eq623 TAesE A TAiE A PAeE APAiE AAk1kE A 8243 K 100250Ak1kE A 6344 K Energy Eq 12VAes 2 AE E hAiE A hAesE A 10048243 6344 19066 kJkg Eq730 12VAeAC 2 AE E 12VAes 2 AE E ηANOZE A 18113 kJkg VeAC A 2 1000 JkJ 18113 kJkgEA 602 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1055 Solve the previous problem using the air tables Solution using air tables A7 CV Turbine hAiE A 13367 sAo TiE A 83940 sAesE A sAiE A then from Eq619 sAo TesE A sAo TiE A R lnPAeE APAiE A 83940 0287 ln 2501500 78798 kJkg K Table A71 TAesE A 796 K hAesE A 8179 kJkg Energy Eq wATsE A hAiE A hAesE A 13367 8179 5188 kJkg Eq727 wATACE A wATsE A ηATE A 441 kJkg hAiE A hAeACE A hAeACE A 8957 TAeACE A 866 K sAo TeE A 79730 kJkg K CV Nozzle hAiE A 8957 kJkg sAo TiE A 79730 kJkgK sAesE A sAiE A then from Eq619 sAo TesE A sAo TiE A R lnPAeE APAiE A 79730 0287 ln 100250 77100 kJkgK Table A71 TAesE A 681 K hAesE A 6931 kJkg Energy Eq 12VAes 2 AE E hAiE A hAesE A 8957 6931 2026 kJkg Eq730 12VAeAC 2 AE E 12VAes 2 AE E ηANOZE A 19247 kJkg VeAC A 2 1000 JkJ 19247 kJkgEA 620 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1056 An afterburner in a jet engine adds fuel after the turbine thus raising the pressure and temperature due to the energy of combustion Assume a standard condition of 800 K 250 kPa after the turbine into the nozzle that exhausts at 95 kPa Assume the afterburner adds 450 kJkg to that state with a rise in pressure for same specific volume and neglect any upstream effects on the turbine Find the nozzle exit velocity before and after the afterburner is turned on Solution Before afterburner is on 1 800 K 250 kPa and 2 95 kPa After afterburner is on 3 v vA1E A and 4 95 kPa 1 3 2 4 1 2 3 4 P 95 kPa T s v1 Assume reversible adiabatic nozzle flow then constant s from Eq623 TA2E A TA1E A PA2E APA1E AAk1kE A 800 K 95250 A02857E A 6068 K Energy Eq 12VA2 2 AE CAPE ATA1E A TA2E A V2 EA 2 CAP ATA1 A TA2 AEA A 2 1004 JkgK 800 6068 KEA 6228 ms Add the qAABE A at assumed constant volume then energy equation gives TA3E A TA1E A qAABE ACAvE A 800 K 450 kJkg0717 kJkgK 14276 K vA3E A vA1E A PA3E A PA1E A TA3E ATA1E A 250 kPa 14276800 4461 kPa Reversible adiabatic expansion again from Eq623 TA4E A TA3E A PA4E APA3E AAk1kE A 14276 K 954461 A02857E A 9177 K V2 EA 2 CAP ATA3 A TA4 AEA A 2 1004 JkgK 14276 9177 KEA 1012 ms Comment The real process adds some fuel that burns releasing energy so the temperature goes up and due to the confined space then pressure goes up As the pressure goes up the exit velocity increases altering the mass flow rate through the nozzle As in most problems the real device is more complicated than we can describe with our simple analysis Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Airstandard refrigeration cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1057 An air standard refrigeration cycle has air into the compressor at 100 kPa 270 K with a compression ratio of 31 The temperature after heat rejection is 300 K Find the COP and the highest cycle temperature From the isentropic compressionexpansion processes TA2E A TA1E A PA2E APA1E AA k1 k E A 3A 02857E A 136874 TA3E ATA4E TA2E A TA1E A 136874 270 K 136874 3696 K The COP Eq105 is β A 1 T2 ET1 1E A 136874 1A 1E A 2712 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1058 A standard air refrigeration cycle has 10AoE AC 100 kPa into the compressor and the ambient cools the air down to 35AoE AC at 400 kPa Find the lowest temperature in the cycle the low T specific heat transfer and the specific compressor work Solution State 3 35AoE AC 3082 K 400 kPa The lowest T is at state 4 which we can relate to state 3 constant s Eq623 TA4E A TA3E APA4E APA3E AA k1 k E A 3082 K A 100 400 02857E A 2049 K 683AoE AC Now the heat rejected is A4E AqA1E A hA1E A hA4E A CApE A TA1E A TA4E A 1004 kJkgK 10 683 K 585 kJkg The isentropic compression TA2E A TA1E APA2E APA1E AA k1 k E A 2632 K A 400 100 02857E A 3913 K from which we get the compressor work wAcE A hA1E A hA2E A CApE A TA1E A TA2E A 1004 kJkgK 2632 3913 K 1286 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1059 The formula for the COP assuming cold air properties is given for the standard refrigeration cycle in Eq105 Develop the similar formula for the cycle variation with a heat exchanger as shown in Fig 1012 Definition of COP β A qL Ewnet E A A qL EqH qL E A A 1 qH EqL 1E A Assuming an ideal heat exchanger hA4E A hA6E A and hA1E A hA3E A so qALE A wAEE From the refrigeration cycle we get the ratio of the heat transfers as A qH EqL E A A CpT2 T3 ECpT6 T5E A A T2 T1 ET4 T5 E A A T1 ET5 E A A T2T1 1 ET4T5 1E The pressure ratios are the same and we have isentropic compressionexpansion A P2 EP1 E A A P4 EP5 E A A T2 T1 kk1E A A T4 T5 kk1E so now we get A T2 ET1 E A A T4 ET5 E A so A qH EqL E A A T1 ET5 E A A T1 ET4 E A A T4 ET5 E A A T1 ET4 E A ArP k1kE and the COP reduces to β A 1 T1 ET5 1E A A 1 T1 T4 rP k1k 1E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1060 Assume a refrigeration cycle as shown in Fig 1012 with reversible adiabatic compressor and expander For this cycle the low pressure is 100 kPa and the high pressure is 14 MPa with constant pressure heat exchangers see Fig 1012 Ts diagram The temperatures are T4 T6 50AoE AC and T1 T3 15AoE AC Find the COP for this refrigeration cycle Solution EXP COMP q H q L 5 4 6 3 2 1 2 s T 1 3 4 5 6 Standard air refrigeration cycle with PA1E A 100 kPa PA2E A 14 MPa We will solve the problem with cold air properties Compressor isentropic sA2E A sA1E A so from Eq623 TA2E A TA1E APA2E APA1E AA k1 k E A 2882 K 1400100A0286E A 613 K wACE A wA12E A CAP0E ATA2E A TA1E A 1004 kJkgK 613 2882 K 326 kJkg Expansion in expander turbine sA5E A sA4E A TA5E A TA4E APA5E APA4E AA k1 k E A 2232 K 1001400A0286E A 1049 K wAEE A CAP0E ATA4E A TA5E A 1004 kJkgK 2232 1049 K 1187 kJkg Net cycle work wANETE A wAEE A wACE A 1187 3260 2073 kJkg qALE A CAP0E ATA6E A TA5E A wAEE A 1187 kJkg Overall cycle performance COP β qALE AwANETE A 1187 2073 0573 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1061 Repeat Problems 1060 but assume that helium is the cycle working fluid instead of air Discuss the significance of the results A heat exchanger is incorporated into an ideal airstandard refrigeration cycle as shown in Fig P1060 It may be assumed that both the compression and the expansion are reversible adiabatic processes in this ideal case Determine the coefficient of performance for the cycle Solution EXP COMP q H q L 5 4 6 3 2 1 2 s T 1 3 4 5 6 Standard air refrigeration cycle with helium and states as TA1E A TA3E A 15 AoE AC 2882 K PA1E A 100 kPa PA2E A 14 MPa TA4E A TA6E A 50 AoE AC 2232 K Compressor isentropic sA2E A sA1E A so from Eq623 TA2E A TA1E APA2E APA1E AA k1 k E A 2882A 1400 100 E A 040E A 8282 K wACE A CAP0E ATA2E A TA1E A 5193 kJkgK 8282 2882 K 28041 kJkg Expansion in expander turbine sA5E A sA4E A TA5E A TA4E APA5E APA4E AA k1 k E A 2232 K A 100 1400 E A 040E A 777 K wAEE A CAP0E ATA4E A TA5E A 15193 kJkgK 2232 777 K 7555 kJkg Net cycle work wANETE A 7555 28041 20486 kJkg qALE A CAP0E ATA6E A TA5E A 5193 kJkgK 2232 777 K 7555 kJkg Overall cycle performance COP β qALE AwANETE A 755520486 0369 Notice that the low temperature is lower and work terms higher than with air It is due to the higher heat capacity CAP0E A and ratio of specific heats k 1 23 The expense is a lower COP requiring more work input per kJ cooling Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1062 Repeat Problem 1060 but assume an isentropic efficiency of 75 for both the compressor and the expander Standard air refrigeration cycle with TA1E A TA3E A 15 AoE AC 2882 K PA1E A 100 kPa PA2E A 14 MPa TA4E A TA6E A 50 AoE AC 2232 K We will solve the problem with cold air properties Ideal compressor isentropic sA2SE A sA1E A so from Eq623 TA2SE A TA1E APA2E APA1E AA k1 k E A 28821400100A0286E A 613 K wASCE A wA12E A CAP0E ATA2SE A TA1E A 1004613 2882 326 kJkg The actual compressor wACE A wASCE A ηASCE A 326075 4346 kJkg Expansion in ideal expander turbine sA5E A sA4E A TA5SE A TA4E APA5E APA4E AA k1 k E A 22321001400A0286E A 1049 K wAEE A CAP0E ATA4E A TA5E A 10042232 1049 1187 kJkg The actual expander turbine wAEE A ηASEE A wASEE A 075 1187 890 kJkg CAP0E ATA4E ATA5E A 10042232 TA5E A TA5E A 1345 K wANETE A 890 4346 3456 kJkg qALE A CAP0E ATA6E A TA5E A 10042232 1345 890 kJkg β qALE AwANETE A 8903456 0258 1 2 3 s T 4 5 6 2S 5S Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Otto Cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1063 The mean effective pressure scales with the net work and thus efficiency Assume the heat transfer per unit mass is a given it depends on the fuelair mixture how does the total power output then vary with the inlet conditions PA1E A TA1E A The power output comes from speed and displacement in Eq 1011 AW E A PAmeffE AVAdisplE A ARPM 60E A A1 2E A m wAnetE A ARPM 60E A A1 2E A m η qAHE A ARPM 60E A A1 2E From the last expression we notice that the power scales with mass and efficiency From our simple analysis we know that the efficiency only depends on compression ratio CR and not on state 1 which then leaves the mass as affected by state 1 The mass of fresh charge is assumed proportional to the mass of air at ambient conditions close to state 1 in the volume VAdisplE A The constant of proportionality is called the volumetric efficiency and depends on the throttling and other restrictions in the intake system which then leads to state 1 m VAdisplE A vA1E A VAdisplE A PA1E ARTA1E A cold charge TA1E A low with high pressure PA1E A a super or turbocharger will give the highest mass and thus the most power Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1064 A 4 stroke gasoline engine runs at 1800 RPM with a total displacement of 3L and a compression ratio of 101 The intake is at 290 K 75 kPa with a mean effective pressure of 600 kPa Find the cycle efficiency and power output Efficiency from the compression ratio η 1 CRA1kE A 1 10A04E A 060 The power output comes from speed and displacement in Eq 1011 AW E A PAmeffE AVAdisplE A ARPM 60E A A1 2E A 600 kPa 0003 mA3E A A1800 60E A 1s A1 2E A 27 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1065 Find the missing pressures and temperatures in the previous cycle Efficiency from the compression ratio η 1 CRA1kE A 1 10A04E A 060 vA1E A RTA1E APA1E A 0287 kJkgK 290 K75 kPa 110973 mA3E Akg Compression TA2E A TA1E A CRAk1E A 290 K 10 A04E A 72845 K PA2E A PA1E A CRAkE A 75 kPa 10 A14E A 18839 kPa qAHE A uA3E A uA2E A CAvE ATA3E A TA2E A wAnetE Aη PAmeffE A vA1E A vA2E Aη 600 1109731 1CR060 99876 kJkg TA3E A TA2E A qAHE A CAvE A 72845 998760717 21214 K PA3E A PA2E A TA3E ATA2E A 18839 21214 72845 5486 kPa TA4E A TA3E ACRA04E A 21214 K 10A04E A 8445 K PA4E A PA3E ACRA14E A 5486 kPa10A14E A 2184 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1066 Air flows into a gasoline engine at 95 kPa 300 K The air is then compressed with a volumetric compression ratio of 101 In the combustion process 1300 kJkg of energy is released as the fuel burns Find the temperature and pressure after combustion using cold air properties Solution Solve the problem with constant specific heat Compression 1 to 2 sA2E A sA1E A From Eq624 and Eq625 TA2E A TA1E A vA1E AvA2E AAk1E A 300 10A04E A 7536 K PA2E A PA1E AvA1E AvA2E AAkE A 95 10A14E A 2386 kPa Combustion 2 to 3 at constant volume uA3E A uA2E A qAHE TA3E A TA2E A qAHE ACAvE A 7536 13000717 2567 K PA3E A PA2E A TA3E ATA2E A 2386 2567 7536 8127 kPa P v 1 2 3 4 s 1 2 3 4 v T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1067 A 4 stroke gasoline 42 L engine running at 2000 RPM has inlet state of 85 kPa 280 K and after combustion it is 2000 K and the highest pressure is 5 MPa Find the compression ratio the cycle efficiency and the exhaust temperature Solution Combustion vA3E A vA2E A Highest T and P are after combustion CR vA1E AvA2E A vA1E AvA3E A TA1E APA3E ATA3E APA1E A A280 5000 2000 85E A 8235 Efficiency from the compression ratio η 1 CRA1kE A 1 8235A04E A 057 Expansion TA4E A TA3E A vA3E AvA4E AAk1E A TA3E A 1 CR Ak1E A 2000 1 8235 A04E A 8605 K P v 1 2 3 4 s 1 2 3 4 v T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1068 Find the power from the engine in Problem 1067 Solution Combustion vA3E A vA2E A Highest T and P are after combustion CR vA1E AvA2E A vA1E AvA3E A TA1E APA3E ATA3E APA1E A A280 5000 2000 85E A 8235 Efficiency from the compression ratio η 1 CRA1kE A 1 8235A04E A 057 Compression TA2E A TA1E ACRAk1E A 280 K 8235 A04E A 6508 K qAHE A uA3E A uA2E A CAvE ATA3E A TA2E A 0717 kJkgK 2000 6508 K 9674 kJkg vA1E A RTA1E APA1E A 0287 kJkgK 280 K85 kPa 09454 mA3E Akg Displacement and then mep from net work vA1E A vA2E A vA1E A vA1E ACR vA1E A1 1CR 08306 mA3E Akg PAmeffE A wAnetE AvA1E A vA2E A η qAHE A vA1E A vA2E A 057 967408306 6639 kPa Total power from Eq 1011 and 4 stroke cycle AW E A PAmeffE AVAdisplE A ARPM 60E A A1 2E A 6639 00042 A2000 60E A A1 2E A 4645 kW P v 1 2 3 4 s 1 2 3 4 v T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1069 A 4 stroke 24 L gasoline engine runs at 2500 RPM has an efficiency of 60 The state before compression is 40 kPa 280 K and after combustion it is at 2400 K Find the highest T and P in the cycle the specific heat transfer added the cycle mep and the total power produced This is a basic 4stroke Otto cycle Compression ratio from the efficiency η 1 CRA1kE A CR 1 ηA104E A 9882 Compression TA2E A TA1E ACRAk1E A 280 9882 A04E A 700 K PA2E A PA1E A CRAkE A 40 9882A14E A 9882 kPa Combustion vA3E A vA2E A Highest T and P are after combustion TA3E A 2400 K PA3E A PA2E A TA3E A TA2E A A9882 2400 700E A 3388 kPa qAHE A uA3E A uA2E A CAvE ATA3E A T A2E A 0717 2400 700 12189 kJkg wAnetE A ηATHE A qAHE A 06 12189 73134 kJkg Displacement and PAmeffE vA1E A RTA1E APA1E A 0287 kJkgK 280 K40 kPa 2009 mA3E Akg vA2E A 19882 vA1E A 02033 mA3E Akg PAmeffE A A wNET Ev1v2 E A A 73134 2009 02033E A 405 kPa Now we can find the power from Eq1011 AW E A PAmeffE A VAdisplE A ARPM 60E A A1 2E A 405 00024 A2500 60E A A1 2E A 203 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1070 Suppose we reconsider the previous problem and instead of the standard ideal cycle we assume the expansion is a polytropic process with n 15 What are the exhaust temperature and the expansion specific work This is a modified 4stroke Otto cycle Assume the same state 123 so only state 4 is changed and thus also the efficiency is less than the original 60 Compression ratio from the original efficiency assuming the standard cycle η 1 CRA1kE A CR 1 ηA104E A 9882 If a polytropic expansion with n 15 instead then TA4modE A TA3E A CRA1nE A 2400 9882A05E A 7635 K This means a heat loss out and thus less work see diagrams w A R 1 nE A TA4modE A TA3E A A 0287 1 15E A 7635 2400 93935 kJkg Comments The heat loss is A3E AqA4E A uA4E A uA3E A A3E AwA4E A CAvE ATA4E A TA3E A A3E AwA4E A 07177635 2400 93935 234 kJkg The original cycle has expansion work as TA4E A TA3E A CRA1kE A 2400 9882A04E A 960 K A3E AwA4E A CAvE ATA3E A TA4E A 0717 2400 960 10325 kJkg P v 1 2 3 4 s 4mod 1 2 3 4 v T s 4mod Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1071 Air flows into a gasoline engine at 95 kPa 300 K The air is then compressed with a volumetric compression ratio of 81 In the combustion process 1300 kJkg of energy is released as the fuel burns Find the temperature and pressure after combustion using cold air properties Solution P v 1 2 3 4 s 1 2 3 4 v T s Compression Isentropic so we use Eqs624625 P2 P1v1v2k 95 kPa 814 1746 kPa T2 T1v1v2k1 300 K 804 6892 K Combustion 2 to 3 T3 T2 qHCv 6892 13000717 25023 kJkg T3 2502 K P3 P2 T3T2 1746 2502 6892 6338 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1072 A gasoline engine has a volumetric compression ratio of 8 and before compression has air at 280 K 85 kPa The combustion generates a peak pressure of 5500 kPa Find the peak temperature the energy added by the combustion process and the exhaust temperature Solution Solve the problem with cold air properties Compression Isentropic so we use Eqs624625 P2 P1v1v2k 85 kPa 814 1562 kPa T2 T1v1v2k1 280 K 804 6433 K Combustion Constant volume T3 T2 P3P2 6433 K 55001562 2265 K qH u3 u2 CvT3 T2 0717 kJkgK 2265 6433 K 1163 kJkg Exhaust Isentropic expansion so from Eq624 T4 T3804 2265 K 22974 986 K P v 1 2 3 4 1 2 3 4 v T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1073 To approximate an actual sparkignition engine consider an airstandard Otto cycle that has a heat addition of 1800 kJkg of air a compression ratio of 7 and a pressure and temperature at the beginning of the compression process of 90 kPa 10C Assuming constant specific heat with the value from Table A5 determine the maximum pressure and temperature of the cycle the thermal efficiency of the cycle and the mean effective pressure Solution P v 1 2 3 4 1 2 3 4 v T s Compression Reversible and adiabatic so constant s from Eq62425 P2 P1v1v2k 90 kPa 714 1372 kPa T2 T1v1v2k1 2832 K 704 6166 K Combustion constant volume T3 T2 qHCV0 6166 18000717 3127 K P3 P2T3T2 1372 kPa 3127 6166 6958 kPa Efficiency and net work ηTH 1 T1T2 1 28326165 0541 wnet ηTH qH 0541 1800 9738 kJkg Displacement and Pmeff v1 RT1P1 0287 kJkgK 2832 K90 kPa 09029 m3kg v2 17 v1 01290 m3kg Pmeff v1v2 wNET 9738 09029 0129 1258 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1074 A 33 L minivan engine runs at 2000 RPM with a compression ratio of 101 The intake is at 50 kPa 280 K and after expansion it is at 750 K Find the highest T in the cycle the specific heat transfer added by combustion and the mean effective pressure P v 1 2 3 4 1 2 3 4 v T s Compression Reversible and adiabatic so constant s from Eq62425 T2 T1v1v2k1 280 1004 7033 K Expansion isentropic T3 is the highest T and T4 was given T3 T4 CRk1 750 K 1004 18839 K qH u3 u2 CV0 T3 T2 0717 18839 7033 8465 kJkg Efficiency and net work ηTH 1 T1T2 1 2807033 0602 wnet ηTH qH 0602 8465 5096 kJkg Displacement and Pmeff v1 RT1P1 0287 kJkgK 280 K50 kPa 16072 m3kg v2 110 v1 016072 m3kg Pmeff v1v2 wNET 5096 16072 016072 3523 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1075 A gasoline engine takes air in at 290 K 90 kPa and then compresses it The combustion adds 1000 kJkg to the air after which the temperature is 2050 K Use cold air properties ie constant heat capacities at 300 K and find the compression ratio the compression specific work and the highest pressure in the cycle Solution Standard Otto Cycle Combustion process T3 2050 K u2 u3 q H T2 T3 qH Cvo 2050 1000 0717 6553 K Compression process P2 P1T2 T1kk1 906553290 35 1561 kPa CR v1 v2 T2 T11k1 6553 290 25 767 1w2 u2 u1 Cvo T2 T1 0717 kJkgK 6553 290 K 262 kJkg Highest pressure is after the combustion where v3 v2 so we get P3 P2T3 T2 1561 kPa 2050 6553 4883 kPa P v 1 2 3 4 1 2 3 4 v T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1076 Answer the same three questions for the previous problem but use variable heat capacities use table A7 A gasoline engine takes air in at 290 K 90 kPa and then compresses it The combustion adds 1000 kJkg to the air after which the temperature is 2050 K Use the cold air properties ie constant heat capacities at 300 K and find the compression ratio the compression specific work and the highest pressure in the cycle Solution Standard Otto cycle solve using Table A71 Combustion process T3 2050 K u3 17257 kJkg u2 u3 qH 17257 1000 7257 kJkg T2 9605 K so T2 80889 kJkg K Compression 1 to 2 s2 s1 From Eq619 0 so T2 so T1 R lnP2P1 so T2 so T1 R lnΤ2v1T1v2 80889 68352 0287 ln9605290 0287 lnv1v2 Solving v1 v2 2378 Comment This is much too high for an actual Otto cycle 1w2 u2 u1 7257 2072 5185 kJkg Highest pressure is after combustion where v3 v2 so we get P3 P2T3 T2 P1T3 T1v1 v3 90 2050 290 2378 15 129 kPa P v 1 2 3 4 1 2 3 4 v T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1077 Redo the previous problem but assume the combustion adds 1225 kJkg Standard Otto cycle solve using Table A71 Combustion process T3 2050 K u3 17257 kJkg u2 u3 qH 17257 1225 5007 kJkg T2 6848 K so T2 771647 kJkg K Compression 1 to 2 s2 s1 From Eq619 0 so T2 so T1 R lnP2P1 so T2 so T1 R lnΤ2v1T1v2 771647 683521 0287 ln6848290 0287 lnv1v2 Solving v1 v2 913 1w2 u2 u1 5007 2072 2935 kJkg Highest pressure is after combustion where v3 v2 so we get P3 P2T3 T2 P1T3 T1v1 v3 90 2050 290 913 5 809 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1078 A four stroke gasoline engine has a compression ratio of 101 with 4 cylinders of total displacement 23 L the inlet state is 280 K 70 kPa and the engine is running at 2100 RPM with the fuel adding 1400 kJkg in the combustion process What is the net work in the cycle and how much power is produced Solution Overall cycle efficiency is from Eq1012 rv v1v2 10 ηTH 1 r1k v 1 1004 0602 wnet ηTH qH 0602 1400 8428 kJkg We also need specific volume to evaluate Eqs109 to 1011 v1 RT1 P1 0287 280 70 1148 m3kg Pmeff v1 v2 wnet v1 1 1 rv wnet 8428 1148 09 8157 kPa Now we can find the power from Eq1011 W Pmeff Vdispl RPM 60 1 2 8157 00023 2100 60 1 2 328 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1079 A gasoline engine has a volumetric compression ratio of 10 and before compression has air at 290 K 85 kPa in the cylinder The combustion peak pressure is 6000 kPa Assume cold air properties What is the highest temperature in the cycle Find the temperature at the beginning of the exhaust heat rejection and the overall cycle efficiency Solution Compression Isentropic so we use Eqs62425 P2 P1v1v2k 85 kPa 1014 21351 kPa T2 T1v1v2k1 290 K 1004 72845 K Combustion Constant volume T3 T2 P3P2 72845 K 600021351 2047 K Exhaust is after the isentropic expansion so from Eq624 T4 T3 v1v2k1 T3 1004 2047 25119 8149 K Overall cycle efficiency is from Eq1012 rv v1v2 η 1 r 1k v 1 1004 0602 Comment No actual gasoline engine has an efficiency that high maybe 35 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1080 Repeat Problem 1073 but assume variable specific heat The ideal gas air tables Table A7 are recommended for this calculation or the specific heat from Fig 326 at high temperature Solution Table A7 T1 2832 K u1 2023 kJkg so T1 68113 kJkg K Compression 1 to 2 s2 s1 From Eq619 0 so T2 so T1 R lnP2P1 so T2 so T1 R lnΤ2v1T1v2 so T2 R lnΤ2T1 so T1 R lnv1v2 68113 0287 ln 7 73698 This becomes trial and error so estimate first at 600 K and use A71 LHS600 75764 0287 ln6002832 73609 too low LHS620 76109 0287 ln6202832 73860 too high Interpolate to get T2 6071 K u2 4405 kJkg 1w2 u2 u1 2382 kJkg u3 4405 1800 22405 T3 25758 K so T3 92859 kJkgK P3 P1 v1v3 T3T1 90 kPa 7 25758 2832 5730 kPa Expansion 3 to 4 s4 s3 From Eq619 as before so T4 R lnΤ4T3 so T3 R lnv3v4 92859 0287 ln17 87274 This becomes trial and error so estimate first at 1400 K and use A71 LHS1400 85289 0287 ln140025758 87039 too low LHS1450 85711 0287 ln145025758 87360 too high Interpolation T4 14366 K u4 11469 kJkg 3w4 u3 u4 22405 11469 10936 kJkg Net work efficiency and mep wnet 3w4 1w2 10936 2382 8554 kJkg ηTH wnet qH 8554 1800 0475 v1 P1 RT1 0287 2832 90 09029 m3kg v2 v1 CR 01290 m3kg Pmeff v1 v2 wnet 8554 09029 0129 1105 kPa Comment Iterations are avoided by the use of the vr function in A72 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1081 Assume a state of 5000 kPa 2100 K after combustion in an Otto cycle with a compression ratio of 101 the intake temperature is 300 K Find the missing temperatures in the cycle using Table A7 and then give the average value of k ratio of specific heats for the compression and expansion processes Compression 1 to 2 s2 s1 From Eq619 0 so T2 so T1 R lnP2P1 so T2 so T1 R lnΤ2v1T1v2 so T2 R lnΤ2T1 so T1 R lnv1v2 686926 0287 ln 10 75301 This becomes trial and error so estimate first at 600 K and use A71 LHS700 77401 0287 ln700300 749692 too low LHS740 780008 0287 ln740300 754096 too high Interpolate to get T2 7301 K u2 5365 kJkg Expansion 3 to 4 s4 s3 From Eq619 as before so T4 R lnΤ4T3 so T3 R lnv3v4 902721 0287 ln 10 836637 This becomes trial and error so estimate first at 1400 K and use A71 LHS1100 824449 0287 ln11002100 843007 too high LHS1050 819081 0287 ln10502100 8389743 still too high LHS1000 813493 0287 ln10002100 8347866 too low Interpolation T4 10221 K u4 77815 kJkg Compression process Cv T2 T2 u2 u1 5365 21436 7301 300 0749 k Cv RCv 138 Expansion process Cv T3 T4 u3 u4 177406 77815 2100 10221 09239 k Cv RCv 131 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1082 An Otto cycle has the lowest T as 290 K and the lowest P is 150 kPa the highest T is 2400 K and combustion adds 1200 kJkg as heat transfer Find the compression ratio and the mean effective pressure Solution Identify states T1 290 K P1 150 kPa T3 2400 K qH 1200 kJkg Combustion qH u3 u2 CvoT3 T2 1200 kJkg T2 T3 qH Cvo 2400 12000717 72636 K Compression CR rv v1v2 T2T11k1 72636 290 25 993 Overall cycle efficiency is from Eq1012 rv v1v2 993 ηTH 1 r1k v 1 T1T2 1 29072636 0601 wnet ηTH qH 0601 1200 7212 kJkg We also need specific volume to evaluate Eqs109 to 1011 v1 RT1 P1 0287 290 150 055487 m3kg v2 v1CR 005588 m3kg Pmeff v1 v2 wnet 7212 055487 005588 kJkg m3kg 1445 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1083 The cycle in the previous problem is used in a 24 L engine running 1800 RPM How much power does it produce Identify states T1 290 K P1 85 kPa T3 2400 K qH 1200 kJkg Combustion qH u3 u2 CvoT3 T2 1200 kJkg T2 T3 qH Cvo 2400 12000717 72636 K Compression CR rv v1v2 T2T11k1 72636 290 25 993 Overall cycle efficiency is from Eq1012 rv v1v2 993 ηTH 1 r1k v 1 T1T2 1 29072636 0601 wnet ηTH qH 0601 1200 7212 kJkg We also need specific volume to evaluate Eqs109 to 1011 v1 RT1 P1 0287 290 150 055487 m3kg v2 v1CR 005588 m3kg Pmeff v1 v2 wnet 7212 055487 005588 kJkg m3kg 1445 kPa Now we can find the power from Eq1011 W Pmeff Vdispl RPM 60 1 2 1445 00024 1800 60 1 2 52 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1084 When methanol produced from coal is considered as an alternative fuel to gasoline for automotive engines it is recognized that the engine can be designed with a higher compression ratio say 10 instead of 7 but that the energy release with combustion for a stoichiometric mixture with air is slightly smaller about 1700 kJkg Repeat Problem 1073 using these values Solution P v 1 2 3 4 1 2 3 4 v T s Compression Reversible and adiabatic so constant s from Eq62425 P2 P1v1v2k 90 kPa 1014 22607 kPa T2 T1v1v2k1 28315 K 1004 7112 K Combustion constant volume T3 T2 qH Cvo 7112 1700 0717 3082 K P3 P2T3 T2 22607 kPa 3082 7112 9797 kPa Efficiency net work displacement and Pmeff ηTH 1 T1T2 1 283157112 0602 wnet ηTH qH 06 1700 10234 kJkg v1 RT1P1 0287 kJkgK 28315 K90 kPa 09029 m3kg v2 v110 00903 m3kg Pmeff v1 v2 wnet 10234 09029 00903 1255 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1085 It is found experimentally that the power stroke expansion in an internal combustion engine can be approximated with a polytropic process with a value of the polytropic exponent n somewhat larger than the specific heat ratio k Repeat Problem 1073 but assume that the expansion process is reversible and polytropic instead of the isentropic expansion in the Otto cycle with n equal to 150 From the average temperature during expansion estimate the actual average k See solution to 1073 except for process 3 to 4 T3 3127 K P3 6958 MPa v3 RT3P3 v2 0129 m3kg v4 v1 09029 m3kg Process Pv15 constant P4 P3v3v415 6958 1715 3757 kPa T4 T3v3v405 31271705 11819 K 1w2 Pdv R 114T2 T1 0287 04 6066 28315 2393 kJkg 3w4 Pdv RT4 T31 15 028711819 312705 11165 kJkg wNET 11165 2393 8772 kJkg ηCYCLE wNETqH 87721800 0487 Pmeff v1 v2 wnet 877209029 0129 1133 kPa Note a smaller wNET ηCYCLE Pmeff compared to an ideal cycle The T changes from 3 to 4 3127 K to 11819 K average is about 2150 K CP0 2200 2100 h2200 h2100 250263 237682 2200 2100 1258 kJkgK k CP0 CP0 R 12581258 0287 1296 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1086 In the Otto cycle all the heat transfer qH occurs at constant volume It is more realistic to assume that part of qH occurs after the piston has started its downward motion in the expansion stroke Therefore consider a cycle identical to the Otto cycle except that the first twothirds of the total qH occurs at constant volume and the last onethird occurs at constant pressure Assume that the total qH is 2100 kJkg that the state at the beginning of the compression process is 90 kPa 20C and that the compression ratio is 9 Calculate the maximum pressure and temperature and the thermal efficiency of this cycle Compare the results with those of a conventional Otto cycle having the same given variables P1 90 kPa T1 20oC rV v1v2 9 q23 23 2100 1400 kJkg q34 21003 700 kJkg Constant s compression Eqs62425 P2 P1v1v2k 90 kPa 914 1951 kPa T2 T1v1v2k1 29315 K 904 706 K Constant v combustion T3 T2 q23CV0 706 14000717 2660 K P3 P2T3T2 1951 kPa 2660706 73508 kPa P4 Constant P combustion T4 T3 q34CP0 2660 7001004 3357 K Remaining expansion v5 v4 v1 v4 P4 P1 T1 T4 73508 90 29315 3357 7131 T5 T4v4v5k1 3357 K 1713104 1530 K qL CV0T5T1 0717 kJkgK 1530 29315 K 8862 kJkg ηTH 1 qLqH 1 88622100 0578 Std Otto Cycle ηTH 1 904 0585 small difference 1 2 3 4 s s P v 5 1 2 3 4 T s s s v v 5 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1087 A gasoline engine has a volumetric compression ratio of 9 The state before compression is 290 K 90 kPa and the peak cycle temperature is 1800 K Find the pressure after expansion the cycle net work and the cycle efficiency using properties from Table A72 Compression 1 to 2 s2 s1 vr2 vr1v2v1 vr2 196379 21819 T2 680 K Pr2 20784 u2 49694 P2 P1Pr2Pr1 90 kPa 20784 0995 1880 kPa 1w2 u1 u2 20719 49694 28975 kJkg Combustion 2 to 3 qH u3 u2 148633 49694 98939 kJkg P3 P2T3T2 1880 kPa 1800 680 4976 kPa Expansion 3 to 4 s4 s3 vr4 vr3 9 1143 9 10278 T4 889 K Pr4 57773 u4 6658 kJkg P4 P3Pr4Pr3 4976 kPa 57773 1051 2735 kPa 3w4 u3 u4 148633 6658 8205 kJkg wNET 3w4 1w2 8205 28975 5308 kJkg η wNETqH 530898939 0536 P v 1 2 3 4 s 1 2 3 4 v T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1088 Solve Problem 1076 using the Pr and vr functions from Table A72 A gasoline engine takes air in at 290 K 90 kPa and then compresses it The combustion adds 1000 kJkg to the air after which the temperature is 2050 K Use the cold air properties ie constant heat capacities at 300 K and find the compression ratio the compression specific work and the highest pressure in the cycle Solution Standard Otto cycle solve using Table A71 and Table A72 Combustion process T3 2050 K u3 17257 kJkg u2 u3 qH 17257 1000 7257 kJkg T2 9605 K vr2 82166 Compression 1 to 2 s2 s1 From the vr function v1v2 vr1vr2 19536 82166 2378 Comment This is much too high for an actual Otto cycle 1w2 u2 u1 7257 2072 5185 kJkg Highest pressure is after combustion P3 P2T3 T2 P1T3 T1v1 v3 90 kPa 2050 290 2378 15 129 kPa P v 1 2 3 4 1 2 3 4 v T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Diesel Cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1089 A diesel engine has an inlet at 95 kPa 300 K and a compression ratio of 201 The combustion releases 1300 kJkg Find the temperature after combustion using cold air properties Solution Compression process isentropic from Eqs62425 T2 T1v1 v2k1 300 2004 9943 K Combustion at constant P which is the maximum pressure 2w3 P v3 v2 2q3 u3 u2 2w3 h3 h2 CpoT3 T2 T3 T2 2q3 Cpo 9943 1300 1004 2289 K P v 1 2 3 4 s s 1 2 3 4 v T s P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1090 A diesel engine has a state before compression of 95 kPa 290 K and a peak pressure of 6000 kPa a maximum temperature of 2400 K Find the volumetric compression ratio and the thermal efficiency Solution Standard Diesel cycle and we will use cold air properties Compression process isentropic from Eqs62425 P2P1 v1v2k CR 14 CR v1v2 P2P11k 600095114 1932 T2 T1P2P1k1k 290 600095 02857 9479 K Combustion and expansion volumes v3 v2 T3T2 v1 T3T2 CR v4 v1 Expansion process isentropic from Eq623 T4 T3 v3v4k1 T3 T3 CR T2k1 2400 K 24001932 9479 04 10646 K Efficiency from Eq1013 η 1 1 k T4 T1 T3 T2 1 1 14 10646 290 2400 9479 0619 P v 1 2 3 4 s s 1 4 v T s 2 3 P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1091 Find the cycle efficiency and mean effective pressure for the cycle in Problem 1089 Solution Compression process isentropic from Eqs62425 T2 T1v1 v2k1 300 2004 9943 K Combustion at constant P which is the maximum pressure 2w3 P v3 v2 2q3 u3 u2 2w3 h3 h2 CpoT3 T2 T3 T2 2q3 Cpo 9943 1300 1004 2289 K Combustion and expansion volumes v3 v2 T3T2 v1 T3T2 CR v4 v1 Expansion process isentropic from Eq623 T4 T3 v3v4k1 T3 T3 CR T2k1 2289 228920 9943 04 964 K Efficiency from Eq1013 η 1 1 k T4 T1 T3 T2 1 1 14 964 300 2289 9943 0634 wnet η qH 0634 1300 8242 kJkg v1 RT1P1 09063 m3kg Pmeff vmax vmin wnet v1 v1CR wnet 8242 09063 1 120 kJkg m3kg 957 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1092 The cut off ratio is the ratio of v3v2 see Fig 1018 which is the expansion while combustion occurs at constant pressure Determine this ratio for the cycle in Problem 1089 Solution Compression process isentropic from Eqs62425 T2 T1v1 v2k1 300 2004 9943 K Combustion at constant P2 P3 which is the maximum pressure 2w3 P v3 v2 2q3 u3 u2 2w3 h3 h2 CpoT3 T2 T3 T2 2q3 Cpo 9943 1300 1004 2289 K Cutoff ratio v3 v2 T3 T2 2289 9943 2302 P v 1 2 3 4 s s 1 2 3 4 v T s P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1093 A diesel engine has a compression ratio of 201 with an inlet of 95 kPa 290 K state 1 with volume 05 L The maximum cycle temperature is 1800 K Find the maximum pressure the net specific work the cut off ratio see problem 1092 and the thermal efficiency Solution Compression process isentropic from Eqs 62425 T2 T1v1 v2k1 290 2004 961 K P2 95 20 14 62975 kPa v2 v120 RT120 P1 0043805 m3kg 1w2 u2 u1 Cvo T2 T1 0717 961 290 4811 kJkg Combustion at constant P which is the maximum pressure P3 P2 6298 kPa v3 v2 T3 T2 0043805 1800961 008205 m3kg Cutoff ratio v3 v2 T3 T2 1800 961 1873 2w3 P v3 v2 6298 008215 0043805 2415 kJkg 2q3 u3 u2 2w3 h3 h2 CpoT3 T2 10041800 961 8424 kJkg Expansion process isentropic from Eq624 T4 T3 v3 v404 1800 008205 08761 04 698 K 3w4 u3 u4 CvoT3 T4 0717 1800 698 7901 kJkg Cycle net work and efficiency wnet 2w3 3w4 1w2 2415 7901 4811 5505 kJkg η wnet qH 5505 8424 0653 P v 1 2 3 4 s s 1 2 3 4 v T s P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1094 A diesel engine has a bore of 01 m a stroke of 011 m and a compression ratio of 191 running at 2000 RPM revolutions per minute Each cycle takes two revolutions and has a mean effective pressure of 1400 kPa With a total of 6 cylinders find the engine power in kW and horsepower hp Solution Work from mean effective pressure Eq109 Pmeff vmax vmin wnet wnet Pmeff vmax vmin The displacement is V πBore2 025 S π 012 025 011 0000864 m3 Work per cylinder per power stroke Eq1010 W PmeffVmax Vmin 1400 0000864 kPa m3 12096 kJcycle Only every second revolution has a power stroke so we can find the power see also Eq1011 W W Ncyl RPM 05 cycles minmin 60 skJ cycle 12096 6 2000 05 160 121 kW 162 hp The conversion factor from kW to hp is from Table A1 under power Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1095 A super charger is used for a two stroke 10 L diesel engine so intake is 200 kPa 320 K and the cycle has compression ratio of 181 and mean effective pressure of 830 kPa The engine is 10 L running at 200 RPM find the power output Solution The power is from Eq1011 W m wnet Ncyl RPM 160 m Pmeff v1 v2 Ncyl RPM60 Pmeff Vdispl RPM60 830 kPa 0010 m3 200 revmin 60 smin 277 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1096 A diesel engine has a state before compression of 95 kPa 290 K and a peak pressure of 6000 kPa a maximum temperature of 2400 K Use the air tables to find the volumetric compression ratio and the thermal efficiency Solution Compression s2 s1 from Eq619 so T2 so T1 R lnP2 P1 68352 0287 ln600095 8025 kJkg K A71 T2 9076 K h2 94116 kJkg h3 27558 kJkg so T3 919586 kJkg K qH h3 h2 27558 94116 18142 kJkg CR v1v2 T1T2P2P1 2909076 6000 95 2018 Expansion process so T4 so T3 R lnP4 P3 so T3 R lnT4 T3 R lnv3v4 v3v4 v3v1 v2v1 T3T2 T3T2 1CR 24009076 12018 013104 so T4 R lnT4 T3 so T3 R lnv3v4 91958 0287 ln 013104 861254 Trial and error on T4 since it appears both in so T4 and the ln function T4 1300 K LHS 84405 0287 ln 13002400 8616 T4 1250 K LHS 83940 0287 ln 12502400 85812 Now Linear interpolation T4 1295 K u4 101826 kJkg qL u4 u1 101826 20719 81108 kJkg η 1 qL qH 1 8110818142 0553 P v 1 2 3 4 s s 1 4 v T s 2 3 P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1097 At the beginning of compression in a diesel cycle T 300 K P 200 kPa and after combustion heat addition is complete T 1500 K and P 70 MPa Find the compression ratio the thermal efficiency and the mean effective pressure Solution Standard Diesel cycle See Pv and Ts diagrams for state numbers Compression process isentropic from Eqs624825 P2 P3 7000 kPa v1 v2 P2P11 k 7000 20007143 1267 T2 T1P2 P1k1 k 300 K 7000 200 02857 8284 K Expansion process isentropic first get the volume ratios v3 v2 T3 T2 1500 8284 181 v4 v3 v1 v3 v1 v2 v2 v3 1267 181 7 The exhaust temperature follows from Eq624 T4 T3v3 v4k1 1500 K 1 7 04 6887 K qL CvoT4 T1 0717 kJkgK 6887 300 K 2785 kJkg qH h3 h2 CpoT3 T2 1004 kJkgK 1500 8284 K 674 kJkg Overall performance η 1 qL qH 1 2785 674 0587 wnet qnet qH qL 674 2785 3955 kJkg vmax v1 R T1 P1 0287 kJkgK 300 K 200 kPa 04305 m3kg vmin vmax v1 v2 04305 1267 0034 m3kg Pmeff vmax vmin wnet 3955 04305 0034 997 kPa P v 1 2 3 4 s s 1 2 3 4 v T s P Remark This is a too low compression ratio for a practical diesel cycle Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1098 The worlds largest diesel engine has displacement of 25 m3 running at 200 RPM in a two stroke cycle producing 100 000 hp Assume an inlet state of 200 kPa 300 K and a compression ratio of 201 What is the mean effective pressure and the flow of air to the engine We have 3 parameters for the cycle T1 P1 and CR we need one more so this comes from the total rate of work W m wnet Ncyl RPM 160 m Pmeff v1 v2 Ncyl RPM60 Pmeff Vdispl RPM60 Pmeff W 60 Vdispl RPM 100 000 0746 kW 60 s 25 m3 200 895 kPa With this information we could now get all the cycle states etc Assume the displacement volume is filled with air at state 1 then m P1Vdispl RT1 200 kPa 25 m3 0287 kJkgK 300 K 5807 kg This is repeated 200 times per minute so the rate is m m RPM 160 5807 200 60 19357 kgs A similar engine is in this container ship shown at dock The engine was build up inside the ship as it is too large to put in finished The cranes are on the port dock Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1099 A diesel engine has air before compression at 280 K 85 kPa The highest temperature is 2200 K and the highest pressure is 6 MPa Find the volumetric compression ratio and the mean effective pressure using cold air properties at 300 K Solution Compression P2P1 v1v2k CR k CR v1v2 P2P11k 600085114 2092 T2 T1P2P1k1k 280 600085 02857 9448 K Combustion Highest temperature is after combustion h2 CPT3 T 2 1004 kJkgK 2200 9448 K 12602 kJkg Expansion v4 CR v2 and P2 P3 T4 T3 v3v4k1 T3 v3 CR v2k1 T3 T3 CR T2 k1 2200 K 22002092 9448 04 9142 K qL u4 u1 CV T4 T1 0717 kJkgK 9142 280 K 4547 kJkg v1 RT1P1 0287 kJkgK 280 K85 kPa 09454 m3kg Displacement and mep from net work v1 v2 v1 v1CR v11 1CR 09002 m3kg Pmeff wnetv1 v2 qH qL v1 v2 12602 454709002 8948 kPa P v 1 2 3 4 s s 1 4 v T s 2 3 P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10100 Consider an ideal airstandard diesel cycle in which the state before the compression process is 95 kPa 290 K and the compression ratio is 20 Find the thermal efficiency for a maximum temperature of 2000 K Solution Diesel cycle P1 95 kPa T1 290 K CR v1v2 20 v1 0287 kJkgK 290 K95 kPa 08761 m3kg v4 CR v2 Compression process isentropic from Eqs62425 T2 T1v1 v2k1 290 K 2004 9612 K Combustion at constant P which is the maximum pressure v3 v2 T3 T2 08761 20 m3kg 2000 9612 009115 m3kg Expansion process isentropic from Eq624 T4 T3 v3 v404 2000 K 009115 08761 04 8089 K Cycle net work and efficiency ηTH 1 kT3 T2 T4 T1 1 8089 290 14 2000 9612 0643 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10101 Solve Problem 1090 using the Pr and vr functions from Table A72 A diesel engine has a state before compression of 95 kPa 290 K and a peak pressure of 6000 kPa a maximum temperature of 2400 K Find the volumetric compression ratio and the thermal efficiency Solution Compression s2 s1 From definition of the Pr function Pr2 Pr1 P2P1 09899 600095 6252 A71 T2 907 K h2 9410 kJkg h3 27558 kJkg vr3 043338 qH h3 h2 27558 9410 18148 kJkg CR v1v2 T1T2P2P1 290907 6000 95 2019 Expansion process vr4 vr3 v4 v3 vr3 v1 v3 vr3 v1 v2 T2T3 vr3 CR T2T3 043338 2019 9072400 330675 Linear interpolation T4 12948 K u4 10181 kJkg qL u4 u1 10181 2072 8109 kJkg η 1 qL qH 1 810918148 0553 P v 1 2 3 4 s s 1 4 v T s 2 3 P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Stirling and Carnot Cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10102 Consider an ideal Stirlingcycle engine in which the state at the beginning of the isothermal compression process is 100 kPa 25C the compression ratio is 6 and the maximum temperature in the cycle is 1100C Calculate the maximum cycle pressure and the thermal efficiency of the cycle with and without regenerators Ideal Stirling cycle T1 T2 25 oC P1 100 kPa CR v1v2 6 T3 T4 1100 oC Isothermal compression heat goes out T1 T2 P2 P1v1v2 100 kPa 6 600 kPa 1w2 1q2 RT1 lnv1v2 0287 2982 ln6 1533 kJkg Constant volume heat addition V2 V3 P3 P2T3T2 600137322982 2763 kPa q23 u3 u2 Cv oT3 T2 0717 1100 25 7708 kJkg Isothermal expansion heat comes in w34 q34 RT3 lnv4v3 0287 13732 ln6 7061 kJkg wnet 7061 1533 5528 kJkg Efficiency without regenerator q23 and q34 are coming in from source ηNO REGEN q23 q34 wnet 5528 7708 7061 0374 Efficiency with regenerator Now only q34 is coming in from source ηWITH REGEN q34 wnet 5528 7061 0783 T T v v 1 2 3 4 P v 1 2 3 4 T T v v s T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10103 An airstandard Stirling cycle uses helium as the working fluid The isothermal compression brings helium from 100 kPa 37C to 600 kPa The expansion takes place at 1200 K and there is no regenerator Find the work and heat transfer in all of the 4 processes per kg helium and the thermal cycle efficiency Helium table A5 R 2077 kJkg K Cvo 31156 kJkg K Compressionexpansion v4 v3 v1 v2 P2 P1 600 100 6 1 2 1w2 q12 P dv R T1lnv1 v2 RT1ln P2 P1 2077 kJkgK 310 K ln 6 11537 kJkg 2 3 2w3 0 q23 CvoT3 T2 311561200 310 2773 kJkg 3 4 3w4 q34 R T3ln v4 v3 20771200 ln 6 44658 kJkg 4 1 4w1 0 q41 CvoT4 T1 2773 kJkg ηcycle q23 q34 1w2 3w4 11537 44658 2773 44658 0458 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10104 Consider an ideal airstandard Stirling cycle with an ideal regenerator The minimum pressure and temperature in the cycle are 100 kPa 25C the compression ratio is 10 and the maximum temperature in the cycle is 1000C Analyze each of the four processes in this cycle for work and heat transfer and determine the overall performance of the engine Ideal Stirling cycle diagram as in Fig 1019 with P1 100 kPa T1 T2 25oC v1v2 10 T3 T4 1000oC From 12 at const T 1w2 1q2 T1s2 s1 RT1lnv1v2 0287 2982 ln10 1971 kJkg From 23 at const V 2w3 0 q23 CV0T3 T2 0717 1000 25 699 kJkg From 34 at const T 3w4 3q4 T3s4 s3 RT3 ln v3 v4 0287 12372 ln10 8414 kJkg From 41 at const V 4w1 0 q41 CV0T1 T4 0717 25 1000 699 kJkg wNET 1971 0 8414 0 6443 kJkg Since q23 is supplied by q41 regenerator qH q34 8414 kJkg ηTH qH wNET 6443 8414 0766 NOTE qH q34 RT3 ln10 qL 1q2 RT1 ln10 ηTH qH qH qL Error Bookmark not defined T3 T3 T1 975 12732 0766 Carnot efficiency Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10105 The airstandard Carnot cycle was not shown in the text show the Ts diagram for this cycle In an airstandard Carnot cycle the low temperature is 320 K and the efficiency is 50 If the pressure before compression and after heat rejection is 100 kPa find the high temperature and the pressure just before heat addition Solution Carnot cycle efficiency from Eq75 η 05 1 THT L TH TL05 640 K Just before heat addition is state 2 and after heat rejection is state 1 so P1 100 kPa and the isentropic compression is from Eq623 P2 P1 TH TL k k1 1131 kPa P v 1 2 3 4 s s T T 1 4 T s 2 3 T T H L Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10106 Air in a pistoncylinder goes through a Carnot cycle in which TL 268C and the total cycle efficiency is η 23 Find TH the specific work and volume ratio in the adiabatic expansion for constant CP Cv Solution Carnot cycle efficiency Eq55 η 1 TLTH 23 TH 3 TL 3 300 900 K Adiabatic expansion 3 to 4 Pvk constant work from Eq629 n k 3w4 P4v4 P3v31 k R 1 kT4 T3 u3 u4 CvT3 T4 0717 kJkgK 900 300 K 4299 kJkg v4v3 T3T41k 1 325 156 P v 1 2 3 4 s s T T 1 4 T s 2 3 T T H L Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10107 Do the problem 10105 using values from Table A71 The airstandard Carnot cycle was not shown in the text show the Ts diagram for this cycle In an airstandard Carnot cycle the low temperature is 320 K and the efficiency is 50 If the pressure before compression and after heat rejection is 100 kPa find the high temperature and the pressure just before heat addition Solution Carnot cycle efficiency Eq75 η 1 TLTH 12 TH 2 TL 2 320 640 K From A71 s T1 693413 kJkg K s T2 764448 kJkg K From the entropy s1 s2 and Eq 619 we get P2 P1 exp s T2 s T1R 100 kPa exp 764448 6934130287 1188 kPa P v 1 2 3 4 s s T T 1 4 T s 2 3 T T H L Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10108 Do Problem 10105 using the Pr vr functions in Table A72 The airstandard Carnot cycle was not shown in the text show the Ts diagram for this cycle In an airstandard Carnot cycle the low temperature is 320 K and the efficiency is 50 If the pressure before compression and after heat rejection is 100 kPa find the high temperature and the pressure just before heat addition Solution Carnot cycle efficiency Eq75 η 1 TLTH 12 TH 2 TL 2 320 640 K From A72 Pr1 13972 Pr2 16598 From the entropy s1 s2 and equation below Table A72 we get P2 P1 Pr2 Pr1 100 kPa 16598 13972 1188 kPa P v 1 2 3 4 s s T T 1 4 T s 2 3 T T H L Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Atkinson and Miller cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10109 An Atkinson cycle has state 1 as 150 kPa 300 K compression ratio of 9 and a heat release of 1000 kJkg Find the needed expansion ratio Solution Solve the problem with constant heat capacity Compression 1 to 2 s2 s1 From Eq624 and Eq625 T2 T1 v1v2k1 300 K 9 04 7225 K P2 P1v1v2k 150 kPa 9 14 3251 kPa Combustion 2 to 3 at constant volume u3 u2 qH T3 T2 qHCv 7225 K 1000 kJkg 0717 kJkgK 2117 K P3 P2 T3T2 3251 kPa 2117 7225 9526 kPa The expansion should now bring pressure down so P4 P 1 P4 P1 P3 v3v4k P3 CRk CR P3 P4 1k CR 9526 150071429 194 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10110 An Atkinson cycle has state 1 as 150 kPa 300 K compression ratio of 9 and expansion ratio of 14 Find the needed heat release in the combustion Solve the problem with constant heat capacity Compression 1 to 2 s2 s1 From Eq624 and Eq625 T2 T1 v1v2k1 300 K 9 04 7225 K P2 P1v1v2k 150 kPa 9 14 3251 kPa The expansion backwards from P4 P1 gives P3 P4 v4v3k P4 CRk 150 kPa 1414 60349 kPa Combustion 2 to 3 at constant volume v3 v2 P3 P2 T3T2 Energy Eq u3 u2 qH qH Cv T3 T2 Solve for T3 from process equation T3 T2 P3 P2 7225 K 60349 3251 13412 K Solve for qH from the energy equation qH Cv T3 T2 0717 13412 7225 4436 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10111 Assume we change the Otto cycle in Problem 1066 to an Atkinson cycle by keeping the same conditions and only increase the expansion to give a different state 4 Find the expansion ratio and the cycle efficiency Solution Solve the problem with constant heat capacity Compression 1 to 2 s2 s1 From Eq624 and Eq625 T2 T1 v1v2k1 300 1004 7536 K P2 P1v1v2k 95 1014 2386 kPa Combustion 2 to 3 at constant volume u3 u2 qH T3 T2 qHCv 7536 13000717 2567 K P3 P2 T3T2 2386 2567 7536 8127 kPa The expansion should now bring pressure down so P4 P 1 P4 P1 P3 v3v4k P3 CRk CR P3 P4 1k CR 8127 95071429 24 So now the efficiency becomes η 1 k CR CR1 CRk CRk 1 1 14 24 10 24 k 10 k 0675 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10112 Repeat Problem 1073 assuming we change the Otto cycle to an Atkinson cycle by keeping the same conditions and only increase the expansion to give a different state 4 Solution P v 1 2 3 4 s s 1 2 3 4 v T s P Compression Reversible and adiabatic so constant s from Eq62425 P2 P1v1v2k 90 kPa 714 1372 kPa T2 T1v1v2k1 2832 Κ 704 6166 K Combustion constant volume T3 T2 qHCv 6166 18000717 3127 K P3 P2T3T2 1372 3127 6166 6958 kPa The expansion should now bring pressure down so P4 P 1 P4 P1 P3 v3v4k P3 CRk CR P3 P4 1k CR 6958 90071429 2232 So now the efficiency and net work become η E A 0654 TH 1 k 1 14 CR CR1 CRk CRk 1 2232 7 2232 k 7 k wAnetE A ηATHE A qAHE A 0654 1800 kJkg 1177 kJkg Displacement we take this as vA4E A vA2E A and PAmeffE vA1E A RTA1E APA1E A 0287 283290 09029 mA3E Akg vA2E A 17 vA1E A 01290 mA3E Akg vA4E A 2232 vA2E A 2879 mA3E Akg PAmeffE A A wNET Ev4 v2 E A A 1177 2879 0129E A 428 kPa Comment The ratio CR CRA1E A 22327 is unrealistic large Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10113 An Atkinson cycle has state 1 as 150 kPa 300 K compression ratio of 9 and expansion ratio of 14 Find the mean effective pressure Solve the problem with constant heat capacity Compression 1 to 2 sA2E A sA1E A From Eq624 and Eq625 TA2E A TA1E A vA1E AvA2E AAk1E A 300 K 9 A04E A 7225 K PA2E A PA1E AvA1E AvA2E AAkE A 150 kPa 9 A14E A 3251 kPa The expansion backwards from PA4E A PA1E A gives PA3E A PA4E A vA4E AvA3E AAkE A PA4E A CRAkE A 150 kPa 14A14E A 60349 kPa Combustion 2 to 3 at constant volume vA3E A vA2E A PA3E A PA2E A TA3E ATA2E A Energy Eq uA3E A uA2E A qAHE A qAHE A CAvE A TA3E A TA2E A Solve for TA3E A from process equation TA3E A TA2E A PA3E A PA2E A 7225 K 60349 3251 13412 K Solve for qAHE A from the energy equation qAHE A CAvE A TA3E A TA2E A 0717 kJkgK 13412 7225 K 4436 kJkg The efficiency and net work become ηATHE A 1 k A CR CR1 CRk AE CRk 1 E 1 14 A 14 9 14 k 9 k E A 0623 wAnetE A ηATHE A qAHE A 0623 4436 kJkg 2764 kJkg Displacement we take this as vA4E A vA2E A and PAmeffE vA1E A RTA1E APA1E A 0287 300150 0574 mA3E Akg vA2E A 19 vA1E A 006378 mA3E Akg vA4E A 14 vA2E A 08929 mA3E Akg PAmeffE A A wNET Ev4 v2 E A A 2764 08929 006378E A 333 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10114 A Miller cycle has state 1 as 150 kPa 300 K compression ratio of 9 and expansion ratio of 14 If PA4E A is 250 kPa find the heat release in the combustion Solution Solve the problem with constant heat capacity Compression 1 to 2 sA2E A sA1E A From Eq624 and Eq625 TA2E A TA1E A vA1E AvA2E AAk1E A 300 K 9 A04E A 7225 K PA2E A PA1E AvA1E AvA2E AAkE A 150 kPa 9 A14E A 3251 kPa The expansion backwards from PA4E A gives PA3E A PA4E A vA4E AvA3E AAkE A PA4E A CRAkE A 250 kPa 14A14E A 10 058 kPa Combustion 2 to 3 at constant volume uA3E A uA2E A qAHE TA3E A TA2E A qAHE ACAvE A and PA3E A PA2E A TA3E ATA2E A TA3E A TA2E A PA3E A PA2E A 7225 K 10 058 3251 22353 K qAHE A CAvE A TA3E A TA2E A 0717 kJkgK 22353 7225 K 1085 kJkg P v 1 2 3 4 s s 5 1 2 3 4 v T s P 5 v Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10115 A Miller cycle has state 1 as 150 kPa 300 K compression ratio of 9 and a heat release of 1000 kJkg Find the needed expansion ratio so PA4E A is 250 kPa Solve the problem with constant heat capacity Compression 1 to 2 sA2E A sA1E A From Eq624 and Eq625 TA2E A TA1E A vA1E AvA2E AAk1E A 300 K 9 A04E A 7225 K PA2E A PA1E AvA1E AvA2E AAkE A 150 kPa 9 A14E A 3251 kPa Combustion 2 to 3 at constant volume uA3E A uA2E A qAHE TA3E A TA2E A qAHE ACAvE A 7225 10000717 2117 K PA3E A PA2E A TA3E ATA2E A 3251 kPa 2117 7225 9526 kPa The expansion should now bring pressure down to PA4E A PA4E A PA3E A vA3E AvA4E AAkE A PA3E A CRAkE A CR PA3E A PA4E A A1kE A CR 9526 250A071429E A 135 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10116 In a Miller cycle assume we know state 1 intake state compression ratios CRA1E A and CR Find an expression for the minimum allowable heat release so PA4E A PA5E A that is it becomes an Atkinson cycle Compression Reversible and adiabatic so constant s from Eq62425 TA2E A TA1E AvA1E AvA2E AAk1E A TA1E ACRA1E Ak1E Combustion constant volume TA3E A TA2E A qAHE ACAvE A and PA3E A PA2E ATA3E ATA2E A The expansion should now bring pressure down so PA4E A PA1E PA4E A PA1E A PA3E A vA3E AvA4E AAkE A PA3E A CRAkE A From the ideal gas law and the above pressure relation we have PA3E AvA3E A PA1E AvA1E A TA3E A TA1E A PA3E A PA1E ACRA1E A CRAkE A CRA1E so now substitute TA3E A TA2E A qAHE ACAvE A TA1E ACRA1E Ak1E A qAHE ACAvE to get TA1E ACRA1E Ak1E A qAHE ACAvE A TA1E A CRAkE A CRA1E A and now solve for qAHE qAHE A CAvE A TA1E A CRAkE A CRA1E A CRA1E Ak1E A CAvE A TA1E A CRAkE A CRA1E AkE A CRA1E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Combined Cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10117 A Rankine steam power plant should operate with a high pressure of 3 MPa a low pressure of 10 kPa and the boiler exit temperature should be 500C The available hightemperature source is the exhaust of 175 kgs air at 600C from a gas turbine If the boiler operates as a counterflowing heat exchanger where the temperature difference at the pinch point is 20C find the maximum water mass flow rate possible and the air exit temperature Solution CV Pump wAPE A hA2E A hA1E A vA1E APA2E A PA1E A 0001013000 10 302 kJkg hA2E A hA1E A wAPE A 19183 302 19485 kJkg Heat exchanger water states State 2a TA2aE A TASATE A 2339 C hA2aE A 100842 kJkg State 3 hA3E A 34565 kJkg Heat exchanger air states inlet hAairinE A 90316 kJkg State 2a hAairE ATA2aE A 20 53128 kJkg HEAT EXCH i e a 2a 3 2 Air temperature should be 2539C at the point where the water is at state 2a CV Section 2a3 ia AmE AH2OhA3E A hA2aE A AmE AairE AhAiE A hAaE A AmE AH2O 175 A 90316 53128 34565 100842E A 26584 kgs Take CV Total AmE AH2OhA3E A hA2E A AmE AairE AhAiE A hAeE A hAeE A hAiE A A mE AH2OhA3E A hA2E AAmE AairE A 9036 2658434565 19485175 40813 kJkg TAeE A 4067 K 1336 C TAeE A TA2E A 465 C OK 1 T 3 2 s 2a Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10118 A simple Rankine cycle with R410A as the working fluid is to be used as a bottoming cycle for an electrical generating facility driven by the exhaust gas from a Diesel engine as the high temperature energy source in the R410A boiler Diesel inlet conditions are 100 kPa 20C the compression ratio is 20 and the maximum temperature in the cycle is 2800 K The R410A leaves the bottoming cycle boiler at 80C 4 MPa and the condenser pressure is 1800 kPa The power output of the Diesel engine is 1 MW Assuming ideal cycles throughout determine a The flow rate required in the diesel engine b The power output of the bottoming cycle assuming that the diesel exhaust is cooled to 200C in the R410A boiler 1 4 v T s 2 3 P AIRSTD DIESEL CYCLE Diesel cycle information given means AW E ADIESELE A 1 MW PA1E A 100 kPa TA1E A 20 AoE AC CR vA1E AvA2E A 20 TA3E A 2800 K Consider the Diesel cycle TA2E A TA1E AvA1E AvA2E AAk1E A 2932 K 20A04E A 9718 K PA2E A PA1E AvA1E AvA2E AAkE A 100 kPa 20A14E A 6629 kPa qAHE A CAP0E ATA3E A TA2E A 1004 kJkgK2800 9718 K 18355 kJkg vA1E A A0287 2932 100E A 08415 mA3E Akg vA2E A A08415 20E A 004208 mA3E Akg vA3E A vA2E ATA3E ATA2E A 004208 mA3E Akg 28009718 012124 mA3E Akg TA4E A TA3E A v3 Ev4 E A k1E A 2800 K A 012124 08415 E A 04E A 1290 K qALE A 071729315 1290 7147 kJkg wANETE A 18355 7147 112076 kJkg AmE AAIRE A AW E ANETE AwANETE A 1000 kW112076 kJkg 0892 kgs T s 6 5 7 8 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Rankine cycle information given means Boiler exit state T7 80oC P7 4 MPa Condenser P8 P5 1800 kPa T5 2822oC h5 hf 1031 kJkg Consider the Rankine cycle s8 s7 10028 Sup vapor T8 3292oC h8 2906 kJkg wT h7 h8 31148 2906 2088 kJkg wP v5P6 P5 00009595 m3kg 4000 1800 kPa 211 kJkg h6 h5 wP 1031 211 10521 kJkg qH h7 h6 31148 10521 20627 kJkg Connecting the two cycles Q H available from Diesel exhaust flow cooled to 200 oC Q H 0892 kgs 1004 kJkgK 1290 4732 K 731 kW m R410a Q HqH 731 kW 20627 kJkg 3544 kgs W R410a 3544 kgs 2088 211 kJkg 665 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10119 A small utility gasoline engine of 250 cc runs at 1500 RPM with a compression ratio of 71 The inlet state is 75 kPa 17oC and the combustion adds 1500 kJkg to the charge This engine runs a heat pump using R410A with high pressure of 4 MPa and an evaporator operating at 0oC Find the rate of heating the heat pump can deliver Overall cycle efficiency is from Eq1012 rv v1v2 7 ηTH 1 r1k 1 704 05408 wnet ηTH qH 05408 1500 81127 kJkg We also need specific volume to evaluate Eqs109 to 1011 v1 RT1 P1 0287 kJkgK 290 K 75 kPa 11097 m3kg v2 v1CR 015853 m3kg Pmeff v1 v2 wnet 81127 1097 015853 kJkg m3kg 8529 kPa Now we can find the power from Eq1011 assume 4 stroke engine W Pmeff Vdispl RPM 60 1 2 8529 25 104 1500 60 1 2 2665 kW For the refrigeration cycle we have State 1 h1 27912 kJkg s1 10368 kJkgK State 2 4 MPa s2 s1 interpolate h2 32381 kJkg State 3 4 MPa interpolate h3 17162 kJkg βHP wC qH h2 h1 h2 h3 32381 17162 32381 27912 3405 The work out of the heat engine equals the input to the heat pump Q H βHP W 3405 2665 kW 907 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10120 Can the combined cycles in the previous problem deliver more heat than what comes from the R410A Find any amounts if so by assuming some conditions The hot exhaust gases from the gasoline engine can also be used and as the heat pump delivers heat at a temperature that equals the saturation T at 4 MPa which is 619oC the exhaust gas can be cooled to that Not all the heat out of the heat engine is exhaust gas part of Q L is heat transfer rejected by the cooling air or coolant So as the engine produces 266 kW with an efficiency of 54 we must have Q H fuel W ηTH 266 kW 054 493 kW input so Q L engine Q H fuel W 493 266 23 kW Generally half of this is exhaust gas and the other half heat transfer So we can potentially get an additional 1 kW to heat with from the exhaust gas alone in addition to the heat pump output Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10121 The power plant shown in Fig 1024 combines a gasturbine cycle and a steam turbine cycle The following data are known for the gasturbine cycle Air enters the compressor at 100 kPa 25C the compressor pressure ratio is 14 the heater input rate is 60 MW the turbine inlet temperature is 1250C the exhaust pressure is 100 kPa the cycle exhaust temperature from the heat exchanger is 200C The following data are known for the steamturbine cycle The pump inlet state is saturated liquid at 10 kPa the pump exit pressure is 125 MPa turbine inlet temperature is 500C Determine a The mass flow rate of air in the gasturbine cycle b The mass flow rate of water in the steam cycle c The overall thermal efficiency of the combined cycle a From Air Tables A7 Pr1 10913 h1 29866 h5 47584 kJkg s2 s1 Pr2S Pr1P2P1 10913 14 152782 T2 629 K h2 63448 kJkg wC h1 h2 29866 63448 33582 kJkg At T3 15232 K Pr3 515493 h3 166391 kJkg m AIR Q Hh3 h2 60 000 166391 63448 5828 kgs b Pr4S Pr3P4P3 515493114 368209 T4 791 K h4 81268 kJkg wT gas h3 h4 166391 81268 85123 kJkg Steam cycle wP 000101 m3kg 12500 10 kPa 12615 kJkg h6 h9 wP 19183 12615 20445 kJkg At 125 MPa 500 oC h7 33411 kJkg s7 64704 kJkg K m H2O m AIR h4 h5 h7 h6 5828 81268 47584 33411 20445 6259 kgs c s8 s7 64704 06492 x8 7501 x8 07761 h8 19181 07761 23928 20489 kJkg wT steam h7 h8 33411 20489 12922 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful W NET m wT wC AIR m wT wP H2O 5828 85123 33582 6259 12922 12615 30 038 8009 38 047 kW 3805 MW ηTH W NETQ H 3804760 0634 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Exergy Concepts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10122 Consider the Brayton cycle in Problem 1028 Find all the flows and fluxes of exergy and find the overall cycle secondlaw efficiency Assume the heat transfers are internally reversible processes and we neglect any external irreversibility Solution Efficiency is from Eq101 η W net Q H wnet qH 1 rk1k p 1 160414 0547 from the required power we can find the needed heat transfer Q H W net η 14 000 0547 25 594 kW m Q H qH 25 594 kW 1160 kJkg 22064 kgs Temperature after compression is T2 T1 rk1k p 290 160414 64035 K The highest temperature is after combustion T3 T2 qHCp 64035 1160 1004 17957 K For the exit flow I need the exhaust temperature T4 T3 rp k1 k 17957 1602857 81325 K ηII W NETΦ H since the low T exergy flow out is lost The high T exergy input from combustion is Φ H m ψ3 ψ2 m h3 h2 Ts3 s2 22064 1160 298 1004 ln 17957 64035 18 784 kW ηII W NETΦ H 14 000 18 784 0745 Φ flow in m ψ1 ψo m h1 ho Tos1 so 22064 100417 25 298 1004 ln 290 298 25 kW Φ flow out m ψ4 ψo m h4 ho Ts4 so 22064 1004813 298 298 1004 ln 813 298 4545 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10123 A Brayton cycle has a compression ratio of 151 with a high temperature of 1600 K and an inlet state of 290 K 100 kPa Use cold air properties to find the specific net work output and the second law efficiency if we neglect the value of the exhaust flow Brayton cycle so this means Minimum T T1 290 K Maximum T T3 1600 K Pressure ratio P2P1 15 Solve using constant CP0 Compression in compressor s2 s1 Implemented in Eq623 T2 T1P2P1 k1 k 290 K 150286 62865 K Energy input is from the combustor qH h3 h2 CP0T3 T2 1004 kJkgK 1600 62865 K 9752 kJkg Do the overall cycle efficiency and net work η W net Q H wnet qH 1 rk1k p 1 150414 05387 wNET η qH 05387 9752 52534 kJkg Notice the qH does not come at a single T so neglecting external irreversibility we get ΦqH increase in flow exergy ψ3 ψ2 h3 h2 Τos3 s2 qH Τo CP lnT3T2 9752 29815 1004 ln 1600 62865 69556 kJkg ηII ψ3 ψ2 wnet 52534 69556 0755 1 2 3 4 P P 100 kPa T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10124 Reconsider the previous problem and find the second law efficiency if you do consider the value of the exhaust flow Brayton cycle so this means Minimum T T1 290 K Maximum T T3 1600 K Pressure ratio P2P1 15 Solve using constant CP0 Compression in compressor s2 s1 Implemented in Eq623 T2 T1P2P1 k1 k 290150286 62865 K Energy input is from the combustor qH h3 h2 CP0T3 T2 1004 1600 62865 9752 kJkg Do the overall cycle thermal efficiency and net work η W net Q H wnet qH 1 rk1k p 1 150414 05387 wNET η qH 05387 9752 52534 kJkg Notice the qH does not come at a single T so neglecting external irreversibility we get ΦqH increase in flow exergy ψ3 ψ2 h3 h2 Τos3 s2 qH Τo CP lnT3T2 9752 29815 1004 ln 1600 62865 69556 kJkg T4 T3 P4P3 k1 k 1600 Κ 1150286 738 K ΦqL ψ4 ψ1 h4 h1 Τos4 s1 CPT4 T1 Τo CP ln T1 T4 1004 kJkgK 738 290 298 ln738 290 K 17033 kJkg ηII ΦqH ΦqL wnet 52534 69556 17033 1 Why is it 1 the cycle is reversible so we could have said that right away 1 2 3 4 P P 100 kPa T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10125 For Problem 10117 determine the change of exergy of the water flow and that of the air flow Use these to determine a second law efficiency for the boiler heat exchanger From solution to 10117 m H2O 26584 kgs h2 19485 kJkg s2 06587 kJkg K h3 34565 kJkg s3 72338 sTi 79820 sTe 71762 kJkg K hi 90316 kJkg he 40813 kJkg ψ3 ψ2 h3 h2 T0s3 s2 130128 kJkg ψi ψe hi he T0sTi sTe 25478 kJkg ηII ψ3 ψ2m H2O ψi ψem air 130128 26584 25478 175 0776 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10126 Determine the second law efficiency of an ideal regenerator in the Brayton cycle The ideal regenerator has no temperature difference between the two flows and is therefore reversible Its second law efficiency is 100 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10127 Assume a regenerator in a Brayton cycle has an efficiency of 75 Find an expression for the second law efficiency To do this assume the pressure is constant in the two flows so we do not have any pressure loss The change in enthalpy and entropy is then only due to changes in temperature for the two flows ηII ψ4 ψ2 ψx ψ2 h4 h2 Tos4 s2 hx h2 Tosx s2 CP Tx T2 To CP lnTxT2 CP T4 T2 To CP lnT4T2 Tx T2 To lnTxT2 T4 T2 To lnT4T2 Finally we have ηII 1 a lnT4T2 ηreg a lnTxT2 where a is defined as a To T4 T2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10128 The Brayton cycle in Problem 1019 had a heat addition of 800 kJkg What is the exergy increase in the heat addition process Solution Combustion h3 h2 qH 2w3 0 and Tmax T3 1400 K T2 T3 qHCP 1400 K 800 kJkg 1004 kJkgK 6032 K Now the exergy increase from 2 to 3 becomes P3 P2 ψ3 ψ2 h3 h2 T0s3 s2 qH T0 CP lnT3T2 800 29815 1004 ln 1400 6032 54796 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10129 The conversion efficiency of the Brayton cycle in Eq101 was done with cold air properties Find a similar formula for the second law efficiency assuming the low T heat rejection is assigned zero exergy value The thermal efficiency first law is Eq101 ηTH W net Q H wnet qH 1 rk1k p The corresponding 2nd law efficiency is ηII wnet ΦH h3 h2 h4 h1 h3 h2 Tos3 s2 where we used ΦH increase in flow exergy ψ3 ψ2 h3 h2 Τos3 s2 Now divide the difference h3 h2 out to get ηII 1 h4 h1 h3 h2 1 Tos3 s2 h3 h2 ηTH 1 Tos3 s2 h3 h2 ηTH 1 To CP lnT3 T2 CP T3 T2 ηTH 1 To lnT3 T2 T3 T2 Comment Due to the temperature sensitivity of ΦH the temperatures do not reduce out from the expression Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10130 Redo the previous problem for a large stationary Brayton cycle where the low T heat rejection is used in a process application and thus has a nonzero exergy The thermal efficiency first law is Eq101 ηTH W net Q H wnet qH 1 rk1k p The corresponding 2nd law efficiency is ηII wnet ΦH ΦL h3 h2 h4 h1 h3 h2 Tos3 s2 where we used ΦH increase in flow exergy ψ3 ψ2 h3 h2 Τos3 s2 ΦL rejection of flow exergy ψ4 ψ1 h4 h1 Τos4 s1 Now divide all terms with the difference h3 h2 to get wnet h3 h2 ηTH ΦHh3 h2 1 Τos3 s2h3 h2 1 Τo lnT3 T2 T3 T2 1 ΦLh3 h2 h3 h2 h4 h1 1 Τo s4 s1 h4 h1 1 ηTH1 Τo T4 T1 ln T4T1 Substitute all terms to get ηII ηTH 1 To lnT3 T2 T3 T2 1 ηTH1 Τo lnT4 T1 T4 T1 Comment Due to the temperature sensitivity of ΦH and ΦL the temperatures do not reduce out from the expression Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10131 Solve Problem 1019 with variable specific heats using Table A7 A Brayton cycle inlet is at 300 K 100 kPa and the combustion adds 800 kJkg The maximum temperature is 1400 K due to material considerations What is the maximum allowed compression ratio For this compression ratio calculate the net work and cycle efficiency using cold air properties Solution Combustion h3 h2 qH 2w3 0 and Tmax T3 1400 K h2 h3 qH 151527 800 71527 kJkg From Table A71 T2 7016 K so T2 774251 T1 300 K so T1 686926 kJkg K Reversible adiabatic compression leads to constant s from Eq619 P2 P1 exp so T2 so T1R exp304268 2096 Reversible adiabatic expansion leads to constant s from Eq619 so T4 so T3 R lnP4 P3 852891 0287 ln 12096 765568 kJkgK From Table A71 by linear interpolation T4 64685 K h4 65681 kJkg wT h3 h4 151527 65681 85846 kJkg wC h2 h1 71527 30047 4148 kJkg wnet wT wC 85846 4148 4437 kJkg η wnet qH 4437 800 055 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10132 Do Problem 1028 with properties from table A71 instead of cold air properties Solution With the variable specific heat we must go through the processes one by one to get net work and the highest temperature T3 From A71 h1 29043 kJkg so T1 683521 kJkg K The compression is reversible and adiabatic so constant s From Eq619 s2 s1 so T2 so T1 RlnP2P1 683521 0287 ln16 763094 T2 6319 K h2 641 kJkg Energy equation with compressor work in wC 1w2 h2 h1 641 29043 35057 kJkg Energy Eq combustor h3 h2 qH 641 1160 1801 kJkg State 3 P h T3 16357 K so T3 871738 kJkg K The expansion is reversible and adiabatic so constant s From Eq619 s4 s3 so T4 so T3 RlnP4P3 871738 0287ln116 792165 T4 8273 K h4 85231 kJkg Energy equation with turbine work out wT h3 h4 1801 85231 94869 kJkg Now the net work is wnet wT wC 94869 35057 59812 kJkg The total required power requires a mass flow rate as m wnet W net 14 000 59812 kW kJkg 2341 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10133 Repeat Problem 1036 but assume that the compressor has an efficiency of 82 that both turbines have efficiencies of 87 and that the regenerator efficiency is 70 a From solution 1036 T2 T1 P2 P1 k1 k 300 K 60286 5008 K wC w12 CP0T2 T1 1004 kJkgK 5008 300 K 2016 kJkg wC wSCηSC 2016082 2458 kJkg w T1 CP0T4 T5 1004 kJkgK 1600 K T5 T5 13552 K wST1 wT1ηST1 2458087 2825 kJkg CP0T4 T5S 10041600 T5S T5S 13186 K s5S s4 P5 P4T5ST4 k k1 60013186 1600 35 3049 kPa b P6 100 kPa s6S s5 T6S T5 P6 P5 k1 k 13552 100 3049 0286 9852K wST2 CP0T5T6S 100413552 9852 3715 kJkg wT2 ηST2 wST2 087 3715 3232 kJkg 3232 CP0T5T6 100413552 T6 T6 10333K m W NETwNET 1503232 0464 kgs c wC 2458 CP0T2 T1 1004T2 300 T2 5448 K ηREG h6 h2 h3 h2 T6 T2 T3 T2 10333 5448 T3 5448 07 T3 8868 K qH CP0T4 T3 10041600 8868 716 kJkg ηTH wNETqH 3232716 0451 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10134 Consider a gas turbine cycle with two stages of compression and two stages of expansion The pressure ratio across each compressor stage and each turbine stage is 8 to 1 The pressure at the entrance of the first compressor is 100 kPa the temperature entering each compressor is 20oC and the temperature entering each turbine is 1100oC A regenerator is also incorporated into the cycle and it has an efficiency of 70 Determine the compressor work the turbine work and the thermal efficiency of the cycle P2P1 P4P3 P6P7 P8P9 80 P1 100 kPa T1 T3 20 oC T6 T8 1100 oC Cold air and s2 s1 and s4 s 3 T4 T2 T1 P1 P2 k1 k 2931580286 531 K 1 3 2 4 5 6 7 8 9 10 T s Total wC 2 w12 2CP0T2 T1 2 1004531 29315 4776 kJkg Also s6 s7 and s8 s9 T7 T9 T6 P6 P7 k1 k 137315 1 8 0286 758 K Total wT 2 w67 2CP0T6 T7 2 1004137315 758 12352 kJkg wNET 12352 4776 7576 kJkg Ideal regenerator T5 T9 T10 T4 so the actual one has ηREG h9 h4 h5 h4 T9 T4 T5 T4 758 531 T5 531 07 T5 6899 K qH h6 h5 h8 h7 CP0T6 T5 CP0T8 T7 1004137315 6899 1004 137315 758 13036 kJkg ηTH wNETqH 757613036 0581 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful REG C O MP TURB TURB C O M P CC CC I C 1 2 4 10 6 7 8 9 5 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10135 A gas turbine cycle has two stages of compression with an intercooler between the stages Air enters the first stage at 100 kPa 300 K The pressure ratio across each compressor stage is 5 to 1 and each stage has an isentropic efficiency of 82 Air exits the intercooler at 330 K The maximum cycle temperature is 1500 K and the cycle has a single turbine stage with an isentropic efficiency of 86 The cycle also includes a regenerator with an efficiency of 80 Calculate the temperature at the exit of each compressor stage the secondlaw efficiency of the turbine and the cycle thermal efficiency State 1 P1 100 kPa T1 300 K State 7 P7 Po 100 kPa State 3 T3 330 K State 6 T6 1500 K P6 P 4 P2 5 P1 500 kPa P4 5 P3 2500 kPa Ideal compression T2s T1 P2P1k1k 4754 K 1st Law q hi he w q 0 wc1 h1 h2 CPT1 T2 wc1 s CPT1 T2s 1760 kJkg wc1 wc1 s η 2146 T2 T1 wc1CP 5139 K T4s T3 P4P3k1k 4754 K wc2 s CPT3 T4s 1936 kJkg wc2 2361 kJkg T4 T3 wc2 CP 5652 K Ideal Turbine reversible and adiabatic T7s T6P7P6k1k 5974 K wTs CPT6 T7s 9058 kJkg 1st Law Turbine q h6 h7 w q 0 wT h6 h7 CPT6 T7 ηTs wTs 086 9058 7790 kJkg T7 T6 wT CP 1500 7791004 7237 K s6 s7 CP ln T7 T6 R ln P7 P6 01925 kJkg K ψ6 ψ7 h6 h7 Tos6 s7 7790 2981501925 8368 kJkg η2nd Law ψ6ψ7 wT 7790 8368 0931 d ηth qH wnet wnet wT wc1 wc2 3283 kJkg 1st Law Combustor q hi he w w 0 qc h6 h5 CPT6 T5 Regenerator ηreg T7 T4 T5 T4 08 T5 6921 K qH qc 8107 kJkg ηth 0405 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10136 Repeat the questions in Problem 1036 when we assume that friction causes pressure drops in the burner and on both sides of the regenerator In each case the pressure drop is estimated to be 2 of the inlet pressure to that component of the system so P3 588 kPa P4 098 P3 and P6 102 kPa Solution From solution Problem 1036 T2 T1 P2 P1 k1 k 300 K 60286 5008 K wC w12 CP0T2 T1 1004 kJkgK 5008 300 K 2016 kJkg P3 098 600 588 kPa P4 098 588 5762 kPa s5 s4 P5 P4T5ST4 k k1 5762 kPa 13992 1600 35 3604 kPa P6 100098 102 kPa s6S s5 T6 T5 P6 P5 k1 k 13992 102 2928 0286 9752 K wST2 CP0T5T6 100413992 9752 4257 kJkg m W NETwNET 150 kW 4257 kJkg 0352 kgs T3 T6 9752 K qH CP0T4 T3 1004 1600 9752 6273 kJkg ηTH wNETqH 42576273 0678 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10137 A gasoline engine has a volumetric compression ratio of 9 The state before compression is 290 K 90 kPa and the peak cycle temperature is 1800 K Find the pressure after expansion the cycle net work and the cycle efficiency using properties from Table A7 Use table A71 and interpolation Compression 1 to 2 s2 s1 From Eq619 0 so T2 so T1 R lnP2P1 so T2 so T1 R lnΤ2v1T1v2 so T2 R lnΤ2T1 so T1 R lnv1v2 683521 0287 ln 9 74658 This becomes trial and error so estimate first at 680 K and use A71 LHS680 77090 0287 ln680290 74644 too low LHS700 77401 0287 ln700290 74872 too high Interpolate to get T2 68123 K u2 4979 kJkg P2 P1 Τ2T1 v1v2 90 68123 290 9 19027 kPa 1w2 u1 u2 2072 4979 2907 kJkg Combustion 2 to 3 constant volume v3 v 2 qH u3 u2 14863 4979 9884 kJkg P3 P2T3T2 19027 18006812 5028 kPa Expansion 3 to 4 s4 s3 From Eq619 as before so T4 R lnΤ4T3 so T3 R lnv3v4 88352 0287 ln19 82046 This becomes trial and error so estimate first at 850 K and use A71 LHS850 77090 0287 ln8501800 81674 too low LHS900 77401 0287 ln9001800 82147 too high Interpolation T4 8893 K u4 666 kJkg P4 P3T4T3v3v4 5028 88931800 19 276 kPa 3w4 u3 u4 14863 6660 8203 kJkg Net work and overall efficiency wNET 3w4 1w2 8203 2907 5296 kJkg η wNETqH 52969884 0536 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10138 Consider an ideal airstandard diesel cycle in which the state before the compression process is 95 kPa 290 K and the compression ratio is 20 Find the maximum temperature by iteration in the cycle to have a thermal efficiency of 60 Solution Diesel cycle P1 95 kPa T1 290 K v1v2 20 ηTH 06 Since the efficiency depends on T3 and T4 which are connected through the expansion process in a nonlinear manner we have an iterative problem T2 T1v1v2 k1 290 K 2004 9612 K v1 0287 kJkgK 290 K95 kPa 0876 m3kg v4 v2 v1CR 0876 20 00438 m3kg v3 v2T3T2 00438 m3kg T39612 00000456 T3 T3 T4 v4v3 k1 0876 00000456 T3 04 T4 0019345 T3 14 Now substitute this into the formula for the efficiency ηTH 060 1 kT3 T2 T4 T1 1 0019345 T3 14 290 14T3 9612 0019345 T3 14 056 T3 248272 0 Trial and error on this nonlinear equation in T3 3050 K LHS 106 3040 K LHS 0036 Linear interpolation T3 3040 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10139 Find the temperature after combustion and the specific energy release by combustion in Problem 1098 using cold air properties This is a difficult problem and it requires iterations The worlds largest diesel engine has displacement of 25 m3 running at 200 RPM in a two stroke cycle producing 100 000 hp Assume an inlet state of 200 kPa 300 K and a compression ratio of 201 What is the mean effective pressure We have 3 parameters for the cycle T1 P1 and CR we need one more so this comes from the total rate of work Eq1011 W m Pmeff v1 v2 Ncyl RPM60 Pmeff Vdispl RPM60 Pmeff W 60 Vdispl RPM 100 000 0746 kW 60 s 25 m3 200 895 kPa v1 RTP 0287 kJkgK 300 K 200 kPa 04305 m3kg wnet Pmeff v1 v2 895 04305 1 1 20 36603 kJkg T2 T1v1v2 k1 300 K 2004 9943 K Combustion and expansion volumes v3 v2 T3T2 v1 T3T2 CR v4 v1 Expansion process isentropic from Eq623 T4 T3 v3v4k1 T3 T3 CRT2k1 T k 3 CRT21k 001908 T k 3 The net work is also given by the heat transfers wnet qH qL CPT3 T2 CvT4 T1 Substitute wnet and T4 into this equation and we get 36603 1004T3 9943 0717001908 T k 3 300 divide by 1004 and combine terms to get 114463 T3 0013626 T k 3 Trial and error guess T3 RHS1400 10541 RHS1600 1183 Interpolate T3 15405 K qH CPT3 T2 100415405 9943 548 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10140 Reevaluate the combined Brayton and Rankine cycles in Problem 10121 For a more realistic case assume the air compressor the air turbine the steam turbine and the pump all have an isentropic efficiency of 87 HEAT EXCH STEAM TURB COND GAS TURB COMP HEAT P AIR 1 2 3 4 5 6 7 8 9 Q 60 MW H T 1250 C 3 o P 100 kPa T 25 C P P 14 η 087 1 1 1 2 SC o T 200 C 5 o P P 125 MPa 6 7 H O 2 η 085 SP P P 10 kPa 8 9 η 087 ST W ST T 500 C 7 o P 100 kPa η 087 4 ST W NET CT a From Air Tables A7 Pr1 10913 h1 29866 h5 47584 kJkg s2 s1 Pr2S Pr1P2P1 10913 14 152782 T2S 629 K h2S 63448 wSC h1 h2S 29866 63448 33582 kJkg wC wSCηSC 33582087 386 h1 h2 h2 68466 kJkg At T3 15232 K Pr3 515493 h3 166391 kJkg m AIR Q Hh3 h2 60 000 166391 68466 6127 kgs b Pr4S Pr3P4P3 515493114 368209 T4S 791 K h4S 81268 kJkg wST h3 h4S 166391 81268 85123 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful wT ηST wST 087 85123 74057 h3 h4 h4 92334 kJkg Steam cycle wSP 00010112500 10 12615 kJkg wP wSPηSP 12615085 1484 kJkg h6 h9 wP 19183 1484 20667 kJkg At 125 MPa 500 oC h7 33417 kJkg s7 64617 kJkg K m H2O m AIR h4 h5 h7 h6 6127 92334 47584 33417 20667 8746 kgs c s8S s7 64617 06492 x8S 7501 x8S 07749 h8S 19181 07749 23928 20460 kJkg wST h7 h8S 33417 20460 12957 kJkg wT ηST wST 087 12957 11273 kJkg W NET m wTwC AIR m wTwP H2O 612774057 3860 874611273 1484 21725 9730 31455 kW 31455 MW ηTH W NETQ H 3145560 0524 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Problems solved using Table A72 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1037 Repeat Problem 1035 when the intercooler brings the air to T3 320 K The corrected formula for the optimal pressure is P2 P1P4 T3T1nn112 see Problem 7245 where n is the exponent in the assumed polytropic process Solution The polytropic process has n k isentropic so nn 1 1404 35 P2 400 kPa 32029035 4752 kPa CV C1 s2 s1 Pr2 Pr1P2P1 09899 4752100 4704 T2 452 K h2 45375 wC1 h2 h1 45375 29043 1633 kJkg CV Cooler qOUT h2 h3 45375 320576 1332 kJkg CV C2 s4 s3 Pr4 Pr3P4P3 13972 16004752 4704 T4 T2 452 K h4 45375 wC2 h4 h3 45375 320576 1332 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1042 A gas turbine with air as the working fluid has two ideal turbine sections as shown in Fig P1042 the first of which drives the ideal compressor with the second producing the power output The compressor input is at 290 K 100 kPa and the exit is at 450 kPa A fraction of flow x bypasses the burner and the rest 1 x goes through the burner where 1200 kJkg is added by combustion The two flows then mix before entering the first turbine and continue through the second turbine with exhaust at 100 kPa If the mixing should result in a temperature of 1000 K into the first turbine find the fraction x Find the required pressure and temperature into the second turbine and its specific power output CVComp wC h2 h1 s2 s1 Pr2 Pr1 P2P1 09899 450100 44545 A72 interpolate T2 445 K h2 44674 kJkg wC 44674 29043 1563 kJkg CVBurner h3 h2 qH 44674 1200 164674 kJkg T3 1509 K CVMixing chamber 1 xh3 xh2 hMIX 104622 kJkg x h3 hMIX h3 h2 164674 104622 164674 44674 050 W T1 W Cin w T1 wC 1563 kJkg h3 h4 h4 104622 1563 8899 kJkg T4 861 K P4 Pr4PrMIX PMIX 519165 450 2504 kPa s4 s5 Pr5 Pr4 P5P4 51 1002504 20367 A72 interpolate h5 6882 kJkg T5 676 K wT2 h4 h5 8899 6882 2017 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1080 Repeat Problem 1073 but assume variable specific heat The ideal gas air tables Table A7 are recommended for this calculation and the specific heat from Fig 326 at high temperature Solution Table A7 is used with interpolation T1 2832 K u1 2023 kJkg vr1 21044 Compression 1 to 2 s2 s1 From definition of the vr function vr2 vr1 v2v1 2104 17 30063 Interpolate to get T2 6039 K u2 4381 kJkg 1w2 u2 u1 2358 kJkg u3 4381 1800 22381 T3 25734 K vr3 034118 P3 90 7 25734 2832 5725 kPa Expansion 3 to 4 s4 s3 From the vr function as before vr4 vr3 v4v3 034118 7 23883 Interpolation T4 14354 K u4 11458 kJkg 3w4 u3 u4 22381 11458 10923 kJkg Net work efficiency and mep wnet 3w4 1w2 10923 2358 8565 kJkg ηTH wnet qH 8565 1800 0476 v1 RT1P1 0287 283290 09029 m3kg v2 17 v1 01290 m3kg Pmeff v1 v2 wnet 8565 09029 0129 1107 kPa UPDATED AUGUST 2013 SOLUTION MANUAL CHAPTER 10 English Units Fundamentals of Thermodynamic Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CHAPTER 10 SUBSECTION PROB NO Brayton Cycles 141150 Otto cycles 151161 Diesel cycles 162166 Stirling and Carnot Cycles 167169 Atkinson Miller cycles 170174 Exergy combined cycles review 175181 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Brayton Cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10141E In a Brayton cycle the inlet is at 540 R 14 psia and the combustion adds 350 Btulbm The maximum temperature is 2500 R due to material considerations Find the maximum permissible compression ratio and for that the cycle efficiency using cold air properties Solution Combustion h3 h2 qH 2w3 0 and Tmax T3 2500 R T2 T3 qHCP 2500 350024 104167 R Reversible adiabatic compression leads to constant s from Eq623 P2 P1 T2T1 k k1 10416754035 99696 We can find the efficiency from the compression ratio Eq101 η 1 T1T2 1 540104167 0482 We can also compute all process work terms Reversible adiabatic expansion leads to constant s from Eq623 T4 T3 P4P3 k1 k T3 T1T2 2500 R 540 104167 1296 R For net work we get wT CP T3 T4 024 2500 1296 28896 Btulbm wC CP T2 T1 024 104167 540 1204 Btulbm wnet wT wC 28896 1204 16856 Btulbm η wnet qH 16856 350 0482 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10142E A Brayton cycle has compression ratio of 151 with a high temperature of 2900 R and the inlet at 520 R 14 psia Use cold air properties and find the specific heat transfer input and the specific net work output Brayton cycle so this means Minimum T T1 520 R Maximum T T3 2900 R Pressure ratio P2P1 15 Compression in compressor s2 s 1 Implemented in Eq623 1 2 3 4 P P 14 psi T s T2 T1 P2P1 k1 k 520 150286 11281 R Energy input is from the combustor qH CP0T3 T2 024 2900 11281 4253 Btulbm Do the overall cycle efficiency and the net work η W net Q H wnet qH 1 rk1k p 1 150414 05387 wNET η qH 05387 4253 2291 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10143E A large stationary Brayton cycle gasturbine power plant delivers a power output of 100 000 hp to an electric generator The minimum temperature in the cycle is 540 R and the exhaust temperature is 1350 R The minimum pressure in the cycle is 1 atm and the compressor pressure ratio is 14 to 1 Calculate the power output of the turbine the fraction of the turbine output required to drive the compressor and the thermal efficiency of the cycle Brayton w NET 100 000 hp P1 1 atm T1 540 R P2P1 14 T3 2900 R Solve using constant CP0 1 2 3 4 P P 1 atm T s Compression in compressor s2 s1 Implemented in Eq623 T2 T1 P2 P1 k1 k 5401402857 11486 R wC h2 h1 CP0T2T1 024 11486 540 1461 Btulbm Expansion in turbine s4 s3 Implemented in Eq623 T3 T4 P3 P4 k1 k 1350 1402857 28693 R wT h3 h4 CP0T3T4 02428693 1350 3646 Btulbm wNET wT wC 3646 1461 2185 Btulbm m W NETwNET 100 000 25442185 1 164 302 lbmh W T m wT 166 865 hp wCwT 040 Energy input is from the combustor qH CP0T3 T2 02428693 11486 413 Btulbm ηTH wNETqH 2185413 0529 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10144E A Brayton cycle produces 14 000 Btus with an inlet state of 60 F 147 psia and a compression ratio of 161 The heat added in the combustion is 480 Btulbm What are the highest temperature and the mass flow rate of air assuming cold air properties Solution Efficiency is from Eq101 η W net Q H wnet qH 1 rk1k p 1 160414 0547 from the required power we can find the needed heat transfer Q H W net η 14 000 0547 25 594 Btus m Q H qH 25 594 Btus 480 Btulbm 5332 lbms Temperature after compression is Eq623 T2 T1 rk1k p 5197 160414 1148 R The highest temperature is after combustion T3 T2 qHCp 1148 480 024 3148 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10145E Do the previous problem with properties from table F5 instead of cold air properties Solution With the variable specific heat we must go through the processes one by one to get net work and the highest temperature T3 From F5 h1 12438 btulbm so T1 163074 Btulbm R The compression is reversible and adiabatic so constant s From Eq619 s2 s1 so T2 so T1 R ln P2 P1 163074 5334 778 ln16 182083 Btulbm R back interpolate in F5 T2 11335 R h2 27458 Btulbm Energy equation with compressor work in wC 1w2 h2 h1 27458 124383 1502 Btulbm Energy Eq combustor h3 h2 qH 27458 480 75458 Btulbm State 3 P h T3 28767 R so T3 207443 Btulbm R The expansion is reversible and adiabatic so constant s From Eq619 s4 s3 so T4 so T3 RlnP4P3 207443 5334 778 ln116 188434 T4 14506 R h4 35627 Btulbm Energy equation with turbine work out wT h3 h4 75458 35627 39831 Btulbm Now the net work is wnet wT wC 39831 1502 24811 Btulbm The total required power requires a mass flow rate as m wnet W net 14 000 24811 Btus Btulbm 564 lbms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10146E Solve Problem 10142E using the air tables F5 instead of cold air properties Brayton cycle so this means Minimum T T1 520 R Maximum T T3 2900 R Pressure ratio P2P1 15 Compression in compressor s2 s 1 Implemented in Eq619 1 2 3 4 P P 14 psi T s s2 s1 so T2 so T1 RlnP2P1 16307 5334 778 ln15 18164 T2 11138 R h2 2696 Btulbm wC h2 h1 2696 1244 1452 Btulbm Energy input is from the combustor qH h3 h2 7614 2696 4918 Btulbm The expansion is reversible and adiabatic so constant s From Eq619 s4 s3 so T4 so T3 RlnP4P3 20768 5334 778 ln115 18911 T4 14885 R h4 3662 Btulbm Energy equation with turbine work out wT h3 h4 7613 3662 3951 Btulbm Now the net work is wnet wT wC 3951 1452 2499 Btulbm Comment Cycle efficiency η W net Q H wnet qH 2499 4918 0508 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10147E An ideal regenerator is incorporated into the ideal airstandard Brayton cycle of Problem 10143E Calculate the cycle thermal efficiency with this modification Solution 1 2 3 4 P v s s 1 2 3 4 P 1 atm T s x y Compression ratio P2 P1 14 Exhaust temperature T4 1350 R Solving for the cycle work terms and heat transfer from problem 10143E The compression is reversible and adiabatic so constant s From Eq623 T2 T1 P2 P1 k1 k 540 14 02857 11486 R wC h2 h1 CP0T2T1 024 11486 540 1461 Btulbm Expansion in turbine s4 s3 Implemented in Eq623 T3 T4 P3 P4 k1 k 1350 1402857 28693 R wT h3 h4 CP0T3T4 02428693 1350 3646 Btulbm wNET wT wC 3646 1461 2185 Btulbm Ideal regenerator TX T4 1350 R qH h3 hX 02428693 1350 3646 Btulbm wT ηTH wNETqH 21853646 060 The short answer from the efficiency of an ideal regenerator T4 Tx from p471 ηTH 1 T3 T1 P1 P2 k1 k 1 T2 T3 1 11486 28693 060 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10148E An airstandard Ericsson cycle has an ideal regenerator as shown in Fig P1044 Heat is supplied at 1800 F and heat is rejected at 150 F Pressure at the beginning of the isothermal compression process is 10 lbfin2 The heat added is 300 Btulbm Find the compressor work the turbine work and the cycle efficiency Identify the states Heat supplied at high temperature T4 T3 1800 F 23497 R Heat rejected at low temperature T1 T2 150 F 6097 R Beginning of the compression P1 10 lbfin 2 Ideal regenerator 2q3 4q1 qH 3q4 wT qH 300 Btulbm ηTH ηCARNOT TH 1 TLTH 1 609723497 07405 wnet ηTH qH 07405 300 Btulbm 222 Btulbm qL wC 300 222 78 Btulbm P v 1 2 3 4 T T P P 1 2 3 4 T T P P s T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10149E The turbine in a jet engine receives air at 2200 R 220 lbfin2 It exhausts to a nozzle at 35 lbfin2 which in turn exhausts to the atmosphere at 147 lbfin2 Find the nozzle inlet temperature and the nozzle exit velocity Assume negligible kinetic energy out of the turbine and reversible processes Solution CV Turbine hi 560588 Btulbm so Ti 199765 Btulbm R ses si Then from Eq619 so Tes so Ti R lnPePi 199765 5334 778 ln 35220 18716 Btu lbm R Table F5 Tes 1382 R hes 33827 Btulbm Energy eq wTs hi hes 560588 33827 2223 Btulbm CV Nozzle hi 33827 Btulbm so Ti 18716 Btulbm R ses si Then from Eq619 so Tes so Ti R lnPePi 18716 5334 778 ln 147 35 181212 Btu lbm R Table F5 Tes 10950 R hes 2649 Btulbm Energy Eq 12Ves 2 hi hes 33827 2649 7337 Btulbm VeAC 2 25037 7337 1917 fts Recall 1 Btulbm 25 037 ft2s 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10150E An air standard refrigeration cycle has air into the compressor at 14 psia 500 R with a compression ratio of 31 The temperature after heat rejection is 540 R Find the COP and the highest cycle temperature From the isentropic compressionexpansion processes Eq623 T2 T1 P2P1 k1 k 3 02857 136874 T3T 4 T2 136874 T1 6844 R The COP Eq105 is β 1 T2 T1 1 136874 1 1 2712 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Otto Diesel Stirling and Carnot Cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12151E A 4 stroke gasoline engine runs at 1800 RPM with a total displacement of 150 in3 and a compression ratio of 101 The intake is at 520 R 10 psia with a mean effective pressure of 90 psia Find the cycle efficiency and power output Efficiency from the compression ratio η 1 CR1k 1 1004 060 The power output comes from speed and displacement in Eq 1011 W PmeffVdispl RPM 60 1 2 90 lbf in2 150 in3 1800 60 1s 1 2 16 875 lbffts 307 hp 217 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10152E Air flows into a gasoline engine at 14 lbfin2 540 R The air is then compressed with a volumetric compression ratio of 101 In the combustion process 560 Btulbm of energy is released as the fuel burns Find the temperature and pressure after combustion Solution Solve the problem with constant heat capacity Compression 1 to 2 s2 s1 From Eq624 and 625 T2 T1 v1v2k1 540 R 1004 13564 R P2 P1v1v2k 14 lbfin2 1014 3517 lbfin 2 Combustion 2 to 3 at constant volume u3 u2 qH T3 T2 qHCv 13564 5600171 4631 R P3 P2 T3T2 3517 lbfin2 4631 13564 1201 lbfin 2 P v 1 2 3 4 s 1 2 3 4 v T s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10153E Find the missing pressures and temperatures in the previous cycle Compression 1 to 2 s2 s1 From Eq624 and 625 T2 T1 v1v2k1 540 R 1004 13564 R P2 P1v1v2k 14 lbfin2 1014 3517 lbfin 2 Combustion 2 to 3 at constant volume u3 u2 qH T3 T2 qHCv 13564 5600171 4631 R P3 P2 T3T2 3517 lbfin2 4631 13564 1201 lbfin 2 T4 T3 v1v2k1 4631 R 1004 18436 R P4 P3 v1v2k 1201 psi 1014 478 lbfin2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10154E To approximate an actual sparkignition engine consider an airstandard Otto cycle that has a heat addition of 800 Btulbm of air a compression ratio of 7 and a pressure and temperature at the beginning of the compression process of 13 lbfin2 50 F Assuming constant specific heat with the value from Table F4 determine the maximum pressure and temperature of the cycle the thermal efficiency of the cycle and the mean effective pressure Solution P v 1 2 3 4 1 2 3 4 v T s State 1 v1 RT1P1 5334510 13144 14532 ft3lbm v2 v17 2076 ft3lbm The compression process reversible adiabatic so then isentropic The constant s is implemented with Eq625 and Eq624 P2 P1v1v2k 13714 1982 lbfin2 T2 T1v1v2k1 510704 11107 R The combustion process with constant volume qH 800 Btulbm T3 T2 qHCV0 11107 8000171 5789 R P3 P2T3T2 1982 578911107 1033 lbfin2 Cycle efficiency from the ideal cycle as in Eq1012 ηTH 1 T1T2 1 51011107 0541 To get the mean effective pressure we need the net work wNET ηTH qH 0541 800 4328 Btulbm Pmeff v1v2 wNET 4328778 145322076144 188 lbfin2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10155E A 4 stroke 85 in3 gasoline engine runs at 2500 RPM has an efficiency of 60 The state before compression is 6 psia 500 R and after combustion it is at 4400 R Find the highest T and P in the cycle the specific heat transfer added the cycle mep and the total power produced This is a basic 4stroke Otto cycle Compression ratio from the efficiency η 1 CR1k CR 1 η104 9882 Compression T2 T1CRk1 500 9882 04 1250 R P2 P1 CRk 6 988214 1482 psia Combustion v3 v2 Highest T and P are after combustion T3 4400 R P3 P2 T3 T2 1482 4400 1250 5217 psia qH u3 u2 CvT3 T 2 0171 4400 1250 53865 Btulbm wnet ηTH qH 06 53865 32319 Btulbm Displacement and Pmeff v1 RT1P1 5334 ftlbflbmR 500 R1446 psia 30868 ft3lbm v2 19882 v1 3124 ft3lbm Pmeff v1v2 wNET 32319778 30868 3124 144 62937 psia Now we can find the power from Eq1011 W Pmeff Vdispl RPM 60 1 2 62937 85 12 2500 60 1 2 92876 lbffts 169 hp Recall 1 hp 550 lbffts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10156E A for stroke gasoline engine has a compression ratio of 101 with 4 cylinders of total displacement 75 in3 the inlet state is 500 R 10 psia and the engine is running at 2100 RPM with the fuel adding 600 Btulbm in the combustion process What is the net work in the cycle and how much power is produced Solution Overall cycle efficiency is from Eq1012 rv v1v2 ηTH 1 r1k v 1 1004 0602 wnet ηTH qH 0602 600 3612 Btulbm We also need specific volume to evaluate Eqs109 to 1011 v1 RT1 P1 5334 500 10 144 1852 ft3lbm Pmeff v1 v2 wnet v1 1 1 rv wnet 3612 1852 09 778 144 1171 psia Now we can find the power from Eq1011 W Pmeff Vdispl RPM 60 1 2 1171 75 12 2100 60 1 2 12 808 lbffts 23 hp Recall 1 hp 550 lbffts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10157E An Otto cycle has the lowest T as 520 R and the lowest P as 22 psia the highest T is 4500 R and combustion adds 500 Btulbm as heat transfer Find the compression ratio and the mean effective pressure Solution Identify states T1 520 R P1 22 psia T3 4500 R qH 500 Btulbm Combustion qH u3 u2 CvoT3 T2 500 Btulbm T2 T3 qH Cvo 4500 5000171 1576 R Compression CR rv v1v2 T2T11k1 1576 520 25 1599 Overall cycle efficiency is from Eq1012 rv v1v2 1599 ηTH 1 r1k v 1 T1T2 1 5201576 0670 wnet ηTH qH 0670 500 335 Btulbm We also need specific volume to evaluate Eqs109 to 1011 v1 RT1 P1 5334144 520 22 8755 ft3lbm v2 v1CR 05475 ft3lbm Pmeff v1 v2 wnet 335 8755 05475 Btulbm ft3lbm 778 lbfftBtu 144 inft2 2205 psia Comment This is too high a compression ratio for a standard Otto cycle Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10158E A gasoline engine has a volumetric compression ratio of 10 and before compression has air at 520 R 122 psia in the cylinder The combustion peak pressure is 900 psia Assume cold air properties What is the highest temperature in the cycle Find the temperature at the beginning of the exhaust heat rejection and the overall cycle efficiency Solution Compression Isentropic so we use Eqs624625 P2 P1v1v2k 122 1014 30645 psia T2 T1v1v2k1 520 1004 13062 R Combustion Constant volume T3 T2 P3P2 13062 90030645 3836 R Exhaust Isentropic expansion so from Eq623 T4 T3 v1v2k1 T3 1004 3836 25119 1527 R Overall cycle efficiency is from Eq1012 rv v1v2 η 1 r 1k v 1 1004 0602 Comment No actual gasoline engine has an efficiency that high maybe 35 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10159E The cycle in Problem 10157E is used in a 150 in3 engine running at 1800 RPM How much power does it produce Identify states T1 520 R P1 12 psia T3 4500 R qH 500 Btulbm Combustion qH u3 u2 CvoT3 T2 500 Btulbm T2 T3 qH Cvo 4500 5000171 1576 R Compression CR rv v1v2 T2T11k1 1576 520 25 1599 Overall cycle efficiency is from Eq1012 rv v1v2 1599 ηTH 1 r1k v 1 T1T2 1 5201576 0670 wnet ηTH qH 0670 500 335 Btulbm We also need specific volume to evaluate Eqs109 to 1011 v1 RT1 P1 5334144 520 22 8755 ft3lbm v2 v1CR 05475 ft3lbm Pmeff v1 v2 wnet 335 8755 05475 Btulbm ft3lbm 778 lbfftBtu 144 inft2 2205 psia Now we can find the power from Eq1011 W Pmeff Vdispl RPM 60 1 2 2205 lbf in2 150 in3 1800 60 1s 1 2 41 344 lbffts 75 hp 53 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10160E It is found experimentally that the power stroke expansion in an internal combustion engine can be approximated with a polytropic process with a value of the polytropic exponent n somewhat larger than the specific heat ratio k Repeat Problem 10154E but assume the expansion process is reversible and polytropic instead of the isentropic expansion in the Otto cycle with n equal to 150 First find states 2 and 3 Based on the inlet state we get v4 v1 RT1P1 533451013144 14532 ft3lbm v3 v2 v17 2076 ft3lbm After compression we have constant s leads to Eq624 and Eq625 P2 P1v1v2k 13714 1982 lbfin2 T2 T1v1v2k1 510704 11107 R Constant volume combustion T3 T2 qHCV0 11107 8000171 5789 R P3 P2T3T2 1982 578911107 1033 lbfin2 Process 3 to 4 Pv15 constant P4 P3v3v415 10331715 5578 lbfin2 T4 T3v3v405 57891705 2188 R For the mean effective pressure we need the net work and therefore the induvidual process work terms 1w2 P dv RT2 T11 14 533411107 51004778 10296 Btulbm 3w4 P dv RT4 T31 15 53342188 578905778 4938 Btulbm wNET 4938 10296 39084 Btulbm ηCYCLE wNETqH 39084700 0488 Pmeff wNETv1v2 3908477814532 2076 1695 lbfin2 Notice a smaller wNET ηCYCLE Pmeff compared to ideal cycle Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10161E In the Otto cycle all the heat transfer qH occurs at constant volume It is more realistic to assume that part of qH occurs after the piston has started its downwards motion in the expansion stroke Therefore consider a cycle identical to the Otto cycle except that the first twothirds of the total qH occurs at constant volume and the last onethird occurs at constant pressure Assume the total qH is 700 Btulbm that the state at the beginning of the compression process is 13 lbfin2 68 F and that the compression ratio is 9 Calculate the maximum pressure and temperature and the thermal efficiency of this cycle Compare the results with those of a conventional Otto cycle having the same given variables P1 13 psia T1 52767 R rV v1v2 7 q23 2 3700 4667 Btu lbm q34 1 3700 2333 Btu lbm Constant s compression Eqs62425 P2 P1v1v2k 13914 2818 lbfin2 T2 T1v1v2k1 52767904 12707 R Constant v combustion T3 T2 q23CV0 12707 46670171 4000 R P3 P2T3T2 2818 400012707 8871 lbfin2 P 4 Constant P combustion T4 T3 q34CP0 4000 2333024 4972 R Remaining expansion v5 v4 v1 v4 P4P1 T1T4 881 13 52767 4972 7242 T5 T4v4v5k1 49721724204 2252 R qL CV0T5T1 01712252 52767 2949 Btulbm ηTH 1 qLqH 1 2949700 0579 Standard Otto cycle ηTH 1 904 0585 1 2 3 4 T s s s v v 5 1 2 3 4 s s P v 5 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10162E A diesel engine has a bore of 4 in a stroke of 43 in and a compression ratio of 191 running at 2000 RPM revolutions per minute Each cycle takes two revolutions and has a mean effective pressure of 200 lbfin2 With a total of 6 cylinders find the engine power in Btus and horsepower hp Solution Work from mean effective pressure Eq109 Pmeff wnet vmax vmin wnet Pmeff vmax vmin The displacement is V πBore2 025 S π 42 025 43 54035 in3 Work per cylinder per power stroke Eq1010 W PmeffVmax Vmin 200 54035 12 778 11575 Btucycle Only every second revolution has a power stroke so we can find the power see also Eq1011 W W Ncyl RPM 05 cycles min min 60 s Btu cycle 11575 6 2000 05 160 11575 Btus 11575 3600254443 hp 164 hp The conversion factor from Btus to hp is from Table A1 under power Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10163E A super charger is used for a diesel engine so intake is 30 psia 580 R The cycle has compression ratio of 181 and the mean effective pressure is 120 psia If the engine is 600 in3 running at 200 RPM find the power output Solution The power is from Eq1011 W m wnet Ncyl RPM 160 m Pmeff v1 v2 Ncyl RPM60 Pmeff Vdispl RPM60 120 lbf in2 600 in3 200 60 1s 20 000 lbffts 257 Btus 366 hp Comment multiply with ½ if this is a four stroke engine Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10164E At the beginning of compression in a diesel cycle T 540 R P 30 lbfin2 and the state after combustion heat addition is 2600 R and 1000 lbfin2 Find the compression ratio the thermal efficiency and the mean effective pressure Solution Compression process isentropic from Eq628 or p295 P2 P3 1000 lbfin2 v1v2 P2P11k 10003007143 1224 T2 T1P2P1k1k 540100030 02857 14706 R Expansion process isentropic first get the volume ratios v3v2 T3T2 260014706 1768 v4v3 v1v3 v1v2v2v3 12241768 6923 The exhaust temperature follows from state 3 and constant s Eq624 T4 T3v3v4k1 2600692304 1199 R qL CVT4 T1 01711199540 1127 Btulbm qH h3 h2 CPT3 T2 0242600 14706 2711 Btulbm Overall performance η 1 qLqH 1 1127 2711 05843 wnet qnet 2711 1127 1584 Btulbm vmax v1 RT1P1 5334 54030 144 66675 ft3lbm vmin vmaxv1v2 66675 1224 0545 ft3lbm Pmeff vmax vmin wnet 1584 66675 0545 778 144 1398 lbfin 2 P v 1 2 3 4 s s 1 2 3 4 v T s P Remark This is a too low compression ratio for a practical diesel cycle Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10165E The cut off ratio is the ratio of v3v2 see Fig 1018 which is the expansion while combustion occurs at constant pressure Determine this ratio for the cycle in Problem 10164E Compression process isentropic from Eqs62325 T2 T1P2P1k1k 540100030 02857 14706 R Cutoff ratio v3 v2 T3 T2 260014706 1768 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10166E A diesel cycle has state 1 as 14 lbfin2 63 F and a compression ratio of 20 For a maximum temperature of 4000 R find the cycle efficiency Diesel cycle P1 14 psia T1 52267 R CR v1v2 20 From the inlet state and the compression we get T2 T1v1v2k1 52267 R 2004 17324 R v1 533452267 14144 13829 ft3lbm v2 13829 20 06915 ft3lbm Constant pressure combustion relates v3 and T 3 v3 v2T3T2 06915 400017324 15966 ft3lbm Expansion process isentropic and v4 v1 from Eq624 T4 T3 v3 v404 4000 R 15966 1382904 16866 R Cycle net work and efficiency ηTH 1 kT3 T2 T4 T1 1 16866 52267 14 4000 17324 0633 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10167E Consider an ideal Stirlingcycle engine in which the pressure and temperature at the beginning of the isothermal compression process are 147 lbfin2 80 F the compression ratio is 6 and the maximum temperature in the cycle is 2000 F Calculate the maximum pressure in the cycle and the thermal efficiency of the cycle with and without regenerators T T v v 1 2 3 4 P v 1 2 3 4 T T v v s T Ideal Stirling cycle T1 T2 80 F P1 147 lbfin 2 v1 v2 6 T3 T4 2000 F Isothermal compression heat goes out T1 T2 P2 P1 v1v2 1476 882 w12 q12 RT1 ln v2 v1 5334 778 540 ln 6 663 Btulbm Constant volume heat addition V2 V3 P3 P2 T3T2 882 2460 540 4018 lbfin 2 q23 CV0T3T2 0171 2460540 3283 Btulbm Isothermal expansion heat comes in w34 q34 RT3 ln v4v3 5334778 2460 ln 6 3022 Btulbm wNET 3022 663 2359 Btulbm Efficiency without regenerator q23 and q34 are coming in from source ηNO REGEN 2359 30223283 0374 Efficiency with regenerator Now only q34 is coming in from source ηWITH REGEN 2359 3022 0781 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10168E Air in a pistoncylinder goes through a Carnot cycle in which TL 560 R and the total cycle efficiency is η 23 Find TH the specific work and volume ratio in the adiabatic expansion for constant Cp Cv Carnot cycle efficiency from Eq55 η 1 TLTH 23 TH 3 TL 3 560 1680 R Adiabatic expansion 3 to 4 Pvk constant work from Eq629 n k 3w4 P4v4 P3v31 k R1kT4 T3 u3 u4 CvT3 T4 0171 BtulbmR 1680 540 R 19494 Btulbm v4v3 T3T41k 1 325 156 P v 1 2 3 4 s s T T 1 4 T s 2 3 T T H L Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10169E Do the previous problem 10168E using Table F5 Air in a pistoncylinder goes through a Carnot cycle in which TL 560 R and the total cycle efficiency is η 23 Find TH the specific work and volume ratio in the adiabatic expansion for constant Cp Cv Carnot cycle η 1 TLTH 23 TH 3 TL 3 560 1680 R 3w4 u3 u4 302124 95589 206535 Btulbm Adiabatic expansion 3 to 4 s4 s3 Eq619 s o T4 s o T3 R ln P4 P3 Table F5 for standard entropy P4 P3 exps o T4 s o T3R exp164852 192338 5334778 0018151 Ideal gas law then gives v4 v3 T4 T3 P3 P4 560 1680 1 0018151 1836 P v 1 2 3 4 s s T T 1 4 T s 2 3 T T H L Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Atkinson and Miller cycles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10170E An Atkinson cycle has state 1 as 20 psia 540 R compression ratio of 9 and expansion ratio of 14 Find the needed heat release in the combustion Solve the problem with constant heat capacity Compression 1 to 2 s2 s1 From Eq 624 and Eq 625 T2 T1 v1v2k1 540 9 04 13004 R P2 P1v1v2k 20 9 14 4335 psi The expansion backwards from P4 P1 gives P3 P4 v4v3k P4 CRk 20 1414 80465 psi Combustion 2 to 3 at constant volume v3 v2 P3 P2 T3T2 Energy Eq u3 u2 qH qH Cv T3 T2 Solve for T3 from process equation T3 T2 P3 P2 13004 R 80465 4335 24138 R Solve for qH from the energy equation qH Cv T3 T2 0171 24138 13004 1904 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10171E Assume we change the Otto cycle in Problem 10152E to an Atkinson cycle by keeping the same conditions and only increase the expansion to give a different state 4 Find the expansion ratio and the cycle efficiency Solution Solve the problem with constant heat capacity Compression 1 to 2 s2 s1 From Eq624 and Eq625 T2 T1 v1v2k1 540 1004 13564 R P2 P1v1v2k 14 1014 3517 psi Combustion 2 to 3 at constant volume u3 u2 qH T3 T2 qHCv 13564 5600171 4631 R P3 P2 T3T2 3517 lbfin2 4631 13564 1201 lbfin 2 The expansion should now bring pressure down so P4 P 1 P4 P1 P3 v3v4k P3 CRk CR P3 P4 1k CR 1201 14071429 2404 So now the efficiency becomes η 1 k CR CR1 CRk CRk 1 1 14 2404 10 2404 k 10 k 0676 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10172E An Atkinson cycle has state 1 as 20 psia 540 R compression ratio of 9 and expansion ratio of 14 Find the mean effective pressure Solve the problem with constant heat capacity Compression 1 to 2 s2 s1 From Eq624 and Eq625 T2 T1 v1v2k1 540 9 04 13004 R P2 P1v1v2k 20 9 14 4335 psia The expansion backwards from P4 P1 gives P3 P4 v4v3k P4 CRk 20 1414 80465 psia Combustion 2 to 3 at constant volume v3 v2 P3 P2 T3T2 Energy Eq u3 u2 qH qH Cv T3 T2 Solve for T3 from process equation T3 T2 P3 P2 13004 80465 4335 24138 R Solve for qH from the energy equation qH Cv T3 T2 0171 24138 13004 1904 Btulbm The efficiency and net work become η E A 0623 TH 1 k 1 14 CR CR1 CRk CRk 1 14 9 14 k 9 k wAnetE A ηATHE A qAHE A 0623 1904 1186 Btulbm Displacement we take this as vA4E A vA2E A and PAmeffE vA1E A RTA1E APA1E A 5334 54020 144 1000 ftA3E Albm vA2E A 19 vA1E A 1111 ftA3E Albm vA4E A 14 vA2E A 155575 ftA3E Albm PAmeffE A A wNET Ev4 v2 E A A 1186 155575 1111E A A778 144E A 4435 psia Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10173E A Miller cycle has state 1 as 20 psia 540 R compression ratio of 9 and expansion ratio of 14 If PA4E A is 30 psia find the heat release in the combustion Solution Solve the problem with constant heat capacity Compression 1 to 2 sA2E A sA1E A From Eq624 and Eq625 TA2E A TA1E A vA1E AvA2E AAk1E A 540 9 A04E A 13004 R PA2E A PA1E AvA1E AvA2E AAkE A 20 9 A14E A 4335 psia The expansion backwards from PA4E A gives PA3E A PA4E A vA4E AvA3E AAkE A PA4E A CRAkE A 30 14A14E A 1207 psia Combustion 2 to 3 at constant volume uA3E A uA2E A qAHE TA3E A TA2E A qAHE ACAvE A and PA3E A PA2E A TA3E ATA2E A TA3E A TA2E A PA3E A PA2E A 13004 1207 4335 36207 R qAHE A CAvE A TA3E A TA2E A 0171 36207 13004 3968 Btulbm P v 1 2 3 4 s s 5 1 2 3 4 v T s P 5 v Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10174E A Miller cycle has state 1 as 20 psia 540 R compression ratio of 9 and a heat release of 430 Btulbm Find the needed expansion ratio so PA4E A is 30 psia Solve the problem with constant heat capacity Compression 1 to 2 sA2E A sA1E A From Eq624 and Eq625 TA2E A TA1E A vA1E AvA2E AAk1E A 540 9 A04E A 13004 R PA2E A PA1E AvA1E AvA2E AAkE A 20 9 A14E A 4335 psia Combustion 2 to 3 at constant volume uA3E A uA2E A qAHE TA3E A TA2E A qAHE ACAvE A 13004 430 0171 3815 R PA3E A PA2E A TA3E ATA2E A 4335 3815 13004 12718 psia The expansion should now bring pressure down to PA4E A PA4E A PA3E A vA3E AvA4E AAkE A PA3E A CRAkE A CR PA3E A PA4E A A1kE A CR 12718 30A071429E A 1453 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Exergy Combined Cycles and Review Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10175E Consider the Brayton cycle in problem 10144E Find all the flows and fluxes of exergy and find the overall cycle secondlaw efficiency Assume the heat transfers are internally reversible processes and we then neglect any external irreversibility Solution Efficiency is from Eq101 η AW E ANET AQ E AH A wnet EqH E A 1 rAk1k EpE A 1 16A0414E A 0547 from the required power we can find the needed heat transfer AQ HE A AW netE A η 14 000 0547 25 594 Btus AmE A AQ HE A qAHE A 25 594 Btus 400 Btulbm 6399 lbms Temperature after compression is TA2E A TA1E A rAk1k EpE A 51967 16A0414E A 1148 R The highest temperature is after combustion TA3E A TA2E A qAHE ACApE A 1148 A400 024E A 2815 R For the exit flow I need the exhaust temperature TA4E A TA3E A rApE Ak1kE A 2815 16A02857E A 12748 R The high T exergy input from combustion is AΦ E AH AmE AψA3E A ψA2E A AmE AhA3E A hA2E A TsA3E A sA2E A 6399 400 53667 024 ln A2815 1148E A 17 895 Btus Since the low T exergy flow out is lost the second law efficiency is ηII AW E ANETAΦ E AH 14 000 17 895 0782 AΦ E Aflow outE A AmE AψA4E A ψAoE A AmE AhA4E A hAoE A TsA4E A sAoE A 6399 02412748 5367 5367 024 ln A12748 5367E A 4205 Btus AΦ E Aflow inE A AmE AψA1E A ψAoE A AmE AhA1E A hAoE A TsA1E A sAoE A 6399 02460 77 5367 024 ln A5197 5367E A 42 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10176E The Brayton cycle in problem 10141E has a heat addition of 350 Btulbm What is the exergy increase in this process Solution Combustion hA3E A hA2E A qAHE A A2E AwA3E A 0 and TAmaxE A TA3E A 2500 R TA2E A TA3E A qAHE ACAP E A 2500 350024 104167 R Now the exergy increase from 2 to 3 becomes PA3E A PA2E A ψ3 ψ2 hA3E A hA2E A TA0E AsA3E A sA2E A qAHE A TA0E A CAP E A lnAT3T2E A 350 5367 024 ln A 2500 104167E A 2372 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10177E Solve Problem 10149E assuming an isentropic turbine efficiency of 85 and a nozzle efficiency of 95 Solution CV Turbine hAiE A 560588 Btulbm sAo TiE A 199765 Btulbm R sAesE A sAiE A Then from Eq619 sAo TesE A sAo TiE A R lnPAeE APAiE A 199765 A5334 778E A ln 35220 18716 A Btu lbm RE Table F5 TAesE A 1382 R hAesE A 33827 Btulbm Energy eq wATsE A hAiE A hAesE A 560588 33827 2223 Btulbm Eq727 wATACE A wATsE A ηATE A 18896 hAiE A hAeACE A hAeACE A 3716 Table F5 TAeACE A 1509 R sAo TeE A 18947 Btulbm R CV Nozzle hAiE A 3716 Btulbm sAo TiE A 18947 Btulbm R sAesE A sAiE A Then from Eq619 sAo TesE A sAo TiE A R lnPAeE APAiE A 18947 A5334 778E A ln A147 35E A 18352 A Btu lbm RE Table F5 TAesE A 11996 R hAesE A 2913 Btulbm Energy Eq 12VAes 2 AE E hAiE A hAesE A 3716 2913 803 Btulbm Eq730 12VAeAC 2 AE E 12VAes 2 AE E ηANOZE A 7629 Btulbm VeAC A 2 25037 7629EA 1954 fts Recall 1 Btulbm 25 037 ftA2E AsA2E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10178E Solve Problem 10141E with variable specific heats using Table F5 In a Brayton cycle the inlet is at 540 R 14 psia and the combustion adds 350 Btulbm The maximum temperature is 2500 R due to material considerations Find the maximum permissible compression ratio and for that the cycle efficiency using cold air properties Solution Combustion hA3E A hA2E A qAHE A A2E AwA3E A 0 and TAmaxE A TA3E A 2500 R hA2E A hA3E A qAHE A 645721 350 29572 Btulbm From Table F5 find TA2E A from hA2E A TA2E A 12168 R sAo T2E A 183883 TA1E A 540 R sAo T1E A 163979 BtulbmR Reversible adiabatic compression leads to constant s from Eq619 PA2E A PA1E A exp sAo T2E A sAo T1E AR exp290313 18231 Reversible adiabatic expansion leads to constant s from Eq619 sAo T4E A sAo T3E A R lnPA4E APA3E A 203391 A5334 778E A ln A 1 18231E A 183487 BtulbmR From Table F51 by linear interpolation TA4E A 11979 R hA4E A 290905 Btulbm wATE A hA3E A hA4E A 645721 290905 35482 Btulbm wACE A hA2E A hA1E A 29572 12918 16654 Btulbm wAnetE A wATE A wACE A 35482 16654 18828 Btulbm η wAnetE A qAHE A 18828 350 0538 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10179E Consider an ideal gasturbine cycle with two stages of compression and two stages of expansion The pressure ratio across each compressor stage and each turbine stage is 8 to 1 The pressure at the entrance to the first compressor is 14 lbfin2 the temperature entering each compressor is 70 F and the temperature entering each turbine is 2000 F An ideal regenerator is also incorporated into the cycle Determine the compressor work the turbine work and the thermal efficiency of the cycle This is as in Fig 106 REG COMP TURB TURB COMP CC CC IC 1 2 4 10 6 7 8 9 5 PA2E APA1E A PA4E APA3E A PA6E APA7E A PA8E APA9E A 80 PA1E A 14 lbfinA2E TA1E A TA3E A 70 F TA6E A TA8E A 2000 F Assume const specific heat sA2E A sA1E A and sA4E A sA3E TA4E A TA2E A TA1E APA2E APA1E AA k1 k E A 529678A02857E A 9594 R 1 2 3 s T 4 5 6 7 8 9 10 Total compressor work wACE A 2 wA12E A 2CAP0E ATA2E A TA1E A 2 0249594 52967 2063 Btulbm Also sA6E A sA7E A and sA8E A sA9E TA7E A TA9E A TA6E A P7 EP6 E A k1 k E A 245967A 1 8 E A 02857E A 13579 R Total turbine work wATE A 2 wA67E A 2CAP0E ATA6E A TA7E A 2 024245967 13579 52885 Btulbm wANETE A 52885 2063 32255 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Ideal regenerator TA5E A TA9E A TA10E A TA4E qAHE A hA6E A hA5E A hA8E A hA7E A 2CAP0E ATA6E A TA5E A 2 024245967 13579 wATE A 52885 Btulbm ηATHE A wANETE AqAHE A 3225552885 061 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10180E Consider an ideal airstandard diesel cycle where the state before the compression process is 14 lbfin2 63 F and the compression ratio is 20 Find the maximum temperature by iteration in the cycle to have a thermal efficiency of 60 Diesel cycle PA1E A 14 TA1E A 52267 R vA1E AvA2E A 20 ηATHE A 060 From the inlet state and the compression we get TA2E A TA1E AvA1E AvA2E AAk1E A 5226720A04E A 17324 R vA1E A A533452267 14144E A 13829 ftA3E Albm vA2E A A13829 20E A 06915 ftA3E Albm Constant pressure combustion relates vA3E A and TA3E vA3E A vA2E ATA3E ATA2E A 06915TA3E A17324 0000399 TA3E The expansion then gives TA4E A interms of TA3E A T3 ET4 E A A v4 Ev3 E AA k1E A A 13829 0000399 T3 E AA 04E A TA4E A 00153 TA 14 3E Now these Ts relate to the given efficiency ηATHE A 060 1 A T4T1 EkT3T2E A 1 A 00153 T 14 3 52267 E14T317324E 00153 TA 14 3E A 056 TA3E A 4475 0 Trial and error on this nonlinear equation 5100 R LHS 3554 5500 R LHS 504 5450 R LHS 05 Linear interpolation TA3E A 5455 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 10181E Repeat Problem 10179E but assume that each compressor stage and each turbine stage has an isentropic efficiency of 85 Also assume that the regenerator has an efficiency of 70 TA4SE A TA2SE A 9594 R wACSE A 2063 TA7SE A TA9SE A 13579 R wATSE A 52885 wACE A wASCE AηASCE A 2427 Btulbm wA12E A wA34E A 24272 12135 Btulbm TA2E A TA4E A TA1E A wA12E ACAP0E 52967 12135024 10353 R 1 2 3 s T 4 5 6 7 8 9 4S 9S 7S 2S wATE A ηATE A wATSE A 085 52885 4495 Btulbm Energy Eq for actual turbines TA7E A TA9E A TA6E A wA67E ACAP0E A 245967 44952024 1523 R Regenerator efficiency ηAREGE A A h5 h4 Eh9 h4 E A A T5 T4 ET9 T4 E A A T5 10353 E1523 10353E A 07 TA5E A 13767 R The two combustors heat addition qAHE A CAP0E ATA6E A TA5E A CAP0E ATA8E A TA7E A 024245967 13767 024245967 1523 4847 Btulbm wANETE A wATE A wACE A 4495 2427 2068 Btulbm ηATHE A wANETE AqAHE A 20684847 0427 Updated June 2013 8e SOLUTION MANUAL CHAPTER 11 Fundamentals of Thermodynamics Borgnakke Sonntag Borgnakke Sonntag Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CONTENT SUBSECTION PROB NO InText concept questions aj Concept Problems 112 Mixture composition and properties 1325 Simple processes 2650 Entropy generation 5166 Airwater vapor mixtures 6783 Tables and formulas or psychrometric chart 84108 Psychrometric chart only 109122 Availability exergy in mixtures 123128 Review Problems 129144 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful InText Concept Questions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11a Are the mass and mole fractions for a mixture ever the same Generally not If the components all had the same molecular mass the mass and mole fractions would be the same 11b For a mixture how many component concentrations are needed A total of N1 concentrations are needed N equals total number of components whether mass or mole fractions They must sum up to one so the last one is by default 11c Are any of the properties P T v for oxygen and nitrogen in air the same In any mixture under equilibrium T is the same for all components Each species has its own pressure equal to its partial pressure Pi The partial volume for a component is vi Vmi and V is the same for all components so vi is not Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11d If I want to heat a flow of a 4 component mixture from 300 to 310 K at constant P how many properties and which ones do I need to know to find the heat transfer You need to know the flow rate the four mass fractions and the component specific heat values or the h values at both temperatures 11e To evaluate the change in entropy between two states at different T and P values for a given mixture do I need to find the partial pressures Not necessarily provided it is an ideal gas If the mixture composition does not change then the mixture can be treated as a pure substance where each of the partial pressures is a constant fraction of the total pressure Eq1110 and the changes in u h and s can be evaluated with the mixture properties as in Eqs 112024 If constant specific heat is an inappropriate model to use then u h and a standard entropy must be evaluated from expressions as in Eqs111112 and 1116 this is precisely what is done to make the air tables A7 from the nitrogen oxygen and argon properties If the substance is not an ideal gas mixture then the properties will depend on the partial pressures Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11f What happens to relative and absolute humidity when moist air is heated Relative humidity decreases while absolute humidity remains constant See Figs 118 and 119 11g If I cool moist air do I reach the dew first in a constantP or constantV process The constantvolume line is steeper than the constantpressure line see Fig 113 Saturation in the constantP process is at a higher T 11h What happens to relative and absolute humidity when moist air is cooled Relative humidity increase while absolute humidity remains constant until we reach the dew point See Figs 118 and 119 If we cool below the dew point the relative humidity stays at 100 and the absolute humidity humidity ratio drops as water condenses to liquid or freezes to solid and drops out of the gas mixture s T v C P C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11i Explain in words what the absolute and relative humidity expresses Absolute humidity is the ratio of the mass of vapor to the mass of dry air It says how much water is there per unit mass of dry air Relative humidity is the ratio of the mole fraction of vapor to that in a saturated mixture at the same T and P It expresses how close to the saturated state the water is 11j An adiabatic saturation process changes Φ ω and T In which direction Relative humidity and absolute humidity increase and temperature decreases Why does the temperature decrease The energy to evaporate some liquid water to go into the gas mixture comes from the immediate surroundings to the liquid water surface where water evaporates look at the dashed curve in Fig 119 The moist air and the liquid water both cool down Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Conceptstudy Guide Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 111 Equal masses of argon and helium are mixed Is the molecular mass of the mixture the linear average of the two individual ones No The individual molecular masses must be combined using the mole fractions as in Mmix yjM j Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 112 A constant flow of pure argon and pure helium are mixed to produce a flow of mixture mole fractions 025 and 075 respectively Explain how to meter the inlet flows to ensure the proper ratio assuming inlet pressures are equal to the total exit pressure and all temperatures are the same The inlet flow rate in terms of mass or moles is the same as the exit rate for each component in the mixture Since the inlet P for each component is the same as the total exit P which is the sum of the partial pressures if ideal gas then the volume flow rates in and out are different for each species P V i m i RiT n i RT P V tot n tot RT We can therefore meter the volume flow rate V i to be proportional to n i for each line of the inlet flows From these two equations we can get the ratio as V i V tot n i n tot yi Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 113 For a gas mixture in a tank are the partial pressures important Yes The sum of the partial pressures equals the total pressure and if they are ideal gases the partial pressures are equal to the mole fraction times the total pressure so Pi yi P and Σ Pi Σ yi P P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 114 An ideal mixture at T P is made from ideal gases at T P by charging them into a steel tank Assume heat is transferred so T stays the same as the supply How do the properties P v and u for each component change up down or constant Solution Ideal gas u uT so constant P drops from P to partial Pi v increases from v at P to v at Pi same T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 115 An ideal mixture at T P is made from ideal gases at T P by flow into a mixing chamber without any external heat transfer and an exit at P How do the properties P v and h for each component change up down or constant Solution Ideal gas hmix Σ cihiout Σ cihiin same function of T so constant T and then also constant hi P drops from P to partial Pi v increases from v at P to v at Pi same T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 116 If a certain mixture is used in a number of different processes do I need to consider partial pressures No If the mixture composition stays the same the pressure for each component which is a partial pressure is the same fraction of the total pressure thus any variation follows the total pressure Recall air is a mixture and we can deal with most processes involving air without knowledge about its composition However to make the air properties we do need to deal with the composition but only once Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 117 Why is it that I can use a set of tables for air which is a mixture without dealing with its composition As long as the composition is fixed any property is a fixed weighted average of the components properties and thus only varies with T and total P A process that will cool air to saturation and condensation can not be handled by the air tables In such a process the composition of the liquid and vapor mixtures are different Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 118 Develop a formula to show how the mass fraction of water vapor is connected to the humidity ratio By definition the mass concentration is c mv ma mv mvma 1 mv ma ω 1 ω and since ω is small then 1 ω 1 and c is close to ω but not equal to Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 119 For air at 110oC and 100 kPa is there any limit on the amount of water it can hold No Since Pg 1433 kPa at 110oC and Pv 100 kPa ω can be infinity ω 0622 Pv Pa 0622 Pv P Pv As Pv approaches P w goes towards infinity Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1110 Can moist air below the freezing point say 5oC have a dew point Yes At the dew point water would begin to appear as a solid It snows Since it is frost forming on surfaces rather than dew you can call it frost point The contrails after the jets are tiny ice particles formed due to the very low 40 C temperature at high altitudes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1111 Why does a car with an airconditioner running often have water dripping out The cold evaporator that cools down an air flow brings it below the dew point temperature and thus condenses water out Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1112 Moist air at 35oC ω 00175 and Φ 50 should be brought to a state of 20oC ω 001 and Φ 70 Do I need to add or subtract water The humidity ratio absolute humidity expresses how much water vapor is present in the mixture ω mv ma so to decrease ω we must subtract water from the mixture The relative humidity expresses how close to the saturated state the vapor is as Φ Pv P g and not about how much water there is Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Mixture composition and properties Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1113 If oxygen is 21 by mole of air what is the oxygen state P T v in a room at 300 K 100 kPa of total volume 60 m3 The temperature is 300 K The partial pressure is P yPtot 021 100 21 kPa At this T P vO2 RTPO2 02598 kJkgK 300 K 21 kPa 3711 m3kg Remark If we found the oxygen mass then mO2vO2 V 60 m3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1114 A 3 L liquid mixture is 13 of each of water ammonia and ethanol by volume Find the mass fractions and total mass of the mixture Each component has a partial volume of 1 L 0001 m3 mwater Vvf 0001 m3 0001 m3kg 1 kg mamm Vvf 0001 m3 604 kgm3 0604 kg methanol Vvf 0001 m3 783 kgm3 0783 kg The vf for water is from Table B11 and vf 1ρ is from A4 Total mass is m 1 0604 0783 2387 kg cwater mwaterm 1 2387 0419 camm mammm 0604 2387 0253 cethanol methanolm 0783 2387 0328 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1115 A flow of oxygen and one of nitrogen both 300 K are mixed to produce 1 kgs air at 300 K 100 kPa What are the mass and volume flow rates of each line For the mixture M 02132 07928013 2885 For O2 c 021 32 2885 02329 For N2 c 079 28013 2885 07671 Since the total flow out is 1 kgs these are the component flows in kgs Volume flow of O2 in is V cm v cm RT P 02329 kgs 02598 kJkgK 300 K 100 kPa 01815 m3s Volume flow of N2 in is V cm v cm RT P 07671 kgs 02968 kJkgK 300 K 100 kPa 06830 m3s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1116 A gas mixture at 20C 125 kPa is 50 N2 30 H2O and 20 O2 on a mole basis Find the mass fractions the mixture gas constant and the volume for 5 kg of mixture Solution The conversion follows the definitions and identities From Eq113 ci yi Mi yjM j From Eq115 Mmix yjMj 0528013 0318015 0231999 140065 54045 63998 25811 cN2 140065 25811 05427 cH2O 54045 25811 02094 cO2 63998 25811 02479 sums to 1 OK From Eq1114 Rmix RMmix 83145 kJkmolK 25811 kgkmol 03221 kJkg K V mRmix TP 5 kg 03221 kJkgK 39315 K125 kPa 5065 m 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1117 A mixture of 60 N2 30 Ar and 10 O2 on a mass basis is in a cylinder at 250 kPa 310 K and volume 05 m3 Find the mole and the mass fractions and the mass of argon Solution Total mixture PV m RmixT Mixture composition cs for N2 Ar O2 06 03 01 From Eq1115 Rmix ciRi 06 02968 03 02081 01 02598 026629 kJkg K m PVRmixT 026649 kJkgK 310 K 250 kPa 05 m3 1513 kg mar 03 m 0454 kg From Eq114 yi ci Mi cjM j ci Mi ciMi yi N2 06 28013 002141 0668 Ar 03 39948 000751 0234 O2 01 31999 0003125 0098 round up 0032055 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1118 A slightly oxygenated air mixture is 69 N2 1 Ar and 30 O2 on a mole basis Assume a total pressure of 101 kPa and find the mass fraction of oxygen and its partial pressure Solution From Eq 113 ci yi Mi yjM j Eq115 Mmix yjMj 06928013 00139948 0331999 29328 cN2 06928013 29328 0659 cAr 00139948 29328 0014 cO2 0331999 29328 0327 sums to 1 OK From Eq1110 PO2 yO2 P 03 101 303 kPa Rmix RMMIX 83145 kJkmolK 29328 kgkmol 02835 kJkg K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1119 A new refrigerant R407 is a mixture of 23 R32 25 R125 and 52 R134a on a mass basis Find the mole fractions the mixture gas constant and the mixture heat capacities for this new refrigerant Solution From the conversion in Eq114 we get ci Mi ciMi yi R32 023 52024 0004421 0381 R125 025 120022 0002083 0180 R134a 052 10203 00050965 0439 00116005 Eq1115 Rmix ciRi 023 01598 025 006927 052 008149 009645 kJkg K Eq1123 CP mix ci CP i 023 0822 025 0791 052 0852 08298 kJkg K Eq1121 Cv mix ciCv i 023 0662 025 0721 052 0771 07334 kJkg K CP mix Rmix Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1120 In a car engine gasoline assume octane C8H18 is evaporated and then mixed with air in a ratio of 115 by mass In the cylinder the mixture is at 750 kPa 650 K when the spark fires For that time find the partial pressure of the octane and the specific volume of the mixture Assuming ideal gas the partial pressure is Pi yi P and cC8H18 116 00625 From Eq 114 yi ci Mi cjM j yC8H18 00625114232 00625114232 093752897 0008186 PC8H18 0008186 750 614 kPa The gas constant from Eq1115 Rmix ciRi 00625 007279 09375 0287 027361 kJkg K vmix RmixTP 027361 650750 02371 m3kg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1121 A 100 m3 storage tank with fuel gases is at 20C 100 kPa containing a mixture of acetylene C2H2 propane C3H8 and butane C4H10 A test shows the partial pressure of the C2H2 is 15 kPa and that of C3H8 is 65 kPa How much mass is there of each component Solution Assume ideal gases then the ratio of partial to total pressure is the mole fraction y PPtot yC2H2 15100 015 yC3H8 65100 065 yC4H10 20100 020 ntot PV RT 100 kPa 100 m3 831451 kJkmolK 29315 K 41027 kmol mC2H2 nMC2H2 yC2H2 ntot MC2H2 01541027 kmol 26038 kgkmol 16024 kg mC3H8 nMC3H8 yC3H8 ntot MC3H8 06541027 kmol 44097 kgkmol 117597 kg mC4H10 nMC4H10 yC4H10 ntot MC4H10 02041027 kmol 58124 kgkmol 47693 kg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1122 A 2 kg mixture of 25 N2 50 O2 and 25 CO2 by mass is at 150 kPa and 300 K Find the mixture gas constant and the total volume Solution From Eq1115 Rmix ciRi 025 02968 05 02598 025 01889 02513 kJkg K Ideal gas law PV mRmixT V mRmixTP 2 kg 02513 kJkgK 300 K 150 kPa 1005 m 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1123 A diesel engine sprays fuel assume nDodecane C12H26 M 17034 kgkmol into the combustion chamber so it becomes filled with an amount of 1 mol fuel per 88 mol air Find the fuel fraction on a mass basis and the fuel mass for a chamber that is 05 L at 800 K and total pressure of 4000 kPa From Eq 113 ci yi Mi yjM j cfuel 189 17034 8889 2897 189 17034 006263 Use ideal gas for the fuel vapor mfuel RfuelT PfuelV 8314517034 kJkgK 800 K 189 4000 kPa 00005 m3 0575 g We could also have done the total mass and then used the mass fraction Eq115 Mmix yjMj 189 17034 8889 2897 30558 Rmix RMmix b 83145 30558 027209 m PV RmixT 4000 00005 027209 800 0009188 kg mfuel cfuel m 006263 0009188 kg 0000575 kg 0575 g Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1124 A new refrigerant R410A is a mixture of R32 and R125 in a 11 mass ratio What are the overall molecular mass the gas constant and the ratio of specific heats for such a mixture Eq115 M yjMj 1 cj Mj 120022 1 05 52024 05 72586 Eq1115 Rmix ciRi 05 01598 05 006927 01145 kJkg K RMMIX 83145 kJkmolK 72586 kgkmol same from Eq1114 Eq1123 CP mix ci CP i 05 0822 05 0791 08065 kJkg K Eq1121 CV mix ciCV i 05 0662 05 0722 0692 kJkg K CP mix Rmix kmix CP mix CV mix 08065 0692 11655 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1125 Do Problem 1124 for R507a which is 11 mass ratio of R125 and R143a The refrigerant R143a has molecular mass of 84041 kgkmol and Cp 0929 kJkg K Refrigerant R143a is not in Table A5 so R RM 83145 kJkmolK 84041 kgkmol 0098934 kJkgK CV Cp R 0929 0098934 08301 kJkgK Eq115 M yjMj 1 cj Mj 84041 1 05 120022 05 98859 Eq1115 Rmix ciRi 05 006927 05 0098934 00841 kJkg K RMMIX 83145 kJkmolK 98859 kgkmol same this is from Eq1114 Eq1123 CP mix ci CP i 05 0791 05 0929 086 kJkg K Eq1121 CV mix ciCV i 05 0722 05 08301 0776 kJkg K CP mix Rmix kmix CP mix CV mix 086 0776 1108 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Simple processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1126 A rigid container has 1 kg CO2 at 300 K and 1 kg argon at 400 K both at 150 kPa Now they are allowed to mix without any heat transfer What is final T P No Q No W so the energy equation gives constant U Energy Eq U2 U1 0 mCO2u2 u1CO2 mAru2 u1Ar mCO2Cv CO2T2 T1CO2 mArCv ArT2 T1Ar 10653 10312 kJK T2 10653300 10312400 kJ T2 3323 K V V1 VCO2 VAr mCO2RCO2TCO2P mArRArTArP 101889300 102081400 kJ 150 kPa 0932 73 m3 Pressure from ideal gas law and Eq1115 for R mR 1 kg 01889 kJkgK 1 kg 02081 kJkgK 0397 kJK P2 mRTV 0397 kJK 3323 K 0932 73 m3 1414 kPa CO Ar 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1127 At a certain point in a coal gasification process a sample of the gas is taken and stored in a 1L cylinder An analysis of the mixture yields the following results Component H2 CO CO2 N2 Percent by mass 2 45 28 25 Determine the mole fractions and total mass in the cylinder at 100 kPa 20C How much heat transfer must be transferred to heat the sample at constant volume from the initial state to 100C Solution Determine mole fractions from Eq114 yi ci Mi cjM j cj Mj 002 2016 045 2801 028 4401 025 28013 0009921 0016065 0006362 000892 0041268 kmolkg Mmix 1 cjMj 10041268 24232 kgkmol From Eq114 yH2 0009921 24232 02404 yCO 0016065 24232 03893 yCO2 0006362 24232 01542 yN2 000892 24232 02161 Rmix RMmix 83145 kJkmolK 24232 kgkmol 034312 kJkgK m PV RT 034312 kJkgK 29315 K 100 kPa 103m3 9942104 kg CV0 MIX ci CV0 i 002 10085 045 0744 028 0653 025 0745 09056 kJkg K 1Q2 U2 U1 mCV0T2T1 9942 104 kg 09056 kJkgK 10020 K 00720 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1128 The mixture in Problem 1122 is heated to 500 K with constant volume Find the final pressure and the total heat transfer needed using Table A5 Solution CV Mixture of constant volume Process V constant 1W2 P dV 0 Energy Eq 1Q2 mu2 u1 m CVmix T2 T1 Ideal gas PV mRT P2 P1T2 T1V1V2 P2 P1T2T1 150 kPa 500300 250 kPa From Eq1121 CVmix ciCV i 025 0745 05 0662 025 0653 06805 kJkg K 1Q2 2 kg 06805 kJkgK 500 300 K 2722 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1129 The mixture in Problem 1122 is heated up to 500 K in a constant pressure process Find the final volume and the total heat transfer using Table A5 Solution CV Mixture Process P constant 1W2 P dV P V2 V1 Energy Eq 1Q2 mu2 u1 1W2 mu2 u1 Pm v2 v1 mh2 h1 m CP mixT2 T1 From Eq1115 Rmix ciRi 025 02968 05 02598 025 01889 02513 kJkg K From Eq1123 CP mix ci CP i 025 1042 05 0922 025 0842 0932 kJkg K V2 m Rmix T2P 2 2 kg 02513 kJkgK 500 K150 kPa 1675 m3 1Q2 2 kg 0932 kJkgK 500 300 K 3728 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1130 A flow of 1 kgs argon at 300 K and another flow of 1 kgs CO2 at 1600 K both at 150 kPa are mixed without any heat transfer What is the exit T P No work implies no pressure change for a simple flow Pe 150 kPa The energy equation becomes m hi m he m hiAr m hiCO2 m heAr m heCO2 m CO2Cp CO2Te TiCO2 m ArCp ArTe TiAr 0 m ArCp ArTi m CO2Cp CO2Ti m ArCp Ar m CO2Cp CO2 Te 10520300 108421600 kW 10520 10842 kWK Te Te 11037 K 1 Ar 2 CO 3 Mix MIXING CHAMBER 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1131 A flow of 1 kgs argon at 300 K and another flow of 1 kgs CO2 at 1600 K both at 150 kPa are mixed without any heat transfer Find the exit T P using variable specific heats No work implies no pressure change for a simple flow Pe 150 kPa The energy equation becomes m hi m he m hiAr m hiCO2 m heAr m heCO2 m CO2 he hiCO2 m ArCp ArTe TiAr 0 1 kgs he 174812 kJkg 1 052 kWK Te 300 K 0 he CO2 052 Te 174812 052 300 190412 kJkg Trial and error on Te using Table A8 for h e CO2 Te 1100 K LHS 109636 052 1100 166836 too small Te 1300 K LHS 135228 052 1300 202828 too large Te 1200 K LHS 122334 052 1200 184734 too small Final interpolation Te 1200 100 190412 184734 202828 184734 12314 K 1 Ar 2 CO 3 Mix MIXING CHAMBER 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1132 A pipe flows 01 kgs of a mixture with mass fractions of 40 CO2 and 60 N2 at 400 kPa 300 K Heating tape is wrapped around a section of pipe with insulation added and 2 kW electrical power is heating the pipe flow Find the mixture exit temperature Solution CV Pipe heating section Assume no heat loss to the outside ideal gases Energy Eq Q m he hi m CP mixTe Ti From Eq1123 CP mix ci CP i 04 0842 06 1042 0962 kJkg K Substitute into energy equation and solve for exit temperature Te Ti Q m CP mix 300 2 kW 01 0962 kWK 3208 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1133 A steady flow of 01 kgs carbon dioxide at 1000 K in one line is mixed with 02 kgs of nitrogen at 400 K from another line both at 100 kPa The mixing chamber is insulated and has constant pressure of 100 kPa Use constant heat capacity to find the mixing chamber exit temperature Take CV around the mixing chamber Continuity Eq49 m 1 m 2 m 3 Concentrations cCO2 m 1m 3 13 cN2 m 2m 3 23 CP mix ci CP i 13 0842 23 1042 097533 kJkg Rmix ciRi 13 01889 23 02968 02608 kJkg Energy Eq m 1h1 m 2h2 m 3h3 m 1h3 CO2 m 2h3 N2 Divide this equation with m 3 and take differences in h as CP T 1 3 CP CO2T1 2 3 CP N2T2 1 3 CP CO2 2 3 CP N2 T3 CP mixT3 T3 1 3 0842 1000 2 3 1042 400 097533 5727 K 1 3 2 Mix Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1134 An insulated gas turbine receives a mixture of 10 CO2 10 H2O and 80 N2 on a mass basis at 1000 K 500 kPa The volume flow rate is 2 m3s and its exhaust is at 700 K 100 kPa Find the power output in kW using constant specific heat from A5 at 300 K Solution CV Turbine Steady 1 inlet 1 exit flow with an ideal gas mixture q 0 Energy Eq W T m hi he m CP mixTi Te Properties From Eqs1115 and 1123 Rmix ciRi 01 01889 01 04615 08 02968 030248 kJkg K CP mix ci CP i 01 0842 01 1872 08 1042 1105 kJkg K PV mRmixT m PV RmixT m 500 kPa 2 m3s 0302481000 kJkg 3306 kgs W T 3306 kgs 1105 kJkgK 1000 700 K 1096 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1135 Solve Problem 1134 using the values of enthalpy from Table A8 An insulated gas turbine receives a mixture of 10 CO2 10 H2O and 80 N2 on a mass basis at 1000 K 500 kPa The volume flow rate is 2 m3s and its exhaust is at 700 K 100 kPa Find the power output in kW using constant specific heat from A5 at 300 K Solution CV Turbine Steady 1 inlet 1 exit flow with an ideal gas mixture q 0 Energy Eq W T m hi he m cj hi he j Properties From Eqs1115 and 1123 Rmix ciRi 01 01889 01 04615 08 02968 030248 kJkg K PV mRmixT m PV RmixT m 500 kPa 2 m3s 030248 1000 kJkg 3306 kgs Now get the h values from Table A8 all in kJkg W T 3306 kgs 01 97167 61622 01 199413 133856 08 107591 73586 kJkg 1234 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1136 Solve Problem 1133 assuming the flows are in kmols A steady flow of 01 kgs carbon dioxide at 1000 K in one line is mixed with 02 kgs of nitrogen at 400 K from another line both at 100 kPa The mixing chamber is insulated and has constant pressure of 100 kPa Use constant heat capacity to find the mixing chamber exit temperature Take CV around the mixing chamber Continuity Eq mole basis n 1 n 2 n 3 Concentrations yCO2 n 1n 3 13 yN2 n 2n 3 23 C P mix yiC P i 1 3 0842 4401 2 3 1042 28013 3181 kJ kmol Mmix yiMi 13 4401 23 28013 33345 kgkmol Energy Eq Mole basis n 1h1 n 2h 2 n 3h 3 n 1h 3 CO2 n 2h 3 N2 Divide this equation with n 3 and take differences in h as C P T 1 3 C P CO2T1 2 3 C P N2T2 1 3 C P CO2 2 3 C P N2 T3 C P mix T3 T3 1 3 37056 1000 2 3 29189 400 3181 633 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1137 Solve Problem 1134 with the percentages on a mole basis and use Table A9 An insulated gas turbine receives a mixture of 10 CO2 10 H2O and 80 N2 on a mole basis at 1000 K 500 kPa The volume flow rate is 2 m3s and its exhaust is at 700 K 100 kPa Find the power output in kW using constant specific heat from A5 at 300 K CV Turbine Steady flow 1 inlet 1 exit flow with an ideal gas mixture and no heat transfer so q 0 Energy Eq W T m hi he n h i h e n yj h i h ej Ideal gas law PV nRT n RT PV 83145 kJkmolK 1000 K 500 kPa 2 m3s 01203 kmols Read the enthalpies from Table A9 they are all in kJkmol W T 012030133397 17754 0126000 14190 0821463 11937 1247 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1138 A mixture of 05 kg nitrogen and 05 kg oxygen is at 100 kPa 300 K in a piston cylinder keeping constant pressure Now 800 kJ is added by heating Find the final temperature and the increase in entropy of the mixture using Table A5 values Solution CV Mixture in the piston cylinder Energy Eq mu2 u1 1Q2 1W 2 Process P constant 1W2 P dV P V2 V1 1Q2 mu2 u1 1W2 mu2 u1 mPv2 v1 mh2 h1 h2 h1 1Q2m CP mix T2 T1 From Eq1123 and Table A5 CP mix 12 0922 12 1042 0982 kJkg K T2 T1 1Q2mCP mix 300 K 800 kJ1 0982 kJK 11147 K From Eq1124 ms2 s1 mCP mix lnT2 T1 R lnP2 P1 1 kg 0982 kJkgK ln 11147300 129 kJK Mixture F C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1139 A rigid insulated vessel contains 12 kg of oxygen at 200 kPa 280 K separated by a membrane from 26 kg carbon dioxide at 400 kPa 360 K The membrane is removed and the mixture comes to a uniform state Find the final temperature and pressure of the mixture Solution CV Total vessel Control mass with two different initial states Mass m mO2 mCO2 12 26 38 kg Process V constant rigid W 0 insulated Q 0 Energy U2 U1 0 0 mO2 CV O2T2 T1 O2 mCO2CV CO2T2 T1 CO2 Initial state from ideal gas Table A5 CV O2 0662 kJkg CV CO2 0653 kJkg K O2 VO2 mRT1P 12 02598 280200 43646 m3 CO2 VCO2 mRT1P 26 01889 360400 44203 m 3 Final state mixture Rmix ciRi 12 02598 26 01889 38 02113 kJkg K The energy equation becomes mO2 CV O2 T2 mCO2CV CO2 T2 mO2 CV O2 T1 O2 mCO2CV CO2 T1 CO2 7944 16978 kJK T2 222432 611208 83364 kJ T2 3345 K From mixture gas constant and total volume V2 VO2 VCO2 43646 m3 44203 m3 87849 m 3 P2 mRmixT2V2 38 kg 02113 kJkgK 3345 K 87849 m3 3057 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1140 A mixture of 05 kg nitrogen and 05 kg oxygen is at 100 kPa 300 K in a piston cylinder keeping constant pressure Now 1200 kJ is added by heating Find the final temperature and the increase in entropy of the mixture using Table A8 values Solution CV Mixture in the piston cylinder Energy Eq mu2 u1 1Q2 1W 2 Process P constant 1W2 P dV P V2 V1 1Q2 mu2 u1 1W2 mu2 u1 mPv2 v1 mh2 h1 h2 h1 1Q2m 1200 kJ 1 kg 1200 kJkg Since T2 is high we use Table A8 values guessing a T 2 h2 h11400K 1 2 155687 31167 1 2 142644 27315 1199245 kJkg low h2 h11500K 1 2 168070 31167 1 2 154023 27315 131806 kJkg too high T2 1400 1001200 1199245131806 1199245 14006 K From Eqs1116 and 1118 s2 s1 1 2 s T2 s T1N2 1 2 s T2 s T1 O2 1 2 85495 68463 1 2 79869 64168 16367 kJkg K Notice that the pressure term drop out composition does not change and total pressure is constant Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1141 New refrigerant R410A is a mixture of R32 and R125 in a 11 mass ratio A process brings 05 kg R410A from 270 K to 320 K at a constant pressure 250 kPa in a piston cylinder Find the work and heat transfer Solution CV R410A Energy Eq mu2 u1 1Q2 1W2 1Q2 P V2 V1 Process P constant 1W2 P V2 V1 mRT2 T1 1Q2 mu2 u1 1W2 mh2 h1 From Eq1115 Rmix ciRi 1 2 01598 1 2 006927 01145 kJkg K FromEq1123 CP mix 1 2 0822 1 2 0791 08065 kJkg K From the process equation 1W2 05 kg 01145 kJkgK 320 270 K 2863 kJ From the energy equation 1Q2 m CP mix T2 T1 05 kg 08065 kJkgK 320 270 K 2016 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1142 A pistoncylinder device contains 01 kg of a mixture of 40 methane and 60 propane gases by mass at 300 K and 100 kPa The gas is now slowly compressed in an isothermal T constant process to a final pressure of 250 kPa Show the process in a PV diagram and find both the work and heat transfer in the process Solution CV Mixture of methane and propane this is a control mass Assume methane propane are ideal gases at these conditions Energy Eq35 mu2 u1 1Q2 1W 2 Property from Eq1115 Rmix 04 RCH4 06 RC3H8 04 05183 06 01886 03205 kJkg K Process T constant ideal gas 1W2 P dV mRmixT 1VdV mRmixT ln V2V1 mRmixT ln P1P2 01 kg 03205 kJkgK 300 K ln 100250 881 kJ Now heat transfer from the energy equation where we notice that u is a constant ideal gas and constant T so 1Q2 mu2 u1 1W2 1W2 881 kJ P v 2 1 T s 2 1 T P C v 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1143 The substance R410A see Problem 1141 is at 100 kPa 290 K It is now brought to 250 kPa 400 K in a reversible polytropic process Find the change in specific volume specific enthalpy and specific entropy for the process Solution Eq1115 Rmix Σ ciRi 1 2 01598 1 2 006927 01145 kJkg K Eq1123 CPmix Σ ciCPi 1 2 0822 1 2 0791 08065 kJkg K v1 RT1P1 01145 290100 033205 m3kg v2 RT2P2 01145 400250 01832 m3kg v2 v1 01832 033205 014885 m3kg h2 h1 CPmix T2 T1 08065 kJkgK 400 290 K 8872 kJkg From Eq1124 s2 s1 CPmix lnT2 T1 Rmix lnP2 P1 08065 ln 400290 01145 ln 250100 0154 kJkg K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1144 Natural gas as a mixture of 75 methane and 25 ethane by mass is flowing to a compressor at 17C 100 kPa The reversible adiabatic compressor brings the flow to 350 kPa Find the exit temperature and the needed work per kg flow Solution CV Compressor Steady adiabatic q 0 reversible sgen 0 Energy Eq413 w hex hin Entropy Eq78 si sgen s e Process reversible sgen 0 se si Assume ideal gas mixture and constant heat capacity so we need k and CP From Eq1115 and 1123 Rmix ciRi 075 05183 025 02765 045785 kJkg K CP mix ciCPi 075 2254 025 1766 2132 kJkg K CV CP mix Rmix 2132 045785 16742 kJkg K Ratio of specific heats k Cp Cv 12734 The isentropic process gives Eq623 from Eq1124 with se si Te Ti Pe Pik1k 290 350100 02147 3795 K Work from the energy equation wc in CP Te Ti 2132 kJkgK 3795 290 K 1908 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1145 A compressor brings R410A see problem 1141 from 10 oC 125 kPa up to 500 kPa in an adiabatic reversible compression Assume ideal gas behavior and find the exit temperature and the specific work Solution CV Compressor Process q 0 adiabatic and reversible Energy Eq413 w hi he Entropy Eq78 se si sgen dqT si 0 0 si From Eq1115 Rmix ciRi 1 2 01598 1 2 006927 01145 kJkg K FromEq1123 CP mix 1 2 0822 1 2 0791 08065 kJkg K Rmix CP mix 0114508065 014197 For constant s ideal gas and use constant specific heat as in Eq623 TeTi PePiRCp Te 26315 K 500125014197 32039 K w CP mix Ti Te 08065 kJkgK 26315 32039 K 46164 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1146 Two insulated tanks A and B are connected by a valve Tank A has a volume of 1 m3 and initially contains argon at 300 kPa 10C Tank B has a volume of 2 m3 and initially contains ethane at 200 kPa 50C The valve is opened and remains open until the resulting gas mixture comes to a uniform state Determine the final pressure and temperature Solution CV Tanks A B Control mass no W no Q Energy Eq35 and 336 U2U1 0 mArCV0T2TA1 mC2H6CVOT2 TB1 mAr PA1VARTA1 300 1 02081 28315 50913 kg mC2H6 PB1VBRTB1 200 2 02765 32315 44767 kg Continuity Eq m2 mAr mC2H6 9568 kg Energy Eq 50913 0312 T2 2832 44767 1490 T2 3232 0 Solving T2 3155 K Rmix Σ ciRi 50913 9568 02081 44767 9568 02765 02401 kJkg K P2 m2RT2VAVB 9568 kg 02401 kJkgK 3155 K 3 m3 242 kPa A B cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1147 A steady flow of 01 kmols carbon dioxide at 1000 K in one line is mixed with 02 kmols of nitrogen at 400 K from another line both at 100 kPa The exit mixture at 100 kPa is compressed by a reversible adiabatic compressor to 500 kPa Use constant specific heat to find the mixing chamber exit temperature and the needed compressor power Take CV around the mixing chamber Continuity Eq mole basis n 1 n 2 n 3 Concentrations yCO2 n 1n 3 13 yN2 n 2n 3 23 C P mix yiC P i 1 3 0842 4401 2 3 1042 28013 3181 kJ kmol RmixCP mix R C P mix 83145 3181 02614 Energy Eq Mole basis n 1h1 n 2h 2 n 3h 3 n 1h 3 CO2 n 2h 3 N2 Divide this equation with n 3 and take differences in h as C P T 1 3 C P CO2T1 2 3 C P N2T2 1 3 C P CO2 2 3 C P N2 T3 C P mix T3 T3 1 3 37056 1000 2 3 29189 400 3181 633 K Now we can do the adiabatic compression T4 T3 P4 P3RCp 633 502614 9641 K w C C P mix T4 T3 3181 kJkmolK 9641 633 K 10 532 kJkmol W m w C 03 kmols 10 532 kJkmol 3160 kW 1 4 3 2 Mix C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1148 A mixture of 2 kg oxygen and 2 kg of argon is in an insulated piston cylinder arrangement at 100 kPa 300 K The piston now compresses the mixture to half its initial volume Find the final pressure temperature and the piston work Solution CV Mixture Control mass boundary work and no Q assume reversible Energy Eq35 u2 u1 1q2 1w2 1w2 Entropy Eq637 s2 s1 0 0 0 Process constant s Pvk constant v2 v12 Assume ideal gases T1 TC and use kmix and Cv mix for properties Eq1115 Rmix Σ ciRi 05 025983 05 020813 0234 kJkg K Eq1123 CPmix Σ ciCPi 05 09216 05 05203 0721 kJkg K Cvmix CPmix Rmix 0487 kJkg K Ratio of specific heats kmix CPmixCvmix 14805 The relations for the polytropic process Eq625 P2 P1v1v2k P12k 100214805 279 kPa Eq624 T2 T1v1v2k1 T12k1 300204805 4186 K Work from the energy equation 1W2 mtot u1 u2 mtot CvT1 T2 4 kg 0487 kJkgK 300 4186 K 231 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1149 The gas mixture from Problem 1127 is compressed in a reversible adiabatic process from the initial state in the sample cylinder to a volume of 02 L Determine the final temperature of the mixture and the work done during the process Solution From Eq1115 Rmix ciRi 002 41243 045 02968 028 01889 025 02968 034314 kJkg K m PVRmixT 100103034314 29315 9941104 kg CV0 MIX ci CV0 i 002 10085 045 0744 028 0653 025 0745 09056 kJkg K CP0 MIX CV0 MIX Rmix 09056 034314 12487 kJkg K k CP0CV0 1248709056 1379 The process adiabatic and reversible is isentropic expressed in Eq632 T2 T1 V1 V2 k1 29315 1 02 0379 5395 K 1W2 U12 mCV0T2T1 9941104 kg 09056 kJkgK 5395 29315 K 022 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1150 A piston cylinder has 01 kg mixture of 25 argon 25 nitrogen and 50 carbon dioxide by mass at total pressure 100 kPa and 290 K Now the piston compresses the gases to a volume 7 times smaller in a polytropic process with n 13 Find the final pressure and temperature the work and the heat transfer for the process Solution Expansion ratio v2 v1 17 Mixture properties Rmix Σ ciRi 025 02081 025 02968 05 01889 0220675 kJkg K Cv mix ci Cvi 025 0312 025 0745 05 0653 059075 kJkg K Process eq Rev adiabatic and ideal gas gives Pvn C with n 13 P2 P1 v1v2n 100 713 12549 kPa T2 T1 v1v2n1 290 703 5199 K Polytropic process work term from Eq321 and ideal gas law 1W2 mR 1 n T2 T1 01 0220675 03 kJK 5199 290 K 1691 kJ Energy Eq 1Q2 U2 U1 1W2 m Cv mix T2 T1 1W 2 01 kg 059075 kJkgK 5199 290 K 1691 kJ 333 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Entropy generation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1151 A flow of gas A and a flow of gas B are mixed in a 12 mole ratio with the same T What is the entropy generation per kmole flow out For this case the total flow is 3 mol units so yA nAntot 13 yB nBntot 23 Eq 1119 S R 13 ln 13 23 ln 23 06365 R 5292 kJkmol K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1152 A rigid container has 1 kg argon at 300 K and 1 kg argon at 400 K both at 150 kPa Now they are allowed to mix without any external heat transfer What is final T P Is any s generated Energy Eq U2 U1 0 2mu2 mu1a mu1b mCv2T2 T1a T1b T2 T1a T1b2 350 K Process Eq V constant P2V 2mRT2 mRT1a T1b P1V1a P1V1b P1V P2 P1 150 kPa S due to temperature changes only not P internally we have a Q over a T S m s2 s1a m s2 s1b mCp ln T2T1a ln T2T1b 1 kg 0520 kJkgK ln 350 300 ln 350 400 00107 kJK Ar Ar cb Why did we not account for partial pressures Since we mix argon with argon there is no entropy generation due to mixing and no partial pressure Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1153 What is the entropy generation in problem 1126 No Q No W so the energy equation gives constant U Energy Eq U2 U1 0 mCO2u2 u1CO2 mAru2 u1Ar mCO2Cv CO2T2 T1CO2 mArCv ArT2 T1Ar 10653 10312 T2 10653300 10312400 T2 3323 K V V1 VCO2 VAr mCO2RCO2TCO2P mArRArTArP 101889300150 102081400150 0932 73 m3 Pressure from ideal gas law and Eq1115 for R P2 101889 102081 33230932 73 1414 kPa S2 S1 0 1S2 gen mCO2s2 s1CO2 mArs2 s1Ar For each component s2 s1 CP ln T1 T2 R ln P1 yP2 Ps are total pressure yCO2 cCO2MCO2 cCO2MCO2 cArMAr 05 4401 05 39948 05 4401 04758 yAr 1 yCO2 05242 1S2 gen 1 kg 0520 ln3323 400 02081 ln 05242 1414 150 kJkgK 1 kg 0842 ln3323 300 01889 ln 04758 1414 150 kJkgK 005027 023756 02878 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1154 A flow of 2 kgs mixture of 50 CO2 and 50 O2 by mass is heated in a constant pressure heat exchanger from 400 K to 1000 K by a radiation source at 1400 K Find the rate of heat transfer and the entropy generation in the process shown in Fig P1154 Solution CV Heat exchanger w 0 Energy Eq412 Q in m he hi Values from Table A8 due to the high T Q in 2 1 2 97167 30376 1 2 98095 36603 12828 kW Entropy Eq78 m ese m isi Q Ts S gen As the pressure is constant the pressure correction in Eq619 drops out to give the generation as S gen m se si Q Ts 2 kgs 1 2 6119 51196 1 2 76121 66838 kJkgK 12828 kW 1400 K 101 kWK 1400 i e Radiation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1155 A flow of 18 kgs steam at 400 kPa 400oC is mixed with 32 kgs oxygen at 400 kPa 400 K in a steady flow mixingchamber without any heat transfer Find the exit temperature and the rate of entropy generation CV Mixing chamber steady flow no work no heat transfer To do the entropies we need the mole fractions n H2O m H2O MH2O 18 18015 01 kmols n O2 MO2 m O2 32 31999 01 kmols yH2O yO2 05 Energy Eq m H2O h1 m O2 h2 m H2O h3 H2O m O2 h 3 O2 Entropy Eq m H2O s1 m O2 s2 S gen m H2O s3 H2O m O2 s 3 O2 Solve for T from the energy equation m H2O h3 H2O h1 m O2 h3 O2 h2 0 m H2O CP H2OT3 T1 m O2 CP O2T3 T2 0 18 1872 T3 400 27315 32 0922T3 400 0 T3 5456 K S gen m H2O s3 H2O s1 m O2 s3 O2 s2 m H2O CP H2O ln T1 T3 R ln yH2O m O2 CP O2 ln T2 T3 R ln yO2 18 kgs 1872 ln 5456 67315 04615 ln 05 kJkgK 32 kgs 0922 ln 5456 400 02598 ln 05 kJkgK 0132 1492 136 kWK 700 C A B Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1156 Carbon dioxide gas at 320 K is mixed with nitrogen at 280 K in an insulated mixing chamber Both flows are at 100 kPa and the mass ratio of carbon dioxide to nitrogen is 21 Find the exit temperature and the total entropy generation per kg of the exit mixture Solution CV mixing chamber The inlet ratio is so m CO2 2 m N2 and assume no external heat transfer no work involved Continuity Eq69 m N2 2m N2 m ex 3m N2 Energy Eq610 m N2hN2 2 hCO2 3m N2hmix ex Take 300 K as reference and write h h300 CPmixT 300 CP N2Ti N2 300 2CP CO2Ti CO2 300 3CP mixTmix ex 300 CP mix ciCP i 2 3 0842 1 3 1042 09087 kJkg K 3CP mixTmix ex CP N2Ti N2 2CP CO2Ti CO2 83064 kJkg Tmix ex 3047 K To find the entropies we need the partial pressures which assuming ideal gas are equal to the mole fractions times the total pressure yi ci Mi cjMj yN2 03333 28013 03333 28013 06666 4401 044 yCO2 1 yN2 056 S gen m exsex m siCO2 m siN2 m N2se siN2 2m N2se siCO2 S gen 3m N2 1 3 CPN2ln Tex TiN2 RN2ln yN2 2 3 CPCO2ln Tex TiCO2 RCO2ln yCO2 1 3 1042 ln3047 280 02968 ln 044 2 3 0842 ln3047 320 01889 ln 056 0110585 0068275 01789 kJkg mix K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1157 Carbon dioxide gas at 320 K is mixed with nitrogen at 280 K in an insulated mixing chamber Both flows are coming in at 100 kPa and the mole ratio of carbon dioxide to nitrogen is 21 Find the exit temperature and the total entropy generation per kmole of the exit mixture CV mixing chamber steady flow The inlet ratio is n CO2 2 n N2 and assume no external heat transfer no work involved Continuity n CO2 2n N2 n ex 3n N2 Energy Eq n N2h N2 2h CO2 3n N2h mix ex Take 300 K as reference and write h h 300 C PmixT 300 C P N2Ti N2 300 2C P CO2Ti CO2 300 3C P mixTmix ex 300 Find the specific heats in Table A5 to get C P mix yiC P i 1042 28013 2 0842 44013 3443 kJkmol K 3C P mixTmix ex C P N2Ti N2 2C P CO2Ti CO2 31889 kJkmol Tmix ex 3087 K Partial pressures are total pressure times molefraction Pex N2 Ptot3 Pex CO2 2Ptot3 S gen n exs ex nsiCO2 nsiN2 n N2s e s iN2 2n N2s e s i CO2 S gen3n N2 C PN2ln Tex TiN2 Rln yN2 2C PCO2ln Tex TiCO2 2 Rln yCO23 28485 91343 2660767423 535 kJkmol mix K 1 N 2 CO 3 Mix MIXING CHAMBER 2 2 S ge Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1158 A flow of 1 kgs carbon dioxide at 1600 K 100 kPa is mixed with a flow of 2 kgs water at 800 K 100 kPa and after the mixing it goes through a heat exchanger where it is cooled to 500 K by a 400 K ambient How much heat transfer is taken out in the heat exchanger What is the entropy generation rate for the whole process Solution CV Total mixing section and heat exchanger Steady flow and no work To do the entropy at the partial pressures we need the mole fractions n H2O m H2OMH2O 2 18015 011102 kmols n CO2 m CO2MCO2 1 4401 0022722 kmols yH2O 011102 011102 0022722 08301 yCO2 1 yH2O 01699 Energy Eq m H2O h1 m CO2 h2 Q cool m H2O h4 H2O m CO2 h 4 CO2 Entropy Eq m H2O s1 m CO2 s2 S gen Q cool Tamb m H2O s4 H2O m CO2 s 4 CO2 As T is fairly high we use Table A8 for properties on a mass basis 1 2 4 H2O 4 CO2 h kJkg 155013 174812 93512 40152 s o T kJkg K 124244 67254 114644 53375 Q cool m H2O h1 h4 H2O m CO2 h2 h4 CO2 2 155013 93512 1 174812 40152 2577 kW S gen m H2O s4 H2O s1 m CO2 s4 CO2 s2 Q coolTamb 2 kgs 114644 124244 04615 ln08301 kJkgK 1 kgs 53375 67254 01889 ln01699 kJkgK 2577 400 kWK 174813 105307 64415 364 kWK 1 2 3 4 Mixing chamber Q cool Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1159 The only known sources of helium are the atmosphere mole fraction approximately 5 106 and natural gas A large unit is being constructed to separate 100 m3s of natural gas assumed to be 0001 He mole fraction and 0999 CH4 The gas enters the unit at 150 kPa 10C Pure helium exits at 100 kPa 20C and pure methane exits at 150 kPa 30C Any heat transfer is with the surroundings at 20C Is an electrical power input of 3000 kW sufficient to drive this unit 0999 CH4 0001 He at 150 kPa 10 oC V 1 100 m3s W CV 3000 kW P2 100 kPa T2 20 oC P3 140 kPa T3 30 oC n 1 P1 V1RT1 150 kPa 100 m3s 831452832 kJkmol 637 kmols n 2 0001 n 1 0006 37 n 3 63636 kmols C P He 4003 kgkmol 5193kJkgK 207876 kJkmol K C P CH4 16043 kgkmol 2254 kJkgK 361609 kJkmol K Energy Eq Q CV n 2h 2 n 3h 3 n 1h 1 W CV n 2C P0 HeT2T1 n 3C P0 CH4T3T1 W CV 00063720787620 10 6363636160930 10 3000 1600 kW Entropy Eq S gen n 2s 2 n 3s 3 n 1s 1 Q CVT0 000637207876 ln 2932 2832 83145 ln 100 0001150 63636361609 ln 3032 2832 83145 ln 140 0999150 16002932 135 kWK 0 Since positive this is possible 1 2 3 He CH 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1160 Repeat Problem 1144 for an isentropic compressor efficiency of 82 Solution CV Compressor Steady adiabatic q 0 reversible sgen 0 Energy Eq413 w hex hin Entropy Eq78 si sgen si s e Process reversible sgen 0 se si Assume ideal gas mixture and constant heat capacity so we need k and CP From Eq1115 and 1123 Rmix ciRi 075 05183 025 02765 045785 kJkg K CP mix ciCPi 075 2254 025 1766 2132 kJkg K CV CP mix Rmix 2132 045785 16742 kJkg K Ratio of specific heats k Cp Cv 12734 The isentropic process gives Eq623 from Eq1124 with se si Te Ti Pe Pik1k 290 350100 02147 3795 K Isentropic work from the energy equation wc in CP Te Ti 2132 kJkgK 3795 290 K 1908 kJkg The actual compressor requires more work wc actual wc inη 1908082 2327 kJkg Cp Te actual Ti Te actual Ti wc actualCP 290 K 2327 kJkg 2132 kJkgK 3991 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1161 A steady flow 03 kgs of 60 carbon dioxide and 40 water mixture by mass at 1200 K and 200 kPa is used in a constant pressure heat exchanger where 300 kW is extracted from the flow Find the exit temperature and rate of change in entropy using Table A5 Solution CV Heat exchanger Steady 1 inlet 1 exit no work Continuity Eq cCO2 cH2O 05 Energy Eq Q m he hi m CP Te Ti Te Ti Q m C P Inlet state Table A5 CP 06 0842 04 1872 1254 kJkgK Exit state Te Ti Q m CP 1200 K 300 kW03 1254 kWK 4026 K The rate of change of entropy for the flow is P is assumed constant m se si m so Te so Ti R lnPePi m so Te so Ti m CP lnTeTi 03 1254 ln4026 1200 0411 kWK The entropy generation rate cannot be estimated unless the average T at which the heat transfer leaves the control volume is known Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1162 A steady flow of 03 kgs of 60 carbon dioxide and 40 water by mass at 1200 K and 200 kPa is used in a heat exchanger where 300 kW is extracted from the flow Find the flow exit temperature and the rate of change of entropy using Table A8 Solution CV Heat exchanger Steady 1 inlet 1 exit no work Continuity Eq cCO2 06 cH2O 04 Energy Eq Q m he hi he hi Q m Inlet state Table A8 hi 06 122334 04 246625 17205 kJkg Exit state he hi Q m 17205 30003 7205 kJkg Trial and error for T with h values from Table A8 500 K he 06 40152 04 93512 61496 kJkg 600 K he 06 50607 04 113367 75711 kJkg 550 K he 06 45303 04 103363 68527 kJkg Interpolate to have the right h T 575 K Entropy Eq78 m se m si Q T S gen The rate of change of entropy for the flow is P is assumed constant m se si m so Te so Ti 03 06548175 63483 041173925 133492 0349 kWK The entropy generation rate cannot be estimated unless the integral dQT can be evaluated If the dQ leaves at the local mixture T assumed uniform over the flow cross section the process is internally reversible and then externally irreversible Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1163 A mixture of 60 helium and 40 nitrogen by mass enters a turbine at 1 MPa 800 K at a rate of 2 kgs The adiabatic turbine has an exit pressure of 100 kPa and an isentropic efficiency of 85 Find the turbine work Solution Assume ideal gas mixture and take CV as turbine Energy Eq413 wT s hi hes Entropy Eq78 ses si adiabatic and reversible Process Eq623 Tes TiPePik1k Properties from Eq1123 1115 and 623 CP mix 06 5193 04 1042 35326 kJkgK Rmix 06 20771 04 02968 1365 kJkgK k1k RCP mix 136535326 03864 Tes 800 K 100100003864 3286 K wTs CPTi Tes 35326 kJkgK 800 3286 K 1665 kJkg wT ac ηwTs 085 1665 kJkg 14155 kJkg W T ac m wT ac 2 kgs 14155 kJkg 2831 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1164 Three steady flows are mixed in an adiabatic chamber at 150 kPa Flow one is 2 kgs of O2 at 340 K flow two is 4 kgs of N2 at 280 K and flow three is 3 kgs of CO2 at 310 K All flows are at 150 kPa the same as the total exit pressure Find the exit temperature and the rate of entropy generation in the process Solution CV Mixing chamber no heat transfer no work Continuity Eq49 m 1 m 2 m 3 m 4 Energy Eq410 m 1h1 m 2h2 m 3h3 m 4h4 Entropy Eq77 m 1s1 m 2s2 m 3s3 S gen m 4s4 Assume ideal gases and since T is close to 300 K use heat capacity from A5 in the energy equation as m 1CP O2T1 T4 m 2CP N2T2 T4 m 3CP CO2T3 T4 0 2 0922 340 4 1042 280 3 0842 310 kW 2 0922 4 1042 3 0842 kWK T4 257706 K 8538 T4 T4 30183 K State 4 is a mixture so the component exit pressure is the partial pressure For each component se si CP lnTe Ti R lnPe Pi and the pressure ratio is Pe Pi y P4 Pi y for each n m M 2 32 4 28013 3 4401 00625 01428 006817 02735 yO2 00625 02735 02285 yN2 01428 02735 05222 yCO2 006817 02735 02493 The entropy generation becomes S gen m 1s4 s1 m 2s4 s2 m 3s4 s3 2 0922 ln30183340 02598 ln02285 4 1042 ln30183280 02968 ln05222 3 0842 ln30183310 01889 ln02493 05475 1084 02399 1871 kWK 1 2 3 O N CO 2 2 2 4 mix MIX Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1165 A tank has two sides initially separated by a diaphragm Side A contains 1 kg of water and side B contains 12 kg of air both at 20C 100 kPa The diaphragm is now broken and the whole tank is heated to 600C by a 700C reservoir Find the final total pressure heat transfer and total entropy generation CV Total tank out to reservoir Energy Eq35 U2 U1 mau2 u1a mvu2 u1v 1Q2 Entropy Eq637 S2 S1 mas2 s1a mvs2 s1v 1Q2Tres Sgen Volume V2 VA VB mvvv1 mava1 0001 1009 101 m 3 vv2 V2mv 101 m3kg T2 P2v 400 kPa va2 V2ma 08417 m3kg T2 P2a mRT2V2 2977 kPa P2tot P2v P2a 6977 kPa Water table B1 u1 8395 kJkg u2 3300 kJkg s1 02966 kJkg K s2 84558 kJkg K Air table A7 u1 293 kJkg u2 6523 kJkg sT1 2492 kJkg K sT2 3628 kJkg K From energy equation we have 1Q2 13300 8395 126526 2094 37479 kJ From the entropy equation we have Sgen 184557 02966 12 79816 6846 0287 ln2977100 37479 9732 53 kJK 700 C A B Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1166 Reconsider the Problem 1146 but let the tanks have a small amount of heat transfer so the final mixture is at 400 K Find the final pressure the heat transfer and the entropy change for the process CV Both tanks Control mass with mixing and heating of two ideal gases nAr PA1VARTA1 3001 831452832 01274 kmol nC2H6 PB1VBRTB1 2002 831453232 01489 kmol Continuity Eq n2 nAr nC2H6 02763 kmol Energy Eq U2U1 nArC V0T2TA1 nC2H6C VOT2TB1 1Q 2 P2 n2RT2VAVB 0276383145400 3 3063 kPa 1Q2 01274399480312400 28315 014893007149400 32315 6983 kJ SSURR 1Q2TSURR SSYS nArS Ar nC2H6S C2H6 yAr 0127402763 04611 S Ar C P Ar ln T2 TA1 R ln PA1 yArP2 399480520 ln 400 28315 83145 ln 046113063 300 13445 kJkmol K S C2H6 C C2H6 ln T2 TB1 R ln PB1 yC2H6P2 30071766 ln 400 32315 83145 ln 053893063 200 129270 kJkmol K Assume the surroundings are at 400 K it heats the gas SNET nArS Ar nC2H6S C2H6 S SURR 0127413445 01489129270 6983400 1892 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Air water vapor mixtures Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1167 Atmospheric air is at 100 kPa 25oC and relative humidity 65 Find the absolute humidity and the dew point of the mixture If the mixture is heated to 30oC what is the new relative humidity Solution Eq1125 Pv φ Pg 065 3169 206 kPa Eq1128 w 0622 PvPtot Pv 0622 206100 206 001308 Tdew is the T such that PgT Pv 206 kPa B11 Tdew 179 C Heating w is constant Pv is constant From Table B11 Pg30C 4246 kPa φ PvPg 2064246 0485 or 49 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1168 A flow of 1 kgs saturated moist air relative humidity 100 at 100 kPa 10oC goes through a heat exchanger and comes out at 25oC What is the exit relative humidity and how much power is needed Solution State 1 φ1 1 Pv Pg 12276 kPa Eq1128 w 0622 PvPa 0622 12276100 12276 000773 State 2 No water added w2 w1 Pv2 P v1 φ2 Pv2Pg2 122763169 0387 or 39 Energy Eq610 Q m 2h2 m 1h1 m a h2 h1air wm a h2 h1vapor m tot m a m v m a1 w1 Energy equation with CP air from A5 and hs from B11 Q m tot 1 w1 CP air 25 10 m tot 1 w1 w hg2 hg1 1 100773 100425 10 1 000773 100773 254717 251974 149445 0210407 1515 kW Q 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1169 If I have air at 100 kPa and a 10oC b 45oC and c 110oC what is the maximum absolute humidity I can have Humidity is related to relative humidity max 100 and the pressures as in Eq1128 where from Eq1125 Pv Φ Pg and Pa Ptot Pv ω 0622 Pv Pa 0622 Φ Pg Ptot ΦPg a Pg 02601 kPa ω 0622 02601 100 026 0001 62 b Pg 9593 kPa ω 0622 9593 100 9593 00660 c Pg 1433 kPa no limit on ω for Ptot 100 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1170 A new highefficiency home heating system includes an airtoair heat exchanger which uses energy from outgoing stale air to heat the fresh incoming air If the outside ambient temperature is 10C and the relative humidity is 50 how much water will have to be added to the incoming air if it flows in at the rate of 1 m3s and must eventually be conditioned to 20C and 45 relative humidity Solution Outside ambient air Pv1 φ1Pg1 050 02602 01301 kPa Assuming P1 P2 100 kPa PA1 100 01301 9987 kPa m A PA1V 1 RAT1 9987 1 0287 2632 13224 kgs From Eq1128 w1 0622 01301 9987 000081 Conditioned to T2 20 oC φ2 045 Eq1125 Pv2 φ2Pg2 045 2339 kPa 10526 kPa Eq1128 w2 0622 10526 100 10526 000662 Continuity equation for water m liq in m Aw2 w1 13224 000662 000081 0007683 kgs 2766 kgh OUTSIDE IN SIDE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1171 Consider 100 m3 of atmospheric air which is an airwater vapor mixture at 100 kPa 15C and 40 relative humidity Find the mass of water and the humidity ratio What is the dew point of the mixture Solution Airvapor P 100 kPa T 15 oC φ 40 Use Table B11 and then Eq1125 Pg Psat15 1705 kPa Pv φ Pg 041705 0682 kPa mv RvT PvV 0682100 046128815 0513 kg Pa Ptot Pv1 100 0682 9932 kPa ma RaT PaV 9932100 028728815 1201 kg w1 ma mv 0513 1201 00043 Tdew is T when Pv Pg 0682 kPa Table B12 gives T 14 oC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1172 A flow of 2 kgs completely dry air at T1 100 kPa is cooled down to 10C by spraying liquid water at 10C 100 kPa into it so it becomes saturated moist air at 10C The process is steady state with no external heat transfer or work Find the exit moist air humidity ratio and the flow rate of liquid water Find also the dry air inlet temperature T1 Solution 2 saturated Pv Pg 12276 kPa and hfg 10C 24777 kJkg Eq1125 w2 0622 12276 100 12276 000773 CV Box Continuity Eq m a m liq m a1 w2 m liq w2 m a 00155 kgs Energy Eq m a ha1 m liq hf m a ha2 w2 hg2 ha1 ha2 Cpa T1 T2 w2 hg2 w2 hf w2 hfg 00073 kgkg dry air 247775 kJkg 915 kJkg dry air T1 T2 ha1 ha2Cpa 10 9151004 291C 1 2 Liquid water Dry air Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1173 The products of combustion are flowing through a heat exchanger with 12 CO2 13 H2O and 75 N2 on a volume basis at the rate 01 kgs and 100 kPa What is the dewpoint temperature If the mixture is cooled 10C below the dewpoint temperature how long will it take to collect 10 kg of liquid water Solution Volume basis is the same as mole fraction yH2O 013 PH2O yH2O P 013100 kPa 13 kPa Table B12 TDEW 5095 oC Cool to 4095 oC TDEW so saturated PG 7805 kPa yH2O PGP 7805100 nH2OvnH2Ov 087 nH2Ov 007365 per kmol mix in nLIQ 013 007365 005635 Eq115 MMIX IN 0124401 01318015 07528013 2863 kgkmol n MIX IN m TOTALMMIX IN 01 kgs 2863 kgkmol 0003493 kmols n LIQ COND 0003 493005635 0000 197 kmols or m LIQ COND 0000 197 kmols 18015 kgkmol 0003 55 kgs For 10 kg it takes Δt mm 10 kg 000355 kgs 47 minutes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1174 Consider a 1 m3s flow of atmospheric air at 100 kPa 25C and 80 relative humidity Assume this flows into a basement room where it cools to 15C 100 kPa Find the rate of water condensing out and the exit mixture volume flow rate Solution State 1 Pg Psat25 3169 kPa Pv φ Pg 08 3169 2535 kPa m v1 PvV RvT 2535 1 0461 29815 00184 kgs w1 m v1 m A1 0622 Pv1 PA1 0622 2535 100 2535 00162 m A1 m v1 w1 00184 00162 1136 kgs m A2 continuity for air Check for state 2 Pg15C 1705 kPa P v1 so liquid water out State 2 is saturated φ2 100 Pv2 Pg2 1705 kPa w2 0622 Pv2 PA2 0622 1705 100 1705 00108 m v2 w2m A2 00108 1136 00123 kgs m liq m v1 m v2 00184 00123 00061 kgs V m A2 RaT2 Pa2 m A2 RaT2 P Pv2 1136 0287 28815 100 1705 0956 m3s Note that the given volume flow rate at the inlet is not that at the exit The mass flow rate of dry air is the quantity that is the same at the inlet and exit Q 1 2 Liquid Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1175 Ambient moist air enters a steadyflow airconditioning unit at 102 kPa 30C with a 60 relative humidity The volume flow rate entering the unit is 100 Ls The moist air leaves the unit at 95 kPa 15C with a relative humidity of 100 Liquid also leaves the unit at 15C Determine the rate of heat transfer for this process Solution State 1 PV1 φ1PG1 060 4246 25476 w1 0622 25476102 25476 001593 m A PA1V 1 RAT1 994501 02873032 01143 kgs Pv2 Pg2 1705 kPa w2 0622 170595 1705 001137 Energy Eq410 Q CV m AhA1 m V1hV1 m AhA2 m V2hA2 m 3hL3 Q CVm A CP0AT2T1 w2hV2 w1hV1 w1w2h L3 10041530 00113725289 00159325562 000456630 26732 kJkg air Q CV 011432673 3055 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1176 A room with air at 40 relative humidity 20oC having 50 kg of dry air is made moist by boiling water to a final state of 20oC and 80 humidity How much water was added to the air The water content is expressed by the absolute humidity humidity ratio from Eq1128 and 1125 w1 0622 04 2339 101325 04 2339 0005797 w2 0622 08 2339 101325 08 2339 0011703 mwater ma w2 w1 50 kg 0011703 0005797 0295 kg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1177 Consider a 500L rigid tank containing an airwater vapor mixture at 100 kPa 35C with a 70 relative humidity The system is cooled until the water just begins to condense Determine the final temperature in the tank and the heat transfer for the process Solution Pv1 φPG1 075628 39396 kPa Since mv const V const also Pv PG2 PG2 Pv1 T2T1 39396 T23082 001278 T 2 Assume T2 30oC 0012783032 3875 4246 P G 30C Assume T2 25oC 0012782982 3811 3169 P G 25C interpolating T2 282 oC w2 w1 0622 39396 10039369 0025 51 ma Pa1VRaT1 1003940502873082 0543 kg Energy Eq 1Q2 U2U1 maua2ua1 mvuv2uv1 ma Cv Ta2Ta1 w1 uv2uv1 0543 0717282 35 002551 24142 24234 0543 kg 511 kJkg 277 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1178 A saturated airwater vapor mixture at 20 oC 100 kPa is contained in a 5m3 closed tank in equilibrium with 1 kg of liquid water The tank is heated to 80oC Is there any liquid water in the final state Find the heat transfer for the process a Since Vliq mliqvF 0001 m3 VGAS V φ1 100 Pv1 PG1 2339 kPa w1 0622 2339 100 2339 00149 ma RaT1 Pa1V 976614999 02872932 5802 kg mv1 w1ma 0086 kg At state 2 Pa2 97661 kPa 3532 2932 4999 5 117623 kPa wMAX 2 0622 4739 117623 02506 But w2 ACTUAL 0086 10 5802 01872 wMAX 2 No liquid at 2 mv2 mv1 mliq 0086 kg 1 kg 1086 kg Q12 ma ua2 ua1 mv2 uv2 mv1uv1 mliq 1uliq 1 ma Cv Ta2 Ta1 mv2 uv2 mv1uv1 mliq 1uliq 1 5802 071780 20 1086 24822 0086 24029 1 840 2496 26957 20665 84 2655 kJ Q 12 AIR VAP LIQ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1179 A flow of 02 kgs liquid water at 80oC is sprayed into a chamber together with 16 kgs dry air at 60oC All the water evaporates and the air leaves at 40oC What is the exit relative humidity and the heat transfer CV Chamber Continuity Eq water m liq wex m a Energy Eq m liq hliq m a ha i Q m a whv ha ex wex m liq m a 02 16 00125 From Eq1125 and 1128 you can get φex Pv Pg w 0622 w P Pg 00125 0622 00125 100 7384 0267 27 Q m a whv haex m liq hliq m a ha i m a ha ex ha i m liqhv hliq m a Cp a Tex Tin m liq hv 40 hf 80 16 1004 40 60 02 257426 33488 32128 44788 1266 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1180 A rigid container 10 m3 in volume contains moist air at 45C 100 kPa φ 40 The container is now cooled to 5C Neglect the volume of any liquid that might be present and find the final mass of water vapor final total pressure and the heat transfer Solution CV container m2 m1 m2u2 m1u1 1Q2 State 1 45C φ 40 w1 00236 Tdew 277C Final state T2 Tdew so condensation φ2 100 Pv1 04 Pg 04 9593 3837 kPa Pa1 Ptot Pv1 96163 kPa ma Pa1VRT1 10532 kg mv1 w1 ma 0248 kg Pv2 Pg2 08721 kPa Pa2 Pa1T2T1 84073 kPa P2 Pa2 Pv2 8495 kPa mv2 Pv2VRvT2 006794 kg Vvg 006797 steam table mf2 mv1 mv2 0180 kg The heat transfer from the energy equation becomes 1Q2 mau2u1a mv2ug2 mf2uf2 mv1ug1 ma CvT2 T1 mv2 23823 mf2 2097 mv1 24368 30206 161853 3775 60433 7408 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1181 A waterfilled reactor of 1 m3 is at 20 MPa 360C and located inside an insulated containment room of 100 m3 that contains air at 100 kPa and 25C Due to a failure the reactor ruptures and the water fills the containment room Find the final pressure CV Total container mv u2 u1 ma u2 u1 1Q2 1W2 0 Initial water v1 00018226 u1 17028 kJkg mv Vv 54867 kg Initial air ma PV RT 100 99 0287 2982 1157 kg Substitute into energy equation 54867 u2 17028 11570717 T2 25 0 u2 01511 T2 17066 kJkg v2 V2mv 018226 m3kg Trial and error 2phase Tguess v2 x2 u2 LHS T 150C LHS 1546 T 160C LHS 18202 T 155C LHS 16781 T 156C LHS 17057 OK x2 05372 Psat 5575 kPa Pa2 Pa1V1T2V2T1 100 99 42915 10029815 1425 kPa P2 Pa2 Psat 700 kPa 100 m 3 1 m 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1182 In the production of ethanol from corn the solids left after fermentation are dried in a continuous flow oven This process generates a flow of 15 kgs moist air 90C with 70 relative humidity which contains some volatile organic compounds and some particles To remove the organic gasses and the particles the flow is send to a thermal oxidicer where natural gas flames brings the mixture to 800C Find the rate of heating by the natural gas burners For this problem we will just consider the heating of the gas mixture and due to the exit temperature we will use the ideal gas tables A7 and A8 Eq1125 Pv φ Pg 070 7014 49098 kPa Eq1128 ω 0622 Pv Ptot Pv 0622 49098 100 49098 060 Flow m tot m a m v m a1 ω so m a 1 ω m tot 9375 kgs m v m tot ω 1 ω 5625 kgs Process Heating ω is constant Pv is constant For ideal gas the enthalpy does not depend on pressure so the energy equation gives the heat transfer as Q m a h2 h1a m v h2 h1v 9375 11302 3640 5625 21643 67275 15 573 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1183 To save on natural gas use in the previous problem it is suggested to take and cool the mixture and condense out some water before heating it back up So the flow is cooled from 90C to 50C and the now dryer mixture is heated to 800C Find the amount of water condensed out and the rate of heating by the natural gas burners for this case Eq1125 Pv1 φ1 Pg 070 7014 49098 kPa Eq1128 ω1 0622 Ptot Pv Pv 0622 49098 100 49098 060 Flow m tot m a m v m a1 ω so m a 1 ω m tot 9375 kgs m v1 m tot ω 1 ω 5625 kgs Now cool to 50C Pg 1235 kPa Pv1 so φ2 100 Pv2 1235 kPa ω2 0622 Ptot Pv Pv 0622 1235 100 1235 00876 m liq m a ω1 ω2 9375 06 00876 480375 kgs The air flow is not changed so the water vapor flow for heating is m v2 m v1 m liq 5625 480375 082125 kgs Now the energy equation becomes Q m a h3 h2a m v2 h3 h2v 9375 11302 32375 082125 21643 59765 8847 kW Comment If you solve the previous problem you find this is only 57 of the heat for the case of no water removal Qcool m liq heat 2 1 3 Q Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Tables and formulas or psychrometric chart Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1184 I want to bring air at 35oC Φ 40 to a state of 25oC ω 0015 do I need to add or subtract water The humidity ratio absolute humidity expresses how much water vapor is present in the mixture ω mv ma Assuming P 100 kPa ω 0622 Pv Pa 0622 Pv P Pv and Pv Φ Pg At 35C 40 ω 0622 0405628 100 0405628 0014 32 To get to ω 0015 it is necessary to add water Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1185 A flow moist air at 100 kPa 40C 40 relative humidity is cooled to 15C in a constant pressure device Find the humidity ratio of the inlet and the exit flow and the heat transfer in the device per kg dry air Solution CV Cooler m v1 m liq m v2 Tables Pg1 7384 kPa Pv1 φ Pg 04 7384 2954 kPa ω1 0622 2954 100 2954 00189 T2 Tdew from PgTdew 2954 Pv2 1705 kPa Pg2 ω2 0622 1705 100 1705 00108 hv1 25743 kJkg hv2 25289 kJkg hf 6298 kJkg qout CPT1 T2 ω1hv1 ω2 hv2 ω1 ω2 hf 100440 15 00189 25743 00108 25289 00073 6298 4598 kJkg dry air Psychrometric chart State 2 T Tdew 23C φ2 100 m v1m a ω1 0018 h 1 106 m v2m a ω2 00107 h2 62 m liqm a ω1 ω2 00073 hf 6298 kJkg m a q out m ah1 m liq hf m a h2 qout h1 ω1 ω2 hf h2 106 00073 6298 62 4354 kJkgdry air w T Φ 100 Φ Φ Φ 80 40 10 dry 1 2 T dew Dew point Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1186 Use the formulas and the steam tables to find the missing property of φ ω and Tdry total pressure is 100 kPa repeat the answers using the psychrometric chart a φ 50 ω 0010 b Tdry 25C Twet 21C Solution a From Eq1128 with Pa P Pv solve for Pv Pv P ω 0622 ω 100 0010632 1582 kPa From Eq1125 Pg Pvφ 158205 3165 kPa T 25C b At 21C Pg 2505 ω2 0622 2505100 2505 0016 From the steam tables B11 hf2 88126 and hfg2 245176 kJkg hv1 254717 From Eq1130 ω1 CpT2T1 ω2 hfg2 hv1 hf2 00143 From Eq1128 with Pa P Pv solve for Pv Pv P ω 0622 ω 2247 kPa From Eq1125 φ 22473169 071 Using the psychrometric chart E4 a Tdry 253 C b ω 00141 φ 7172 w T Φ 100 Φ Φ 21 70 50 25 dry a T wet b Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1187 The discharge moist air from a clothes dryer is at 40oC 80 relative humidity The flow is guided through a pipe up through the roof and a vent to the atmosphere Due to heat transfer in the pipe the flow is cooled to 24oC by the time it reaches the vent Find the humidity ratio in the flow out of the clothes dryer and at the vent Find the heat transfer and any amount of liquid that may be forming per kg dry air for the flow Solution State 1 Just outside chart Pa 08 7384 5907 kPa ω1 0622 5907 100 5907 00390 Tdew 358oC State 2 24oC Tdew so it is saturated ω2 0019 m liqm a ω1 ω2 00099 kgkg dry air 1 2 Energy Eq Q m a ha2 ω2 hv2 ha1 ω1hv1 ω1 ω2 hf Cpa T2 T1 ω2 hv2 ω1hv1 ω1 ω2 hf 100424 40 001925451 003925743 00099 1007 67 kJkg dry air Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1188 A flow 02 kgs dry air of moist air at 40C 50 relative humidity flows from the outside state 1 down into a basement where it cools to 16C state 2 Then it flows up to the living room where it is heated to 25C state 3 Find the dew point for state 1 any amount of liquid that may appear the heat transfer that takes place in the basement and the relative humidity in the living room at state 3 Solve using psychrometric chart a Tdew 272 w w1 φ 100 w1 00232 h1 1182 kJkg air b T2 Tdew so we have φ2 100 liquid water appear in the basement w2 00114 h2 644 and from steam tbl hf 6717 m liq m airw1w2 020023200114 000236 kgs c Energy equation m air h1 m liq hf m air h2 Q out Q out 021182 644 001186717 106 kW d w3 w2 00114 25C φ3 58 If you solve by the formulas and the tables the numbers are Pg40 7384 kPa Pv1 φ Pg40 05 7384 3692 kPa w1 0622 3692 100 3692 002384 Pv1 Pg Tdew Tdew 1 275 C 2 φ 100 Pv2 Pg2 1832 kPa w2 0622183298168 00116 m liq m air w1w2 02001223 000245 kgs 3 w3 w2 Pv3 Pv2 1832 kPa Pg3 3169 kPa φ3 PvPg 18323169 578 w T Φ 100 Φ Φ Φ 50 40 10 dry 1 2 T dew Dew point 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1189 A steady supply of 10 m3s air at 25C 100 kPa 50 relative humidity is needed to heat a building in the winter The outdoor ambient is at 10C 100 kPa 50 relative humidity What are the required liquid water input and heat transfer rates for this purpose Solution Air Ra 0287 kJkg K Cp 1004 kJkgK State 1 T1 10C φ1 50 P1 100 kPa Pg1 12276 kPa Pv1 φ1Pg1 06138 kPa Pa1 P1 Pv1 9939 kPa ω1 0622 Pv1Pa1 0003841 State 2 T2 25C P2 100 kPa φ2 50 V 2 1 m3s Pg2 3169 kPa Pv2 φ2Pg2 15845 kPa Pa2 P2 Pv2 98415 kPa ω2 0622 Pv2Pa2 0010014 m a2 Pa2 V 2RaT2 98415 10287 29815 115 kgs Steam tables B11 hv1 25197 kJkg hv2 25472 kJkg State 3 Assume liq water at T3 25C hf3 1049 kJkg Conservation of Mass m a1 m a2 m f3 m v2 m v1 m f3 m a2ω2 ω1 115 0006173 00071 kgs Energy Eq Q m a1ha1 m v1hv1 m f3hf3 m a2ha2 m v2h v2 q Q m a CpT2 T1 ω2hv2 ω1hv1 m f3 m a hf3 100425 10 0010014 25472 0003841 25197 0006173 1049 3024 kJkg dry air Q m a1 q 115 kgs 3024 kJkg 3478 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1190 In a ventilation system inside air at 34oC and 70 relative humidity is blown through a channel where it cools to 25oC with a flow rate of 075 kgs dry air Find the dew point of the inside air the relative humidity at the end of the channel and the heat transfer in the channel CV Cooler m v1 m liq m v2 Tables Pg1 5352 kPa Pv1 φ Pg 07 5352 375 kPa ω1 0622 375 101 375 0024 T2 Tdew 27C from PgTdew 375 Pv2 3169 kPa Pg2 ω2 0622 3169 101 3169 002015 hv1 256347 kJkg hv2 254717 kJkg hf 10487 kJkg q out CPT1 T2 ω1hv1 ω2 hv2 ω1 ω2 hf 100434 25 0024 256347 002015 254717 000385 10487 1883 kJkg dry air Q m a q out 075 kgs 1883 kJkg 141 kW Psychrometric chart State 2 T Tdew 275C φ2 100 m v1m a ω1 00234 h 1 1137 kJkg air m v2m a ω2 00202 h2 96 kJkg air m liqm a ω1 ω2 00032 hf 10487 kJkg m a q out m ah1 m liq hf m a h2 qout h1 ω1 ω2 hf h2 1137 00032 10487 96 1736 kJkgair Q m a qout 130 kW w T Φ 100 Φ Φ Φ 70 40 10 dry 1 2 T dew Dew point Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1191 Two moist air streams with 85 relative humidity both flowing at a rate of 01 kgs of dry air are mixed in a steady setup One inlet flowstream is at 325C and the other at 16C Find the exit relative humidity Solution CV mixing chamber Continuity Eq water m air w1 m air w2 2m air wex Energy Eq m air h1 m air h2 2m air h ex Properties from the tables and formulas Pg325 4937 kPa Pv1 0854937 kPa 4196 kPa w1 0622 4196 100 4196 00272 Pg16 1831 kPa Pv2 0851831 kPa 1556 kPa w2 0622 1556 100 1556 000983 Continuity Eq water wex w1 w22 00185 For the energy equation we have h ha whv so 2 hex h1 h2 0 2ha ex ha 1 ha 2 2wexhv ex w1hv 1 wh v 2 we will use constant specific heat to avoid an iteration on Tex Cp air2Tex T1 T2 Cp H2O2wexTex w1T1 w2T2 0 Tex Cp airT1 T2 Cp H2Ow1T1 w2T2 2Cp air 2wexCp H2O 1004 325 16 187200272 325 000983 1620773 244C Pv ex 0622 wex wex Ptot 00185 0622 00185 100 2888 kPa Pg ex 3069 kPa φ 2888 3069 094 or 94 Properties taken from the psychrometric chart State 1 w1 00266 h1 120 State 2 w2 00094 h2 60 Continuity Eq water wex w1 w22 0018 Energy Eq hex h1 h22 90 kJkg dry air exit wex hex Tex 245C φ 94 Notice how the energy in terms of temperature is close to the average of the two flows but the relative humidity is not Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1192 A combination air cooler and dehumidification unit receives outside ambient air at 35C 100 kPa 90 relative humidity The moist air is first cooled to a low temperature T2 to condense the proper amount of water assume all the liquid leaves at T2 The moist air is then heated and leaves the unit at 20C 100 kPa relative humidity 30 with volume flow rate of 001 m3s Find the temperature T2 the mass of liquid per kilogram of dry air and the overall heat transfer rate Solution MIX OUT MIX IN 2 LIQ OUT Q H Q C 1 2 HEAT COOL CV Pv1 φ1PG1 09 5628 50652 kPa w1 0622 50652 10050652 0033 19 Pv3 φ3PG3 03 2339 07017 kPa w2 w3 0622 07017 10007017 00044 m LIQ 2m a w1 w2 0033 19 00044 0028 79 kgkg air PG2 Pv3 07017 kPa T2 17 oC For a CV around the entire unit Q CV Q H Q C Net heat transfer Energy Eq Q CVm a ha3ha1 w3hv3 w1hv1 m L2 hL2m a 10042035 0004425381 0033 1925653 0028 79728 8882 kJkg air m a Pa3V3 RaT3 10007017001 02872932 00118 kgs Q CV 00118 kgs 8882 kJkg air 105 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1193 To make dry coffee powder we spray 02 kgs coffee assume liquid water at 80o C into a chamber where we add 10 kgs dry air at T All the water should evaporate and the air should leave with a minimum 40oC and we neglect the powder How high should T in the inlet air flow be CV Chamber We assume it is adiabatic Continuity Eq water alone m liq wex m a Energy Eq m liq hf 80 m a ha Ti m liq hv 40 m a h a 40 wex m liq m a 02 8 0025 From the energy equation you get ha Ti ha 40 Cp a Tin Tex m liq hv 40 hf 80 m a 1004 kJkgK T 02 257426 33488 10 44788 kJkg T 446oC Tin 40 446 846oC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1194 An insulated tank has an air inlet ω1 00084 and an outlet T2 22C φ2 90 both at 100 kPa A third line sprays 025 kgs of water at 80C 100 kPa For steady operation find the outlet specific humidity the mass flow rate of air needed and the required air inlet temperature T1 Solution Take CV tank in steady state Continuity and energy equations are Continuity Eq water m 3 m a w1 m a w 2 Energy Eq m 3hf m a h1 m a h 2 All state properties are known except T1 From the psychrometric chart we get State 2 w2 0015 h2 795 State 3 hf 33491 steam tbl m a m 3w2 w1 025001500084 3788 kgs h1 h2 w2 w1hf 795 00066 33491 773 Chart w1 h1 T1 365C Using the tables and formulas we get State 2 Pg2 2671 Pv2 09 2671 24039 kPa w2 0622 24039 100 24039 00153 m a m 3w2 w1 02500153 00084 3623 kgs To avoid iterations on T1 we use specific heat values also for water vapor by writing hv1 hv2 Cp h2oT1 T2 so the energy equation is Cp a T1 w1Cp h2oT1 T2 w1hv2 Cp a T2 w2hv2 w2 w1 h f The equation now becomes 1004 00084 1872T1 00084 1872 1004 22 00153 0008425417 33491 37219 T1 365C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1195 An air flow 2 kgs at 30C and relative humidity 80 is conditioned by taking half the flow cooling it and mix it with the other half Assume the outlet flow should have a water content that is 75 of the original flow Find the temperature to cool to the rate of cooling and the final exit flow temperature FIGURE P1195 CV Total setup Continuity water 0 m a ω1 m liq m a ω3 m liq m a ω1 ω3 State 1 Pg1 4246 Pv1 08 4246 33968 kPa ω1 0622 33968 100 33968 002187 CV Junction Continuity water 0 05 m a ω1 05 m a ω2 m a ω3 ω3 075 ω 1 Energy Eq 0 05 m aha1 ω1hv1 05 m aha2 ω2hv2 m aha3 ω3 hv3 Solve for ω2 ω2 2 ω3 ω1 15 ω1 ω1 05 ω1 0010935 State 2 φ2 100 ω2 0010935 Pv2 100 ω2 0622 ω2 1728 kPa Pv2 Pg2 T2 152C hf 6374 kJkg CV Cooler Q 05 m a h2 h1a 05 m a ω2hv2 ω1hv1 m liq hf 05 m a Cp a T2 T1 ω2hv2 ω1hv1 2ω1 ω3 hf 05 2 1 ω1 kgs 1004 152 30 0010935 25293 002187 255625 2 025 002187 6374 kJkg 415 kW From the energy equation for the junction ha3 ω3 hv3 05 ha1 ω1hv1 ha2 ω2hv2 051004 30 152 002187 255625 0010935 25293 64472 kJkg This is now trial and error on T3 T3 20C LHS 1004 20 0016403 253806 6171 kJkg T3 25C LHS 1004 25 0016403 254717 6688 kJkg Interpolate for final ans T3 227C Cooler 3 2 1 liquid Q cool Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1196 A watercooling tower for a power plant cools 45C liquid water by evaporation The tower receives air at 195C φ 30 100 kPa that is blown throughover the water such that it leaves the tower at 25C φ 70 The remaining liquid water flows back to the condenser at 30C having given off 1 MW Find the mass flow rate of air and the amount of water that evaporates Solution CV Total cooling tower steady state Continuity Eq for water in air win m evapm a wex Energy Eq m a hin m 1 h45 m a hex m 1 m evap h30 Inlet 195C 30 rel hum win 00041 hin 50 kJkg dry air Exit 25C 70 rel hum wex 00138 hex 80 kJkg dry air Take the two water flow difference to mean the 1 MW Q m 1 h45 m 1 m evap h30 1 MW Substitute this into the energy equation above and we get m ahex hin m a80 50 1000 kW m a 3333 kgs m evap wex win m a 00097 3333 0323 kgs The needed makeup water flow could be added to give a slightly different meaning to the 1 MW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1197 Moist air at 31oC and 50 relative humidity flows over a large surface of liquid water Find the adiabatic saturation temperature by trial and error Hint it is around 225oC For adiabatic saturation Φ2 1 and assume P 100 kPa energy Eq1130 ω1 hv1 hf2 CpT2 T1 ω2 hfg2 State 1 ω1 0622 P1 ΦPg1 ΦPg1 0622 05 45 100 05 45 001432 φ2 1 ω2 0622 Pg2P2 Pg2 Only one unknown in energy Eq T2 Trial and error on energy equation CpT2 ω2 hfg2 ω1 hf2 CpT1 ω1hv1 1004 31 001432 2558 677546 kJkg T2 20 oC Pg2 2339 kPa hf2 8394 kJkg hfg2 245412 kJkg ω2 0622 2339 97661 00149 LHS 1004 20 00149 245412 001432 8394 57848 T2 25 oC Pg2 3169 kPa hf2 10487 kJkg hfg2 24423 kJkg ω2 0622 3169 96831 002036 LHS 1004 25 002036 24423 001432 10487 76327 Linear interpolation to match RHS 677546 T2 20 25 20 677546 57848 76327 57848 227 oC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1198 A flow of air at 10C φ 90 is brought into a house where it is conditioned to 25C 60 relative humidity This is done with a combined heaterevaporator where any liquid water is at 10C Find any flow of liquid and the necessary heat transfer both per kilogram dry air flowing Find the dew point for the final mixture CV heater and evaporator Use psychrometric chart Inlet ω1 00069 h1 47 kJkg dry air hf 4199 kJkg Exit ω2 00118 h2 75 kJkg dry air Tdew 165C From these numbers we see that water and heat must be added Continuity eq for water m LIQ INm A ω2 ω1 00049 kgkg dry air Energy equation per kg dry air q h2 h1 ω2 ω1 hf 278 kJkg dry air Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1199 An air conditioner for an airport receives desert air at 45oC 10 relative humidity and must deliver this to the buildings at 20oC 50 relative humidity They have a cooling system with R410A running with high pressure of 3000 kPa and low pressure of 1000 kPa and their tap water is 18oC What should be done to the air Find the needed heatingcooling per kg dry air Check out the psychrometric chart State 1 w1 00056 h1 79 State 2 w2 00072 h2 58 kJkg Liquid tap hliq 75556 kJkg from B11 Now we know the following We must add water w2 w1 and then cool Twet 1 20oC Water continuity equation m liq m air w2 w1 Energy equation h1 w2 w1 hliq q h2 q h2 h1 w2 w1 hliq 58 79 00072 00056 75556 2112 kJkg dry air For the refrigeration cycle we can find from table B31 Plow 200 kPa Tevaporator 12oC which is cold enough Phigh 1500 kPa Tcondenser 59oC 45oC so it is hot enough No absolute scaling was provided the mass flow rates or W so we do not know if the motorcompressor combination is big enough Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11100 A flow of moist air from a domestic furnace state 1 is at 45oC 10 relative humidity with a flow rate of 005 kgs dry air A small electric heater adds steam at 100oC 100 kPa generated from tap water at 15oC Up in the living room the flow comes out at state 4 30oC 60 relative humidity Find the power needed for the electric heater and the heat transfer to the flow from state 1 to state 4 1 3 4 Liquid cb 2 Properties from the psychrometric chart State 1 w1 00056 h1 79 kJkg dry air State 4 w4 00160 h4 905 kJkg dry air Continuity equation for water from 1 to 4 m liq m a ω4 ω1 005 0016 00056 000052 kgs Energy Eq for heater Q heater m liq hout hin 000052 267605 6298 136 kW Energy Eq for line Q line m a h4 h1 m liq hvap 005905 79 000052 267605 0816 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11101 One means of airconditioning hot summer air is by evaporative cooling which is a process similar to the adiabatic saturation process Consider outdoor ambient air at 35C 100 kPa 30 relative humidity What is the maximum amount of cooling that can be achieved by such a technique What disadvantage is there to this approach Solve the problem using a first law analysis and repeat it using the psychrometric chart Fig E4 Ambient Air Air 1 2 3 Liquid Cooled P1 P2 100 kPa Pv1 φ1Pg1 0305628 16884 ω1 0622168849831 001068 For adiabatic saturation Max cooling is for φ2 1 Energy Eq Eq1130 ω1 hv1 hf2 CpT2 T1 ω2 hfg2 φ2 1 ω2 0622 PG2P2 PG2 Only one unknown T2 Trial and error on energy equation CpT2 ω2 hfg2 ω1 hf2 CpT1 ω1hv1 1004 35 001068 25653 62537 kJkg T2 20 oC PG2 2339 kPa hf2 8394 hfg2 245412 kJkg ω2 0622 2339 97661 00149 LHS 1004 20 00149 24541 001068 8394 57543 kJkg T2 25 oC PG2 3169 kPa hf2 10487 hfg2 24423 kJkg ω2 0622 3169 96831 002036 LHS 1004 25 002036 24423 001068 10487 75945 kJkg linear interpolation T2 214 oC This method lowers the temperature but the relative and absolute humidity becomes very high and the slightest cooling like on a wall results in condensation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Solution 11101 Continued b chart E4 Adiabatic saturation T2 WetBulbTemperature 215 oC w T Φ 100 Φ Φ Φ 80 30 10 dry 1 2 35 215 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11102 A flow out of a clothes dryer of 005 kgs dry air is at 40oC and relative humidity 60 It flows through a heat exchanger where it exits at 20oC After heat exchanger the flow combines with another flow of 003 kgs dry air at 30oC and relative humidity 30 Find the dew point of state 1 see Fig P11102 the heat transfer per kg dry air and the final exit state humidity ratio and relative humidity Use the psychrometric chart to solve the problem State 1 w 00286 h1 1315 Tdew 31oC State 2 20oC Tdew so it is saturated w2 00148 h2 778 kJkg air State 3 w3 00078 h3 698 kJkg air m liqm a ω1 ω2 00138 kgkg dry air Energy Eq Q m a h1 h2 ω1 ω2 h f 1315 778 00138 8394 525 kJkg dry air Do CV around the junction where flow 2 and 3 combines to give exit at 4 Continuity water m a1 ω2 m a3 ω3 m a4 ω4 ω4 m a1m a4 ω2 m a3m a4 ω3 005 008 00148 003 008 00078 0012175 Energy Equation m a1 h2 m a3 h3 m a4 h 4 h4 m a1m a4 h2 m a3m a4 h 3 005 008 778 003 008 698 748 kJkg air From chart given ω4h4 we get Φ4 65 and T4 24oC 1 4 2 3 liq q Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11103 Atmospheric air at 35C relative humidity of 10 is too warm and also too dry An air conditioner should deliver air at 21C and 50 relative humidity in the amount of 3600 m3 per hour Sketch a setup to accomplish this find any amount of liquid at 20C that is needed or discarded and any heat transfer CV air conditioner First we must check if water should be added or subtracted We can know this from the absolute humidity ratio Properties from the tables and formulas State 1 Pg35 5628 kPa Pv1 0105628 kPa 05628 kPa w1 0622 05628 101325 05628 0003474 State 2 Pg21 2505 kPa Pv2 052505 kPa 1253 kPa w2 0622 1253 101325 1253 0007785 As w goes up we must add liquid water Now we get Continuity Eq m air 1 w1 m liq m air 1 w2 Energy Eq m airh1 m liqhf Q CV m airh2 For the liquid flow we need the air mass flowrate out 3600 m3h 1 m3s m air Pa 2V RT 101325 12531028729415 1185 kgs m liq m air w2 w1 000511 kgs 184 kgh Q CV m air Cp aT2 T1 w2hv2 w1hv1 m liqh f 1185 1004 21 35 0007785 25399 0003474 25653 000511 8396 421 kW If from psychrometric chart Inlet w1 00030 h1 630 hf20 8396 kJkg Exit w2 00076 h2 602 kJkg dry air Pv2 m air same as above Q CV m air h2 h1 m liqhf 1185602 63 000511 8396 374 kW Liquid water Cooler Inlet Exit 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11104 In a cars defrostdefog system atmospheric air 21C relative humidity 80 is taken in and cooled such that liquid water drips out The now dryer air is heated to 41C and then blown onto the windshield where it should have a maximum of 10 relative humidity to remove water from the windshield Find the dew point of the atmospheric air specific humidity of air onto the windshield the lowest temperature and the specific heat transfer in the cooler Solution Solve using the psychrometric chart 1 2 3 Liquid Q cool Q heat w T Φ 100 Φ Φ 80 10 dry 1 2 T dew 1 Dew point 3 for 1 T dew 3 Air inlet 21C φ 80 w1 00124 Tdew 173C h1 72 Air exit 41C φ 10 w3 00044 Tdew 19C To remove enough water we must cool to the exit Tdew followed by heating to Tex The enthalpy from chart h2 325 and from B11 hf19C 8 kJkg CV cooler m liqm air w1 w3 00124 00044 0008 kg liqkg air q Q CVm air h2 w1 w3 hf h1 325 00088 72 394 kJkg dry air If the steam and air tables are used the numbers are State 1 Pg1 2505 kPa Pv1 2004 kPa w1 001259 hg1 25399 ha1 2943 h1 3263 kJkg State 3 Pg3 7826 Pv3 0783 w3 000486 State 2 wg3 w3 T2 T3dew 33C hf2 1377 kJkg hg2 25074 ha2 27656 h2 28875 kJkg m liqm air 000773 q 28875 000773 1377 3263 3745 kJkg air Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11105 A commercial laundry runs a dryer that has an exit flow of 05 kgs moist air at 48 oC 70 relative humidity To save on the heating cost a counterflow stack heat exchanger is used to heat a cold water line at 10oC for the washers with the exit flow as shown in Fig P11105 Assume the outgoing flow can be cooled to 25oC Is there a missing flow in the figure Find the rate of energy recovered by this heat exchanger and the amount of cold water that can be heated to 30oC The flow out of the dryer is outside the range for the psychrometric chart so the solution is done with the tables and the formulas Dryer oulet 1 48C Φ 70 hg1 258851 kJkg Pg1 11247 kPa Pv1 07 11247 kPa 7873 kPa ω1 0622 7873 101325 7873 005241 w T Φ 100 Φ 70 dry 1 2 T dew 1 Dew point for 1 T 2 State 2 T2 Tdew 1 42oC so Φ2 100 Pg2 3169 kPa hg2 254717 kJkg hf2 10487 kJkg ω2 0622 3169101325 3169 002008 Continuity Eq water 12 line m a ω1 m a ω2 m liq m liq m aω1 ω2 The mass flow rate of dry air is m a m moist air 1 ω1 05 kgs 1 005241 04751 kgs The heat out of the exhaust air which also equals the energy recovered becomes Q CV m a Cp aT1 T2 ω1 hg1 ω1 ω2 hf2 ω2 hg2 04751 100448 25 005241 258851 003233 10487 002008 254717 04751 kgs 1042 kJkg 4951 kW m liq Q CV CP liq ΔTliq 4951 418 30 10 0592 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11106 A flow of moist air at 45oC 10 relative humidity with a flow rate of 02 kgs dry air is mixed with a flow of moist air at 25oC and absolute humidity of w 0018 with a rate of 03 kgs dry air The mixing takes place in an air duct at 100 kPa and there is no significant heat transfer After the mixing there is heat transfer to a final temperature of 25oC Find the temperature and relative humidity after mixing Find the heat transfer and the final exit relative humidity Solution CV Total Setup state 3 is internal to CV Continuity Eq m a1 w1 m a2 w2 m a1 m a2 w3 m a1 m a2 w 4 Energy Eq m a1 h1 m a2 h2 m a1 m a2 h 3 State 1 From Psychrometric chart w1 00056 h1 79 kJkg dry air State 2 From Psychrometric chart Φ2 90 h2 905 kJkg dry air w3 w4 m a1w1 m a2w2 m a1 m a2 02 05 00056 03 05 0018 001304 h3 m a1h1 m a2h2 m a1 m a2 02 05 79 03 05 905 859 kJkg dry air State 3 From Psychrometric chart T3 325oC Φ3 45 State 4 25oC w4 001304 Read from Psychrometric chart h4 78 Φ4 66 Now do the energy equation for the whole setup Energy Eq m a1 h1 m a2 h2 Q m a1 m a2 h4 Q m a1 m a2 h4 m a1 h1 m a2 h2 05 78 02 79 03 905 395 kW Q heat cool 1 2 4 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11107 An indoor pool evaporates 1512 kgh of water which is removed by a dehumidifier to maintain 21C φ 70 in the room The dehumidifier shown in Fig P11107 is a refrigeration cycle in which air flowing over the evaporator cools such that liquid water drops out and the air continues flowing over the condenser For an air flow rate of 01 kgs the unit requires 14 kW input to a motor driving a fan and the compressor and it has a coefficient of performance β QLWc 20 Find the state of the air as it returns to the room and the compressor work input Solution The unit must remove 1512 kgh liquid to keep steady state in the room As water condenses out state 2 is saturated State 1 21C 70 w1 00108 h1 685 CV 1 to 2 m liq m aw1 w2 w2 w1 m liqm a qL h1 h2 w1 w2 hf2 w2 00108 1512360001 00066 State 2 w2 100 T2 8C h2 45 hf2 336 qL 685 45 00042 336 2336 kJkg dry air CV Total system h3 h1 W elm a w1w2 hf 685 14 014 8236 kJkg dry air State 3 w3 w2 h3 T3 46C φ3 1112 W c m a qL β 1165 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11108 A moist air flow of 5 kgmin at 30oC Φ 60 100 kPa goes through a dehumidifier in a setup shown in Problem 11107 The air is cooled down to 15oC and then blown over the condenser The refrigeration cycle runs with R134a with a low pressure of 200 kPa and a high pressure of 1000 kPa Find the COP of the refrigeration cycle the ratio m R134am air and the outgoing T3 and Φ3 Standard Refrigeration Cycle Table B5 h1 39215 kJkg s1 1732 kJkg K h4 h3 25556 kJkg CV Compressor assume ideal m 1 m 2 wC h2 h1 s2 s1 s gen P2 s s1 h2 42571 kJkg wC 3356 kJkg CV Evaporator qL h1 h4 39215 25556 13659 kJkg CV Condenser qH h2 h3 42571 25556 17015 kJkg COP β qL wC 13659 3356 407 C 3 4 1 2 W C Air in Air sat 15 C Air ex R134a Evaporator R134a Condenser o liquid P1 200 kPa P2 1000 kPa For the air processes let us use the psychrometric chart Air inlet win 0016 hin 905 kJkg dry air Tdew 21oC 15oC Air 15oC φ 100 w7 00107 h7 62 hf 6298 B11 Now do the continuity for water and energy equations for the cooling process m liqm air win w7 0016 00107 00053 kgkg air qcool hin h7 m liqhfm air 905 62 00053 6298 2817 kJkg air T s 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Solution 11108 continued Now the cooling of the air is done by the R134a so Q cool m air qcool m R134a qL m R134am air qcool qL 2817 13659 02062 Energy eq for the air flow being heated Q heat m air hex h7 hex h7 Q heat m air h7 qH m R134am air hex 62 17015 02062 9708 kJkg dry air and wex w 7 Locate state in the psychrometric chart just outside edge of chart Tex 493oC and φex 15 w T Φ 60 Φ 30 dry 1 15 21 30 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Psychrometric chart only Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11109 Use the psychrometric chart to find the missing property of φ ω Twet T dry a Tdry 25C φ 80 b Tdry 15C φ 100 cTdry 20C and ω 0010 d Tdry 25C Twet 23C Solution a 25C φ 80 ω 0016 Twet 223C b 15C φ 100 ω 00106 Twet 15C c 20C ω 0010 φ 70 Twet 163C d 25C Twet 23C ω 0017 φ 86 w T Φ 100 Φ Φ Φ 80 40 10 dry a d b c Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11110 Use the psychrometric chart to find the missing property of φ ω Twet T dry a φ 50 ω 0014 b Twet 15C φ 60 c ω 0008 and Twet 15C d Tdry 10C ω 0006 Solution a φ 50 ω 0014 Tdry 31C Twet 226 C b Twet 15C φ 60 Tdry 202C ω 00086 c ω 0008 Twet 15C Tdry 218C φ 50 d Tdry 10C ω 0006 φ 80 Twet 82C w T Φ 100 Φ Φ Φ 80 40 10 dry a d b c Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11111 For each of the states in Problem 11110 find the dew point temperature Solution The dew point is the state with the same humidity ratio abs humidity ω and completely saturated φ 100 From psychrometric chart a Tdew 192C c Tdew 108C b Tdew 12C d Tdew 65C Finding the solution from the tables is done for cases ac and d as Eq1128 solve Pv Pg ωPtot ω 0622 Pg Tdew in B11 For case b use energy Eq 1330 to find ω1 first from Tad sat Twet w T Φ 100 Φ Φ Φ 80 40 10 dry a d b c Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11112 Use the formulas and the steam tables to find the missing property of φ ω and Tdry total pressure is 100 kPa repeat the answers using the psychrometric chart a φ 50 ω 0010 b Twet 15C φ 50 c Tdry 25C Twet 21C a From Eq1128 with Pa P Pv solve for Pv Pv P ω 0622 ω 100 0010632 1582 kPa From Eq1125 Pg Pvφ 158205 3165 kPa T 25C b Assume Twet is adiabatic saturation T and use energy Eq1130 At 15C Pg 1705 kPa ω 0622 1705 100 1705 001079 LHS ω1 hv1 hf2 CpT1 RHS CpT2 ω2 hfg2 RHS 100415 001079 246593 41667 kJkg ω1 0622 φPg100 φPg where Pg is at T1 Trial and error LHS25C 4998 LHS20C 383 T 214C ω1 0008 c At 21C Pg 2505 kPa ω2 0622 2505 100 2505 0016 hf2 88126 and hfg2 245176 kJkg hv1 254717 kJkg From Eq1130 ω1 CpT2T1 ω2 hfg2 hv1 hf2 00143 Pv P ω 0622 ω 2247 φ 22473169 071 Using the psychrometric chart E4 a Tdry 253 C b Tdry 216C ω 0008 c ω 00141 φ 7172 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11113 An airconditioner should cool a flow of ambient moist air at 40C 40 relative humidity with 02 kgs flow of dry air The exit temperature should be 20C and the pressure is 100 kPa Find the rate of heat transfer needed and check for the formation of liquid water Solution Before we know if we should have a liquid water flow term we need to check for condensation the dew point Using the psychrometric chart i ωi 0018 hi 106 kJkg air Tdew 23C Since Tdew Te then condensation occurs and the exit is φ 100 Exit state 20C he 775 kJkg air ωe 00148 hf 8394 kJkg CV heat exchanger Water m a ωi m liq m a ωe m liq m a ωi ωe Energy ha ωhvi q ha ωhve ωi ωe hf q he hi ωi ωe hf 775 106 0018 00148 8394 2823 kJkg dry air Q m a q 02 kgs 2823 kJkg 565 kW it goes out Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11114 A flow of moist air at 21C 60 relative humidity should be produced from mixing of two different moist air flows Flow 1 is at 10C relative humidity 80 and flow 2 is at 32C and has Twet 27C The mixing chamber can be followed by a heater or a cooler No liquid water is added and P 100 kPa Find the two controls one is the ratio of the two mass flow rates m a1m a2 and the other is the heat transfer in the heatercooler per kg dry air Solution CV Total Setup state 3 is internal to CV Continuity Eq m a1 w1 m a2 w2 m a1 m a2 w4 Energy Eq m a1 h 1 m a2 h 2 Q a1 m a1 m a2 h 4 Define x m a1m a2 and substitute into continuity equation x w1 w2 1x w4 x w4 w2 w1 w4 3773 Energy equation scaled to total flow of dry air q Q a1m a1 m a2 h 4 x1x h 1 11x h2 64 07905 45 02095 105 643 kJkgdry air w T Φ 80 Φ Φ 60 40 dry 1 2 4 State 1 w1 0006 h 1 45 State 2 w2 00208 h 2 105 State 4 w4 00091 h 4 64 Tdew 4 125C Q heat cool 1 2 4 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11115 Consider a mixing process as in Fig P11114 where flow 1 comes in as cold and moist 10C φ 90 It is mixed with a flow 2 at 42C Twet 31C If the exit flow should be at 21C φ 50 find the ratio m a1m a2 and the heat transfer per kg dry air out CV Total Setup state 3 is internal to CV Assume no liquid flow then the balance equations are Continuity Eq m a1 w1 m a2 w2 m a1 m a2 w4 Energy Eq m a1 h1 m a2 h2 Q m a1 m a2 h4 The continuity equation is solved for the ratio x m a1m a2 x m a1m a2 w2 w4 w4 w1 00245 00076 00076 00067 187778 Then solve the energy eq for the heat transfer Q m a4 h4 x 1 x h1 1 1 x h2 60 187778 197778 47 1 197778 124 9107 kJkg air out w T Φ 90 Φ Φ 50 40 dry 1 2 4 State 1 w1 00067 h 1 47 State 2 w2 00245 h 2 124 State 4 w4 00076 h 4 60 Q heat cool 1 2 4 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11116 In a hot and dry climate air enters an airconditioner unit at 100 kPa 40C and 5 relative humidity at the steady rate of 10 m3s Liquid water at 20C is sprayed into the air in the AC unit at the rate 20 kghour and heat is rejected from the unit at at the rate 20 kW The exit pressure is 100 kPa What are the exit temperature and relative humidity State 1 T1 40C P1 100 kPa φ1 5 V a1 1 m3s Pg1 73837 kPa Pv1 φ1Pg1 0369 kPa Pa1 P Pv1 9963 kPa ω1 0622 Pv1 Pa1 00023 m a1 Pa1V a1 RTa1 1108 kgs hv1 25743 kJkg State 2 Liq Water 20C m f2 20 kghr 000556 kgs hf2 839 kJkg Conservation of Mass m a1 m a3 m v1 m 12 m v3 ω3 m f2 m a1 ω1 0005561108 00023 00073 State 3 P3 100 kPa and Pv3 P3ω30622 ω3 116 kPa Energy Eq with Q 20 kW Q m a1ha1 m v1hv1 m f2hf2 m a3ha3 m v3hv3 ha3ha1 ω3hv3 ω1hv1 m f2hf2 Q m a1 00023 25743 000556 839 201108 117 Unknowns ha3 hv3 implicitly given by a single unknown T3 Trial and Error for T EA PAv3 A EPAg3 AE A 094 3 T3 10C Pg3 123 kPa φ3 If we solved with the psychrometric chart we would get State 1 Am v1E AAm aE A ωA1E A 0002 AhE A1E A 65 kJkg dry air State 3 ωA3E A Am E Af2E A Am E Aa1E A ωA1E A 0005561108 0002 0007 Now the energy equation becomes AhE A3E A AhE A1E A Am E Af2E AhAf2E A AQ E A Am E Aa1E A 65 000556839 201108 474 Given ωA3E A we find the state around 10C and φA3E A 90 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11117 Compare the weather two places where it is cloudy and breezy At beach A it is 20C 1035 kPa relative humidity 90 and beach B has 25C 99 kPa relative humidity 40 Suppose you just took a swim and came out of the water Where would you feel more comfortable and why Solution As your skin is wet and air is flowing over it you will feel TAwetE A With the small difference in pressure from 100 kPa we can use the psychrometric chart A 20C φ 90 TAwetE A 187C B 25C φ 40 TAwetE A 16C At beach A it is comfortable at beach B it feels chilly w T Φ 100 Φ Φ Φ 80 40 10 dry A B Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11118 Ambient air at 100 kPa 30C 40 relative humidity goes through a constant pressure heat exchanger as a steady flow In one case it is heated to 45C and in another case it is cooled until it reaches saturation For both cases find the exit relative humidity and the amount of heat transfer per kilogram dry air Solution CV heat exchanger AmE AAiE A AmE AAeE A AmE AviE A AmE AveE A wAeE A wAiE hAaE A whAvE AAiE A q hAaE A whAvE AAeE A AhE AeE A q AhE AeE A AhE AiE Using the psychrometric chart i wAiE A 00104 AhE AiE A 76 Case I e TAeE A 45 AoE AC wAeE A wAiE A AhE AeE A 92 φAeE A 17 q 9276 16 kJkg dry air Case II e wAeE A wAiE A φAeE A 100 AhE AeE A 61 TAeE A 145AoE AC q 6176 15 kJkg dry air w T Φ 100 Φ Φ Φ 80 40 10 dry i CASE II e e CASE I Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11119 A flow of moist air at 100 kPa 35AoE AC 40 relative humidity is cooled by adiabatic evaporation of liquid 20AoE AC water to reach a saturated state Find the amount of water added per kg dry air and the exit temperature Since the liquid is not necessarily at the adiabatic saturation temperature the exit may be close to but not exactly the wet bulb temperature We will use that as a good guess and check the energy equation Chart E4 TAwetE A 236AoE AC wA1E A 00138 AhE A1E A mix 90 kJkg wA2E A 00186 AhE A2mixE A 906 kJkg Continuity Eq AmE AAE A1 wA1E A AmE AliqE A AmE AAE A1 wA2E A Energy Eq q 0 AmE AAE AhE A1mix m liqhAfE A AmE AAE AhE A2mixE Divide the energy equation by the mass flow rate of air m liqAmE AAE A wA2E A wA1E A 00186 00138 00048 kg waterkg air so AhE A1mix wA2E A wA1E A hAfE A AhE A2mixE A LHS 90 00048 8394 904 RHS 906 The temperature should be a little lower which will lower wA2E A also so TA2E A 235AoE AC wA2E A 00185 m liqAmE AAE A 00047 kg waterkg air This is close to the accuracy by which we can read the chart and the first answer is nearly as good Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11120 A flow out of a clothes dryer of 01 kgs dry air is at 60AoE AC and relative humidity 60 It flows through a heat exchanger where it exits at 20AoE AC After the heat exchanger the flow combines with another flow of 003 kgs dry air at 30AoE AC and relative humidity 40 Find the dew point of state 1 see Fig P11120 the heat transfer per kg dry air and the final exit state humidity ratio and relative humidity Use the psychrometric chart to solve the problem except for state 1 State 1 PAg1E A 19941 kPa PAv1E A ΦA1E A PAg1E A 11965 kPa TAdewE A TAsatE A PAv1E A 493AoE AC ωA1E A 0622 A 11965 100 11965E A 0084534 AhE A1E A hAa1E A hAa 20CE A ωA1E AhAa1E A 33381 25345 0084534 260959 301 kJkg air State 2 20AoE AC TAdewE A so it is saturated ωA2E A 00148 AhE A2E A 778 kJkg air State 3 ωA3E A 00078 AhE A3E A 698 kJkg air Am E AliqE AAm E AaE A ωA1E A ωA2E A 006973 kgkg dry air Energy Eq AQ E AAm E AaE A AhE A1E A AhE A2E A ωA1E A ωA2E A hAfE A 301 778 006973 8394 21735 kJkg dry air Do CV around the junction where flow 2 and 3 combines to give exit at 4 Continuity water Am E Aa1E A ωA2E A Am E Aa3E A ωA3E A Am E Aa4E A ωA4E ωA4E A Am E Aa1E AAm E Aa4E A ωA2E A Am E Aa3E AAm E Aa4E A ωA3E A A005 008E A 00148 A003 008E A 00078 0012175 Energy Equation Am E Aa1E A AhE A2E A Am E Aa3E A AhE A3E A Am E Aa4E A AhE A4E AhE A4E A Am E Aa1E AAm E Aa4E A AhE A2E A Am E Aa3E AAm E Aa4E A AhE A3E A005 008E A 778 A003 008E A 698 748 kJkg air From chart given ωA4E AAhE A4E A we get ΦA4E A 65 and TA4E A 24AoE AC 1 4 2 3 liq q Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11121 Consider two states of atmospheric air 1 40C φ 50 and 2 25C TAwetE A 16C Suggest a system of devices that will allow air in a steady flow process to change from 1 to 2 and from 2 to 1 Heaters coolers dehumidifiers liquid traps etc are available and any liquidsolid flowing is assumed to be at the lowest temperature seen in the process Find the specific and relative humidity for state 1 dew point for state 2 and the heat transfer per kilogram dry air in each component in the systems Use the psychrometric chart E4 1 w1 00232 AhE A1 1185 ΦA1E A 50 Tdew 272C AhE Adew 106 2 w2 00077 AhE A2 65 Φ2 40 Tdew 102C AhE Adew 50 Since w2 w1 water must be removed in process I to II and added in the process II to I Water can only be removed by cooling below dew point temperature so I to II Cool to dew point 3 then cool to state 4 dew pt for state 2 while liquid water is removed then heat to state 2 II to I Heat while adding water once w2 reached only heat Am E AliqE AAm E AaE A ωA1E A ωA2E A 00232 00077 00155 I to II qcool AhE A1E A AhE A4E A ωA1E A ωA2E A hfA4E 1185 50 00155 4283 678 kJkg air qheat AhE A2 AhE A4E A 65 50 15 kJkg air II to I qheat AhE A1E A AhE A2 ωA1E A ωA2E A hf2 1185 65 00155 10487 519 kJkg air w T Φ 100 Φ Φ Φ 80 50 10 dry 1 3 4 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11122 To refresh air in a room a counterflow heat exchanger see Fig P11122 is mounted in the wall drawing in outside air at 05C 80 relative humidity and pushing out room air 40C 50 relative humidity Assume an exchange of 3 kgmin dry air in a steady flow device and also that the room air exits the heat exchanger at 23C to the outside Find the net amount of water removed from the room any liquid flow in the heat exchanger and T φ for the fresh air entering the room We will use the psychrometric chart to solve this problem State 3 room w3 00232 AhE A3 1192 Tdew 3 27C State 1outside 05C φ 80 w1 00032 AhE A1 292 kJkg dry air CV room AmE Avout AmE Aa w3 w2 AmE Aa w3 w1 30023200032 006 kgmin The room air is cooled to 23C Tdew 3 so liquid will form in the exit flow channel and state 4 is saturated State 4 23C φ 100 w4 00178 AhE A4 88 hf4 9652 kJkg CV 3 to 4 flow cooled below Tdew 3 so liquid forms AmE Aliq 4 AmE Aa w3 w4 3 00232 00178 00162 kgmin CV Heat exchanger with no external heat transfer maAhE A2 AhE A1 maAhE A3 AhE A4 mliqhf4 AhE A2 AhE A1 AhE A3 AhE A4 w3w4 hf4 292 1192 88 000549652 599 kJkg dry air State 2 w2 w1 AhE A2 T2 325C φ 12 w T Φ 100 Φ Φ Φ 80 50 10 dry 3 4 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Exergy in mixtures Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11123 Consider several flow processes with ideal gasses 1 A compression of a gas mixture from 100 kPa to 200 kPa 2 Cooling a gas mixture from 50C to ambient 20C using ambient air 3 Mixing two different gasses at 100 kPa 4 Throttle a gas mixture from 125 kPa to 100 kPa For each case explain What happens to the exergy is there any exergy destruction and is the composition needed 1 Exergy increases mixture specific heat and gas constant needed 2 Exergy decreases of mixture and some exergy is destroyed due to heat transfer over T Specific heat for mixture needed 3 Composition needed exergy is destroyed 4 Exergy is destroyed mixture gas constant needed Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11124 Find the second law efficiency of the heat exchanger in Problem 1154 A flow of 2 kgs mixture of 50 COA2E A and 50 OA2E A by mass is heated in a constant pressure heat exchanger from 400 K to 1000 K by a radiation source at 1400 K Find the rate of heat transfer and the entropy generation in the process Solution The second law efficiency follows Eq1132 where the wanted term is the flow increase of exergy and the source is the radiation AΦ E AflowE A AmE AψAexE A ψAinE A AΦ E AsourceE A AQ E AinE A 1 To Tsource Heat exchanger Energy Eq412 AQ E AinE A AmE AhAeE A hAiE A Values from Table A8 due to the high T AQ E AinE A 2 A1 2E A 97167 30376 A1 2E A 98095 36603 12828 kW AΦ E AsourceE A AQ E AinE A 1 To Tsource 12828 1 A29815 1400E A 10096 kW Entropy Eq78 AmE AeE AsAeE A AmE AiE AsAiE A AQ E ATAsE A S gen As P C the pressure correction in Eq628 drops out to give generation as S gen AmE AsAeE A sAiE A AQ E ATAsE A 2 05 6119 51196 05 76121 66838 128281400 101 kWK AΦ E AflowE A AΦ E AsourceE A AΦ E AdestructionE A AΦ E AsourceE A T S gen 10096 29815 101 7085 η A Φ flow EΦ source E A A 7085 10096E A 070 Remark We could also explicitly have found the flow exergy increase 1400 K i e Radiation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11125 Consider the mixing of a steam flow with an oxygen flow in Problem 1155 Find the rate of total inflowing exergy and the rate of exergy destruction in the process A flow of 18 kgs steam at 400 kPa 400AoE AC is mixed with 32 kgs oxygen at 400 kPa 400 K in a steady flow mixingchamber without any heat transfer Find the exit temperature and the rate of entropy generation Exergy Flow AΦ E AinE A Am E A ψAinE A AmE AH2OE A ψA1E A AmE AO2E A ψA2E ψA1E A hA1E A hAoE A TAoE AsA1E A sAoE A CAP H2OE ATA1E A TAoE A TAoE A CAP H2OE A lnTA1E ATAoE A R lnPA1E APAoE A 1872 400 25 29815 1872 ln A67315 29815E A 04615 ln A400 100E A 702 29815 15245 063978 4382 kJkg ψA2E A hA2E A hAoE A TAoE AsA2E A sAoE A CAP O2E ATA2E A TAoE A TAoE A CAP O2E A lnTA2E ATAoE A R lnPA2E APAoE A 09221268525 298150922 ln A 400 29815E A 02598 ln A400 100E A 93906 29815 027095 036016 1205 kJkg AΦ E AinE A AmE AH2OE A ψA1E A AmE AO2E A ψA2E A 18 4382 32 1205 11744 kW CV Mixing chamber steady flow no work no heat transfer To do the entropies we need the mole fractions AnE AH2OE A A m H2O EMH2O E A A 18 18015E A 01 kmols AnE AO2E A A m O2 EMO2 E A A 32 31999E A 01 kmols yAH2OE A yAO2E A 05 Energy Eq AmE AH2OE A hA1E A AmE AO2E A hA2E A AmE AH2OE A hA3 H2OE A AmE AO2E A hA3 O2E Entropy Eq AmE AH2OE A sA1E A AmE AO2E A sA2E A AS E AgenE A AmE AH2OE A sA3 H2OE A AmE AO2E A sA3 O2E Solve for T from the energy equation AmE AH2OE A hA3 H2OE A hA1E A AmE AO2E A hA3 O2E A hA2E A 0 AmE AH2OE A CAP H2OE ATA3E A TA1E A AmE AO2E A CAP O2E ATA3E A TA2E A 0 18 1872 TA3E A 400 27315 32 0922TA3E A 400 0 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful TA3E A 5456 K AS E AgenE A AmE AH2OE A sA3 H2OE A sA1E A AmE AO2E A sA3 O2E A sA2E A AmE AH2OE A CAP H2OE A ln A T3 ET1 E A R ln yAH2OE A AmE AO2E A CAP O2E A ln A T3 ET2 E A R ln yAO2E A 18 1872 ln A 5456 67315E A 04615 ln 05 32 0922 ln A5456 400E A 02598 ln 05 0132 1492 136 kWK The exergy destruction is proportional to the entropy generation AΦ E AinE A TAoE A AS E AgenE A 29815 136 4055 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11126 A mixture of 75 carbon dioxide and 25 water by mole is flowing at 1600 K 100 kPa into a heat exchanger where it is used to deliver energy to a heat engine The mixture leave the heat exchanger at 500 K with a mass flow rate of 2 kgmin Find the rate of energy and the rate of exergy delivered to the heat engine CV Heat exchanger steady flow and no work From Table A8 CO2 hAinE A 174812 kJkg sA o T inE A 67254 kJkg K CO2 hAexE A 40152 kJkg sA o T exE A 53375 kJkg K H2O hAinE A 348769 kJkg sA o T inE A 140822 kJkg K H2O hAexE A 93512 kJkg sA o T exE A 114644 kJkg K Energy Eq AQ E A Am E A hAinE A hAexE A Am E A ci hin hexi A 2 60E A 075 174812 40152 025348769 93512 A 1 30E A 100995 63814 5494 kW Entropy change P does not change so partial pressures are constant sAinE A sAexE A 07567254 53375 025140822 119644 16954 kJkg K Exergy Flux AΦ E A Am E A ψAinE A ψAexE A AQ E A TAoE A Am E A sAinE A sAexE A 5494 29815 A 1 30E A 16954 3809 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11127 For flows with moist air where the water content is changed either by evaporation or by condensation what happens to the exergy Is the water vapor in air flowing over a lake in equilibrium with the liquid water If the liquid water and the water vapor in the air were in thermodynamic equilibrium then there would be no tendency to move from one phase to the other The water vapor is at its partial pressure and the air mixture temperature whereas the liquid water is at 1 atm pressure and at the liquid water temperature In such situations equilibrium is attained when the partial Gibbs function g h Ts of the water is the same in the two phases see Chapter 14 Notice if T TAoE A the Gibbs function becomes the same expression as for the flow exergy The water will transport from the higher Gibbs function exergy towards the lower Gibbs function exergy if there is a difference between the two The difference then determines if evaporation or condensation occurs If the temperature of the liquid and vapor is the same then the vapor should be at the saturated pressure PAgE A plus a very small amount ε to be in equilibrium with the liquid When the conditions are close to equilibrium the exergy of the water in the two phases are nearly equal and there is then no difference in the exergy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11128 A semipermeable membrane is used for the partial removal of oxygen from air that is blown through a grain elevator storage facility Ambient air 79 nitrogen 21 oxygen on a mole basis is compressed to an appropriate pressure cooled to ambient temperature 25C and then fed through a bundle of hollow polymer fibers that selectively absorb oxygen so the mixture leaving at 120 kPa 25C contains only 5 oxygen The absorbed oxygen is bled off through the fiber walls at 40 kPa 25C to a vacuum pump Assume the process to be reversible and adiabatic and determine the minimum inlet air pressure to the fiber bundle State 1 is the air inlet state 2 is the 5 00416079 00416 oxygen exit mix state 3 pure OA2E A exit 079 NA2E 021 OA2E 079 NA2E A 00416 OA2E A 01684 OA2E Let reference be at state 1 so s 0 at T 25AoE AC PA1E AsE AMIX 1E A 0 0 yAA1E ARE A ln yAA1E A yAB1E ARE A ln yAB1E AsE AMIX 2E A 0 0 ARE A ln PA2E APA1E A yAA2E ARE A ln yAA2E A yAB2E ARE A ln yAB2E Pure OA2E A AsE A3E A 0 ARE A ln PA3E APA1E A Entropy equation AnE A1E AsE A1E A AnE A2E AsE A2E A AnE A3E AsE A3E A All Ts the same so only partial pressure terms ARE A08316 ln PA2E APA1E A 079 ln 095 00416 ln 005 01684 ln PA3E APA1E A 079 ln 079 021 ln 021 0 08316 ln PA2E APA1E A 01684 ln PA3E APA1E A 03488 46025 ln PA1E A 03488 PA1 minE A 141 kPa For PA1E A PA1 minE A we would have entropy generation AS E A 0 1 2 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11129 Weighing of masses gives a mixture at 60C 225 kPa with 05 kg O2 15 kg N2 and 05 kg CH4 Find the partial pressures of each component the mixture specific volume mass basis mixture molecular weight and the total volume Solution From Eq114 yi mi Mi mjM j ntot mjMj 0531999 1528013 051604 0015625 0053546 0031172 0100343 yO2 00156250100343 01557 yN2 00535460100343 05336 yCH4 00311720100343 03107 From Eq1110 PO2 yO2 Ptot 01557225 35 kPa PN2 yN2 Ptot 05336225 120 kPa PCH4 yCH4 Ptot 03107225 70 kPa Vtot ntot RTP 0100343 831451 33315 225 1235 m 3 v Vtotmtot 1235 05 15 05 0494 m3kg From Eq115 Mmix yjMj mtotntot 250100343 24914 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11130 A carbureted internal combustion engine is converted to run on methane gas natural gas The airfuel ratio in the cylinder is to be 20 to 1 on a mass basis How many moles of oxygen per mole of methane are there in the cylinder Solution The mass ratio mAIRmCH4 20 so relate mass and mole n mM nAIR nCH4 mAIR mCH4 MCH4MAIR 20 16042897 110735 nO2 nCH4 nO2 nAIR nAIR nCH4 021110735 2325 mole O2mole CH 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11131 A mixture of 50 carbon dioxide and 50 water by mass is brought from 1500 K 1 MPa to 500 K 200 kPa in a polytropic process through a steady state device Find the necessary heat transfer and work involved using values from Table A5 Solution Process Pvn constant leading to n lnv2v1 lnP1P2 v RTP n ln 1000 200 ln 500 1000 200 1500 31507 Eq1115 Rmix Σ ciRi 05 01889 05 04615 03252 kJkg K Eq1123 CP mix Σ ciCPi 05 08418 05 1872 13569 kJkg K Work is from Eq718 w vdP n n1 Peve Pivi nR n1 Te Ti 4764 kJkg Heat transfer from the energy equation q he hi w CPTe Ti w 8805 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11132 The accuracy of calculations can be improved by using a better estimate for the specific heat Reconsider the previous problem and use CP hT from Table A8 centered at 1000 K A mixture of 50 carbon dioxide and 50 water by mass is brought from 1500 K 1 MPa to 500 K 200 kPa in a polytropic process through a steady state device Find the necessary heat transfer and work involved using values from Table A5 Solution Using values from Table A8 we estimate the heat capacities CEAP COA2 AE A A109636 84972 1100 900E A 12332 kJkg K CEAP HA2 AOE A A222673 17686 1100 900E A 22906 kJkg K Eq1123 CAP mixE A Σ cAiE ACAPiE A 05 12332 05 22906 17619 kJkg K Eq1115 RAmixE A Σ cAiE ARAiE A 05 01889 05 04615 03252 kJkg K Process PvAnE A C n lnPA1E APA2E A lnvA2E AvA1E A and use Pv RT n ln A 1000 200 E A ln A 500 1000 200 1500 E A 31507 Work is from Eq718 w AvEAdP A n n1E A PAeE AvAeE A PAiE AvAiE A AnR n1E A TAeE A TAiE A 4764 kJkg Heat transfer from energy equation q hAeE A hAiE A w 17619500 1500 4764 12855 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11133 A large air separation plant takes in ambient air 79 NA2E A 21 OA2E A by mole at 100 kPa 20C at a rate of 25 kgs It discharges a stream of pure OA2E A gas at 200 kPa 100C and a stream of pure NA2E A gas at 100 kPa 20C The plant operates on an electrical power input of 2000 kW Calculate the net rate of entropy change for the process Air 79 NA2E 21 OA2E PA1E A 100 kPa TA1E A 20 AoE AC AmE A1E A 25 kgs AW E AINE A 2000 kW PA2E A 200 kPa TA2E A 100 AoE AC PA3E A 100 kPa TA3E A 20 AoE AC Solution To have the flow terms on a mass basis let us find the mass fractions From Eq 113 cAiE A yAiE A MAiE A A yEAjE AMAjE cAO2 E A 021 32 021 32 079 28013 023293 cAN2 E A 1 cAO2 E A 076707 AmE A2E A cAO2 E AmE A1E A 5823 kgs AmE A3E A cAN2 E AmE A1E A 19177 kgs The energy equation Eq410 gives the heat transfer rate as AQ E ACVE A Σ AmE AhAiE A AW E ACVE A AmE AO2 E ACAP0 O2 E ATA2E ATA1E A AmE AN2 E ACAP0 N2 E ATA3E A TA1E A AW E ACVE 5823 0922 10020 0 2000 15705 kW The entropy equation Eq77 gives the generation rates as AS E AgenE A Σ AmE AiE AsAiE A AQ E ACVE ATA0E A AmE A2E AsA2E A AmE A3E AsA3E A AmE A1E AsA1E A AQ E ACVE ATA0E A Use Eq616 for the entropy change Σ AmE AiE AsAiE A 5823 0922 ln A3732 2932E A 02598 ln A200 21E A 19177 0 02968 ln 10079 3456 kWK AS E AgenE A 157052932 3456 190 kWK pure O 2 pure N 2 1 2 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11134 Take Problem 1157 with inlet temperature of 1400 K for the carbon dioxide and 300 K for the nitrogen First estimate the exit temperature with the specific heats from Table A5 and use this to start iterations using A9 to find the exit temperature CV mixing chamber steady flow The inlet ratio is n CO2 2 n N2 and assume no external heat transfer no work involved C P CO2 4401 0842 3706 C P N2 28013 1042 29189 kJkmol K Continuity Equation 0 Σn in Σn ex Energy Equation 0 Σn in h in Σn ex h ex 0 2n N2 C P CO2Tin Tex CO2 n N2 C P N2 Tin Tex N2 0 2 3706 1400Tex 29189 300Tex 0 103768 87567 103309 Tex Tex 1089 K From Table A9 Σn in h in n N2 2 55895 1 54 n N2 111844 1000K Σn ex h ex n N2 2 33397 21463 n N2 88257 1100K Σn ex h ex n N2 2 38885 24760 n N2 102530 1200K Σn ex h ex n N2 2 44473 28109 n N2 117055 Now linear interpolation between 1100 K and 1200 K Tex 1100 100 111844102530 117055102530 1164 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11135 A pistoncylinder has 100 kg of saturated moist air at 100 kPa 5C If it is heated to 45C in an isobaric process find 1Q2 and the final relative humidity If it is compressed from the initial state to 200 kPa in an isothermal process find the mass of water condensing Solution Energy Eq mu2 u1 1Q2 1W2 Initial state 1 φ1 100 Table B11 Pv1 08721 kPa hv1 251054 Eq1128 w1 0622 Ptot Pv1 Pv1 08721 100 08721 0005472 Eq1126 with ma mtot mv1 mtot w1ma gives ma mtot1 w1 99456 kg Eq1126 mv1 w1ma 0544 kg Case a P constant 1W2 mPv2v1 1Q2 mu2 u1 1W2 mh2 h1 maCpT2 T1 mhv2 hv1 State 2 w2 w1 T2 Pv2 Pv1 and Table B11 hv2 258319 kJkg Pg2 9593 kPa Eq1125 φ2 Pg2 Pv2 08721 9593 0091 or φ2 91 From the energy equation 1Q2 99456 100445 5 0544258319 251054 4034 kJ Case b As P is raised Pv yv P would be higher than Pg condensation T constant φ2 100 Pv Pg 08721 kPa w2 0622 Pa2 Pv2 0622 Ptot2 Pv2 Pv2 08721 200 08721 0002724 mv2 w2 ma 0271 kg mliq mv1 mv2 0273 kg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11136 A pistoncylinder contains helium at 110 kPa at ambient temperature 20C and initial volume of 20 L as shown in Fig P11136 The stops are mounted to give a maximum volume of 25 L and the nitrogen line conditions are 300 kPa 30C The valve is now opened which allows nitrogen to flow in and mix with the helium The valve is closed when the pressure inside reaches 200 kPa at which point the temperature inside is 40C Is this process consistent with the second law of thermodynamics P1 110 kPa T1 20 oC V1 20 L Vmax 25 L V 2 P2 200 kPa T2 40 oC Pi 300 kPa Ti 30 oC Constant P to stops then constant V Vmax Wcv P1V2 V1 Qcv U2 U1 Wcv nih i n2h 2 n1h 1 nih i P2 P1V 2 nAh A2 h Ai nBh B2 h B1 P2 P1V2 nB n1 P1V1R T1 110002831452932 00009 kmol n2 nA nB P2V2R T2 2000025831453132 000192 kmol nA n2 nB 000102 kmol Mole fractions yA2 000102000192 05313 yB2 04687 Qcv 00010228013104240 30 000094003519340 20 200 110 0025 0298 0374 225 1578 kJ Sgen n2s 2 n1s 1 nis i QcvT 0 nAs A2 s Ai nBs B2 s B1 QcvT 0 s A2 s Ai 29189 ln 3132 3032 R ln 05313200 300 95763 s B2 s B1 207876 ln 3132 2932 R ln 04687200 110 27015 Sgen 00010295763 0000927015 15782932 00176 kJK 0 Satisfies 2nd law Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11137 A spherical balloon has an initial diameter of 1 m and contains argon gas at 200 kPa 40C The balloon is connected by a valve to a 500L rigid tank containing carbon dioxide at 100 kPa 100C The valve is opened and eventually the balloon and tank reach a uniform state in which the pressure is 185 kPa The balloon pressure is directly proportional to its diameter Take the balloon and tank as a control volume and calculate the final temperature and the heat transfer for the process CO2 A B VA1 π 6 13 05236 mA1 RTA1 PA1VA1 20005236 0208 133132 1606 kg mB1 PB1VB1RTB1 1000500188923732 0709 kg Process P C D CV13 polytropic n 13 VA2 VA1 P2 PA1 3 05236185 200 3 04144 m 3 2 Uniform ideal gas mixture P2VA2 VB mARA mBRBT 2 T2 P2VA2 VB mARA mBRB 185 04144050 1606020813 0709018892 3613 K Boundary work in a polytropic process from Eq629 W12 1 13 P2VA2 PA1VA1 18504144 20005236 43 kPa m3 210 kJ The heat transfer from the energy equation Q mACV0AT2 TA1 mBCV0BT2 TB1 W12 160603123613 3132 070906533613 3732 210 186 210 24 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11138 An insulated rigid 2 m3 tank A contains CO2 gas at 200C 1MPa An uninsulated rigid 1 m3 tank B contains ethane C2H6 gas at 200 kPa room temperature 20C The two are connected by a oneway check valve that will allow gas from A to B but not from B to A The valve is opened and gas flows from A to B until the pressure in B reaches 500 kPa and the valve is closed The mixture in B is kept at room temperature due to heat transfer Find the total number of moles and the ethane mole fraction at the final state in B Find the final temperature and pressure in tank A and the heat transfer tofrom tank B Solution A B CO C H 2 2 6 Tank A VA 2 m3 state A1 CO2 TA1 200C 4732 K PA1 1 MPa C v0 CO2 0653 4401 2874 C P0 CO2 0842 4401 3706 kJkmol K Tank B VB 1 m3 state B1 C2H6 TB1 20C 2932 K PB1 200 kPa Slow Flow A to B to PB2 500 kPa and assume TB2 TB1 T 0 Total moles scales to pressure so with same V and T we have PB1VB nB1R TB1 PB2VB nB2 mix R T B2 Mole fraction yEEACA2 AHA6 A BA2 AE A EA nAB1 A EnAB2 AE A EA PAB1 A EPAB2 AE A A200 500E A 0400 nAB1E A EA PAB1 AVAB A ERTAB1 AE A A 200 1 R 2932E A 008204 kmol nAB2 mixE A EA PAB2 AVAB A ERTAB2 AE A A 500 1 R 2932E A 02051 kmol nEACOA2 A B2E A 02051 008201 012306 kmol Now we can work backwards to final state in A nAA1E A EA PAA1 AVAA A ERTAA1 AE A A 10002 R4732E A 050833 kmol nAA2E A nAA1E A nEACOA2 A B2E A 038527 kmol Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CV A All COA2E A Transient with flow out and adiabatic Energy Eq QACV AE A 0 nAA2E A AuE A AA2E A nAA1E A AuE A AA1E A nAaveE A AhE A AaveE 0 nAA2E A ACE AEAv0 COA2 AE ATAA2E A nAA1E A ACE AEAv0 COA2 AE ATAA1E A nAaveE A ACE AEAP0 COA2 AE A TAA1E A TAA2E A2 0 2874 038527 TAA2E A 0508334732 012306 37064732 TAA2E A2 TAA2E A 4369 K PAA2E A EA nAA2 A RTAA2 A E VAA AE A A038527 83145 4369 2E A 700 kPa CV B Transient with flow in and nonadiabatic QACV BE A nABiE A AhE A ABi aveE A nAuE A AB2E A nAuE A AB1E A nAuE AEACOA2 A B2E A nAuE AEACA2 AHA6 A B2E A nAuE AEACA2 AHA6 A B1E QACV BE A 012306 2874 2932 0 012306 3706 4732 43692 1038 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11139 You have just washed your hair and now blow dry it in a room with 23C φ 60 1 The dryer 500 W heats the air to 49C 2 blows it through your hair where the air becomes saturated 3 and then flows on to hit a window where it cools to 15C 4 Find the relative humidity at state 2 the heat transfer per kilogram of dry air in the dryer the air flow rate and the amount of water condensed on the window if any The blowdryer heats the air at constant specific humidity to 2 and it then goes through an adiabatic saturation process to state 3 finally cooling to 4 1 23C 60 rel hum w1 00104 AhE A1 69 kJkg dry air 2 wA2E A w1 TA2E A φ2 15 AhE A2E A 95 kJkg dry air CV 1 to 2 wA2E A w1 q AhE A2 AhE A1 95 69 26 kJkg dry air AmE Aa Qq 0526 001923 kgs CV 2 to 3 w3 wA2E A AmE AliqAmE Aa AmE Aa AhE A2E A AmE Aliq hf AmE Aa AhE A3 3 φ 100 T3 Twet2 248C w3 00198 4 φ 100 T4 w4 001065 AmE Aliq w3 w4 AmE Aa 00198 001065 001923 0176 gs If the steam tables and formulas are used then we get hAg1E A 25435 hAg2E A 25903 PAg1E A 2837 kPa Pv1 17022 kPa PAg2E A 118 kPa w1 001077 wA2E A w1 PAv2E A PAv1E φA2E A PAv2E APAg2E A 144 hf3 114 kJkg Trial and error for adiabatic saturation temperature T3 25C w3 002 PAv4E A PAg4E A 1705 kPa wA4E A 062217051001705 00108 w T Φ 100 Φ Φ Φ 60 15 10 dry 3 4 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11140 A 02 mA3E A insulated rigid vessel is divided into two equal parts A and B by an insulated partition as shown in Fig P11140 The partition will support a pressure difference of 400 kPa before breaking Side A contains methane and side B contains carbon dioxide Both sides are initially at 1 MPa 30C A valve on side B is opened and carbon dioxide flows out The carbon dioxide that remains in B is assumed to undergo a reversible adiabatic expansion while there is flow out Eventually the partition breaks and the valve is closed Calculate the net entropy change for the process that begins when the valve is closed A B CH 4 CO 2 PAMAXE A 400 kPa PAA1E A PAB1E A 1 MPa VAA1E A VAB1E A 01 mA3E TAA1E A TAB1E A 30AoE AC 3032 K COA2E A inside B sAB2E A sAB1E A to PAB2E A 600 kPa PAA2E A 1000 kPa For COA2E A k 1289 TAB2E A 3032A 600 1000E AA 0289 1289 E A 2704 K nAB2E A PAB2E AVAB2E AARE ATAB2E A 60001831452704 0026 688 nAA2E A nAA1E A 100001831453032 0039 668 kmol The process 2 to 3 is adiabatic but irreversible with no work QA23E A 0 nA3E AuE A3E A Ai EA nAi2E AuE Ai2E A 0 nAA2E ACE Avo AE ATA3E ATAA2E A nAB2E ACE Avo BE ATA3E ATAB2E A 0 0039 66816041736TA3E A3032 0026 68844010653TA3E A2704 0 Solve TA3E A 2898 K Get total and partial pressures for the entropy change PA3E A nARE ATV 006635683145289802 7994 kPa PAA3E A 059787994 4779 kPa PAB3E A PA3E A PAA3E A 3215 kPa AsE AA3E A AsE AA2E A 16042254 lnA2898 3032E A 83145 ln A4779 1000E A 4505 kJkmol K AsE AB3E A AsE AB2E A 44010842 lnA2898 2704E A 83145 ln A3215 600E A 77546 kJkmol K SANETE A nAA2E AAsE AA3E A AsE AA2E A nAB2E A AsE AB3E A AsE AB2E A 0039668 4505 0026688 77546 03857 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11141 Ambient air is at a condition of 100 kPa 35C 50 relative humidity A steady stream of air at 100 kPa 23C 70 relative humidity is to be produced by first cooling one stream to an appropriate temperature to condense out the proper amount of water and then mix this stream adiabatically with the second one at ambient conditions What is the ratio of the two flow rates To what temperature must the first stream be cooled PA1E A PA2E A 100 kPa TA1E A TA2E A 35 AoE AC φA1E A φA2E A 050 φA4E A 10 PA5E A 100 TA5E A 23 AoE AC φA5E A 070 PAv1E A PAv2E A 055628 2814 kPa wA1E A wA2E A 0622A 2814 1002814E A 00180 PAv5E A 072837 19859 kPa wA5E A 0622A 19859 10019859E A 00126 CV Mixing chamber Call the mass flow ratio r mAa2E AmAa1E A cons mass wA1E A r wA4E A 1 rwA5E Energy Eq hAa1E A wA1E AhAv1E A rhAa4E A rwA4E AhAv4E A 1rhAa5E A 1rwA5E AhAv5E 0018 rwA4E A 1r 00126 or r A ma2 Ema1 E A A001800126 00126w4 E A with wA4E A 0622 A PG4 E100PG4 E 10043082 001825653 r1004TA4E A r wA4E AhAv4E 1r1004 2962 1r0012625436 or r1004TA4E A wA4E AhAG4E A 3293 262 0 Assume TA4E A 5 AoE AC PAG4E A 08721 hAG4E A 25105 wA4E A 06220872110008721 00055 r mAa2E AmAa1E A A 001800126 0012600055E A 07606 0760610042782 0005525105 3296 262 142 0 OK COOL LIQ H O 2 Q 0 MIX QCOOL MIX 1 2 3 4 5 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11142 An airwater vapor mixture enters a steady flow heater humidifier unit at state 1 10C 10 relative humidity at the rate of 1 mA3E As A second airvapor stream enters the unit at state 2 20C 20 relative humidity at the rate of 2 mA3E As Liquid water enters at state 3 10C at the rate of 400 kg per hour A single air vapor flow exits the unit at state 4 40C as shown in Fig P11142 Calculate the relative humidity of the exit flow and the rate of heat transfer to the unit State 1 TA1E A 10C φA1E A 10 AV E Aa1E A 1 mA3E As PAg1E A 12276 kPa PAv1E A φA1E APAg1E A 01228 kPa PAa1E A P PAv1E A 99877 kPa ωA1E A 0622 EA PAv1 A E PAa1 AE A 0000765 Am E Aa1E A EA PAa1 AV E Aa1 A RTAa1 AE A 12288 kgs Am E Av1E A ωA1E Am E Aa1E A 000094 kgs hAv1E A hAg1E A 25197 kJkg State 2 TA2E A 20C φA2E A 20 AV E Aa2E A 2 mA3E As PAg2E A 23385 kPa PAv2E A φ PAg2E A 04677 kPa PAa2E A P PAv2E A 99532 kPa ωA2E A 0622 EA PAv2 A E PAa2 AE A 0002923 Am E Aa2E A EA PAa2 A AV E A Aa2 A RTAa2 AE A 23656 kgs Am E Av2E A ωA2E Am E Aa2E A 000691 kgs hAv2E A hAg2E A 25381 kJkg State 3 Liquid TA3E A 10C Am E Af3E A 400 kghr 01111 kgs hAf3E A 42 kJkg Continuity Eq air Am E Aa4E A Am E Aa2E A Am E Aa1E A 35944 kgs Continuity Eq water Am E Av4E A Am E Av1E A Am E Av2E A Am E Af3E A 011896 kgs ωA4E A Am E Av4E AAm E Aa4E A 00331 0622 EA PAv4 A EPPAv4 AE A PAv4E A 5052 kPa PAg4E A 7384 kPa φA4E A PAv4E A PAg4E A 0684 hAv4E A hAg4E A 25743 kJkg Energy Eq AQ E A Am E Aa1E AhAa1E A Am E Av1E AhAv1E A Am E Aa2E AhAa2E A Am E Av2E AhAv2E A Am E Af3E AhAf3E A Am E Aa4E AhAa4E A Am E Av4E AhAv4E AQ E A 100435944 40 12288 10 23656 20 011896 25743 000094 25197 000691 25381 01111 420 366 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11143 A dehumidifier receives a flow of 025 kgs dry air at 35AoE AC 90 relative humidity as shown in figure P11107 It is cooled down to 20AoE AC as it flows over the evaporator and then heated up again as it flows over the condenser The standard refrigeration cycle uses R410A with an evaporator temperature of 5AoE AC and a condensation pressure of 3000 kPa Find the amount of liquid water removed and the heat transfer in the cooling process How much compressor work is needed What is the final air exit temperature and relative humidity Solution This setup has a standard refrigeration cycle with R410A This cycle and the air flow interacts through the two heat transfer processes The cooling of the air is provided by the refrigeration cycle and thus requires an amount of work that depends on the cycle COP Refrigeration cycle State 1 x 1 hA1E A 27753 kJkg sA1E A 10466 kJkg K State 2 sA2E A sA1E A hA2E A 31897 kJkg TA2E A 729AoE AC State 3 xA3E A 00 hA3E A hAfE A 14178 kJkg TA3E A 4907C State 4 hA4E A hA3E A and PA4E A PA1E C 3 4 1 2 W C Air in Air sat 20 C Air ex R410a Evaporator R410a Condenser o T s 1 2 3 4 TA1E A 5AoE AC PA2E A 3000 kPa Now for the cycle we get wACE A hA2E A hA1E A 31897 27753 4144 kJkg qAHE A hA2E A hA3E A 31897 14178 17719 kJkg qALE A hA1E A hA4E A 27753 14178 13575 kJkg For the air processes let us use the psychrometric chart Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Air inlet wAinE A 0019 AhE AinE A 96 kJkg dry air TAdewE A 24AoE AC 15AoE AC Air 15 φ 100 wA20E A 00148 AhE A20E A 775 hAfE A 8394 B11 Now do the continuity for water and energy equations for the cooling process AmE AliqE A m air wAinE A wA20E A 025 0019 00148 000105 kgs AQ E AcoolE A m air AhE AinE A AhE A20E A AmE AliqE AhAfE A 02596 775 000105 8394 4537 kW Now the heater from the R410A cycle has AQ E AheatE A AQ E AcoolE A qAHE A qALE A 4537 17719 13575 5922 kW so the compressor work is the balance of the two AW E ACE A AQ E AheatE A AQ E AcoolE A 5922 4537 1385 kW Energy eq for the air flow being heated AQ E AheatE A m air AhE AexE A AhE A20E A AhE AexE A AhE A20E A AQ E AheatE A m air AhE AexE A 775 5922 025 1012 kJkg dry air and wAexE A wA20E Locate state in the psychrometric chart T 435AoE AC and φ 27 w T Φ 80 Φ 27 dry in 20 24 435 ex sat Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11144 The airconditioning by evaporative cooling in Problem 11101 is modified by adding a dehumidification process before the water spray cooling process This dehumidification is achieved as shown in Fig P11144 by using a desiccant material which absorbs water on one side of a rotating drum heat exchanger The desiccant is regenerated by heating on the other side of the drum to drive the water out The pressure is 100 kPa everywhere and other properties are on the diagram Calculate the relative humidity of the cool air supplied to the room at state 4 and the heat transfer per unit mass of air that needs to be supplied to the heater unit States as noted on Fig P11144 text page 552 At state 1 35 AoE AC PAv1E A φA1E APAg1E A 030 5628 16884 kPa wA1E A 0622 168849831 0010 68 At TA3E A 25 AoE AC wA3E A wA2E A wA1E A2 000534 Evaporative cooling process to state 4 where TA4E A 20AoE AC Energy as in Eq1130 wA3E AhAv3E A hAf4E A CAP0AE ATA4E A TA3E A wA4E A hAfg4E 0005 34 25472 839 100420 25 wA4E A 24542 wA4E A 00074 0622 PAv4E A 100 PAv4E A PAv4E A 1176 kPa φA4E A 1176 2339 0503 Following now the flow back we have TA5E A 25 AoE AC wA5E A wA4E A 00074 Evaporative cooling process to state 6 where TA6E A 20AoE AC wA5E AhAv5E A hAf6E A CAP0AE ATA6E A TA5E A wA6E A hAfg6E 0007425472 839 100420 25 wA6E A 24542 wA6E A 0009 47 For adiabatic heat exchanger AmE AA2E A AmE AA3E A AmE AA6E A AmE AA7E A AmE AAE A Also wA2E A wA3E A wA6E A wA7E So now only TA7E A is unknown in the energy equation hAA2E A wA2E AhAv2E A hAA6E A wA6E AhAv6E A hAA3E A wA3E AhAv3E A hAA7E A wA7E AhAv7E or CAP0AE ATA7E A wA6E AhAv7E A hAv6E A CAP0AE ATA2E A TA6E A TA3E A wA2E AhAv2E A hAv3E A 1004 TA7E A 0009 47hAv7E A 25381 100460 20 25 0005 3426096 25472 55526 By trial and error TA7E A 547 AoE AC hAv7E A 26003 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful For the heater 78 wA8E A wA7E A AQE AAmE AAE A CAP0AE ATA8E A TA7E A wA7E AhAv8E A hAv7E A 100480 547 0009 4726437 26003 258 kJkg dry air ENGLISH UNIT PROBLEMS Updated August 2013 SOLUTION MANUAL CHAPTER 11 English Units Fundamentals of Thermodynamics Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CHAPTER 11 SUBSECTION PROB NO Concept problem 145 Mixture Composition and Properties 146151 Simple Processes 152160 Entropy Generation 161169 Air Water vapor Mixtures 170183 Review Problems 184187 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Concept Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11145E If oxygen is 21 by mole of air what is the oxygen state P T v in a room at 540 R 15 psia of total volume 2000 ft3 The temperature is 540 R The partial pressure is PO2 yPtot 021 15 psia 315 psia At this T P v RTP 4828 540 315 144 ftlbflbm R R lbfin2 inft2 5748 ft3lbm Remark If we found the oxygen mass then mO2vO2 V 2000 ft3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Mixture Composition and Properties Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11146E A flow of oxygen and one of nitrogen both 540 R are mixed to produce 1 lbms air at 540 R 15 psia What are the mass and volume flow rates of each line We assume air has mole fraction of oxygen 021 and nitrogen 079 For the mixture M 021 32 079 28013 2885 For O2 c 021 32 2885 02329 For N2 c 079 28013 2885 07671 Since the total flow out is 1 lbms these are the component flows in lbms Volume flow of O2 in is V cm v cm RT P 02329 1 lbms 4828 lbfftlbmR 540 R 15 psi 144 in2ft2 281 ft3s Volume flow of N2 in is V cm v cm RT P 07671 1 lbms 5515 lbfftlbmR 540 R 15 psi 144 in2ft2 1058 ft3s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11147E A gas mixture at 250 F 18 lbfin2 is 50 N2 30 H2O and 20 O2 on a mole basis Find the mass fractions the mixture gas constant and the volume for 10 lbm of mixture Solution The conversion follows the definitions and identities From Eq113 ci yi Mi yjM j From Eq115 Mmix yjMj 05 28013 03 18015 02 31999 140065 54045 63998 25811 cN2 140065 25811 05427 cH2O 54045 25811 02094 cO2 63998 25811 02479 sums to 1 OK From Eq1114 and R 154536 lbfftlbmolR Rmix RMmix 154536 25811 5987 lbf ftlbm R V mRmix TP 10 lbm 5987 lbfftlbmR 710 R 18 psi 144 in2ft2 164 ft3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11148E In a car engine gasoline assume octane C8H18 is evaporated and then mixed with air in a ratio of 115 by mass In the cylinder the mixture is at 110 psia 1200 R when the spark fires For that time find the partial pressure of the octane and the specific volume of the mixture Assuming ideal gas the partial pressure is Pi yi P and cC8H18 116 00625 From Eq 114 yi ci Mi cjM j yC8H18 00625114232 00625114232 093752897 0008186 PC8H18 0008186 110 090 psia The gas constant from Eq1115 Rmix ciRi 00625 1353 09375 5334 50852 ftlbflbmR vmix RmixTP 50852 lbfftlbmR 1200 R 110 psi 144 in2ft2 3852 ft3lbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11149E A diesel engine sprays fuel assume nDodecane C12H26 M 17034 lbmlbmol into the combustion chamber already filled with in an amount of 1 mol fuel per 88 mol air Find the fuel fraction on a mass basis and the fuel mass for a chamber that is 007 ft3 at 1400 R and total pressure of 600 psia From Eq 113 ci yi Mi yjM j cfuel 189 17034 8889 2897 189 17034 006263 Use ideal gas for the fuel vapor mfuel RfuelT PfuelV 1545417034 ftlbflbmR 1400 R 189 600 psi 007 ft3 144 inft2 000535 lbm We could also have done the total mass and then used the mass fraction Eq115 Mmix yjMj 189 17034 8889 2897 30558 Rmix RMmix 15454 30558 505727 ftlbflbmR m PV RmixT 505727 ftlbflbmR 1400 R 600 psi 007 ft3 144 inft2 008542 lbm mfuel cfuel m 006263 008542 lbm 000535 lbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11150E A new refrigerant R410A is a mixture of R32 and R125 in a 11 mass ratio What is the overall molecular weight the gas constant and the ratio of specific heats for such a mixture Eq1115 Rmix ciRi 05 297 05 1287 21285 ftlbflbm R Eq1123 CP mix ci CP i 05 0196 05 0189 01925 Btulbm R Eq1121 CV mix ciCV i 05 0158 05 0172 0165 Btulbm R CP mix Rmix kmix CP mix CV mix 01925 0165 11667 M yjMj 1 cj Mj 120022 1 05 52024 05 72586 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11151E Do the previous problem for R507a which is a 11 mass ratio of R125 and R 143a The refrigerant R143a has molecular mass of 84041 lbmlbmol and CP 0222 BtulbmR For R143a R 15453684041 18388 ftlbflbmR CV CP R 0222 1838877817 01984 BtulbmR Eq1115 Rmix ciRi 05 1287 05 18388 15629 ftlbflbm R Eq1123 CP mix ci CP i 05 0189 05 0222 02055 Btulbm R Eq1121 CV mix ciCV i 05 0172 05 01984 01852 Btulbm R CP mix Rmix kmix CP mix CV mix 02055 01852 11096 M yjMj 1 cj Mj 120022 1 05 84041 05 9886 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Simple Processes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11152E A rigid container has 1 lbm CO2 at 540 R and 1 lbm argon at 720 R both at 20 psia Now they are allowed to mix without any heat transfer What is final T P No Q No W so the energy equation gives constant U Energy Eq U2 U1 0 mCO2u2 u1CO2 mAru2 u1Ar mCO2Cv CO2T2 T1CO2 mArCv ArT2 T1Ar 10201 10124 BtuR T2 10201540 10124720 Btu T2 6087 R Volume from the beginning state V V1 VCO2 VAr mCO2RCO2TCO2P mArRArTArP 1351054020 1386872020 144 1625 ft3 Pressure from ideal gas law and Eq1115 for R P2 mRTV 13510 13868 60871625 144 192 psia CO Ar cb 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11153E A flow of 1 lbms argon at 540 R and another flow of 1 lbms CO2 at 2800 R both at 20 psia are mixed without any heat transfer What is the exit T P No work implies no pressure change for a simple flow Pe 20 psia The energy equation becomes m hi m he m hiAr m hiCO2 m heAr m heCO2 m CO2Cp CO2Te TiCO2 m ArCp ArTe TiAr 0 m ArCp ArTi m CO2Cp CO2Ti m ArCp Ar m CO2Cp CO2 Te 1 0124 540 1 0201 2800 Btus 1 0124 1 0201 BtusR Te Te 19377 R 1 Ar 2 CO 3 Mix MIXING CHAMBER cb 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11154E Repeat the previous problem using variable specific heats A flow of 1 lbms argon at 540 R and another flow of 1 lbms CO2 at 2800 R both at 20 psia are mixed without any heat transfer What is the exit T P No work implies no pressure change for a simple flow Pe 20 psia The energy equation becomes m hi m he m hiAr m hiCO2 m heAr m heCO2 m CO2he hiCO2 m ArCp ArTe TiAr 0 m ArCp ArTi m CO2hi m ArCp Ar Te m CO2 heCO2 1 0124 540 1 27 9264401 7015 Btus 1 lbms 0124 BtulbmR Te heCO2 Trial and error on RHS using Table F6 for heCO2 Te 2000 R RHS 169824401 0124 2000 63387 too small Te 2200 R RHS 196594401 0124 2200 719494 too large Final interpolation Te 2000 200 7015 63387 719494 63387 2158 R 1 Ar 2 CO 3 Mix MIXING CHAMBER cb 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11155E A pipe flows 01 lbms mixture with mass fractions of 40 CO2 and 60 N2 at 60 lbfin2 540 R Heating tape is wrapped around a section of pipe with insulation added and 2 Btus electrical power is heating the pipe flow Find the mixture exit temperature Solution CV Pipe heating section Assume no heat loss to the outside ideal gases Energy Eq Q m he hi m CP mixTe Ti From Eq1123 CP mix ci CP i 04 0201 06 0249 023 Btulbm R Substitute into energy equation and solve for exit temperature Te Ti Q m CP MIX 540 R 2 Btus 01 lbms 023 Btulbm R 6269 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11156E An insulated gas turbine receives a mixture of 10 CO2 10 H2O and 80 N2 on a mass basis at 1800 R 75 lbfin2 The volume flow rate is 70 ft3s and its exhaust is at 1300 R 15 lbfin2 Find the power output in Btus using constant specific heat from F4 at 540 R CV Turbine Steady 1 inlet 1 exit flow with an ideal gas mixture q 0 Energy Eq W T m hi he n h i h e n C P mixTi Te PV nRT n PV RT 75 psi 144 in2ft2 70 ft3s 15454 lbfftlbmolR 1800 R 0272 lbmols C P mix yi C i 01 4401 0201 01 18015 0447 08 28013 0249 727 Btulbmol R W T 0272 lbmols 727 BtulbmolR 1800 1300 R 9887 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11157E Solve Problem 11156 using the values of enthalpy from Table F6 CV Turbine Steady 1 inlet 1 exit flow with an ideal gas mixture q 0 Energy Eq W T m hi he n h i h e PV nRT n PV RT 75 psi 144 in2ft2 70 ft3s 15454 lbfftlbmolR 1800 R 0272 lbmols W T 0272 lbmols 0114 358 8121 0111 178 64685 089227 5431 Btulbmol 11237 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11158E A piston cylinder device contains 03 lbm of a mixture of 40 methane and 60 propane by mass at 540 R and 15 psia The gas is now slowly compressed in an isothermal T constant process to a final pressure of 40 psia Show the process in a PV diagram and find both the work and heat transfer in the process Solution CV Mixture of methane and propane this is a control mass Assume methane propane are ideal gases at these conditions Energy Eq35 mu2 u1 1Q2 1W 2 Property from Eq1115 Rmix 04 RCH4 06 RC3H8 04 9635 06 3504 59564 ftlbf lbm R 007656 Btu lbm R Process T constant ideal gas 1W2 P dV mRmixT 1VdV mRmixT ln V2V1 mRmixT ln P1P2 03 lbm 007656 Btu lbm R 540 R ln 1540 1216 Btu Now heat transfer from the energy equation where we notice that u is a constant ideal gas and constant T so 1Q2 mu2 u1 1W2 1W2 1216 Btu P v 2 1 T s 2 1 T C P C v 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11159E Two insulated tanks A and B are connected by a valve Tank A has a volume of 30 ft3 and initially contains argon at 50 lbfin2 50 F Tank B has a volume of 60 ft3 and initially contains ethane at 30 lbfin2 120 F The valve is opened and remains open until the resulting gas mixture comes to a uniform state Find the final pressure and temperature Energy eq U2U1 0 nArC V0T2TA1 nC2H6 C VOT2TB1 nAr PA1VAR TA1 50 psi144 in2ft2 30 ft3 1545 lbfftlbmolR5097 R 02743 lbmol nC2H6 PB1VBR TB1 30 psi144 in2ft2 60 ft3 1545 lbfftlbmolR5797 R 02894 lbmol n2 nAr nC2H6 05637 lbmol Substitute this into the energy equation 02743 39948 00756 T2 5097 02894 3007 0361 T2 5097 0 Solving T2 5651 R P2 n2R T2VAVB 0563715455651 90144 38 lbfin 2 A B cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11160E A mixture of 4 lbm oxygen and 4 lbm of argon is in an insulated piston cylinder arrangement at 147 lbfin2 540 R The piston now compresses the mixture to half its initial volume Find the final pressure temperature and the piston work Since T1 TC assume ideal gases Energy Eq u2 u1 1q2 1w2 1w2 Entropy Eq s2 s1 0 Process Eq Pvk constant v2 v12 P2 P1v1v2k P12k T2 T1v1v2k1 T12 k1 Find kmix to get P2 T2 and Cv mix for u2 u 1 Rmix ΣciRi 05 4828 05 3868778 0055887 Btulbm R CPmix ΣciCPi 05 0219 05 01253 017215 Btulbm R Cvmix CPmix Rmix 011626 BtulbmR kmix CPmixCvmix 14807 Eq625 P2 147214805 4103 lbfin2 Eq624 T2 540 204805 7535 R Work from the energy equation 1w2 u1 u2 CvT1T2 011626 540 7535 2482 Btulbm 1W2 mtot 1w2 8 2482 1986 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11161E A flow of gas A and a flow of gas B are mixed in a 12 mole ratio with same T What is the entropy generation per lbmole flow out For this the mole fractions are yA nAntot 13 yB nBntot 23 Eq 1119 S R 13 ln13 23 ln23 063651 R 063651 198589 1264 BtulbmolR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11162E A rigid container has 1 lbm argon at 540 R and 1 lbm argon at 720 R both at 20 psia Now they are allowed to mix without any external heat transfer What is final T P Is any s generated Energy Eq U2 U1 0 2mu2 mu1a mu1b mCv2T2 T1a T1b T2 T1a T1b2 630 R Process Eq V constant P2V 2mRT2 mRT1a T1b P1V1a P1V1b P1V P2 P1 20 psia S due to temp changes only not P S m s2 s1a m s2 s1b mC ln T2T1a ln T2T1b 1 0124 ln 630 540 ln 630 720 000256 BtuR Ar Ar cb Why did we not account for partial pressures Since we mix argon with argon there is no entropy generation due to mixing and no partial pressure Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11163E What is the rate of entropy increase in problem 11153E Using Eq 114 the mole fraction of CO2 in the mixture is yCO2 cCO2MCO2 cCO2MCO2 cArMAr 05 4401 05 39948 05 4401 04758 yAr 1 yCO2 05242 The energy equation becomes m hi m he m hiAr m hiCO2 m heAr m heCO2 m CO2Cp CO2Te TiCO2 m ArCp ArTe TiAr 0 m ArCp ArTi m CO2Cp CO2Ti m ArCp Ar m CO2Cp CO2 Te 1 0124 540 1 0201 2800 Btus 1 0124 1 0201 BtusR Te Te 19377 R From Eqs 1116 and 1117 from the two inlet states to state 2 m se si 1 0124 ln19377 540 3868 778 ln 0524220 20 1 0201 ln19377 2800 3510 778 ln 0475820 20 015 Btus R The increase is also the entropy generation as there is no heat transfer Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Entropy Generation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11164E Find the entropy generation for the process in Problem 11159E Energy eq U2U1 0 nArC V0T2TA1 nC2H6 C VOT2TB1 nAr PA1VAR TA1 50 psi144 in2ft2 30 ft3 1545 lbfftlbmolR5097 R 02743 lbmol nC2H6 PB1VBR TB1 30 psi144 in2ft2 60 ft3 1545 lbfftlbmolR5797 R 02894 lbmol n2 nAr nC2H6 05637 lbmol Substitute into energy equation 02743 39948 00756 T2 5097 02894 3007 0361 T2 5097 0 Solving T2 5651 R P2 n2R T2VAVB 0563715455651 90144 38 lbfin 2 SSURR 0 Sgen SSYS nArS Ar nC2H6 S C2H6 yAr 0274305637 04866 S Ar C P Ar ln T2 TA1 R ln PA1 yArP2 3994801253 ln 5651 5097 1545 778 ln 0486638 50 24919 Btulbmol R S C2H6 C C2H6 ln T2 TB1 R ln PB1 yC2H6P2 30070427 ln 5651 5797 1545 778 ln 0513438 30 05270 Btulbmol R Sgen 0274324919 0289405270 0836 BtuR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11165E Carbon dioxide gas at 580 R is mixed with nitrogen at 500 R in an insulated mixing chamber Both flows are at 147 lbfin2 and the mole ratio of carbon dioxide to nitrogen is 21 Find the exit temperature and the total entropy generation per mole of the exit mixture CV mixing chamber steady flow The inlet ratio is n CO2 2 n N2 and assume no external heat transfer no work involved Continuity n CO2 2n N2 n ex 3n N2 Energy Eq n N2h N2 2h CO2 3n N2h mix ex Take 540 R as reference and write h h 540 C PmixT540 C P N2 Ti N2 540 2C P CO2 Ti CO2 540 3C P mixTmix ex 540 C P mix yiC P i 29178 237053 82718 Btulbmol R 3C P mixTmix ex C P N2 Ti N2 2C P CO2 Ti CO2 13 837 Btulbmol Tmix ex 5576 R Partial pressures are total pressure times molefraction Pex N2 Ptot3 Pex CO2 2Ptot3 S gen n exs exnsiCO2 nsiN2 n N2 s e s iN2 2n N2 s e s i CO2 S genn N2 C PN2 ln Tex TiN2 R ln yN2 2C PCO2 ln Tex TiCO2 2 R ln y CO2 07575 21817 07038 16104 3846 Btulbmol N2 R 1 N 2 CO 3 Mix MIXING CHAMBER cb 2 2 Sgen Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11166E A steady flow 06 lbms of 60 carbon dioxide and 40 water mixture by mass at 2200 R and 30 psia is used in a constant pressure heat exchanger where 300 Btus is extracted from the flow Find the exit temperature and rate of change in entropy using Table F4 Solution CV Heat exchanger Steady 1 inlet 1 exit no work Continuity Eq cCO2 06 cH2O 04 Energy Eq Q m he hi m CP Te Ti Te Ti Q m C P Inlet state Table F4 CP 06 0201 04 0447 02994 BtulbmR Exit state Te Ti Q m CP 2200 R 300 Btus 06 lbms 02994 BtulbmR 530 R The rate of change of entropy for the flow is P is assumed constant m se si m so Te so Ti R lnPePi m so Te so Ti m CP lnTeTi 06 02994 ln530 2200 0256 BtusR The entropy generation rate cannot be estimated unless the average T at which the heat transfer leaves the control volume is known Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11167E A steady flow 06 lbms of 60 carbon dioxide and 40 water mixture by mass at 2200 R and 30 psia is used in a constant pressure heat exchanger where 300 Btus is extracted from the flow Find the exit temperature and rate of change in entropy using Table F6 Solution CV Heat exchanger Steady 1 inlet 1 exit no work Continuity Eq cCO2 06 cH2O 04 Energy Eq Q m he hi he hi Q m Inlet state Table F6 hi 06 196594401 04 1525418015 60671 Btulbm Exit state he hi Q m 60671 30006 10671 Btulbm Trial and error for T with h values from Table F6 800 R he 06 25254401 04 214218015 81984 Btulbm 1000 R he 06 46554401 04 382418015 14837 Btulbm Interpolate to have the right h T 8745 R Entropy Eq78 m se m si Q T S gen The rate of change of entropy for the flow is P is assumed constant m se si m so Te so Ti 06 0655732 669524401 0449019 5760518015 0344 BtusR The entropy generation rate cannot be estimated unless the average T at which the heat transfer leaves the control volume is known Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11168E A mixture of 60 helium and 40 nitrogen by mole enters a turbine at 150 lbfin2 1500 R at a rate of 4 lbms The adiabatic turbine has an exit pressure of 15 lbfin2 and an isentropic efficiency of 85 Find the turbine work Assume ideal gas mixture and take CV as turbine Energy Eq ideal turbine wT s hi hes Entropy Eq ideal turbine ses si Tes TiPePik1k Properties from Eq1123 1115 and 623 C P mix 06 125 4003 04 0248 28013 57811 Btulbmol R k1k R C P mix 154557811778 03435 Mmix 06 4003 04 28013 13607 CP C PMmix 04249 Btulbm R Tes 1500 R 1515003435 680 R wTs CPTiTes 3484 Btulbm Then do the actual turbine wT ac ηwTs 2961 Btulbm W m wTs 1184 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11169E A tank has two sides initially separated by a diaphragm Side A contains 2 lbm of water and side B contains 24 lbm of air both at 68 F 147 lbfin2 The diaphragm is now broken and the whole tank is heated to 1100 F by a 1300 F reservoir Find the final total pressure heat transfer and total entropy generation CV Total tank out to reservoir Energy Eq35 U2 U1 mau2 u1a mvu2 u1v 1Q2 Entropy Eq637 S2 S1 mas2 s1a mvs2 s1v 1Q2Tres Sgen V2 VA VB mvvv1 mava1 00321 31911 31944 ft 3 Water u1 3608 Btulbm s1 00708 Btulbm R vv2 V2mv 159718 T2 Table F72 P2v 587 lbfin 2 u2 14143 Btulbm s2 2011 Btulbm R Air u1 9005 Btulbm u2 27823 Btulbm sT1 16342 Btulbm R sT2 19036 Btulbm R va2 V2ma 133098 T2 P2a mRT2V2 43415 lbfin 2 P2tot P2v P2a 102 lbfin 2 1Q2 214143 3608 2427823 9005 3208 Btu Sgen 22011 00708 2419036 16342 5334 778 ln43415 147 3208 1760 38804 04684 1823 2526 BtuR 1300 F Q 1 2 A B Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Air Water vapor Mixtures Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11170E A 1 lbms flow of saturated moist air relative humidity 100 at 147 psia and 50 F goes through a heat exchanger and comes out at 77 F What is the exit relative humidity and the how much power is needed Solution State 1 φ1 1 Pv Pg 0178 psia Eq1128 w 0622 PvPa 0622 0178147 0178 000762 State 2 No water added w2 w1 Pv2 P v1 φ2 Pv2Pg2 01780464 0384 or 38 Energy Eq410 Q m 2h2 m 1h1 m a h2 h1air wm a h2 h1vapor m tot m a m v m a1 w1 Energy equation with CP air from F4 and hs from F71 Q m tot 1 w1 CP air 77 50 m tot 1 w1 w hg2 hg1 1 100762 024 77 50 1 000762 100762 109073 108329 6431 00563 649 Btus Q 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11171E If I have air at 147 psia and a 15 F b 115 F and c 230 F what is the maximum absolute humidity I can have Humidity is related to relative humidity max 100 and the pressures as in Eq1128 where from Eq1125 Pv Φ Pg and Pa Ptot Pv ω 0622 Pv Pa 0622 Φ Pg Ptot ΦPg a Pg 003963 psia ω 0622 003963 147 003963 0001 68 b Pg 14855 psia ω 0622 14855 147 14855 0070 c Pg 20781 psia no limit on ω for P 147 psia Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11172E Consider a volume of 2000 ft3 that contains an airwater vapor mixture at 147 lbfin2 60 F and 40 relative humidity Find the mass of water and the humidity ratio What is the dew point of the mixture Airvap P 147 lbfin2 T 60 F φ 40 Pg Psat60 0256 lbfin2 Pv φ Pg 04 0256 01024 lbfin2 mv1 RvT PvV 01024 psi 144 in2ft2 2000 ft3 8576 lbfftlbmR 520 R 0661 lbm Pa Ptot Pv1 147 01024 14598 lbfin 2 ma RaT PaV 14598 psi 144 in2ft2 2000 ft3 5334 lbfftlbmR 520 R 151576 lbm w1 ma mv 0661 151576 000436 Tdew is T when PgTdew 01024 lbfin2 T 355 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11173E Consider a 35 ft3s flow of atmospheric air at 147 psia 77 F and 80 relative humidity Assume this flows into a basement room where it cools to 60 F at 147 psia How much liquid will condense out Solution State 1 Pg Psat77 0464 psia Pv φ Pg 08 0464 0371 psia m v1 PvV RvT 0371 psi 144 in2ft2 35 ft3s 8576 lbfftlbmR 53667 R 00406 lbms w1 m v1 m A1 0622 Pv1 PA1 0622 0371 147 0371 00161 m A1 m v1 w1 00406 00161 2522 lbms m A2 continuity for air Check for state 2 Pg60F 0256 psia P v1 so liquid water out Q 1 2 Liquid State 2 is saturated φ2 100 Pv2 Pg2 0256 psia w2 0622 Pv2 PA2 0622 0256 147 0256 00110 m v2 w2m A2 00110 2522 00277 lbms m liq m v1 m v2 00406 00277 00129 lbms Note that the given volume flow rate at the inlet is not that at the exit The mass flow rate of dry air is the quantity that is the same at the inlet and exit Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11174E Consider a 10ft3 rigid tank containing an airwater vapor mixture at 147 lbfin2 90 F with a 70 relative humidity The system is cooled until the water just begins to condense Determine the final temperature in the tank and the heat transfer for the process Pv1 φPG1 07 06988 0489 lbfin2 Since mv const V const also Pv PG2 PG2 Pv1T2T1 0489T25497 For T2 80 F 048953975497 04801 05073 PG at 80 F For T2 70 F 048952975497 04712 03632 PG at 70 F interpolating T2 780 F w2 w1 0622 0489 1470489 00214 ma RaT1 Pa1V 14211 psi 144 in2ft2 10 ft3 5334 lbfftlbmR 5497 R 0698 lbm Energy Eq 1Q2 U2 U1 maua2 ua1 mvuv2 uv1 ma Cv Ta2 Ta1 w1 uv2 uv1 0698 lbm 017178 90 0021410363 10402 Btulbm 0698 lbm 2135 Btulbm 149 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11175E A waterfilled reactor of 50 ft3 is at 2000 lbfin2 550 F and located inside an insulated containment room of 5000 ft3 that has air at 1 atm and 77 F Due to a failure the reactor ruptures and the water fills the containment room Find the final pressure CV Total container Energy mvu2 u1 mau2 u1 1Q2 1W2 0 Initial water v1 0021407 ft3lbm u1 53924 mv Vv 23357 lbm Initial air ma PVRT 147 4950 1445334 53667 36604 lbm Substitute into energy equation 23357 u2 53924 36604 0171 T2 77 0 u2 00268 T2 5413 v2 V2mv 21407 ft3lbm Trial and error 2phase Tguess v2 x2 u2 LHS T 300 x2 21407 00174564537 0329 u2 54273 Btulbm LHS 550789 Btulbm too large T 290 x2 21407 00173574486 028507 u2 49827 Btulbm LHS 50605 Btulbm too small T2 298 F x2 03198 Psat 65 lbfin2 LHS 5415 OK Pa2 Pa1V1T2V2T1 147 psi 4950 7577 5000 53667 2055 lbfin2 P2 Pa2 Psat 8555 lbfin2 5000 ft 3 50 ft 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11176E In the production of ethanol from corn the solids left after fermentation are dried in a continuous flow oven This process generates a flow of 35 lbms moist air 200 F with 70 relative humidity which contains some volatile organic compounds and some particles To remove the organic gasses and the particles the flow is send to a thermal oxidicer see Fig P1182 where natural gas flames brings the mixture to 1500 F Find the rate of heating by the natural gas burners For this problem we will just consider the heating of the gas mixture and due to the exit temperature we will use the ideal gas tables F5 and F6 Eq1125 Pv φ Pg 070 1153 psia 807 psia Eq1128 ω 0622 Pv Ptot Pv 0622 807 147 807 0757 Flow m tot m a m v m a1 ω so m a 1 ω m tot 1992 lbms m v m tot ω 1 ω 1508 lbms Process Heating ω is constant Pv is constant For ideal gas the enthalpy does not depend on pressure so the energy equation gives the heat transfer as Q m a h2 h1a m v h2 h1v 1992 49361 15796 1508 12779 9964518015 6686 9863 16 549 Btus Comment Notice how much energy is spend on heating the water vapor Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11177E To save on natural gas use in the previous problem it is suggested to take and cool the mixture and condense out some water before heating it back up So the flow is cooled from 200 F to 120 F as shown in Fig P1183 and the now dryer mixture is heated to 1500 F Find the amount of water condensed out and the rate of heating by the natural gas burners for this case Eq1125 Pv φ Pg 070 1153 psia 807 psia Eq1128 ω 0622 Pv Ptot Pv 0622 807 147 807 0757 Flow m tot m a m v m a1 ω so m a 1 ω m tot 1992 lbms m v m tot ω 1 ω 1508 lbms Now cool to 120 F Pg 1695 psia Pv1 so φ2 100 Pv2 1695 psia ω2 0622 Ptot Pv Pv 0622 1695 147 1695 008107 m liq m a ω1 ω2 1992 0757 008107 13464 lbms The air flow is not changed so the water vapor flow for heating is m v2 m v1 m liq 1508 13464 1616 lbms Now the energy equation becomes Q m a h3 h2a m v2 h3 h2v 1992 49361 15796 1616 12779 9964518015 7743 Btus Comment If you solve the previous problem you find this is only 47 of the heat for the case of no water removal Qcool m liq heat 2 1 3 Q Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11178E Two moist air streams with 85 relative humidity both flowing at a rate of 02 lbms of dry air are mixed in a steady flow setup One inlet flowstream is at 90 F and the other at 61 F Find the exit relative humidity Solution CV mixing chamber Continuity Eq water m air w1 m air w2 2m air wex Energy Eq m air h1 m air h2 2m air h ex Properties from the tables and formulas Pg90 0699 Pv1 085 0699 0594 psia w1 0622 0594 147 0594 00262 Pg61 02667 Pv2 085 02667 02267 psia w2 0622 02267 147 02267 000974 Continuity Eq water wex w1 w22 0018 For the energy equation we have h ha whv so 2 hex h1 h2 0 2ha ex ha 1 ha 2 2wexhv ex w1hv 1 wh v 2 we will use constant heat capacity to avoid an iteration on Tex Cp air2Tex T1 T2 Cp H2O2wexTex w1T1 w2T2 0 Tex Cp airT1 T2 Cp H2Ow1T1 w2T2 2Cp air 2wexCp H2O 024 90 61 044700262 90 000974 6104961 757 F Pv ex 0622 wex wex Ptot 0018 0622 0018 147 0413 psia Pg ex 0445 psia φ 0413 0445 093 or 93 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11179E A flow of moist air from a domestic furnace state 1 in Figure P11100 is at 120 F 10 relative humidity with a flow rate of 01 lbms dry air A small electric heater adds steam at 212 F 147 psia generated from tap water at 60 F Up in the living room the flow comes out at state 4 90 F 60 relative humidity Find the power needed for the electric heater and the heat transfer to the flow from state 1 to state 4 1 3 4 Liquid cb 2 State 1 F71 Pg1 1695 psia hg1 111354 Btulbm Pv1 φ Pg1 01 1695 01695 psia w1 0622 Ptot Pv1 Pv1 0622 01695 147 01695 000726 Starte 2 hf 2808 Btulbm State 2a hg 212 115049 Btulbm State 4 Pg4 0699 psia hg4 110072 Btulbm Pv4 φ Pg4 06 0699 04194 psia w4 0622 Ptot Pv4 Pv4 0622 04194 147 04194 00183 m liq m a ω1 ω4 01 00183 000726 00011 lbms Energy Eq for heater Q heater m liq hout hin 00011 115049 2808 1235 Btus 117 kW Energy Eq for line excluding the heater Q line m a ha4 w4hg4 ha1 w1hg1 m liq hg 212 01 02490 120 00183 110072 000726 111354 00011 115049 078 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11180E Atmospheric air at 95 F relative humidity of 10 is too warm and also too dry An air conditioner should deliver air at 70 F and 50 relative humidity in the amount of 3600 ft3 per hour Sketch a setup to accomplish this find any amount of liquid at 68 F that is needed or discarded and any heat transfer CV air conditioner Check first the two states inlet 1 exit 2 In Pg1 08246 psia hg1 11029 Btulbm hf68 3608 Btulbm Pv1 φ1 Pg1 008246 psia w1 0622 Pv1PtotPv1 00035 Ex Pg2 036324 psia hg2 1092 Btulbm Pv2 φ2 Pg2 01816 psia w2 0622 Pv2PtotPv2 000778 Water must be added w2 w1 Continuity and energy equations m A1 w1 m liq m A1 w2 m Ah1mix m liqhf Q CV m Ah2mix m tot PV totRT 1473600144533452967 270 lbmh m A m tot1 w2 26791 lbmh m liq m Aw2 w1 26791000778 00035 1147 lbmh Q CV m ACp a T2 T1 w2hg2 w1hg1 m liqhf68 26791 02470 95 000778 1092 00035 11029 1147 3608 4068 Btuh Liquid water Cooler Inlet Exit 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11181E A commercial laundry runs a dryer that has an exit flow of 1 lbms moist air at 120 F 70 relative humidity To save on the heating cost a counterflow stack heat exchanger is used to heat a cold water line at 50 F for the washers with the exit flow as shown in Fig P11105 Assume the outgoing flow can be cooled to 70 F Is there a missing flow in the figure Find the rate of energy recovered by this heat exchanger and the amount of cold water that can be heated to 85 F Dryer oulet 1 120 F Φ 70 hg1 111354 Btulbm Pg1 1695 psia Pv1 07 1695 psia 11865 psia ω1 0622 11865 147 11865 00546 w T Φ 100 Φ 70 dry 1 2 T dew 1 Dew point for 1 T 2 State 2 T2 Tdew 1 107 F so Φ2 100 Pg2 0363 psia hg2 109204 Btulbm hf2 3809 Btulbm ω2 0622 0363147 0363 001575 Continuity Eq water 12 line m a ω1 m a ω2 m liq m liq m aω1 ω2 The mass flow rate of dry air is m a m moist air 1 ω1 1 lbms 1 00546 0948 lbms The heat out of the exhaust air which also equals the energy recovered becomes Q CV m a Cp aT1 T2 ω1 hg1 ω1 ω2 hf2 ω2 hg2 0948 024120 70 00546 111354 003885 3809 001575 109204 0948 lbms 5412 Btulbm 513 Btus m liq Q CV CP liq ΔTliq 513 10 85 50 1466 lbms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11182E An indoor pool evaporates 3 lbmh of water which is removed by a dehumidifier to maintain 70 F Φ 70 in the room The dehumidifier is a refrigeration cycle in which air flowing over the evaporator cools such that liquid water drops out and the air continues flowing over the condenser as shown in Fig P11107 For an air flow rate of 02 lbms the unit requires 12 Btus input to a motor driving a fan and the compressor and it has a coefficient of performance β QL WC 20 Find the state of the air after the evaporator T2 ω2 Φ2 and the heat rejected Find the state of the air as it returns to the room and the compressor work input The unit must remove 3 lbmh liquid to keep steady state in the room As water condenses out state 2 is saturated 1 70 F 70 Pg1 0363 psia hg1 10920 Btulbm Pv1 φ1 Pg1 02541 psia w1 0622 Pv1PtotPv1 001094 CV 1 to 2 m liq m aw1 w2 w2 w1 m liqm a qL h1 h2 w1 w2 hf2 w2 001094 33600 02 0006774 Pv2 Pg2 Ptot w2 0622 w2 147 0006774 0628774 01584 psia Table F71 T2 468 F hf2 1488 btulbm hg2 1081905 Btulbm qL 02470 468 001094 1092 0006774 1081905 000417 1488 1012 Btulbm dry air W c m a qL β 1 Btus CV Total system h3 h1 W elm a w1w2 hf 1202 0062 5938 Btulbm dry air Cp a T3 T1 w2hv3 w1hv1 Trial and error on T3 3 w3 w2 h3 T3 112 F Pg3 136 psia Pv3 Pv2 01584 φ3 Pv3Pg3 012 or φ3 12 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11183E To refresh air in a room a counterflow heat exchanger is mounted in the wall as shown in Fig P11122 It draws in outside air at 33 F 80 relative humidity and draws room air 104 F 50 relative humidity out Assume an exchange of 6 lbmmin dry air in a steady flow device and also that the room air exits the heat exchanger to the atmosphere at 72 F Find the net amount of water removed from room any liquid flow in the heat exchanger and T φ for the fresh air entering the room State 3 Pg3 10804 psia hg3 110673 Btulbm Pv3 φ3 Pg3 05402 w3 0622 Pv3PtotPv3 002373 The room air is cooled to 72 F Tdew1 82 F so liquid will form in the exit flow channel and state 4 is saturated 4 72 F φ 100 Pg4 03918 psia hg4 109291 Btulbm w4 0017 hf4 4009 Btulbm 1 33 F φ 80 Pg1 00925 psia hg1 107583 Btulbm Pv1 φ1 Pg1 0074 psia w1 000315 CV 3 to 4 m liq4 m a w3 w4 6 002373 0017 004 lbmmin CV room m vout m a w3 w2 m a w3 w1 6002373 000315 01235 lbmmin CV Heat exchanger m ah2 h1 m ah3 h4 m liqhf4 Cp aT2T1 w2hv2 w1hv1 Cp aT3T4 w3hv3 w4hv4 w3w4 hf4 024T233 w2hv2 33888 02410472 262627 185795 02698 024 T2 000315 hv2 26402 btulbm Trial and error on T2 T2 955 F Pg2 0837 psia Pv2 Pv1 φ Pv2 Pg2 0074 0837 0088 or φ 9 w T Φ 100 Φ Φ Φ 80 50 10 dry 3 4 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11184E Weighing of masses gives a mixture at 80 F 35 lbfin2 with 1 lbm O2 3 lbm N2 and 1 lbm CH4 Find the partial pressures of each component the mixture specific volume mass basis mixture molecular weight and the total volume From Eq 114 yi mi Mi mjM j ntot mjMj 131999 328013 11604 0031251 0107093 0062344 0200688 yO2 00312510200688 01557 yN2 01070930200688 05336 yCH4 00623440200688 03107 From Eq1110 PO2 yO2 Ptot 01557 35 545 lbfin2 PN2 yN2 Ptot 05336 35 18676 lbfin2 PCH4 yCH4 Ptot 03107 35 10875 lbfin 2 Vtot ntot RTP 0200688 1545 5397 35 144 332 ft 3 v Vtotmtot 332 1 3 1 664 ft3lbm From Eq115 Mmix yjMj mtotntot 50200688 24914 lbmlbmol Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11185E A mixture of 50 carbon dioxide and 50 water by mass is brought from 2800 R 150 lbfin2 to 900 R 30 lbfin2 in a polytropic process through a steady flow device Find the necessary heat transfer and work involved using values from F4 Process Pvn constant leading to n lnv2v1 lnP1P2 v RTP n ln15030 ln900 15030 2800 33922 Rmix ΣciRi 05 351 05 8576778 007767 Btulbm R CP mix ΣciCPi 05 0203 05 0445 0324 Btulbm R Work is from Eq718 w vdP n n1 Peve Pivi nR n1 Te Ti 33922 007767 23922 900 2800 2093 Btu lbm Heat transfer from the energy equation q he hi w CPTe Ti w 4063 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11186E A large air separation plant takes in ambient air 79 N2 21 O2 by volume at 147 lbfin2 70 F at a rate of 2 lb mols It discharges a stream of pure O2 gas at 30 lbfin2 200 F and a stream of pure N2 gas at 147 lbfin2 70 F The plant operates on an electrical power input of 2000 kW Calculate the net rate of entropy change for the process Air 79 N2 21 O2 P1 147 psia T1 70 F n 1 2 lbmols W IN 2000 kW P2 30 psia T2 200 F P3 147 psia T3 70 F S gen Q CV T0 i n is i T0 Q CV n 2s 2 n 3s 3 n 1s 1 The energy equation Eq410 gives the heat transfer rate as Q CV Σnh i W CV n O2 C P0 O2 T2T1 n N2 C P0 N2 T3T1 W CV 021232 0213 200 70 0 2000 34123600 3826 18956 1513 Btus Use Eq616 see page 295 also for the entropy change Σn is i 0212320219 ln 660 530 1545 778 ln 30 021147 07920 1545 778 ln 147 079147 19906 BtuR s S gen 1513 530 19906 0864 BtuR s pure O 2 pure N 2 1 2 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 11187E Ambient air is at a condition of 147 lbfin2 95 F 50 relative humidity A steady stream of air at 147 lbfin2 73 F 70 relative humidity is to be produced by first cooling one stream to an appropriate temperature to condense out the proper amount of water and then mix this stream adiabatically with the second one at ambient conditions What is the ratio of the two flow rates To what temperature must the first stream be cooled COOL LIQ H O 2 Q 0 MIX Q COOL MIX 1 2 3 4 5 P1 P2 P5 147 lbfin 2 T1 T2 95 F φ1 φ2 050 φ4 10 T5 73 F φ5 070 Pv1 Pv2 0508246 04123 w1 w2 0622 04123 14704123 00179 Pv5 0704064 02845 w5 0622 02845 14702845 00123 MIX Call the mass flow ratio r ma2ma1 Conservation of water mass w1 r w4 1 rw5 Energy Eq ha1 w1hv1 rha4 rw4hv4 1 rha5 1 rw5hv5 00179 rw4 1 r 00123 or r 0017900123 00123w4 with w4 0622 PG4 147PG4 024555 00179 11072 r 024 T4 rw4hv4 1 r 024 533 1 r 00123 10933 or r024 T4 w4hG4 1414 1166 0 Assume T4 40 F PG4 0121 66 psia hG4 10789 Btulbm w4 0622 0121 66 1470121 66 00052 ma2 ma1 0017900123 0012300052 07887 07887024500 0005210789 1414 1166 029 0 OK T4 40 F Updated June 2013 8e SOLUTION MANUAL CHAPTER 12 Fundamentals of Thermodynamics Borgnakke Sonntag Borgnakke Sonntag Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Fundamentals of Thermodynamics 8th Edition Borgnakke and Sonntag CONTENT CHAPTER 12 SUBSECTION PROB NO InText concept questions af Study guide problems 115 Clapeyron equation 1633 Property Relations Maxwell and those for Enthalpy internal Energy and Entropy 3443 Volume Expansivity and Compressibility 4459 Equations of State 6081 Generalized Charts 82120 Mixtures 121133 Helmholtz EOS 134138 Review problems 139148 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful The following table gives the values for the compressibility enthalpy departure and the entropy departure along the saturated liquidvapor boundary These are used for all the problems using generalized charts as the figures are very difficult to read accurately consistently along the saturated liquid line It is suggested that the instructor hands out copies of this page or let the students use the computer for homework solutions Tr Pr Zf Zg dhRTf dhRTg dsRf dsRg 096 078 014 054 365 139 345 110 094 069 012 059 381 119 374 094 092 061 010 064 395 103 400 082 090 053 009 067 407 090 425 072 088 046 008 070 417 078 449 064 086 040 007 073 426 069 473 057 084 035 006 076 435 060 497 050 082 030 005 079 443 052 522 045 080 025 004 081 451 046 546 039 078 021 0035 083 458 040 572 035 076 018 003 085 465 034 598 031 074 015 0025 087 472 029 626 027 072 012 002 088 479 025 654 023 070 010 0017 090 485 021 683 020 068 008 0014 091 492 018 714 017 066 006 001 092 498 015 747 015 064 005 0009 094 504 012 781 012 060 003 0005 095 516 008 856 008 058 002 0004 096 522 006 897 007 054 001 0002 098 534 003 987 004 052 00007 00014 098 541 002 1038 003 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful InText Concept Questions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12a Mention two uses of the Clapeyron equation If you have experimental information about saturation properties down to a certain temperature Clapeyron equation will allow you to make an intelligent curve extrapolation of the saturated pressure versus temperature function PsatT for lower temperatures From Clapeyrons equation we can calculate a heat of evaporation heat of sublimation or heat of fusion based on measurable properties P T and v The similar changes in entropy are also obtained since hfg Tsfg hif Tsif hig Ts ig Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12b If I raise the temperature in a constant pressure process does g go up or down From the definition and variation in Gibbs function see Eq1215 and Maxwells relation Eq1221 last one we get dg s dT so Gibbs function decreases as temperature increases Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12c If I raise the pressure in an isentropic process does h go up or down Is that independent upon the phase Tds 0 dh vdP so h increases as P increases for any phase The magnitude is proportional to v ie large for vapor and small for liquid and solid phases Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12d If I raise the pressure in a solid at constant T does s go up or down In Example 124 it is found that change in s with P at constant T is negatively related to volume expansivity a positive value for a solid dsT v α P dPT so raising P decreases s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12e What does it imply if the compressibility factor is larger than 1 A compressibility factor that is greater than one comes from domination of intermolecular forces of repulsion short range over forces of attraction long range either high temperature or very high density This implies that the density is lower than what is predicted by the ideal gas law the ideal gas law assumes the molecules atoms can be pressed closer together 12f What is the benefit of the generalized charts Which properties must be known besides the charts themselves The generalized charts allow for the approximate calculations of enthalpy and entropy changes and PvT behavior for processes in cases where specific data or equation of state are not known They also allow for approximate phase boundary determinations It is necessary to know the critical pressure and temperature as well as idealgas specific heat Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful ConceptStudy Guide Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 121 The slope dPdT of the vaporization line is finite as you approach the critical point yet hfg and vfg both approach zero How can that be The slope is dP dT sat hfg Tvfg Recall the math problem what is the limit of fxgx when x goes towards a point where both functions f and g goes towards zero A finite limit for the ratio is obtained if both first derivatives are different from zero so we have dPdT dhfg dT dTvfgdT as T Tc Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 122 In view of Clapeyrons equation and Fig 24 is there something special about ice I versus the other forms of ice Yes The slope of the phase boundary dPdT is negative for ice I to liquid whereas it is positive for all the other ice to liquid interphases This also means that these other forms of ice are all heavier than liquid water The pressure must be more than 200 MPa 2000 atm so even the deepest ocean cannot reach that pressure recall about 1 atm per 10 meters down Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 123 If we take a derivative as PTv in the twophase region see Figs 27 and 28 does it matter what v is How about T In the twophase region P is a function only of T and not dependent on v The slope is the same at a given T regardless of v The slope becomes higher with higher T and generally is the highest near the critical point P T v V L S CP Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 124 Sketch on a PT diagram how a constant v line behaves in the compressed liquid region the twophase LV region and the superheated vapor region P T V L CrP S v v c v large v medium v v c small P T v V L S CP v large v v v c medium v v c The figures are for water where liquid is denser than solid Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 125 If I raise the pressure in an isothermal process does h go up or down for a liquid or solid What do you need to know if it is a gas phase Eq 1225 h PT v T v TP v1 TαP Liquid or solid αP is very small h increases with P For a gas we need to know the equation of state 126 The equation of state in Example 123 was used as explicit in v Is it explicit in P Yes the equation can be written explicitly in P P RT v C T3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 127 Over what range of states are the various coefficients in Section 125 most useful For solids or liquids where the coefficients are essentially constant over a wide range of Ps and Ts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 128 For a liquid or a solid is v more sensitive to T or P How about an ideal gas For a liquid or solid v is much more sensitive to T than P For an ideal gas v RTP varies directly with T inversely with P 129 Most equations of state are developed to cover which range of states Most equations of state are developed to cover the gaseous phase from low to moderate densities Many cover highdensity regions as well including the compressed liquid region To cover a wider region the EOS must be more complex and usually has many terms so it is only useful on a computer Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1210 Is an equation of state valid in the twophase regions No In a twophase region P depends only on T There is a discontinuity at each phase boundary It is actually difficult to determine the phase boundary from the EOS 1211 As P 0 the specific volume v For P does v 0 At very low P the substance will be essentially an ideal gas Pv RT so that v becomes very large However at very high P the substance eventually must become a solid which cannot be compressed to a volume approaching zero Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1212 Must an equation of state satisfy the two conditions in Eqs 1249 and 1250 It has been observed from experimental measurements that substances do behave in that manner If an equation of state is to be accurate in the nearcritical region it would have to satisfy these two conditions If the equation is simple it may be overly restrictive to impose these as it may lead to larger inaccuracies in other regions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1213 At which states are the departure terms for h and s small What is Z there Departure terms for h and s are small at very low pressure or at very high temperature In both cases Z is close to 1 and this is the ideal gas region Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1214 The departure functions for h and s as defined are always positive What does that imply for the real substance h and s values relative to ideal gas values Realsubstance h and s are less than the corresponding idealgas values This is true for the range shown in the figures Pr 10 For higher P the isotherms do bend down and negative values are possible Generally this means that there are slightly attractive forces between the molecules leading to some binding energy that shows up as a negative potential energy thus smaller h and u and it also give a little less chaos smaller s due to the stronger binding as compared to ideal gas When the pressure becomes large the molecules are so close together that the attractive forces turns into repulsive forces and the departure terms are negative Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1215 What is the benefit of Kays rule versus a mixture equation of state Kays rule for a mixture is not nearly as accurate as an equation of state for the mixture but it is very simple to use and it is general For common mixtures new mixture EOS are becoming available in which case they are preferable for a greater accuracy in the calculation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Clapeyron Equation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1216 An approximation for the saturation pressure can be ln Psat A BT where A and B are constants Which phase transition is that suitable for and what kind of property variations are assumed Clapeyron Equation expressed for the three phase transitions are shown in Eqs 125127 The last two leads to a natural log function if integrated and ideal gas for the vapor is assumed dPsat dT Psat hevap RT2 where hevap is either hfg or hig Separate the variables and integrate P 1 sat dPsat hevap R1 T2 dT ln Psat A BT B hevap R1 if we also assume hevap is constant and A is an integration constant The function then applies to the liquidvapor and the solidvapor interphases with different values of A and B As hevap is not excactly constant over a wide interval in T it means that the equation cannot be used for the total domain Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1217 Verify that Clapeyrons equation is satisfied for R410A at 10oC in Table B4 Clapeyron Eq dPsat dT dPg dT hfg Tvfg B4 P 10857 kPa hfg 20857 kJkg vfg 002295 m3kg Slope around 10oC best approximated by cord from 5oC to 15oC dPg dT 12554 9339 15 5 3215 kPaK hfg Tvfg 20857 28315 002295 32096 kPaK This fits very well Use CATT3 to do from 9 to 11oC for better approximation as the saturated pressure is very nonlinear in T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1218 In a Carnot heat engine the heat addition changes the working fluid from saturated liquid to saturated vapor at T P The heat rejection process occurs at lower temperature and pressure T T P P The cycle takes place in a piston cylinder arrangement where the work is boundary work Apply both the first and second law with simple approximations for the integral equal to work Then show that the relation between P and T results in the Clapeyron equation in the limit T dT qH Tsfg qL TTsfg wnet qH qL Ts fg The boundary movement work w Pdv wNET Pv2v1 2 3 Pdv P Pv4 v3 1 4 Pdv Approximating 2 3 Pdv P P 2 v3 v2 1 4 Pdv P P 2 v1 v4 Collecting terms wNET P 2 v2v3 2 v1v4 the smaller the P the better the approximation P T 1 2v2 v3 1 2v1 v4 sfg In the limit as T 0 v3 v2 vg v4 v1 v f lim T0 P T dPsat dT sfg vfg s P v P T T T 1 2 4 3 P P P P P s at T v at T fg fg 4 3 1 2 T T T T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1219 Verify that Clapeyrons equation is satisfied for carbon dioxide at 6oC in Table B3 Clapeyron Eq dPsat dT dPg dT hfg Tvfg B3 P 40720 kPa hfg 21159 kJkg vfg 000732 m3kg Slope around 6oC best approximated by cord from 4oC to 8oC dPg dT 42831 38688 8 4 103575 kPaK hfg Tvfg 21159 27915 000732 kJkg K m3kg 103549 kPaK This fits very well Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1220 Use the approximation given in problem 1216 and Table B1 to determine A and B for steam from properties at 25oC only Use the equation to predict the saturation pressure at 30oC and compare to table value ln Psat A BT dPsat dT Psat BT2 so we notice from Eq127 and Table values from B11 and A5 that B hfg R 24423 04615 kJkg kJkgK 5292 K Now the constant A comes from the saturation pressure as A ln Psat BT ln 3169 5292 27315 25 189032 Use the equation to predict the saturation pressure at 30oC as ln Psat A BT 189032 5292 27315 30 14462 Psat 42469 kPa compare this with the table value of Psat 4246 kPa and we have a very accurate approximation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1221 A certain refrigerant vapor enters a steady flow constant pressure condenser at 150 kPa 70C at a rate of 15 kgs and it exits as saturated liquid Calculate the rate of heat transfer from the condenser It may be assumed that the vapor is an ideal gas and also that at saturation vf vg The following quantities are known for this refrigerant ln Pg 815 1000T CP 07 kJkg K with pressure in kPa and temperature in K The molecular mass is 100 Solution Refrigerant State 1 T1 70oC P1 150 kPa State 2 P2 150 kPa x2 10 State 3 P3 150 kPa x3 00 Energy Eq 1q3 h3 h1 h3 h2 h2 h1 hfg T3 CP0T2 T1 Get the saturation temperature at the given pressure ln 150 815 1000T2 T2 3185 K 453oC T3 Now get the enthalpy of evaporation hfg T3 dPg dT hfg Tvfg vfg vg RT Pg dT dPg Pg d ln Pg dT hfg RT2 Pg d ln Pg dT 1000T2 hfgRT2 hfg 1000 R 1000 83145100 8315 kJkg Substitute into the energy equation 1q3 8315 07453 70 10044 kJkg Q COND 1510044 1506 kW v P s T 1 2 3 1 2 3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1222 Calculate the values hfg and sfg for nitrogen at 70 K and at 110 K from the Clapeyron equation using the necessary pressure and specific volume values from Table B61 Clapeyron equation Eq127 dPg dT hfg Tvfg sfg vfg For N2 at 70 K using values for Pg from Table B6 at 75 K and 65 K and also vfg at 70 K hfg Tvgvf Τ Pg 700525 0157611741 7565 2157 kJkg sfg hfgT 3081 kJkg K Table B61 hfg 2078 kJkg and sfg 297 kJkg K Comparison not very close because Pg not linear function of T Using 71 K 69 K from the software we can then get hfg 700525 01544563324 7169 2080 kJkg At 110 K hfg 1100014 3421938810842 115105 13482 kJkg sfg hfgT 13482 110 1226 kJkg K Table B61 hfg 13417 kJkg and sfg 122 kJkg K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1223 Find the saturation pressure for refrigerant R410A at 80oC assuming it is higher than the triple point temperature The lowest temperature in Table B4 for R410A is 60oC so it must be extended to 80oC using the Clapeyron Eq 127 integrated as in example 121 Table B4 at T1 60oC 21315 K P1 641 kPa R 01145 kJkgK ln P P1 R hfg T T1 T T1 27996 01145 19315 21315 19315 21315 11878 P 641 exp11878 1954 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1224 Ammonia at 70oC is used in a special application at a quality of 50 Assume the only table available is B2 that goes down to 50oC To size a tank to hold 05 kg with x 05 give your best estimate for the saturated pressure and the tank volume To size the tank we need the volume and thus the specific volume If we do not have the table values for vf and vg we must estimate those at the lower T We therefore use Clapeyron equation to extrapolate from 50oC to 70oC to get the saturation pressure and thus vg assuming ideal gas for the vapor The values for vf and hfg do not change significantly so we estimate Between 50oC and 70oC hfg 1430 kJkg and at 70oC we get vf 0001375 m3kg The integration of Eq127 is the same as in Example 121 so we get ln P2 P1 hfg R T2 T1 T2T1 1430 04882 70 50 20315 22315 12923 P2 P1 exp12923 409 exp12923 112 kPa vg RT2P2 04882 20315 112 8855 m3kg v2 1x vf x vg 05 0001375 05 8855 4428 m3kg V2 mv2 2214 m3 A straight line extrapolation will give a negative pressure P T 70 50 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1225 Use the approximation given in problem 1216 and Table B4 to determine A and B for refrigerant R410A from properties at 0oC only Use the equation to predict the saturation pressure at 5oC and compare to table value ln Psat A BT dPsat dT Psat BT2 so we notice from Eq127 and Table values from B41 and A5 that B hfg R 22137 01145 kJkg kJkgK 19334 K Now the constant A comes from the saturation pressure as A ln Psat BT ln 7987 19334 27315 137611 Use the equation to predict the saturation pressure at 5oC as ln Psat A BT 137611 19334 27315 5 68102 Psat 907 kPa compare this with the table value of Psat 9339 kPa and we have an approximation 3 low Notice hfg decreases so we could have used a lower value for the average in the interval Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1226 The triple point of CO2 is 564oC Predict the saturation pressure at that point using Table B3 The lowest temperature in Table B3 for CO2 is 50oC so it must be extended to 564oC 21675 K using the Clapeyron Eq 127 integrated as in Ex 121 Table B3 at T1 50oC 22315 K P1 6823 kPa hfg 33973 kJkg Table A5 R 01889 kJkgK ln P P1 R hfg T T1 T T1 33973 01889 21675 22315 21675 22315 023797 P 6823 exp023797 5378 kPa Notice from Table 32 P 5208 kPa so we are 3 high As hfg becomes larger for lower Ts we could have estimated a more suitable value for the interval from 50 to 564oC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1227 Helium boils at 422 K at atmospheric pressure 1013 kPa with hfg 833 kJkmol By pumping a vacuum over liquid helium the pressure can be lowered and it may then boil at a lower temperature Estimate the necessary pressure to produce a boiling temperature of 1 K and one of 05 K Solution Helium at 422 K P1 01013 MPa h FG 833 kJkmol dPSAT dT hFG TvFG hFGPSAT RT2 ln P2 P1 hFG R 1 T1 1 T2 For T2 10 K ln P2 1013 833 83145 1 422 1 10 P2 0048 kPa 48 Pa For T2 05 K ln P2 1013 833 83145 1 422 1 05 P2 21601106 kPa 21601 103 Pa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1228 Using the properties of water at the triple point develop an equation for the saturation pressure along the fusion line as a function of temperature Solution The fusion line is shown in Fig 23 as the SL interphase From Eq125 we have dPfusion dT hif Tvif Assume hif and vif are constant over a range of Ts We do not have any simple models for these as function of T other than curve fitting Then we can integrate the above equation from the triple point T1 P1 to get the pressure PT as P P1 hif vif ln T T1 Now take the properties at the triple point from B11 and B15 P1 06113 kPa T1 27316 K vif vf vi 0001 00010908 908 105 m3kg hif hf hi 00 3334 3334 kJkg The function that approximates the pressure becomes P 06113 3672 106 ln T T1 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1229 Using thermodynamic data for water from Tables B11 and B15 estimate the freezing temperature of liquid water at a pressure of 30 MPa H2O dT dPif Tvif hif const At the triple point vif vf vi 0001 000 0001 090 8 0000 090 8 m3kg hif hf hi 001 33340 33341 kJkg dPif dT 33341 273160000 090 8 13 442 kPaK at P 30 MPa T 001 30 00006 13 442 22 oC TP T P 30 MPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1230 Ice solid water at 3C 100 kPa is compressed isothermally until it becomes liquid Find the required pressure Water triple point T 001oC P 06113 kPa Table B11 vf 0001 m3kg hf 001 kJkg Tabel B15 vi 0001 0908 m3kg hi 3334 kJkg Clapeyron dPif dT hf hi vf viT 3334 00000908 27316 13 442 kPaK P dPif dT T 13 4423 001 40 460 kPa P Ptp P 40 461 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1231 From the phase diagram for carbon dioxide in Fig 25 and 24 for water what can you infer for the specific volume change during melting assuming the liquid has a higher h than the solid phase for those two substances The saturated pressure versus temperature has a positive slope for carbon dioxide and a negative slope for water Clapeyron dPif dT hf hi vf viT So if we assume hf hi 0 then we notice that the volume change in the melting gives Water vf vi 0 so vf vi Carbon dioxide vf vi 0 so vf v i Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1232 A container has a double wall where the wall cavity is filled with carbon dioxide at room temperature and pressure When the container is filled with a cryogenic liquid at 100 K the carbon dioxide will freeze so that the wall cavity has a mixture of solid and vapor carbon dioxide at the sublimation pressure Assume that we do not have data for CO2 at 100 K but it is known that at 90C Psat 381 kPa hIG 5745 kJkg Estimate the pressure in the wall cavity at 100 K Solution For CO2 space at T1 90 oC 1832 K P1 381 kPa hIG 5745 kJkg For T2 TcO2 100 K Clapeyron dPSUB dT hIG TvIG hIGPSUB RT2 ln P2 P1 hIG R 1 1832 1 100 5745 0188 92 1 1832 1 100 1381 or P2 P1 1005106 P2 383105 kPa 383102 Pa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1233 Small solid particles formed in combustion should be investigated We would like to know the sublimation pressure as a function of temperature The only information available is T hFG for boiling at 1013 kPa and T hIF for melting at 1013 kPa Develop a procedure that will allow a determination of the sublimation pressure PsatT TNBP normal boiling pt T TNMP normal melting pt T TTP triple point T 1 TTP T NMP 2 01013 MPa PTP 1PSAT dPSAT TNMP TTP hFG RT2 dT Since hFG const hFG NBP the integral over temperature becomes ln PTP 01013 hFG NBP R 1 TNBP 1 TTP get PTP 3 hIG at TP hG hI hG hF hF hI hFG NBP hIF NMP Assume hIG const again we can evaluate the integral ln PSUB PTP PTP PSUB 1PSUB dPSUB TTP T hIG RT2 dT hIG R 1 TTP 1 T or PSUB fnT P TP TP NMP NBP 1013 kPa T Solid Liquid Vap T T T P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Property Relations Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1234 Use Gibbs relation du Tds Pdv and one of Maxwells relations to find an expression for uPT that only has properties P v and T involved What is the value of that partial derivative if you have an ideal gas du Tds Pdv divide this by dP so we get u P T T s P T P v P T T v T P P v P T where we have used Maxwell Eq1219 Now for an ideal gas we get Ideal gas Pv RT v RT P then the derivatives are v T P R P and v P T RTP 2 and the derivative of u is u P T T v T P P v P T T R P P RTP2 0 This confirms that u is not sensitive to P and only a function of T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1235 The JouleThomson coefficient µJ is a measure of the direction and magnitude of the temperature change with pressure in a throttling process For any three properties xyz use the mathematical relation x y z y z x z x y 1 to show the following relations for the JouleThomson coefficient µJ T P h T v T P v CP RT2 PCP Z T P Let x T y P and z h and substitute into the relations as T P h P h T h T P 1 Then we have the definition of specific heat as CP h T P so solve for the first term µJ T P h 1 CP P h T 1 CP h P T The last derivative is substituted with Eq1225 so we get µJ T P h T v T P v CP If we use the compressibility factor then we get Pv ZRT v T P ZR P RT P Z T P v T RT P Z T P so then T v T P v v RT2 P Z T P v RT2 P Z T P and we have shown the last expression also µJ T P h T v T P v CP RT2 PCP Z T P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1236 Find the JouleThomson coefficient for an ideal gas from the expression given in Problem 1235 µJ T P h T v T P v CP RT2 PCP Z T P For an ideal gas v RTP so then the partial derivative v T P R P T v T P v RT P v v v 0 For an ideal gas Z 1 so the very last derivative of Z is also zero Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1237 Start from Gibbs relation dh Tds vdP and use one of Maxwells equation to get hvT in terms of properties P v and T Then use Eq1224 to also find an expression for hTv Find h vT and h Tv dh Tds vdP and use Eq1218 h vT T s vT vP vT T P Tv vP vT Also for the second first derivative use Eq1224 h Tv Ts Tv vP Tv Cv vP Tv Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1238 From Eqs 1223 and 1224 and the knowledge that Cp Cv what can you conclude about the slopes of constant v and constant P curves in a Ts diagram Notice that we are looking at functions Ts P or v given Solution The functions and their slopes are Constant v Ts at that v with slope T s v Constant P Ts at that P with slope T s P Slopes of these functions are now evaluated using Eq1223 and Eq1224 as T s P s T P 1 T Cp T s v s T v 1 T Cv Since we know Cp Cv then it follows that TCv TCp and therefore T s v T s P which means that constant vlines are steeper than constant P lines in a Ts diagram Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1239 Derive expressions for Tvu and for hsv that do not contain the properties h u or s Use Eq 1230 with du 0 T vu u vTu Tv Cv P TP Tv see Eqs 1233 and 1234 As dh Tds vdP h sv T vP sv T vT vs Eq1220 But T vs s vTs Tv Cv TP Tv Eq1222 h sv T vT Cv P Tv Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1240 Evaluate the isothermal changes in the internal energy the enthalpy and the entropy for an ideal gas Confirm the results in Chapters 3 and 6 We need to evaluate duT dhT and dsT for an ideal gas P RTv From Eq1231 we get duT T P Tv P dvT T R v P dvT P P dvT 0 From Eq1227 we get using v RTP dhT v T v TP dPT v T R P dPT v v dPT 0 These two equations confirms the statements in chapter 5 that u and h are functions of T only for an ideal gas From Eq1232 or Eq1234 we get dsT v TP dPT P Tv dvT R P dPT R v dvT so the change in s can be integrated to find s2 s1 R ln P2 P1 R ln v2 v1 when T2 T1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1241 Develop an expression for the variation in temperature with pressure in a constant entropy process TPs that only includes the properties PvT and the specific heat Cp Follow the development for Eq1232 T Ps s PT s TP v TP CPT T CP v TP s PT v TP Maxwell relation Eq 1223 and the other is Eq1227 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1242 Use Eq 1234 to get an expression for the derivative Tvs What is the general shape of a constant s process curve in a Tv diagram For an ideal gas can you say a little more about the shape Equation 1234 says ds Cv dT T P Tv dv so then in a constant s process we have ds 0 and we find T vs T Cv P T v As T is higher the slope is steeper but negative unless the last term PTv counteracts If we have an ideal gas this last term can be determined P RTv P Tv R v T vs T Cv R v P Cv and we see the slope is steeper for higher P and a little lower for higher T as Cv is an increasing function of T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1243 Show that the PvT relation as Pv b RT satisfies the mathematical relation in Problem 1235 x y z y z x z x y 1 Let x y z be P v T so we have P v T v T P T P v 1 The first derivative becomes P RTv b P v T RT v b2 Pv b The second derivative v b RTP v T P RP The third derivative T PRv b T P v v bR Substitute all three derivatives into the relation P v T v T P T P v RT v b2 R P v b R RT v b 1 P 1 with the last one recognized as a rewrite of the original EOS Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Volume Expansivity and Compressibility Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1244 What are the volume expansivity αp the isothermal compressibility βT and the adiabatic compressibility βs for an ideal gas The volume expansivity from Eq1237 and ideal gas v RTP gives αp 1 vv TP 1 v R P 1 T The isothermal compressibility from Eq1238 and ideal gas gives βT 1 vv PT 1 v RT P2 1 P The adiabatic compressibility βs from Eq1240 and ideal gas βs 1 vv Ps From Eq1232 we get for constant s ds 0 T Ps T Cp v TP T Cp R P v Cp and from Eq1234 we get v Ts Cv T P Tv Cv T v R Cv P Finally we can form the desired derivative v Ps v Ts T Ps Cv P v Cp v kP βs 1 vv Ps 1 v v kP 1 kP 1 k βT Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1245 Assume a substance has uniform properties in all directions with V LxLyLz and show that volume expansivity αp 3δT Hint differentiate with respect to T and divide by V V LxLyLz From Eq1237 αp 1 VV TP 1 LxLyLz T LxLyLz P LyLz LxLyLz Lx T P LxLyLz LxLz T LyP LxLy LxLyLz Lz T P 1 Lx T LxP 1 Ly T LyP 1 Lz T Lz P 3 δT This of course assumes isotropic properties the same in all directions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1246 Determine the volume expansivity αP and the isothermal compressibility βT for water at 20C 5 MPa and at 300C and 15 MPa using the steam tables Water at 20oC 5 MPa compressed liquid αP 1 vv TP 1 vv TP βT 1 vv PT 1 vv P T Estimate by finite difference using values at 0oC 20oC and 40oC αP 1 0000 9995 0001 0056 0000 9977 40 0 0000 1976 oC1 Using values at saturation 5 MPa and 10 MPa βT 1 0000 9995 0000 9972 0001 0022 10 00023 0000 50 MPa1 Water at 300oC 15 MPa compressed liquid αP 1 0001 377 0001 4724 0001 3084 320 280 0002 977 oC1 βT 1 0001 377 0001 3596 0001 3972 20 10 0002 731 MPa1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1247 Use the CATT3 software to solve the previous problem The benefit of the software to solve for the partial derivatives is that we can narrow the interval over which we determine the slope Water at 20oC 5 MPa compressed liquid αP 1 vv TP 1 vv TP βT 1 vv PT 1 vv P T Estimate by finite difference using values at 19oC 20oC and 21oC αP 1 0000 9995 0000 9997 0000 9993 21 19 0000 40 oC1 Using values at saturation 45 MPa and 55 MPa βT 1 0000 9995 0000 9993 0000 9997 55 45 0000 40 MPa1 Water at 300oC 15 MPa compressed liquid αP 1 0001 377 0001 385 0001 369 302 298 0011 619 oC1 βT 1 0001 377 0001 373 0001 381 16 14 0002 905 MPa1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1248 A cylinder fitted with a piston contains liquid methanol at 20C 100 kPa and volume 10 L The piston is moved compressing the methanol to 20 MPa at constant temperature Calculate the work required for this process The isothermal compressibility of liquid methanol at 20C is 122 109 m2N 1w2 1 2 Pdv Pv PT dPT 1 2 vβT PdPT For v constant βT constant the integral can be evaluated 1w2 vβT 2 P 2 2 P 2 1 For liquid methanol from Table A4 ρ 787 m3kg V1 10 L m 001 787 787 kg 1W2 0011220 2 202 012 2440 J 244 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1249 For commercial copper at 25oC see table A3 the speed of sound is about 4800 ms What is the adiabatic compressibility βs From Eq1241 and Eq1240 c2 P ρs v2P vs 1 1 vv Ps ρ 1 βsρ Then we get using density from Table A3 βs 1 c2ρ 1 48002 8300 m2 kg s2 m3 1000 48002 8300 1 kPa 523 109 kPa 1 Cu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1250 Use Eq 1232 to solve for TPs in terms of T v Cp and αp How large a temperature change does 25oC water αp 21 104 K1 have when compressed from 100 kPa to 1000 kPa in an isentropic process From Eq1232 we get for constant s ds 0 and Eq1237 T Ps T Cp v TP T Cp αp v Assuming the derivative is constant for the isentropic compression we estimate with heat capacity from Table A3 and v from B11 Ts T Ps Ps T Cp αp v P s 27315 25 418 21 104 0001003 1000 100 0013 K barely measurable Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1251 Sound waves propagate through a media as pressure waves that cause the media to go through isentropic compression and expansion processes The speed of sound c is defined by c2 Pρs and it can be related to the adiabatic compressibility which for liquid ethanol at 20C is 94 1010 m2N Find the speed of sound at this temperature c2 P ρs v2P vs 1 1 vv Ps ρ 1 βsρ From Table A4 for ethanol ρ 783 kgm3 c 1 9401012783 12 1166 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1252 Use Table B3 to find the speed of sound for carbon dioxide at 2500 kPa near 100oC Approximate the partial derivative numerically c2 P ρs v2P vs We will use the 2000 kPa and 3000 kPa table entries We need to find the change in v between two states with the same s at those two pressures At 100oC 2500 kPa s 16843 159542 163985 kJkgK v 003359 0021822 0027705 m3kg 2000 kPa s 163985 kJkgK v 0031822 m3kg 3000 kPa s 163985 kJkgK v 00230556 m3kg c2 v2P vs 00277052 3000 2000 002305560031822 kJ kg 87 5578 J kg c 87 5578 2959 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1253 Use the CATT3 software to solve the previous problem At 100oC 2500 kPa s 1636 kJkgK v 002653 m3kg 101oC s 1636 kJkgK v 002627 m3kg P 2531 MPa 99oC s 1636 kJkgK v 002679 m3kg P 2469 MPa c2 v2P vs 0026532 2531 2469 002627 002679 kJ kg 83 9195 J kg c 83 9195 2897 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1254 Consider the speed of sound as defined in Eq 1241 Calculate the speed of sound for liquid water at 20C 25 MPa and for water vapor at 200C 300 kPa using the steam tables From Eq 1241 c2 P ρs v2P v s Liquid water at 20oC 25 MPa assume P vs P v T Using saturated liquid at 20oC and compressed liquid at 20oC 5 MPa c2 0001 0020000 9995 2 2 500023 0000 99950001 002 MJ kg 2002106 J kg c 1415 ms Superheated vapor water at 200oC 300 kPa v 07163 m3kg s 73115 kJkg K At P 200 kPa s 73115 kJkg K T 157oC v 09766 m3kg At P 400 kPa s 73115 kJkg K T 2338oC v 05754 m3kg c2 071632 04000200 0575409766 MJ kg 02558 106 m2s2 c 506 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1255 Use the CATT3 software to solve the previous problem From Eq 1241 c2 P ρs v2P v s Liquid water at 20oC 25 MPa assume P vs P vs and CATT3 v 0001001 m3kg s 02961 kJkgK Using liquid at 3 MPa and 2 MPa at the same s 02961 kJkgK c2 00010012 3 2 0001 0001 001 MJ kg 1002106 J kg c 1001 ms Superheated vapor water at 200oC 300 kPa CATT3 v 07163 m3kg s 7311 kJkg K At P 290 kPa s 7311 kJkg K T 1962oC v 07351 m3kg At P 310 kPa s 7311 kJkg K T 2037oC v 06986 m3kg c2 071632 0310 0290 06986 07351 MJ kg 028114 106 m2s2 c 530 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1256 Soft rubber is used as a part of a motor mounting Its adiabatic bulk modulus is Bs 282 106 kPa and the volume expansivity is αp 486 104 K1 What is the speed of sound vibrations through the rubber and what is the relative volume change for a pressure change of 1 MPa From Eq1241 and Eq1240 c2 P ρs v2P vs 1 1 vv Ps ρ 1 βsρ ρ Bs 282 106 1000 Pa 1100 kgm3 2564 106 m2s2 c 1601 ms If the volume change is fast it is isentropic and if it is slow it is isothermal We will assume it is isentropic 1 VV Ps βs 1 Bs then V V P Bs 1000 282 106 355 104 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1257 Liquid methanol at 25oC has an adiabatic compressibility of 105 109 m2N What is the speed of sound If it is compressed from 100 kPa to 10 MPa in an insulated pistoncylinder what is the specific work From Eq1241 and Eq1240 and the density from table A4 c2 P ρs v2P vs 1 βsρ 1 105 109 787 1210 106 m2s2 c 1100 ms The specific work becomes w P dv P βsv dP βsv P dP βs v 1 2 P dP βs v 05 P2 2 P2 1 105 109 m2N 05 787 m3kg 10 0002 1002 10002 Pa 2 667 Jkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1258 Use Eq 1232 to solve for TPs in terms of T v Cp and αp How much higher does the temperature become for the compression of the methanol in Problem 1257 Use αp 24 104 K1 for methanol at 25oC From Eq1232 we get for constant s ds 0 and Eq1237 T Ps T Cp v TP T Cp αp v Assuming the derivative is constant for the isentropic compression we estimate with heat capacity and density v 1ρ from Table A4 Ts T Ps Ps T Cp αp v P s 29815 255 K kg K kJ 24 104 K1 1 787 kg m3 10 000 100 kPa 0353 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1259 Find the speed of sound for air at 20C 100 kPa using the definition in Eq 1241 and relations for polytropic processes in ideal gases From problem 1241 c2 P ρs v2P v s For ideal gas and isentropic process Pvk constant P Cvk P v kCvk1 kPv 1 c2 v2kPv1 kPv kRT c kRT 140287293151000 3432 ms For every 3 seconds after the lightning the sound travels about 1 km Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Equations of State Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1260 Use Table B3 and find the compressibility of carbon dioxide at the critical point Pv Z RT At the critical point from B3 P 73773 kPa T 31C 30415 K v 0002139 m3kg from A5 R 01889 kJkgK Z Pv RT 73773 0002139 01889 30415 027 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1261 Use the equation of state as shown in Example 123 where changes in enthalpy and entropy were found Find the isothermal change in internal energy in a similar fashion do not compute it from enthalpy The equation of state is Pv RT 1 C P T4 and to integrate for changes in u from Eq1231 we make it explicit in P as P T4 v R T3 C 1 Now perform the partial derivative of P P Tv 4 T3 v R T3 C 1 T4 v R T3 C 2 3 v R T 2 4 P T T4 P2 3 v R T2 4 P T 3 P T Pv RT P T 4 3 Pv RT Substitute into Eq1231 duT T P Tv P dvT P 4 3 Pv RT P dvT 3 P 1 Pv RT dvT 3 P C P T4 dvT The P must be eliminated in terms of v or the opposite we do the latter as from the equation of state v RT P C R 1 T3 dvT RT P2 dPT so now duT 3 C P2 T4 dvT 3 C R 1 T3 dPT and the integration becomes u2 u1 3 C R T3 P2 P1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1262 Use Table B4 to find the compressibility of R410A at 60oC and a saturated liquid b saturated vapor and c 3000 kPa Table A2 R 831451 72585 01145 kJkgK a Table B41 P 38369 kPa v 0001227 m3kg Z Pv RT 38369 0001227 01145 33315 01234 b Table B41 P 38369 kPa v 000497 m3kg Z Pv RT 38369 000497 01145 33315 05 c Table B42 P 3000 kPa v 000858 m3kg Z Pv RT 3000 000858 01145 33315 0675 The R410A is not an ideal gas at any of these states Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1263 Use a truncated virial EOS that includes the term with B for carbon dioxide at 20o C 1 MPa for which B 0128 m3kmol and TdBdT 0266 m3kmol Find the difference between the idealgas value and the realgas value of the internal energy virial eq P RT v BRT v2 P T v R v BR v2 RT v2 dB dT uu EA v ART2 E v2 A AdB dT AdvEA ART vE A T AdB dTE A v P T v Pdv ESolution of virial equation quadratic formula AvE A A1 2E A EA AR AT EPE A 1 EA 1 4BPAR ATEA where EA AR AT EPE A A8314529315 1000E A 243737 AvE A A1 2E A 243737 1 A 1 40128243737EA 23018 mA3E Akmol Using the minussign root of the quadratic formula results in a compressibility factor 05 which is not consistent with such a truncated equation of state u uAE A A83145 29315 23018E A 0266 4401 64 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1264 Solve the previous Problem with Table B3 values and find the compressibility of the carbon dioxide at that state B3 v 005236 mA3E Akg u 32727 kJkg A5 R 01889 kJkgK Z APv RTE A A1000 005236 01889 29315E A 09455 close to ideal gas To get uAE A let us look at the lowest pressure 400 kPa 20AoE AC v 013551 mA3E Akg and u 33157 kJkg Z PvRT 400 01355101889 29315 097883 It is not very close to ideal gas but this is the lowest P in the printed table u uAE A 32727 33157 43 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1265 A gas is represented by the virial EOS with the first two terms B and C Find an expression for the work in an isothermal expansion process in a pistoncylinder Virial EOS P ART vE A ABT RT E v2 E A ACT RT E v3 E A RT vA1E A BT vA2E A CT vA3E A The work is w A P dvEA RT A v1 BT v2 CT v3 dvE With just the first two terms we get w RT ln A v2 Ev1 E A BT vA1 2E A vA1 1E A A1 2E A CT vA2 2E A vA2 1E A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1266 Extend problem 1263 to find the difference between the idealgas value and the realgas value of the entropy and compare to table B3 Calculate the difference in entropy of the idealgas value and the realgas value for carbon dioxide at the state 20C 1 MPa as determined using the virial equation of state Use numerical values given in Problem 1263 COA2E A at T 20AoE AC P 1 MPa sA PE A sAPE A A vP ERTP P Tv dvE A ID Gas sA PE A sAPE A A vP ERTP R v dvEA R ln A P P E Therefore at P sA PE A sAPE A R ln A P P E A A vP ERTP P Tv dvE A virial P ART vE A ABRT v2 E A and AP TE AAvE A AR vE A ABR v2 E A ART v2 E AAdB dTE A Integrating sA PE A sAPE A R ln A P P E A R ln ART PvE A RB TAdB dTE AA1 vE A A P ERTE A Rln ART PvE A B TAdB dTE AA1 vE A Using values for COA2E A from solution 1263 and R 01889 kJkgK sA PE A sAPE A 01889ln A2437 37 23018E A 0128 0266A 1 23018E A 002214 kJkg K From Table B3 take the ideal as the lowest P 400 kPa sA PE A sAPE A 17904 16025 01889 ln4001000 00148 kJkgK The lowest P 400 kPa in B3 is not exactly ideal gas Z PvRT 09788 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1267 Two uninsulated tanks of equal volume are connected by a valve One tank contains a gas at a moderate pressure PA1E A and the other tank is evacuated The valve is opened and remains open for a long time Is the final pressure PA2E A greater than equal to or less than PA1E A2 Hint Recall Fig 125 Assume the temperature stays constant then for an ideal gas the pressure will be reduced to half the original pressure For the real gas the compressibility factor maybe different from 1 and then changes towards one as the pressure drops VAAE A VABE A VA2E A 2VA1E A TA2E A TA1E A T A P2 EP1 E A A V1 EV2 E A A Z2 EZ1 E A AmRT mRTE A A1 2E A A Z2 EZ1 E If T TABE A ZA2E A ZA1E A A P2 EP1 E A A1 2E If T TABE A ZA2E A ZA1E A A P2 EP1 E A A1 2E A B GAS EVAC P Z 1 1 2 2 10 T T B T T B P 1 P 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1268 Show how to get the constants in Eq1252 for van der Waals EOS van der Waals EOS P A RT v bE A A a v2 E The conditions at the critical point relate to these derivatives AP vE AATE A A RT v b2 E A A2a v3 E A A2P Ev2 E AATE A A 2RT v b 3 E A A6a v4 E Set both derivatives to zero at the critical point RTc vc b2 A2a v 3 c E A 0 1 2RTc vc b3 A6a v 4 c E A 0 2 we also have from the EOS Pc RTc vc b A a v 2 c E A 3 Now we need to solve theses three equations for vc a and b Solve the first equation for a and substitute into the second equation to give A2a v 3 c E A RTc vc b2 A6a v 4 c E A 3RTc vc vc b2 substitute into Eq2 3RTc vc vc b2 2RTc vc b3 now solve to get vc 3b Substitute back into the first equation to get 2 a RTc vc b2 v3 c RTc A27 4E A b Now finally substitute a and vc into the EOS Eq3 to get b Pc RTc vc b A a v 2 c E A RTc 2b RTc 27 8 b 9 b2 RTc 05 38 b The result is as in Eq1252 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1269 Show that the van der Waals equation can be written as a cubic equation in the compressibility factor involving the reduced pressure and reduced temperature as ZA3E A EA PAr A E8TAr AE A 1 ZA2E A A 27 Pr E64 T 2 r E A Z EA 27 PAr A2 E 512 TAr A 3 E A 0 van der Waals equation Eq1255 P ART vbE A A a v2 E a A27 64E A EA R2TAC A2 EPAC AE A b EA RTAC A E8PAC AE multiply equation by Av2vb EPE Get vA3E A b ART PE A vA2E A Aa PE A v Aab PE A 0 Multiply by A P3 E R3 T3 E A and substitute Z APv RTE Get ZA3E A AbP RTE A 1 ZA2E A A aP R2T2 E A Z A abP2 ER3 T3 E A 0 Substitute for a and b get ZA3E A EA PAr A E8TAr AE A 1 ZA2E A A 27 Pr E64 T 2 r E A Z EA 27 PAr A2 E 512 TAr A 3 E A 0 Where PArE A EA P PAc AE A TArE A EA T TAc AE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1270 Evaluate changes in an isothermal process for u h and s for a gas with an equation of state as P v b RT From Eq1231 we get duT T AP TE AAvE A P dvT T A R v bE A P dvT P P dvT 0 From Eq1227 we get using v b RTP dhT v T Av TE AAPE A dPT v T AR PE A dPT b dPT From Eq1232 or Eq1234 we get dsT Av TE AAPE A dPT AP TE AAvE A dvT AR PE A dPT A R v bE A dvT Now the changes in u h and s can be integrated to find u2 u1 0 h2 h1 A b dPEA bP2 P1 s2 s1 R ln P2 P1 R ln v2 b v1 b Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1271 Develop expressions for isothermal changes in internal energy enthalpy and entropy for a gas obeying the van der Waals equation of state van der Waals equation of state P ART vbE A A a v2 E AP TE AAvE A A R vbE Au vE AATE A TAP TE AAvE A P ART vbE A ART vbE A A a v2 E uA2E AuA1E AATE A A 1 2 TP Tv PdvE A A 1 2 a v2dvEA aA 1 v1 E A A 1 v2 E A hA2E AhA1E AATE A uA2E AuA1E AATE A PA2E AvA2E A PA1E AvA1E A PA2E AvA2E A PA1E AvA1E A aA 1 v1 E A A 1 v2 E A sA2E AsA1E AATE A A 1 2 P Tv dvEA A 1 2 R vb dvEA R lnA v2b Ev1bE A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1272 Consider the following equation of state expressed in terms of reduced pressure and temperature Z 1 PArE A14TArE A1 6TA2 rE A What does this predict for the reduced Boyle temperature Z APv RTE A 1 A Pr E14 Tr E A1 A 6 Tr 2 E A Z P T A 1 14PcTr E A1 A 6 Tr 2 E A To find the Boyle temperature we must have ALim P0E A Z P T 0 at Tboyle The above derivative is then zero if 1 A 6 Tr 2 E A 0 TArE A A 6EA 245 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1273 Use the result of Problem 1235 to find the reduced temperature at which the JouleThomson coefficient is zero for a gas that follows the EOS given in Problem 1272 Z APv RTE A 1 A Pr E14 Tr E A1 A 6 Tr 2 E A From Problem 1235 µJ T P h T v T P v CP ART2 AE PCP E Z T P Z T P Pr 14Tc T 2 r 1 6 AT 2 rE A A Pr E14 Tr E A 12 AT 3 rE A Tc Pr 14Tc T 2 r 18 AT 2 rE A 1 So this is zero for 18 AT 2 rE A 1 or TArE A A 18E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1274 What is the Boyle temperature for the following equation of state P ART vbE A A a v2TE where a and b are constants P ART vbE A A a v2TE Multiplying by Avb PE A gives v b ART PE A Aa1bv EPvTE Using for TABoyleE A Alim P0E AAZ PE AATE A Alim P0E A AZ1 P0E A A 1 RTE A Alim P0E Av ART PE A Alim P0E Av ART PE A b Aa10 ERTTE A b A a RT2 E A 0 at TABoyleE or TABoyleE A A a Rb EA A 27 64 R2T 3 EC PC 1 R 8PC RTC EA A 27 8 EA TACE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1275 Determine the reduced Boyle temperature as predicted by an equation of state the experimentally observed value for most substances is about 25 using the van der Waals equation and the RedlichKwong equation Note It is helpful to use Eqs 1247 and 1248 in addition to Eq 1246 The Boyle temp is that T at which Alim P0E AAZ PE AATE A 0 But Alim P0E AAZ PE AATE A Alim P0E A AZ1 P0E A A 1 RTE A Alim P0E Av ART PE A van der Waals P ART vbE A A a v2 E multiply by Avb PE A get vb ART PE A Aavb EPv2 E A or v ART PE A b Aa1bv EPvE RT Alim P0E AAZ PE AATE A b Aa10 ERTE A 0 only at TABoyleE or TABoyleE A A a RbE A A27 8E A TACE A 3375 TACE RedlichKwong P ART vbE A A a vvbT12 E as in the first part get v ART PE A b A a1bv EPv1bvT12 E RT Alim P0E AAZ PE AATE A b A a10 EPv10T12 E A 0 only at TABoyleE or TA 32 BoyleE A A a RbE A A 0427 48 R2 T 52 EC RPC E AA PC E008 664 R TC E TABoyleE A A0427 48 0086 64E AA 23E ATACE A 29 TACE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1276 One early attempt to improve on the van der Waals equation of state was an expression of the form P ART vbE A A a v2TE Solve for the constants a b and vACE A using the same procedure as for the van der Waals equation From the equation of state take the first two derivatives of P with v AP vE AATE A A RT vb2 E A A 2a v3TE A and A2P Ev2 E AATE A A 2RT vb3 E A A 6a v4TE Since both these derivatives are zero at the critical point A RT vb2 E A A 2a v3TE A 0 and A 2RT vb3 E A A 6a v4TE A 0 Also PACE A A RTC EvCbE A A a v 2 C TC E solving these three equations vACE A 3b a A27 64E A A R2T 3 EC PC E A b A RTC E8PC E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1277 Develop expressions for isothermal changes in internal energy enthalpy and entropy for a gas obeying RedlichKwong equation of state RedlichKwong equation of state P A RT v bE A A a vv bT12 E AP TE AAvE A A R v bE A A a 2vv bT32 E From Eq1231 uA2E A uA1E AATE A A 1 2 3a 2vv bT12 dvEA A 3a 2bT12 E A lnA v2 b Ev2 E AA v1 Ev1 bE A We find change in h from change in u so we do not do the derivative in Eq1227 This is due to the form of the EOS hA2E A hA1E AATE A PA2E AvA2E A PA1E AvA1E A A 3a 2bT12 E A lnA v2 b Ev2 E AA v1 Ev1 bE A Entropy follows from Eq1235 sA2E A sA1E AATE A A 1 2 R v b a2 Evv bT32dvE R lnA v2 b Ev1 bE A A a 2bT32 E A lnA v2 b Ev2 E AA v1 Ev1 bE A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1278 Determine the second virial coefficient BT using the van der Waals equation of state Also find its value at the critical temperature where the experimentally observed value is about 034 RTAcE APAcE A From Eq1248 BT Alim P0E A α where Eq 1244 α ART P Ev From Eq 1251 van der Waals P ART vbE A A a v2 E A which we can multiply by Avb PE A get v b ART PE A Aavb EPv2 E A or v ART PE A b Aa1bv EPvE Taking the limit for P 0 then Pv RT and v we get BT b aRT RTC PC A1 8E A 27 TC 64 T where ab are from Eq1252 At T TC then we have BTACE A RTC PC A19 64E A 0297 RTC PC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1279 Determine the second virial coefficient BT using the RedlichKwong equation of state Also find its value at the critical temperature where the experimentally observed value is about 034 RTAcE APAcE A From Eq1248 BT Alim P0E A α where Eq1244 α ART P Ev For Redlich Kwong the result becomes v ART PE A b A a1 bv Pv1 bv T12 E A Taking the limit for P 0 then Pv RT and v we get BT b A a RT32 E Now substitute Eqs 1254 and 1255 for a and b BT RTC PC 008664 042748 A TC T 32 AEE and evaluated at TC it becomes BTC RTC PC 008664 042748 0341 RTC PC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1280 Oxygen in a rigid tank with 1 kg is at 160 K 4 MPa Find the volume of the tank by iterations using the RedlichKwong EOS Compare the result with the ideal gas law For the ideal gas law Pv RT so v RTP v 02598 160 4000 00104 mA3E Akg V mv 00104 mA3E For RedlichKwong Eq1253 and oxygen PAcE A 5040 kPa TAcE A 1546 K R 02598 kJkg K b 008664 A RTc EPc E A 008664 A02598 1546 5040E A 0000 690 5 mA3E Akg a 0427 48 A R2T 52 Ec Pc E A 0427 48 A025982 154652 E5040E A 17013 P A RT v bE A A a vv bT12 E A trial and error to get v due to nonlinearity v 001 mA3E Akg P 44651 12799 31852 kPa too low v 0008 mA3E Akg P 568685 19681 37188 kPa too low v 00075 mA3E Akg P 610441 222743 387698 kPa v 0007 mA3E Akg P 658816 254170 404646 kPa Now we interpolate between the last two entries and check v 000714 mA3E Akg P 644515 24473 39978 kPa OK V mv 000714 mA3E A 69 of the ideal gas value Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1281 A flow of oxygen at 230 K 5 MPa is throttled to 100 kPa in a steady flow process Find the exit temperature and the specific entropy generation using RedlichKwong equation of state and ideal gas heat capacity Notice this becomes iterative due to the nonlinearity coupling h P v and T CV Throttle Steady single flow no heat transfer and no work Energy eq hA1E A 0 hA2E A 0 so constant h Entropy Eq sA1E A sAgenE A sA2E A so entropy generation Find the change in h from Eq1226 assuming CApE A is constant RedlichKwong equation of state P A RT v bE A A a vv bT12 E AP TE AAvE A A R v bE A A a 2vv bT32 E From Eq1231 uA2E A uA1E AATE A A 1 2 3a 2vv bT12 dvEA A 3a 2bT12 E A lnA v2 b Ev2 E AA v1 Ev1 bE A We find change in h from change in u so we do not do the derivative in Eq1227 This is due to the form of the EOS hA2E A hA1E AATE A PA2E AvA2E A PA1E AvA1E A A 3a 2bT12 E A lnA v2 b Ev2 E AA v1 Ev1 bE A Entropy follows from Eq1235 sA2E A sA1E AATE A A 1 2 R v b a2 Evv bT32dvE R lnA v2 b Ev1 bE A A a 2bT32 E A lnA v2 b Ev2 E AA v1 Ev1 bE A PAcE A 5040 kPa TAcE A 1546 K R 02598 kJkg K b 008664 A RTc EPc E A 008664 A02598 1546 5040E A 0000 690 5 mA3E Akg a 0427 48 A R2T 52 Ec Pc E A 0427 48 A025982 154652 E5040E A 17013 We need to find TA2E A so the energy equation is satisfied Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful hA2E A hA1E A hA2E A hAxE A hAxE A hA1E A CApE ATA2E A TA1E A hA2E A hA1E AATE A 0 and we will evaluate it similar to Fig 124 where the first term is done from state x to 2 and the second term is done from state 1 to state x at TA1E A 230 K We do this as we assume state 2 is close to ideal gas but we do not know TA2E A We first need to find vA1E A from the EOS so guess v and find P vA1E A 0011 mA3E Akg P 57960 87235 4924 too low vA1E A 001082 mA3E Akg P 58990 9007 49983 OK Now evaluate the change in h along the 230 K from state 1 to state x that requires a value for vAxE A Guess ideal gas at TAxE A 230 K vAxE A RTAxE APA2E A 02598 230100 059754 mA3E Akg From the EOS PA2E A 1001157 03138 99802 kPa close A few more guesses and adjustments gives vAxE A 059635 mA3E Akg PA2E A 1003157 03151 1000006 kPa OK hAxE A hA1E AATE A PAxE AvAxE A PA1E AvA1E A A 3a 2bT12 E A lnA vx b Evx E AA v1 Ev1 bE A 59635 5000 001082 243694 ln A059704 059635E A A001082 001151E A 59635 541 1478335 20318 kJkg From energy eq TA2E A TA1E A hAxE A hA1E AATE ACApE A 230 20318 0922 208 K Now the change in s is done in a similar fashion sAgenE A sA2E A sA1E A sAxE A sA1E AATE A sA2E A sAxE R lnA vx b Ev1 bE A A a 2bT32 E A lnA vx b Evx E AA v1 Ev1 bE A CApE A ln A T2 ETx E A 02598 lnA 059566 00101295E A 035318 ln 094114 0922 lnA208 230E A 105848 0021425 0092699 0987 kJkg K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Generalized Charts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1282 How low should the pressure be so that nitrous oxide NA2E AO gas at 2786 K can be treated as an ideal gas with 5 accuracy or better From Table A2 TAcE A 3096 K PAcE A 724 MPa TAr1E A A2786 3096E A 09 Look in Fig D1 following the curve TAr1E A 09 to the point where Z 095 PAr1E A 0125 so P 0125 724 MPa 09 MPA 900 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1283 Nitrous oxide NA2E AO at 2786 K is at a pressure so that it can be in a twophase state Find the generalized enthalpy departure for the two saturated states of liquid and vapor From Table A2 TAcE A 3096 K PAcE A 724 MPa TAr1E A A2786 3096E A 09 From Fig D2 Saturated liquid hAE A hAfE ARTAcE A 408 Saturated vapor hAE A hAgE ARTAcE A 0875 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1284 Find the heat of evaporation hAfgE A for R134a at 0C from the generalized charts and compare to the value in Table B5 From Table A2 TAcE A 3742 K PAcE A 406 MPa TAr1E A A27315 3742E A 073 From Fig D2 Saturated liquid hAE A hAfE ARTAcE A 47 Saturated vapor hAE A hAgE ARTAcE A 03 hAfgE A hAgE A hAfE A RTAcE A 03 47 44 RTAcE A 44 008149 3742 1342 kJkg Table B51 hAfgE A 19836 kJkg The approximation is not very good and can be improved by using the accentric factor Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1285 A 200L rigid tank contains propane at 9 MPa 280C The propane is then allowed to cool to 50C as heat is transferred with the surroundings Determine the quality at the final state and the mass of liquid in the tank using the generalized compressibility chart Fig D1 Propane CA3E AHA8E A V 02 mA3E A PA1E A 9 MPa TA1E A 280AoE AC 5532 K cool to TA2E A 50 AoE AC 3232 K From Table A2 TAcE A 3698 K PAcE A 425 MPa PAr1E A A 9 425E A 2118 TAr1E A A5532 3698E A 1496 From Fig D1 ZA1E A 0825 vA2E A vA1E A A Z1RT1 EP1 E A A08250188 555532 9 000E A 000956 mA3E Akg From Fig D1 at TAr2E A 0874 PAG2E A 045 4250 1912 kPa vAG2E A 071 0188 55 32321912 002263 mA3E Akg vAF2E A 0075 0188 55 32321912 000239 mA3E Akg 000956 0002 39 xA2E A002263 000239 xA2E A 0354 mALIQ 2E A 1035402000956 1351 kg These tanks contain liquid propane Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1286 A rigid tank contains 5 kg of ethylene at 3 MPa 30C It is cooled until the ethylene reaches the saturated vapor curve What is the final temperature V const m 5 kg PA1E A 3 MPa TA1E A 30 AoE AC 3032 K cool to xA2E A 10 PAr1E A A 3 504E A 0595 TAr1E A A3032 2824E A 1074 Fig D1 ZA1E A 082 Final state x 1 vA2E A vA1E Pv ZRT so take the ratio between the two states PAr2E A PAr1E A A Z2Tr2 EZ1Tr1 E A 0595 A ZG2Tr2 E0821074E A 06756 ZAG2E ATAr2E Trial error Table D4 may be easier to use than Fig D1 TAr2E A ZAG2E A PAr2E A PAr2 CALCE 0866 072 042 0421 OK TA2E A 2446 K 2 4 C H T v 1 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1287 The new refrigerant R152a is used in a refrigerator with an evaporator temperature of 20AoE AC and a condensing temperature of 30AoE AC What are the high and low pressures in this cycle Since we do not have the printed tables for R152a we will use generalized charts The critical properties are TAcE A 3864 K PAcE A 452 MPa Evaporator TAr1E A TTAcE A 27315 203864 0655 Fig D1 PAG T1E A PAr1 satE A PAcE A 006 452 0271 MPa Condenser TAr2E A TTAcE A 27315 303864 0785 Fig D1 PAG T2E A PAr2 satE A PAcE A 022 452 0994 MPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1288 A 4mA3E A storage tank contains ethane gas at 10 MPa 100AoE AC Use LeeKesler EOS and find the mass of the ethane The LeeKesler EOS is shown as the generalized charts Table A2 TAcE A 3054 K PAcE A 488 MPa Table A5 R 02765 kJkgK The reduced properties are PAr1E A A 10 488E A 205 TAr1E A A37315 3054E A 122 Fig D1 Z 056 m A PV ZRTE A A 10 000 4 056 02765 37315E A 6923 kg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1289 The ethane gas in the storage tank from the previous problem is cooled to 0AoE AC Find the new pressure The new final state is given by TA2E A vA2E A vA1E A TAr2E A A27315 3054E A 08944 Since Z and P are unknown this becomes trial and error solution PA2E A ZA2E A mRTA2E A V 6923 02765 27315 4 130717 kPa Assume it is saturated PAr2E A 053 see Fig D1 ZAgE A 067 ZAfE A 009 PA2E A 053 4880 2586 kPa and ZA2E A PA2E A 130717 0198 two phase OK Ans PA2E A 2586 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1290 Use CATT3 to solve the previous two problems when the acentric factor is used to improve the accuracy Problem 1288 The LeeKesler EOS is shown as the generalized charts Table A2 TAcE A 3054 K PAcE A 488 MPa Table A5 R 02765 kJkgK Table D4 ω 0099 The reduced properties are PAr1E A A 10 488E A 205 TAr1E A A37315 3054E A 122 CATT3 Z 0605 m A PV ZRTE A A 10 000 4 0605 02765 37315E A 6408 kg Problem 1289 The new final state is given by TA2E A vA2E A vA1E A TAr2E A A27315 3054E A 08944 Since Z and P are unknown this becomes trial and error solution PA2E A ZA2E A mRTA2E A V 6408 02765 27315 4 120993 kPa Assume it is saturated vapor PAr2E A 0483 CATT3 ZAgE A 069 ZAfE A 0078 PA2E A 0483 4880 2357 kPa and ZA2E A PA2E A 12 0993 01948 two phase OK Ans PA2E A 2357 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1291 A geothermal power plant uses butane as saturated vapor 80AoE AC into the turbine and the condenser operates at 30AoE AC Find the reversible specific turbine work CA4E AHA10E A cycle TA1E A 80 AoE AC xA1E A 10 TA3E A 30AoE AC xA3E A 00 TAr1E A A3532 4252E A 0831 From D1 D2 and D3 PA1E A 0325 3800 1235 kPa hA 1E AhA1E A 0143 044252056 341 sA 1E AsA1E A 0143 040475 00680 TAr3E A A3032 4252E A 0713 From D1 D2 and D3 PA3E A 01133800 429 kPa sat liq hAE A hAfE A RTAcE A481 2925 sAE A sAfE A R664 0950 sat vap hAE A hAgE A RTAcE A0235 143 sAE A sAgE A R022 0031 Because of the combination of properties of CA4E AHA10E A particularly the large CAP0E AR sA1E A is larger than sAgE A at TA3E A To demonstrate sA 1E AsA g3E A 17164 ln A3532 3032E A 0143 04 ln A1235 429E A 01107 sA1E AsAg3E A 00680 01107 0031 00737 kJkg K so that TA2SE A will be TA3E A as shown in the Ts diagram A number of other heavy hydrocarbons also exhibit this behavior Turbine Cond Ht Exch P 3 1 4 2 Q H W T WP 3 2 2s 1 s T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Solution 1291 Continued Assume T2S 315 K Tr2S 315 4252 0741 From D2 and D3 h 2S h2S RTc021 128 and s 2S s2S R019 0027 s 1 s 2S 17164 ln 3532 315 0143 04 ln 1235 429 00453 s1 s2S 00680 00453 0027 0 T2S 315 K h 1 h 2S 171643532315 656 wST h1 h2S 341 656 128 443 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1292 Consider the following EOS expressed in terms of reduced pressure and temperature Z 1 Pr 14 Tr 1 6 Tr 2 What does this equation predict for enthalpy departure from the ideal gas value at the state Pr 04 Tr 09 What is it from the generalized charts Z Pv RT 1 14 Tr Pr 1 6 Tr 2 v RT P 14Pc RTc 1 T2 6Tc 2 v T p R P 12RT 3 c 14PcT3 v T v T p RTc 14Pc 18RT 3 c 14PcT2 Now Eq1227 is integrated with limits similar to Eq1262 h h 0 P v T v T p dP RTc 14 1 18 Tr 2 Pr 0606 RT c Evaluate at Pr 04 Tr 09 from Fig D2 to get h h h h 06 RTc This result matched with in the accuracy the figure can be read Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1293 Consider the following equation of state expressed in terms of reduced pressure and temperature Z 1 Pr 14 Tr 1 6 Tr 2 What does this equation predict for entropy departure from the ideal gas value at the state Pr 04 Tr 09 The entropy departure is the change in s for a real gas minus the change in s for an ideal gas so from Eq1232 and eq614 we get ds s Cp dT T v T p dP Cp dT T R P dP R P v T p dP Solve now for v from the compressibility factor Z PvRT to get Z Pv RT 1 14 Tr Pr 1 6 Tr 2 v RT P 14Pc RTc 1 T2 6Tc 2 v T p R P 12RT 3 c 14PcT3 s s 0 P R P v T p dP 0 P 12RT 3 c 14PcT3 dP 6 7 R Pr T3 r Evaluate at Pr 04 Tr 09 to get s s 04703 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1294 A very low temperature refrigerator uses neon From the compressor the neon at 15 MPa 80 K goes through the condenser and comes out at saturated liquid 40 K Find the specific heat transfer using generalized charts State 1 80 K 15 MPa Tr1 80 444 1802 Pr1 15 276 0543 State 2 40 K x 0 Tr2 090 Pr2 0532 The enthalpy departure chart Fig D2 h h1 022 RTc h h2 410 RT c h 2 h 1 103 40 80 412 kJkg h2 h1 h 2 h 1 h h2 h h 1 412 0412 444 410 022 1122 kJkg q h2 h1 1122 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1295 Repeat the previous problem using CATT3 software for the neon properties From CATT3 h1 1383 kJkg h2 3024 kJkg P 146 MPa q h2 h1 3024 1383 1081 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1296 A pistoncylinder contains 5 kg of butane gas at 500 K 5 MPa The butane expands in a reversible polytropic process to 3 MPa 460 K Determine the polytropic exponent n and the work done during the process C4H10 m 5 kg T1 500 K P1 5 MPa Rev polytropic process P1V n 1 P2V n 2 Tr1 500 4252 1176 Pr1 5 38 1316 From Fig D1 Z1 068 Tr2 460 4252 1082 Pr2 3 38 0789 From Fig D1 Z2 074 V1 mZRT P 5 068 01430 500 5000 00486 m 3 V2 mZRT P 5 074 01430 460 3000 00811 m 3 Solve for the polytropic exponent n as n lnP1P2 lnV2V1 ln 5 3 ln 00811 00486 09976 1W2 1 2 PdV 1n P2V2 P1V1 300000811 500000486 1 09976 125 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1297 Calculate the heat transfer during the process described in Problem 1296 From solution 1296 V1 00486 m3 V2 00811 m3 1W2 125 kJ Tr1 500 4252 1176 Pr1 5 38 1316 From Fig D1 Z1 068 Tr2 1082 Pr2 0789 T2 460 K From Fig D2 h h1 130 RTC h h2 090 RT C h 2 h 1 1716460 500 831 kJkg h2 h1 831 831454252 58124 090 130 588 kJkg U2 U1 mh2 h1 P2V2 P1V 1 5588 3000 00811 5000 00486 2883 kJ 1Q2 U2 U1 1W2 1743 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1298 An ordinary lighter is nearly full of liquid propane with a small amount of vapor the volume is 5 cm3 and temperature is 23C The propane is now discharged slowly such that heat transfer keeps the propane and valve flow at 23C Find the initial pressure and mass of propane and the total heat transfer to empty the lighter Propane C3H8 T1 23oC 2962 K constant x1 00 V1 5 cm3 5106 m3 Tr1 29623698 0804 From Figs D1 and D2 P1 PG T1 025425 1063 MPa Z1 004 h 1h1 0188 553698451 3145 m1 Z1RT1 P1V1 0040188 552962 10635106 000238 kg State 2 Assume vapor at 100 kPa 23oC ideal gas so no corrections Therefore m2 much smaller than m1 90 106 kg QCV m2u2 m1u1 mehe m2h2 m1h1 P2P1V m1m2he m2h2he m1heh1 P2P1V he h1 0 0 3145 QCV 0 0002383145 100 10635106 0753 kJ Actual lighters use butane and some propane Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1299 250L tank contains propane at 30C 90 quality The tank is heated to 300C Calculate the heat transfer during the process V 250 L 025 m3 T1 30 oC 3032 K x1 090 Heat to T2 300 oC 5732 K M 44094 Tc 3698 K Pc 425 MPa R 0188 55 CP0 16794 Tr1 082 Fig D1 Z1 1 x1 Zf1 x1 Zg1 01 005 09 0785 0711 Fig D2 EA hA 1 AhA1 A E RTAc AE A 01 443 09 052 0911 PA SAT rE A 030 PA SAT 1E A 1275 MPa m A 1275025 07110188 553032E A 7842 kg PAr2E A A 7842Z20188 555732 E0254250E A A Z2 E1254E at TAr2E A 155 Trial and error on PAr2E PAr2E A 0743 PA2E A 3158 MPa ZA2E A 094 AhE A hA2E A 035 RTACE hA 2E AhA 1E A 1679430030 4534 kJkg hA 1E AhA1E A 09110188 553698 635 kJkg hA 2E AhA2E A 0350188 553698 244 kJkg QA12E A mhA2E AhA1E A PA2E APA1E AV 78422444534635 31581275025 3862 471 3391 kJ T v 1 2 C H 3 8 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12100 Find the heat of evaporation hAfgE A for isobutane TAcE A 4082 K PAcE A 365 MPa M 58124 at 126C from the generalized charts and compare to the values in the CATT3 computerized tables To read the charts we need the reduced temperature TAr1E A TTAcE A 126 273154082 070 AhE A hAgE A 02 RTAcE A AhE A hAfE A 485 RTAcE hAfgE A hAgE A hAfE A AhE A hAgE A AhE A hAfE A 02 485 RTAcE 465 RTAcE A 465 8314558124 4082 2715 kJkg CATT3 hAfgE A 2562 3687 343 kJkg The generalized charts are not super accurate some improvement can be done using the accentric factor Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12101 A cylinder contains ethylene CA2E AHA4E A at 1536 MPa 13C It is now compressed isothermally in a reversible process to 512 MPa Find the specific work and heat transfer Ethylene CA2E AHA4E A PA1E A 1536 MPa TA2E A TA1E A 13AoE AC 2602 K TAr2E A TAr1E A 2602 2824 0921 PAr1E A 1536 504 0305 From D1 D2 and D3 ZA1E A 085 hA 1E AhA1E A 029642824040 335 and sA 1E AsA1E A 02964030 00889 From D1 D2 and D3 ZA2E A 017 PAr2E A 512504 1016 comp liquid hA 2E AhA2E A 02964282440 3348 and sA 2E AsA2E A 0296436 1067 Ideal gas hA 2E AhA 1E A 0 and sA 2E AsA 1E A 0 02964 ln A 512 1536E A 03568 A1E AqA2E A TsA2E AsA1E A 26021067 03568 00889 3473 kJkg hA2E A hA1E A 3348 0 335 3013 kJkg uA2E A uA1E A hA2E AhA1E A RTZA2E AZA1E A 3013 029642602017085 2489 A1E AwA2E A A1E AqA2E A uA2E A uA1E A 3473 2489 984 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12102 Saturated vapor R410A at 30C is throttled to 200 kPa in a steady flow process Calculate the exit temperature assuming no changes in the kinetic energy using the generalized charts Fig D2 and the R410A tables Table B4 R410A throttling process Energy Eq hA2E A hA1E A 0 hA2E A hA 2E A hA 2E A hA 1E A hA 1E A hA1E A Generalized Chart Fig D2 R 83145172585 011455 kJkgK TAr1E A A3032 3445E A 088 hA 1E AhA1E A 011455 3445 085 3354 kJkg For CAP0E A use h values from Table B4 at low pressure CAP0E A 33083 31440 40 20 08215 kJkg K Substituting hA2E AhA 2E A 08215 TA2E A30 3354 0 at PAr2E A 2004900 0041 Assume TA2E A 10AoE AC TAr2E A 26323445 0764 hA 2E AhA2E A RT 01 011455 3445 01 395 Substituting 395 08215 10 30 3354 327 Assume TA2E A 5AoE AC TAr2E A 26823445 0778 hA 2E AhA2E A RT 01 011455 3445 01 395 Substituting 395 08215 5 30 3354 084 TA2E A 60 AoE AC R410A tables B4 at TA1E A 30AoE AC xA1E A 10 hA1E A 28416 kJkg hA2E A hA1E A 28416 PA2E A 02 MPa TA2E A 134 AoE AC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12103 Repeat Problem 1291 using CATT3 and include the acentric factor for butane to improve the accuracy CA4E AHA10E A cycle TA1E A 80 AoE AC xA1E A 10 TA3E A 30AoE AC xA3E A 00 TAr1E A A3532 4252E A 0831 From CATT3 with ω 0199 PA1E A 02646 3800 1005 kPa hA 1E AhA1E A 01430 425205685 346 sA 1E AsA1E A 01430 04996 00714 TAr3E A A3032 4252E A 0713 From CATT3 with ω 0199 PA3E A 007443 3800 2828 kPa sat liq hAE A hAfE A RTAcE A6048 36774 sAE A sAfE A R8399 1201 sat vap hAE A hAgE A RTAcE A0202 1228 sAE A sAgE A R0201 00287 Because of the combination of properties of CA4E AHA10E A particularly the large CAP0E AR sA1E A is larger than sAgE A at TA3E A To demonstrate sA 1E AsA g3E A 1716 ln A3532 3032E A 01430 ln A1005 2828E A 00806 sA1E AsAg3E A 00714 00806 00287 00379 kJkg K so that TA2SE A will be TA3E A as shown in the Ts diagram A number of other heavy hydrocarbons also exhibit this behavior Turbine Cond Ht Exch P 3 1 4 2 Q H W T WP 3 2 2s 1 s T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Solution 12103 continued Assume T2S 315 K Tr2S 315 4252 0741 From CATT3 h 2S h2S RTc0183 1113 and s 2S s2S R01746 0025 s 1 s 2S 1716 ln 3532 315 01430 ln 1005 2828 001509 s1 s2S 00714 001509 0025 0031 Repeat at T2S 310 K to get Tr2S 0729 h 2S h2S RTc01907 11595 and s 2S s2S R01853 00265 s 1 s 2S 1716 ln 3532 310 01430 ln 1005 2828 004255 s1 s2S 00714 004255 00265 00023 very close to 0 OK T2S 310 K h 1 h 2S 1716 3532 310 7413 wST h1 h2S 346 7413 116 511 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12104 A cylinder contains ethylene C2H4 at 1536 MPa 13C It is now compressed in a reversible isobaric constant P process to saturated liquid Find the specific work and heat transfer Ethylene C2H4 P1 1536 MPa P2 T1 13oC 2602 K State 2 saturated liquid x2 00 Tr1 2602 2824 0921 Pr1 Pr2 1536 504 0305 From Figs D1 D2 Z1 085 h 1 h1RTc 040 v1 P1 Z1RT1 0850296372602 1536 0042675 h 1 h1 0296 37 2824 040 335 From Figs D1 D2 T2 08242824 2327 K Z2 005 h 2 h2RTc 442 v2 P2 Z2RT2 0050296372327 1536 0002245 m3kg h 2 h2 0296 37 2824 442 3699 h 2 h 1 CP0T2 T1 154822327 2602 426 w12 Pdv Pv2 v1 1536 kPa 0002 245 0042 675 m3kg 621 kJkg q12 u2 u1 w12 h2 h1 3699 426 335 379 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12105 Refrigerant123 dichlorotrifluoroethane which is currently under development as a potential replacement for environmentally hazardous refrigerants undergoes an isothermal steady flow process in which the R123 enters a heat exchanger as saturated liquid at 40C and exits at 100 kPa Calculate the heat transfer per kilogram of R123 using the generalized charts Fig D2 R123 M 15293 TC 4569 K PC 367 MPa T1 T2 40 oC x1 0 P2 100 kPa Tr1 Tr2 31324569 0685 Pr2 01367 0027 From Fig D2 Pr1 0084 h h1RTC 49 From D1 saturated P1 00843670 308 kPa P2 P1 with no work done so process is irreversibel Energy Eq q h1 h2 Entropy Eq s1 dqT sgen s2 sgen 0 From Fig D2 h h2RTC 0056 q h2 h1 83145 4569 0056 0 49015293 1204 kJkg 1 2 Heat Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12106 Carbon dioxide collected from a fermentation process at 5C 100 kPa should be brought to 243 K 4 MPa in a steady flow process Find the minimum amount of work required and the heat transfer What devices are needed to accomplish this change of state Tri 2782 3041 0915 Pri 100 7380 00136 From D2 and D3 hhriRTC 002 ssriR 001 Tre 243 3041 080 Pre 4 738 0542 From D2 and D3 hhreRTC 45 ssreR 474 hihe h i hi h i h e h ehe 0188 923041001 084182782243 0188 92304145 2876 kJkg sise s i si s i s e s ese 0188 92001 08418 ln2782243 0188 92 ln014 0188 92474 17044 kJkg K wrev hihe T0sise 2876 278217044 1866 kJkg qrev hehi wrev 2876 1866 4742 kJkg We need a compressor to bring the pressure up and a cooler to bring the temperature down Cooling it before compression and intercooling between stages in the compressor lowers the compressor work In an actual setup we require more work than the above reversible limit Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12107 Determine how accurate the generalized chart is for the carbon dioxide process in Problem 12106 by using the CATT3 software for the carbon dioxide properties From the CATT3 software Inlet superheated vapor hi 3761 kJkg and si 2015 kJkgK Exit compressed liquid he 2021 kJkg and se 007349 kJkgK wrev hi he T0si se 3761 2021 2782 2015 007349 1841 kJkg qrev he hi wrev 2021 3761 1841 540 kJkg The work is very accurate but the heat transfer a little less so The enthalpy and entropy differences are both underestimated by the generalized charts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12108 A geothermal power plant on the Raft River uses isobutane as the working fluid The fluid enters the reversible adiabatic turbine at 160C 5475 MPa and the condenser exit condition is saturated liquid at 33C Isobutane has the properties Tc 40814 K Pc 365 MPa CP0 1664 kJkg K and ratio of specific heats k 1094 with a molecular weight as 58124 Find the specific turbine work and the specific pump work Turbine inlet T1 160oC P1 5475 MPa Condenser exit T3 33oC x3 00 Tr3 3062 4081 075 From Fig D1 Pr3 016 Z3 003 P2 P3 016 365 0584 MPa Tr1 4332 4081 1061 Pr1 5475 365 150 From Fig D2 D3 h 1 h1 0143 054081284 1658 s 1 s1 0143 05215 03076 s 2 s 1 1664 ln 3062 4332 0143 05 ln 0584 5475 02572 s 2 s2 s 2 sF2 x2 s FG2 0143 05612 x20143 05612029 08755 x208340 s2 s1 0 08755 x208340 02572 03076 x2 099 h 2 h 1 CP0T2 T1 16643062 4332 2113 From Fig D2 h 2 h2 h 2 hF2 x2hFG2 0143 054081469099469032 2738 099 2551 213 Turbine wT h1 h2 1658 2113 213 668 kJkg Pump vF3 P3 ZF3RT3 0030143 053062 584 000225 wP v dP vF3P4 P3 000225 5475584 110 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12109 A flow of oxygen at 230 K 5 MPa is throttled to 100 kPa in a steady flow process Find the exit temperature and the entropy generation Process Throttling Small surface area Q 0 No shaft W 0 Irreversible S gen 0 We will solve the problem using generalized charts Tri 230 1546 1488 Pri 5 504 0992 Pre 01 504 002 From D2 h i hi RTc h 02598 1546 050 201 kJkg Energy Eq he hi 0 h ehe h eh i h i hi Assume Te 208 K Tre 1345 h eh i Cp Te Ti 0922 208 230 203 kJkg From D2 h ehe RTc h 02598 1546 001 04 Check first law he hi 04 203 201 0 OK Te 208 K From D3 s i si 02598025 00649 and s ese 02598001 00026 s es i 09216 ln 208 230 02598 ln 01 5 09238 kJkg K sgen se si 00026 09238 00649 09861 kJkg K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12110 An uninsulated pistoncylinder contains propene C3H6 at ambient temperature 19C with a quality of 50 and a volume of 10 L The propene now expands very slowly until the pressure in the cylinder drops to 460 kPa Calculate the mass of propene the work and heat transfer for this process Propene C3H6 T1 19oC 2922 K x1 050 V1 10 L From Fig D1 Tr1 29223649 080 Pr1 Pr sat 025 P1 025 46 115 MPa From D1 Z1 05 004 05 0805 04225 m P1V1 Z1RT1 11500010 042250197 582922 0471 kg Assume reversible and isothermal process slow no friction not insulated 1Q2 mu2u1 1W2 1W2 1 2 PdV cannot integrate 1Q2 1 2 TdS Tms2s1 From Figs D2 and D3 h 1 h1 019758 364905 451 05 046 1792 kJkg s 1 s1 0197 58 05 546 05 039 05779 kJkg K The ideal gas change in h and s are h 2 h 1 0 and s 2 s 1 0 0197 58 ln 460 1161 01829 kJkg K At Tr2 080 Pr2 010 from D1 D2 and D3 Z2 093 h 2 h2 0197 58 3649 016 115 kJkg s 2 s2 0197 58 013 00257 kJkg K Now we can do the change in s and h from state 1 to state 2 s2 s1 s 2 s2 s 2 s 1 s 1 s1 00257 01829 05779 07351 kJkg K h2 h1 h 2 h2 h 2 h 1 h 1 h 1 115 0 1792 1677 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Solution 12110 continued The heat transfer is found from the second law 1q2 2922 07351 2148 kJkg 1Q2 m 1q2 1012 kJ We need the internal energy in the energy equation u2 u1 h2 h1 RTZ1 Z2 1677 0197 58 2922 04225 093 1384 kJkg 1w2 1q2 u2 u1 2148 1384 764 kJkg 1W2 m 1w2 360 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12111 An alternative energy power plant has carbon dioxide at 6 MPa 100oC flowing into a turbine with an exit as saturated vapor at 1 MPa Find the specific turbine work using generalized charts and repeat using Table B3 From Table A5 R 01889 kJkgK CP 0842 kJkg From Table A2 TC 3041 K PC 738 MPa Pri 6 738 0813 Tri 3732 3041 1227 Δ h 070 From D2 and D3 h i hi RTC Δ h 01889 3041 070 468 kJkg Pre 1 738 01355 x 1 so Tre 073 Te 073 3041 222 K From D2 and D3 h e he RTC Δ h 01889 3041 025 468 kJkg w hi he h i h e h i hi h e he CP hi he RTC Δ h i Δ h e 0842 37315 222 01889 03041 07 025 10145 kJkg From Table B3 w hi he 42169 32239 993 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12112 A distributor of bottled propane C3H8 needs to bring propane from 350 K 100 kPa to saturated liquid at 290 K in a steady flow process If this should be accomplished in a reversible setup given the surroundings at 300 K find the ratio of the volume flow rates V inV out the heat transfer and the work involved in the process From Table A2 Tri 350 3698 0946 Pri 01 425 0024 From D1 D2 and D3 Zi 099 h i hi 01886 3698 003 21 kJkg s i si 01886 002 00038 kJkg K Tre 290 3698 0784 and x 0 From D1 D2 and D3 Pre 022 Pe 022 425 0935 MPa and Ze 0036 h ehe 01886 3698 457 3186 kJkg s ese 01886 566 10672 kJkg K h eh i 1679290 350 1008 kJkg s es i 1679 ln 290 350 01886 ln 0935 01 07373 kJkg K hehi 3186 1008 21 4173 kJkg sesi 10672 07373 00038 18007 kJkg K V in V out ZiTiPi ZeTePe 099 0036 350 290 0935 01 3103 wrev hihe T0sise 4173 30018007 1229 kJkg qrev hehi wrev 4173 1229 5402 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12113 An insulated cylinder fitted with a frictionless piston contains saturatedvapor carbon dioxide at 0oC at which point the cylinder volume is 20 L The external force on the piston is now slowly decreased allowing the carbon dioxide to expand until the temperature reaches 30oC Calculate the work done by the CO2 during this process CO2 TC 3041 K Pc 738 MPa Cp 0842 kJkgK R 01889 kJkg K State 1 T1 0oC sat vap x1 10 V1 20 L Tr1 09 P1 Pr1PC 053 7380 3911 kPa Z1 Zg 067 h 1 h1g 09 RTC s 1 s1gR 072 m P1V1 Z1RT1 2262 kg State 2 T2 30oC Tr2 08 P2 Pr2Pc 025 7380 1845 kPa Entropy Eq Sgen ms2 s1 1Q2T 1Q2 0 Sgen 0 s2 s1 s2 s 2 s 2 s 1 s 1 s1 0 s 2 s 1 CP ln T1 T2 R ln P1 P2 0044 kJkgK s 1 s1 0136 kJkgK s 2 s2 0180 kJkg K s 2 s2f 546 R s 2 s2g 039 R s 2 s2 1x2s 2 s2f x2 s 2 s2g x2 0889 Energy Eq 1Q2 mu2 u1 1W2 1Q2 0 u h Pv Z2 1 x2Zf x2Zg 0111 004 0889 081 0725 h2 h1 h2 h 2 h 2 h 1 h 1 h1 h 2 h 1 CpT2 T1 253 kJkg h 1 h1 517 kJkg h 2 h2f 451 RTC h 2 h2g 046 RT C h 2 h2 1 x2h 2 h2f x2 h 2 h2g 522 kJkg h2 h1 522 253 517 258 kJkg u2 u1 h2 h1 Z2RT2 Z1RT1 258 0725 018892 2432 067 018892 2732 245 kJkg 1W2 554 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12114 A control mass of 10 kg butane gas initially at 80C 500 kPa is compressed in a reversible isothermal process to onefifth of its initial volume What is the heat transfer in the process Butane C4H10 m 10 kg T1 80 oC P1 500 kPa Compressed reversible T const to V2 V15 Tr1 3532 4252 0831 Pr1 500 3800 0132 From D1 and D3 Z1 092 s 1 s1 0143016 00230 v1 P1 Z1RT1 09201433532 500 009296 m3kg State 2 v2 v15 001859 m3kg Tr2 Tr1 0831 From D1 PG 03253800 1235 kPa sat liq ZF 005 ssF R508 07266 sat vap ZG 0775 ssG R0475 00680 Therefore vF 00501433532 1235 000205 m3kg vG 077501433532 1235 00317 m3kg Since vF v2 vG x2 v2vFvGvF 05578 s 2 s2 1 x2s 2 sF2 x2s 2 sG2 04422 07266 05578 00680 03592 kJkg K s 2 s 1 CP0 ln T2T1 R ln P2P1 0 0143 ln 1235 500 01293 s2 s1 03592 01293 00230 04655 kJkg K 1Q2 Tms2 s1 3532 10 04655 1644 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12115 A line with a steady supply of octane C8H18 is at 400C 3 MPa What is your best estimate for the availability in a steady flow setup where changes in potential and kinetic energies may be neglected Availability of Octane at Ti 400 oC Pi 3 MPa From Table A5 R 007279 kJkgK CP 1711 kJkg From Table A2 TC 5688 K PC 249 MPa Pri 3 249 1205 Tri 6732 5688 1184 From D2 and D3 h i hi RTC Δh 0072 79 5688 113 468 kJkg s i si R Δs 0072 79 069 005 kJkgK This is relative to the dead ambient state assume T0 2982 K P0 100 kPa Tr0 2982 5688 0524 Pr0 01 249 0040 From D2 and D3 The s correction is outside chart extrapolate or use CATT3 h 0 h0 RTC 54 2236 and s 0 s0 R 9 0655 kJkgK h i h 0 CPTi T0 1711 6732 2982 6417 kJkg s i s 0 1711 ln 6732 2982 0072 79 ln 3 01 11459 kJkgK hi h0 468 6417 2236 8185 kJkg si s0 005 11459 0655 17509 kJkgK ϕi wrev hi h0 T0si s0 8185 298217509 2965 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12116 The environmentally safe refrigerant R152a is to be evaluated as the working fluid for a heat pump system that will heat a house It uses an evaporator temperature of 20oC and a condensing temperature of 30oC Assume all processes are ideal and R152a has a heat capacity of Cp 0996 kJkg K Determine the cycle coefficient of performance Ideal Heat Pump TH 30 oC From A2 M 6605 R 0125 88 TC 3864 K PC 452 MPa Tr3 3032 3864 0785 Pr3 Pr2 022 P3 P2 994 kPa Satliq h 3 h3 456RTC 2218 T1 20 oC 2532 K Tr1 0655 Pr1 0058 P1 262 kPa h 1 h1 014RTC 68 and s 1 s1 014R 00176 Assume T2 307 K Tr2 0795 given Pr2 022 From D2 D3 s 2 s2 034R 00428 h 2 h2 040RTc 195 s 2 s 1 0996 ln 307 2532 0125 88 ln 994 262 00241 s2 s1 00428 00241 00176 0001 0 OK h2 h1 195 09963072532 68 409 h2 h3 195 09963073032 2218 2061 β qH wIN h2 h3 h2 h1 2061 409 504 T v 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12117 Rework the previous problem using an evaporator temperature of 0oC Ideal Heat Pump TH 30 oC From A2 M 6605 R 0125 88 TC 3864 K PC 452 MPa Tr3 3032 3864 0785 Pr3 Pr2 022 P3 P2 994 kPa Satliq h 3 h3 456RTC 2218 T1 0 oC 2732 K Tr1 0707 Pr1 0106 P1 479 kPa h 1 h1 022RTC 107 and s 1 s1 021R 00264 Assume T2 305 K Tr2 0789 s 2 s2 035R 00441 and h 2 h2 038RTC 185 s 2 s 1 0996 ln 3050 2732 0125 88 ln 994 479 00178 s2 s1 00441 00178 00264 00001 0 OK h2 h1 185 099630502732 107 239 h2 h3 185 099630503032 2218 2051 β h2 h3 h2 h1 2051 239 858 T v 1 2 3 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12118 An uninsulated compressor delivers ethylene C2H4 to a pipe D 10 cm at 1024 MPa 94C and velocity 30 ms The ethylene enters the compressor at 64 MPa 205C and the work input required is 300 kJkg Find the mass flow rate the total heat transfer and entropy generation assuming the surroundings are at 25C Tri 2937 2824 1040 Pri 64 504 1270 From D2 and D3 h i hi 0296 37 2824 265 2218 kJkg s i si 0296 37 208 06164 kJkg K Tre 3672 2824 130 Pre 1024 504 2032 From D1 Ze 069 ve Pe ZeRTe 0690296 373672 10 240 00073 m3kg Ae π 4 D 2 e 0007 85 m2 m ve AeVe 0007 8530 00073 3226 kgs From D2 and D3 h ehe 0296 37 2824 16 1339 kJkg s ese 0296 37 090 02667 kJkg K h eh i 1548236722937 1138 s es i 15482 ln 3672 2937 0296 37 ln 1024 64 02065 hehi 1339 1138 2218 2017 kJkg sesi 02667 02065 06164 05562 kJkg K Energy Eq q hehi KEe w 2017 21000 302 300 979 kJkg Q cv m q 3226979 3158 kW S gen Q cv To m se si 3158 2982 322605562 2853 kWK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12119 The new refrigerant fluid R123 see Table A2 is used in a refrigeration system that operates in the ideal refrigeration cycle except the compressor is neither reversible nor adiabatic Saturated vapor at 265C enters the compressor and superheated vapor exits at 65C Heat is rejected from the compressor as 1 kW and the R123 flow rate is 01 kgs Saturated liquid exits the condenser at 375C Specific heat for R123 is CP 06 kJkg Find the coefficient of performance R123 Tc 4569 K Pc 367 MPa M 15293 kgkmol R 005438 kJkg K State 1 T1 265oC 2467 K sat vap x1 10 Tr1 054 Fig D1 Pr1 001 P1 Pr1Pc 37 kPa Fig D2 h 1h1 003 RTC 08 kJkg State 2 T2 65oC 3382 K State 3 T3 375oC 3107 K sat liq x3 0 Tr3 068 Fig D1 Pr3 008 P3 Pr3Pc 294 kPa P2 P3 294 kPa Pr2 0080 Tr2 074 Fig D2 h 2h2 025 RTC 62 kJkg h 3h3 492 RTC 1222 kJkg State 4 T4 T1 2467 K h4 h 3 Energy Eq Evaporator qL h4 h1 w w 0 h4 h3 qL h1 h3 h1 h 1 h 1 h 3 h 3 h3 h 1 h 3 CPT1 T3 384 kJkg qL 08 384 1222 830 kJkg Energy Eq Compressor q h1 h2 wc Q 10 kW m 01 kgs wc h1 h2 q h1 h2 h1 h 1 h 1 h 2 h 2 h2 h 1 h 2 CPT1 T2 549 kJkg wc 08 549 62 100 595 kJkg β qLwc 830595 1395 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12120 An evacuated 100L rigid tank is connected to a line flowing R142b gas chlorodifluoroethane at 2 MPa 100C The valve is opened allowing the gas to flow into the tank for a period of time and then it is closed Eventually the tank cools to ambient temperature 20C at which point it contains 50 liquid 50 vapor by volume Calculate the quality at the final state and the heat transfer for the process The idealgas specific heat of R142b is Cp 0787 kJkg K Rigid tank V 100 L m1 0 Line R142b CH3CClF2 M 100495 TC 4103 K PC 425 MPa CP0 0787 kJkg K R RM 831451 100495 0082 73 kJkg K Line Pi 2 MPa Ti 100 oC Flow in to T2 T0 20oC VLIQ 2 VVAP 2 50 L Continuity mi m2 Energy QCV mihi m2u2 m2h2 P2V From D2 at i Pri 2 425 0471 Tri 37315 4103 091 h i hi 0082 734103072 244 h 2h i CP0T2Ti 078720100 630 From D2 Tr2 2932 4103 0715 P2 01154250 489 kPa sat liq ZF 002 hhF RTC485 1646 sat vap ZG 088 hhG RTC025 85 mLIQ 2 ZFRT2 P2VLIQ 2 4890050 0020082 732932 504 kg mVAP 2 ZGRT2 P2VVAP 2 115 kg m2 5155 kg x2 mVAP 2m2 00223 h 2h2 1x2h 2hF2 x2h 2hG2 09777 1646 00223 85 1611 QCV m2h2 hi P2V 51551611 630 244 489 010 10 343 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Mixtures Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12121 A 2 kg mixture of 50 argon and 50 nitrogen by mole is in a tank at 2 MPa 180 K How large is the volume using a model of a ideal gas and b Kays rule with generalized compressibility charts a Ideal gas mixture Eq115 Mmix yi Mi 05 39948 05 28013 33981 V mRT MmixP 2 83145 180 33981 2000 0044 m 3 b Kays rule Eq1284 Pc mix 05 487 05 339 413 MPa Tc mix 05 1508 05 1262 1385 K Reduced properties Pr 2 413 0484 Tr 180 1385 130 Fig D1 Z 0925 V Z mRT MmixP 0925 0044 00407 m3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12122 A 2 kg mixture of 50 argon and 50 nitrogen by mass is in a tank at 2 MPa 180 K How large is the volume using a model of a ideal gas and b van der Waals equation of state with a b for a mixture a Ideal gas mixture Eq1115 Rmix ci Ri 05 02081 05 02968 025245 kJkg K V mRmixT P 2 025245 180 2000 00454 m 3 b van der Waals equation of state before we can do the parameters a b for the mixture we need the individual component parameters aAr E A A02081 15082 E4870E A 008531 27 64 R2T2 c Pc 27 64 aN2 A27 64E A A R2 AET2 c Pc E A27 64E A A02968 12622 E3390E A 017459 bAr RTc 8Pc A02081 1508 8 4870E A 0000 805 bN2 RTc 8Pc A02968 1262 8 3390E A 0001 381 Now the mixture parameters are from Eq1284 amix A ci a12 i 2 AEE 05 A 008531EA 05 A 017459EAA2E A 0126 bmix ci bi 05 0000 805 05 0001 381 0001 093 Using now Eq1252 P A RT v bE A A a v2 E A 2000 A025245 180 v 0001 093E A A0126 v2 E By trial and error we find the specific volume v 002097 mA3E Akg V mv 004194 mA3E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12123 R410A is a 11 mass ratio mixture of R32 and R125 Find the specific volume at 20AoE AC 1200 kPa using Kays rule and the generalized charts and compare to Table B4 Kays rule Eq1283 Pc mix 05 578 05 362 470 MPa Tc mix 05 3513 05 3392 34525 K Reduced properties Pr A 12 470E A 0255 Tr A29315 34525E A 0849 Table A5 R 01145 kJkgK or compute from mix Fig D1 Z 085 v ZRTP 085 01145 29315 1200 00238 mA3E Akg Table B4 v 002260 mA3E Akg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12124 The R410A in Problem 12123 is flowing through a heat exchanger with an exit at 120AoE AC 1200 kPa Find the specific heat transfer using Kays rule and the generalized charts and compare to solution using Table B4 Rmix 05 01598 05 006927 01145 kJkgK CP mix ci CP i 05 0822 05 0791 08065 kJkg K Kays rule Eq1284 Pc mix 05 578 05 362 470 MPa Tc mix 05 3513 05 3392 34525 K Reduced properties 1 Pr1 A 12 470E A 0255 Tr1 A29315 34525E A 0849 Fig D1 hA 1E A hA1E A 04 RTAcE A 04 01145 34525 1581 kJkg Reduced properties 1 Pr2 A 12 470E A 0255 Tr2 A39315 34525E A 1139 Fig D1 hA 2E A hA2E A 02 RTAcE A 02 01145 34525 7906 kJkg The energy equation gives A1E AqA2E A hA2E A hA1E A hA2E A hA 2E A hA 2E A hA 1E A hA 1E A hA1E A 7906 08065 120 20 1581 8855 kJkg mix Table B42 q hA2E A hA1E A 39313 29051 10262 kJkg The main difference is in the value of specific heat about 1 kJkgK at the avg T whereas it is 08 kJkgK at 25AoE AC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12125 A 2 kg mixture of 50 argon and 50 nitrogen by mass is in a tank at 2 MPa 180 K How large is the volume using a model of a ideal gas and b Redlich Kwong equation of state with a b for a mixture a Ideal gas mixture Eq1115 Rmix ci Ri 05 02081 05 02968 025245 kJkg K V mRmixT P A2 025245 180 2000E A 00454 mA3E b Redlich Kwong equation of state Before we can do the parameters a b for the mixture we need the individual component parameters Eq1254 1355 aAr 042748 A R2 AET52 c Pc E 042748 A020812 150825 E4870E A 106154 aN2 042748 A R2 AET52 c Pc E 042748 A029682 126225 E3390E A 198743 bAr 008664 RTc Pc 008664 A02081 1508 4870E A 0000 558 bN2 008664 RTc Pc 008664 A02968 1262 3390E A 0000 957 Now the mixture parameters are from Eq1284 amix A ci a12 i 2 AEE 05 A 106154EA 05 A 198743EAA2E A 14885 bmix ci bi 05 0000 558 05 0000 957 0000 758 Using now Eq1253 P A RT v bE A A a vv bT12 E A 2000 A025245 180 v 0000 758E A A 14885 vv 0000 758 18012 E By trial and error we find the specific volume v 002102 mA3E Akg V mv 004204 mA3E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12126 A modern jet engine operates so that the fuel is sprayed into air at a P T higher than the fuel critical point Assume we have a rich mixture of 50 noctane and 50 air by mole at 500 K and 35 MPa near the nozzle exit Do I need to treat this as a real gas mixture or is an ideal gas assumption reasonable To answer find Z and the enthalpy departure for the mixture assuming Kays rule and the generalized charts The mole fractions are yC8H18 05 yN2 05 079 0395 yO2 05 021 0105 Eq115 Mmix yi Mi 05 114232 0395 28013 0105 31999 71541 Kays rule Eq1284 Pc mix 05 249 0395 339 0105 504 3113 MPa Tc mix 05 5688 0395 1262 0105 1546 3505 K Reduced properties Pr A 35 3113E A 1124 Tr A 500 3505E A 1427 Fig D1 Z 087 I must treat it as a real gas mixture Fig D2 AhE A h 070 RTAcE A 070 A83145 71541E A 3505 2851 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12127 A mixture of 60 ethylene and 40 acetylene by moles is at 6 MPa 300 K The mixture flows through a preheater where it is heated to 400 K at constant P Using the Redlich Kwong equation of state with a b for a mixture find the inlet specific volume Repeat using Kays rule and the generalized charts To do the EOS we need the gas constant so from Eq115 we get Mmix yi Mi 06 28054 04 26068 2726 Rmix 831452726 0305 kJkg K Redlich Kwong EOS the individual component parameters Eq1254 1255 aC2H4 042748 A R2 AET52 c Pc E 042748 A029642 282425 E5040E A 99863 aC2H2 042748 A R2 AET52 c Pc E 042748 A031932 308325 E6140E A 118462 bC2H4 008664 RTc Pc 008664 A02964 2824 5040E A 0001 439 bC2H2 008664 RTc Pc 008664 A03193 3083 6140E A 0001 389 Now the mixture parameters are from Eq1284 so we need the mass fractions cC2H4 y M Mmix A06 28054 2726E A 06175 cC2H4 1 cC2H4 03825 amix A ci a12 i 2 AEE 06175 A 99863EA 03825 A 118462EAA2E A 10679 bmix ci bi 06175 0001 439 03825 0001 389 0001 42 Using now Eq1253 P A RT v bE A A a vv bT12 E A 6000 A0305 300 v 0001 42E A A 10679 vv 0001 42 30012 E By trial and error we find the specific volume v 0006683 mA3E Akg Kays rule Eq1284 Pc mix 06 504 04 614 548 MPa Tc mix 06 2824 04 3083 2928 K Reduced properties Pr A 6 548E A 1095 Tr A 300 2928E A 1025 Fig D1 Z 04 difficult to read v ZRTP 04 0305 300 6000 00061 mA3E Akg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12128 For the previous problem find the specific heat transfer using Kays rule and the generalized charts To do the EOS we need the gas constant so from Eq115 we get Mmix yi Mi 06 28054 04 26068 2726 Rmix 831452726 0305 kJkg K cC2H4 y M Mmix A06 28054 2726E A 06175 cC2H4 1 cC2H4 03825 CP mix ci CP i 06175 1548 03825 1699 1606 kJkg K Kays rule Eq1284 Pc mix 06 504 04 614 548 MPa Tc mix 06 2824 04 3083 2928 K Reduced properties 1 Pr1 A 6 548E A 1095 Tr1 A 300 2928E A 1025 Fig D1 hA 1E A hA1E A 21 RTAcE A 21 0305 2928 1875 kJkg Reduced properties 2 Pr2 A 6 548E A 1095 Tr2 A 400 2928E A 1366 Fig D1 hA 2E A hA2E A 07 RTAcE A 07 0305 2928 625 kJkg The energy equation gives A1E AqA2E A hA2E A hA1E A hA2E A hA 2E A hA 2E A hA 1E A hA 1E A hA1E A 625 1606 400 300 1875 2856 kJkg mix Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12129 A gas mixture of a known composition is frequently required for different purposes eg in the calibration of gas analyzers It is desired to prepare a gas mixture of 80 ethylene and 20 carbon dioxide mole basis at 10 MPa 25C in an uninsulated rigid 50L tank The tank is initially to contain CO2 at 25C and some pressure P1 The valve to a line flowing C2H4 at 25C 10 MPa is now opened slightly and remains open until the tank reaches 10 MPa at which point the temperature can be assumed to be 25C Assume that the gas mixture so prepared can be represented by Kays rule and the generalized charts Given the desired final state what is the initial pressure of the carbon dioxide P1 A CA2E AHA4E A B COA2E TA1E A 25 AoE AC PA2E A 10 MPa TA2E A 25 AoE AC yAA2E A 08 yAB2E A 02 Mixture at 2 PAC2E A 08 504 02 738 5508 MPa TAC2E A 08 2824 02 3041 2867 K TAr2E A 298152867 1040 PAr2E A 105508 1816 D1 ZA2E A 032 nA2E A A P2V EZ2R T2 E A A 10 000005 032831452982E A 06302 kmol nAA2E A nAiE A 08 nA2E A 05042 kmol CA2E AHA4E nAB2E A nA1E A 02 nA2E A 01260 kmol COA2E TAr1E A A2982 3041E A 0981 PAr1E A A n1ZB1R E T1 PCBVE A A 0126 ZB1 831452982 E7380005E A 08466 ZAB1E By trial error PAr1E A 0618 ZAB1E A 073 PA1E A 0618 738 456 MPa B P 10 MPa i T 25 C o i A V005 m3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12130 One kmols of saturated liquid methane CH4 at 1 MPa and 2 kmols of ethane C2H6 at 250C 1 MPa are fed to a mixing chamber with the resultant mixture exiting at 50C 1 MPa Assume that Kays rule applies to the mixture and determine the heat transfer in the process Control volume the mixing chamber inlet CH4 is 1 inlet C2H6 is 2 and the exit state is 3 Energy equation is AQ E ACVE A AnE A3E A AhE A3 AnE A1E A AhE A1 AnE A2E A AhE A2 Select the ideal gas reference temperature to be T3 and use the generalized charts for all three states Pr1 Prsat 1460 02174 Trsat 0783 T1 0783 1904 1491 K h1 457 Pr2 1488 0205 Tr2 5233054 1713 h2 008 AhE A1 ACE A1T1 T3 h1 ARE ATc 36151491 3232 457 83145 1904 13528 kJkmol AhE A2 ACE A2T2 T3 h2 ARE ATc 5311250 50 008 83145 3054 10 419 kJkmol Kays rule Eq1284 Tcmix 1 1904 2 30543 2671 K Pcmix 1 460 2 4883 479 MPa Tr3 32322671 121 Pr3 1479 021 h3 015 AhE A3 0 015 2671 83145 333 kJkmol AQ E ACVE A 3333 113528 210 419 8309 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12131 Saturatedliquid ethane at TA1E A 14C is throttled into a steady flow mixing chamber at the rate of 025 kmols Argon gas at TA2E A 25C PA2E A 800 kPa enters the chamber at the rate of 075 kmols Heat is transferred to the chamber from a heat source at a constant temperature of 150AoE AC at a rate such that a gas mixture exits the chamber at TA3E A 120AoE AC PA3E A 800 kPa Find the rate of heat transfer and the rate of entropy generation Argon Ta2 25oC P2 800 kPa n 2 075 kmols Tca 150 K Pca 487 MPa Ma 39948 kgkmol Cpa 052 kJkg K h a3 h a2 MaCpaT3 Ta2 19734 kJkmol Inlet Ethane Tb1 14oC sat liq xb1 0 n 1 025 kmols Tcb 3054 K Pcb 488 MPa Mb 3007 kgkmol Cpb 1766 kJkgK Tr1 094 Pb1 Pr1Pcb 069 4880 3367 kPa h b1 h b1 381 RTcb 96745 kJkmol s b1 s b1 374 ARE A 311 h b3 h b1 MbCpbT3 Tb1 56296 kJkmol Exit Mix T3 120oC P3 800 kPa consider this an ideal gas mixture Energy Eq n 1h b1 n 2h a2 AQ E A n 3h 3 n 1h b3 n 2h a3 AQ E A n 1h b3 h b1 n 2h a3 h a2 025 56296 96745 07519734 5306 kW Entropy Eq S gen n 1s b3 s b1 n 2s a3 s a2 AQ E ATH TH 150oC ya n 2n tot 075 yb n 1n tot 025 s a3 s a2 MaCpaln T3 Ta2 ARE A ln yaP3 Pa2 814 kJkmolK s b3 s b1 MbCpbln T3 Tb1 ARE A ln ybP3 Pb1 s b1 s b1 40172 311 7127 kJkmol K S gen 025 7127 075 814 5306 423 1138 kWK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12132 A cylinderpiston contains a gas mixture 50 CO2 and 50 C2H6 mole basis at 700 kPa 35C at which point the cylinder volume is 5 L The mixture is now compressed to 55 MPa in a reversible isothermal process Calculate the heat transfer and work for the process using the following model for the gas mixture a Ideal gas mixture b Kays rule and the generalized charts a Ideal gas mixture U2 U1 mCAv mixE AT2 T1 0 Q12 W12 A EA P dV P1V1 lnV2V1 P1V1 lnP2P1 700 0005 ln5500700 771 kJ b Kays rule Tcmix 05 3041 05 3054 30475 K Pcmix 05 738 05 488 613 MPa Tr1 3081530475 1011 Pr1 07613 01142 Z1 096 h1 012 s1 008 n P1V1Z1ARE A T1 A 7000005 09628314530815E A 000142 kmol Tr2 Tr1 Pr2 55613 0897 Z2 058 h2 135 s2 10 AhE A2 AhE A1 AhE A2 AhE A1 ARE A Tch2 h1 0 83145 30475135 012 3117 AuE A2 AuE A1 AhE A2 AhE A1 ARE ATZ1 Z2 3117 83145 30815096 058 2143 kJkmol Q12 nTAsE A2 AsE A1T 000142 30815 83145 0 ln5507 10 008 1085 kJ W12 Q12 nAuE A2 AuE A1 1085 0001422143 781 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12133 A cylinderpiston contains a gas mixture 50 CO2 and 50 C2H6 mole basis at 700 kPa 35C at which point the cylinder volume is 5 L The mixture is now compressed to 55 MPa in a reversible isothermal process Calculate the heat transfer and work for the process using the following model for the gas mixture a Ideal gas mixture b The van der Waals equation of state a Ideal gas mixture U2 U1 mCAv mixE AT2 T1 0 Q12 W12 A EA P dV P1V1 lnV2V1 P1V1 lnP2P1 700 0005 ln5500700 771 kJ b van der waals equation For CO2 b ARE A Tc8Pc 83145 30418 7380 004282 a 27 Pc b2 27 7380 0042822 36545 For C2H6 b ARE A Tc8Pc 83145 30548 4880 006504 a 27 Pc b2 27 4880 0065042 55741 amix 05A 36545EA 05A 55741EA2 456384 bmix 05 004282 05 006504 005393 A831453082 v1 005393E A A456384 v12E A 700 0 By trial and error AvE A1 35329 mA3E Akmol A831453082 v2 005393E A A456384 v22E A 5500 0 By trial and error AvE A2 02815 mA3E Akmol n V1AvE A1 000535329 000142 Q12 nTAsE A2 AsE A1T n ARE A T ln A v2 b Ev1 bE 000142 83145 3082 ln A02815 005392 35329 005392E A 993 kJ U2 U1 000142 45639353291 028151 212 kJ Q12 U2 U1 W12 W12 993 212 781 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Helmholtz EOS Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12134 Verify that the ideal gas part of Helmholtz function substituted in Eq1289 does lead to the ideal gas law as in note after Eq1297 The ideal gas Helmholtz function is from its definition see Eq1212 aAE A uAE A TsAE A hAE A RT T sAE We have hAE A hA oE A CPo dT sAE A sA oE A CPo T dT R ln ρT ρoTo Now from Eq1289 we need to look at A a Eρ TE A so following 1292 A a Eρ TE A A h Eρ TE A A RT ρ TE A T A s Eρ TE 0 0 TR A ρE A ln ρT ρoTo T RT ρ So then ρA2E A a Eρ TE A P ρRT Ideal gas OK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12135 Gases like argon and neon have constant specific heats Develop an expression for the ideal gas contribution to Helmholtz function in Eq1292 for these cases The ideal gas Helmholtz function is from its definition see Eq1292 aAE A uAE A TsAE A hAE A RT T sAE We have hAE A hA oE A CPo dT hA oE A CPoT To sAE A sA oE A CPo T dT R ln ρT ρoTo sA oE A CPo lnT To R ln ρT ρoTo So now we get aAE A hA oE A CPoT To RT T sAE hA oE A T sA oE A CPoT To CPoT ln T To RT RT ln ρT ρoTo Co C1 T C2 T ln T To RT ln ρ ρo where Co hA oE A CPoTo C1 CPo R sA oE A C2 CPo R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12136 Find an expression for the change in Helmholtz function for a gas with an EOS as Pv b RT From Eq1231 we get duT T AP TE AAvE A P dvT T A R v bE A P dvT P P dvT 0 From Eq1232 or Eq1234 we get dsT Av TE AAPE A dPT AP TE AAvE A dvT AR PE A dPT A R v bE A dvT Now the changes in u and s can be integrated to find u2 u1T 0 s2 s1T R ln P2 P1 R ln v2 b v1 b to this we now need to add the variation due to T For this we get u2 u1v 1 2 Cv dT and s2 s1v 1 2 Cv T dT Finally since the Helmholtz function contains the product Ts we need the absolute value of the entropy so s1 so o 1 Cv T dT R ln v1 b vo b Then the change in Helmholtz function becomes a2 a1 u2 u1 T2s2 T1s1 u2 u1 T2s2 s1 T1 T2 s1 1 2 Cv dT T2 1 2 Cv T dT R ln v2 b v1 b T1 T2 s1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12137 Use the equation of state in Example 123 and find an expression for isothermal changes in Helmholtz function between two states The EOS is APv RTE A 1 C A P T4 E A or v ART PE A A C T3 E and we use the isothermal changes found in Ex143 as h2 h1T A4C T3 E A P2 P1T s2 s1T R ln P2 P1 T A3C T4 E A P2 P1T As Helmholtz function is a u Ts we get a2 a1T u2 u1T Ts2 s1T h2 h1T P2v2 P1v1 Ts2 s1T A4C T3 E A P2 P1T RT2 RT1 A C T3 E AP2 P1T RT ln P2 P1 T A3C T3 E A P2 P1T A4C T3 E A P2 P1T A C T3 E AP2 P1T RT ln P2 P1 T A3C T3 E A P2 P1T This now reduces to the final answer a2 a1T RT ln P2 P1 T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12138 Assume a Helmholtz equation as aAE A Co C1 T C2 T ln T To RT ln ρ ρo where Co C1 C2 are constants and To and ρo reference values for temperature and density see Eqs 1292 95 Find the properties P u and s from this expression Is anything assumed for this particular form Given Helmholtz function we can find the pressure and entropy from Eq 1221 and then u from the definition a u Ts aAE A Co C1 T C2 T ln T To RT ln v vo A a Ev TE A RT A vE A ln v vo RT v P A a Ev TE A RT v ie Ideal gas s A a ET vE A C1 C2 ln T To C2 R ln v vo Notice how it looks like Eq617 u a Ts Co C1 T C2 T ln T To RT ln v vo C1T C2T ln T To C2T RT ln v vo Co C2T We find that u is linear in T Not only is it ideal gas but it also has constant specific heats Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12139 Saturated liquid ethane at 244 MPa enters a heat exchanger and is brought to 611 K at constant pressure after which it enters a reversible adiabatic turbine where it expands to 100 kPa Find the heat transfer in the heat exchanger the turbine exit temperature and turbine work From D2 PAr1E A 244488 050 TAr1E A 089 TA1E A 0893054 2718 K hA 1E AhA1E A 027653054412 3479 hA 2E AhA 1E A 1766 611 2718 5990 PAr2E A 050 TAr2E A 6113054 200 From D2 hA 2E AhA2E A RTAcE A 014 027653054014 118 q hA2E AhA1E A 118 5990 3479 9351 kJkg From D3 sA 2E AsA2E A 02765005 00138 sA 3E AsA 2E A 1766 ln A T3 E611E A 02765 ln A 100 2440E Assume TA3E A 368 K TAr3E A 1205 at PAr3E A 0020 sA 3E AsA 2E A 08954 08833 00121 From D3 sA 3E AsA3E A 02765001 00028 sA3E AsA2E A 00028 00121 00138 0 ΟΚ Therefore TA3E A 368 K From D2 hA 3E AhA3E A 027653054001 08 w hA2E AhA3E A 118 1766 611 368 08 4181 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12140 A pistoncylinder initially contains propane at T 7C quality 50 and volume 10L A valve connecting the cylinder to a line flowing nitrogen gas at T 20C P 1 MPa is opened and nitrogen flows in When the valve is closed the cylinder contains a gas mixture of 50 nitrogen 50 propane on a mole basis at T 20C P 500 kPa What is the cylinder volume at the final state and how much heat transfer took place State 1 Propane TA1E A 7oC xA1E A 05 VA1E A 10 L Tc 3698 K Pc 425 kPa CP 1679 kJkgK M 44097 kgkmol Fig D1 Tr1 072 Pr1 012 PA1E A Pr1Pc 510 kPa Fig D1 Zf1 0020 Zg1 088 ZA1E A 1 x1Zf1 x1Zg1 045 nA1E A PA1E AVA1E AZA1E ARE ATA1E A 510 001045 83145 2662 000512 kmol h 1 h 1o AC PE ATA1E A To h 1 h 1 h 1o 0 Ah E1E Ah 1f RTc 479 Ah E1E Ah 1g RTc 025 Ah E1E A h 1 1 x1 Ah E1E A h 1f x1 Ah E1E A h 1g 7748 kJkmol h 1 0 1679 440947 20 7748 9747 kJkmol Inlet Nitrogen Ti 20oC Pi 10 MPa Tc 1262 K Pc 339 MPa Cpn 1042 kJkgK M 28013 kgkmol Tri 2323 Pri 0295 Ah EiE Ah i 006 83145 1262 6296 kJkmol h i h io AC PnE ATi To h i h i h io 0 Ti To 0 State 2 50 Propane 50 Nitrogen by mol T2 20oC P2 500 kPa TAcmixE A yAiE ATAciE A 248 K PAcmixE A yAiE APAciE A 382 MPa Tr2 1182 Pr2 0131 Z2 097 Ah E2E A h 2RTc 006 h 2 h 2o AC PmixE AT2 To h 2 h 2 h 2o 0 T2 To 0 a ni n1 n2 n1 ni 01024 V2 n2Z2RT2P2 00484 m3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful b Energy Eq Qcv nih i n2u 2 n21u 21 Wcv AuE A Ah Pv EE Wcv P1 P2V2 V12 1988 kJ Qcv n2h 2 n1h 1 nih i P2V2 P1V1 Wcv h i 6296 kJkmol h 2 1237 kJkmol Qcv 5003 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12141 A newly developed compound is being considered for use as the working fluid in a small Rankinecycle power plant driven by a supply of waste heat Assume the cycle is ideal with saturated vapor at 200C entering the turbine and saturated liquid at 20C exiting the condenser The only properties known for this compound are molecular weight of 80 kgkmol ideal gas heat capacity CAPOE A 080 kJkg K and TACE A 500 K PACE A 5 MPa Calculate the work input per kilogram to the pump and the cycle thermal efficiency TA1E A 200AoE AC 4732 K xA1E A 10 TA3E A 20AoE AC 2932 K xA3E A 00 Properties known M 80 CAPOE A 08 kJkg K TACE A 500 K PACE A 50 MPa TAr1E A A4732 500E A 0946 TAr3E A A2932 500E A 0586 R RM 83145180 010393 kJkg K From Fig D1 PAr1E A 072 PA1E A 072 5 36 MPa PA4E A PAr3E A 0023 PA3E A 0115 MPa PA2E A ZAF3E A 0004 vAF3E A A ZF3RT3 EP3 E A A0004 010393 2932 115E A 000106 mA3E Akg wAPE A A 3 4 vdPEA vAF3E APA4E A PA3E A 0001063600115 37 kJkg qAHE A hA4E A hA1E A but hA3E A hA4E A wAPE A qAHE A hA1E AhA3E A wAPE A From Fig D2 hA 1E AhA1E A RTACE A 125 0103 93 500 125 649 kJkg hA 3E AhA3E A 0103 93 500 52 2702 kJkg hA 1E AhA 3E A CAP0E ATA1E ATA3E A 08020020 1440 kJkg hA1E AhA3E A 649 1440 2702 3493 kJkg qAHE A 3493 37 3456 kJkg Turbine Cond Ht Exch P 3 1 4 2 Q H W T WP Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Turbine sA2E A sA1E A 0 sA 2E A sA2E AsA 2E A sA 1E A sA 1E A sA1E A From Fig D3 sA 1E AsA1E A 010393099 01029 kJkg K sA 2E AsA 1E A 080 ln A2932 4732E A 0103 93 ln A 115 3600E A 00250 Substituting sA 2E AsA2E A 01029 00250 00779 sA 2E AsAF2E A xA2E AsAFG2E 00779 0103 93885 xA2E A0103 93885006 xA2E A 0922 hA 2E AhA2E A hA 2E AhAF2E A xA2E AhAFG2E From Fig D2 hAFG2E A 010393 500 52007 2666 hA 2E AhA2E A 2702 0922 2666 250 wATE A hA1E AhA2E A 649 1440 250 1041 kJkg ηATHE A A wNET EqH E A A104137 3456E A 029 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12142 A 200L rigid tank contains propane at 400 K 35 MPa A valve is opened and propane flows out until half the initial mass has escaped at which point the valve is closed During this process the mass remaining inside the tank expands according to the relation PvA14E A constant Calculate the heat transfer to the tank during the process CA3E AHA8E A V 200 L TA1E A 400 K PA1E A 35 MPa Flow out to mA2E A mA1E A2 PvA14E A const inside TAr1E A A 400 3698E A 1082 PAr1E A A 35 425E A 0824 Fig D1 ZA1E A 074 vA1E A A0740188 55400 3500E A 001594 vA2E A 2vA1E A 003188 mA1E A A 02 0015 94E A 1255 kg mA2E A A1 2E A mA1E A 6275 kg PA2E A PA1E AA v1 Ev2 E AA 14E A A3500 214 E A 1326 kPa A Pr2 1326 E425 0312 P2v2 Z2RT2 E A A Trial error saturated with T2 08263698 3055 K EZ2 1326003188 0188 553055 0734E ZA2E A ZAF2E A xA2E AZAG2E A ZAF2E A 0734 005 xA2E A078005 xA2E A 0937 hA 1E AhA1E A 0188 55369809 628 hA 2E AhA 1E A 167943055400 1587 hA 2E AhA2E A hA 2E AhAF2E A xA2E AhAFG2E A 0188 553698441 0937441055 553 Energy Eq QACVE A mA2E AhA2E A mA1E AhA1E A PA1E APA2E AV mAeE AhAe AVEE Let hA 1E A 0 then hA1E A 0 hA1E AhA 1E A 628 hA2E A hA 1E A hA 2E AhA 1E A hA2E AhA 2E A 0 1587 553 2140 hAe AVEE A hA1E AhA2E A2 1384 QACVE A 62752140 1255628 3500132602 62751384 9814 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12143 One kilogram per second water enters a solar collector at 40C and exits at 190C as shown in Fig P12143 The hot water is sprayed into a directcontact heat exchanger no mixing of the two fluids used to boil the liquid butane Pure saturatedvapor butane exits at the top at 80C and is fed to the turbine If the butane condenser temperature is 30C and the turbine and pump isentropic efficiencies are each 80 determine the net power output of the cycle HA2E AO cycle solar energy input raises 1 kgs of liquid HA2E AO from 40AoE AC to 190AoE AC Therefore corresponding heat input to the butane in the heat exchanger is AQ E AHE A AmE AhAF 190 CE AhAF 40 CE AAH2OE A 18076216757 64005 kW CA4E AHA10E A cycle TA1E A 80 AoE AC xA1E A 10 TA3E A 30 AoE AC xA3E A 00 ηASTE A ηASPE A 080 TAr1E A A3532 4252E A 0831 From D1 D2 and D3 PA1E A 03253800 1235 kPa hA 1E AhA1E A 0143 044252056 341 sA 1E AsA1E A 0143 040475 00680 TAr3E A A3032 4252E A 0713 From D1 D2 and D3 PA3E A 01133800 429 kPa sat liq hAE AhAFE A RTACE A481 2925 sAE AsAFE A R664 0950 sat vap hAE AhAGE A RTACE A0235 143 sAE AsAGE A R022 0031 Because of the combination of properties of CA4E AHA10E A particularly the large CAP0E AR sA1E A is larger than sAGE A at TA3E A To demonstrate sA 1E AsA G3E A 17164 ln A3532 3032E A 0143 04 ln A1235 429E A 01107 sA1E AsAG3E A 00680 01107 0031 00737 kJkg K Turbine Cond Ht Exch P 3 1 4 2 Q H W T WP Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful so that TA2SE A will be TA3E A as shown in the Ts diagram A number of other heavy hydrocarbons also exhibit this behavior Assume TA2SE A 315 K TAr2SE A 0741 From D2 and D3 hA 2SE AhA2SE A RTACE A021 128 and sA 2SE AsA2SE A R019 0027 sA 1E AsA 2SE A 17164 ln A3532 315E A 0143 04 ln A1235 429E A 00453 sA1E AsA2SE A 00680 00453 0027 0 TA2SE A 315 K hA 1E AhA 2SE A 171643532315 656 wASTE A hA1E AhA2SE A 341 656 128 443 kJkg wATE A ηASE AwASTE A 080443 354 kJkg At state 3 vA3E A A00190143 043032 429E A 0001 92 mA3E Akg wASPE A vA3E APA4E APA3E A 0001 921235429 155 kJkg wAPE A A wSP EηSP E A A155 08E A 194 kJkg wANETE A wATE A wAPE A 354 194 3346 kJkg For the heat exchanger AQ E AHE A 64005 AmE AC4H10E AhA1E AhA4E A But hA1E AhA4E A hA1E AhA3E AwAPE A hA1E AhA3E A hA1E AhA 1E A hA 1E AhA 3E A hA 3E AhA3E A 341 171680 30 2925 3442 kJkg Therefore AmE AC4H10E A A 64005 3442194E A 187 kgs AW E ANETE A AmE AC4H10E AwANETE A 187 3346 6257 kW 3 2 2s 1 s T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12144 A pistoncylinder contains ethane gas initially at 500 kPa 100 L and at ambient temperature 0C The piston is now moved compressing the ethane until it is at 20C with a quality of 50 The work required is 25 more than would have been required for a reversible polytropic process between the same initial and final states Calculate the heat transfer and the entropy generation for the process Ethane Tc 3054 K Pc 488 MPa R 02765 kJkgK Cp 1766 kJkg K State 1 Tr1 0895 Pr1 0102 Z1 095 v1 Z1RT1P1 01435 m3kg m1 V1v1 0697 kg h 1 h1 013RTc 110 kJkg s 1 s1 009 R 0025 kJkg K State 2 T2 20oC x2 05 1W2 125Wrev Tr2 096 Pr2 078 P2 Pr2Pc 3806 kPa Zf2 014 Zg2 054 Z2 1 x2Zf x2Zg 034 h 2 h2 1 x2 365 RTc x2 139 RTc 2128 kJkg s 2 s2 1 x2 345 R x2 110 R 0629 kJkg K v2 Z2RT2P2 00072 m3kg V2 mv2 0005 m3 Polytropic process P1Vn 1 P2Vn 2 ln P2 P1 n ln V1 V2 n 06783 Wrev P dV P2V2 P1V1 1 n 963 kJ 1W2 125Wrev 1204 kJ a Energy Eq 1Q2 mu2 u1 1W2 u h Pv h2 h1 h2 h 2 h 2 h 1 h 1 h1 2128 176620 0 110 1665 kJkg u2 u1 h2 h1 P2v2 P1v1 1222 kJkg 1Q2 0697 kg 1222 kJkg 1204 kJ 2056 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful b Entropy Eq Sgen ms2 s1 1Q2 To To 0oC 27315 K s2 s1 s2 s 2 s 2 s 1 s 1 s1 s 2 s 1 Cp lnT2 T1 R lnP2 P1 0436 kJkg K Sgen 06970629 0436 0025 2056 27315 0028 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12145 Carbon dioxide gas enters a turbine at 5 MPa 100C and exits at 1 MPa If the isentropic efficiency of the turbine is 75 determine the exit temperature and the secondlaw efficiency using the generalized charts CO2 turbine ηS wwS 075 inlet T1 100oC P1 5 MPa exhaust P2 1 MPa a Pr1 5 738 0678 Tr1 3732 3041 1227 Pr2 1 738 0136 From D2 and D3 h 1 h1 0188 923041052 299 s 1 s1 0188 92030 00567 Assume T2S 253 K Tr2S 0832 From D2 and D3 h 2Sh2S RTC020 115 s 2S s2S R017 00321 s 2S s 1 08418 ln 253 3732 0188 92 ln 1 5 00232 s2S s1 00321 00232 00567 0 T2S 253 K h 2S h 1 084182533732 1012 wS h1 h2S 299 1012 115 828 kJkg w ηSwS 075828 621 kJkg h1h 1 h 1h 2 h 2h2 Assume T2 275 K Tr2 0904 h 1 h 2 08418 3732 275 827 From D2 and D3 h 2 h2 RTC017 98 s 2 s2 R013 00245 Substituting w 299 827 98 627 621 T2 275 K For the exergies we need the entropy changes Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful s 2 s 1 08418 ln 275 3732 0188 92 ln 1 5 00470 s2 s1 00245 00470 00567 00792 Assuming T0 25 oC ϕ1 ϕ2 h1 h2 T0s1 s2 621 298200792 857 kJkg η2nd Law w ϕ1ϕ2 621 857 0725 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12146 A 10 m3 storage tank contains methane at low temperature The pressure inside is 700 kPa and the tank contains 25 liquid and 75 vapor on a volume basis The tank warms very slowly because heat is transferred from the ambient a What is the temperature of the methane when the pressure reaches 10 MPa b Calculate the heat transferred in the process using the generalized charts c Repeat parts a and b using the methane tables Table B7 Discuss the differences in the results CH4 V 10 m3 P1 700 kPa VLIQ 1 25 m3 VVAP 1 75 m 3 a Pr1 070 460 0152 Pr2 10 460 2174 From D1 ZF1 0025 ZG1 087 T1 074 1904 1409 K vF1 0025051831409 700 000261 vG1 087051831409 700 00908 mLIQ 1 25 000261 9579 kg mVAP 1 75 00908 826 kg Total m 10403 kg v2 v1 V m 10 10405 000961 Z2051831904Tr2 10 000 or Z2Tr2 09737 at Pr2 2174 By trial and error Tr2 1334 Z2 073 T2 13341904 2540 K b Energy Eq Q12 mu2u1 mh2h1 VP2P1 Using D2 x1 826 10405 00794 h 1h1 h 1hf1 x1h fg1 0518 35190447200794472029 4311 h 2 h 1 CP T2 T1 2254 2540 1409 2549 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful h 2 h2 05183 1904147 1451 h2 h1 1451 2549 4311 5409 kJkg Q12 10405 kg 5409 kJkg 10 m3 10 000700 kPa 469 806 kJ c Using Table B7 for CH4 T1 TSAT 1 1417 K vF1 0002 675 uF1 17847 vG1 0090 45 uG1 19984 mLIQ 1 25 0002 675 9346 mVAP 1 75 0090 45 829 Total mass m 10175 kg and v2 10 10175 0009 828 m3kg At v2 P2 10 MPa T2 2591 K u2 29611 Q12 mu2u1 1017529611 934617847 82919984 451 523 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12147 Consider the following reference state conditions the entropy of real saturated liquid methane at 100C is to be taken as 100 kJkmol K and the entropy of hypothetical ideal gas ethane at 100C is to be taken as 200 kJkmol K Calculate the entropy per kmol of a real gas mixture of 50 methane 50 ethane mole basis at 20C 4 MPa in terms of the specified reference state values and assuming Kays rule for the real mixture behavior CH4 T0 100 oC s LIQ 0 100 kJkmol K C2H6 T0 100 oC P0 1 MPa s 0 200 kJkmol K Also for CH4 TC 1904 K PC 460 MPa For a 50 mixture Kays rule Eq1284 Tcmix 05 1904 05 3054 2479 K Pcmix 05 460 05 488 474 MPa IG MIX at T0100 oC P01 MPa CH4 Tr0 091 PG 057 460 2622 MPa s 0 CH4 s LIQ 0 PG ss LIQat PG R ln P0PG 100 40183145 83145 ln 12622 14136 s 0 MIX 0514136 05200 8314505 ln 05 05 ln 05 17644 C P0 MIX 0516042254 0530071766 44629 s TP MIX 17644 44629 ln 2932 1732 83145 ln 4 1 18841 kJkmol K For the mixture at T P Tr 1183 Pr 0844 Entropy departure s TP MIX s TP MIX 0436383145 363 kJkmol K Therefore s TP MIX 18841 363 18478 kJkmol K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful An alternative is to form the ideal gas mixture at T P instead of at T0 P0 s TP CH4 s LIQ 0 ss LIQ C P0 CH4 ln T T0 R ln P PG PG T0 at PG T 0 100 3334 16042254 ln 2932 1732 83145 ln 4 26 100 3334 1903 353 14884 kJkmol K s TP C2H6 200 30071766 ln 2932 1732 83145 ln 4 1 200 2796 1153 21643 kJkmol K s TP MIX 0514884 0521643 8314505 ln 05 05 ln 05 18841 kJkmol K s TP MIX 18841 363 18478 kJkmol K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12148 Determine the heat transfer and the net entropy change in problem 12129 Use the initial pressure of the carbon dioxide to be 456 MPa before the ethylene is flowing into the tank A gas mixture of a known composition is frequently required for different purposes eg in the calibration of gas analyzers It is desired to prepare a gas mixture of 80 ethylene and 20 carbon dioxide mole basis at 10 MPa 25C in an uninsulated rigid 50L tank The tank is initially to contain CO2 at 25C and some pressure P1 The valve to a line flowing C2H4 at 25C 10 MPa is now opened slightly and remains open until the tank reaches 10 MPa at which point the temperature can be assumed to be 25C Assume that the gas mixture so prepared can be represented by Kays rule and the generalized charts Given the desired final state what is the initial pressure of the carbon dioxide P1 A C2H4 B CO2 T1 25 oC P2 10 MPa T2 25 oC yA2 08 yB2 02 Mixture at 2 PC2 08 504 02 738 5508 MPa TC2 08 2824 02 3041 2867 K Tr2 298152867 1040 Pr2 105508 1816 D1 Z2 032 n2 Z2R T2 P2V 10 000005 032831452982 06302 kmol nA2 ni 08 n2 05042 kmol C2H4 nB2 n1 02 n2 01260 kmol CO 2 Tr1 2982 3041 0981 and Pr1 4560 7380 0618 Energy Eq QCV nih i n2u 2 n1u 1 n2h 2 n1h 1 P2P1V or QCV n2h 2h 2 n1h 1h 1 nih ih i P2P1V since Ti T1 T2 h i h 1 h 2 h 1h 1 083 83145 3041 2099 kJkmol B P 10 MPa i T 25 C o i A V005 m3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful h 2h 2 340 83145 2867 8105 kJkmol Tri 2982 2824 1056 Pri 10 504 1984 h i h i 335831452824 7866 kJkmol QCV 063028105 01262099 050427866 10 0004560005 1149 kJ SCV n2s 2 n1s 1 SSURR QCVT0 nis i Sgen n2s 2 n1s 1 QCVT0 nis i Let s A0 s B0 0 at T0 25 oC P0 01 MPa Then s MIX 0 83145 08 ln 08 02 ln 02 4161 kJkmol K s 1 s B0 s P1 T1s P0 T0B s 1s P1 T1 B 0 083145 ln 456 01 060 83145 3675 kJkmol K s i s A0 s Pi Tis P0 T0A s is Pi Ti A 0 083145 ln 10 01 24483145 5858 kJkmol K s 2 s MIX 0 s P2 T2s P0 T0MIX s 2s P2 T2 MIX 4161 083145 ln 10 01 255183145 5534 kJkmol K Sgen 063025533 01263675 050425858 11492982 315 kJK UPDATED JULY 2013 SOLUTION MANUAL CHAPTER 12 English Units Fundamentals of Thermodynamics Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CHAPTER 12 CONTENT CHAPTER 12 SUBSECTION PROB NO Clapeyron equation 149153 Volume Expansivity and Compressibility 154160 Equations of State 161162 Generalized Charts 163178 Mixtures 179181 Review problem 182 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Clapeyron Equation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12149E Verify that Clapeyrons equation is satisfied for R410A at 30 F in Table F9 Clapeyron Eq dPsat dT dPg dT hfg Tvfg F9 P 111796 psia hfg 9575 Btulbm vfg 05289 ft3lbm Slope around 30 F best approximated by cord from 20 F to 40 F dPg dT 133163 93128 40 20 200 psiaR hfg Tvfg 9575 4897 05289 778 144 1997 psiaR This fits very well Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12150E Use the approximation given in Problem 1216 and Table F7 to determine A and B for steam from properties at 70 F only Use the equation to predict the saturation pressure at 80 F and compare to table value ln Psat A BT dPsat dT Psat BT2 so we notice from Eq127 and Table values from F71 and F4 that B hfg R 105395 8576 778 95613 R Now the constant A comes from the saturation pressure as A ln Psat BT ln 0363 95613 45967 70 17038 Use the equation to predict the saturation pressure at 80 F as ln Psat A BT 17038 95613 45967 80 06789 Psat 05071 psia compare this with the table value of Psat 0507 psia and we have a very accurate approximation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12151E Find the saturation pressure for refrigerant R410A at 100 F assuming it is higher than the triple point temperature The lowest temperature in Table F9 for R410A is 80 F so it must be extended to 100 F using the Clapeyron Eq 127 integrated as in example 121 At T1 80 F 3797 R P1 8196 lbfin2 and R 19859 72585 0027 36 Btulbm R ln P P1 R hfg T T1 TT1 12114 0027 36 35973797 3597 3797 06484 P P1 exp06484 4286 lbfin 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12152E Using thermodynamic data for water from Tables F71 and F73 estimate the freezing temperature of liquid water at a pressure of 5000 lbfin2 H2O dT dPif Tvif hif constant At the triple point vif vf vi 0016 022 0017 473 0001 451 ft3lbm hif hf hi 00 14334 14334 Btulbm dPif dT 14334 491690001 451 7782 144 10858 lbfin2 R at P 5000 lbfin2 T 3202 5000009 10858 274 F TP T P 5000 psia Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12153E Ice solid water at 27 F 1 atm is compressed isothermally until it becomes liquid Find the required pressure Water triple point T 3202 F 49169 R P 0088 67 lbfin2 vf 0016 022 ft3lbm vi 0017 473 ft3lbm hf 000 Btulbm hi 14334 Btulbm Clapeyron dPif dT hf hi vf viT 143347782 0001 45149169144 10858 psiaR P dPif dT T 10858 27 3202 54507 lbfin2 P Ptp P 5451 lbfin2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Volume Expansivity and Compressibility Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12154E Determine the volume expansivity αP and the isothermal compressibility βT for water at 50 F 500 lbfin2 and at 400 F 1500 lbfin2 using the steam tables Water at 50 F 500 lbfin2 compressed liquid αP 1 vv TP 1 vv T P Using values at 32 F 50 F and 100 F αP 1 0015 99 0016 1 0015 99 100 32 0000 101 F1 βT 1 vv PT 1 vv PT Using values at saturation 500 and 1000 lbfin2 βT 1 0015 99 0015 99 0016 02 1000 0178 0000 0019 in2lbf Water at 400 F 1500 lbfin2 compressed liquid αP 1 0018 45 0019 25 0017 85 450 350 0000 76 F1 βT 1 0018 45 00184 00185 2000 1000 0000 0054 in2lbf Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12155E Use the CATT3 software to solve the previous problem The benefit of the software to solve for the partial derivatives is that we can narrow the interval over which we determine the slope Water at 50 F 500 lbfin2 compressed liquid αP 1 vv TP 1 vv TP βT 1 vv PT 1 vv P T Estimate by finite difference using values at 40 F 50 F and 60 F αP 1 001600 001601 001599 60 40 0000 0625 F1 Using values at saturation 500 psia and 1000 psia βT 1 00160 001597 001602 1000 0178 0000 0031 psi1 Water at 400 F 1500 lbfin2 compressed liquid Estimate by finite difference using values at 350 F 400 F and 450 F αP 1 001 849 0019 26 0017 86 450 350 0000 757 F1 Using values at 1000 psia 1500 psia and 2000 psia βT 1 0018 49 001 844 001 855 2000 1000 0000 0059 psi1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12156E A cylinder fitted with a piston contains liquid methanol at 70 F 15 lbfin2 and volume 1 ft3 The piston is moved compressing the methanol to 3000 lbfin2 at constant temperature Calculate the work required for this process The isothermal compressibility of liquid methanol at 70 F is 83 106 in2lbf 1w2 1 2 Pdv Pv PT dPT 1 2 vβT PdPT For v const βT const 1w2 2 vβTP 2 2 P 2 1 For liquid methanol from Table F3 ρ 491 lbmft3 V1 10 ft3 m 10 491 491 lbm 1W2 10 83 106 2 30002 152144 53784 ft lbf 69 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12157E Sound waves propagate through a media as pressure waves that causes the media to go through isentropic compression and expansion processes The speed of sound c is defined by c2 Pρs and it can be related to the adiabatic compressibility which for liquid ethanol at 70 F is 64 106 in2lbf Find the speed of sound at this temperature c2 P ρs v2P vs 1 1 vv Ps ρ 1 βsρ From Table F3 for ethanol ρ 489 lbmft3 c 32174144 64106489 12 3848 fts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12158E Consider the speed of sound as defined in Eq 1242 Calculate the speed of sound for liquid water at 50 F 250 lbfin2 and for water vapor at 400 F 80 lbfin2 using the steam tables From Eq 1242 c2 P ρs v2P v s Liquid water at 50 F 250 lbfin2 Assume P vs P v T Using saturated liquid at 50 F and compressed liquid at 50 F 500 lbfin2 c2 00160240015998 2 2 00159980016024 50001814432174 2283210 6 c 4778 fts Superheated vapor water at 400 F 80 lbfin2 v 6217 ft3lbm s 16790 Btulbm R At P 60 lbfin2 s 16790 T 3438 F v 77471 ft3lbm At P 100 lbfin2 s 16790 T 4462 F v 52394 ft3lbm c2 62172 1006014432174 5239477471 2856 106 c 1690 fts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12159E Liquid methanol at 77 F has an adiabatic compressibility of 71 106 in2lbf What is the speed of sound If it is compressed from 15 psia to 1500 psia in an insulated pistoncylinder what is the specific work From Eq1241 and Eq1242 and the density from table F3 c2 P ρs v2P vs 1 βsρ 1 71 106 491 144 32174 ft2s 2 13290 106 ft2s2 c 3645 fts The specific work becomes w P dv P βsv dP βsv P dP βs v 1 2 P dP βs v 05 P2 2 P2 1 71 106 in2lbf 05 491 ft3lbm 15002 152 psi2 0163 ftlbflbm ftin2 234 ftlbflbm 003 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Equations of State Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12160E Use Table F9 to find the compressibility of R410A at 140 F and a saturated liquid b saturated vapor and c 400 psia Table F1 R 154536 72585 2129 lbfftlbmR a Table F91 P 556488 psia v 001966 ft3lbm Z Pv RT 556488 001966 144 2129 5997 01234 b Table F91 P 556488 psia v 00796 ft3lbm Z Pv RT 556488 00796 144 2129 5997 05 c Table F92 P 400 psia v 01574 ft3lbm Z Pv RT 400 01574 144 2129 5997 071 The R410A is not an ideal gas at any of these states Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12161E Calculate the difference in internal energy of the idealgas value and the realgas value for carbon dioxide at the state 70 F 150 lbfin2 as determined using the virial EOS At this state B 2036 ft3lb mol TdBdT 4236 ft3lb mol Solution CO2 at 70 F 150 lbfin 2 virial P RT v BRT v2 P Tv R v BR v2 RT v2dB dT uu v TP Tv Pdv v RT2 v2 dB dT dv RT2 v dB dT B 2036 ft3lbmol TdB dT 4236 ft3lbmol Solution of virial equation quadratic formula v 1 2 R T P 1 14BPR T But R T P 15455297 150144 378883 v 053788831 142036378883 357294 ft3lbmol Using the minussign root of the quadratic formula results in a compressibility factor 05 which is not consistent with such a truncated equation of state u u RT v T dB dT 198595297 357294 4236 1239 Btulbmol Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Generalized Charts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12162E How low should the pressure be so that nitrous oxide N2O gas at 5016 R can be treated as an ideal gas with 5 accuracy or better From Table F1 Tc 5573 R Pc 1050 psi Tr1 5016 5573 09 Look in Fig D1 following the curve Tr1 09 to the point where Z 095 Pr1 0125 so P 0125 1050 psi 131 psi Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12163E Nitrous oxide N2O at 5016 R is at a pressure so that it can be in a twophase state Find the generalized enthalpy departure for the two saturated states of liquid and vapor From Table F1 Tc 5573 R Pc 1050 psi Tr1 5016 5573 09 From Fig D2 Saturated liquid h hfRTc 408 Saturated vapor h hgRTc 0875 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12164E Find the heat of evaporation hfg for R134a at 30 F from the generalized charts and compare to the value in Table F10 From Table F1 Tc 6736 R Pc 589 psia Tr1 4897 6736 0727 R 19858910203 001946 BtulbmR From Fig D2 Saturated liquid h hfRTc 47 Saturated vapor h hgRTc 03 hfg hg hf RTc 03 47 44 RTc 44 001946 6736 5767 Btulbm Table F 10 hfg 8563 Btulbm The approximation is not very good and can be improved by using the accentric factor Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12165E A 7ft3 rigid tank contains propane at 1300 lbfin2 540 F The propane is then allowed to cool to 120 F as heat is transferred with the surroundings Determine the quality at the final state and the mass of liquid in the tank using the generalized compressibility charts Propane C3H8 V 70 ft3 P1 1300 lbfin2 T1 540 F 1000 R cool to T2 120 F 580 R From Table F1 TC 6656 R PC 616 lbfin2 Pr1 1300 616 2110 Tr1 1000 6656 1502 From D1 Z1 083 v2 v1 Z1RT1 P1 083 3504 1000 1300 144 01554 ft3lbm From D1 at Tr2 0871 saturated PG2 043 616 265 lbfin2 vG2 0715 3504 580 265 144 03808 ft3lbm vF2 0075 3504 580 265 144 00399 ft3lbm 01554 00399 x20378100399 x2 03388 mLIQ 2 1 03388 70 01554 298 lbm These tanks contain liquid propane Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12166E A rigid tank contains 5 lbm of ethylene at 450 lbfin2 90 F It is cooled until the ethylene reaches the saturated vapor curve What is the final temperature 2 4 C H T v 1 2 C2H4 m 5 lbm P1 450 lbfin2 T1 90 F 2497 R Pr1 450 731 0616 Tr1 5497 5083 1082 Fig D1 Z1 082 Pr2 Pr1 Z2Tr2 Z1Tr1 0616 ZG2Tr2 082 1082 06943 ZG2T r2 Trial error to match a saturated Pr2 Tr2 and the ZG2 so Eq is satisfied Guess a Tr2 and find the rest and compare with computed Pr2 from Eq Tr2 ZG2 Pr2 P r2 CALC 0871 0715 043 0432 OK T2 4427 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12167E A pistoncylinder contains 10 lbm of butane gas at 900 R 750 lbfin2 The butane expands in a reversible polytropic process with polytropic exponent n 105 until the final pressure is 450 lbfin2 Determine the final temperature and the work done during the process C4H10 m 10 lbm T1 900 R P1 750 lbfin2 Rev polytropic process n 105 P2 450 lbfin 2 Tr1 900 7654 1176 Pr1 750 551 1361 From Fig D1 Z1 067 V1 mZRTP 10 067 2658 900 750 144 1484 ft3 P1V n 1 P2V n 2 V2 1484 750 450 1 105 2414 ft 3 Z2Tr2 P2V2 mRTC 450 144 2414 10 2658 7654 07688 at Pr2 450551 0817 Trial error Tr2 1068 Z2 072 T2 8174 R 1W2 1 2 PdV 1n P2V2 P1V1 450 2414 750 1484 1 105 144 778 988 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12168E Calculate the heat transfer during the process described in Problem 12167E From solution 12167 V1 1473 ft3 V2 2396 ft3 1W2 988 Btu Tr1 1176 Pr1 1361 Tr2 1068 Pr2 0817 T2 8174 R From D1 hh RTC1 136 RTC hh2 095 h 2 h 1 0415 8174 900 343 Btulbm h2 h1 343 26587654 778 095 136 236 Btulbm U2 U1 mh2 h1 P2V2 P1V 1 10236 4501442414 778 7501441484 778 2311 Btu 1Q2 U2 U1 1W2 1323 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12169E The new refrigerant R152a is used in a refrigerator with an evaporator temperature of 10 F and a condensing temperature of 90 F What are the high and low pressures in this cycle Since we do not have the printed tables for R152a we will use generalized charts The critical properties are Tc 6955 R Pc 656 psi Tr1 TTc 45967 10 6955 0646 Fig D1 PG T1 Pr1 sat Pc 006 656 394 psi Tr2 TTc 45967 90 6955 079 Fig D1 PG T2 Pr2 sat Pc 025 656 164 psi Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12170E A cylinder contains ethylene C2H4 at 2226 lbfin2 8 F It is now compressed in a reversible isobaric constant P process to saturated liquid Find the specific work and heat transfer Ethylene C2H4 P1 2226 lbfin2 P2 T1 8 F 4677 R State 2 saturated liquid x2 00 R 5507 ft lbflbm R 0070 78 Btulbm R Tr1 4677 5083 0920 Pr1 Pr2 2226 731 0305 From D1 and D2 Z1 085 RTC hh1 040 v1 P1 Z1RT1 085 5507 4677 2226 144 0683 h 1h1 0070 78 5083 040 144 From D1 and D2 T2 0822 5083 4178 R Z2 005 RTC hh2 442 v2 P2 Z2RT2 005 5507 4178 2226 144 0035 89 ft3lbm h 2h2 0070 78 5083 442 1590 Btulbm h 2h 1 CP0T2T1 04114178 4677 205 Btulbm 1w2 1 2 Pdv Pv2v1 22260035 89 0683 144 778 267 Btulbm 1q2 u2 u1 1w2 h2h1 1590 205 144 1651 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12171E Saturated vapor R410A at 80 F is throttled to 30 lbfin2 in a steady flow process Find the exit temperature neglecting kinetic energy using the generalized charts Fig D2 and repeat using Table F9 R410A throttling process R 19858972585 002736 BtulbmR T1 80 F x1 100 P2 30 lbfin2 Energy Eq h2h1 h2h 2 h 2h 1 h 1h1 0 Generalized charts Tr1 5397 6201 087 From D2 h 1h1 002736 6201 075 1272 Btulbm To get CP0 use h values from Table F9 at low pressure 5 psia CP0 137613375 8060 01925 Btulbm R Substituting into energy Eq h2h 2 01925 T2 30 1272 0 at Pr2 30 711 0042 Assume T2 10 F 4697 R Tr2 4697 6201 0757 h 2h2 002736 6201 007 119 Substituting 119 0192510 80 1272 1945 Guess T2 20 F 4797 R Tr2 4797 6201 0774 h 2h2 002736 6201 007 119 Substituting 119 0192520 80 1272 002 nearly OK T2 20 F maybe reestimate CP0 at avg T R410A tables F9 T1 80 F x1 10 h1 12214 Btulbm h2 h1 12214 Btulbm P2 30 lbfin2 T2 815 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12172E Find the heat of evaporation hfg for isobutane Tc 7358 R Pc 5294 psia M 58124 at 547 F from the generalized charts and compare to the values in the CATT3 computerized tables To read the charts we need the reduced temperature Tr1 TTc 547 45977358 070 R 198589 58124 003417 BtulbmR From Fig D2 h hg 02 RTc h hf 485 RT c hfg hg hf h hg h hf 02 485 RT c 465 RTc 465 003417 BtulbmR 7358 R 1169 kJkg CATT3 hfg 1101 1585 1475 Btulbm The generalized charts are not super accurate some improvement can be done using the accentric factor Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12173E A 10ft3 tank contains propane at 90 F 90 quality The tank is heated to 600 F Calculate the heat transfer during the process V P 1 2 V 10 ft3 T1 90 F 5497 R x1 090 Heat to T2 600 F 10597 R M 44094 TC 6656 R PC 616 lbfin 2 R 3504 CP0 0407 Btulbm R Tr1 0826 Figs D1 and D2 Z1 01 0053 09 078 0707 h 1h1 RTc 01 44 09 055 0935 P SAT r 031 P SAT 1 031 616 191 lbfin2 m PV ZRT 191 144 10 0707 3504 5497 202 lbm Pr2 616 144 10 202 Z2 3504 10597 1183 Z2 at Tr2 1592 Trial error Pr2 079 and P2 490 lbfin2 Z2 094 and h 2h2 RTc 036 h 2h 1 040760090 2076 Btulbm h 1h1 0935 3504 6659 778 280 h 2h2 036 3504 6659 778 108 From the energy equation Q12 mh2h1 P2P1V 202 108 2076 280 490 191 144 10 778 4541 553 3988 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12174E Carbon dioxide collected from a fermentation process at 40 F 15 lbfin2 should be brought to 438 R 590 lbfin2 in a steady flow process Find the minimum amount of work required and the heat transfer What devices are needed to accomplish this change of state R 351 778 0045 12 Btulbm R Tri 500 5474 0913 Pri 15 1070 0014 From D2 and D3 hh RTC 002 ss R 001 R Tre 438 5474 080 Pre 590 1070 0551 From D2 and D3 h ehe 450 RTc s ese 470 R hihe h ihi h ih e h ehe 0045 12 5474 002 0203500 438 0045 12 5474 450 1232 Btulbm sise s isi s is e s ese 0045 12001 0203 ln 500 438 0045 12 ln 15 590 0045 12470 04042 Btulbm R wrev hihe T0sise 1232 50004042 784 Btulbm qrev hehi wrev 1232 789 2021 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12175E A cylinder contains ethylene C2H4 at 2226 lbfin2 8 F It is now compressed isothermally in a reversible process to 742 lbfin2 Find the specific work and heat transfer Ethylene C2H4 R 5507 ft lbflbm R 0070 78 Btulbm R State 1 P1 2226 lbfin2 T2 T1 8 F 4677 R State 2 P2 742 lbfin2 Tr2 Tr1 4677 5083 0920 Pr1 2226 731 0305 From D1 D2 and D3 Z1 085 RTC hh1 040 h 1h1 0070 78 5083 040 144 Btulbm s 1s1 0070 78 030 00212 Btulbm R Pr2 742 731 1015 comp liquid From D1 D2 and D3 Z2 017 h 2h2 0070 78 5083 40 1439 s 2s2 0070 78 36 02548 h 2h 1 0 s 2s 1 0 0070 78 ln 742 2226 00852 1q2 Ts2s1 467702548 00852 00212 1491 Btulbm h2h1 1439 0 144 1295 u2u1 h2h1 RTZ2Z1 1295 0070 78 4677 017 085 1070 1w2 1q2 u2u1 1491 1070 421 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12176E A geothermal power plant on the Raft river uses isobutane as the working fluid The fluid enters the reversible adiabatic turbine at 320 F 805 lbfin2 and the condenser exit condition is saturated liquid at 91 F Isobutane has the properties Tc 73465 R Pc 537 lbfin2 Cpo 03974 Btulbm R and ratio of specific heats k 1094 with a molecular weight as 58124 Find the specific turbine work and the specific pump work R 2658 ft lbflbm R 0034 166 Btulbm R Turbine inlet Tr1 7797 7347 1061 Pr1 805 537 1499 Condenser exit T3 91 F x3 00 Tr3 5507 7347 075 From D1 Pr3 0165 Z3 00275 P2 P3 0165 537 886 lbfin2 From D2 and D3 h 1h1 0034 166 7347 285 715 Btulbm s 1s1 0034 166 215 00735 Btulbm R s 2s 1 03974 ln 5507 7797 0034 166 ln 886 805 00628 Btulbm R s 2s2 s 2sF2 x2sFG2 0034 166 612 x2 0034 166612 029 02090 x2 01992 s2s1 0 02090 x2 0199200628 00735 x2 09955 h 2h 1 CP0T2T1 0397455077797 910 Btulbm From D2 h 2h2 h 2hF2 x2hFG2 0034 1667347469 09955469 032 1177 09955 1097 85 Btulbm Turbine wT h1h2 715 910 85 280 Btulbm Pump vF3 P3 ZF3RT3 00275 2658 5507 886 144 0031 55 ft3lbm wP 3 4 vdP vF3P4 P3 0031 55805886 144 778 42 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12177E A line with a steady supply of octane C8H18 is at 750 F 440 lbfin2 What is your best estimate for the availability in an steady flow setup where changes in potential and kinetic energies may be neglected Availability of Octane at Ti 750 F Pi 440 lbfin2 R 1353 ft lbflbmR 0017 39 BtulbmR Pri 440 361 1219 Tri 12097 10238 1182 From D2 and D3 h 1h1 0017 39 10238 115 205 Btulbm s 1s1 0017 39 071 00123 Btulbm R Exit state in equilibrium with the surroundings Assume T0 77 F P0 147 lbfin2 Tr0 5367 10238 0524 Pr0 147 361 0041 From D2 and D3 h 0h0 RTC 541 963 and s 0s0 R 1038 01805 h ih 0 040912097 5367 2753 Btulbm s is 0 0409 ln 12097 5367 0017 39 ln 440 147 02733 Btulbm R hih0 205 2753 963 3511 Btulbm sis0 00123 02733 01805 04415 Btulbm R ψi wrev hih0 T0sis0 3511 5367 04415 1141 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12178E A distributor of bottled propane C3H8 needs to bring propane from 630 R 147 lbfin2 to saturated liquid at 520 R in a steady flow process If this should be accomplished in a reversible setup given the surroundings at 540 R find the ratio of the volume flow rates VinVout the heat transfer and the work involved in the process R 3504778 0045 04 Btulbm R Tri 630 6656 0947 Pri 147 616 0024 From D1 D2 and D3 Zi 099 h ihi 0045 04 6656 003 09 Btulbm s isi 0045 04 002 00009 Btulbm R Tre 5206656 0781 From D1 D2 and D3 Pre 021 Pe 021 616 129 lbfin2 Ze 0035 h ehe 0045 04 6656 458 1373 Btulbm s ese 0045 04 572 02576 Btulbm R h eh i 0407 520 630 448 Btulbm s es i 0407 ln 520 630 0045 04 ln 132 147 01770 Btulbm R hehi 1373 448 09 1812 Btulbm sesi 02576 01759 00009 04326 Btulbm R V in V out ZiTiPi ZeTePe 099 0035 630 520 129 147 3007 wrev hi he T0si se 1812 54004326 524 Btulbm qrev he hi wrev 1812 524 2336 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Mixtures Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12179E A 4 lbm mixture of 50 argon and 50 nitrogen by mole is in a tank at 300 psia 320 R How large is the volume using a model of a ideal gas and b Kays rule with generalized compressibility charts a Ideal gas mixture Eq115 Mmix yi Mi 05 39948 05 28013 33981 R 154536 33981 45477 lbfftlbmR V mRT P 4 45477 320 300 144 lbfft psi in2ft2 1347 ft 3 b Kays rule Eq1284 Pc mix 05 706 05 492 599 psia Tc mix 05 2714 05 2272 2493 R Reduced properties Pr 300 599 050 Tr 320 2493 1284 Fig D1 Z 092 V Z mRT MmixP Z VID gas 092 1347 124 ft3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12180E R410A is a 11 mass ratio mixture of R32 and R125 Find the specific volume at 80 F 200 psia using Kays rule and the generalized charts and compare to Table F9 Kays rule Eq1284 Pc mix 05 838 05 525 6815 psia Tc mix 05 6323 05 6106 62145 R Reduced properties Pr 200 6815 0293 Tr 53967 62145 0868 Table F1 R 154536 72585 2129 lbfftlbmR Fig D1 Z 082 v ZRT P 082 2129 53967 200 144 lbfftlbm psi in2ft2 0327 ft3lbm Table F9 v 03174 ft3lbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12181E The R410A in Problem 12180E flows through a heat exchanger and exits at 280 F 200 psia Find the specific heat transfer using Kays rule and the generalized charts and compare this to solution found using Table F9 Kays rule Eq1284 Pc mix 05 838 05 525 6815 psia Tc mix 05 6323 05 6106 62145 R Table F1 R 154536 72585 2129 lbfftlbmR 002736 BtulbmR CP mix ci CP i 05 0196 05 0189 01925 BtulbmR Reduced properties 1 Pr1 200 6815 0293 Tr1 53967 62145 0868 Fig D1 h 1 h1 045 RTc 045 002736 62145 765 Btulbm Reduced properties 2 Pr2 200 6815 0293 Tr2 73967 62145 119 Fig D1 h 2 h2 025 RTc 025 002736 62145 425 Btulbm The energy equation gives 1q2 h2 h1 h2 h 2 h 2 h 1 h 1 h1 425 01925 280 80 765 419 Btulbm mix Table F92 q h2 h1 17626 12634 4992 Btulbm The main difference is in the value of specific heat about 025 BtulbmR at the avg T whereas it is 01925 BtulbmR at 77 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problem Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 12182E A new compound is used in an ideal Rankine cycle where saturated vapor at 400 F enters the turbine and saturated liquid at 70 F exits the condenser The only properties known for this compound are a molecular mass of 80 lbmlbmol an ideal gas heat capacity of Cpo 020 BtulbmR and Tc 900 R Pc 750 psia Find the specific work input to the pump and the cycle thermal efficiency using the generalized charts Turbine Cond Ht Exch P 3 1 4 2 Q H W T W P T1 400 F 860 R x1 10 T3 70 F 530 R x3 00 Properties known M 80 CPO 02 Btulbm R TC 900 R PC 750 lbfin2 Tr1 860 900 0956 Tr3 530 900 0589 From D1 Pr1 076 P1 076 750 570 lbfin2 P4 Pr3 0025 P3 19 lbfin2 P2 Zf3 00045 vf3 Zf3RT3P3 00045 1545 530 19 144 80 00168 ft3lbm wP 3 4 vdP vf3P4 P3 00168 570 19 144 778 171 Btulbm qH h4 h1 but h3 h4 wP qH h1 h3 wP From D2 h 1 h1 1985980 900 134 300 Btulbm h 3 h3 1985980 900 52 1161 Btulbm h 1 h 3 CP0T1T3 0240070 660 Btulbm h1 h3 300 660 1161 1521 Btulbm qH 1521 171 1504 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Turbine s2 s1 0 s 2 s2 s 2 s 1 s 1 s1 From D3 s 1 s1 1985980 106 00263 Btulbm R s 2 s 1 020 ln 530 860 0024 82 ln 19 570 00124 Btulbm R Substituting s 2s2 00124 00263 00139 s 2sf2 x2s fg2 00139 0024 82 877 x2 0024 82 877 0075 x2 09444 h 2h2 h 2hf2 x2h fg2 From D2 hfg2 0024 82 900 52 007 1146 Btulbm h 2h2 1161 09444 1146 79 Btulbm wT h1h2 300 660 79 439 Btulbm ηTH qH wNET 439 17 1504 0281 Updated June 2013 SOLUTION MANUAL CHAPTER 13 Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CONTENT CHAPTER 13 SUBSECTION PROB NO InText concept questions ak ConceptStudy Guide Problems 119 Fuels and the Combustion Process 2035 Energy Equation Enthalpy of Formation 3653 Enthalpy of Combustion and heating Value 5480 Adiabatic Flame Temperature 81101 Second Law for the Combustion Process 102115 Problems Involving Generalized Charts or Real Mixtures 116121 Fuel cells 122133 Combustion Applications and Efficiency 134146 Review Problems 147162 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful InText Concept Questions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13a How many kmol of air are needed to burn 1 kmol of carbon Combustion Eq C O2 1 CO2 One kmol of O2 is required to form one kmol CO2 Since air is 21 O2 this means 476 kmol of air Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13b If I burn 1 kmol of hydrogen H2 with 6 kmol of air what is the airfuel ratio on a mole basis and what is the percent theoretical air Combustion Eq stoichiometric H2 νO2O2 376 N2 1 H2O 376 νO2 N 2 νO2 05 AFS νO2 1 376 1 238 Six kmol of air is 126 O2 474 N2 The AF mole ratio is 6 so the percent theoretical air is Theoretical air AFac AFS 100 6 238 100 252 Hydrogen flames are nearly colorless but even small traces of other substances can give color Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13c For the 110 theoretical air in Eq138 what is the equivalence ratio Is that mixture rich or lean 110 Theoretical air means also AF 11 AFS so from the definition in Eq136 Φ AFS AF 1 110 0909 a lean mixture 13d In most cases combustion products are exhausted above the dew point Why If any water in the products condenses it will be acidic due to the other gases in the products There are always minute amounts of unburned or partially burned fuel and intermediate species in the products that can combine with water and create a very corrosive mixture Corrosion on a metal surface Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13e How is a fuel enthalpy of combustion connected to its enthalpy of formation The enthalpy of combustion of a fuel is the difference in enthalpy of the products and reactants for the combustion involving the fuel these enthalpies include the various enthalpies of formation At reference condition H RP H P H R H P h 0 f fuel vap or liq The value is negative so talk about the heating value as HV H RP Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13f What is the higher and lower heating value HHV LHV of nbutane The heating value is the negative of the enthalpy of combustion From Table 133 the HHV of gaseous nbutane at 25C is HHV 49 500 kJkg and the corresponding LHV is LHV 45 714 kJkg Notice the table is on a mass basis per kg fuel A butane flame Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 15g What is the value of hfg for noctane This can be obtained from two places From Table A10 we get hfg h0 f vap h0 f liq M 208 600 250 105 114232 363 kJkg The hfg of a fuel listed in Table 133 is the difference between the first two columns in the table or the third and fourth For noctane this value is hfg 47 893 48 256 363 kJkg To see this remember H RP H P H R H P h 0 f fuel vap or liq so when we take the difference between fuel as gas and fuel as liquid all other terms will cancel out leaving hfg for the fuel when numbers are on mass basis Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13h What happens to the adiabatic flame temperature if I burn rich or lean The higher the percent theoretical air used in combustion the leaner the larger the number of moles of products especially nitrogen which decreases the adiabatic flame temperature Burning rich causes incomplete combustion however with a smaller release of energy Experimentally the highest temperature is reached for slightly rich Heavy molecules show up as yellow Oxygen diffuses in from the air and the fuel evaporates from the wick As air mixes in the flame cools 13i Is the irreversibility in a combustion process significant Why is that A combustion process is highly irreversible with a large increase in entropy It takes place at a rapid rate due to large driving forces and results in stable products of combustion that have little or no tendency to return to their former constituents and states C Borgnakke Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13j If the airfuel ratio is larger than stoichiometric is it more or less reversible Less reversible more irreversible The excess oxidizer air is being heated up Q over a finite temperature difference is an irreversible process The same is true for AF smaller than one where the excess fuel is heated up Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13k What makes the fuel cell attractive from a powergenerating point of view Fuel cells are attractive for power generation because their direct output is electrical energy They also have a much higher power density as power per unit volume or power per unit mass and thus can be used in mobile applications Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful ConceptStudy Guide Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 131 Is mass conserved in combustion Is the number of moles constant Yes mass is conserved as we do not consider the quantum physic effect of mass and energy conversion Moles are not conserved what is conserved are the atoms so each atom in the reactants must be found in the products 132 Does all combustion take place with air Most combustion takes place with air as the oxidizer special cases do occur where a different oxidizer is used To reach very high temperatures a gas welder uses a bottle of pure oxygen and one with the fuel instead of air and then the flame can cut through steel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 133 Why would I sometimes need AF on a mole basis on a mass basis If you want to meter measure the fuel and air flows it can be done as a volume flowrate which is proportional to moles PV nRT in which case concentrations on a mole basis are needed The fuel and air flows can also be measured with a method that measures mass flow rate m or if you are filling up tanks to store the fuel and oxidizer as in a rocket in both cases the concentrations on a mass basis are needed 134 Why is there no significant difference between the number of moles of reactants versus products in combustion of hydrocarbon fuels with air In most hydrocarbon fuels there are approximately twice as many hydrogen atoms as carbon atoms so the numbers of moles of CO2 and H2O in the products are roughly equal the total of which is not too different from the O2 required in the reactants The number of excess O2 is the same in reactants and products The total number of moles is dominated by the N2 in each especially with excess air Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 135 Why are products measured on a dry basis Combustion products have traditionally been measured by passing the gas mixture through a series of solutions that selectively absorb the components onebyone and measuring the resulting gas volume decreases The water component is condensed out in these processes leaving the others that is a dry basis Other and newer instruments measure the concentrations by optical means and these are sensitive to moisture content which can corrode the surfaces and destroy the sensors If the water stays in the mixture it typically has to stay hot to prevent condensation at undesirable locations where that would alter the concentrations of the remaining gas mixture components Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 136 What is the dew point of hydrogen burned with stoichiometric pure oxygen air For H2 burned with pure oxygen H2 05 O2 1 H2O with the only product as water so the dewpoint at 100 kPa pressure is 996C For H2 burned with stoichiometric air H2 05 O2 376 N2 1 H2O 188 N2 the product mixture is water and nitrogen The partial pressure of the water at a total pressure of 100 kPa is Pv Ptot yv 100 1 1 188 347 kPa corresponding to a dewpoint of 723C A hydrogen flame is nearly colorless If other species are present they can give rise to some color of the flame Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 137 How does the dew point change as equivalence ratio goes from 09 to 1 to 11 For a given amount of water in the products the smaller the total number of moles of other gases is as Φ increases the higher the partial pressure of the water and therefore the dewpoint temperature As Φ becomes greater than 10 there will be incomplete combustion and the resulting composition will be affected to have some unburned fuel and therefore relative less water The relative maximum amount of water is then at a stoichiometric mixture Φ 1 and this is also maximum dew point temperature 138 Why does combustion contribute to global warming Any combustion of a hydrocarbon fuel produces carbon dioxide which in the atmosphere is a contributing factor to global warming Carbon dioxide absorbs radiation over a wide spectrum and thus heats the atmosphere This is not just manmade but nature has forest fires and volcanic action that liberate gases into the atmosphere Natural gas mainly methane CH4 has the highest hydrogen to carbon ratio of the hydrocarbons we find on earth very little hydrogen can be found Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 139 What is the enthalpy of formation for oxygen as O2 If O For CO2 The enthalpy of formation of O2 is zero by choice of the reference base Relative to this base the value for the monatomic form O is h f O 249 170 kJkmol Table A9 and the value for CO2 is h f CO2 393 522 kJkmol Table A9 or A10 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1310 If the nitrogen content of air could be lowered will the adiabatic flame temperature increase or decrease Since that would reduce the number of nitrogen molecules in the products it will increase the adiabatic flame temperature more energy per molecule Energy Eq HP HV HR given HP νCO2 h CO2 νH2O h H2O νN2 h N2 at Tad so a lower νN2 means the hs must be larger 1311 Does the enthalpy of combustion depend on AF NO HRP HP HR depends on T P in general Assuming ideal gas it does not depend on P It is assumed the number is scaled to 1 kmol or 1 kg of fuel burned Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1312 Why do some fuels not have entries for liquid fuel in Table 133 Those fuels cannot exist as liquids at 25C above their critical temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1313 Is a heating value a fixed number for a fuel Yes The heating value is the negative of the enthalpy of combustion and is only useful if it is a well defined number It is fixed but there are normally 4 values quoted as the higher water liquid and lower water vapor heating value for the fuel as a liquid or vapor if applicable It should also be clear whether it is the enthalpies or the internal energies that are used ie do you have a flow or a non flow situation So these qualifiers must accompany the number for it to be unique Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1314 Is an adiabatic flame temperature a fixed number for a fuel Qualified Yes It is a single number if we assume reactants are supplied at the reference conditions in a stoichiometric ratio and we know the phase of the fuel liquid or gas and the type of oxidizer air or pure oxygen Also you need to know if the combustion takes place at constant pressure the most common or it may take place at constant volume unusual If anyone of these conditions is changed the adiabatic flame temperature is changed This is less fixed than the heating value more variables can change the adiabatic flame temperature say adding an inert gas will lower the flame temperature but keep the heating value the same Chemical equilibrium reactions and incomplete combustion makes the real flame temperature lower than the theoretical adiabatic flame temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1315 Does it make a difference for the enthalpy of combustion whether I burn with pure oxygen or air What about the adiabatic flame temperature No difference in the enthalpy of combustion the nitrogen in the air is the same in the reactants and products and its enthalpy cancels out nitrogen and oxygen formation enthalpy are both zero The adiabatic flame temperature is much lower for combustion with air because a significant part of the energy release from the combustion goes into heating the nitrogen as well as the other products to the flame temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1316 A welder uses a bottle with acetylene and a bottle with oxygen Why does he use oxygen instead of air The oxygen in the air comes with nitrogen that is also being heated up to the flame temperature The heating value is being shared among all the product components so when they increase having the nitrogen also the adiabatic flame temperature drops correspondingly Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1317 Some gaswelding is done using bottles of fuel oxygen and argon Why do you think argon is used Argon is used to shield the welding spot from the oxygen in the surrounding air If oxygen is on the surface of the liquid metal it may form metaloxides creating a weak location in the weld 1318 Is combustion a reversible process NO It is a highly irreversible process 1319 Is combustion with more than 100 theoretical air more or less reversible It is less reversible Any deviation from a stoichiometric mixture lean or rich will increase the irreversibility You may also think of the adiabatic flame temperature which is highest for a stoichiometric mixture like adding the heating value at that T So with any other mixture the adiabatic flame temperature is lower so the heating value is added at a lower T a more irreversible process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Fuels and the Combustion Process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1320 In a picnic grill gaseous propane is fed to a burner together with stoichiometric air Find the airfuel ratio on a mass basis and the total amount of reactant mass for 1 kg propane burned The reaction equation for stoichiometric mixture is C3H8 νO2 O2 376 N2 a H2O b CO2 c N2 C balance 3 b H balance 8 2a a 4 O balance 2 νO2 a 2b 4 2 3 νO2 5 N2 balance 376 νO2 c 188 The reaction is C3H8 5 O2 376 N2 4 H2O 3 CO2 188 N2 AFmol 5 188 1 238 AFmass AFmol Mair Mfuel 238 2897 44097 1564 Total mass of reactants 1 1564 1664 kg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1321 A mixture of fuels is E85 which is 85 ethanol and 15 gasoline assume octane by mass Find the AF ratio on a mass basis for stoichiometric combustion Since the fuel mixture is specified on a mass basis we need to find the mole fractions for the combustion equation From Eq134 we get yethanol 08546069 08546069 015114232 093356 yoctane 1 yethanol 006644 The reaction equation is 093356 C2H5OH 006644 C8H18 νO2 O2 376 N2 νCO2 CO2 νH2OH2O 376 νO2 N 2 C balance 2 093356 8 006644 νCO2 239864 H balance 6 093356 18 006644 2νH2O 679728 νH2O 339864 O balance 093356 2 νO2 2 νCO2 νH2O 819592 νO2 363118 The AF ratio is AF 476 νO2 2897 093356 46069 006644 114232 9896 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1322 Calculate the theoretical airfuel ratio on a mass and mole basis for the combustion of ethanol C2H5OH Reaction Eq C2H5OH νO2O2 376N2 aCO2 bH2O cN 2 Do the atom balance Balance C 2 a Balance H 6 2b b 3 Balance O 1 2νO2 2a b 4 3 7 νO2 3 AFmol νO21 3761 3 476 1428 AFmass νO2MO2 νN2 MN2M Fuel 3 31999 1128 28013 46069 8943 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1323 Methane is burned with 200 theoretical air Find the composition and the dew point of the products The reaction equation for stoichiometric mixture is CH4 νO2 O2 376 N2 a H2O b CO2 c N2 C balance 1 b H balance 4 2a O balance 2 νO2 a 2b 2 2 1 νO2 2 N2 balance 376 νO2 c 752 200 theoretical air νO2 2 2 4 so now more O2 and N2 CH4 νO2 O2 376 N2 a H2O b CO2 c N2 d O 2 N2 balance 376 νO2 c 1504 Extra oxygen d 4 1 1 2 Products 2 H2O 1 CO2 1504 N2 2 O 2 Water vapor mole fraction yv 2 1 2 2 1504 00998 Partial water vapor pressure Pv yv Po 00998 101 998 kPa PgTdew Pv 998 kPa Tdew 458oC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1324 Natural gas B from Table 132 is burned with 20 excess air Determine the composition of the products The reaction equation stoichiometric and complete combustion with the fuel composition is 601 CH4 148 C2H6 134 C3H8 42 C4H10 75 N2 νO2 O2 376 N2 a H2O b CO2 c N2 C balance 601 2148 3134 442 b 1467 H balance 4601 6148 8134 1042 2a 4784 a 2392 O balance 2 νO2 a 2b 2392 21467 νO2 2663 N2 balance 75 376 νO2 c 10088 20 excess air νO2 122663 31956 so now more O2 and N2 Extra oxygen d 31956 2663 5326 c 75 37631956 1209 Products 2392 H2O 1467 CO2 1209 N2 5326 O 2 To the expert the color of the flame can tell about the composition It can tell about specific gases present if they have distinct color emission Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1325 A certain fuel oil has the composition C10H22 If this fuel is burned with 150 theoretical air what is the composition of the products of combustion C10H22 1φ νO2 O2 376 N2 a H2O b CO2 c N2 d O 2 Stoichiometric combustion φ 1 d 0 C balance b 10 H balance a 222 11 O balance 2 νO2 a 2b 11 20 31 νO2 155 Actual case 1φ 15 νO2 15 155 2325 H balance a 11 C balance b 10 N balance c 2325 376 8742 O2 balance d 2325 10 112 775 excess oxygen The combustion process is C10H22 2325 O2 376 N2 11 H2O 10 CO2 8742 N2 775 O 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1326 A Pennsylvania coal contains 742 C 51 H 67 O dry basis mass percent plus ash and small percentages of N and S This coal is fed into a gasifier along with oxygen and steam as shown in Fig P1326 The exiting product gas composition is measured on a mole basis to 399 CO 308 H2 114 CO2 164 H2O plus small percentages of CH4 N2 and H2S How many kilograms of coal are required to produce 100 kmol of product gas How much oxygen and steam are required Convert the mass concentrations to number of kmol per 100 kg coal C n 7421201 6178 H2 n 512016 2530 O2 n 6731999 0209 Now the combustion equation reads x6178 C 253 H2 0209 O2 y H2O z O2 in and 399 CO 308 H2 114 CO2 164 H2O out in 100 kmol of mix out Now we can do the atom balance to find x y z C balance 6178 x 399 114 x 8304 H2 balance 2538304 y 308 164 y 26191 O2 balance 0209 8304 26191 2 z 399 2 114 164 2 z 24719 Therefore for 100 kmol of mixture out require 8304 kg of coal 26191 kmol of steam 24719 kmol of oxygen Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1327 For complete stoichiometric combustion of gasoline C7H17 determine the fuel molecular weight the combustion products and the mass of carbon dioxide produced per kg of fuel burned Stoichiometric combustion C7H17 νO2 O2 376 N2 a H2O b CO2 c N2 C balance 7 b H balance 17 2a a 85 O balance 2 νO2 a 2b 85 14 225 νO2 1125 N balance c 376 νO2 376 1125 423 MFUEL 7 MC 17 MH 7 12011 17 1008 101213 mCO2 mFUEL 7 MCO2 MFUEL 7 4401 101213 3044 kg CO2 per kg fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1328 A sample of pine bark has the following ultimate analysis on a dry basis percent by mass 56 H 534 C 01 S 01 N 379 O and 29 ash This bark will be used as a fuel by burning it with 100 theoretical air in a furnace Determine the airfuel ratio on a mass basis Converting the Bark Analysis from a mass basis Substance S H2 C O2 N2 cM 0132 562 53412 37932 0128 kmol 100 kg coal 0003 280 445 1184 0004 Product SO2 H2O CO2 oxygen required 0003 140 445 Combustion requires 0003 140 445 5853 kmol O2 there is in the bark 1184 kmol O2 so the net from air is 4669 kmol O2 AF 4669 4669 376 2897 100 644 kg air kg bark Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1329 Liquid propane is burned with dry air A volumetric analysis of the products of combustion yields the following volume percent composition on a dry basis 86 CO2 06 CO 72 O2 and 836 N2 Determine the percent of theoretical air used in this combustion process a C3H8 b O2 c N2 86 CO2 06 CO d H2O 72 O2 836 N 2 C balance 3a 86 06 92 a 3067 H2 balance 4a d d 12267 N2 balance c 836 O2 balance b 86 06 2 12267 2 72 22234 AirFuel ratio 22234 836 3067 3451 Theoretical C3H8 5 O2 188 N2 3 CO2 4 H2O 188 N2 theo AF ratio 5 188 1 238 theoretical air 3451 238 100 145 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1330 The coal gasifier in an integrated gasification combined cycle IGCC power plant produces a gas mixture with the following volumetric percent composition Product CH4 H2 CO CO2 N2 H2O H2S NH3 vol 03 296 410 100 08 170 11 02 This gas is cooled to 40C 3 MPa and the H2S and NH3 are removed in water scrubbers Assuming that the resulting mixture which is sent to the combustors is saturated with water determine the mixture composition and the theoretical airfuel ratio in the combustors CH4 H2 CO CO2 N2 n 03 296 410 100 08 817 yH2O nv 817 nv where nv number of moles of water vapor Cool to 40C PG 7384 P 3000 kPa yH2O MAX 7384 3000 nv nv 817 nv 02016 a Mixture composition CH4 H2 CO CO2 N2 H2Ov 03 kmol 296 410 100 08 02016 819016 kmol from 100 kmol of the original gas mixture 03 CH4 06 O2 03 CO2 06 H2O 296 H2 148 O2 296 H2O 41 CO 205 O2 41 CO 2 Number of moles of O2 06 148 205 359 Number of moles of air 359 376 359 N2 AF 2897359 376359 0316 2962 4128 1044 0828 0201618 295 kg airkg fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1331 In a combustion process with decane C10H22 and air the dry product mole fractions are 8361 N2 491 O2 1056 CO2 and 092 CO Find the equivalence ratio and the percent theoretical air of the reactants x C10H22 1φ νO2 O2 376 N2 a H2O b CO2 c CO d N2 e O 2 Stoichiometric combustion φ 1 c 0 e 0 C balance b 10x H balance a 22x2 11x O balance 2 νO2 a 2b 11x 20x 31x νO2 155x νN2 5828x AFs νO2 νN2x 7378 Actual combustion d 8361 N balance 1φ νO2 376 8361 1φ νO2 2224 C balance 10x 1056 092 1148 x 1148 AFac 1φ νO2 4761148 92215 φ FAac FAs AFs AFac 7378 92215 080 or φ 08 Percent theoretical air 100 1φ 125 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1332 The output gas mixture of a certain airblown coal gasifier has the composition of producer gas as listed in Table 132 Consider the combustion of this gas with 120 theoretical air at 100 kPa pressure Determine the dew point of the products and find how many kilograms of water will be condensed per kilogram of fuel if the products are cooled 10C below the dewpoint temperature 3 CH4 14 H2 509 N2 06 O2 27 CO 45 CO2 311 O2 1169 N2 345 CO2 20 H2O 52 O2 1678 N 2 Products yH2O yH2O MAX PG100 20 345 20 52 1678 PG 879 kPa TDEW PT 432C At T 332C PG 513 kPa yH2O max 513 100 nH2O nH2O345521678 nH2O 1122 mH2O LIQ 316 142 50928 0632 2728 4544 87818 00639 kgkg fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1333 Methanol CH3OH is burned with 200 theoretical air in an engine and the products are brought to 100 kPa 30C How much water is condensed per kilogram of fuel CH3OH νO2O2 376 N2 CO2 2 H2O 376 νO2N2 Stoichiometric νO2 S 15 νO2 AC 3 Actual products CO2 2 H2O 15 O2 1128 N2 Psat30C 4246 kPa yH2O 004246 νH2O 1 νH2O 15 1128 νH2O 0611 νH2O cond 2 0611 1389 MFu 32042 MH2O MFu 1389 18 32042 0781 kg H2O kg fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1334 Butane is burned with dry air at 40oC 100 kPa with AF 22 on a mass basis For complete combustion find the equivalence ratio theoretical air and the dew point of the products How much water kgkg fuel is condensed out if any when the products are cooled down to ambient temperature 40oC Solution C4H10 νO2O2 376 N2 4 CO2 5 H2O 376 νO2N2 Stoichiometric νO2 S 4 52 65 376 νO2 2444 AFS 6531999 376 2801358124 153574 Actual νO2ac AFac AFs νO2 S 22 153574 65 931 Theoretical air 22 153574 100 1433 Equivalence ratio Φ 114325 07 Actual products 4 CO2 5 H2O 281 O2 350 N2 The water partial pressure becomes Pv yv Ptot 5 4 5 281 350 100 1068 kPa Tdew 469oC Pg 40 7348 kPa yv max 7384 100 νH2O 4 νH2O 281 350 Solve for νH2O vap νH2O vap 3333 still vapor νH2O LIQ 5 3333 1667 is liquid mH2O LIQ mFuel 1667 18015 58124 0517 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1335 The hot exhaust gas from an internal combustion engine is analyzed and found to have the following percent composition on a volumetric basis at the engine exhaust manifold 10 CO2 2 CO 13 H2O 3 O2 and 72 N2 This gas is fed to an exhaust gas reactor and mixed with a certain amount of air to eliminate the carbon monoxide as shown in Fig P1335 It has been determined that a mole fraction of 10 oxygen in the mixture at state 3 will ensure that no CO remains What must the ratio of flows be entering the reactor Exhaust gas at state 1 CO2 10 H2O 13 CO 2 O2 3 N2 72 Exhaust gas at state 3 CO 0 O2 10 Air Exh gas gas out Reactor 1 2 3 Reaction equation for the carbon monoxide 002 CO x O2 376x N2 002 CO2 x001 O2 376x N2 At 3 νCO2 010 002 012 νH2O 013 νO2 x001 003 x 002 νN2 072 376x or nTOT 012 013 x 002 072 376x 099 476x yO2 010 x 002 099 476x x 0151 or air 2 Exh Gas 1 476x 1 0718 kmol air kmol Exh gas Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Energy Equation Enthalpy of Formation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1336 Acetylene gas C2H2 is burned with stoichiometric air in a torch The reactants are supplied at the reference conditions Po To The products come out from the flame at 2500 K after some heat loss by radiation Find the heat loss per kmol fuel The reaction equation is C2H2 25 O2 376 N2 2 CO2 1 H2O 94 N2 Energy Eq HP Qout HR Ho P HP Qout Ho R HP 2 h CO2 h H2O 94 h N2 2 121 9265 99 1305 94 74 3015 1 041 418 kJkmol Qout Ho R Ho P HP f fuel h h f H2O 2 h f CO2 HP Qout 226 731 241 826 2393 522 1 041 418 214 183 kJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1337 Hydrogen is burned with stoichiometric air in a steady flow process where the reactants are supplied at 100 kPa 298 K The products are cooled to 800 K in a heat exchanger Find the heat transfer per kmol hydrogen The combustion equation is H2 05 O2 376 N2 1 H2O 188 N2 The reactants are at 298 K HR 0 0 0 0 The products are at 800 K so water is gaseous and from A9 HP 1241 826 18 002 188 0 15 046 195 538 kJkmol fuel QCV HP HR 195 538 kJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1338 Natural gas methane is burned with stoichiometric air reactants supplied at reference Po To in a steady flow burner The products come out at 800 K if the burner should deliver 10 kW what is the need flowrate of natural gas in kgs The reaction equation for stoichiometric mixture is CH4 νO2 O2 376 N2 2 H2O 1 CO2 c N2 O balance 2 νO2 2 2 νO2 2 c 2 376 752 The products are cooled to 800 K so we do not consider condensation and the energy equation is Energy Eq HR Q HP H P HP H R Q Q H P HP H R 2 h f H2O h f CO2 HP h f fuel HP 2 h H2O h CO2 752 h N2 2 18 002 22 806 752 15 046 171 956 Q 2241 826 393 522 171 956 74 873 630 345 kJkmol fuel Q n Qout Qout m M so m Q MQout 10 kW 16043 kgkmol 630 345 kJkmol 00002545 kgs 0916 kgh Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1339 Butane gas and 200 theoretical air both at 25C enter a steady flow combustor The products of combustion exits at 1000 K Calculate the heat transfer from the combustor per kmol of butane burned C4H10 1φ νO2O2 376 N2 a CO2 b H2O c N2 d O2 First we need to find the stoichiometric air φ 1 d 0 C balance 4 a H balance 10 2b b 5 O balance 2νO2 2a b 8 5 13 νO2 65 Now we can do the actual air 1φ 2 νO2 2 65 13 N balance c 376 νO2 4888 O balance d 13 65 65 Energy Eq q HR HP Ho R Ho P HP Table A10 Ho R 126 200 0 0 126 200 kJkmol fuel Ho P 4 393 522 5241 826 0 0 2 783 218 kJkmol fuel The rest of the values are from Table A9 at 1000 K h CO2 33397 h N2 21463 h O2 22703 h H2O 26000 kJkmol HP 4 33 397 5 26 000 4888 21 463 65 22 703 1 460 269 kJkmol fuel From the energy equation we get q 126 200 2 783 218 1 460 269 1 196 749 kJkmol butane Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1340 One alternative to using petroleum or natural gas as fuels is ethanol C2H5OH which is commonly produced from grain by fermentation Consider a combustion process in which liquid ethanol is burned with 110 theoretical air in a steady flow process The reactants enter the combustion chamber at 25C and the products exit at 60C 100 kPa Calculate the heat transfer per kilomole of ethanol C2H5OH 11 3 O2 376 N2 2CO2 3H2O 03O2 12408N 2 Fuel h0 f 277 380 kJkmol for liquid from Table A10 Products at 60C 100 kPa check for condensation of water yH2O MIX 1994 100 nV MAX nV MAX 2 03 12408 nV MAX 366 3 No liquid HR 1277 380 0 0 277 380 kJkmol fuel HP 2393 522 1373 3241 826 1185 030 10395 124080 1021 1 493 241 kJkmol fuel QCV HP HR 1 215 860 kJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1341 Do the previous problem with the ethanol fuel delivered as a vapor One alternative to using petroleum or natural gas as fuels is ethanol C2H5OH which is commonly produced from grain by fermentation Consider a combustion process in which liquid ethanol is burned with 120 theoretical air in a steady flow process The reactants enter the combustion chamber at 25C and the products exit at 60C 100 kPa Calculate the heat transfer per kilomole of ethanol C2H5OH 11 3 O2 376 N2 2CO2 3H2O 03O2 12408N 2 Fuel h0 f 235 000 kJkmol for IG from Table A10 Products at 60C 100 kPa check for condensation of water yH2O MIX 1994 100 nV MAX nV MAX 2 03 12408 nV MAX 366 3 No liquid The mixture can then hold 366 mol of vapor 3 present No liquid HR 1235 000 0 0 235 000 kJkmol fuel HP 2393 522 1373 3241 826 1185 030 10395 124080 1021 1 493 241 kJkmol fuel QCV HP HR 1 258 241 kJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1342 As an alternative fuel consider liquid methanol burned with stoichiometric air both supplied at Po To in a constant pressure process exhausting the products at 900 K What is the heat transfer per kmol of fuel CH3OH 15 O2 376 N2 1 CO2 2 H2O 564 N2 Reactants at 25oC products are at 900 K 100 kPa CH3OH ho f 239 220 kJkmol from table A10 for the liquid state HR 1 h LIQ 239 220 kJkmol fuel HP 1393 522 28 030 2241 826 21 937 56418 223 702 492 kJkmol fuel Q HP HR 702 492 239 220 463 272 kJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1343 The combustion of heptane C7H16 takes place in a steady flow burner where fuel and air are added as gases at P0 T0 The mixture has 125 theoretical air and the products are going through a heat exchanger where they are cooled to 600 KFind the heat transfer from the heat exchanger per kmol of heptane burned The reaction equation for stoichiometric ratio is C7H16 vO2 O2 376 N2 7CO2 8 H2O vO2 376 N2 So the balance C and H was done in equation of oxygen gives vO2 7 4 11 vO2 ac 125 11 1375 Now the actual reaction equation is C7H16 1375 O2 517 N2 7CO2 8 H2O 517 N2 275 O 2 To find the heat transfer take a control volume as combustion chamber and heat exchanger HR Q HP Q H o P HP H o R Take the enthalpies from Tables A9 for the products and A10 for the fuel Q 7393 522 12 906 8 241 826 10 499 5178894 2759245 187 900 3 841 784 kJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1344 Pentene C5H10 is burned with stoichiometric pure oxygen in a steady flow process After giving some heat out the products are at 700 K and used in a heat exchanger where they are cooled to 35C Find the specific heat transfer in the heat exchanger C5H10 νO2O2 5 CO2 5 H2O νO2 75 Check for condensation amount Find x yH2O max Pg35 Ptot 00555 x 5 x x 02938 Out of the 5 H2O only 02938 are still vapor at the 35C exit temperature The heat exchanger cools the products so the energy equation is 5 n F h CO2 5 n F h H2O Q 5 n F h CO2 5 x n F h liq H2O x n F h vap H2O Find the enthalpy at 700 K all gas A9 and at 35C 30815 K part of water is a gas in table A9 and use B11 and A10 for the liquid water h liq H2O hf liq h 308 285 830 1801514666 10487 285 077 h vap H2O hf vap h 241 826 h Q n F 5 h 308 h 700CO2 5x h liq H2O hf vap h 700 H2O x E A AEf vap E AhE A700 H2O vapor h h 308 f vap h 53896 17 754 4706 285 077 241 826 14 190 02938 241 826 3381 241 826 14 190 361 209 kJkmol Fu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1345 Another alternative fuel to be seriously considered is hydrogen It can be produced from water by various techniques that are under extensive study Its biggest problem at the present time are cost storage and safety Repeat Problem 1340 using hydrogen gas as the fuel instead of ethanol HA2E A 11 05 OA2E A 11 376 05 NA2E A 1 HA2E AO 005 OA2E A 2068 NA2E Products at 60C 100 kPa check for condensation of water yAH2O MAXE A A1994 100E A A nV MAX EnV MAX 005 2068E Solving nAV MAXE A 05275 1 nAVE A 05275 nALIQE A 04725 HARE A 0 0 0 0 Notice the products are at 60C so add for water liquid from steam tables HAPE A 04725 285 830 18015 25111 10487 05275 241 826 1185 0050 10395 20680 1021 258 585 kJ QACVE A HAPE A HARE A 258 585 kJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1346 In a new highefficiency furnace natural gas assumed to be 90 methane and 10 ethane by volume and 110 theoretical air each enter at 25C 100 kPa and the products assumed to be 100 gaseous exit the furnace at 40C 100 kPa What is the heat transfer for this process Compare this to an older furnace where the products exit at 250C 100 kPa 09 CHA4E A 01 CA2E AHA6E A 11 215 OA2E A 376 2365 NA2E 11 COA2E A 21 HA2E AO 0215 OA2E A 8892 NA2E Fuel values from table A10 and the rest from Table A9 HARE A 0974 873 0184 740 75 860 kJkmol fuel HAPE A 11393 522 586 21241 826 5075 0215445 88924376 935 012 kJkmol fuel assuming all gas QACVE A HAPE A HARE A 859 152 kJkmol fuel b TAPE A 250 AoE AC HAPE A 11393 522 9370 21241 826 7750 02156817 88926602 853 956 kJ QACVE A HAPE A HARE A 778 096 kJkmol fuel Products 40 C 100 kPa 110 Air 25 C 100 kPa 09 CH 01 C H 25 C 100 kPa 4 2 6 Heat transfer Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1347 Repeat the previous problem but take into account the actual phase behavior of the products exiting the furnace 09 CH4 01 C2H6 11 215 O2 376 2365 N2 11 CO2 21 H2O 0215 O2 8892 N 2 Same as 1541 except check products for saturation at 40oC 100 kPa yv max 7384 100 nV MAX nV MAX10207 Solving nv max 0814 nV 0814 nliq 21 0814 1286 Fuel values from table A10 and the rest from Table A9 HR 0974 873 0184 740 75 860 kJkmol fuel For the liquid water add difference 40oC 25oC from steam tables Hliq 1286285 830 180151676 1049 366 125 kJkmol fuel HGAS 11393 522 586 0814241 826 5075 0215445 88924376 624 676 kJkmol fuel QCV HP HR 366 125 624 676 75 860 914 941 kJkmol fuel b TP 250 oC HP 11393 522 9370 21241 826 7750 02156817 88926602 853 956 kJkmol QCV HP HR 778 096 kJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1348 Methane CH4 is burned in a steady adiabatic flow process with two different oxidizers Case A Pure oxygen O2 and case B A mixture of O2 x Ar The reactants are supplied at T0 P0 and the products for both cases should be at 2100 K Find the required equivalence ratio in case A and the amount of Argon x for a stoichiometric ratio in case B a Stoichiometric has ν 2 actual has CH4 νO2 CO2 2H2O ν 2O2 Energy eq H R H P HP 2100 HP 2100 H R H P ho f fuel 0 ho f CO2 2ho f H2O 0 RHS 74 873 393 522 2241 826 802 301 kJkmol h CO2 97 500 h H2O 77 970 h O2 62 973 all in kJkmol HP 2100 h CO2 2h H2O ν 2 h O2 97 500 2 77 970 ν 2 62 973 127 494 ν 62 973 802 301 kJkmol fuel ν 1072 Φ AFS AF 2 1072 01866 b CH4 2 O2 2x Ar CO2 2H2O 2x Ar h Ar C P Ar T CP Ar MAr T from table A5 HP 2100 97 500 2 77 970 2x 052 39948 2100 298 253 440 x 74 866 Now the energy equation becomes 802 301 253 440 x 74 866 x 733 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1349 A rigid container has a 11 mole ratio of propane and butane gas together with a stoichiometric ratio of air at Po To The charge burns and there is heat transfer to a final temperature of 1000 K Find the final pressure and the heat transfer per kmol of fuel mixture C3H8 C4H10 νO2O2 376 N2 7 CO2 9 H2O 376 νO2 N2 The carbon and hydrogen balance has been used so oxygen gives νO2 7 92 115 νN2 4324 The energy equation is UP UR Q 0 HP HR nPRTP nRRT R H o P HP H o R nPRTP nRRTR nR 1 1 115 476 5674 nP 7 9 4324 5924 From Table A9 HP 7 33 397 9 26 000 4324 21 463 1 395 839 kJ From Table 133 convert kJkg to kJkmol with M H o P H o R 44094 46 352 58124 45 714 4 700 926 kJ Substitute into energy equation Q 4 700 926 1 395 839 831455924 1000 5674 29815 3 656 981 kJ2 kmol fuel 1 828 490 kJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1350 A closed insulated container is charged with a stoichiometric ratio of oxygen and hydrogen at 25C and 150 kPa After combustion liquid water at 25C is sprayed in such that the final temperature is 1200 K What is the final pressure Combustion reaction H2 1 2 O2 H2O Products 1 H2O xiH2O Energy eq per 1 kmol hydrogen remember flow in U2 U1 xih i xih f liq 1 xi HP HR 1 xi R TP 3 2R TR Solve for xi xi h f liq HP R TP HP HR R TP 3 2R TR From Table A9 HR 0 HP 241 826 34 506 207 320 kJkmol From Table A10 h f liq 285 830 kJkmol Substitute into the energy equation xi 285830 207320 83145 1200 207 320 83145 1200 3 229815 213 579 xi 3116 Volume is constant P1V1 nRR T1 P2V1 npR Tp nR 1 1 2 15 np 1 xi P2 P1 1 xi TP 3 2 T1 150 4116 1200 3 2 29815 1657 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1351 In a gas turbine natural gas methane and stoichiometric air flows into the combustion chamber at 1000 kPa 500 K Secondary air see FigP1351 also at 1000 kPa 500 K is added right after the combustion to result in a product mixture temperature of 1500 K Find the AF ratio mass basis for the primary reactant flow and the ratio of the secondary air to the primary air mass flow rates ratio CH4 νO2 376 N2 CO2 2H2O 376 ν Ν2 O balance ν 2 νN2 752 For the primary flow AF mass 2 32 752 28013 16043 1712 Do a CV around the entire setup then the energy equation becomes Energy Eq HR1 HR2 HP 1500 H P HP 1500 HR1 H R HR1 f fuel ho HFuel Hair1 Hair1 2HO2 376 HN2 26086 376 5911 56 623 kJkmol HR2 0 Hair2 νO2 add HO2 376 HN2 νO2 add 28 311 kJkmol HFuel M CP T 16043 2254 500 298 73045 kJkmol ho f fuel H P HV 16043 50 010 802 310 kJkmol HP 1500 h CO2 2h H2O 752 h N2 νO2 add h O2 376 h N2 61 705 2 48 149 752 38 405 νO2 add 40 600 376 38 405 446 809 νO2 add 185 003 kJkmol Now substitute everything into energy equation 802 310 73045 56 623 νO2 add 28 311 446 809 νO2 add 185 003 νO2 add 2677 mair2mair1 nair2nair1 νO2 add ν 26772 134 Fuel Air1 Air2 1500 K T ad Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1352 A rigid vessel initially contains 2 kmol of carbon and 2 kmol of oxygen at 25C 200 kPa Combustion occurs and the resulting products consist of 1 kmol of carbon dioxide 1 kmol of carbon monoxide and excess oxygen at a temperature of 1000 K Determine the final pressure in the vessel and the heat transfer from the vessel during the process 2 C 2 O2 1 CO2 1 CO 1 2 O 2 Process V constant C solid n1GAS 2 n2GAS 25 P2 P1 n2T2 n1T1 200 25 1000 2 2982 8384 kPa H1 0 and we neglect the Pv for carbon when doing the U1 H2 1393 522 33 397 1110 527 21 686 120 22 703 437 615 kJ Now find heat transfer from energy equation 1Q2 U2 U1 H2 H1 n2RT2 n1RT 1 437 615 0 8314525 1000 2 2982 453 442 kJ Combustion gas Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1353 Gaseous propane mixes with air both supplied at 500 K 01 MPa The mixture goes into a combustion chamber and products of combustion exit at 1300 K 01 MPa The products analyzed on a dry basis are 1142 CO2 079 CO 268 O2 and 8511 N2 on a volume basis Find the equivalence ratio and the heat transfer per kmol of fuel C3H8 α O2 376 α N2 β CO2 γ H2O 376 α N2 β 3 γ 4 α β γ2 5 AF S 476α 238 The actual combustion reaction is xC3H8 α O2 376 α N2 1142 CO2 y H2O 8511 N2 079 CO 268 O2 C balance 3x 1142 079 x 407 H balance 8x 2y y 4x 1628 O balance 2α 2 1142 y 079 2 268 4527 α 22635 N balance 376 α 8511 α 226356 checks close enough Rescale the equation by dividing with x to give C3H8 55614 O2 376 N2 2806 CO2 4 H2O 2091 N2 0194 CO 06584 O2 AF 55614 1 376 1 26472 φ AF S AF 238 26472 0899 Theo air 1φ 111 q hP hR ho P νih1300 K hR hR ho f fuel hfuel 55614 hO2 2091 hN2 103 900 1679 44094 500 298 55614 6086 2091 5911 68 500 kJkmol fuel hP ho P νih1300 K 2806 393 522 50 148 4241 826 38 941 2091 0 31 503 0194 110 527 31 867 06584 0 33 345 1 109 625 kJkmol fuel q 1 109 625 68 500 1 178 125 kJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Enthalpy of Combustion and Heating Value Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1354 Find the enthalpy of combustion and the heating value for pure carbon Reaction C νO2 O2 376 N2 CO2 νO2376 N2 oxygen balance νO2 1 H RP Ho P Ho R h f CO2 f C h 393 522 0 393 522 kJkmol 32 791 kJkg M 12011 HV 393 522 kJkmol 32 791 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1355 Phenol has an entry in Table 133 but it does not have a corresponding value of the enthalpy of formation in Table A10 Can you calculate it C6H5OH νO2 O2 376 N2 3 H2O 6 CO2 376νO2 N2 The C and H balance was introduced 6 Cs and 6 Hs At the reference condition the oxygen and nitrogen have zero enthalpy of formation Energy Eq HP HR Ho P Ho R since ref T is assumed H RP HP HR Ho P Ho R 3 h f H2O 6 h f CO2 h f fuel Table 133 is on a mass basis and let us chose liquid fuel so we get the molecular weight from the composition M 6 12011 3 2016 16 94114 H RP 94114 31 117 2 928 545 kJkmol Solve the energy equation for fuel formation enthalpy h f fuel 3 h f H2O 6 h f CO2 H RP 3 241 826 6393 522 2 928 545 158 065 kJkmol For fuel as vapor we get H RP 94114 31 774 2 990 378 kJkmol h f fuel 3 h f H2O 6 h f CO2 H RP 3 241 826 6393 522 2 990 378 96 232 kJkmol Notice if I took liquid water in products to do H RP then I must use liquid value for h f H2O 285 830 kJkmol and the final result is the same Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1356 Acetylene gas C2H2 is burned with stoichiometric air in a torch The reactants are supplied at the reference conditions Po To The products come out from the flame at 2800 K after a heat loss by radiation Find the lower heating value for the fuel as it is not listed in Table 133 and the heat loss per kmol of fuel The reaction equation is C2H2 25 O2 376 N2 2 CO2 1 H2O 94 N2 Definition of the heating value LHV H RP Ho R Ho P h f fuel h f H2O 2 h f CO2 226 731 241 826 2393 522 1 255 601 kJkmol 48 222 kJkg Energy Eq HP Qout HR Ho P HP Qout Ho R Qout Ho R Ho P HP LHV HP HP 2 140 435 115 463 94 85 323 1 198 369 kJkmol Qout 1 255 601 1 198 369 57 232 kJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1357 Some type of wood can be characterized as C1H15O07 with a lower heating value of 19 500 kJkg Find its formation enthalpy C1H15O07 νO2 O2 376 N2 CO2 075 H2O 376νO2 N2 O balance 07 2νO2 2 075 νO2 1025 νN2 3854 M 1 120111 15 1008 07 16 24712 H RP Ho P Ho R M HV 24712 19 500 481 884 kJkmol 075 h f H2O 1 h f CO2 h f fuel h f fuel 075 h f H2O 1 h f CO2 M HV 075 241 826 393 522 481 884 93 008 kJkmol Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1358 Do problem 1340 using table 133 instead of Table A10 for the solution One alternative to using petroleum or natural gas as fuels is ethanol C2H5OH which is commonly produced from grain by fermentation Consider a combustion process in which liquid ethanol is burned with 120 theoretical air in a steady flow process The reactants enter the combustion chamber at 25C and the products exit at 60C 100 kPa Calculate the heat transfer per kilomole of ethanol C2H5OH 11 3 O2 376 N2 2 CO2 3 H2O 03 O2 12408 N 2 Products at 60C 100 kPa so check for condensation of water yH2O MIX 1994 100 nV MAX nV MAX 2 03 12408 nV MAX 366 3 No liquid Fuel table 133 select liquid fuel water vapor and convert to mole basis H RP 46069 26 811 1 235 156 kJkmol Since the reactants enter at the reference state the energy equation becomes QCV HP HR H o P HP H o R H RP H P HP 2 h CO2 3 h H2O 03 h O2 12408 h N2 21373 31185 0310395 124081021 19 281 kJkmol QCV 1 235 156 19 281 1 215 874 kJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1359 Agriculturally derived butanol C4H10O molecular mass 7412 also called bio butanol has a lower heating value LHV 33 075 kJkg for liquid fuel Find its formation enthalpy The reaction equation for stiochiometric ratio is C4H10O νO2 O2 376 N2 4CO2 5 H2O νO2 376 N2 where the carbon and hydrogen balance have been done Now the oxygen O 1 2νO2 4 2 5 νO2 6 νN2 2256 The lower heating value per kmol fuel is LHV H RP Ho R Ho P h f fuel 4 h f CO2 5 h f H2O so the fuel formation enthalpy becomes h f fuel LHV 4 h f CO2 5 h f H2O 7412 33 075 4 393 522 5241 826 2 451 519 1 574 088 1 209 130 331 699 kJkmol Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1360 In a picnic grill gaseous propane and stoichiometric air are mixed and fed to the burner both at ambient Po and To After combustion the products cool down and at some point exit at 500 K How much heat transfer was given out for 1 kg of propane The reaction equation for stoichiometric mixture is C3H8 νO2 O2 376 N2 a H2O b CO2 c N2 C balance 3 b H balance 8 2a a 4 O balance 2 νO2 a 2b 4 2 3 νO2 5 N2 balance 376 νO2 c 188 AF 5 476 2897 44094 15637 The reaction is C3H8 5 O2 376 N2 4 H2O 3 CO2 188 N2 All the water is in the vapor phase 500 K products Energy Eq HR Qout HP Ho R Qout Ho P ΔH P A9 ΔHP 4 6922 3 8305 188 5911 174 682 kJkmol fuel 174 682 44094 kJkg 39616 kJkg Table 133 H RP HV 46 352 kJkg Qout Ho R Ho P ΔHP H RP ΔHP HV ΔHP 46 352 39616 kJkg fuel 42 390 kJkg fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1361 Propylbenzene C9H12 is listed in Table 133 but not in table A10 No molecular weight is listed in the book Find the molecular weight the enthalpy of formation for the liquid fuel and the enthalpy of evaporation C9H12 12 O2 9 CO2 6 H2O M 9 12011 6 2016 120195 h RP H o P H o R P νih fi h fFu h fFu P νih fi h RP Formation enthalpies from Table A10 and enthalpy of combustion from Table 133 h fFu 9h fCO2 6h fH2O g M 41 219liq Fu H2O vap 9393 522 6241 826 12019541 219 38 336 kJkmol Take the enthalpy of combustion from Table 133 for fuel as a gas and as a vapor the difference is the enthalpy of evaporation hfg h RP gas h RP liq 41 603 41 219 384 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1362 Liquid pentane is burned with dry air and the products are measured on a dry basis as 101 CO2 02 CO 59 O2 remainder N2 Find the enthalpy of formation for the fuel and the actual equivalence ratio νFuC5H12 νO2O2 376 νO2N2 x H2O 101 CO2 02 CO 59 O2 838 N2 Balance of C 5 νFu 101 02 νFu 206 Balance of H 12 νFu 2 x x 6 νFu 1236 Balance of O 2 νO2 x 202 02 2 59 νO2 2228 Balance of N 2 376 νO2 838 2 νO2 22287 OK νO2 for 1 kmol fuel 10816 φ 1 C5H12 8 O2 8 376 N2 6 H2O 5 CO2 3008 N2 H RP H P H R 6 h f H2O 5 h f CO2 f fuel h Table 133 H RP 44 983 72151 3 245 568 kJkmol fuel h f fuel 6 h f H2O 5 h f CO2 H RP 6 241 826 5 393 522 3 245 568 172 998 kJkmol φ AFs AF νO2 stoichνO2 AC 810816 074 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1363 Wet biomass waste from a foodprocessing plant is fed to a catalytic reactor where in a steady flow process it is converted into a lowenergy fuel gas suitable for firing the processing plant boilers The fuel gas has a composition of 50 methane 45 carbon dioxide and 5 hydrogen on a volumetric basis Determine the lower heating value of this fuel gas mixture per unit volume For 1 kmol fuel gas 05 CH4 045 CO2 005 H2 1025 O 2 05 045 CO2 105 H2O The lower heating value is with water vapor in the products Since the 045 CO2 cancels h RP 05393 522 105241 826 0574 873 0050 413 242 kJkmol fuel gas With n V PR T 100 83145 2982 004033 kmolm 3 LHV 413 242 004033 16 666 kJm3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1364 Determine the lower heating value of the gas generated from coal as described in Problem 1330 Do not include the components removed by the water scrubbers The gas from problem 1330 is saturated with water vapor Lower heating value LHV has water as vapor LHV H RP H P H R Only CH4 H2 and CO contributes find gas mixture after the scrubbers CH4 H2 CO CO2 N2 n 03 296 410 100 08 817 Cool to 40C PG 7384 P 3000 kPa yH2O MAX 7384 3000 nv nv 817 nv 02016 Mixture composition CH4 H2 CO CO2 N2 H2Ov 03 kmol 296 410 100 08 02016 819016 kmol from 100 kmol of the original gas mixture 03 CH4 296 H2 41 CO 10 CO2 08 N2 02016 H2O LHV 03H RPCH4 296H RPH2 41H RPCO819 0350 010 16043 296241 826 41393 522 110 527819 232 009 kJ kmol gas Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1365 Do Problem 1342 using Table 133 instead of Table A10 for the solution As an alternative fuel consider liquid methanol burned with stoichiometric air both supplied at Po To in a constant pressure process exhausting the products at 900 K What is the heat transfer per kmol of fuel CH3OH 15 O2 376 N2 1 CO2 2 H2O 564 N2 Reactants at 25oC products are at 900 K 100 kPa CH3OH From table 133 for liquid fuel water vapor H RP 19 910 kJkg 19 910 32042 637 956 kJkmol fuel A9 ΔHP 1 28 030 2 21 937 564 18 223 174 682 kJkmol fuel Q HPo HP HRo H RP HP 637 956 174 682 463 274 kJkmol fuel Various species added to the methanol changes the flame color Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1366 Do problem 1343 using table 133 instead of Table A10 for the solution The reaction equation for stiochiometric ratio is C7H16 νO2 O2 376 N2 7CO2 8 H2O νO2 376 N2 So the balance C and H was done in equation of oxygen gives vO2 7 4 11 and actual one is 11125 1375 Now the actual reaction equation is C7H16 1375 O2 517 N2 7CO2 8 H2O 517 N2 275 O 2 To find the heat transfer take a control volume as combustion chamber and heat exchanger HR Q HP Q Ho P HP Ho R H RP HP Now we get the enthalpy of combustion from table 133 which is per kg so scale it with the molecular weight for the fuel Add all the HP from A9 HP 7 h CO2 8 h H2O 517 h N2 275 h O2 712 906 810 499 5178894 2759245 659 578 kJ kmol Q M H RP HP 10020544 922 659 578 3 841 831 kJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1367 E85 is a liquid mixture of 85 ethanol and 15 gasoline assume octane by mass Find the lower heating value for this blend The heating value of the blend becomes LHV 085 LHVethanol 015 LHVoctane 085 26 811 015 44 425 kJkg 29 453 kJkg As this is lower than gasoline a larger amount of fuel is being used for the same energy release in the engine If the mixture was given on a mole basis we would have used LHV on a mole basis instead multiplying Table 133 entries by M Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1368 Assume the products of combustion in Problem 1367 are sent out of the tailpipe and cool to ambient 20C Find the fraction of the product water that will condense Since the fuel mixture is specified on a mass basis we need to find the mole fractions for the combustion equation From Eq134 we get yethanol 08546069 08546069 015114232 093356 yoctane 1 yethanol 006644 The reaction equation is 093356 C2H5OH 006644 C8H18 νO2 O2 376 N2 νCO2 CO2 νH2OH2O 376 νO2 N 2 C balance 2 093356 8 006644 νCO2 239864 H2 balance 3 093356 9 006644 νH2O 339864 O balance 093356 2 νO2 2 νCO2 νH2O 819592 νO2 363118 Now the products are 239864 CO2 339864 H2O 1365324 N2 Check for condensation amount Find x yH2O max Pg20 Ptot 2339 101325 0023084 x 239864 x 1365324 x 03793 mol of water can stay vapor Condensing water 339864 03793 301934 or 89 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1369 Gaseous propane and stoichiometric air are mixed and fed to a burner both at ambient Po and To After combustion the products cool down and eventually reach ambient To How much heat transfer was given out for 1 kg of propane The reaction equation for stoichiometric mixture is C3H8 νO2 O2 376 N2 a H2O b CO2 c N2 C balance 3 b H balance 8 2a a 4 O balance 2 νO2 a 2b 4 2 3 νO2 5 N2 balance 376 νO2 c 188 AF 5 476 2897 44094 15637 The reaction is C3H8 5 O2 376 N2 4 H2O 3 CO2 188 N2 First we solve the problem assuming all the water is in the vapor phase and then we can solve with the proper split of water into liquid and vapor masses Energy Eq HR Qout HP Ho R Qout Ho P Qout Ho R Ho P H RP HV 46 352 kJkg fuel At To we need to check for condensation of water partial P limited to Pg yv max Pg P 3169 100 nv max nv max 3 188 nv max 07135 and then nliq 4 nv max 32865 So now we know the split of the 4 moles of water into vapor and liquid The change in the heating value becomes HVliq water 50 343 kJkg fuel Qout HV 07135 4 46 352 32865 4 50 343 49 631 kJkg fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1370 In an experiment a 11 mole ratio propane and butane is burned in a steady flow with stoichiometric air Both fuels and air are supplied as gases at 298 K and 100 kPa The products are cooled to 1000 K as they give heat to some application Find the lower heating value per kg fuel mixture and the total heat transfer for 1 kmol of fuel mixture used Combustion C3H8 C4H10 115O2 376 N 7 CO2 9 H2O 4324 N 2 νO2 7 92 115 νN2 376 115 4324 The enthalpy of combustion for the 2 kmol of fuel becomes from Tbl 133 Ho P Ho R 46 352 44094 45 714 58124 4 700 9256 kJ2 kmol fuel HV H RP 4 700 9256 44094 58124 45 989 kJkg fuel Convert to mass basis HP 7 33 397 9 26 000 4324 21 463 1 395 839 kJ2 kmol fuel The heat transfer for the two kmol of fuel is Q HR HP H RP HP 4 700 9256 1 395 839 3 305 087 kJ2 kmol fuel 1 652 544 kJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1371 Blast furnace gas in a steel mill is available at 250C to be burned for the generation of steam The composition of this gas is on a volumetric basis Component CH4 H2 CO CO2 N2 H2O Percent by volume 01 24 233 144 564 34 Find the lower heating value kJm3 of this gas at 250C and ambient pressure Of the six components in the gas mixture only the first 3 contribute to the heating value These are per kmol of mixture 0024 H2 0001 CH4 0233 CO For these components 0024 H2 0001 CH4 0233 CO 01305 O2 0026 H2O 0234 CO 2 The remainder need not be included in the calculation as the contributions to reactants and products cancel For the lower HVwater as vapor at 250C h RP 0026241 826 7750 0234393 522 9370 00240 6558 000174 873 2254 160425025 0233110 527 6629 013050 6817 72 749 kJ kmol fuel v 0 R ToPo 83145 5232100 435015 m3kmol LHV 72 749 435015 1672 kJm3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1372 Consider natural gas A which are listed in Table 132 Calculate the enthalpy of combustion of each gas at 25C assuming that the products include vapor water Repeat the answer for liquid water in the products Natural Gas A 0939 CH4 0036 C2H6 0012 C3H8 0013 C4H10 21485 O2 376 N2 1099 CO2 2099 H2O 80784 N2 HR 093974 878 003684 740 0012103 900 0013126 200 76244 kJ a vapor H2O HP 1099393 522 2099241 826 940 074 kJkmol h RP HP HR 863 830 kJkmol b Liq H2O HP 1099393 522 2099285 830 1 032 438 h RP 956 194 kJkmol These values can be compared to the pure methane Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1373 Consider natural gas D which are listed in Table 132 Calculate the enthalpy of combustion of each gas at 25C assuming that the products include vapor water Repeat the answer for liquid water in the products Natural Gas D 0543 CH4 0163 C2H6 0162 C3H8 0074 C4H10 0058 N2 νO2 O2 376 N2 1651 CO2 2593 H2O νO2 376 N 2 HR 054374 873 016384 740 0162130 900 0074126 200 80 639 kJ a vapor H2O HP 1651393 522 2593241 826 1 276 760 kJ h RP 1 196 121 kJkmol b Liq H2O HP 1651393 522 2593285 830 1 390 862 kJ h RP 1 310 223 kJkmol Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1374 A burner receives a mixture of two fuels with mass fraction 40 nbutane and 60 methanol both vapor The fuel is burned with stoichiometric air Find the product composition and the lower heating value of this fuel mixture kJkg fuel mix Since the fuel mixture is specified on a mass basis we need to find the mole fractions for the combustion equation From Eq134 we get ybutane 0458124 0458124 0632042 026875 ymethanol 1 ybutane 073125 The reaction equation is 073125 CH3OH 026875 C4H10 νO2 O2 376 N2 νCO2 CO2 νH2OH2O 376 νO2 N 2 C balance 073125 4 026875 νCO2 180625 H2 balance 2 073125 5 026875 νH2O 280625 O balance 073125 2 νO2 2 νCO2 νH2O 641875 νO2 284375 Now the products are 180625 CO2 280625 H2O 106925 N2 Since the enthalpy of combustion is on a mass basis in table 133 this is also the negative of the heating value we get LHV 04 45 714 06 21 093 30 941 kJkg fuel mixture Notice we took fuel vapor and water as vapor lower heating value Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1375 Natural gas we assume methane is burned with 200 theoretical air and the reactants are supplied as gases at the reference temperature and pressure The products are flowing through a heat exchanger where they give off energy to some water flowing in at 20oC 500 kPa and out at 700oC 500 kPa The products exit at 400 K to the chimney How much energy per kmole fuel can the products deliver and how many kg water per kg fuel can they heat The reaction equation for stoichiometric mixture is CH4 νO2 O2 376 N2 2 H2O 1 CO2 c N2 O balance 2 νO2 2 2 νO2 2 200 theoretical air νO2 2 2 4 so now more O2 and N2 CH4 4 O2 376 N2 2 H2O 1 CO2 1504 N2 2 O2 The products are cooled to 400 K so we do not consider condensation and the energy equation is Energy Eq HR Q HP H P HP H R Q Q H P H R HP H RP HP From Table 133 H RP 1604 50 010 802 160 kJkmol HP h CO2 2 h H2O 2 h O2 1504 h N2 From Table A9 HP 400 4003 2 3450 2 3027 1504 2971 61 641 kJkmol Q H RP HP 802 160 61 641 740 519 kJkmol qprod Q M 740 519 1604 46 167 kJkg fuel The water flow has a required heat transfer using B13 and B14 as qH2O hout hin 392597 8381 38422 kJkg water The mass of water becomes mH2O mfuel qprod qH2O 120 kg water kg fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1376 Liquid nitromethane is added to the air in a carburetor to make a stoichiometric mixture where both fuel and air are added at 298 K 100 kPa After combustion a constant pressure heat exchanger brings the products to 600 K before being exhausted Assume the nitrogen in the fuel becomes N2 gas Find the total heat transfer per kmole fuel in the whole process CH3NO2 νO2 O2 376 N2 15 H2O 1 CO2 a N2 C and H balances done in equation The remaining O balance 2 2 νO2 15 2 νO2 075 N balance 1 376 νO2 2 2a a 332 Energy eq HR Q HP Q HP HR H P H R HP HR The reactants enter at the reference state HR 0 and the products at 600 K from table A9 HP 15 h H2O h CO2 332 h N2 15 10 499 1 12 906 332 8894 58 183 kJkmol fuel H P H R H RP 6104 10 537 643 178 kJkmol Q 643 178 58 183 584 995 kJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1377 Gasoline C7H17 is burned in a steady state burner with stoichiometric air at Po To The gasoline is flowing as a liquid at To to a carburetor where it is mixed with air to produce a fuel air gas mixture at To The carburetor takes some heat transfer from the hot products to do the heating After the combustion the products go through a heat exchanger which they leave at 600 K The gasoline consumption is 10 kg per hour How much power is given out in the heat exchanger and how much power does the carburetor need Stoichiometric combustion C7H17 νO2 O2 376 N2 85 H2O 7 CO2 c N2 O balance 2 νO2 85 14 225 νO2 1125 N balance c 376 νO2 376 1125 423 MFUEL 7 MC 17 MH 7 12011 85 2016 101213 CV Total heat exchanger and carburetor included Q out Energy Eq HR H R H P HP Qout From Table A9 HP 85 10 499 7 12 906 423 8894 555 800 kJkmol From energy equation and Table 133 Qout H R H P HP H RP HP 101213 44 506 555 800 3 948 786 kJkmol Now the power output is Q n Qout Qout m M 3 948 786 10 3600 101213 1084 kW The carburetor air comes in and leaves at the same T so no change in energy all we need is to evaporate the fuel hfg so Q m hfg 10 3600 44 886 44 506 1 360 380 106 kW Here we used Table 133 for fuel liquid and fuel vapor to get hfg and any phase of the water as long as it is the same for the two Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1378 An isobaric combustion process receives gaseous benzene C6H6 and air in a stoichiometric ratio at Po To To limit the product temperature to 2000 K liquid water is sprayed in after the combustion Find the kmol of liquid water added per kmol of fuel and the dew point of the combined products The reaction equation for stoichiometric mixture with C and H balance done is C6H6 νO2 O2 376 N2 3 H2O 6 CO2 c N2 O balance 2 νO2 3 6 2 15 νO2 75 N balance c 376 νO2 376 75 282 With x kmol of water added per kmol fuel the products are Products 3 x H2O 6 CO2 282 N2 Energy Eq HR H R x h f H2O liq H P HP H P x h f H2O vap 3 x h H2O 6 h CO2 282 h N2 Where the extra water is shown explicitly Rearrange to get H R H P 6h CO2 282h N2 3h H2O x h f H2O vap h f H2O liq h H2O 40 576 78114 6 91 439 282 56 137 3 72 788 x 241 826 285 830 72 788 819 493 x 116 792 x 7017 kmolkmol fuel Dew point yv 3 x 6 282 3 x 02265 Pv yvP 02265 101325 2295 kPa B12 Tdew 63oC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1379 A liquid fuel similar to diesel ndodecane C12H26 is sent into a carburetor where it is vaporized and mixed with air in a stoichiometric ratio The liquid fuel and air are supplied at To and 100 kPa and the heat to the carburetor Q1 is taken from the combustion products as in Fig P1377 After mixing the reactant gas goes through a combustor and after complete combustion the products enter a heat exchanger where they deliver Q1 to the carburetor and Q2 to an application after which the products exit at 800 K Find the heat transfer to the carburetor per kmol fuel Q1 and the second heat transfer Q2 both in kJkmol fuel First do the combustion equation for nDodecane C12H 26 C12H26 νO2O2 376 N2 12 CO2 13 H2O 6956 N2 O balance 2νO2 12 2 13 37 νO2 372 185 N2 balance νO2 376 νO2 376 185 6956 MFUEL 12 MC 26 MH 12 12011 26 1008 17034 CV Carburetor here the air is unchanged liquid fuel vaporizes Q1 HVf vap HVf liq 1703444 467 44 109 60 982 kJkmol CV Total system notice Q1 is internal to this CV Energy Eq 0 HR1 HP2 Q2 Table 133 HV H RP 17034 44 109 7 513 527 kJkmol Table A9 HP2 12 h CO2 13 h H2O 6956 h N2 12 22 806 13 18 002 6956 15 046 1 554 298 kJkmol Q2 HR1 HP2 Ho R Ho P HP2 HV HP2 7 513 527 1 554 298 5 959 229 kJkmol Carburetor Combustion chamber Heat exch Q2 Q1 Q1 Fuel air gas mix 800 K To To Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1380 A mixture of fuels is E85 which is 85 ethanol and 15 gasoline assume octane by mass Assume we put the fuel and air both at To Po into a carburetor and vaporize the fuel as we mix it with stoichiometric air before it flows to an engine Assume the engine has an efficiency as work divided by the lower heating value of 30 and we want it to deliver 40 kW We use heat from the exhaust flow 500 K for the carburetor Find the lower heating value of this fuel kJkg the rate of fuel consumption the heating rate needed in the carburetor and the rate of entropy generation in the carburetor The heating value of the liquid fuel blend becomes LHV 085 LHVethanol 015 LHVoctane 085 26 811 015 44 425 kJkg 29 453 kJkg W 03 m fuel LHV m fuel W 03 LHV 40 kW 03 29 453 kJkg 0004527 kgs 163 kgh The difference in heating value based on liquid versus vapor fuel is ΔLHV 085 27 731 26 811 015 44 788 44 425 085 920 015 363 83645 kJkg Q m fuel ΔLHV 0004527 83645 379 kW Entropy generation is from vaporizing the fuel by transfer of heat from 500 K exhaust gas to the intake system at To and then mixing it with air The mole fractions are Ethanol M 46069 Octane M 114232 DIV 085 46069 015 114232 001845 0001313 0019764 yethanol 001845 0019764 09336 yoctane 1 yethanol 006644 093356 C2H5OH 006644 C8H18 363118 O2 376 N2 239864 CO2 339864 H2O 1365324 N2 C balance 093356 2 006644 8 239864 vCO2 H balance 093356 6 006644 18 679728 2vH2O So the AF ratio on a mole and mass basis are AFmole 363118 476 1 172844 AFmass 172844 2897505996 9896 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful The fuel was mixed before the carburetor so afterwards we have a 1 172844 mix sfg ethanol 282444 16055446069 26458 kJkgK sfg octane 466514 360575114232 0927402 kJkgK sfg fuel mix 085 26458 015 0927 238798 kJkgK Mfuel 09336 46069 006644 114232 505996 Entropy Eq S gen out in m fuel sfg R lnyfuel m air R lnyair Q Texhaust 0004527 238798 83145 505996 ln 1 182844 0004527 9896 0287 ln172844 182844 379 500 kWK 001297 0000723 000758 00061 kWK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Adiabatic Flame Temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1381 In a rocket hydrogen is burned with air both reactants supplied as gases at Po To The combustion is adiabatic and the mixture is stoichiometeric 100 theoretical air Find the products dew point and the adiabatic flame temperature 2500 K The reaction equation is H2 νO2 O2 376 N2 H2O 376 vO2 N2 The balance of hydrogen is done now for oxygen we need vO2 05 and thus we have 188 for nitrogen yv 11188 03472 Pv 101325 03472 3518 kPa Pg Table B12 Tdew 726 C Energy Eq HR HP 0 241 826 hwater 188 hnitrogen HP hwater 188 hnitrogen 241 826 kJkmol fuel Find now from table A9 the two enthalpy terms At 2400 K HP 93741 188 70640 226544 kJkmol fuel At 2600 K HP 104520 188 77963 251090 kJkmol fuel Then interpolate to hit 241 826 to give T 2525 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1382 Hydrogen gas is burned with pure oxygen in a steady flow burner where both reactants are supplied in a stoichiometric ratio at the reference pressure and temperature What is the adiabatic flame temperature The reaction equation is H2 νO2 O2 H2O The balance of hydrogen is done now for oxygen we need vO2 05 Energy Eq HR HP 0 241 826 h H2O h H2O 241 826 kJkmol Interpolate now in table A9 for the temperature to give this enthalpy T 4991 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1383 Some type of wood can be characterized as C1H15O07 with a lower heating value of 19 500 kJkg Find its adiabatic flame temperature when burned with stoichiometric air at 100 kPa 298 K C1H15O07 νO2 O2 376 N2 CO2 075 H2O 376νO2 N2 O balance 07 2νO2 2 075 νO2 1025 νN2 3854 M 1 120111 15 1008 07 16 24712 H RP Ho P Ho R M HV 24712 19 500 481 884 kJkmol Energy Eq HP H P HP HR H R HP H R H P H RP HP h CO2 075 h H2O 3854 h N2 481 884 kJkmol at 2400 K HP 115 779 075 93 741 3854 70 640 458 331 kJ at 2600 K HP 128 074 075 104 520 3854 77 963 506 933 kJ T 2400 200 481 884 458 331 506 933 458 331 2497 K Comment Most wood has some water and some noncombustible solids material so the actual flame temperature will be much lower C Borgnakke Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1384 A gas turbine burns methane with 200 theoretical air The air and fuel comes in through two separate compressors bringing them from 100 kPa 298 K to 1400 kPa and after mixing enters the combustion chamber at 600 K Find the adiabatic flame temperature using constant specific heat for the HP terms The reaction equation for stoichiometric mixture is CH4 νO2 O2 376 N2 2 H2O 1 CO2 c N2 O balance 2 νO2 2 2 νO2 2 200 theoretical air νO2 2 2 4 so now more O2 and N2 CH4 4 O2 376 N2 2 H2O 1 CO2 1504 N2 2 O2 The energy equation around the combustion chamber becomes Energy Eq HP HR 0 HP H R HR H P H RP HR HR HFuel Hair M CP T 4h O2 376 h N2 16043 2254 600298 49245 376 8894 181 666 kJkmol H RP 16043 50 010 802 310 kJkmol HP h CO2 2 h H2O 1504 h N2 2 h O2 Τ νiC Pi 802 310 181 666 983 976 kJkmol from energy Eq νiC Pi 0842 4401 2 1872 18015 1504 1042 28013 2 0922 31999 60252 kJkmolK Τ HP νiC Pi 983 976 60252 16331 K T 298 1633 1931 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1385 Extend the solution to the previous problem by using Table A9 for the HP terms The reaction equation for stoichiometric mixture is CH4 νO2 O2 376 N2 2 H2O 1 CO2 c N2 O balance 2 νO2 2 2 νO2 2 200 theoretical air νO2 2 2 4 so now more O2 and N2 CH4 4 O2 376 N2 2 H2O 1 CO2 1504 N2 2 O2 The energy equation around the combustion chamber becomes Energy Eq HP HR 0 HP H R HR H P H RP HR HR HFuel Hair M CP T 4h O2 376 h N2 16043 2254 600298 49245 376 8894 181 666 kJkmol H RP 16043 50 010 802 310 kJkmol HP h CO2 2 h H2O 1504 h N2 2 h O2 802 310 181 666 983 976 kJkmol from energy Eq Trial and error with h from Table A9 At 1800 K HP 79 432 2 62 693 1504 48 979 2 51 674 1 044 810 kJkmol At 1700 K HP 73 480 2 57 757 1504 45 430 2 47 959 968 179 kJkmol Linear interpolation T 1700 100 983 976 968 179 1 044 810 968 179 1721 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1386 Carbon is burned with air in a furnace with 150 theoretical air and both reactants are supplied at the reference pressure and temperature What is the adiabatic flame temperature C νO2O2 376 νO2N2 1 CO2 376 νO2N2 From this we find νO2 1 and the actual combustion reaction is C 15 O2 564 N2 1 CO2 564 N2 05 O2 HP H P HP HR H R HP H R H P 0 393 522 393 522 kJkmol HP h CO2 564 h N2 05 h O2 Find T so HP takes on the required value To start guessing assume all products are nitrogen 1 564 05 714 that gives 1900 T 2000 K from Table A9 HP 1900 85 420 564 52 549 05 55 414 409 503 too high HP 1800 79 432 564 48 979 05 51 674 381 511 Linear interpolation to find T 1800 100 393 522 381 511 409 503 381 511 1843 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1387 Acetylene gas at 25C 100 kPa is fed to the head of a cutting torch Calculate the adiabatic flame temperature if the acetylene is burned with a 100 theoretical air at 25C b 100 theoretical oxygen at 25C a C2H2 25 O2 25 376 N2 2 CO2 1 H2O 94 N2 HR ho f C2H2 226 731 kJkmol from table A10 HP 2393 522 h CO2 1241 826 h H2O 94 h N2 QCV HP HR 0 2 h CO2 1 h H2O 94 h N2 1 255 601 kJ Trial and Error A9 LHS2800 1 198 369 LHS3000 1 303 775 Linear interpolation TPROD 2909 K b C2H2 25 O2 2 CO2 H2O HR 226 731 kJ HP 2393 522 h CO2 1241 826 h H2O 2 h CO2 1 h H2O 1 255 601 kJkmol fuel At 6000 K limit of A9 2 343 782 302 295 989 859 At 5600 K 2 317 870 278 161 913 901 Slope 75 958400 K change Extrapolate to cover the difference above 989 859 kJkmol fuel TPROD 6000 400265 74275 958 7400 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1388 Hydrogen gas is burned with 200 theoretical air in a steady flow burner where both reactants are supplied at the reference pressure and temperature What is the adiabatic flame temperature The stoichiometric reaction equation is H2 νO2 O2 376 N2 H2O 376 νO2 N2 The balance of hydrogen is done now for oxygen we need νO2 05 and thus we have for the actual mixture νO2 1 The actual reaction is H2 1 O2 376 N2 1 H2O 376 N2 05 O2 The energy equation with formation enthalpy from A9 or A10 for water is HR HP 0 241 826 HP HP hH2O 376 hN2 05 hO2 241 826 kJkmol Find now from table A9 the three enthalpy terms At 2000 K HP 72 788 376 56 137 05 59 176 313 451 At 1800 K HP 62 693 376 48 979 05 51 674 272 691 At 1600 K HP 52 907 376 41 904 05 44 267 232 600 At 1700 K HP 57 757 376 45 430 05 47 959 252 553 Then interpolate to match 241 826 to give T 1600 100 241 826 232 600 252 553 232 600 1646 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1389 Butane gas at 25C is mixed with 150 theoretical air at 600 K and is burned in an adiabatic steady flow combustor What is the temperature of the products exiting the combustor C4H10 1565 O2 376 N2 4 CO2 5 H2O 325 O2 3666 N 2 Energy Eq HP HR 0 HP H R HR H P Reactants HR 9759245 36668894 416 193 kJkmol H R h C4H10 ho f IG 126 200 kJkmol HR 289 993 kJkmol H P 4393522 5241826 2 783 218 kJkmol HP 4 h CO2 5 h H2O 325 h O2 3666 h N2 From the energy equation we then get HP 126 200 416 193 2 783 218 3 073 211 kJkmol Trial and Error LHS2000 K 2 980 000 LHS2200 K 3 369 866 Linear interpolation to match RHS TP 2048 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1390 A stoichiometric mixture of benzene C6H6 and air is mixed from the reactants flowing at 25C 100 kPa Find the adiabatic flame temperature What is the error if constant specific heat at T0 for the products from Table A5 are used C6H6 νO2O2 376 νO2N2 6CO2 3H2O 376 νO2N2 νO2 6 32 75 νN2 282 HP H P HP HR H R HP H RP 40576 78114 3 169 554 kJkmol HP 6 h CO2 3 h H2O 282 h N2 HP 2600K 6128074 3104 520 28277 963 3 280 600 HP 2400K 6115 779 393 741 28270 640 2 968 000 Linear interpolation TAD 2529 K νiC Pi 6 0842 4401 3 1872 18015 282 1042 28013 114666 kJkmol K T HPνiC Pi 3 169 554 114666 2764 TAD 3062 K 21 high Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1391 What is the adiabatic flame temperature before the secondary air is added in Problem 1351 CH4 νO2 376 N2 CO2 2H2O 376 ν Ν2 O balance ν 2 νN2 752 Do a CV around the combustion chamber then the energy equation becomes Energy Eq HR1 HP 1 H P HP 1 HR1 H R HR1 f fuel ho HFuel Hair1 Hair1 2HO2 376 HN2 26086 376 5911 56 623 kJkmol HFuel M CP T 16043 2254 500 298 73045 kJkmol ho f fuel H P HV 16043 50 010 802 310 kJkmol HP 1 h CO2 075 h H2O 3854 h N2 ho f fuel HFuel Hair1 H P from energy Eq 802 310 73045 56 623 866 237 kJkmol fuel at 2400 K HP 1 115 779 2 93 741 752 70 640 834 474 kJkmol at 2600 K HP 1 128 074 2 104 520 752 77 963 923 396 kJkmol T 2400 200 866 237 834 474 923 396 834 474 2471 K Fuel Air1 Air2 1500 K T ad Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1392 Ethene C2H4 burns with 150 theoretical air in a steady flow constantpressure process with reactants entering at P0 T0 Find the adiabatic flame temperature Stoichiometric C2H4 3O2 376N2 2CO2 2H2O 1128N2 Actual C2H4 45O2 376N2 2CO2 2H2O 15 O2 1692N2 HP H P 2h CO2 2h H2O 15h O2 1692h N2 H R f Fu h HP H P H R HP H RP 28054 47 158 1 322 9705 kJ kmol Fu HP 2h CO2 2h H2O 15h O2 1692h N2 Initial guess based on 22151692 N2 from A9 T1 2100 K HP2000 1 366 982 HP1900 1 278 398 TAD 1950 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1393 A special coal burner uses a stoichiometric mixture of coal and an oxygen argon mixture 11 mole ratio with the reactants supplied at Po To Find the adiabatic flame temperature assuming complete combustion Combustion of carbon C O2 Ar CO2 Ar CV Combustion chamber Energy Eq HR H R HP H P HP h f CO2 h CO2 h Ar Table A9 or A10 h f CO2 393 522 kJkmol reference H R 0 Table A5 C P Ar 052 39948 20773 kJkmolK HP H R H P 0 h f CO2 393 522 kJkmol fuel HP 5200 292 112 20773 5200 298 393 938 kJkmol fuel HP 4800 266 488 20773 4800 298 360 005 kJkmol fuel interpolate T4 Tadflame 5195 K Comment At this temperature some chemical equilibrium reactions will be important see Chapter 14 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1394 A gasturbine burns natural gas assume methane where the air is supplied to the combustor at 1000 kPa 500 K and the fuel is at 298 K 1000 kPa What is the equivalence ratio and the percent theoretical air if the adiabatic flame temperature should be limited to 1800 K The reaction equation for a mixture with excess air is CH4 νO2 O2 376 N2 2 H2O 1 CO2 376νO2 N2 νO2 2O 2 HP H P HP HR H R HR From table A9 at 500 K notice fuel is at 298 K HR 0 νO2hO2 376 hN2 νO26086 376 5911 28 3114 νO2 From table A9 at 1800 K HP 2 hH2O hCO2 376 νO2 hN2 νO2 2 hO2 2 62 693 79432 376 νO2 48 979 νO2 2 51 674 101 470 235 835 νO2 From table 133 H P H R H RP 160450 010 802 160 kJkmol Now substitute all terms into the energy equation 802 160 101 470 235 835 νO2 28 3114 ν O2 Solve for νO2 νO2 802 160 101 470 235 835 28 3114 3376 Theoretical air 100 3376 2 1688 Φ AFs AF 2 3376 0592 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1395 Liquid butane at 25oC is mixed with 150 theoretical air at 600 K and is burned in a steady flow burner Use the enthalpy of combustion from Table 133 to find the adiabatic flame temperature out of the burner C4H10 1565 O2 376 N2 4 CO2 5 H2O 325 O2 3666 N 2 Energy Eq HP HR 0 HP H R HR H P H RP HR Reactants HR 9759245 36668894 416 193 kJkmol H RP M HV 58124 45 344 2 635 575 kJkmol HP 4h CO2 5h H2O 325 h O2 3666 h N2 So the energy equation becomes HP 2 635 575 416 193 3 051 768 kJkmol Trial and Error HP 2000 K 4 91439 5 72788 325 59176 3666 56137 2 980 000 HP 2200 K 4 103562 5 83153 325 66770 3666 63362 3 369 866 Linear interpolation to match RHS TP 2037 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1396 Natural gas we assume methane is burned with 200 theoretical air and the reactants are supplied as gases at the reference temperature and pressure The products are flowing through a heat exchanger and then out the exhaust as in Fig P1396 What is the adiabatic flame temperature right after combustion before the heat exchanger The reaction equation for stoichiometric mixture is CH4 νO2 O2 376 N2 1 CO2 2 H2O c N2 O balance 2 νO2 2 2 νO2 2 200 theoretical air νO2 2 2 4 so now more O2 and N2 CH4 4 O2 376 N2 1 CO2 2 H2O 1504 N2 2 O2 Energy Eq Hair Hfuel HR HP H P HP H R HR HP H R HR H P H RP 0 From Table 133 H RP 1604 50 010 802 160 kJkmol HP h CO2 2 h H2O 2 h O2 1504 h N2 From Table A9 HP 1600 67 659 2 52 907 2 44 267 1504 41 904 892 243 HP 1500 61 705 2 48 149 2 40 600 1504 38 405 816 814 HP 1400 55 895 2 43 491 2 36 958 1504 34 936 742 230 Linear interpolation to get 802 160 T 1400 100 802 160 742 230 816 814 742 230 1480 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1397 Solid carbon is burned with stoichiometric air in a steady flow process The reactants at T0 P0 are heated in a preheater to T2 500 K as shown in Fig P1397 with the energy given by the product gases before flowing to a second heat exchanger which they leave at T0 Find the temperature of the products T4 and the heat transfer per kmol of fuel 4 to 5 in the second heat exchanger Combustion of carbon C O2 376 N2 CO2 376 N2 CV Combustion chamber and preheater from 1 to 4 no external Q For this CV states 2 and 3 are internal and do not appear in equations Energy Eq HR H R HP4 H P HP4 h f CO2 h CO2 376h N2 Table A9 or A10 h f CO2 393 522 kJkmol reference H R 0 HP4 H R H P 0 h f CO2 393 522 kJkmol fuel HP4 2400 115 779 376 70 640 381 385 kJkmol fuel HP4 2600 128 074 376 77 963 421 215 kJkmol fuel interpolate T4 Tadflame 2461 K Control volume Total Then energy equation H R Q H P Q RP H h f CO2 0 393 522 kJ kmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1398 Gaseous ethanol C2H5OH is burned with pure oxygen in a constant volume combustion bomb The reactants are charged in a stoichiometric ratio at the reference condition Assume no heat transfer and find the final temperature 5000 K C2H5OH 3 O2 2 CO2 3 H2O Energy Eq UP UR H R HR nRR TR H P HP nPR T P Solve for the properties that depends on TP and recall HR 0 HP nPR TP H R H P nRR TR h0 f fuel 2 h0 f CO2 3 h0 f H2O 4R TR Fuel h0 f fuel 235 000 kJkmol for IG from Table A10 so HP nPR TP 235 000 2393 522 3241 826 4 831451 29815 1 267 606 kJkmol LHS HP nPR TP 2 h CO2 3 h H2O 5 831451 TP From Table A9 we find LHS5600 2 317 870 3 278 161 415726 5600 1 237 417 LHS6000 2 343 782 3 302 295 415726 6000 1 345 014 Tadflame 5712 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1399 Liquid nbutane at T0 is sprayed into a gas turbine with primary air flowing at 10 MPa 400 K in a stoichiometric ratio After complete combustion the products are at the adiabatic flame temperature which is too high so secondary air at 10 MPa 400 K is added with the resulting mixture being at 1400 K Show that Tad 1400 K and find the ratio of secondary to primary air flow CV Combustion Chamber C4H10 65 O2 65 376 N2 5 H2O 4 CO2 2444 N2 Energy Eq Hair Hfuel HR HP H P HP H R HR HP H R HR H P H RP HR HP 45344 58124 65376 2971 3027 2 727 861 kJkmol HP 1400 5 43491 4 55895 2444 34936 1 294 871 HP Remark Try TAD 1400 HP 2 658 263 2400 K HP 2 940 312 2600 K CV Mixing Chamber Air Second νO2 sO2 376 N2 ΗP νO2 second Hair HP 1400 νO2 second Hair 1400 νO2 second HP HP 1400 Hair 1400 Hair 400 1432990 168317 14198 93 ratio νO2 secνO2 prim 9365 143 A Mixing Combustion Chambe To 1400 Fue Air Ai Primary Secondary T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13100 The enthalpy of formation of magnesium oxide MgOs is 601 827 kJkmol at 25C The melting point of magnesium oxide is approximately 3000 K and the increase in enthalpy between 298 and 3000 K is 128 449 kJkmol The enthalpy of sublimation at 3000 K is estimated at 418 000 kJkmol and the specific heat of magnesium oxide vapor above 3000 K is estimated at 3724 kJkmol K a Determine the enthalpy of combustion per kilogram of magnesium b Estimate the adiabatic flame temperature when magnesium is burned with theoretical oxygen a Mg 1 2 O2 MgOs hCOMB M h COMB f M h 601 827 2432 24 746 kJkg b assume TR 25C and also that TP 3000 K MgO vapor phase 1st law QCV HP HR 0 but HR 0 HP h f h 3000 h 298SOL h SUB C P VAPTP 3000 601 827 128 449 418 000 3724TP 3000 0 Solving TP 4487 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13101 In a test of rocket propellant performance liquid hydrazine N2H4 at 100 kPa 25C and oxygen gas at 100 kPa 25C are fed to a combustion chamber in the ratio of 05 kg O2kg N2H4 The heat transfer from the chamber to the surroundings is estimated to be 100 kJkg N2H4 Determine the temperature of the products exiting the chamber Assume that only H2O H2 and N2 are present The enthalpy of formation of liquid hydrazine is 50 417 kJkmol Liq N2H4 100 kPa 25oC Gas O2 100 kPa 25oC Products m O2m N2H4 05 32n O232n N2H4 and Q m N2H4 100 kJkg Energy Eq QCV HP HR 100 32045 3205 kJkmol fuel Combustion eq 1 N2H4 1 2 O2 H2O H2 N2 HR 150417 1 20 50417 kJ HP 241 826 h H2O h H2 h N2 Energy Eq now reads HP HR QCV H o P HP HP h H2O h H2 h N2 H o P HR Q CV 241 826 50 417 3205 289 038 kJkmol fuel Table A9 Guess T and read for water hydrogen and nitrogen 2800 K HP 115 463 81 355 85 323 282 141 too low 3000 K HP 126 548 88 725 92 715 307 988 too high Interpolate to get TP 2854 K Comb chamber Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Second Law for the Combustion Process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13102 Consider the combustion of hydrogen with pure oxygen in a stoichiometric ratio under steady flow adiabatic conditions The reactants enter separately at 298 K 100 kPa and the products exit at a pressure of 100 kPa What is the exit temperature and what is the irreversibility The reaction equation is H2 νO2 O2 H2O The balance of hydrogen is done now for oxygen we need vO2 05 Energy Eq HR HP 0 241 826 h H2O h H2O 241 826 kJkmol Interpolate now in table A9 for the temperature to give this enthalpy T 4991 K For this temperature we find from Table A9 in this case P Po so we do not need any pressure correction for the entropy SP S P s H2O 315848 kJkmol K For the reactants we have again no pressure correction SR s H2 05 s O2 130678 05 205148 233252 kJkmol K Sgen SP SR 315848 233252 82596 kJkmol H2 K I To Sgen 29815 82596 24 626 kJkmol H2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13103 Consider the combustion of methanol CH3OH with 25 excess air The combustion products are passed through a heat exchanger and exit at 200 kPa 400 K Calculate the absolute entropy of the products exiting the heat exchanger assuming all the water is vapor CH3OH 125 15 O2 376 N2 CO2 2 H2O 0375 O2 705 N2 We need to find the mole fractions to do the partial pressures n 1 2 0375 705 10425 yi ni n Gas mixture ni yi si R ln P0 yiP S i CO2 10 00959 225314 13730 239044 H2O 2 01918 198787 7967 206754 O2 0375 00360 213873 20876 234749 N2 705 06763 200181 2511 197670 SGAS MIX niS i 21345 kJK kmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13104 Consider the combustion of methanol CH3OH with 25 excess air The combustion products are passed through a heat exchanger and exit at 200 kPa 40C Calculate the absolute entropy of the products exiting the heat exchanger per kilomole of methanol burned using the proper amounts of liquid and vapor for the water CH3OH 125 15 O2 376 N2 CO2 2 H2O 0375 O2 705 N2 Products exit at 40 oC 200 kPa check for saturation yV MAX P PG 7384 200 nV MAX nV MAX 1 0375 705 nV nV MAX 0323 nLIQ 1677 Gas mixture ni yi si R ln P0 yiP S i CO2 10 01143 215633 12270 227903 H2O 0323 00369 190485 21671 212156 O2 0375 00429 206592 20418 22701 N2 705 08059 193039 3969 18907 SGAS MIX niS i 171450 kJK kmol fuel s LIQ 69950 1801505725 03674 73645 kJkmol SLIQ 1677 73645 12350 kJK kmol fuel SPROD 171450 12350 1838 kJK kmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13105 An inventor claims to have built a device that will take 0001 kgs of water from the faucet at 10C 100 kPa and produce separate streams of hydrogen and oxygen gas each at 400 K 175 kPa It is stated that this device operates in a 25C room on 10kW electrical power input How do you evaluate this claim Liq H2O 10oC 100 kPa 0001 kgs W CV 10 kW H2 gas O2 gas each at 400 K 175 kPa T0 25 oC H2O H2 1 2 O 2 Hi He 285830 180154201 10489 2961 1 2 3027 291 437 kJkmol Si Se 69950 180150151 03674 139219 83145 ln 175 1 2 213873 83145 ln 175 173124 kJkmol K WREV Hi He T0Si Se 291 437 29815173124 239820 kJkmol W REV 000118015239 820 1331 kW I W REV W CV 1331 10 0 Impossible Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13106 Propene C3H6 is burned with air in a steady flow burner with reactants at Po To The mixture is lean so the adiabatic flame temperature is 1800 K Find the entropy generation per kmol fuel neglecting all the partial pressure corrections The reaction equation for a mixture with excess air is C3H6 νO2 O2 376 N2 3 H2O 3 CO2 376νO2 N2 νO2 45O2 Energy Eq HR H R HR H R HP H P HP The entropy equation SR Sgen SP Sgen SP SR SP S R From table A9 at reference T HR hFu νO2hO2 376 hN2 0 From table A9 at 1800 K HP 3 hH2O 3 hCO2 376 νO2 hN2 νO2 45 hO2 3 62 693 3 79432 376 νO2 48 979 νO2 45 51 674 193 842 235 835 νO2 From table 133 H P H R H RP 4208145 780 1 926 468 kJkmol Now substitute all terms into the energy equation 1 926 468 193 842 235 835 νO2 0 Solve for νO2 νO2 1 926 468 193 842 235 835 73468 νN2 27624 Table A910 contains the entropies at 100 kPa so we get SP 3 259452 3 302969 73468 45 264797 27624 248304 930024 kJkmolK SR 267066 73468 205148 27624 191609 706725 kJkmol K Sgen 930024 706725 2233 kJkmolK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13107 Hydrogen peroxide H2O2 enters a gas generator at 25C 500 kPa at the rate of 01 kgs and is decomposed to steam and oxygen exiting at 800 K 500 kPa The resulting mixture is expanded through a turbine to atmospheric pressure 100 kPa as shown in Fig P13107 Determine the power output of the turbine and the heat transfer rate in the gas generator The enthalpy of formation of liquid H2O2 is 187 583 kJkmol H2O2 H2O 1 2 O2 n H2O2 M m H2O2 01 34015 000294 kmols n exmix n H2O2 15 000441 kmols C p mix 2 3 1872 18015 1 3 0922 31999 32317 C v mix 32317 83145 240 kmix 32317240 13464 CV turbine Assume reversible s3 s 2 T3 T2 P3 P2 k1 k 800100 500 02573 5288 K w C p mix T2 T3 32317800 5288 8765 kJkmol W CV n exmix w 000441 8765 3866 kW CV Gas Generator H 1 000294187 583 0 55149 H 2 000294241 826 18002 0001470 15836 63476 Q CV H 2 H 1 63476 55149 8327 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13108 Graphite C at P0 T0 is burned with air coming in at P0 500 K in a ratio so the products exit at P0 1200 K Find the equivalence ratio the percent theoretical air and the total irreversibility C 1φbO2 376 N2 CO2 1φ 1 O2 376 1φ N2 Stoichiometric b 1 Energy Eq HP HR HP 1200 HR H R H P 44 473 1φ 1 29 761 3761φ28 109 1φ 6086 3765911 0 393 522 1φ 3536 Sgen SP SR PR ν s R ln y Reactants yO2 021 yN2 079 carbon is solid Products yO2 01507 yN2 079 yCO2 00593 S P 27939 2536 250011 13295 234227 40275 S R 574 3536220693 376 20674 35348 For the pressure correction the term with the nitrogen drops out same y R PR ν ln y R lnyCO2 2536 lnyO2 prod 3536 lnyO2 reac R ln00593 2536 ln01507 3536 ln021 8314528235 2536 18927 3536 15606 17503 Sgen 40275 35348 17503 5102 kJkmol carbonK I T0 Sgen 152 117 kJ kmol C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13109 Calculate the irreversibility for the process described in Problem 1352 2 C 2 O2 1 CO2 1 CO 1 2 O 2 Process V constant C solid n1GAS 2 n2GAS 25 P2 P1 n2T2 n1T1 200 25 1000 2 2982 8384 kPa H1 HR 0 H2 HP 1393522 33397 1110527 21686 120 22703 437 615 kJ 1Q2 U2 U1 H2 H1 n2R T2 n1R T 1 437 615 0 8314525 1000 2 2982 453 442 kJ GAS COMBUSTION cb Reactants SR 25740 2205148 831451 ln 200 100 410250 kJK Products ni yi s i R ln P0 yiP S i CO2 10 040 269299 10061 259238 CO 10 040 234538 10061 224477 O2 05 020 243579 4298 239281 SP 10259238 10224477 05239281 603355 kJK I T0SP SR 1Q2 29815603355 410250 453 442 511 016 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13110 Two kilomoles of ammonia are burned in a steady flow process with x kmol of oxygen The products consisting of H2O N2 and the excess O2 exit at 200C 7 MPa a Calculate x if half the water in the products is condensed b Calculate the absolute entropy of the products at the exit conditions 2NH3 xO2 3H2O N2 x 15O2 Products at 200 oC 7 MPa with nH2O LIQ nH2O VAP 15 a yH2O VAP PGP 15538 7 15 15 1 x 15 x 5757 b SPROD SGAS MIX SH2O LIQ Gas mixture ni yi si R lnyiPP0 S i H2O 15 0222 204595 22810 181785 O2 4257 0630 218985 31482 187503 N2 10 0148 205110 19439 185671 SGAS MIX 15181785 4257187503 1018567 125655 kJK SH2O LIQ 1569950 1801523223 03674 15775 kJK SPROD 125655 15775 14143 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13111 A flow of 002 kmols methane CH4 and 200 theoretical air both at reference conditions are compressed separately to P3 P4 2 MPa then mixed and then burned in a steady flow setup like a gasturbine After combustion state 6 heat transfer goes out so the exhaust state 7 is at 600 K a Find T3 T4 and T5 b Find the total rate of irreversibility from inlet to state 5 c Find the rate of heat transfer minus the work terms Q W 1 W 2 The reaction equation for stoichiometric ratio is CH4 νO2 O2 376 N2 CO2 2 H2O νO2 376 N 2 where the carbon and hydrogen balance have been done Now the oxygen O 2νO2 2 2 νO2 2 νN2 752 The actual combustion becomes νO2 2 2 4 νN2 1504 CH4 4 O2 376 N2 CO2 2 H2O 1504 N2 2 O2 The two compressors assumed adiabatic and reversible Air T3 T1 P3P1 k1 k 29815 2000100 141 14 7017 K CH4 T4 T2 P4P2 k1 k 29815 2000100 12991 1299 5942 K n air 4 476 n Fu 1904 n Fu 03808 kmols C Pair 2897 1004 290859 kJkmolK C PFu 16043 2254 361609 kJkmolK Mix Comb Heat exch C1 C2 W1 1 2 3 4 5 6 7 W2 Q Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Energy Eq mixing chamber n air h 3 n Fu h 4 n air h 5 air n Fu h 5 Fu use constant specific heats to evaluate and divide with n Fu to get 1904 290859 7017 361609 5942 1904 290859 361609 T5 solve T5 6951 K CV From the inlet 12 to the exit state 5 no Q so y5 air 1904 2004 y5 Fu 1 2004 S gen n airs 5 air s 1 n Fu s 5 Fu s 2 03808 290859 ln 6951 29815 83145 ln 100 1904 2004 2000 002 361609 ln 6951 29815 83145 ln 100 1 2004 2000 03808 01376 002 30625 06649 kWK I To S gen 29815 06649 1982 kW CV Total control volume from inlet 1 2 to final exit at state 7 Q W 1 W 2 n air h 1 n Fu h 2 n P h 7 n Fu HR HP HR HP Ho R Ho P HP H RP H P The lower heating value per kmol fuel is LHV H RP Ho R Ho P 16043 50 010 802 3104 kJkmol HP h CO2 2 h H2O 1504 h N2 2 h O2 12 906 2 10 499 1504 8894 2 9245 186 1598 kJkmol fuel Q W 1 W 2 002 802 3104 186 1598 12 323 kW 123 MW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13112 A flow of hydrogen gas is mixed with a flow of oxygen in a stoichiometric ratio both at 298 K and 50 kPa The mixture burns without any heat transfer in complete combustion Find the adiabatic flame temperature and the amount of entropy generated per kmole hydrogen in the process The reaction equation is H2 νO2 O2 H2O The balance of hydrogen is done now for oxygen we need vO2 05 Energy Eq HR HP 0 241 826 h H2O h H2O 241 826 kJkmol Interpolate now in table A9 for the temperature to give this enthalpy T 4991 K For this temperature we find from Table A9 SP s H2O R lnPPo 315848 831451 ln05 321611 kJkmol K For the reactants we have SR s H2 R lnPPo 05 s O2 R lnPPo 130678 05 205148 15 831451 ln05 241897 kJkmol K Sgen SP SR 321611 241897 79714 kJkmol H2 K Recall that this entropy generation includes the mixing process we did not use the partial pressures after mixing but the total pressure 50 kPa before mixing Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13113 Methane is burned with air both of which are supplied at the reference conditions There is enough excess air to give a flame temperature of 1800 K What are the percent theoretical air and the irreversibility in the process The combustion equation with X times theoretical air is CH4 2XO2 376 N2 CO2 2H2O 2X1O2 752X N2 Energy Eq Hair Hfuel HR HP H P HP H R HR HP H R HR H P H RP 0 From Table 133 H RP 1604 50 010 802 160 kJkmol HP h CO2 2 h H2O 2X1 h O2 752X h N2 From Table A9 and the energy equation HP 1800 79 432 2 62 693 2X1 51 674 752X 48 979 802 160 so 101 470 471 670 X 802 160 X 14856 Theoretical air 1486 The products are Products CO2 2H2O 09712 O2 11172 N 2 The second law Sgen SP SR and I To Sgen Reactants Pi 100 kPa Po 100 kPa so f from Table A9 ni yi so f R ln yiPi Po S i kJ kmol K CH4 1 1 186251 0 186251 O2 2X 021 205148 12976 218124 N2 752 X 079 191609 196 193569 SR niS i 299684 kJK kmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Products Pe 100 kPa Po 100 kPa ni yi so 1800 R ln yiPe Po S i kJ kmol K CO2 1 006604 302969 22595 325564 H2O 2 013208 259452 16831 276283 O2 09712 006413 264797 22838 287635 N2 11172 073775 248304 2529 250833 SP niS i 395972 kJK kmol fuel I ToSP SR 29815395972 299684 287 MJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13114 Pentane gas at 25C 150 kPa enters an insulated steady flow combustion chamber Sufficient excess air to hold the combustion products temperature to 1800 K enters separately at 500 K 150 kPa Calculate the percent theoretical air required and the irreversibility of the process per kmol of pentane burned C5H12 8X O2 376 N2 5 CO2 6 H2O 8X1 O2 3008X N 2 Energy Eq Qcv HR HP WCV WCV 0 Qcv 0 Reactants C5H12 ho f from A9 and h 500 for O2 and N2 from A9 HR f hoC5H12 8X h O2 3008X h N2 146 500 8X 6086 3008 X 5911 226 491 X 146 500 HP 5 f ho hCO2 6 f ho hH2O 8X1 h O2 3008 X h N2 5393 522 79 432 6241 826 62 693 8X1 51 674 3008 X 48 979 1 886 680 X 3 058 640 Energy Eq solve for X HR HP 226 491 X 146 500 1 886 680 X 3 058 640 X 1754 b Reactants Pin 150 kPa Po 100 kPa so f ni yi so f so 500 R ln yiPin Po S i kJ kmol K C5H 12 1 1 348945 3371 345574 O2 8X 021 220693 9605 230298 N2 3008 X 079 20674 1411 205329 SR niS i 1441034 kJK kmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Products Pe 150 kPa Po 100 kPa ni yi so 1800 R ln yiPe Po S i kJ kmol K CO2 5 00716 302969 18550 321519 H2O 6 0086 259452 17027 276479 O2 8X1 00864 264797 16988 281785 N2 3008X 0756 248304 1045 247259 SP niS i 17 732073 kJK kmol fuel I ToSP SR 2981517 73207 14 41034 990 MJkmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13115 A closed rigid container is charged with propene C3H6 and 150 theoretical air at 100 kPa 298 K The mixture is ignited and burns with complete combustion Heat is transferred to a reservoir at 500 K so the final temperature of the products is 700 K Find the final pressure the heat transfer per kmole fuel and the total entropy generated per kmol fuel in the process C3H6 νO2 O2 376 N2 3 CO2 3 H2O x N2 Oxygen O2 balance 2 νO2 6 3 9 νO2 45 Actual Combustion φ 15 νO2 ac 15 45 675 C3H6 675 O2 2538 N2 3 CO2 3 H2O 2538 N2 225 O2 P2 P1 npT2 nRT1 100 3363 700 3313 29815 2383 kPa Enthalpies from Table A9 HP 700 317 754 314 190 253811 937 22512 499 426 916 kJ kmol fuel Enthalpy of combustion from table 133 converted to mole basis H RP 45 780 42081 1 926 468 kJkmol fuel U2 U1 1Q2 0 H2 H1 n2RT2 n1RT1 1Q2 H RP HP 700 nPRT2 n1RT1 1 926 468 426 916 3363 83145 700 3313 83145 29815 1613106 kJ kmol fuel Entropies from Table A9 and pressure correction Reactants ni yi s i R lnyiPP0 S i C3H 8 10 00302 267066 29104 29617 O2 675 02037 205143 13228 218376 N2 2538 07661 191609 2216 189393 S1 29617 675 218376 2538 189393 6577 kJ kmol fuel K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Products ni yi s i R lnyiPP0 S i CO2 3 00892 250752 12875 263627 H2O 3 00892 218739 12875 231614 O2 225 00669 231465 15266 246731 N2 2538 07547 216865 488 211985 S2 3263627 231614 225 246731 2538 211985 7421 kJkmol fuel K 1S2 gen S2 S1 1Q2Tres 7421 6577 1613106 500 4070 kJ kmol fuel K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Problems Involving Generalized Charts or Real Mixtures Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13116 Liquid butane at 25C is mixed with 150 theoretical air at 600 K and is burned in an adiabatic steady state combustor Use the generalized charts for the liquid fuel and find the temperature of the products exiting the combustor Prod at Tp 25 C LIQ C H 4 10 o 150 Air 600 K Adiab Comb Q 0 CV TC 4252 K Tr 0701 h C4H10 ho f IG h LIQ h see Fig D2 126200 485 83145 4252 143 346 kJ C4H10 15 65 O2 376 975 N 2 4 CO2 5 H2O 325 O2 3666 N 2 h AIR 9759245 36668894 416 193 kJ HR 416 193 143 346 272 847 kJ HP 4393522 h CO2 5241826 h H2O 325 h O2 3666 h N2 Energy Eq HP HR 0 4 h CO2 5 h H2O 325 h O2 3666 h N2 3 056 065 Trial and Error LHS2000 K 2 980 000 LHS2200 K 3 369 866 Linear interpolation to match RHS TP 2039 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13117 Repeat Problem 13101 but assume that saturatedliquid oxygen at 90 K is used instead of 25C oxygen gas in the combustion process Use the generalized charts to determine the properties of liquid oxygen Problem same as 13101 except oxygen enters at 2 as saturated liquid at 90 K m O2m N2H4 05 32n O232n N2H4 and Q m N2H4 100 kJkg Energy Eq QCV HP HR 100 32045 3205 kJkmol fuel Reaction equation 1 N2H4 1 2 O2 H2O H2 N2 At 90 K Tr2 901546 0582 h f 52 Figure D2 h h 83145 1546 52 6684 kJkmol h AT 2 6684 0922 32 90 29815 12825 kJkmol HR 50417 1 20 12825 44005 kJ H P 241826 Energy Eq h P h H2O h H2 h N2 Qcv HR H 282 626 kJkmol P From Table A9 HP 2800K 282 141 HP 3000K 307 988 kJkmol Therefore TP 2804 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13118 A gas mixture of 50 ethane and 50 propane by volume enters a combustion chamber at 350 K 10 MPa Determine the enthalpy per kilomole of this mixture relative to the thermochemical base of enthalpy using Kays rule h MIX O 0584740 05103900 94320 kJkmol C P0 MIX 05 3007 17662 05 44097 167 63583 h 350 h 298 63583 350 2982 3294 kJkmol Kays rule TC MIX 05 3054 05 3698 3376 K PC MIX 05 488 05 425 4565 MPa Tr 3503376 1037 Pr 104565 219 From Fig D2 h h 83145 3376 353 9909 kJkmol h MIX 350K10MPa 94320 3294 9909 100 935 kJkmol Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13119 A mixture of 80 ethane and 20 methane on a mole basis is throttled from 10 MPa 65C to 100 kPa and is fed to a combustion chamber where it undergoes complete combustion with air which enters at 100 kPa 600 K The amount of air is such that the products of combustion exit at 100 kPa 1200 K Assume that the combustion process is adiabatic and that all components behave as ideal gases except the fuel mixture which behaves according to the generalized charts with Kays rule for the pseudocritical constants Determine the percentage of theoretical air used in the process and the dew point temperature of the products Reaction equation Fuel mix h0 f FUEL 0274873 0884740 82767 kJkmol C P0 FUEL 02 22537 1604 08 17662 3007 49718 h FUEL 4971865 25 1989 kJkmol TCA 3054 K TCB 1904 K Tc mix2824 K PCA 488 PCB 460 Pc mix 4824 MPa Tr 33822824 1198 Pr 104824 2073 h hFUEL IN 831451 2824 218 5119 h FUEL IN 82767 1989 5119 85897 kJ kmol Energy Eq 18393522 44473 28241826 34506 32x 129761 1203x28109 85897 32x9245 1203x8894 0 a x 4104 or 4104 b nP 18 28 324104 1 1203 4104 63904 yH2O 2863904 PV100 PV 438 kPa T 305C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13120 Saturated liquid butane enters an insulated constant pressure combustion chamber at 25C and x times theoretical oxygen gas enters at the same P and T The combustion products exit at 3400 K With complete combustion find x What is the pressure at the chamber exit and what is the irreversibility of the process Butane T1 To 25oC sat liq x1 0 Tc 4252 K Pc 38 MPa Do the properties from the generalized charts Fig D1 Tr1 07 Pr1 01 P1 Pr1Pc 380 kPa Figs D2 and D3 h 1 h 1 f 485 RTc 1 s s1f 68 R Oxygen T2 To 25oC X Theoretical O2 Products T3 3400 K Assumes complete combustion C4H10 65X O2 4 CO2 5 H2O 65X1 O 2 Energy Eq Qcv HR HP Wcv Qcv 0 Wcv 0 HR nho f hC4H10 1126 200 17 146 143 346 kJ Products CO2 n f ho hCO2 4393 522 177 836 862 744 kJ H2O n f ho hH2O 5241 826 149 073 463 765 kJ O2 n f ho hO2 65X10 114 101 X1741 657 kJ HP ni ho f hi 741 657X 2 068 166 Energy Eq HP HR solve for X X 2594 Assume that the exit pressure equals the inlet pressure Pe Pi 380 kPa sC4H10 so f R ln P1 Po 1 s s1f sO2 so R ln P1 Po SR SC4H10 SO2 306647 1110 56539 20548 1110 65 2594 351645 kJK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Products ni yi so i R ln yiPe Po S i kJ kmol K CO2 4 02065 341988 2016 344004 H2O 5 02582 293550 0158 293708 O2 10368 05353 289499 5904 283595 SP niS i 578487 kJK I ToSP SR 29815 578487 351645 676 329 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13121 Liquid hexane enters a combustion chamber at 31C 200 kPa at the rate 1 kmols 200 theoretical air enters separately at 500 K 200 kPa and the combustion products exit at 1000 K 200 kPa The specific heat of ideal gas hexane is Cp 143 kJkmol K Calculate the rate of irreversibility of the process Hexane Tc 5075 K Pc 3010 kPa Tr1 06 Fig D1 Prg 0028 Pg1 Pr1Pc 8447 kPa Figs D2 and D3 h 1 h 1 f 516 RTc s 1 s1 f 856 R Air T2 500 K P2 200 kPa 200 theoretical air Products T3 1000 K P3 200 kPa a h C6H14 ho f h 1 h 1f 1 h 0 h 0 h h 0 h 0 h 0 0 h 1 h 0 C P T1 To 858 kJkmol ho f 167300 kJkmol h 1 h 1 516 83145 5075 21773 kJkmol h C6H14 188215 kJkmol sC6H14 so To C P ln T1 To R ln P1 Po s1 1 s so To C P ln T1 To R ln P1 Po 387979 285 5763 385066 kJkmolK s 1 s1 85683145 71172 kJkmolK sC6H14 313894 kJkmolK b C6H14 19O2 7144N2 6CO2 7H2O 95O2 7144N2 Tc prod yiTci 1793 K Tr3 T3 Tc prod 558 Ideal Gas c 1st Law Q HR HP W W 0 Q HP HR HR hC6H14 19h O2 7144 h N2 188 215 19 6086 7144 5911 349701 kJkmol fuel HP 6 f ho hCO2 7 f ho hH2O 95 f ho hO2 7144 f ho hN2 CO2 ho f h 393522 33397 360125 kJkmol H2O ho f h 241826 26000 215826 kJkmol Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful O2 ho f h 0 22703 22703 kJkmol N2 ho f h 0 21463 21463 kJkmol HP 1 922 537 kJ Q 2 272 238 kW d I To n SP SR Q To 25oC SR sC6H14 19 500 so Rln yO2P2 Po O2 7144 500 so Rln yN2P2 Po N2 sC6H14 313894 kJkmol K so 500O2 220693 kJkmol K so 500N2 206740 kJkmol K yO2 021 yN2 079 S R 191419 kWK Products ni yi so i R ln yiPe Po S i kJkmolK CO2 6 00639 269299 17105 286404 H2O 7 00745 232739 15829 248568 O2 95 01011 243579 13291 25687 N2 7144 07605 228171 3487 224684 SP nisi 219501 kJK I To n SP SR Q 3 109 628 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Fuel Cells Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13122 In Example 1315 a basic hydrogenoxygen fuel cell reaction was analyzed at 25C 100 kPa Repeat this calculation assuming that the fuel cell operates on air at 25C 100 kPa instead of on pure oxygen at this state Anode 2 H2 4 e 4 H Cathode 4 H 4 e 1 O2 2 H2O Overall 2 H2 1 O2 2 H2O Example 1316 G25C 474 283 kJ for pure O2 For PO2 021 01 MPa S O2 205148 83145 ln 021 218124 kJkmol S 269950 2130678 1218124 33958 kJkmol K G25C 571 660 2981533958 470 414 kJkmol E 470414 96487 4 1219 V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13123 Assume that the basic hydrogenoxygen fuel cell operates at 600 K instead of 298 K as in example 1315 Find the change in the Gibbs function and the reversible EMF it can generate Reaction 2 H2 O2 2 H2O At a temperature of 600 K the water formed is in a vapor state We can thus find the change in the enthalpy as H 0 600 K 2ho f hH2O g 2ho f hH2 ho f h O2 2241 826 10 499 20 8799 0 9245 489 497 kJ4 kmol e S 0 600 K 2 so f H2O g 2 fso H2 so f O2 2 213051 2 151078 22645 102504 kJ4 kmol e K G 0 600 K H 0 600 K TS 0 600 K 489 497 600102504 427 995 kJ4 kmol e Wrev G 0 427 995 kJ4 kmol e E 0 96485 8 G 0 427 995 96 485 4 1109 V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13124 A reversible fuel cell operating with hydrogen and pure oxygen produces water at the reference conditions Po To this is described in Example 1315 Find the work output and any heat transfer both per kmol of hydrogen Assume an actual fuel cell operates with a secondlaw efficiency of 70 and enough heat transfer takes place to keep it at 25C How much heat transfer is that per kmol of hydrogen The basic fuel cell analysis from Example 1315 Reaction 2 H2 O2 2 H2O Wrev ΔG 474 283 kJ2 kmol 237 142 kJkmol This assumed equilibrium with the surroundings so the heat transfer must be Q To SP SR 29815 3266 2 48 688 kJkmol that is the heat transfer goes out A 70 2nd law efficiency must mean the actual work is W 07 Wrev 165 999 kJkmol From the energy equation we get Q HP HR W 571 6602 165 999 119 831 kJkmol All this heat transfer means that the fuel cell will have a hard time staying at the room temperature it would tend to be warmer Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13125 Consider a methaneoxygen fuel cell in which the reaction at the anode is CH4 2H2O CO2 8e 8H The electrons produced by the reaction flow through the external load and the positive ions migrate through the electrolyte to the cathode where the reaction is 8 e 8 H 2 O2 4 H2O Calculate the reversible work and the reversible EMF for the fuel cell operating at 25C 100 kPa CH4 2H2O CO2 8e 8H and 8e 8H 2CO2 4H2O Overall CH4 2O2 CO2 2H2O a 25 oC assume all liquid H2O and all comp at 100 kPa H 0 25 C 393 522 2285 830 74 873 0 890 309 kJ S 0 25 C 213795 269950 186251 2205148 242852 kJK G 0 25 C 890 309 29815242852 817 903 kJ Wrev G 0 817903 kJ E 0 96485 8 G 0 817903 96485 8 106 V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13126 Consider a methaneoxygen fuel cell in which the reaction at the anode is CH4 2H2O CO2 8e 8H The electrons produced by the reaction flow through the external load and the positive ions migrate through the electrolyte to the cathode where the reaction is 8 e 8 H 2 O2 4 H2O Assume that the fuel cell operates at 1200 K instead of at room temperature CH4 2H2O CO2 8e 8H and 8e 8H 2CO2 4H2O Overall CH4 2O2 CO2 2H2O At 1200 K assume all gas H2O and all comp at 100 kPa H 0 1200 K 1393522 44473 2241826 34506 20 29761 174873 16043 22541200 2982 780 948 kJ S 0 1200 K 1279390 2240485 1186251 16043 2254 ln 1200 2982 2250011 237397 kJK G 0 1200 K H 0 1200 K TS 0 1200 K 780 948 1200237397 809 436 kJ Wrev 809 436 kJ E 0 809 436 96 485 8 1049 V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13127 For a PEC fuel cell operating at 350 K the constants in Eq1329 are ileak 001 iL 2 io 0013 all Acm2 b 008 V c 01 V ASR 001 Ω cm2 and EMF 122 V Find the voltage and the power density for the current density i 025 075 and 10 Acm2 The fuel cell output voltage is from Eq1329 V EMF b ln i ileak io i ASRohmic c ln iL iL i ileak i 025 V 122 023966 00025 001393 09639 V p Vi 0241 Wcm2 i 075 V 122 032547 00075 00478 08392 V p Vi 06294 Wcm2 i 10 V 122 034822 001 00703 07915 V p Vi 07915 Wcm2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13128 Assume the PEC fuel cell in the previous problem How large an area does the fuel cell have to deliver 1 kW with a current density of 1 A cm2 The fuel cell output voltage is from Eq1329 V EMF b ln i ileak io i ASRohmic c ln iL iL i ileak i 10 V 122 034822 001 00703 07915 V p Vi 07915 Wcm2 Total power P p A 07915 Wcm2 A 1000 W A 1000 07915 cm2 1263 cm 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13129 A SOC fuel cell at 900 K can be described by an EMF 106 V and the constants in Eq1329 as b 0 V c 01 V ASR 004 Ω cm2 ileak 001 iL 2 io 013 all Acm2 Find the voltage and the power density for the current density i 025 075 and 10 Acm2 The fuel cell output voltage is from Eq1329 V EMF b ln i ileak io i ASRohmic c ln iL iL i ileak i 025 V 106 00 001 001393 1036 V p Vi 0259 Wcm2 i 075 V 106 00 003 00478 09822 V p Vi 0737 Wcm2 i 10 V 106 00 004 00703 095 V p Vi 095 Wcm2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13130 Assume the SOC fuel cell in the previous problem How large an area does the fuel cell have to deliver 1 kW with a current density of 1 A cm2 The fuel cell output voltage is from Eq1329 V EMF b ln i ileak io i ASRohmic c ln iL iL i ileak i 10 V 106 00 004 00703 095 V p Vi 095 Wcm2 Total power P p A 095 Wcm2 A 1000 W A 1000 095 cm2 1053 cm 2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13131 A PEC fuel cell operating at 25oC generates 10 V that also account for losses For a total power of 1 kW what is the hydrogen mass flow rate The work term for the fuel cell is W G 0 298 K so W n G 0 298 K m Mn MW G 0 298 K 2016 kgkmol 1 kW 05 474 283 kJkmol 85 106 kgs Since the actual output was 1 V rather than the ideal 1229 V we then infer an efficiency of 11229 0814 so m ac m 0814 1045 106 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13132 A basic hydrogenoxygen fuel cell operates at 600 K instead of 298 K as in example 1315 For a total power of 5 kW find the hydrogen mass flow rate and the exergy in the exhaust flow Reaction 2 H2 O2 2 H2O At a temperature of 600 K the water formed is in a vapor state We can thus find the change in the enthalpy as H 0 600 K 2ho f hH2O g 2ho f hH2 ho f h O2 2241 826 10 499 20 8799 0 9245 489 497 kJ4 kmol e S 0 600 K 2 so f H2O g 2 fso H2 so f O2 2 213051 2 151078 22645 102504 kJ4 kmol e K G 0 600 K H 0 600 K TS 0 600 K 489 497 600102504 427 995 kJ4 kmol e Wrev G 0 427 995 kJ4 kmol e This was for 2 kmol hydrogen m Mn MW G 0 600 K 2016 kgkmol 5 kW 05 427 995 kJkmol 471 106 kgs The exhaust flow is water vapor at 600 K 100 kPa so use Table B1 m ψ m h ho Tos so 471 106 3129 10487 29815 83037 03673 0031 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13133 Consider the fuel cell with methane in Problem 13125 Find the work output and any heat transfer both per kmol of methane Assume an actual fuel cell operates with a secondlaw efficiency of 75 and enough heat transfer takes place to keep it at 25C How much heat transfer is that per kmol of methane CH4 2H2O CO2 8e 8H and 8e 8H 2CO2 4H2O Overall reaction CH4 2O2 CO2 2H2O T 25 oC 298 K assume all liquid H2O and all comp at 100 kPa H 0 25 C 393 522 2285 830 74 873 0 890 309 kJkmol S 0 25 C 213795 269950 186251 2205148 242852 kJK G 0 25 C 890 309 29815242852 817 903 kJkmol Wrev G 0 817 903 kJkmol This assumed equilibrium with the surroundings so the heat transfer must be Q To SP SR 29815 242852 72 406 kJkmol A 75 2nd law efficiency must mean the actual work is W 075 Wrev 613 427 kJkmol From the energy equation we get Q HP HR W 890 309 613 427 276 881 kJkmol All this heat transfer means that the fuel cell will have a hard time staying at the room temperature it would tend to be warmer Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Combustion applications and efficiency Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13134 For the combustion of methane 150 theoretical air is used at 25oC 100 kPa and relative humidity of 70 Find the composition and dew point of the products The reaction equation for stoichiometric mixture is CH4 νO2 O2 376 N2 2 H2O 1 CO2 376 νO2 N2 C balance νCO2 1 H balance 2νH2O 4 N2 balance νN2 376 νO2 O balance 2 νO2 νH2O 2νCO2 2 2 1 νO2 2 150 theoretical air νO2 15 2 3 so now more O2 and N2 CH4 3 O2 376 N2 2 H2O 1 CO2 1128 N2 1 O2 Add water to the dry air from Eq1128 w 0622 φPg Ptot φPg 0622 07 3169 100 07 3169 00141 So the number of moles to add is from Eq1332 x 7655 w 7655 00141 0108 and the added number of moles is νO2 x 0324 the products are then Products 2324 H2O 1 CO2 1128 N2 1 O 2 The water partial pressure becomes Pv yv Ptot 2324 2324 1 1128 1 100 14894 kPa Tdew 538oC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13135 Pentane is burned with 120 theoretical air in a constant pressure process at 100 kPa The products are cooled to ambient temperature 20C How much mass of water is condensed per kilogram of fuel Repeat the answer assuming that the air used in the combustion has a relative humidity of 90 C5H12 12 8 O2 376 N2 5 CO2 6 H2O 096 O2 361 N 2 Products cooled to 20oC 100 kPa so for H2O at 20C Pg 2339 kPa yH2O MAX PgP 2339 100 nH2O MAX 4206 nH2O MAX nH2O MAX 1007 nH2O Therefore nH2O VAP 1007 nH2O LIQ 6 1007 4993 mH2O LIQ 4993 18015 72151 1247 kgkg fuel Pv1 09 2339 2105 kPa w1 0622 2105 97895 0013375 Water from moist air Eq 1332 nH2O IN 0013375 2897 18015 96 1 376 0983 kmol nH2O OUT 0983 6 6983 nH2O LIQ 6983 1007 5976 kmol mH2O LIQ 5976 18015 72151 1492 kgkg fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13136 A gas turbine burns methane with 150 theoretical air Assume the air is 25oC 100 kPa and relative humidity of 80 How large a fraction of the product mixture water comes from the moist inlet air The reaction equation for stoichiometric mixture is CH4 νO2 O2 376 N2 2 H2O 1 CO2 376 νO2 N2 C balance νCO2 1 H balance 2νH2O 4 N2 balance νN2 376 νO2 O balance 2 νO2 νH2O 2νCO2 2 2 1 νO2 2 150 theoretical air νO2 15 2 3 so now more O2 and N2 CH4 3 O2 376 N2 2 H2O 1 CO2 1128 N2 1 O2 Add water to the dry air from Eq112829 w 0622 φPg Ptot φPg 0622 07 3169 100 07 3169 00141 So the number of moles to add is from Eq1332 x 7655 w 7655 00141 0108 and the added number of moles is νO2 x 0324 the products are then Products 2324 H2O 1 CO2 1128 N2 1 O 2 Fraction of water is 03242324 01394 so 139 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13137 In an engine a mixture of liquid octane and ethanol mole ratio 91 and stoichiometric air are taken in at T0 P0 In the engine the enthalpy of combustion is used so that 30 goes out as work 30 goes out as heat loss and the rest goes out the exhaust Find the work and heat transfer per kilogram of fuel mixture and also the exhaust temperature 09 C8H18 01 C2H5OH 1155 O2 43428 N2 84 H2O 74 CO2 43428 N2 For 09 octane 01 ethanol convert to mole basis H RP mix 09 H RP C8H18 01 H RP C2H5OH 09 44 425 114232 01 26 811 46069 4 690 797 kJ kmol M mix 09 M oct 01 M alc 107414 Energy h in qin h ex ωex h ex h ex ωex h ex h in H RP mix ωex h ex qin H RP mix ωex qin 03 H RP 1 407 239 kJ kmol 13 101 kJ kg Fu h prod h ex 04 H RP 1 876 319 kJ kmol Fu h prod 84 h H2O 74h CO2 43428 h N2 h prod 1300 84 38 941 74 50 148 43428 31 503 2 066 312 h prod 1200 84 34 506 74 44 473 43428 28 109 1 839 668 Linear interpolation to get the value of h prod 1 876 319 satisfied for T 1216 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13138 The gas turbine cycle in problem 1028 has a qH 1160 kJkg air added by combustion Assume the fuel is methane gas and the qH equals the heating value at To Find the AF ratio on a mass basis The heating value and the qH are scaled with the fuel mass and total mixture mass respectively so the relation is see Eq1334 qH HV mmixmfuel HV AF 1 1160 kJkgmix AF HV qH 1 50 010 1160 1 4311 Comment This is about 300 theoretical air Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13139 An oven heated by natural gas burners has the combustion take place inside a U shaped steel pipe so the heating is done from the outside surface of the pipe by radiation Each burner delivers 15 kW of radiation burning 110 theoretical air with methane The products leave the pipe at 800 K Find the flowkgs of methane The burner is now switched to oxygenenriched air 30 O2 and 70 N2 so assume the same conditions as before with the same exit T Find the new flow kgs of methane needed The reaction equation for stoichiometric mixture is CH4 νO2 O2 376 N2 2 H2O 1 CO2 c N2 O balance 2 νO2 2 2 νO2 2 110 theoretical air νO2 11 2 22 so now more O2 and N2 CH4 22 O2 376 N2 2 H2O 1 CO2 8272 N2 02 O2 Energy Eq assume reactant enter at reference T QCV HP HR H P HP H R H RP H P HP 2 h H2O h CO2 8272 h N2 02 h O2 2 18002 22806 8272 15046 02 15836 186 438 kJkmol H RP 16043 50 010 802 310 kJkmol QCV 802 310 186 438 615 872 kJkmol m M Q CVQCV 16043 15 615 872 kgs 000039 kgs The new air just changes the nitrogen to oxygen ratio to 7030 2333 so the actual reaction is changed to CH4 22 O2 2333 N2 2 H2O 1 CO2 51333 N2 02 O2 HP 2 h H2O h CO2 51333 h N2 02 h O2 2 18002 22806 51333 15046 02 15836 139 213 kJkmol QCV 802 310 139 213 663 097 kJkmol m M Q CVQCV 16043 15 663 097 kgs 000036 kgs This is an 8 reduction in the fuel flow rate Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13140 A slight disadvantage of the oxygenenriched air for combustion is an increase in flame temperature which tends to increase NOx Find the flame temperature for the previous problem for both cases standard air and oxygenenriched air Energy Eq assume reactant enter at reference T 0 HP HR H P HP H R H RP HP HP H RP HV 16043 50 010 802 310 kJkmol Reaction std air CH4 22 O2 376 N2 2 H2O 1 CO2 8272 N2 02 O2 HP 2 h H2O h CO2 8272 h N2 02 h O2 The heating value should be spread over 1147 moles of products which is about 70 000 each so look in A9 for oxygen or nitrogen T 2200 2400 K HP 2200K 2 83153 103562 8272 63362 02 66770 807 352 HP 2000K 2 72788 91 439 8272 56137 02 59176 713 215 Interpolate T 2189 K Reaction std air CH4 22 O2 2333 N2 2 H2O 1 CO2 51333 N2 02 O2 HP 2 h H2O h CO2 51333 h N2 02 h O2 The heating value should be spread over 8333 moles of products which is about 96 000 each so look in A9 for oxygen or nitrogen T 3000 K HP 3000K 2 126 548 152 853 51333 92715 02 98 013 901 485 HP 2800K 2 115 463 140 435 51333 85 323 02 90 080 827 366 HP 2600K 2 104 520 128 074 51333 77 963 02 82 225 753 766 Interpolate T 2732 K a significant higher flame temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13141 A gas turbine burns methane with 200 theoretical air The air and fuel comes in through two separate compressors bringing them from 100 kPa 298 K to 1400 kPa and enters a mixing chamber and combustion chamber What are the specific compressor work and qH to be used in Brayton cycle calculation Use constant specific heat to solve The reaction equation for stoichiometric mixture is CH4 νO2 O2 376 N2 2 H2O 1 CO2 c N2 O balance 2 νO2 2 2 νO2 2 200 theoretical air νO2 2 2 4 so now more O2 and N2 CH4 4 O2 376 N2 2 H2O 1 CO2 1504 N2 2 O2 For 1 kmol fuel m 16043 4 476 2897 16043 55159 56763 kg reactants The temperatures after compression are T2 air T1 P2P1 k1k 29815 14 02857 6337 K k 14 T2 fuel T1 P2P1 k1k 29815 14 0230177 5473 K k 1299 wc air Cp T2 T1 10046337 29815 33689 kJkg air wc fuel Cp T2 T1 22545473 29815 56168 kJkg fuel wc total m mfuel wc fuel mair m wc air 1604 56763 56158 55159 56763 33689 34324 kJkg mix q H RP HV but scaled to be per kg reactants q 50 010 16043 56763 14134 kJkg HV AF 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13142 Find the equivalent heat transfer qH to be used in a cycle calculation for constant pressure combustion when the fuel is a methane and b gaseous octane in both cases use water vapor in the products and a stoichiometric mixture Methane combustion becomes CH4 νO2 O2 376 N2 1 CO2 2 H2O 752 N2 Oxygen balance 2νO2 2 2 4 νO2 2 νN2 752 AF 2 31999 752 28013 16043 1712 mairmfuel q HV mmixmfuel HV AF 1 50 010 1712 1 2760 kJkg Octane combustion see reaction and AF in Example 131 HV from Table 133 q HV mmixmfuel HV AF 1 44 788 15 1 2799 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13143 Consider the steady combustion of propane at 25C with air at 400 K The products exit the combustion chamber at 1200 K It may be assumed that the combustion efficiency is 90 and that 95 of the carbon in the propane burns to form carbon dioxide the remaining 5 forms carbon monoxide Determine the ideal fuelair ratio and the heat transfer from the combustion chamber Ideal combustion process assumed adiabatic excess air to keep 1200 K out C3H8 5x O2 188x N2 3 CO2 4 H2O 5x 1 O2 188x N 2 HR 103900 5x0 3027 188x0 2971 103900 70990x HP 3393522 44473 4241826 34506 5x 10 29761 188x0 28109 2025232 677254x Energy Eq HP HR 0 Solving x 3169 FAIDEAL 1238 3169 001326 FAACTUAL 001326090 001473 C3H8 1426 O2 5362 N 2 285 CO2 015 CO 4 H2O 9335 O2 5362 N 2 HR 103900 14260 3027 53620 2971 98 570 kJ HP 285393522 44473 015110527 28427 4241826 34506 93350 29761 53620 28109 51 361 kJ QCV HP HR 149 931 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13144 A gasoline engine is converted to run on propane Assume the propane enters the engine at 25C at the rate 40 kgh Only 90 theoretical air enters at 25C such that 90 of the C burns to form CO2 and 10 of the C burns to form CO The combustion products also include H2O H2 and N2 exit the exhaust at 1000 K Heat loss from the engine primarily to the cooling water is 120 kW What is the power output of the engine What is the thermal efficiency Propane T1 25oC m 40 kghr M 44094 kgkmol Air T2 25oC 90 theoretical Air produces 90 CO2 10 CO Products T3 1000 K CO2 CO H2O H2 N2 C3H8 45O2 1692N2 27 CO2 03CO 33H2O 07H2 1692N 2 n C3H8 m M3600 0000252 kmols Energy eq Q HR HP W Q 120 kW HR nC3H8 ho f 103 900 kJ Products CO2 nCO2ho f h 27393522 33397 9723375 kJ CO nCOho f h 03110527 21686 26652 kJ H2O nH2Oho f h 33241826 26000 712226 kJ H2 nH2 f ho h 070 20663 144641 kJ N2 nN2ho f h 16920 21463 363154 kJ HP ni ho f hi 1 333 598 kJ W Q nHR HP 1899 kW C3H8 Table 133 HRPo 50343 kJkg HHV n C3H8 MHRPo 5594 kW ηth W HHV 0339 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13145 A small aircooled gasoline engine is tested and the output is found to be 10 kW The temperature of the products is measured to 600 K The products are analyzed on a dry volumetric basis with the result 114 CO2 29 CO 16 O2 and 841 N2 The fuel may be considered to be liquid octane The fuel and air enter the engine at 25C and the flow rate of fuel to the engine is 15 104 kgs Determine the rate of heat transfer from the engine and its thermal efficiency a C8H18 b O2 376b N 2 114 CO2 29 CO c H2O 16 O2 841 N 2 b 841 376 2237 a 1 8 114 29 1788 c 9a 16088 C8H18 125 O2 471 N 2 638 CO2 162 CO 9 H2O 089 O2 471 N 2 HR h0 f C8H18 250 105 kJkmol HP 638393 522 15 788 162110527 10 781 9241 826 12 700 0890 11187 4710 10712 4 119 174 kJkmol HP HR 4 119 174 250 105 3 869 069 kJkmol H P H R 000015114233 869 069 5081 kW Mfuel 11423 Q CV 5081 10 4081 kW Fuel heating value from table 133 Q H m HV 000015 47 893 7184 kW ηTH W NETQ H 107184 0139 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13146 A gasoline engine uses liquid octane and air both supplied at P0 T0 in a stoichiometric ratio The products complete combustion flow out of the exhaust valve at 1100 K Assume that the heat loss carried away by the cooling water at 100C is equal to the work output Find the efficiency of the engine expressed as worklower heating value and the second law efficiency C8H18 νO2 O2 376 N2 8 CO2 9 H2O 47 N2 2 νO2 16 9 νO2 125 LHV 44 425 kJ kg fuel LHV 507476106 kJ kmol fuel HP 1100 8 38885 9 30190 47 24760 1746510 CV Total engine Hin Hex W Qloss Hex 2 W 2 W Hin Hex HR Hν H RP HR HP 1100 507476106 0 1746510 3328250 W 1664106 kJ kmol fuel ηth W LHV 507476106 1664106 0328 Find entropies in and out inlet S 360575 Fu E A 21812 S 205148 83145 ln O2 1 476 ASE A AE N2 E 191609 83145 ln A376 476E A 19357 ASE A AEin E 360575 125 21812 47 19357 12185 exit ASE A AE CO2 E 275528 83145 ln A 8 64E A 29282 ASE A AE H2O E 236732 83145 ln A 9 64E A 25304 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful ASE A AE N2 E 231314 81345 ln A47 64E A 23388 ASE A AEex E 8 29282 9 25304 47 23388 15612 Assume the same Qloss out to 100C reservoir in the reversible case and compute QA rev 0E A ASE Ain QA rev 0E AT0 ASE Aex QlossTres QA rev 0E A T0 S ex S in Qloss T0Tres 2981515612 12185 166410A6E A 2981537315 235110A6E A A kJ kmol fuelE Hin QA rev 0E A Hex WArevE A Qloss WArevE A Hin Hex Qloss QA rev 0E A Wac QA rev 0E A 401510A6E A A kJ kmol fuelE ηII Wac WArevE A 166410A6E A401510A6E A 0414 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13147 Repeat Problem 1326 for a certain Utah coal that contains according to the coal analysis 682 C 48 H 157 O on a mass basis The exiting product gas contains 309 CO 267 HA2E A 159 COA2E A and 257 HA2E AO on a mole basis Convert the mass concentrations to number of kmol per 100 kg coal C 6821201 5679 HA2E A 482016 2381 OA2E A 1573200 0491 Now the combustion equation reads x5679 C 2381 HA2E A 0491 OA2E A y HA2E AO z OA2E A in 309 CO 267 HA2E A 159 COA2E A 257 HA2E AO out in 100 kmol of mix out Now we can do the atom balance to find x y z C 5679x 309 159 x 8241 HA2E A 2381 8241 y 267 257 y 32778 OA2E A 0491 8241 A32778 2E A z A309 2E A 159 A257 2E z 23765 Therefore for 100 kmol of mixture out require 8241 kg of coal 32778 kmol of steam 23765 kmol of oxygen Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13148 Many coals from the western United States have a high moisture content Consider the following sample of Wyoming coal for which the ultimate analysis on an asreceived basis is by mass Component Moisture H C S N O Ash mass 289 35 486 05 07 120 58 This coal is burned in the steam generator of a large power plant with 150 theoretical air Determine the airfuel ratio on a mass basis Converting from mass analysis Substance S HA2E C OA2E NA2E cM 0532 352 48612 1232 0728 kmol 100 kg coal 00156 175 405 0375 0025 Product SOA2E HA2E AO COA2E oxygen required 00156 0875 405 Combustion requires then oxygen as 00156 0875 405 49406 The coal does include 0375 OA2E A so only 45656 OA2E A from air100 kg coal AF 15 45656 45656 376 2897100 9444 kg airkg coal Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13149 A fuel CAxE AHAyE A is burned with dry air and the product composition is measured on a dry mole basis to be 96 COA2E A 73 OA2E A and 831 NA2E A Find the fuel composition xy and the percent theoretical air used νAFuE ACAxE AHAyE A νAOE A2E AOA2E A 376νO2N2 96 CO2 73 O2 831 N2 νH2OH2O NA2E A balance 376νO2 831 νO2 22101 OA2E A balance νO2 96 73 A 1 2E A νH2O νH2O 10402 H balance νFu y 2 νH2O 20804 C balance νFu x 96 Fuel composition ratio xy 9620804 0461 Theoretical air νO2AC νO2stoich 100 A 22101 96 1 4 29804E A 100 1493 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13150 In an engine liquid octane and ethanol mole ration 91 and stoichiometric air are taken in at 298 K 100 kPa After complete combustion the products run out of the exhaust system where they are cooled to 10AoE AC Find the dew point of the products and the mass of water condensed per kilogram of fuel mixture Reaction equation with 09 octane and 01 ethanol is 09 C8H18 01 C2H5OH 1155 O2 43428 N2 84 H2O 74 CO2 43428 N2 yH2O A 84 84 74 43428E A 01418 PH2O yH2OPtot 143 kPa Tdew 529 C 10 C PH2O 12276 yH2O 0012276 A x x 74 43428E x 06317 νH2O 777 A kmol kmol Fu mixE Mmix yiMi 09 11423 01 46069 107414 mH2O cond νH2O 18015 107414 1303 A kg kg Fu mixE Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13151 Find the lower heating value for the fuel blend in problem 15134 with a scaling as in Table 133 Reaction equation with 09 octane and 01 ethanol is 09 C8H18 01 C2H5OH 1155 O2 43428 N2 84 H2O 74 CO2 43428 N2 We can solve several ways use Table A10 to find HA RPE A and do it on a mole basis and then divide with M for mixture or we can convert the mole fractions to mass fractions and use Table 133 values Let us use Table 133 Mmix yiMi 09 11423 01 46069 107414 cAethanolE A yMMmix A01 46069 107414E A 0042889 cAoctaneE A 0957111 LHV cAethanolE A LHVAethanolE A cAoctaneE A LHVAoctaneE A 0042889 26 811 0957111 44 425 kJkg 43 670 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13152 Determine the higher heating value of the sample Wyoming coal as specified in Problem 13148 The higher heating value is for liquid water in the products We need the composition of the products Converting from mass analysis Substance S HA2E C OA2E NA2E cM 0532 352 48612 1232 0728 kmol 100 kg coal 00156 175 405 0375 0025 Product SOA2E HA2E AO COA2E So the combustion equation becomes for 100 kg coal Fuel Air 175 HA2E AO 405 COA2E A 00156 SOA2E The formation enthalpies are from Table A10 Therefore AhE ARP0E A HA o PE A HA o RE A 405393 522 175285 830 00156296 842 2 098 597 kJ100 kg coal So that HHV 20 986 kJkg coal Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13153 Ethene CA2E AHA4E A and propane C3H8 in a 11 mole ratio as gases are burned with 120 theoretical air in a gas turbine Fuel is added at 25C 1 MPa and the air comes from the atmosphere 25C 100 kPa through a compressor to 1 MPa and mixed with the fuel The turbine work is such that the exit temperature is 800 K with an exit pressure of 100 kPa Find the mixture temperature before combustion and also the work assuming an adiabatic turbine φ 1 C2H4 C3H8 8 O2 3008 N2 5 CO2 6 H2O 3008 N2 φ 12 C2H4 C3H8 96 O2 36096 N2 5 CO2 6 H2O 16 O2 36096 N2 45696 kmol air per 2 kmol fuel CV Compressor air flow Energy Eq wc h2 h1 Entropy Eq s2 s1 Pr2 Pr1 P2P1 13573 T2 air 5708 K wc 57644 29834 2781 kJkg 80566 kJkmol air CV Mixing Chanber no change in composition n airh air in n Fu1h 1 in n Fu2h 2 in SAMEexit C P F1 C P F2 Texit T0 45696 C P air T2 air Texit C2H4 C P F1 4343 C3H8 C P F2 7406 C P air 2907 Texit 45696C P airT2 4343 7406 T0 4343 7406 45696 C P air 5487 K Dew Point Products yH2O A 6 5 6 16 36096E A 01232 PH2O yH2OPtot 1232 kPa TAdewE A 1055C CV Turb combustor mixer compressor no Q wnet Hin Hout HR HP 800 800K out so no liquid H2O h fC2H4 h fC3H8 5 h CO2 6 h H2O 16 h O2 36096 h N2 2 576 541 A kJ 2 kmol FuE wT wnet wcomp 2 944 695 A kJ 2 kmol FuE A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13154 Phenol CA6E AHA5E AOH M 9411 kgkmol Cp 17 kJkgK is used in a combustion process being burned with stoichiometric air The reactants are supplied at 12 MPa the air at TAoE A and the phenol as liquid at 100C Enough heat transfer QA1E A takes place after the combustion so the products reach 1500 K After the heat transfer the products are sent through a turbine where they expand to 120 kPa a Find the heat transfer QA1E A kJkmol fuel b Find the enthalpy of formation of the fuel as vapor c Find the specific turbine work output W kJkmol fuel First do the combustion equation for Phenol CA6E AHA5E AOH CA6E AHA5E AOH νAOE A2E A OA2E A 376 NA2E A 6 COA2E A 3 HA2E AO 2632 NA2E O balance 1 2νAOE A2E A 6 2 3 15 νAOE A2E A 7 NA2E A balance νANE A2E A 376 νAOE A2E A 376 7 2632 Energy Eq 0 QA1E A HARE A1E A HAPE A2E A QA1E A HA RPE A HARE A1E A HAPE A2E Table 133 HA RPE A 9411 31 117 2 928 421 kJkmol fuel as liquid HARE A1E A M Cp TA1E A TA0E A 9411 17 100 25 11 999 kJkmol Table A9 HAPE A2E A 6 AhE ACO2E A 3 AhE AH2OE A 2632 AhE AN2E 6 61705 3 481549 2632 38405 1 525 497 kJkmol QA1E A HA RPE A HARE A1E A HAPE A2E A 2 928 421 11 999 1 525 497 1 414 923 kJkmol HA RPE A 9411 31 774 2 990 251 6 AhE A f CO2 3 AhE A f H2O Ah Ef fuelE Ah Ef fuelE A 3 AhE A f H2O 6 AhE A f CO2 HA RPE 3 241 826 6393 522 2 990 251 96 359 kJkmol For WATE A we need exit T and as product is mainly NA2E A use air approximation same k Constant CAPE A k TA3E A TA2E A PA3E APA2E A A k1 k E A 1500 1201200 A02857E A 77695 K Variable CAPE A k PAr3E A PAr2E A PA3E APA2E A 4831610 48316 TA3E A 850 K See Table A72 for this or use the standard entropy for this calculation HAPE A3E A 6 25418 3 199695 2632 166345 650 237 kJkmol W HAPE A2E A HAPE A3E A 1 525 497 650 237 875 260 kJkmol Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13155 The phenol in Problem 13154 is used in a gasturbine cycle where we need the heating value or the negative of the enthalpy of combustion at the temperature after the compressor a What is the T after the compressor assuming a simple adiabatic compression from 100 kPa to 12 MPa b Find the HARPE A HAPE A HARE A both at that T except phenol is liquid at 100C Assume the reactants are similar to air Constant CAPE A k TA2E A TA1E A PA2E APA1E A A k1 k E A 29815 1200100 A02857E A 6064 K Variable CAPE A k PAr2E A PAr1E A PA2E APA1E A 10907 12 13088 TA3E A 600 K See Table A72 for this or use the standard entropy for this calculation First do the combustion equation for Phenol CA6E AHA5E AOH CA6E AHA5E AOH νAOE A2E A OA2E A 376 NA2E A 6 COA2E A 3 HA2E AO 2632 NA2E O balance 1 2νAOE A2E A 6 2 3 15 νAOE A2E A 7 NA2E A balance νANE A2E A 376 νAOE A2E A 376 7 2632 For enthalpy of fuel use CAPE A 17 kJkgK similar to other fuels HARE A h fuel 7AhE AO2E A 376AhE AN2E A 17 9411 600 29815 79245 376 8894 347 097 kJkmol HAPE A 6 12906 3 10499 2632 8894 343 023 kJkmol HA RPE A 9411 31 774 2 990 251 Gaseous fuel value at 298 K which is then a hypothetical state as the reference so since the 600 K is higher than the 100C then it does not matter Now we get HA RPE A HA RPE A HAPE A HARE A 2 990 251 343 023 347 097 2 994 325 kJkmol For a cycle calculation we would then use qAHE A HA RPE A 9411 1 AF in kJkg mix Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13156 Consider the gas mixture fed to the combustors in the integrated gasification combined cycle power plant as described in Problem 1330 If the adiabatic flame temperature should be limited to 1500 K what percent theoretical air should be used in the combustors Product CHA4E A HA2E A CO COA2E A NA2E A HA2E AO HA2E AS NHA3E vol 03 296 410 100 08 170 11 02 Mixture may be saturated with water so the gases are HA2E AS and NHA3E A out CHA4E HA2E COA E COA2E NA2E n 03 296 410 100 08 817 yAv maxE A 73843000 nAvE AnAvE A 817 Solving nAvE A 02 kmol the rest is condensed 03 CHA4E A 296 HA2E A 410 CO 100 COA2E A 08 NA2E 02 HA2E AO 359x OA2E A 376 359x NA2E A 513 COA2E A 304 HA2E AO 359x 1 OA2E A 1350x 08 NA2E For the fuel gas mixture at 40C nACE AP0 mixE A 03 1604 22537 296 2016 142091 410 2801 10413 100 4401 08418 08 28013 10416 02 18015 18723 2455157 nAhE A 0 f mixE A 0374873 2960 410110527 100393522 080 02241826 8537654 kJ HAmixE A 8 537 654 245515740 25 8 500 827 kJ Assume air enters at 25C AhE AairE A 0 Products at 1500 K HAPE A 513393522 61705 304241826 48149 359x 10 40600 135x 080 38405 24 336 806 6 642 215x Energy Eq HAPE A HARE A HAmixE x A24336809 8500827 6642215E A 2384 or 238 theo air Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13157 Carbon monoxide CO is burned with 150 theoretical air and both gases are supplied at 150 kPa and 600 K Find the heating value and the adiabatic flame temperature CO νO2 O2 376 N2 CO2 νN2 N2 O balance 1 2νO2 1 νO2 05 νO2 actual 075 Now the actual reaction equation has excess oxygen as CO 075 O2 376 N2 CO2 282 N2 025 O2 From the definition of heating value negative of enthalpy of combustion Eq1314 or 1315 HV HA RPE A HA PE A HA RE A ho f CO2 0 Aho Ef COE 393 522 110 527 282 995 kJkmol CO 10 103 kJkg CO as for Table 133 Actual energy Eq HR HP HA PE A HP HA RE A HR HP HA RE A HR HA PE A HA RPE A h CO 075 h O2 282 h N2 HV h CO 075 h O2 282 h N2 282 995 8942 075 9245 282 8894 323 952 kJkmol The left hand side is HP h CO2 025 h O2 282 h N2 HP 2600 128 074 025 82 225 282 77 963 368 486 HP 2400 115 779 025 74 453 282 70 640 333 597 HP 2200 103 562 025 66 770 282 63 362 298 935 Now we can do a linear interpolation for the adiabatic flame temperature T 2200 200 A323 952 298 935 333 597 298 935E A 2344 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13158 A rigid container is charged with butene C4H8 and air in a stoichiometric ratio at P0 T0 The charge burns in a short time with no heat transfer to state 2 The products then cool with time to 1200 K state 3 Find the final pressure P3 the total heat transfer 1Q3 and the temperature immediately after combustion T2 The reaction equation is having used C and H atom balances C4H8 νO2 O2 376 N2 4 CO2 4 H2O 376 νO2N2 Counting now the oxygen atoms we require νO2 6 CV analysis gives U3 U1 Q W Q H3 H1 P2V2 P1V1 H3 H1 ARE An2T3 n1T1 H3 H1 HP 1200 HA RE A HA PE A HA RE A HP AME A HA RPE A HP 56108 45 316 950055 1 592 535 kJkmol Where AME A 56108 and n1 1 6 476 2956 n2 4 4 6 376 3056 Τable A9 at 1200 K hCO2 44473 hH2O34506 hN228109 Now solving for the heat transfer Q 1592535 831453056 1200 2956 29815 1824164 A kJ kmol fuelE To get the pressure assume ideal gases P3 n2R T3 V2 P1 n2T3 n1T1 101325 kPa A 3056 1200 2956 29815E A 4216 kPa Before heat transfer takes place we have constant U so U2 U1 0 H2 H1 n2 ARE AT2 n1 ARE AT1 Now split the enthalpy H2 HA PE A HPT2 and arrange things with the unknowns on LHS and knowns on RHS HP n2 ARE AT2 HR HA PE A n1 ARE AT1 2 542 590 73278 2 469 312 Trial and error leads to LHS 3000 K 3 209 254 3056 831451 3000 2 446 980 LHS 3200 K 3 471 331 3056 831451 3200 2 658 238 linear interpolation T 3021 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13159 Natural gas approximate it as methane at a ratio of 03 kgs is burned with 250 theoretical air in a combustor at 1 MPa where the reactants are supplied at T0 Steam at 1 MPa 450C at a rate of 25 kgs is added to the products before they enter an adiabatic turbine with an exhaust pressure of 150 kPa Determine the turbine inlet temperature and the turbine work assuming the turbine is reversible CH4 νO2 O2 376 N2 CO2 2 H2O 752 N2 2 νO2 2 2 νO2 2 Actual νO2 2 25 5 CH4 5 O2 188 N2 CO2 2 H2O 3 O2 188 N2 CV combustor and mixing chamber HR nH2O AhE AH2O in HP ex nH2O n H2O n Fu m H2OMFu m FuMH2O A25 16043 03 18015E A 7421 A kmol steam kmol fuelE A Energy equation becomes nH2O h ex h in H2O h CO2 2h H2O 3h O2 188h N2 ex HA RPE A 50 010 16043 802 310 h ex h in H2O AhE AH2O ex 150725 so then h CO2 9421h H2O 3h O2 188h N2 ex 914 163 A kJ kmol fuelE A Trial and error on Tex Tex 1000 K LHS 749 956 Tex 1100 K LHS 867429 Tex 1200 K LHS 987 286 Tex 1139 K Tin turbine If air then TAex turbineE A 700 K and Tavg 920 K Find ACE AP mix between 900 and 1000 K From Table A9 ACE AP mix nAiE ACE APi nAiE A A5367 94214063 33462 188324 E32221E A 35673 kJkmol K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful ACE AV mix ACE AP mix ARE A 273587 kJkmol kmix 1304 Tex turbine 1139 150 1000A02331E A 732 K H732 193706 942115410 313567 18812932 448 371 kJkmol wT Hin Hex Hin Hex 914 163 448 371 465 792 A kJ kmol fuelE W T n FuwT m FuwTM Fu 03 465 79216043 8710 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13160 The turbine in Problem 13153 is adiabatic Is it reversible irreversible or impossible Inlet to the turbine is the exit from the mixing of air and fuel at 1 MPa From solution to 13153 we have ACE AP C2H2 4343 ACE AP C3H8 7406 Tturbinein 5487 K C2H4 C3H8 96 O2 36096 N2 5 CO2 6 H2O 16 O2 36096 N2 Sex Sin dQT Sgen Sgen φ Inlet 1 MPa 5487 K ASE AFu ASE A iE A ACE AP Fu ln TT0 nAiE yAiE AsE A i ARE AlnA yiP EP0 E ASE Ai C2H4 1 002097 24582 12989 258809 C3H8 1 002097 31509 12989 328079 OA2E 96 02013 223497 5816 217681 NA2E 36096 07568 209388 16828 19256 Sin 258809 328079 96 217681 36096 19256 96273 nAiE yAiE AsE A i ARE AlnA yiP EP0 E AS iE COA2E 5 01027 257496 18925 276421 HA2E AO 6 01232 223826 17409 241235 OA2E 16 00329 23592 28399 264319 NA2E 36096 07413 221016 2489 223505 Sex 5 276421 6 241235 16 264319 36096 223505 11320 A kJ 2kmol Fu KE Sgen Sex Sin 1693 A kJ 2kmol Fu KE A 0 Possible but one should check the state after combustion to account for generation by combustion alone and then the turbine expansion separately Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13161 Consider the combustion process described in Problem 13119 a Calculate the absolute entropy of the fuel mixture before it is throttled into the combustion chamber b Calculate the irreversibility for the overall process From solution to 13119 fuel mixture 08 C2H6 02 CH4 at 65C 10 MPa ACE AP0 FUELE A 49718 kJkmol K Using Kays rule TAr1E A 1198 PAr1E A 2073 and x 4104 theoretical air or 1313 OA2E A 4936 NA2E A in at 600 K 100 kPa and 18COA2E A 28HA2E AO 993OA2E A 4936NA2E A out at 100 kPa 1200 K AsE A 0 FUELE A 02186251 08229597 8314502 ln 02 08 ln 08 225088 sA TP E A 49718 ln A3382 2982E A 83145 ln A10 01E A 32031 From Fig D3 AsE A E A AsE AAFUELE A 137 83145 11391 AsE AFUELE A 225088 32031 11391 18166 kJkmol K Air at 600 K 100 kPa nAiE yAiE AsE A i ARE AlnAyiPP0E A AS iE OA2E 1313 021 22645 12976 239426 NA2E 4936 079 212177 196 214137 SAAIRE A nAiE ASE AiE A 1371347 kJK SARE A 18166 1371347 138951 kJK Products at 1200 K 100 kPa PROD nAiE yAiE AsE A o iE ARE AlnAyiPP0E A AS iE COA2E 18 00282 279390 29669 309059 HA2E AO 28 00438 240485 26008 266493 OA2E 993 01554 250011 15479 265490 NA2E 4936 07726 234227 2145 236372 SAPE A nAiE ASE AiE A 156061 kJK I TA0E ASAPE A SARE A QACVE A 2981515 6061 13 8951 0 510 132 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13162 Liquid acetylene CA2E AHA2E A is stored in a highpressure storage tank at ambient temperature 25C The liquid is fed to an insulated combustorsteam boiler at the steady rate of 1 kgs along with 140 theoretical oxygen OA2E A which enters at 500 K as shown in Fig P13162 The combustion products exit the unit at 500 kPa 350 K Liquid water enters the boiler at 10C at the rate of 15 kgs and superheated steam exits at 200 kPa aCalculate the absolute entropy per kmol of liquid acetylene at the storage tank state b Determine the phases of the combustion products exiting the combustor boiler unit and the amount of each if more than one c Determine the temperature of the steam at the boiler exit a C2H2 ASE A IG 25CE A 200958 kJkmolK from A10 TR1 29823083 0967 From Fig D1 PR1 082 P1 082 614 503 MPa A S AE S 1 E 333ARE A 27687 kJkmolK ASE Aliq T1 P1 ASE AE AT0 P0 T ARE A ln P1P A S S E AP1 T1 140695 A kJ kmol KE b 1 C2H2 14 25 O2 2 CO2 1 H2O 1 O2 H1 226731 356 ARE A 3083 217605 kJ H2 350 6086 21301 kJ Products T3 350 K 768C PG 418 kPa yV max PG P A418 500E A 00836 nV max nV max 2 1 nV max 02737 nV gas mix nliq 1 02737 07263 Gas Mix 2 CO2 02737 H2O 1 O2 c Hliq3 07263285830 180153215 1049 204764 kJ Hgas mix3 2393522 2036 02737241826 1756 1541 847138 kJ H3 Hliq3 Hgas mix3 204 764 847 138 1 051 902 kJ H3 H1 H2 1 290 808 kJ or H 3 H 1 H 2 1 290 80826038 49 574 kW m H2O h4 h5 h5 4201 A49574 15E A 33469 T5 4334C UPDATED JULY 2013 SOLUTION MANUAL CHAPTER 13 English Units Fundamentals of Thermodynamics Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CHAPTER 13 CONTENT CHAPTER 13 SUBSECTION PROB NO ConceptStudy Guide Problems 163 Energy Equation Enthalpy of Formation 164174 Enthalpy of Combustion and heating Value 175180 Adiabatic Flame Temperature 181190 Second Law for the Combustion Process 191195 Fuel cells 196 Efficiency and review 197199 Mixtures and generalized charts 200201 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Concept Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13163E The output gas mixture of a certain airblown coal gasifier has the composition of producer gas as listed in Table 132 Consider the combustion of this gas with 120 theoretical air at 147 lbfin2 pressure Find the dew point of the products and the mass of water condensed per poundmass of fuel if the products are cooled 20 F below the dew point temperature 3 CH4 14 H2 509 N2 06 O2 27 CO 45 CO2 311 O2 1169 N2 345 CO2 20 H2O 52 O2 1678 N 2 Products yH2O yH2O MAX PG 147 20 345 20 52 1678 PG 12923 lbfin2 TDEW PT 1104 F At T 904 F PG 07089 lbfin 2 yH2O max 07089 147 nH2O nH2O 345 52 1678 nH2O 1051 nH2O LIQ 20 1051 949 lb mol mH2O LIQ 31614250928063227284544 94918 0069 lbmlbm fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Energy and Enthalpy of Formation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13164E What is the enthalpy of formation for oxygen as O2 If O For CO2 From Table F6 h f O2 0 h f O 107 124 Btulbmol h f CO2 169 184 Btulbmol or Table F11 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13165E One alternative to using petroleum or natural gas as fuels is ethanol C2H5OH which is commonly produced from grain by fermentation Consider a combustion process in which liquid ethanol is burned with 110 theoretical air in a steady flow process The reactants enter the combustion chamber at 77 F and the products exit at 140 F 147 lbfin2 Calculate the heat transfer per lbmol fuel C2H5OH 11 3 O2 376 N2 2CO2 3H2O 03O2 12408N 2 Products at 140 F 147 lbfin2 Psat 2892 psia yH2O 2892147 nv2 03 nv 12408 nv 36023 3 no condensation h 119 252 Btulbmol as liquid from F11 f HR 119 252 0 0 119 252 Btu lbmol fuel HP 2169184 570 347518 5065 030 4437 124080 4384 472 690 Btu lbmol fuel QCV HP HR 353 438 Btulbmol fuel An ethanol flame Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13166E Liquid methanol is burned with stoichiometric air both supplied at Po To in a constant pressure process and the products exits a heat exchanger at 1600 R Find the heat transfer per lbmol fuel CH3OH 15 O2 376 N2 1 CO2 2 H2O 564 N2 Reactants at 77 F products are at 1600 R 147 psia CH3OH ho f 102 846 Btulbmol from table F11 for the liquid state HR 1 h LIQ 0 0 102 846 Btulbmol fuel HP 1169 184 11 798 2103 966 9241 5640 7681 303 515 Btulbmol fuel Q HP HR 303 515 102 846 200 669 Btulbmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13167E In a new highefficiency furnace natural gas assumed to be 90 methane and 10 ethane by volume and 110 theoretical air each enter at 77 F 147 lbfin2 and the products assumed to be 100 gaseous exit the furnace at 100 F 147 lbfin2 What is the heat transfer for this process Compare this to an older furnace where the products exit at 450 F 147 lbfin2 09 CH4 01 C2H6 11 215 O2 376 2365 N2 11 CO2 21 H2O 0215 O2 8892 N 2 a TP 100 F HR 0932190 0136432 32614 Btu HP 11169184 206 21103966 185 02150 162 88920 160 402 360 Btulbmol assuming all gas QCV HP HR 369 746 Btulb mol fuel b TP 450 F HP 11169184 3674 21103966 3057 02152688 88922610 370 184 Btu QCV HP HR 337 570 Btulb mol fuel Products 100 F 1 atm 110 Air 77 F 1 atm 09 CH 01 C H 77 F 1 atm 4 2 6 Heat transfer Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13168E Repeat the previous problem but take into account the actual phase behavior of the products exiting the furnace Same as 13167 except possible condensation 09 CH4 01 C2H6 11 215 O2 376 2365 N2 11 CO2 21 H2O 0215 O2 8892 N 2 a 100 F 147 lbfin2 Psat 095 psia yv max 095 147 nv maxnv max 10207 nv max 0705 nv 0705 nliq 21 0705 1395 Hliq 1395122 885 180156805 4509 170 847 Btulbmol Hgas 11169 184 206 0705103 966 185 0215162 8892160 257 584 Btulbmol HP Hliq Hgas 428 431 Btulbmol QCV HP HR 395817 Btulbmol fuel b TP 450 F no condensation HP 11169 184 3674 21103 966 3057 02152688 88922610 370 184 Btulbmol QCV HP HR 337 570 Btulbmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13169E Pentene C5H10 is burned with pure oxygen in a steady state process The products at one point are brought to 1300 R and used in a heat exchanger where they are cooled to 77 F Find the specific heat transfer in the heat exchanger C5H10 νO2O2 5 CO2 5 H2O stoichiometric νO2 75 Heat exchanger in at 1300 R out at 77 F so some water will condense 5 H2O 5 xH2Oliq x H2Ovap Check for condensation amount yH2Omax Pg 77 Ptot 0464 14696 003158 x 5 x x 0163 q Q n fuel 5 h ex h in CO2 5 h ex h in H2Ovap 5 xh fg H2O 5 8121 5 64685 5 0163 18919 164 459 Btu lb mol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13170E Methane CH4 is burned in a steady state process with two different oxidizers A Pure oxygen O2 and B a mixture of O2 x Ar The reactants are supplied at T0 P0 and the products in are at 3200 R both cases Find the required equivalence ratio in case A and the amount of Argon x for a stoichiometric ratio in case B a CH4 νO2 CO2 2H2O ν 2O2 νs 2 for stoichiometric mixture HP 3200 H R H P HP 3200 h CO2 33 579 Btulbmol h H2O 26 479 h O2 21 860 HP H R H P H RP 50 0102326 1604 344 867 Btulbmol 33 579 2 26 479 ν 221 860 42 817 ν 21 860 ν 138175 φ νsν 2138175 01447 b CH4 2O2 2x Ar CO2 2H2O 2x Ar h Ar C P Ar T CP Ar MAr T HP H R H P H RP 50 0102326 1604 344 867 Btulbmol 33579 2 26479 2 01253 39948 3200 53667 Now the energy equation becomes 344 867 86537 x 266625 x 9689 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13171E A rigid vessel initially contains 2 pound mole of carbon and 2 pound mole of oxygen at 77 F 30 lbfin2 Combustion occurs and the resulting products consist of 1 pound mole of carbon dioxide 1 pound mole of carbon monoxide and excess oxygen at a temperature of 1800 R Determine the final pressure in the vessel and the heat transfer from the vessel during the process 2 C 2 O2 1 CO2 1 CO 1 2 O 2 V constant C solid n1GAS 2 n2GAS 25 P2 P1 n2T2 n1T1 30 25 1800 2 5367 1258 lbf in2 H1 0 and we neglect the Pv for carbon when doing the U1 H2 1169184 14358 147518 9323 1 20 9761 188 141 Btu 1Q2 U2U1 H2H1 n2R T2 n1R T 1 188 141 0 19858925 1800 2 53667 194 945 Btu GAS COMBUSTION cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13172E A closed insulated container is charged with a stoichiometric ratio of oxygen and hydrogen at 77 F and 20 lbfin2 After combustion liquid water at 77 F is sprayed in such that the final temperature is 2100 R What is the final pressure Combustion reaction H2 1 2 O2 H2O Products 1 H2O xi H2O Energy eq per 1 lbmol hydrogen remember flow in U2 U1 xih i xih f liq 1 xi HP HR 1 xi R TP 3 2R TR Solve for xi xi h f liq HP R TP HP HR R TP 3 2R TR Table F6 HR φ HP 103 966 14 2185 89 7475 Btulbmol Table F11 h f liq 122 885 Btulbmol Substitute xi122885 897475 198588 2100 897475 1985882100 3 2 53667 92 3192 Btulbmol xi 3187 Volume is constant P1V1 nRR T1 P2V1 npR Tp nR 1 1 2 15 np 1 xi P2 P11 xiTP 15 T1 2041872100 1553667 2185 lbfin2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13173E Acetylene gas C2H2 is burned with stoichiometric air in a torch The reactants are supplied at the reference conditions Po To The products come out from the flame at 4500 R after some heat loss by radiation Find the heat loss per lbmol of fuel The reaction equation is C2H2 25 O2 376 N2 2 CO2 1 H2O 94 N2 Energy Eq HP Qout HR Ho P HP Qout Ho R Qout Ho R Ho P HP Ho R Ho P f fuel h h f H2O 2 h f CO2 97 477 103 966 2169 184 539 811 Btulbmol HP 2 52 4165 42 612 94 31 9425 447 705 Btulbmol Qout 539 811 447 705 92 106 Btulbmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13174E Natural gas methane is burned with stoichiometric air with reactants supplied at the reference conditions Po To in a steady flow burner The products come out at 1400 R If the burner should deliver 10 Btus what is the needed flowrate of natural gas in lbms The reaction equation for stoichiometric mixture is CH4 νO2 O2 376 N2 2 H2O 1 CO2 c N2 O balance 2 νO2 2 2 νO2 2 c 2 376 752 The products are cooled to 1400 R so we do not consider condensation and the energy equation is Energy Eq HR Q HP H P HP H R Q Q H P HP H R 2 h f H2O h f CO2 HP h f fuel HP 2 h H2O h CO2 752 h N2 2 7371 9315 752 6169 70 448 Btulbmol Q 2103 966 169 184 70 448 32 190 274 478 Btulbmol fuel Q n Qout Qout m M so m Q MQout 10 Btus 16043 lbmlbmol 274 478 Btulbmol 0000584 lbms 21 lbmh Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Enthalpy of combustion and heating value Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13175E What is the higher heating value HHV of nButane Either convert units from Table 133 or compute from the enthalpy of formation From Table F11 h f C4H10 54 256 Btulbmol M 58124 h f H2O liq 122 885 Btulbmol we need liquid for higher heating value h f CO2 169 184 Btulbmol HHV H RP f C4H10 h 4 f CO2 h 5 f H2O liq h 54 256 4169 184 5122 885 1 236 905 Btulbmol 21 280 Btulbm 49 500 kJkg 21 280 Btulbm from Table 133 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13176E Find the enthalpy of combustion and heating value for pure carbon Reaction C νO2 O2 376 N2 CO2 νO2376 N2 oxygen balance νO2 1 H RP Ho P Ho R h f CO2 f C h 169 184 0 169 184 Btulbmol 14 086 Btulbm M 12011 HV 169 184 Btulbmol 14 086 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13177E Acetylene gas C2H2 is burned with stoichiometric air in a torch The reactants are supplied at the reference conditions Po To The products come out from the flame at 5000 R after some heat loss by radiation Find the lower heating value for the fuel as it is not listed in Table 133 and the heat loss per kmol of fuel The reaction equation is C2H2 25 O2 376 N2 2 CO2 1 H2O 94 N2 Definition of the heating value and values from F11 LHV H RP Ho R Ho P h f fuel h f H2O 2 h f CO2 97 477 103 966 2169 184 539 811 Btulbmol 20 731 Btulbm Energy Eq HP Qout HR Ho P HP Qout Ho R Qout Ho R Ho P HP LHV HP HP 2 59 784 49 114 94 36 330 510 184 Btulbmol Qout 539 811 510 184 29 627 Btulbmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13178E Blast furnace gas in a steel mill is available at 500 F to be burned for the generation of steam The composition of this gas is on a volumetric basis Component CH4 H2 CO CO2 N2 H2O Percent by volume 01 24 233 144 564 34 Find the lower heating value Btuft3 of this gas at 500 F and P0 Of the six components in the gas mixture only the first 3 contribute to the heating value These are per lb mol of mixture 0024 H2 0001 CH4 0233 CO For these components 0024 H2 0001 CH4 0233 CO 01305 O2 0026 H2 0234 CO2 The remainder need not be included in the calculation as the contributions to reactants and products cancel For the lower HVwater vapor at 500 F h RP 0026103 966 3488 0234169 184 4229 00240 2101 000132 190 0538 160450077 023347 518 2981 013050 3069 31 257 Btulb mol fuel v 0 R T0 P0 1545 5367 147 144 39147 ft3lb mol LHV 31 680 39147 7985 Btuft3 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13179E A burner receives a mixture of two fuels with mass fraction 40 nbutane and 60 methanol both vapor The fuel is burned with stoichiometric air Find the product composition and the lower heating value of this fuel mixture Btulbm fuel mix Since the fuel mixture is specified on a mass basis we need to find the mole fractions for the combustion equation From Eq134 we get ybutane 0458124 0458124 0632042 026875 ymethanol 1 ybutane 073125 The reaction equation is 073125 CH3OH 026875 C4H10 νO2 O2 376 N2 νCO2 CO2 νH2OH2O 376 νO2 N 2 C balance 073125 4 026875 νCO2 180625 H2 balance 2 073125 5 026875 νH2O 280625 O balance 073125 2 νO2 2 νCO2 νH2O 641875 νO2 284375 Now the products are 180625 CO2 280625 H2O 106925 N2 Since the enthalpy of combustion is on a mass basis in table 133 this is also the negative of the heating value we get LHV 04 45 714 06 21 0932326 13 302 Btulbm fuel mixture Notice we took fuel vapor and water as vapor lower heating value Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13180E A mixture of fuels is E85 which is 85 ethanol and 15 gasoline assume octane by mass Assume we put the fuel and air both at To Po into a carburetor and vaporize the fuel as we mix it with stoichiometric air before it flows to an engine Assume the engine has an efficiency as work divided by the lower heating value of 30 and we want it to deliver 55 hp We use heat from the exhaust flow 900 R for the carburetor Find the lower heating value of this fuel Btulbm the rate of fuel consumption the heating rate needed in the carburetor and the rate of entropy generation in the carburetor The heating value of the liquid fuel blend values from Tbl 133 converted LHV 085 LHVethanol 015 LHVoctane 085 26 811 015 44 425 042992 Btulbm 12 662 Btulbm W 03 m fuel LHV m fuel W 03 LHV 55 hp 2544 Btuhhp 03 12 662 Btulbm 36835 lbmh 00102 lbms The difference in heating value based on liquid versus vapor fuel is ΔLHV 085 27 731 26 811 015 44 788 44 425 042992 085 920 015 363 042992 3596 Btulbm Q m fuel ΔLHV 00102 lbms 3596 Btulbm 368 Btus Entropy generation is from vaporizing the fuel by transfer of heat from 900 R exhaust gas to the intake system at To and then mixing it with air The mole fractions are Ethanol M 46069 Octane M 114232 DIV 085 46069 015 114232 001845 0001313 0019764 yethanol 001845 0019764 09336 yoctane 1 yethanol 006644 093356 C2H5OH 006644 C8H18 363118 O2 376 N2 239864 CO2 339864 H2O 1365324 N2 C balance 093356 2 006644 8 239864 vCO2 H balance 093356 6 006644 18 679728 2vH2O So the AF ratio on a mole and mass basis are Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful AFmole 363118 476 1 172844 AFmass 172844 2897505996 9896 The fuel was mixed before the carburetor so afterwards we have a 1 172844 mix sfg ethanol 67434 3832146069 063194 BtulbmR sfg octane 111399 86122114232 022128 BtulbmR sfg fuel mix 085 063194 015 022128 057034 BtulbmR Mfuel 09336 46069 006644 114232 505996 Entropy Eq S gen out in m fuel sfg R lnyfuel m air R lnyair Q Texhaust 00102 057034 198589 505996 ln 1 182844BtusR 00102 9896 006855 ln172844 182844 BtusR 368 900 BtusR 000698 0000389 0004089 000328 BtusR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Adiabatic flame temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13181E Hydrogen gas is burned with pure oxygen in a steady flow burner where both reactants are supplied in a stoichiometric ratio at the reference pressure and temperature What is the adiabatic flame temperature The reaction equation is H2 νO2 O2 H2O The balance of hydrogen is done now for oxygen we need vO2 05 Energy Eq HR HP 0 103 966 h H2O h H2O 103 966 Btulbmol Interpolate now in table F6 for the temperature to give this enthalpy T 8985 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13182E Some type of wood can be characterized as C1H15O07 with a lower heating value of 8380 Btulbm Find its adiabatic flame temperature when burned with stoichiometric air at 1 atm 77 F C1H15O07 νO2 O2 376 N2 CO2 075 H2O 376νO2 N2 O balance 07 2νO2 2 075 νO2 1025 νN2 3854 M 1 120111 15 1008 07 16 24712 H RP Ho P Ho R M HV 24712 8380 207 086 Btulbmol Energy Eq HP H P HP HR H R HP H R H P H RP HP h CO2 075 h H2O 3854 h N2 207 086 Btulbmol at 4400 R HP 50 948 075 41 325 3854 31 068 201 678 Btu at 4600 R HP 53 885 075 43 899 3854 32 817 213 286 Btu T 4400 200 207 086 201 678 213 286 201 678 4493 R Comment Most wood has some water and some noncombustible solids material so the actual flame temperature will be much lower Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13183E Carbon is burned with air in a furnace with 150 theoretical air and both reactants are supplied at the reference pressure and temperature What is the adiabatic flame temperature C νO2O2 376 νO2N2 1 CO2 376 νO2N2 From this we find νO2 1 and the actual combustion reaction is C 15 O2 564 N2 1 CO2 564 N2 05 O2 HP H P HP HR H R HP H R H P 0 169 184 169 184 Btulbmol HP h CO2 564 h N2 05 h O2 Find T so HP takes on the required value To start guessing assume all products are nitrogen 1 564 05 714 that gives 3400 T 3600 R from Table F6 HP 3400 36 437 564 22 421 05 23 644 174 713 too high HP 3200 33 579 564 20 717 05 21 860 161 353 Linear interpolation to find T 3200 200 169 184 161 353 174 713 161 353 3317 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13184E Acetylene gas at 77 F 147 lbfin2 is fed to the head of a cutting torch Calculate the adiabatic flame temperature if the acetylene is burned with 100 theoretical air at 77 F Repeat the answer for 100 theoretical oxygen at 77 F a C2H2 25 O2 25 376 N2 2 CO2 1 H2O 94 N2 HR ho f C2H2 97 477 Btu HP 2169 184 h CO2 1103 966 h H2O 94 h N2 QCV HP HR 0 2 h CO2 1 h H2O 94 h N2 539 811 Btulbmol Trial and Error TPROD 5236 R 2 147 196 121 488 94 89 348 1 255 751 OK b C2H2 25 O2 2 CO2 H2O HR 97 477 Btu HP 2169 184 h CO2 1103 966 h H2O 2 h CO2 1 h H2O 539 811 At 10 000 R limit of F6 2 135 426 118 440 389 292 At 9500 R 2 127 734 111 289 366 757 or 4507100 R change Difference extrapolating TPROD 10 000 150519 4507 13 340 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13185E Butane gas at 77 F is mixed with 150 theoretical air at 1000 R and is burned in an adiabatic steady state combustor What is the temperature of the products exiting the combustor C4H10 15 65O2 376N2 4CO2 5H2O 325 O2 3666N2 HR H R Hairin HP H P 4h CO2 5h H2O 325h O2 3666h N2 HP HR HP H R H P Hairin HP H RP Hairin 45714 2326 58124 975 3366 3666 3251 1 294 339 Btulbmol fuel 4h CO2 5h H2O 325h O2 3666h N2 at Tad Trial and Error Find the enthalpies from Table F6 HP3600R 1 281 185 HP3800R 1 374 068 Linear interpolation to match LHS Tad 3628 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13186E A special coal burner uses a stoichiometric mixture of coal and an oxygen argon mixture 11 mole ratio with the reactants supplied at Po To Find the adiabatic flame temperature assuming complete combustion Combustion of carbon C O2 Ar CO2 Ar CV Combustion chamber Energy Eq HR H R HP H P HP h f CO2 h CO2 h Ar Table F6 or F11 h f CO2 169 184 Btulbmol reference H R 0 Table F4 C P Ar 0124 39948 49536 BtulbmolR HP H R H P 0 h f CO2 169 184 Btulbmol HP 9000 120 071 49536 9000 4597 162 376 Btulbmol HP 9500 127 734 49536 9500 4597 172 516 Btulbmol interpolate T4 Tadflame 9336 R Comment At this temperature some chemical equilibrium reactions will be important see Chapter 14 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13187E Liquid nbutane at T0 is sprayed into a gas turbine with primary air flowing at 150 lbfin2 700 R in a stoichiometric ratio After complete combustion the products are at the adiabatic flame temperature which is too high so secondary air at 150 lbfin2 700 R is added with the resulting mixture being at 2500 R Show that Tad 2500 R and find the ratio of secondary to primary air flow CV Combustion Chamber C4H10 65 O2 65 376 N2 5 H2O 4 CO2 2444 N2 HR Hair HFu HP H P HP H R HR HP H R H P HR H RP HR 45344 581242326 651158 376 1138 1 168 433 Btulbmol fuel HP2500R 4 23755 5 18478 2444 14855 550 466 Btulbmol fuel HP HP2500R Tad 2500 R If iteration Tad 4400 R CV Mixing chamber HP νO2 2ndHair700 HP2500R νO2 2ndHair2500R νO2 2nd HP HP2500 Hair2500 Hair700 1168433 550466 71571 5437 9344 Ratio νO2 2ndνO2 Prim 934465 144 AD Mixing Combustion Chamber To turbine 2500 R Fuel Air Air Primary Secondary T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13188E Ethene C2H4 burns with 150 theoretical air in a steady flow constantpressure process with reactants entering at P0 T0 Find the adiabatic flame temperature Stoichiometric C2H4 3O2 376N2 2CO2 2H2O 1128N2 Actual C2H4 45O2 376N2 2 CO2 2 H2O 15 O2 1692 N2 HP HR H P HP H R From Table 133 units converted HP H R H P H RP 28054 47 1582326 568 775 Btulbmol HP 2h CO2 2h H2O 15h O2 1692h N2 Trial and error on Tad HP3400R 2 36 437 2 28 867 15 23 644 1692 22 421 545 437 Btulbmol HP3600R 2 39 312 2 31 293 15 25 441 1692 24 135 587 736 Btulbmol Linear interpolation Tad 3510 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13189E Solid carbon is burned with stoichiometric air in a steady state process as shown in Fig P1397 The reactants at T0 P0 are heated in a preheater to T2 900 R with the energy given by the products before flowing to a second heat exchanger which they leave at T0 Find the temperature of the products T4 and the heat transfer per lb mol of fuel 4 to 5 in the second heat exchanger a Following the flow we have Inlet T1 after preheater T2 after mixing and combustion chamber T3 after preheater T4 after last heat exchanger T5 T1 b Products out of preheater T4 Control volume Total minus last heat exchanger C O2 376 N2 CO2 376 N2 Energy Eq HR H R HP3 H P HP3 h f CO2 h CO2 376h N2 hf CO2 169 184 Btulbmol HP3 4400 50 948 376 31068 167 764 Btulbmol HP3 4600 53 885 376 32 817 177 277 Btulbmol Linear interpolation T3 Tadflame 4430 R c Control volume total Then energy equation H R q H P E q H h f CO2 0 169 184 RP Btu lbmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13190E In a test of rocket propellant performance liquid hydrazine N2HA4E A at 147 lbfin2 77 F and oxygen gas at 147 lbfin2 77 F are fed to a combustion chamber in the ratio of 05 lbm O2lbm N2HA4E A The heat transfer from the chamber to the surroundings is estimated to be 45 Btulbm N2HA4E A Determine the temperature of the products exiting the chamber Assume that only H2O H2 and N2 are present The enthalpy of formation of liquid hydrazine is 21 647 Btulb mole NA2E AHA4E OA2E 1 2 3 Comb Chamber Products NA2E AHA4E A A1 2E A OA2E A HA2E AO HA2E A NA2E AmE AO2E AAmE AN2H4E A 05 32AnE AO2E A32AnE AN2H4E A and AQ E AAmE AN2H4E A 45 Btulbm QCV 45 32045 1442 A Btu lb mol fuE CV combustion chamber n Fuh 1 n O2h 2 Q CV n toth 3 or H1 H2 QCV HP3 HA RE A QCV HA P E A HP3 HP3 HA RE A HA P E A QCV 21 647 103 966 1442 124 171 A Btu lb mol fuelE Trial and error on T3 HAP 5000RE A 49 114 34 627 36 330 120 071 Btulbmol HAP 5500RE A 55 739 39 032 40 745 135 516 Btulbmol Interpolate T3 5133 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Second law for the combustion process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13191E Two pound moles of ammonia are burned in a steady state process with x lb mol of oxygen The products consisting of HA2E AO NA2E A and the excess OA2E A exit at 400 F 1000 lbfinA2E A a Calculate x if half the water in the products is condensed b Calculate the absolute entropy of the products at the exit conditions 2NHA3E A xOA2E A 3HA2E AO NA2E A x 15OA2E Products at 400 F 1000 lbfinA2E A with nAH2O LIQE A nAH2O VAP E A 15 a yAH2O VAP E A A PG EPE A A2471 1000E A A 15 15 1 x 15E x 5070 b SAPRODE A SAGAS MIXE A SAH2O LIQE Gas mixture nAiE yAiE AsE A i ARE AlnA yiP EP0 E ASE Ai HA2E AO 15 02471 48939 5604 43335 OA2E 357 05881 52366 7326 45040 NA2E 10 01648 49049 4800 44249 SAGAS MIXE A 1543335 35745040 1044249 27004 BtuR SAH2O LIQE A 1516707 1801505647 00877 3795 BtuR SAPRODE A 27004 3795 30799 BtuR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13192E Propene C3H6 is burned with air in a steady flow burner with reactants at Po To The mixture is lean so the adiabatic flame temperature is 3200 R Find the entropy generation per lbmol fuel neglecting all the partial pressure corrections The reaction equation for a mixture with excess air is CA3E AHA6E A νAO2E A OA2E A 376 NA2E A 3 HA2E AO 3 COA2E A 376νAO2E A NA2E A νAO2E A 45OA2E Energy Eq HR HA RE A HR HA RE A HP HA P E A HP The entropy equation SR Sgen SP Sgen SP SR SP SA RE From table F6 at reference T HR hFu νAO2E AhO2 376 hN2 0 From table F6 at 3200 R HP 3 hH2O 3 hCO2 376 νAO2E A hN2 νAO2E A 45 hO2 3 26 479 3 33579 376 νAO2E A 20 717 νAO2E A 45 21 860 81 804 99 756 νAO2E From table F11 HA P E A HA RE A 3103966 3169184 8783 828 233 Btulbmol Now substitute all terms into the energy equation 828 233 81 804 99 756 νAO2E A 0 Solve for νAO2E A νAO2E A A828 233 81 804 99 756E A 748255 νAN2E A 28134 Table F6 contains the entropies at 147 psia so we get SP 3 61796 3 72160 748255 45 63109 28134 59175 225492 BtulbmolR SR 63761 748255 48973 28134 45739 171702 Btulbmol R Sgen 225492 171702 5379 BtulbmolR Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13193E Graphite C at Po To is burned with air coming in at Po 900 R in a ratio so the products exit at Po 2200 R Find the equivalence ratio the percent theoretical air and the total irreversibility C xOA2E A 376NA2E A COA2E A x 1OA2E A 376 x NA2E HAP E A HARE A HAP2200E A HARE A HA AER E HA AEP E HA AE RP 19659 x 113136 376 x 12407 x2616 376 2541 0 169184 6523 x 476162 169184 x 3416 or 342 theoretical air Equivalence ratio φ 1x 0293 Sgen sP sR P νisA AEi E ARE Aln yi R νisA AEi E ARE Aln yi R yAO2 E A 021 yAN2 E A 079 P yAO2 E A 01507 yAN2 E A 079 yACO2 E A 00593 SAgenE A 66952 5610 241659844 3758 376 341656066 0468 1371 341652686 3099 37649353 0468 1205 Btulbmol C R I ToSgen 64 677 Btulbmol C Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13194E Hydrogen peroxide H2O2 enters a gas generator at 77 F 75 lbfin2 at the rate of 02 lbms and is decomposed to steam and oxygen exiting at 1500 R 75 lbfin2 The resulting mixture is expanded through a turbine to atmospheric pressure 147 lbfin2 as shown in Fig P13107 Determine the power output of the turbine and the heat transfer rate in the gas generator The enthalpy of formation of liquid HA2E AOA2E A is 80 541 Btulb mol H2O2 H2O A1 2E A O2 AnE AFu AmE AFuMFu 0234015 000588 lbmols AnE Aexmix 15 AnE AFu 000882 lbmols ACE Ap mix A 2 3E A 0445 18015 A 1 3E A 0219 31999 76804 ACE Av mix ACE Ap mix 198588 56945 kmix ACE Ap mixACE Av mix 13487 Reversible turbine T3 T2 P3P2Ak1kE A 1500 14775A02585E A 9843 R AwE A ACE Ap mixT2 T3 768041500 9843 39608 Btulbmol AWE ACV AnE Amix AwE A 000882 39608 349 Btus CV Gas generator AQE ACV AHE A2 AHE A1 000588 103 966 8306 00029472975 00058880 541 6745 Btus Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13195E Methane is burned with air both supplied at the reference conditions There is enough excess air to give a flame temperature of 3200 R What are the percent theoretical air and the irreversibility in the process The combustion equation with X times theoretical air is CHA4E A 2XOA2E A 376 NA2E A COA2E A 2HA2E AO 2X1OA2E A 752X NA2E Energy Eq Hair Hfuel HR HP HA P E A HP HA RE A HR HP HA RE A HR HA P E A HA RP E A 0 From Table 133 HA RP E A 1604 50 0102326 344 867 Btulbmol HP AhE A CO2E A 2 AhE A H2OE A 2X1 AhE A O2E A 752X AhE A N2E From Table F6 and the energy equation HP 3200 33 579 2 26 479 2X1 21 860 752X 20 717 344 867 so 42 817 199 512 X 344 867 X 1514 Theoretical air 1514 The products are Products COA2E A 2HA2E AO 09712 OA2E A 11172 NA2E The second law SAgenE A SAPE A SARE A and I TAoE A SAgenE Reactants Pi 147 psia Po 147 psia Aso EfE A from Table F6 and F11 ni yi Aso EfE A ARE A ln yiPi Po ASE Ai A Btu lbmol RE CHA4E 1 1 44459 0 44459 OA2E 2X 021 48973 3099 52072 NA2E 752 X 079 45739 0468 46207 SR ni ASE Ai 72821 BtuR lbmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Products Pe 147 psia Po 147 psia From Table F6 ni yi Aso E3200E ARE A ln yiPe Po ASE Ai A Btu lbmol RE COA2E 1 006604 72160 53966 77557 HA2E AO 2 013208 61796 40201 65816 OA2E 09712 006413 61109 54549 66564 NA2E 11172 073775 59175 06040 59779 SP ni ASE Ai 94169 BtuR lbmol fuel I ToSAPE A SARE A 5366794169 72821 114 568 Btulbmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Fuel Cells Efficiency and Review Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13196E In Example 1315 a basic hydrogenoxygen fuel cell reaction was analyzed at 25C 100 kPa Repeat this calculation assuming that the fuel cell operates on air at 77 F 147 lbfin2 instead of on pure oxygen at this state Anode 2 HA2E A 4 eAE A 4 HAE Cathode 4 HAE A 4 eAE A 1 OA2E A 2 HA2E AO Overall 2 HA2E A 1 OA2E A 2 HA2E AO Example A 1315E A GA25CE A 474 283 kJkmol Or GA77 FE A 203 904 Btulbmol PEAOA2 AE A yAO2E A P 021 147 3087 lbfin2 AsE AEAOA2 AE A 48973 198589 ln 021 52072 S 216707 231186 152072 8103 BtuR H 2122 885 20 10 245 770 Btulbmol GA77 FE A 245 770 536678103 202 284 Btulbmol EAE A GNA0E Ae nAeE A 202 284 232696 485 4 1219 V Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13197E Pentane is burned with 120 theoretical air in a constant pressure process at 147 lbfinA2E A The products are cooled to ambient temperature 70 F How much mass of water is condensed per poundmass of fuel Repeat the answer assuming that the air used in the combustion has a relative humidity of 90 CA5E AHA12E A 12 8 OA2E A 376 NA2E A 5 COA2E A 6 HA2E AO 096 OA2E A 361 NA2E Products cooled to 70 F 147 lbfinA2E a for HA2E AO at 70 F PAGE A 03632 lbfinA2E yAH2O MAXE A A PG EPE A A03632 147E A A nH2O MAX EnH2O MAX 4206E Solving nAH2O MAXE A 1066 nAH2OE Therefore nAH2O VAP E A 1066 nAH2O LIQE A 6 1066 4934 mAH2O LIQE A A4934 18015 72151E A 1232 lbmlbm fuel b PAv1E A 09 03632 03269 lbfinA2E wA1E A 0622 A03269 14373E A 0014 147 nAH2O INE A 0014147 A 2897 18015E A 96 361 1040 lbmol nAH2O OUTE A 104 6 704 nAH2O LIQE A 704 1066 5974 lb mol nAH2O LIQE A A5974 18015 72151E A 1492 lbmlbm fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13198E A small aircooled gasoline engine is tested and the output is found to be 20 hp The temperature of the products is measured and found to be 730 F The products are analyzed on a dry volumetric basis with the following result 114 COA2E A 29 CO 16 OA2E A and 841 NA2E A The fuel may be considered to be liquid octane The fuel and air enter the engine at 77 F and the flow rate of fuel to the engine is 18 lbmh Determine the rate of heat transfer from the engine and its thermal efficiency a CA8E AHA18E A b OA2E A 376b NA2E A 114 COA2E A 29 CO c HA2E AO 16 OA2E A 841 NA2E b 841376 2237 a 18114 29 1788 c 9a 16088 CA8E AHA18E A 125 OA2E A 471 NA2E 638 COA2E A 162 CO 9 HA2E AO 089 OA2E A 471 NA2E a HARE A Ah Ef C8H18E A 107 526 Btulbmol HAP E A 638169 184 6807 16247 518 4647 9103 966 5475 08904822 4710 4617 1 770 092 Btulbmol HAP E A HARE A 1 770 092 107 526 1 662 566 Btulbmol AH E AP E A AH E ARE A A 18 11423E A 1 662 566 26 198 Btuh MAfuelE A 11423 AQ E ACVE A 26 198 20 2544 21 110 Btuh b Fuel heating value from table 153 converted to Btulbm AQ E AHE A 18 47 8932326 37 062 Btuh AWE ANETE A 20 2544 5088 Btuh ηATHE A A 5088 37062E A 0137 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13199E A gasoline engine uses liquid octane and air both supplied at Po To in a stoichiometric ratio The products complete combustion flow out of the exhaust valve at 2000 R Assume that the heat loss carried away by the cooling water at 200 F is equal to the work output Find the efficiency of the engine expressed as worklower heating value and the second law efficiency C8H18 125O2 376N2 8 CO2 9 H2O 47 N2 Table 133 LHV convert units kJkg to Btulbm LHV 44 425 114232 2326 2 181 753 Btulbmol fuel Table F6 HP2000 8 16 982 9 13 183 47 10 804 643 470 CV Total engine Hin Hex W Qloss Hex 2W W Hin Hex2 HR HP2 HA AERP E HP20002 2 181 753 643 4702 769 142 Btulbmol fuel ηTH WLHV 769 1422 181 753 0353 For 2nd law efficiency we must find reversible work ASE Ain AsE Afuel 125AsE AO2 376AsE AN2 86122 12548973 198589 ln 1 476 4745739 198589 ln 376 476 29088 Btulbmol fuel R ASE Aex 8AsE ACO2 9AsE AH2O 47AsE AN2 865677 198589 ln 8 64 956619 198589 ln A 9 64E A 4755302 198589 ln A47 64E A 37311 Btulbmol fuel R Assume the same Qloss out to 200 F 65967 R reservoir and compute QA0 rev AE ASE Ain QA0 rev AET0 ASE Aex QlossTres QA0 rev AE T0ASE Aex ASE Ain QlossT0Tres 5366737311 29088 769 1425366765967 1 067 034 Btulbmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful WArevE A Hin Hex Qloss QA0 rev AE Wac QA0 rev AE 769 142 1 067 034 1 836 176 Btulbmol fuel ηII WacWArevE A 769 1421 836 176 0419 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13200E Ethene CA2E AHA4E A and propane CA3E AHA8E A in a 11 mole ratio as gases are burned with 120 theoretical air in a gas turbine Fuel is added at 77 F 150 lbfin2 and the air comes from the atmosphere 77 F 15 lbfin2 through a compressor to 150 lbfin2 and mixed with the fuel The turbine work is such that the exit temperature is 1500 R with an exit pressure of 147 lbfin2 Find the mixture temperature before combustion and also the work assuming an adiabatic turbine C2H4 C3H8 νO2O2376N2 5CO2 6H2O νN2N2 φ 1 νO2 8 φ 112 νO2 96 so we have 45696 lbmol air per 2 lbmol fuel C2H4 C3H8 96O2 376N2 5CO2 6H2O 16 O2 36096N2 CV Compressor air flow wcin h2 h1 s2 s1 Pr1 10907 Pr2 Pr1P2P1 10907 T2 air 10273 R wcin 24781 12838 11953 Btulbm 34624 Btulbmol air CV Mixing chamber AnE Aair AhE Aair in AnE Afu1 AhE Afu1 AnE Afu2 AhE Afu2 sameexit ACE APF1 ACE APF2Tex T0 45696 ACE AP airT2 air Tex ACE APF1 1153 ACE APF2 1795 ACE AP air 6953 BtulbmolR Tex 45696C P airT2 C PF1 C PF2T0 C PF1 C PF2 45696C air 9856 R Tin combust Turbine work take CV total and subtract compressor work Wtotal Hin Hout HR HP1500 AhE A AEf E F1 AhE A AEf E F2 5AhE ACO2 6AhE AH2O 36096AhE AN2 16AhE AO2 22557 44669 510557 169184 68306 103966 36096 6925 16 72975 1 083 342 Btu2 lbmol Fuel wT wtot wcin 1 083 342 34624 45696 1 241 560 Btu2 lbmol fuel Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 13201E Repeat problem 13190E but assume that saturatedliquid oxygen at 170 R is used instead of 77 F oxygen gas in the combustion process Use the generalized charts to determine the properties of liquid oxygen Problem the same as 13190E except oxygen enters at state 2 as saturated liquid at 170 R At 170 R TAr2E A A 170 2786E A 061 AhE Af 51 From Fig D2 AhE A E A AhE A 198589 2786 51 2822 Btulbmol HP3 HA RE A HR HA P E A QCV 21 647 052822 050219170 5366732 103 966 1442 121 475 Btulbmol With HAP 5000RE A 49 114 34 627 36 330 120 071 Btulbmol HAP 5500RE A 55 739 39 032 40 745 135 516 Btulbmol Linear interpolation T3 5045 R Updated June 2013 8e SOLUTION MANUAL CHAPTER 14 Fundamentals of Thermodynamics Borgnakke Sonntag Borgnakke Sonntag Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful CONTENT CHAPTER 14 SUBSECTION PROB NO InTextConceptQuestions ag ConceptStudy Guide Problems 117 Equilibrium and Phase Equilibrium 1821 Chemical Equilibrium Equilibrium Constant 2274 Simultaneous Reaction 7586 Gasification 8793 Ionization 9499 Applications 100108 Review Problems 109117 English Unit Problems 118144 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful InText Concept Questions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14a For a mixture of O2 and O the pressure is increased at constant T what happens to the composition An increase in pressure causes the reaction to go toward the side of smaller total number of moles in this case toward the O2 14b For a mixture of O2 and O the temperature is increased at constant P what happens to the composition A temperature increase causes more O2 to dissociate to O 14c For a mixture of O2 and O I add some argon keeping constant T P what happens to the moles of O Diluting the mixture with a nonreacting gas has the same effect as decreasing the pressure causing the reaction to shift toward the side of larger total number of moles in this case the O Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14d When dissociations occur after combustion does T go up or down Dissociation reactions of combustion products lower the temperature 14e For nearly all the dissociations and ionization reactions what happens with the composition when the pressure is raised The reactions move towards the side with fewer moles of particles that is ions and electrons towards the monatomic side and monatomic species combine to form the molecules diatomic or more Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14f How does the time scale for NO formation change when P is lower at same T Look at the expression for the time scale in Eq1440 As P is lowered the time scale becomes larger The formation rates drops and the effect is explained by the larger distance between the moleculesatoms density lower the same T essentially means that they have the same characteristic velocity 14g Which atom in air ionizes first as T increases What is the explanation Using Fig 1411 we note that as temperature increases atomic N ionizes to N becoming significant at about 68000 K N has a lower ionization potential compared to O or Ar Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful ConceptStudy Guide Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 141 Is the concept of equilibrium limited to thermodynamics Equilibrium is a condition in which the driving forces present are balanced with no tendency for a change to occur spontaneously This concept applies to many diverse fields of study one no doubt familiar to the student being that of mechanical equilibrium in statics or engineering mechanics 142 How does Gibbs function vary with quality as you move from liquid to vapor There is no change in Gibbs function between liquid and vapor For equilibrium we have gg gf Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 143 How is a chemical equilibrium process different from a combustion process Chemical equilibrium occurs at a given state T and P following a chemical reaction process possibly a combustion which is a chemical reaction process together with several reactions amongst the combustion products Whereas the combustion is a oneway process irreversible the chemical equilibrium is a reversible process that can proceed in both directions 144 Must P and T be held fixed to obtain chemical equilibrium No but we commonly evaluate the condition of chemical equilibrium at a state corresponding to a given temperature and pressure If T and P changes in a process it means that the chemical composition adjusts towards equilibrium and the composition changes along with the process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 145 The change in Gibbs function Go for a reaction is a function of which property The change in Gibbs function for a reaction G is a function of T and P The change in standardstate Gibbs function Go is a function only of T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 146 In a steady flow burner T is not controlled which properties are The pressure tends to be constant only minor pressure changes due to acceleration of the products as density decreases velocity must increase to have the same mass flow rate The product temperature depends on heat losses radiation etc and any chemical reactions that may take place generally lowering the temperature below the standard adiabatic flame temperature 147 In a closed rigid combustion bomb which properties are held fixed The volume is constant The number of atoms of each element is conserved although the amounts of various chemical species change As the products have more internal energy but cannot expand the pressure increases significantly Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 148 Is the dissociation of water pressure sensitive Yes since the total number of moles on the left and right sides of the reaction equations is not the same 149 At 298 K K exp184 for the water dissociation what does that imply This is an extremely small number meaning that the reaction tends to go strongly from right to left in other words does not tend to go from left to right dissociation of water at all Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1410 If a reaction is insensitive to pressure prove that it is also insensitive to dilution effects at a given T Assume the standard reaction we used to develop the expression for the equilibrium constant νA A νB B νC C νD D let us assume we add an inert component E so the total moles become ntot nA nB nC nD n E This will now lower the mole fractions of A B C and D If the reaction is pressure insensitive then νA νB νC νD and the equilibrium equation becomes K yνC C yνD D yνA A yνB B P Po νC νD νA νB C yνD D yνA A yνB B yνC Since each yi ni ntot we get K yνC C yνD D yνA A yνB B nCntotνC nDntotνD nAntotνA nBntotνB nνC C nνD D nνA A nνB B ntot νA νB νC νD C nνD D nνA A nνB B nνC Now we see that the total number of moles that includes nE does not enter the equation and thus will not affect any progress of the reaction Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1411 For a pressure sensitive reaction an inert gas is added dilution how does the reaction shift Assume the standard reaction we used to develop the expression for the equilibrium constant νA A νB B νC C νD D let us assume we add an inert component E so the total moles become ntot nA nB nC nD n E This will now lower the mole fractions of A B C and D If the reaction is pressure sensitive then νA νB νC νD and the equilibrium equation becomes K yνC C yνD D yνA A yνB B P Po νC νD νA νB Since each yi ni ntot we get K nCntotνC nDntotνD nAntotνA nBntotνB P Po νC νD νA νB nνC C nνD D nνA A nνB B P Po ntot νC νD νA νB As ntot is raised due to nE it acts as if the pressure P is lowered thus pushing the reaction towards the side with a larger number of moles Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1412 In a combustion process is the adiabatic flame temperature affected by reactions The adiabatic flame temperature is lower due to dissociation reactions of the products as those products absorb some of the energy ie 2 O atoms have more energy h 2249 170 h than a single O2 molecule h h see Table A9 The temperature is also influenced by other reactions like the water gas reaction Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1413 In equilibrium Gibbs function of the reactants and the products is the same how about the energy The chemical equilibrium mixture at a given T P has a certain total internal energy There is no restriction on its division among the constituents The conservation of energy from the reactants to the products will determine the temperature so if it takes place in a fixed volume combustion bomb then U is constant whereas if it is in a flow like a steady flow burner then H is constant When this is combined with chemical equilibrium it is actually a lengthy procedure to determine both the composition and the temperature in the actual process Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1414 Does a dissociation process require energy or does it give out energy Dissociation reactions require energy and is thus endothermic Notice from Table A9 that all the atoms N O H has a much higher formation enthalpy than the diatomic molecules which have formation enthalpy equal to zero Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1415 If I consider the nonfrozen composition can vary specific heat but still assume all components are ideal gases does that C become a function of temperature of pressure The nonfrozen mixture heat capacity will be a function of both T and P because the mixture composition depends on T and P while the individual component specific heat capacities depend only on T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1416 What is K for the water gas reaction in Example 144 at 1200 K Using the result of Example 144 and Table A11 ln K 1 2 ln KI ln KII 05 35736 36363 03135 K 13682 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1417 What would happen to the concentrations of the monatomic species like O N if the pressure is higher in Fig 1411 Since those reaction are pressure sensitive more moles on RHS than on LHS the higher pressure will push these reactions to the left and reduce the concentrations of the monatomic species Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful Equilibrium and Phase Equilibrium Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1418 Carbon dioxide at 15 MPa is injected into the top of a 5km deep well in connection with an enhanced oilrecovery process The fluid column standing in the well is at a uniform temperature of 40C What is the pressure at the bottom of the well assuming ideal gas behavior Z 1 Z 2 CO 2 cb Z1 Z2 5000 m P1 15 MPa T 40 oC constant Equilibrium at constant T wREV 0 g PE RT ln P2P1 gZ2 Z1 0 ln P2P1 98075000 10000188 923132 08287 P2 15 MPa exp08287 3436 MPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1419 Consider a 2kmdeep gas well containing a gas mixture of methane and ethane at a uniform temperature of 30oC The pressure at the top of the well is 14 MPa and the composition on a mole basis is 90 methane 10 ethane Each component is in equilibrium top to bottom with dG g dZ 0 and assume ideal gas so for each component Eq1410 applies Determine the pressure and composition at the bottom of the well Z 1 Z 2 mixture cb Gas A B Z1 Z2 2000 m Let A CH4 B C2H6 P1 14 MPa yA1 090 yB1 010 T 30 oC constant From section 141 for A to be at equilibrium between 1 and 2 WREV 0 nAG A1 G A2 nA MA g Z1 Z2 Similarly for B WREV 0 nBG B1 G B2 nB MB g Z1 Z2 Using eq 1410 for A R T ln PA2PA1 MA g Z1 Z2 with a similar expression for B Now ideal gas mixture PA1 yA1P etc Substituting ln yA2P2 yA1P1 MAgZ1Z2 R T and ln yB2P2 yB1P1 MBgZ1Z2 R T ln yA2 P2 ln0914 160498072000 1000831453032 26585 yA2 P2 142748 ln yB2P2 ln0114 300798072000 1000831453032 0570 43 yB2 P2 1 yA2 P2 176903 Solving P2 16044 MPa yA2 08897 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1420 A container has liquid water at 20oC 100 kPa in equilibrium with a mixture of water vapor and dry air also at 20oC 100 kPa How much is the water vapor pressure and what is the saturated water vapor pressure From the steam tables we have for saturated liquid Pg 2339 kPa vf 0001002 m3kg The liquid is at 100 kPa so it is compressed liquid still at 20oC so from Eq1415 at constant T gliq gf v dP vf P Pg The vapor in the moist air is at the partial pressure Pv also at 20oC so we assume ideal gas for the vapor gvap gg v dP RT ln Pg Pv We have the two saturated phases so gf gg q hfg Tsfg and now for equilibrium the two Gibbs function must be the same as gvap gliq RT ln Pv Pg gg vf P Pg g f leaving us with ln Pv Pg vf P Pg RT 0001002 100 2339 04615 29315 0000723 Pv Pg exp0000723 23407 kPa This is only a minute amount above the saturation pressure For the moist air applications in Chapter 11 we neglected such differences and assumed the partial water vapor pressure at equilibrium 100 relative humidity is Pg The pressure has to be much higher for this to be a significant difference Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1421 Using the same assumptions as those in developing Eq d in Example 141 develop an expression for pressure at the bottom of a deep column of liquid in terms of the isothermal compressibility βT For liquid water at 20oC βT 00005 1MPa Use the result of the first question to estimate the pressure in the Pacific ocean at the depth of 3 km d gT v 1 βTP dPT d gT g dz 0 v 1 β EAv1βAT AP dPAT AEA g AdzE TP dPT g dz 0 and integrate EA PA0 A EP 1βAT AP dPAT AEA A g vE A A0 HdzEA P PA0E A βATE A A1 2E A PA2E A PA0E A2E A A g vE A H P 1 A1 2E A βATE A P PA0E A A1 2E AβATE A PA0E A2E A A g vE A H v vAf 20CE A 0001002 H 3000 m g 980665 msA2E A βATE A 00005 1MPa P 1 A1 2E A 00005 P 0101 A1 2E A 00005 0101A2E A 980665 30000001002 10A6E 29462 MPa which is close to P Solve by iteration or solve the quadratic equation P 29682 MPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful Chemical Equilibrium Equilibrium Constant Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1422 Which of the reactions listed in Table A11 are pressure sensitive Check if νAAE A νABE A νACE A νADE Reaction Check P sensitive HA2E A 2H 1 2 yes OA2E A 2O 1 2 yes NA2E A 2N 1 2 yes 2 HA2E AO 2 HA2E A 1 OA2E A 2 3 yes 2 HA2E AO 2 HA2E A 2 OH 2 4 yes 2 COA2E A 2 CO 1 OA2E A 2 3 yes NA2E A OA2E A 2 NO 2 2 no NA2E A 2OA2E A 2 NOA2E A 3 2 yes Most of them have more moles on RHS and thus will move towards the RHS if the pressure is lowered Only the last one has the opposite and will move towards the LHS if the pressure is lowered Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1423 Calculate the equilibrium constant for the reaction OA2E A 2O at temperatures of 298 K and 6000 K Verify the result with Table A11 Reaction OA2E A 2O At 25 AoE AC 29815 K HA0E A 2AhE A 0 f OE A 1AhE A 0 f O2E A 2249 170 10 498 340 kJkmol SA0E A 2AsE A 0 OE A 1AsE A 0 O2E A 2161059 1205148 11697 kJkmol K GA0E A HA0E A TSA0E A 498 340 2981511697 463 465 kJkmol ln K AG0 ER TE A A 463 465 8314529815E A 186961 At 6000 K HA0E A 2249 170 121 264 0 224 210 516 658 kJkmol SA0E A 2224597 1313457 135737 kJkmol K GA0E A 516 658 6000135737 297 764 kJkmol ln K A 297 764 831456000E A 5969 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1424 Calculate the equilibrium constant for the reaction HA2E A 2H at a temperature of 2000 K using properties from Table A9 Compare the result with the value listed in Table A11 From Table A9 at 2000 K we find Eh HA2 A 52 942 kJkmol E s HA2 A 188419 kJkmol K Aho EfE A 0 h H 35 375 kJkmol s H 154279 kJkmol K Aho EfE A 217 999 kJkmol GA0E A H TS HARHSE A HALHSE A T SA0E ARHSE A SA0E ALHSE A 2 35 375 217 999 52943 20002154279 182419 213 528 kJkmol ln K GA0E AARE AT 213 528 83145 2000 128407 Table A11 ln K 12841 OK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1425 For the dissociation of oxygen OA2E A 2O around 2000 K we want a mathematical expression for the equilibrium constant KT Assume constant heat capacity at 2000 K for OA2E A and O from Table A9 and develop the expression from Eqs 1412 and 1415 From Eq1415 the equilibrium constant is K exp AG0 ERTE A GA0E A HA0E A T SA0E and the shift is GA0E A 2 h O h O2 T2so O so O2 Substitute the first order approximation to the functions AhE A and AsoE A as AhE A h 2000 K C p T 2000 AsoE A so 2000 K C p ln A T 2000E The properties are from Table A9 and ARE A 83145 kJkmol K Oxygen OA2E A h 2000 K 59 176 kJkmol so 2000 K 268748 kJkmol K C p h 2200 K h 2200 K 2200 1800 A66 770 51 674 400E A 3774 kJkmol K Oxygen O h 2000 K 35 713 249 170 284 883 kJkmol so 2000 K 201247 kJkmol K C p h 2200 K h 2200 K 2200 1800 A39 878 31 547 400E A 208275 kJkmol K Substitute and collect terms AG0 ERTE A AΗ0 ERTE A AS0 ER E A A Η 0 2000 ERTE A A C p 2000 ER E A AT 2000 TE A ln A T 2000E A A S 0 2000 ER E Now we have HA 0 2000E AARE A 2 284 883 59 17683145 61 4096 K AC p 2000E AARE A 2 208275 377483145 0470864 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful SA 0 2000E AARE A 2 201247 26874883145 1608587 so we get AG0 ERTE A A61 4096 TE A 0470864 AT 2000 TE A ln A T 2000E A 1608587 A60 4679 TE A 15615 0470864 ln A T 2000E Now the equilibrium constant KT is approximated as KT exp 15615 A60 4679 TE A 0470864 ln A T 2000E A Remark We could have chosen to expand the function GA0E A ARE AT as a linear expression instead or even expand the whole expGA0E A ARE AT in a linear function Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1426 Find K for COA2E A CO 12OA2E A at 3000 K using A11 The elementary reaction in A11 is 2COA2E A 2CO OA2E so the wanted reaction is 12 times that so K K12 A11 A exp2217EA A 0108935EA 033 or ln K 05 ln KA11 05 2217 11085 K exp11085 033 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1427 Plot to scale the values of ln K versus 1T for the reaction 2 COA2E A 2 CO OA2E A Write an equation for ln K as a function of temperature 2 COA2E A 2 CO 1 OA2E TK 10A4E A A1 TE ln K TK 10A4E A A1 TE ln K 2000 5000 13266 4000 2500 3204 2400 4167 7715 4500 2222 4985 2800 3571 3781 5000 2000 6397 3200 3125 0853 5500 1818 7542 3600 2778 1408 6000 1667 8488 For the range below 5000 K ln K A BT Using values at 2000 K 5000 K A 195056 B 65 543 K 8 4 0 4 8 12 1 2 3 4 5 0 1 almost linear 10 T x 4 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1428 Consider the reaction 2 COA2E A 2 CO OA2E A obtained after heating 1 kmol COA2E A to 3000 K Find the equilibrium constant from the shift in Gibbs function and verify its value with the entry in Table A11 What is the mole fraction of CO at 3000 K 100 kPa From Table A9 we get h CO 93 504 AhE A 0 f COE A 110 527 s CO 273607 Eh COA2 A 152 853 AhE AEA 0 f COA2 AEA 393 522E s COA2 A 33417 Eh OA2 A 98 013 E s OA2 A 284466 GA0E A H TS 2 HACOE A HEAOA2 AE A 2 HEACOA2 AE A T 2s CO E s OA2 A E2s COA2 A 2 93 504 110 527 98 013 0 2152 853 393 522 30002273607 284466 233417 55 285 ln K GA0E AARE AT 55 285 8314513000 22164 Table A11 ln K 2217 OK At 3000 K 2 COA2E A 2 CO 1 OA2E ln K 2217 Initial 1 0 0 K 0108935 Change 2z 2z z Equil 12z 2z z We have P PAoE A 01 MPa and nAtotE A 1 z so from Eq1429 K A yCO E2 yO2 yCO2 2 E A A P P0 E A A 2z 1 2z 2E A A z 1 z E A 1 0108935 4 zA3E A 0108935 1 2zA2E A1 z z 022 yACOE A 2z 1 z 036 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1429 Carbon dioxide is heated at 100 kPa What should the temperature be to see a mole fraction of CO as 025 For that temperature what will the mole fraction of CO be if the pressure is 200 kPa 2 COA2E A 2 CO 1 OA2E Initial 1 0 0 Change 2z 2z z Equil 12z 2z z We have P PAoE A 01 MPa and nAtotE A 1 z so from Eq1429 yACOE A A 2z 1 zE A 025 z 17 K A yCO E2 yO2 yCO2 2 E A A P P0 E A A 2z 1 2z 2E A A z 1 z E A 1 A 2 7 2 2E A A 1 7 1 E A 002 ln K 3912 from A11 T 2785 K K 002 A yCO E2 yO2 yCO2 2 E A A P P0 E A A 2z 1 2z 2E A A z 1 z E A 2 zA3E A 00025 1 2zA2E A1 z z 011776 yACOE A A 2z 1 zE A 02107 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1430 Assume a diatomic gas like OA2E A or NA2E A dissociate at a pressure different from PAoE A Find an expression for the fraction of the original gas that has dissociated at any T assuming equilibrium Look at initially 1 mol Oxygen and shift reaction with x OA2E A 2 O Species OA2E O Initial 1 0 Change x 2x Equil 1x 2x nAtotE A 1 x 2x 1 x yAOE A A 2x 1 xE A and yAO2E A A1 x 1 xE Substitute this into the equilibrium equation as KATE A A yO E2 y02 E A A P Po E AA 21E A A 4x2 E1 x2 E A A1 x 1 xE A A P Po E A A 4x2 E1 x2 E A A P Po E A Now solve for x as xA2E A 1 xA2E A KTPo 4P x KT 4PPo KT Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1431 Hydrogen gas is heated from room temperature to 4000 K 500 kPa at which state the diatomic species has partially dissociated to the monatomic form Determine the equilibrium composition at this state HA2E A 2 H Equil nAH2E A 1 x x 2x nAH E A 0 2x n 1 x K A 2x2 E1x1xE A A P P0 E AA 21E A at 4000 K ln K 0934 K 2545 A 2545 4500100E A 0127 25 A x2 E1x2 E A Solving x 03360 nAH2E A 0664 nAH E A 0672 ntot 1336 yAH2E A 0497 yAH E A 0503 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1432 Consider the dissociation of oxygen OA2E A 2 O starting with 1 kmol oxygen at 298 K and heating it at constant pressure 100 kPa At which temperature will we reach a concentration of monatomic oxygen of 10 Look at initially 1 mol Oxygen and shift reaction with x OA2E A 2 O Species OA2E O Initial 1 0 Change x 2x Equil 1x 2x nAtotE A 1 x 2x 1 x yAOE A A 2x 1 xE A 01 x 012 01 00526 yAO2E A 09 K A yO E2 y02 E A A P Po E AA 21E A A012 E09E A 1 001111 ln K 44998 Now look in Table A11 T 2980 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1433 Redo Problem 1432 for a total pressure of 40 kPa Look at initially 1 mol Oxygen and shift reaction with x OA2E A 2 O Species OA2E O Initial 1 0 Change x 2x Equil 1x 2x nAtotE A 1 x 2x 1 x yAOE A A 2x 1 xE A 01 x 012 01 00526 yAO2E A 09 K A yO E2 y02 E A A P Po E AA 21E A A012 E09E A 04 0004444 ln K 54161 Now look in Table A11 T 2856 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1434 Redo Problem 1432 but start with 1 kmol oxygen and 1 kmol helium at 298 K 100 kPa Look at initially 1 mol Oxygen and shift reaction with x OA2E A 2 O Species OA2E O He Initial 1 0 1 Change x 2x Equil 1x 2x 1 nAtotE A 1 x 2x 1 2 x yAOE A A 2x 2 xE A 01 x 022 01 010526 yAO2E A 1 x2 x 0425 K A yO E2 y02 E A A P Po E AA 21E A A 012 E0425E A 1 0023529 ln K 37495 Now look in Table A11 T 3094 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1435 Calculate the equilibrium constant for the reaction2COA2E A 2CO OA2E A at 3000 K using values from Table A9 and compare the result to Table A11 From Table A9 we get kJkmol kJkmol kJkmol K h CO 93 504 AhE A o f COE A 110 527 s CO 273607 Eh COA2 A 152 853 AhE AEA o f COA2 AEA 393 522E s COA2 A 33417 Eh OA2 A 98 013 AhE AEA o f OA2 AEA 0 E s OA2 A 284466 GA0E A H TS 2 HACOE A HEAOA2 AE A 2 HEACOA2 AE A T 2s CO E s OA2 A E2s COA2 A 2 93 504 110 527 98 013 0 2152 853 393 522 30002273607 284466 233417 55 285 kJkmol ln K GA0E AARE AT 55 285 8314513000 22164 Table A11 ln K 2217 OK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1436 Find the equilibrium constant for CO 12OA2E A COA2E A at 2200 K using Table A11 The elementary reaction in A11 is 2 COA2E A 2CO OA2E The wanted reaction is therefore 05 times that so K K12 A11 1 A exp10232E A 1 A 0000036EA 16667 or ln K 05 ln KA11 05 10232 5116 K exp5116 16667 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1437 Pure oxygen is heated from 25C to 3200 K in an steady flow process at a constant pressure of 200 kPa Find the exit composition and the heat transfer The only reaction will be the dissociation of the oxygen OA2E A 2O From A11 K3200 exp3069 0046467 Look at initially 1 mol Oxygen and shift reaction with x nAO2E A 1 x nAOE A 2x nAtotE A 1 x yAiE A nAiE AnAtotE K A yO E2 y02 E A A P Po E AA 21E A A 4x2 E1 x2 E A A1 x 1 xE A 2 A 8x2 E1 x2 E xA2E A A K8 1 K8E A x 007599 yA02 EA A1 x 1 xE A 0859 yA0 EA 1 yA02 EA 0141 AqE A nA02ex EAhE A02ex EA nA0ex EAhE AOex EA AhE A02in EA 1 xyA02 EAhE A02 EA yA0 EAhE AO EA 0 AhE A02 EA 106 022 kJkmol AhE AO EA 249 170 60 767 309 937 kJkmol AqE A 145 015 kJkmol OA2E q AqE A32 4532 kJkg 33165 if no reaction Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1438 Nitrogen gas N2 is heated to 4000 K 10 kPa What fraction of the N2 is dissociated to N at this state NA2E A 2 N T 4000 K lnK 12671 Initial 1 0 K 314x106 Change x 2x Equil 1x 2x ntot 1 x 2x 1 x yAN2E A A1 x 1 xE A yANE A A 2x 1 xE A K y2 N yN2 A P Po 21 AE 314x106 A 4x2 E1 x2 E A A 10 100 E A x 00028 yN2 A1 x 1 xE A 09944 yN A 2x 1 xE A 00056 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1440 One kilomole Ar and one kilomole O2 are heated up at a constant pressure of 100 kPa to 3200 K where it comes to equilibrium Find the final mole fractions for Ar OA2E A and O The only equilibrium reaction listed in the book is dissociation of OA2E A So assuming that we see in Table A11 lnK 3072 Ar OA2E A Ar 1 x OA2E A 2x O The atom balance already shown in above equation can also be done as Species Ar OA2E A O Start 1 1 0 Change 0 x 2x Total 1 1x 2x The total number of moles is nAtotE A 1 1x 2x 2 x so yAArE A 12 x yAO2 E A 1 x2 x yAOE A 2x2 x and the definition of the equilibrium constant PAtotE A PAoE A becomes K eA3072E A 004633 A yO E2 y02 E A A 4x2 E2 x1 xE The equation to solve becomes from the last expression K 4xA2E A Kx 2K 0 If that is solved we get x 00057 01514 01457 x must be positive yAOE A 01358 yA02 E A 03981 yAArE A 04661 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1441 Air assumed to be 79 nitrogen and 21 oxygen is heated in a steady state process at a constant pressure of 100 kPa and some NO is formed At what temperature will the mole fraction of NO be 0001 079 NA2E A 021 OA2E A heated at 100 kPa forms NO NA2E A OA2E A 2 NO nAN2E A 079 x x x 2x nAO2E A 021 x nANOE A 0 2x ntot 10 At exit yANOE A 0001 A2x 10E A x 00005 nAN2E A 07895 nAO2E A 02095 K A y 2 NO EyN2yO2 E A A P P0 E AA0E A A 106 E0789502095E A 604610A6E A or ln K 12016 From Table A11 T 1444 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1442 Pure oxygen is heated from 25C 100 kPa to 3200 K in a constant volume container Find the final pressure composition and the heat transfer As oxygen is heated it dissociates OA2E A 2O ln KAeqE A 3069 from table A11 C V Heater UA2E A UA1E A A1E AQA2E A HA2E A HA1E A PA2E Av PA1E Av Per mole OA2E A A1E AqE A2E A AhE A2 EA AhE A1 EA ARE ATA1E A nA2E AnA1E ATA2E A Shift x in reaction 1 to have final composition 1 xOA2E A 2xO nA1E A 1 nA2E A 1 x 2x 1 x yAO2 E A 1 x1 x yAOE A 2x1 x Ideal gas and VA2E A VA1E A PA2E A PA1E AnA2E ATA2E AnA1E ATA1E A PA2E APAoE A 1 xTA2E ATA1E Substitute the molefractions and the pressure into the equilibrium equation KAeqE A eA3069E A A yO E2 y02 E A A P2 EPo E A A 2x 1 xE AA2E A A1 x 1 xE A A1 x 1E A A T2 ET1 E A A 4x2 E1 xE A A T1 ET2 E A eA3069E A 000433 x 00324 The final pressure is then PA2E A PAoE A1 xA T2 ET1 E A 100 1 00324 A3200 2982E A 1108 kPa nAO2 E A 09676 nAOE A 00648 nA2E A 10324 A1E AqE A2E A 09676 106022 00648 249170 60767 0 83145 29815 10324 3200 97681 kJkmolOA2E yAO2 E A A09676 10324 E A 0937 yAOE A A00648 10324 E A 00628 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1443 Find the equilibrium constant for the reaction 2NO OA2E A 2NOA2E A from the elementary reactions in Table A11 to answer which of the nitrogen oxides NO or NOA2E A is the more stable at ambient conditions What about at 2000 K 2 NO OA2E A 2 NOA2E A 1 But NA2E A OA2E A 2 NO 2 NA2E A 2 OA2E A 2 NOA2E A 3 Reaction 1 Reaction 3 Reaction 2 GA 0 1E A GA 0 3E A GA 0 2E A ln KA1E A ln KA3E A ln KA2E At 25 AoE AC from Table A11 ln KA1E A 41355 69868 28513 or KA1E A 241610A12E an extremely large number which means reaction 1 tends to go very strongly from left to right At 2000 K ln KA1E A 19136 7825 11311 or KA1E A 1224 10A5E meaning that reaction 1 tends to go quite strongly from right to left Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1444 Assume the equilibrium mole fractions of oxygen and nitrogen are close to those in air find the equilibrium mole fraction for NO at 3000 K 500 kPa disregarding dissociations Assume the simple reaction to make NO as NA2E A OA2E A 2 NO Ar nAN2E A 078 x 078 021 0 001 nAO2E A 021 x x x 2x nANOE A 2x nAarE A 001 078x 021x 2x 001 ntot 10 From A11 at 3000 K ln K 4205 K 0014921 K A 4x2 E078 x021 xE A A P P0 E AA 0E A A x2 E078 x021 xE A A0014921 4E A 000373 and 0 x 021 Solve for x x 00230 yANOE A A2x 10E A 0046 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1445 The combustion products from burning pentane CA5E AHA12E A with pure oxygen in a stoichiometric ratio exists at 2400 K 100 kPa Consider the dissociation of only COA2E A and find the equilibrium mole fraction of CO CA5E AHA12E A 8 OA2E A 5 COA2E A 6 HA2E AO At 2400 K 2 COA2E A 2 CO 1 OA2E ln K 7715 Initial 5 0 0 K 4461 10A4E Change 2z 2z z Equil 52z 2z z Assuming P PAoE A 01 MPa and ntot 5 z 6 11 z K A yCO E2 yO2 yCO2 2 E A A P P0 E A A 2z 5 2z 2E A A z 11 z E A 1 4461 10A4E A Trial Error compute LHS for various values of z z 0291 nACO2 E A 4418 nACOE A 0582 nAO2 E A 0291 yACOE A 00515 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1446 A mixture flows with 2 kmols COA2E A 1 kmols argon and 1 kmols CO at 298 K and it is heated to 3000 K at constant 100 kPa Assume the dissociation of carbon dioxide is the only equilibrium process to be considered Find the exit equilibrium composition and the heat transfer rate Reaction 2 COA2E 2 CO OA2E Ar initial 2 1 0 1 change 2x 2x x equil 2 2x 1 2x x 1 From Table A11 K exp 2217 0108935 A y 2 CO yO2 Ey 2 CO2 E A A P Po E AA 32E A A1 2x2 E4 x2 E A A x 4 xE A A4 x2 E2 2x2 E A 1A 1E A Ax 4E A A1 2x2 E4 xE A A 1 1 x2 E then 043574 4 x 1 xA2E A x 1 2xA2E trial and error solution gives x 032136 The outlet has 135728 COA2E A 164272 CO 1 Ar 032136 OA2E Energy equation gives from Table A9 and A5 for argon CApE AΔT AQ E A n ex h ex n in h in 135728 152853 393522 164272 93504 110527 1 39948 052 3000 298 032136 98013 2 393522 1110527 1 0 630 578 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1447 A mixture of 1 kmol carbon dioxide 2 kmol carbon monoxide and 2 kmol oxygen at 25C 150 kPa is heated in a constant pressure steady state process to 3000 K Assuming that only these same substances are present in the exiting chemical equilibrium mixture determine the composition of that mixture initial mix 1 COA2E A 2 CO 2 OA2E Constant pressure reactor Q Equil mix COA2E A CO OA2E A at T 3000 K P 150 kPa Reaction 2 COA2E 2 CO OA2E initial 1 2 2 change 2x 2x x equil 1 2x 2 2x 2 x nAtotE A 1 2x 2 2x 2 x 5 x so y n nAtotE From A11 at 3000 K K exp2217 0108935 For each n 0 1 x A 1 2E K A y 2 COyO2 Ey 2 CO2 E A A P P0 E AA 1E A 4 A1 x 1 2xE AA 2E A A2 x 5 xE A A150 100E A 0108935 or A1 x 1 2xE AA 2E A A2 x 5 xE A 0018 156 Trial error x 0521 A nCO2 2042 EnCO 0958 nO2 1479 ntot 4479 E A A yCO2 04559 EyCO 02139 yO2 03302 E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1448 Acetylene gas Acetylene gas CA2E AHA2E A is burned with stoichiometric air in a torch The reactants are supplied at the reference conditions PA0E A TA0E A The products come out from the flame at 2800 K after a small heat loss by radiation Consider the dissociation of COA2E A into CO and OA2E A and no others Find the equilibrium composition of the products Are there any other reactions that should be considered Combustion CA2E AHA2E A 25 OA2E A 376NA2E A 2 COA2E A 1 HA2E AO 94 NA2E At 2800 K 2 COA2E A 2 CO 1 OA2E A HA2E AO NA2E ln K 3781 Initial 2 0 0 1 94 K 00228 Change 2z 2z z 0 0 Equil 22z 2z z 1 94 Assuming P PAoE A 01 MPa and ntot 2 z 1 94 124 z K A yCO E2 yO2 yCO2 2 E A A P P0 E A A 2z 2 2z 2E A A z 124 z E A 1 00228 Solve zA3E A 00228 1 zA2E A124 z z 04472 yACOE A A 2z 124 zE A 00696 yACO2E A A 2 2z 124 zE A 0086 yAO2E A A z 124 zE A 00348 yAH2OE A A 1 124 zE A 00778 yAN2E A A 94 124 zE A 07317 Looking in A11 there will be no dissociation of nitrogen and only a small dissociation of water and oxygen which could be included Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1449 Consider combustion of methane with pure oxygen forming carbon dioxide and water as the products Find the equilibrium constant for the reaction at 1000 K Use an average heat capacity of Cp 52 kJkmol K for the fuel and Table A9 for the other components For the reaction equation CHA4E A 2 OA2E A COA2E A 2 HA2E AO At 1000 K from Table A9 and A10 for the fuel at 298 K HA 0 1000 KE A 1393 522 33 397 2241 826 26 000 174 873 521000 2982 20 22 703 798 804 kJkmol SA 0 1000 KE A 1269299 2232739 1186251 lnA1000 2982E A 2243579 487158 kJkmol K GA 0 1000 KE A HA 0 1000 KE A T SA 0 1000 KE 798 804 1000 487158 1 285 962 kJkmol ln K AG0 ER TE A A 1 285 962 831451000E A 154665 K 14796 E 67 This means the reaction is shifted totally to the right Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1450 Water from the combustion of hydrogen and pure oxygen is at 3800 K and 50 kPa Assume we only have HA2E AO OA2E A and HA2E A as gases find the equilibrium composition With only the given components we have the reaction 2 HA2E AO 2HA2E A OA2E which at 3800 K has an equilibrium constant from A11 as ln K 1906 Assume we start with 2 kmol water and let it dissociate x to the left then Species HA2E AO HA2E A OA2E A Initial 2 0 0 Change 2x 2x x Final 2 2x 2x x Tot 2 x Then we have K exp1906 A yH2 E2 yO2 yH2O 2 E A A P P0 212E A A 2x 2 x 2 x 2 x 2 2x 2 x 2 50 E100E which reduces to 0148674 A 1 1 x2 4x3 E2 x 1 4 1 2E A or xA3E A 0297348 1 xA2E A 2 x Trial and error to solve for x 054 then the concentrations are yAH2OE A A2 2x 2 xE A 0362 yAO2E A A x 2 xE A 0213 yAH2E A A 2x 2 xE A 0425 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1451 Repeat problem 1447 for an initial mixture that also includes 2 kmol of nitrogen which does not dissociate during the process This problem has a dilution of the reactant with nitrogen initial mix 1 COA2E A 2 CO 2 OA2E A 2 NA2E Constant pressure reactor Q Equilibrium mix COA2E A CO OA2E A and NA2E A at T 3000 K P 150 kPa Reaction 2 COA2E 2 CO OA2E initial 1 2 2 change 2x 2x x equil 12x 22x 2x From A11 at 3000 K K exp2217 0108935 For each n 0 1 x A 1 2E Equilibrium nACO2E A 1 2x nACOE A 2 2x nAO2E A 2 x nAN2E A 2 so then nAtotE A 7 x K A y 2 COyO2 Ey 2 CO2 E A A P P0 E AA 1E A 4 A1 x 1 2xE AA 2E A A2 x 7 xE A A150 100E A 0108935 or A1 x 1 2xE AA 2E A A2 x 7 xE A 0018167 Trial error x 0464 A nCO2 1928 EnCO 1072 nO2 1536 nN2 20 nTOT 6536 E A A yCO2 0295 EyCO 0164 yO2 0235 yN2 0306 E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1452 Catalytic gas generators are frequently used to decompose a liquid providing a desired gas mixture spacecraft control systems fuel cell gas supply and so forth Consider feeding pure liquid hydrazine NA2E AHA4E A to a gas generator from which exits a gas mixture of NA2E A HA2E A and NHA3E A in chemical equilibrium at 100C 350 kPa Calculate the mole fractions of the species in the equilibrium mixture Initially 2 NA2E AHA4E A 1 NA2E A 1 HA2E A 2 NHA3E Reaction NA2E 3 HA2E 2 NHA3E initial 1 1 2 change x 3x 2x equil 1x 13x 22x nATOTALE A 42x K A y 2 NH3 EyN2y 3 H2 E A A P P0 E AA 2E A A2 2x24 2x2 E1 x1 3x3 E A A350 100E AA 2E At 100 AoE AC 3732 K for NHA3E A use A5 ACE AP0E A 17032130 36276 AhE A 0 NH3E A 45 720 362763732 2982 42 999 kJkmol AsE A 0 NH3E A 192572 36276 ln A3732 2982E A 20071 kJkmol K Using A9 HA 0 100 CE A 242 999 102188 302179 94 723 kJ SA 0 100 CE A 2200711 1198155 3137196 208321 kJK GA 0 100 CE A HA0E A TSA0E A 94 723 3732208321 16 978 kJ ln K AG0 ERTE A A 16 978 831453732E A 54716 K 23784 Therefore A1 x2 x E1 3xE AA 2E A A 1 1 x1 3xE A A23784352 E16E A 182096 By trial and error x 0226 A nN2 0774 EnH2 0322 nNH3 2452 nTOT 3518 E A A yN2 02181 EyH2 00908 yNH3 06911 E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1453 Complete combustion of hydrogen and pure oxygen in a stoichiometric ratio at PAoE A TAoE A to form water would result in a computed adiabatic flame temperature of 4990 K for a steady state setup How should the adiabatic flame temperature be found if the equilibrium reaction 2HA2E A OA2E A 2 HA2E AO is considered Disregard all other possible reactions dissociations and show the final equations to be solved 2HA2E A OA2E A 2HA2E AO Species HA2E A OA2E A HA2E AO Initial 2 1 0 Shift 2x x 2x Final 2 2x 1 x 2x Keq A yH2O E2 yH2 2 yO2 E A A P P0 E AA1E A nAtotE A 2 2x 1 x 2x 3 x Energy Eq HAPE A HARE A HAP Eo E A HAPE A HAR Eo E A 0 HAPE A 1 x2AhE AH2 EA AhE AO2 EA 2xAhE AfH2O Eo E A AhE AH2OE A 0 1 Equilibrium constant KAeqE A A 4x2 E3x2 E A A 3 x2 E2 2x2 E A A3 x 1 xE A Ax23 x E1 x3 E A KATE A 2 A10 AhE AfH2O Eo E A 241 826 kJkmol A11 ln KATE A A9 AhE AH2 EAT AhE AO2 EAT AhE AH2O EAT Trial and error solve for x T using Eqs 1 and 2 Guess T 3600 K K exp1996 735956 x 053986 HAPE A 1 0539862 111 367 122 245 2 053986 241 826 160 484 70 912 high Guess T 3400 K K exp3128 228283 x 0648723 HAPE A 1 06487232 103 736 114 101 2 0648723 241 826 149 073 7381 low Now do a linear interpolation to get HAPE A 0 T 3400 200 738170 912 7381 3419 K K 2050 x 063905 yAO2 E A 0153 yAH2 E A 0306 yAH2OE A 0541 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1454 Consider the water gas reaction in Example 144 Find the equilibrium constant at 500 1000 1200 and 1400 K What can you infer from the result As in Example 144 III HA2E A COA2E A HA2E AO CO I 2 COA2E A 2 CO OA2E II 2 HA2E AO 2 HA2E A OA2E A Then ln KAIIIE A 05 ln KAIE A ln KAIIE A At 500 K ln KAIIIE A 05 115234 105385 49245 K 0007 266 At 1000 K ln KAIIIE A 05 47052 46321 03655 K 0693 85 At 1200 K ln KAIIIE A 05 35736 36363 03135 K 13682 At 1400 K ln KAIIIE A 05 27679 29222 07715 K 2163 It is seen that at lower temperature reaction III tends to go strongly from right to left but as the temperature increases the reaction tends to go more strongly from left to right If the goal of the reaction is to produce more hydrogen then it is desirable to operate at lower temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1455 A pistoncylinder contains 01 kmol hydrogen and 01 kmol Ar gas at 25C 200 kPa It is heated up in a constant pressure process so the mole fraction of atomic hydrogen is 10 Find the final temperature and the heat transfer needed When gas is heated up HA2E A splits partly into H as HA2E A 2H and the gas is diluted with Ar Component HA2E A Ar H Initial 01 01 0 Shift x 0 2x Final 01x 01 2x Total 02 x yAHE A 01 2x 02 x 2x 002 01x x 0010526 nAtotE A 021053 yAH2E A 0425 01 x02 x yAArE A 1 rest 0475 Do the equilibrium constant KT A y 2 H EyH2 E A A P P0 E AA21E A A 001 0425E A A200 100E A 0047059 ln K 3056 so from Table A11 interpolate to get T 3110 K To do the energy eq we look up the enthalpies in Table A9 at 3110 K hAH2E A 92 8291 hAHE A hAfE A h 217 999 58 4474 276 4454 hAArE A 0 CAPE A3110 29815 207863 311029813 58 4479 same as h for H Now get the total number of moles to get nAHE A 0021053 nAH2E A nAtotE A A1x 2xE A 008947 nAArE A 01 Since pressure is constant W PV and Q becomes differences in h Q nh 008947 92 8291 0 0021053 276 4464 0 01 58 4479 19 970 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1456 The vant Hoff equation d ln K AHo RT2 E A dTAPoE relates the chemical equilibrium constant K to the enthalpy of reaction Ho From the value of K in Table A11 for the dissociation of hydrogen at 2000 K and the value of Ho calculated from Table A9 at 2000 K use vant Hoff equation to predict the constant at 2400 K HA2E A 2H H 2 35 375 217 999 52 942 453 806 kJkmol ln KA2000E A 12841 Assume H is constant and integrate the Vant Hoff equation lnKA2400E A lnKA2000E A A 2400 2000 Ho RT2 dTEA EAH AR AE A EA 1 TA2400 AE A EA 1 TA2000 AE A lnKA2400E A lnKA2000E A H EA 1 TA2400 AE A EA 1 TA2000 AE A ARE 12841 453 806 A 65 12000E A 831451 12841 4548 8293 Table A11 lists 8280 H not exactly constant Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1457 A gas mixture of 1 kmol carbon monoxide 1 kmol nitrogen and 1 kmol oxygen at 25C 150 kPa is heated in a constant pressure process The exit mixture can be assumed to be in chemical equilibrium with COA2E A CO OA2E A and NA2E A present The mole fraction of COA2E A at this point is 0176 Calculate the heat transfer for the process initial mix 1 CO 1 OA2E A 1 NA2E Constant pressure reactor Q Equil mix COA2E A CO OA2E A NA2E yACO2E A 0176 P 150 kPa reaction 2 COA2E 2 CO OA2E also NA2E initial 0 1 1 1 change 2x 2x x 0 equil 2x 12x 1x 1 yACO2E A 0176 A 2x 3xE A x 0242 65 A nCO2 04853 EnCO 05147 nO2 07574 nN2 1 E A A yCO2 0176 EyCO 01867 yO2 02747 E K A y 2 COyO2 Ey 2 CO2 E A A P P0 E AA 1E A A01867202747 E01762 E A A150 100E A 04635 From A11 TAPRODE A 3213 K From A10 HARE A 110 527 kJ HAPE A 04853393 522 166 134 05147110 527 101 447 075740 106 545 10 100 617 66 284 kJ QACVE A HAPE A HARE A 66 284 110 527 176 811 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1458 A tank contains 01 kmol hydrogen and 01 kmol of argon gas at 25AoE AC 200 kPa and the tank keeps constant volume To what T should it be heated to have a mole fraction of atomic hydrogen H of 10 For the reaction HA2E A 2H K A yH E2 yH2 E A A P Po E AA 21E Assume the dissociation shifts right with an amount x then we get reaction HA2E 2 H also Ar initial 01 0 01 change x 2x 0 equil 01 x 2x 01 Tot 02 x yAHE A A 2x 02 xE A 010 x 0010526 We need to find T so K will take on the proper value since K depends on P we need to evaluate P first PA1E AV nA1E ARE ATA1E A PA2E AV nA2E ARE ATA2E A PA2E A PA1E A A n2T2 En1T1 E where we have nA1E A 02 and nA2E A 02 x 0210526 K A yH E2 yH2 E A A P Po E AA 21E A A 2x2 AE 01 x n2 E 200 100 n2T2 02 29815 00001661 TA2E Now it is trial and error to get TA2E A so the above equation is satisfied with K from A11 at TA2E A 3600 K ln K 0611 K 05428 RHS 059796 error 005516 3800 K ln K 0201 K 122262 RHS 063118 error 059144 Linear interpolation between the two to make zero error T 3600 200 A 005516 005516 059144E A 3617 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1459 A local stoichiometric combustion of butene CA4E AHA8E A results in only half the C atoms burns to COA2E A and the other half generates CO This means the products should contain a mixture of HA2E AO CO COA2E A OA2E A and NA2E A which after heat transfer are at 1000 K a Write the combustion equation assuming no hydrogen in the products b Use the water gas reaction to estimate the amount of hydrogen HA2E A present Select the amount of air for stoichiometric combustion Combustion CA4E AHA8E A 6OA2E A 376NA2E A 2COA2E A 2 CO 4 HA2E AO 1 OA2E A 2256 NA2E A Since we have 2 CO we did not use 1 OA2E A which is then left in the products Watergas reaction HA2E A COA2E A HA2E AO CO COA2E A CO HA2E AO HA2E A OA2E A NA2E A Start 2 2 4 0 1 2256 Change x x x x 0 0 Final 2x 2 x 4x x 1 2256 tot 3156 Notice the total number of moles is constant See Example 144 for the equilibrium K At 1000 K ln K 05ln KA1E A ln KA2E A 0547052 46321 03655 K exp03655 yACOE A yAH2OE A yACO2E A yAH2E A 1 A2 x 2 xE A A4 x xE Reduce and solve 2 x x exp03655 2 x 4 x 2x xA2E A exp03655 8 6x xA2E A 2 x 0 for all pos concentrations x 1136 nAH2E A 1136 molmol fuel or 36 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1460 A liquid fuel can be produced from a lighter fuel in a catalytic reactor according to CA2E AHA4E A HA2E AO CA2E AHA5E AOH Show the equilibrium constant is ln K 6691 at 700 K using CAPE A 63 kJkmol K for ethylene and CAPE A 115 kJkmol K for ethanol at 500 K 25AoE AC 5 MPa 300AoE AC 5 MPa 2 H O 2 1 C H 2 4 IG chem equil mixture CA2E AHA5E AOH CA2E AHA4E A HA2E AO 700 K 5 MPa 1 CA2E AHA4 E A 1 HA2E AO 1 CA2E AHA5E AOH HA 0 700 KE A 1235 000 1157002982 152 467 63 7002982 1241 826 14 190 38 935 kJ SA 0 700 KE A 1282444 115 ln A 700 2982E A 1219330 63 ln A 700 2982E A 1218739 111253 kJK GA 0 700 KE A HA0E A TSA0E A 38 935 700111253 38 942 kJ ln K AG0 ER TE A A 38 942 831451 700E A 6691 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1461 A rigid container initially contains 2 kmol of carbon monoxide and 2 kmol of oxygen at 25C 100 kPa The content is then heated to 3000 K at which point an equilibrium mixture of COA2E A CO and OA2E A exists Disregard other possible species and determine the final pressure the equilibrium composition and the heat transfer for the process Equilibrium process 2 CO 2 OA2E A 2 COA2E A OA2E A Species CO OA2E A COA2E Initial 2 2 0 Shift 2x 2xx 2x Final 22x 2x 2x nAtotE A 2 2x 2 x 2x 4 x yACOE A A22x 4xE A yAO2 E A A2x 4xE A yACO2 E A A 2x 4xE Energy equation UA2E A UA1E A A1E AQA2E A HA2E A HA1E A PA2E Av PA1E Av 2 2xAhE ACO 2E A 2 xAhE AO22E A 2xAhE ACO2 E A 2AhE AfCO2 Eo E A 2AhE AfO2 Eo E A ARE A 4 xTA2E A 4ARE ATA1E Notice PA2E A is unknown so write it in terms of TA2E A and the number of moles We flipped the reaction relative to the one in A11 so then KAeqE A 1 KAeq A11E KAeqE A eA2217E A A yCO2 E2 y02yCO 2 E A A P2 EPo E AA1E A A 4x2 E41 x2 E A A4 x 2 xE A A 4T1 E4 xT2 E A A x 1 xE AA2E A A 1 2 xE A A1 4E A A T2 ET1 E A eA2217E A 23092 x 08382 yACOE A 0102 yAO2 E A 0368 yACO2 E A 053 Constant volume and ideal gas approximation PV nARE AT PA2E A PA1E A4 x TA2E A 4TA1E A 100 A4 08382 3000 E4 29815E A 7954 kPa A1E AQA2E A 03236110527 93504 1161898013 16764393522 152853 2110527 2Ø 83145429815 300031618 142991 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1462 Use the information in Problem 1490 to estimate the enthalpy of reaction HAoE A at 700 K using Vant Hoff equation see problem 1456 with finite differences for the derivatives d lnK HARTA 2 EA dT or solve for HAoE A H ARTA 2 EA Ad lnK dTE A ARTA 2 EA AlnK TE 831451 700A 2E A A03362 4607 E800 600E A 86 998 kJkmol Remark compare this to A9 values A5 A10 HAoE A HACE A 2HAH2 E A HACH4 E A 061 12 700298 2 11730 2254 1604 700298 74873 86 739 kJkmol Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1463 One kilomole of carbon dioxide CO2 and 1 kmol of hydrogen H2 at room temperature 200 kPa is heated to 1200 K at 200 kPa Use the water gas reaction to determine the mole fraction of CO Neglect dissociations of H2 and O2 1 COA2E A 1 HA2E A 1 CO 1 HA2E AO Initial 1 1 0 0 Shift x x x x Total 1x 1x x x ntot 2 yH2O yCO x2 yH2 yCO2 1x2 The water gas reaction see Example 144 At 1200 K ln KAIIIE A 05 35736 36363 03135 K 13682 Ax2x2 E1x 2 1x 2 E A K A x2 1x2E A A x 1xE A 11697 x 11697 21697 05391 yH2O yCO x2 027 yH2 yCO2 1x2 023 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1464 A step in the production of a synthetic liquid fuel from organic waste material is the following conversion process at 5 MPa 1 kmol ethylene gas converted from the waste at 25AoE AC and 2 kmol steam at 300AoE AC enter a catalytic reactor An ideal gas mixture of ethanol ethylene and water in equilibrium see previous problem leaves the reactor at 700 K 5 MPa Determine the composition of the mixture 25AoE AC 5 MPa 300AoE AC 5 MPa 2 H O 2 1 C H 2 4 IG chem equil mixture CA2E AHA5E AOH CA2E AHA4E A HA2E AO 700 K 5 MPa 1 CA2E AHA4 E A 1 HA2E AO 1 CA2E AHA5E AOH initial 1 2 0 change x x x equil 1x 2x x total 3 x The reaction rate from the previous problem statement is ln K AG0 ER TE A 6691 K 0001 242 A yC2H5OH EyC2H4 yH2O E AA P P0 E AA 1E A x 1xE AA3x 2xE A 0001242 A50 01E A 00621 By trial and error x 00404 CA2E AHA5E AOH n 00404 y 001371 CA2E AHA4E A n 09596 y 03242 HA2E AO n 19596 y 06621 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1465 A special coal burner uses a stoichiometric mixture of coal and an oxygen argon mixture 11 mole ratio with the reactants supplied at Po To Consider the dissociation of CO2 into CO and O2 and no others Find the equilibrium composition of the products for T 4800 K Is the final temperature including dissociations higher or lower than 4800 K Combustion C OA2E A Ar COA2E A Ar Reaction 2 COA2E A 2 CO OA2E A COA2E A CO OA2E A Ar Start 1 0 0 1 Change 2x 2x x 0 Final 12x 2x x 1 tot 2 x At 4800 K K exp58322 yACOE A2E A yAO2E A yACO2E A2E A 1 A 2x 2 xE AA 2E A x 2 xE A A1 2x 2 xE AA 2E A Reduce and solve xA3E A exp583224 2 x 1 2xA2E A xA3E A 085277 2 x 1 2xA2E A x 04885 close nACOE A 0977 nAO2E A 04885 nACO2E A 0023 nAArE A 1 Now we can do the energy equation HAP 4800 KE A 0023 266 488 393 522 1 20773 4800 298 0977 161 285 110 527 04885 172 240 224 325 kJkmol Since HAP 4800 KE A 0 then T 4800 K We should have had HAPE A 0 if we match the energy equation to find the T we have to redo the equilibrium equation as K is different This is a larger trial and error problem to find the actual T and we even neglected other possible reactions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1466 Acetylene gas at 25C is burned with 140 theoretical air which enters the burner at 25C 100 kPa 80 relative humidity The combustion products form a mixture of COA2E A HA2E AO NA2E A OA2E A and NO in chemical equilibrium at 2200 K 100 kPa This mixture is then cooled to 1000 K very rapidly so that the composition does not change Determine the mole fraction of NO in the products and the heat transfer for the overall process CA2E AHA2E A 35 OA2E A 1316 NA2E A water 2 COA2E A 1 HA2E AO 1 OA2E A 1316 NA2E A water water PAVE A 083169 2535 kPa nAVE A nAAE A PAVE APAAE A 35 1316 253597465 0433 So total HA2E AO in products is 1433 Reaction NA2E A OA2E A 2 NO change x x 2x at 2200 K from A11 K exp6836 0001 074 Equil products nACO2E A 2 nAH2OE A 1433 nAO2 E A 1 x nAN2 E A 1316 x nANOE A 0 2x nATOTE A 17593 K A 2x2 E1 x1316 xE A 0001 074 x 00576 yANOE A A200576 17593E A 0006 55 Final products same composition at 1000 K reactants at 25C HARE A 1226 731 0 0433241 826 0 122 020 kJ HAPE A 2393 522 33 397 1433241 82626 000 094240 22 703 1310240 21 463 0115290 291 22 229 713 954 kJ QACVE A HAPE A HARE A 835 974 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1467 Coal is burned with stoichiometric air with the reactants supplied at the reference conditions Po To If no dissociations are considered the adiabatic flame temperature is found to be 2461 K What is it if the dissociation of COA2E A is included Combustion C 1OA2E A 376 NA2E A COA2E A 376 NA2E The dissociation of COA2E A is 2COA2E A 2 CO OA2E COA2E A CO NA2E A OA2E Start 1 0 376 0 Change 2e 2e e Final 12e 2e 376 e Total 476 e Reaction constant K y2 CO yO2 y2 CO2 A P Po 32 AEE A 2e2 e E 1 2e2476 eE A 1 Products at equilibrium 12e COA2E A 2e CO 376 NA2E A e OA2E Energy Eq HARE A HAPE A HAo PE A ΔHAPE A 0 HAPE A 12e AhE A 0 fE A ΔHACO2E A 2e AhE A 0 fE A ΔHACOE A 376 ΔHAN2E A e ΔHAO2E 0 12e 393 522 ΔHACO2E A 2e 110 527 ΔHACOE A 376 ΔHAN2E A e ΔHAO2E A 2 We need to satisfy equations 1 and 2 both functions of T and e We know T must be less than 2461 K as the dissociation requires energy and that e is small Guess 2400 K K exp7715 0000446 Eq1 e 007322 HAPE A 085356393522 115779 014644110 527 71326 376 70640 007322 74453 28 247 kJkmol This shows that HAPE A is too high so also then e too high Guess 2200 K K exp10232 0000036 Eq1 e 003349 HAPE A 093302393522 103562 006698110 527 64012 376 63362 003349 66770 331768 kJkmol Now HAPE A is too small so linear interpolation to hit HAPE A 0 T 2200 200 0 33 176828 247 33 1768 2308 K To confirm try T 2300 K K exp89735 00001267 e 00498 HAPE A 09004393522 1096705 00996110 527 67669 376 67001 003349 706115 5560 kJkmol Final linear interpolation T 2300 100 556028 247 5560 2316 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1468 An important step in the manufacture of chemical fertilizer is the production of ammonia according to the reaction NA2E A 3HA2E A 2NHA3E A Show that the equilibrium constant is K 6202 at 150C 1 NA2E A 3 HA2E A 2 NHA3E A at 150AoE AC Assume ideal gas and constant specific heat for ammonia we could be more accurate if we used Table B2 realizing properties depend also on P We must use the reference h s values from Table A10 AhE A o NH3 150 CE A 45 720 21317031150 25 41 186 kJkmol AsE A o NH3 150 CE A 192572 21317031 ln A4232 2982E A 205272 kJkmolK Use Table A9 for nitrogen and hydrogen gases HA o 150 CE A 241 186 103649 30 3636 96 929 kJkmol SA 0 150 CE A 2205272 1201829 3140860 213865 kJkmolK GA 0 150 CE A 96 929 4232213865 6421 kJkmol ln K A 6421 831451 42315E A 18248 K 6202 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1469 Consider the previous reaction in equilibrium at 150C 5 MPa For an initial composition of 25 nitrogen 75 hydrogen on a mole basis calculate the equilibrium composition Reaction 1 NA2E A 3 HA2E A 2 NHA3E A at 150AoE AC progress variable x Beginning 1 3 0 Change x 3x 2x Final 1x 33x 2x Total n 4 2x The reaction constant is K 6202 so the reaction constant equation is K A y 2 NH3 EyN2y 3 H2 E AA P P0 E AA 2E A A2x24 2x2 E331 x4 E A A P P0 E AA 2E or A x 1xE AA 2E AA2x 1xE AA 2E A A27 16E A 6202 A 5 01E AA 2E A 26165 or A x 1xE AA2x 1xE A 161755 n y Trial Error NHA3E 1843 08544 x 09215 NA2E 00785 00364 HA2E 02355 01092 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1470 At high temperature NO can form from oxygen and nitrogen Natural gas methane is burned with 150 theoretical air at 100 kPa and the product temperature is 2000 K Neglect other reactions and find the equilibrium concentration of NO Does the formation of the NO change the temperature Combustion CHA4E A 15 2OA2E A 376 NA2E A COA2E A 2HA2E AO 1OA2E A 1128 NA2E Reaction NA2E A OA2E A 2 NO progress x NA2E A OA2E 2 NO HA2E AOCOA2E initial 1128 1 0 21 change x x 2x 0 equilibrium 1x 1x 2x 3 nATOTE A 5 K exp7825 00003996 y2 NO yN2 yO2 A P Po 22 AE A 2x2 E1 x 1 xE A A 2x 1 x 2E Solve 2x1x A 00003996EA 001999 x 0009896 yANOE A 2x5 000396 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1471 Methane at 25C 100 kPa is burned with 200 theoretical oxygen at 400 K 100 kPa in an adiabatic steady state process and the products of combustion exit at 100 kPa Assume that the only significant dissociation reaction in the products is that of carbon dioxide going to carbon monoxide and oxygen Determine the equilibrium composition of the products and also their temperature at the combustor exit Combustion CHA4E A 4OA2E A COA2E A 2HA2E AO 2OA2E Dissociation 2 COA2E 2 CO OA2E HA2E AO inert initial 1 0 2 2 change 2x 2x x 0 equilibrium 12x 2x 2x 2 nATOTE A 5x Equil Eqn K A y 2 COyO2 Ey 2 CO2 E A A P P0 E A A x 05 xE AA 2E AA2 x 5 xE AA P P0 E A or A x 05 xE AA 2E AA2 x 5 xE A A K PP0E A 1 Energy Eq HAPE A HARE A 0 12x393 522 AhE ACO2E A 2x110 527 AhE ACOE A 2241 826 AhE AH2OE A 2xAhE AO2E A 174 873 43027 0 or 12xAhE ACO2E A 2xAhE ACOE A 2AhE AH2OE A 2xAhE AO2E A 565 990x 814 409 0 Assume TAPE A 3256 K From A11 K 06053 Solving 1 by trial error x 02712 Substituting x and the AhE A values from A9 at 3256 K into 2 04576168 821 05424103 054 2140 914 22712108 278 565 99002712 814 409 0 OK TAPE A 3256 K x 02712 nACO2E A 04576 nACOE A 05424 nAH2OE A 20 nAO2E A 22712 yACO2E A 00868 yACOE A 01029 yAH2OE A 03794 yAO2E A 04309 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1472 Calculate the irreversibility for the adiabatic combustion process described in the previous problem From solution of Prob 1471 it is found that the product mixture consists of 04576 COA2E A 05424 CO 20 HA2E AO 22712 OA2E A at 3256 K 100 kPa The reactants include 1 CHA4E A at 25 AoE AC 100 kPa and 4 OA2E A at 400 K 100 kPa Reactants SARE A 1186251 4213873 104174 kJK Products nAiE yAiE AsE A 0 iE ARE A ln A yiP EP0 E ASE A iE COA2E 04576 00868 339278 20322 359600 CO 05424 01029 276660 18907 295567 HA2E AO 20 03794 291099 8058 299157 OA2E 22712 04309 287749 7000 294749 SAPE A 04576359600 05424295567 20299157 22712294749 159262 kJK I TA0E ASAPE ASARE A QCV 29815159262 104174 0 164 245 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1473 Consider the stoichiometric combustion of pure carbon with air in a constant pressure process at 100 kPa Find the adiabatic flame temperature no equilibrium reactions Then find the temperature the mixture should be heatedcooled to so that the concentrations of CO and COA2E A are the same Conbustion C 1OA2E A 376 NA2E A COA2E A 376 NA2E Energy Eq HARE A HAPE A HAo PE A ΔHAPE ΔHAPE A HARE A HAo PE A HAo RPE A HV 393 522 kJkmole ΔHACO2E A 376 ΔHAN2E ΔHAP 3000 KE A 152 853 376 92 715 501 461 too big ΔHAP 2400 KE A 115 779 376 70 640 381 385 too small ΔHAP 2600 KE A 128 074 376 77 963 421 215 too big Interpolate T 2461 K Assume now the dissociation of COA2E A as 2COA2E A 2 CO OA2E COA2E A CO NA2E A OA2E Start 1 0 376 0 Change 2e 2e e Final 12e 2e 376 e Total 476 e When the two concentrations are the same then yACO2E A A 1 2e 476 eE A yACOE A A 2e 476 eE A 1 2e 2e e ¼ Reaction constant K y2 CO yO2 y2 CO2 A P Po 32 AE yAO2E A A 025 476 025E A 00499 exp29977 Now look in Table A11 ln KATE A 29977 Interpolate T 2900 K Check for other reactions possible NA2E A 2N KATE A exp235 no progress to RHS OA2E A 2O KATE A exp509 minor progress to RHS NA2E A OA2E A 2NO KATE A exp45 minor progress to RHS To be very accurate we should look at the last two reactions also Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1474 Hydrides are rare earth metals M that have the ability to react with hydrogen to form a different substance MHx with a release of energy The hydrogen can then be released the reaction reversed by heat addition to the MHx In this reaction only the hydrogen is a gas so the formula developed for the chemical equilibrium is inappropriate Show that the proper expression to be used instead of Eq 1414 is ln PAH2E APAoE A GAoE ART when the reaction is scaled to 1 kmol of HA2E A M A1 2E A x HA2E A MHAxE A At equilibrium GP GR assume g of the solid is a function of T only AgE AMHxE A AhE A 0 MHxE A TAsE A 0 MHxE A AgE A 0 MHxE A AgE AME A AhE A 0 ME A TAsE A 0 ME A AgE A 0 ME A AgE AH2E A AhE A 0 H2E A TAsE A 0 H2E A ARE AT lnPH2Po AgE A 0 H2E A ARE AT lnPH2Po GP GR AgE AMHxE A AgE AME A A1 2E A x AgE AH2E A AgE A 0 ME A A1 2E A xAgE A 0 H2E A ARE AT lnPH2Po AGE A0E A AgE A 0 MHxE A AgE A 0 ME A x AgE A 0 H2E A2 AgE A 0 MHxE A AgE A 0 ME A Scale to 1 mole of hydrogen AG E A0E A AgE A 0 MHxE A AgE A 0 ME Ax2 ARE AT lnPAH2E APAoE A which is the desired result Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful Simultaneous Reactions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1475 For the process in Problem 1446 should the dissociation of oxygen also be considered Verbal answer but supported by numbers The dissociation of oxygen Reaction OA2E A 2 O From A11 KATE A exp4356 001283 Equilibrium KATE A 001283 A yO E2 y02 E A A P P0 E A A yO E2 y02 E So we need to solve two simultaneous reaction equations Due to the small equilibrium constant the concentration of O is going to be small but it will have the effect of requiring a larger heat transfer to reach 3000 K From problem 1446 yAO2E A 0074 so assume this is nearly the same yAOE A A yO2KT EA A 0074 001283EA 003 so about half of the oxygen is dissociated and we do need to solve the proper equations Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1476 Which other reactions should be considered in Problem 1453 and which components will be present in the final mixture 2HA2E A OA2E A 2HA2E AO Species HA2E A OA2E A HA2E AO Other possible reactions from table A11 HA2E A 2 H OA2E A 2 O 2 HA2E AO HA2E A 2 OH So the final component list most likely has species as HA2E A OA2E A HA2E AO H O OH Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1477 Redo Problem 1446 and include the oxygen dissociation Reaction 1 2 COA2E A 2 CO OA2E A Progress variable x Reaction 2 OA2E A 2 O Progress variable y Component COA2E CO OA2E O Ar initial 2 1 0 0 1 change 2x 2x xy 2y equil 2 2x 1 2x xy 2y 1 The total number of moles is nATOTE A 4 x y From Table A11 at 3000 K KA1E A exp 2217 0108935 A y 2 CO yO2 Ey 2 CO2 E A A P Po E AA 32E A A 1 2x2 E4 x y2 E A A x y 4 x yE A A4 x y2 E2 2x2 E A 1A 1E A Ax y 4E A A1 2x2 E4 x yE A A 1 1 x2 E then 043574 4 x y 1 xA2E A x y 1 2xA2E A 1 KA2E A exp 4356 00128296 A y 2 O EyO2 E A A P Po E AA 21E A A 2y 2 E4 x y 2 E A A4 x y x yE A 1A 1E A 4 A y2 E4 x yE A A 1 x yE 00032074 4 x y x y yA2E A 2 The two eq 1 and 2 must be solved for the two variables xy with the restrictions x y 0 and 0 x 1 Solution is x 0347805 y 006343 The outlet has 130439 COA2E A 169561 CO 1 Ar 028438 OA2E A 012686 O Energy equation gives from Table A9 and A5 for argon CApE AΔT plus the enthalpy of formation as the composition changes from inlet to exit AQ E A n ex h ex n in h in 130439 152853 393522 169561 93504 110527 1 39948 052 3000 298 028438 98013 012686 56574 249170 2 393522 1110527 1 0 677 568 kW Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1478 Ethane is burned with 150 theoretical air in a gas turbine combustor The products exiting consist of a mixture of COA2E A HA2E AO OA2E A NA2E A and NO in chemical equilibrium at 1800 K 1 MPa Determine the mole fraction of NO in the products Is it reasonable to ignore CO in the products Combustion CA2E AHA6E A 525 OA2E A 1974 NA2E A 2 COA2E A 3 HA2E AO 175 OA2E A 1974 NA2E Products at 1800 K 1 MPa Equilibrium mixture COA2E A HA2E AO OA2E A NA2E A NO NA2E OA2E 2 NO initial 1974 175 0 change x x 2x equil 1974x 175x 2x Equil comp nACO2E A 2 nAO2 E A 175x nANO E A 2x nAH2OE A 3 nAN2 E A 1974x K 119210A4E A A y 2 NO EyN2yO2 E A A P P0 E AA 0E A A 4x2 E1974x175xE Solving x 0031 75 yANOE A A20031 75 2649E A 00024 b 2 COA2E A 2 CO OA2E initial 2 0 0 change 2a 2a 2x equil 22a 2a 2x K 419410A8E A A y 2 COyO2 Ey 2 CO2 E AA P P0 E AA 1E A A 2a 22aE AA 2E AA175xa 2649aE AA 1 01E This equation should be solved simultaneously with the equation solved in part a modified to include the unknown a Since x was found to be small and also a will be very small the two are practically independent Therefore use the value x 0031 75 in the equation above and solve for a A a 1aE AA 2E AA1750031 75a 2649aE A A01 10E A 4194 10A8E Solving a 0000 254 or yACOE A 19210A5E A negligible for most applications Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1479 A mixture of 1 kmol water and 1 kmol oxygen at 400 K is heated to 3000 K 200 kPa in a steady flow process Determine the equilibrium composition at the outlet of the heat exchanger assuming that the mixture consists of HA2E AO HA2E A OA2E A and OH Reactions and equilibrium eqns the same as in example 147 but different initial composition At equil nAH2OE A 1 2a 2b nAH2E A 2a b nAO2E A 1 a nAOHE A 2b nATOTE A 2 a b Since T 3000 K is the same the two equilibrium constants are the same From Table A11 KA1E A 0002 062 KA2E A 0002 893 The two equilibrium equations are KA1E A A 2a b 1 2a 2bE AA 2E A A 1 a 2 a bE AA P P0 E A KA2E A A 2a b 2 a bE AA 2b 1 2a 2bE AA 2E AA P P0 E A which must be solved simultaneously for a b If solving manually it simplifies the solution to divide the first by the second which leaves a quadratic equation in a b can solve for one in terms of the other using the quadratic formula with the root that gives all positive moles This reduces the problem to solving one equation in one unknown by trial error Solving b 0116 a 0038 nAH2OE A 0844 nAH2E A 00398 nAO2E A 0962 nAOHE A 0232 nATOTE A 20778 yAH2OE A 04062 yAH2E A 00191 yAO2E A 04630 yAOHE A 01117 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1480 Assume dry air 79 NA2E A and 21 OA2E A is heated to 2000 K in a steady flow process at 200 kPa and only the reactions listed in Table A11 are possible and their linear combinations Find the final composition anything smaller than 1 ppm parts per million is neglected and the heat transfer needed for 1 kmol of air in We could have NA2E A N OA2E A O NO NOA2E The possible reactions are NA2E A 2 N KT exp41655 0 OA2E A 2 O KT exp14619 small NA2E A OA2E A 2 NO KT exp7825 NA2E A 2 OA2E A 2 NOA2E A KT exp19136 small So from this we have only NO extra NA2E OA2E 2 NO initial 079 021 0 change x x 2x equil 079 x 021 x 2x K exp7825 00004 A y 2 NO EyN2yO2 E A A P P0 E AA 0E A A 4x2 E079 x021 xE Solving x 0004 yAN2E A 0786 yAO2E A 0206 yANOE A 0008 Q HAexE A HAin E A 0786 ΔAhE AN2E A 0206 ΔAhE AO2E A 0008 AhE ANOE 0786 56137 0206 59176 0008 57859 90291 57 499 kJkmol If we had solved also for the oxygen O and NOA2E A formation we would get approximately neglect the effect on total moles yA 2 OE A yAO2E A PPAoE A exp14619 yAOE A 00002 yA 2 NO2E A yAN2E AyA 2 O2E A PPAoE AA 1E A exp19136 yANO2E A 0000018 and the O energy addition would be 0000235713 249170 57 kJkmol very insignificant Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1481 One kilomole of water vapor at 100 kPa 400 K is heated to 3000 K in a constant pressure steady flow process Determine the final composition assuming that HA2E AO HA2E A H OA2E A and OH are present at equilibrium Reactions 1 2 HA2E AO 2 HA2E A OA2E A 2 2 HA2E AO HA2E A 2 OH change 2a 2a a change 2b b 2b 3 HA2E A 2 H change c 2c At equilibrium 3000 K 100 kPa nAH2OE A 12a2b nAO2 E A a nAH E A 2c nAH2 E A 2abc nAOH E A 2b nATOTE A 1abc A K1 EPP0E A A2062103 E1E A A 2abc 12a2bE AA 2E A A a 1abcE A A K2 EPP0E A A2893103 E1E A A 2abc 1abcE AA 2b 12a2bE AA 2E A K3 EPP0E A A2496102 E1E A A 2c2 E2abc1abcE These three equations must be solved simultaneously for a b c a 00622 b 00570 c 00327 and nAH2OE A 07616 yAH2OE A 06611 nAH2 E A 01487 yAH2 E A 01291 nAO2 E A 00622 yAO2 E A 00540 nAOH E A 01140 yAOH E A 00990 nAH E A 00654 yAH E A 00568 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1482 Water from the combustion of hydrogen and pure oxygen is at 3800 K and 50 kPa Assume we only have HA2E AO OA2E A OH and HA2E A as gases with the two simple water dissociation reactions active find the equilibrium composition This problem is very similar to Example 147 in the text The only difference is that we have T 3800 K and P 50 kPa From table A11 we have ln KA1E A 1906 KA1E A 014867 ln KA2E A 0984 KA2E A 03738 KA1E A A 2a b 1 2a 2bE AA 2E A A a 1 a bE AA P P0 E A KA2E A A 2a b 1 a bE AA 2b 1 2a 2bE AA 2E AA P P0 E A So we have two equations as A 2a b 1 2a 2bE AA 2E A A a 1 a bE A KA1E A A P P0 E A 029734 1 A 2a b 1 a bE AA 2b 1 2a 2bE AA 2E A KA2E A A P P0 E A 07476 2 Divide the second equation by the first to give A 4b2 E2a b aE A A 07476 029734E A 25143 or 2aA2E A ba 15909 bA2E A 0 a b4 14 A b2 4 2 15909 b2EA 0676256 b Now we can do trial and error on equation 1 for only one variable say b a 014228 b 02104 nAH2OE A 1 2a 2b 029464 nAH2E A 2a b 049496 nAO2E A a 014228 nAOHE A 2b 04208 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1483 Methane is burned with theoretical oxygen in a steady flow process and the products exit the combustion chamber at 3200 K 700 kPa Calculate the equilibrium composition at this state assuming that only COA2E A CO HA2E AO HA2E A OA2E A and OH are present Combustion CHA4E A 2 OA2E A COA2E A 2 HA2E AO Dissociation reactions 1 2 HA2E AO 2 HA2E A OA2E A 2 2 HA2E AO HA2E A 2 OH change 2a 2a a change 2b b 2b 3 2 COA2E A 2 CO OA2E change 2c 2c c At equilibrium nAH2OE A 22a2b nAO2 E A ac nACO2E A 12c nAH2 E A 2ab nAOH E A 2b nACO E A 2c nATOTE A 3abc Products at 3200 K 700 kPa from A11 KA1E A exp4916 0007 328 A 2ab 22a2bE AA 2E A A ac 3abcE A A700 100E A KA2E A exp4401 0012 265 A 2b 22a2bE AA 2E A A 2ab 3abcE A A700 100E A KA3E A exp0853 0426 135 A 2c 12cE AA 2E A A ac 3abcE A A700 100E A These 3 equations must be solved simultaneously for a b c If solving by hand divide the first equation by the second and solve for c fctab This reduces the solution to 2 equations in 2 unknowns Solving a 0024 b 01455 c 0236 Substance HA2E AO HA2E OA2E OH COA2E CO n 1661 01935 0260 0291 0528 0472 y 04877 00568 00764 00855 01550 01386 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1484 Butane is burned with 200 theoretical air and the products of combustion an equilibrium mixture containing only COA2E A HA2E AO OA2E A NA2E A NO and NOA2E A exit from the combustion chamber at 1400 K 2 MPa Determine the equilibrium composition at this state Combustion CA4E AHA10E A 13 OA2E A 489 NA2E A 4 COA2E A 5 HA2E AO 65 OA2E A 489 NA2E Dissociation 1 NA2E A OA2E A 2 NO 2 NA2E A 2OA2E A 2 NOA2E change a a 2a change b 2b 2b At equilibrium nAH2OE A 5 nAN2 E A 489ab nANO E A 2a nACO2E A 4 nAO2 E A 65a2b nANO2E A 2b nATOTE A 644b At 1400 K from A11 KA1E A 376110A6E A KA2E A 902610A10E KA1E A A 2a2 E489ab65a2bE A KA2E A A 2b2644b E65a2b2489abE AA P P0 E AA 1E As KA1E A and KA2E A are both very small with KA2E A KA1E A the unknowns a b will both be very small with b a From the equilibrium eqs for a first trial a A1 2E A A K148965EA 00173 b A1 2E A65A K2 2 E01489 644 EA 0000 38 Then by trial error A a2 E489ab65a2bE A A3761106 E4E A 0940 2510A6E A b2644b E65a2b2489abE A A 90261010 2 E01 4E A 451310A10E Solving a 0017 27 b 0000 379 nACO2E A 4 nAH2OE A 5 nAN2 E A 48882 nAO2 E A 6482 yACO2E A 0062 11 yAH2OE A 0077 64 yAN2 E A 0759 04 yAO2 E A 0100 65 nANO E A 0034 54 nANO2E A 0000 76 yANO E A 0000 55 yANO2E A 0000 01 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1485 One kilomole of air assumed to be 78 nitrogen 21 oxygen and 1 argon at room temperature is heated to 4000 K 200 kPa Find the equilibrium composition at this state assuming that only N2 O2 NO O and Ar are present 1 kmol air 078 NA2E A 021 OA2E A 001 Ar heated to 4000 K 200 kPa Equil 1 NA2E A OA2E A 2 NO nAN2E A 078a change a a 2a nAO2E A 021ab nAArE A 001 2 OA2E A 2 O nAO E A 2b change b 2b nANOE A 2a nAtotE A 1b KA1E A 00895 A 4a2 E078a021abE A A200 100E AA 0E KA2E A 2221 A 4b2 E1b021abE A A200 100E A Divide 1st eqn by 2nd and solve for a as functionb using X A K1 EK2 E A A P P0 E A 00806 Get a A Xb2 E21bE A1A 140781b EXb2 E A 1 Also A b2 E1b021abE A A K2 E4PP0E A 0277 63 2 Assume b 01280 From 1 get a 00296 Then check a b in 2 OK Therefore nAN2E A 07504 nAO E A 02560 yAN2E A 06652 yAO E A 02269 nAO2E A 00524 nANOE A 00592 yAO2E A 00465 yANOE A 00525 nAArE A 001 yAArE A 00089 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1486 Acetylene gas and x times theoretical air x 1 at room temperature and 500 kPa are burned at constant pressure in an adiabatic steady flow process The flame temperature is 2600 K and the combustion products are assumed to consist of N2 O2 CO2 H2O CO and NO Determine the value of x Combustion CA2E AHA2E A 25x OA2E A 94x NA2E A 2 COA2E A HA2E AO 25x1OA2E A 94x NA2E Eq products 2600 K 500 kPa NA2E A OA2E A COA2E A HA2E AO CO NO 2 Reactions 1 2 COA2E A 2 CO OA2E A 2 NA2E A OA2E A 2 NO change 2a 2a a change b b 2b Equil Comp nAN2 E A 94xb nAH2OE A 1 nACO E A 2a nANO E A 2b nAO2 E A 25x 25 a b nACO2E A 2 2a nATOTE A 119x 05 a At 2600 K from A11 KA1E A 3721 10A3E A KA2E A 4913 10A3E EQ1 A K1 EPPoE A A3721103 E5E A A a 1 aE AA 2E A A25x 25 a b 119x 05 aE A EQ2 KA2E A 491310A3E A A 2b2 E94 b25x 25 a bE Also from the energy Eq HAPE A HARE A 0 where HARE A 1226 731 0 0 226 731 kJ HAPE A 94x b0 77 963 25x 25 a b0 82 225 2 2a393 522 128 074 1241 826 104 520 2a110 527 78 679 2b90 291 80 034 Substituting HARE A and HAPE A into the energy equation EQ3 988 415x 549 425a 180 462b 1 100 496 0 which results in 3 equations in the 3 unknowns x a b Assume x 107 then EQ1 7442 10 A2E A A a 1aE AA 2E AA0175 a b 13233 aE A EQ2 12283 10 A3E A A b2 E10058 b0175 a bE A Solving a 01595 b 00585 Then checking in EQ3 988 415107 549 42501595 180 46200585 1 100 496 0 Therefore x 107 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful Gasification Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1487 One approach to using hydrocarbon fuels in a fuel cell is to reform the hydrocarbon to obtain hydrogen which is then fed to the fuel cell As a part of the analysis of such a procedure consider the reforming reaction CHA4E A HA2E AO 3HA2E A CO Determine the equilibrium constant for this reaction at a temperature of 800 K CHA4E A HA2E AO 3HA2E A CO The equilibrium constant from Eq1415 depends on GA0E A and T For CHA4E A use CAP0E A at average temperature 550 K A 1 2E A 298 800 K Table A6 ACE AP0E A 49316 kJkmol K AhE A 0 800 KE A AhE A 0 fE A ACE AP0E AT 74 873 49316 800 2982 50 126 kJkmol AsE A 0 800 KE A 186251 49316 ln A 800 2982E A 234918 kJkmol K The rest of the properties are from Table A9 at 800 K HA 0 800 KE A 30 14 681 1110 527 15 174 150 126 1241 826 18 002 222 640 kJkmol SA 0 800 KE A 3159554 1227277 1234918 1223826 247195 kJkmol K GA0E A HA0E A TSA0E A 222 640 800247195 24 884 kJkmol ln K AG0 ER TE A A 24 884 83145800E A 37411 K 00237 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1488 A coal gasifier produces a mixture of 1 CO and 2HA2E A that is fed to a catalytic converter to produce methane The reaction is CO 3HA2E A CHA4E A HA2E AO The equilibrium constant at 600 K is K 183 10A6E A What is the composition of the exit flow assuming a pressure of 600 kPa The reaction equation is CO 3 HA2E CHA4E HA2E AO initial 1 2 0 0 change x 3x x x equil 1 x 2 3x x x nATOTALE A 3 2x K A yCH4 yH2O E y 3 H2 yCO E A A P Po E AA 1113E A A x2 3 2x2 E1x2 3x3 E A A P Po E AA 2E 183 10A6E A A600 100E AA 2E A 6588 10A7E A A x2 3 2x2 E1x2 3x3 E Trial and error to solve for x x 06654 LHS 6719 10A7E x 066538 LHS 641 10A7E x 066539 LHS 6562 10A7E A close enough nACH4E A 066539 nAH2OE A 066539 nACOE A 066539 nAH2E A 000383 so we used up nearly all the hydrogen gas Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1489 Gasification of char primarily carbon with steam following coal pyrolysis yields a gas mixture of 1 kmol CO and 1 kmol HA2E A We wish to upgrade the hydrogen content of this syngas fuel mixture so it is fed to an appropriate catalytic reactor along with 1 kmol of HA2E AO Exiting the reactor is a chemical equilibrium gas mixture of CO HA2E A HA2E AO and COA2E A at 600 K 500 kPa Determine the equilibrium composition Note see Example 144 1 CO 1 HA2E 1 HA2E AO Constant pressure reactor Chem Equil Mix CO HA2E A HA2E AO COA2E 600 K 500 kPa 1 1 CO 1 HA2E AO 1 COA2E A 1 HA2E A Water gas reaction x x x x 2 2 HA2E AO 2 HA2E A 1 OA2E A 3 2 COA2E A 2 CO 1 OA2E Reaction 1 A1 2E A 2 A1 2E A 3 From Table A11 at 600 K ln KA1E A A1 2E A8579 9249 335 KA1E A 28503 Equilibrium nACOE A 1x nAH2OE A 1x nACO2E A 0 x nAH2E A 1 x A nEA 3 notice the reaction is pressure insensitive K A yCO2yH2 EyCOyH2O E A A P P0 E AA0E A A yCO2yH2 EyCOyH2O E A Ax1 x E1 x2 E A 28503 Solve for x x 07794 n y CO 02206 00735 735 HA2E AO 02206 00735 735 COA2E 07794 02598 260 HA2E 17794 05932 593 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1490 The equilibrium reaction as CH4 C 2HA2E A has ln K 03362 at 800 K and lnK 4607 at 600 K By noting the relation of K to temperature show how you would interpolate ln K in 1T to find K at 700 K and compare that to a linear interpolation A11 ln K 03362 at 800K ln K 4607 at 600K lnKA700E A lnKA800E A EA A 1 700 A A 1 E800 A A 1 600 A A 1 800 AE A 4607 03362 03362 EA A800 700 A 1 EA800 600 A 1E A 42708 21665 Linear interpolation lnKA700E A lnKA600E A A700 600 800 600E A lnKA800E A lnKA600E A 4607 A1 2E A 03362 4607 24716 Comment Look at the vant Hoff equation in Problem 1456 and integrate it with temperature assuming HA0E A is constant That gives ln K C HA0E AT Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1491 One approach to using hydrocarbon fuels in a fuel cell is to reform the hydrocarbon to obtain hydrogen which is then fed to the fuel cell As a part of the analysis of such a procedure consider the reaction CHA4E A HA2E AO CO 3HA2E A One kilomole each of methane and water are fed to a catalytic reformer A mixture of CH4 HA2E AO HA2E A and CO exits in chemical equilibrium at 800 K 100 kPa determine the equilibrium composition of this mixture using an equilibrium constant of K 00237 The reaction equation is CHA4E HA2E AO 3 HA2E CO initial 1 1 0 0 change x x 3x x equil 1x 1x 3x x nATOTALE A 1 x 1 x 3x x 2 2x K A y 3 H2yCO EyCH4yH2O E A A P P0 E AA 2E A A 3x3x E1 x1 x2 2x2 E A A100 100E AA 2E or A x 1 xE AA 2E AA x 1 xE AA 2E A A400237 271E A 0003 51 or A x2 E1 x2 E A A 0003 51EA 0059 25 Solving x 02365 A nCH4 07635 EnH2O 07635 nH2 07095 nCO 02365 nTOT 2473 E A A yCH4 03087 EyH2O 03087 yH2 02870 yCO 00956 E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1492 Consider a gasifier that receives 4 kmol of carbonmonoxide 3 kmol hydrogen and 376 kmol nitrogen and brings the mixture to equilibrium at 900 K 1 MPa with the following reaction 2 CO 2 HA2E A CHA4E A COA2E which is the sum of Eq1432 and 1433 If the equilibrium constant is K 2679 find the exit composition The reaction takes place with the nitrogen as a dilutant reaction 2 CO 2 HA2E CHA4E COA2E also NA2E initial 4 3 0 0 376 change 2x 2x x x 0 equil 42x 32x x x 376 nATOTE A 1076 2x K A yCH4 yCO2 Ey 2 CO y 2 H2 E A A P P0 E AA 2E A Ax x 1076 2x2 E4 2x23 2x2 E A A P P0 E AA 2E or take the squareroot to get A x1076 2x E4 2x3 2xE A A P P0 E A A KEA A 1 01E A A 2679EA 16368 By trial error or solve the quadratic eq in x x 12781 nACOE A 1444 nAH2E A 0444 nACH4E A nACO2E A 1278 nAN2E A 376 yACOE A 0176 yAH2E A 0054 yACH4E A yACO2E A 0156 yAN2E A 0458 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1493 Consider the production of a synthetic fuel methanol from coal A gas mixture of 50 CO and 50 H2 leaves a coal gasifier at 500 K 1 MPa and enters a catalytic converter A gas mixture of methanol CO and H2 in chemical equilibrium with the reaction CO 2H2 CH3OH leaves the converter at the same temperature and pressure where it is known that ln K 5119 a Calculate the equilibrium composition of the mixture leaving the converter b Would it be more desirable to operate the converter at ambient pressure 1 CO 1 HA2E Converter Equil Mix CHA3E AOH CO HA2E 500 K 1 MPa Reaction CO 2 HA2E A CHA3E AOH initial 1 1 0 change x 2x x equil 1x 12x x a K A yCH3OH EyCOy 2 H2 E AA P P0 E AA 2E A A x 1xE AA22x 12xE AA 2E AA P P0 E AA 2E A A x1x E12x2 E A AK 4E AA P P0 E AA 2E ln K 5119 K 0005 98 A x1x E12x2 E A A0005 98 4E A A 1 01E AA 2E A 01495 x 01045 nACH3OHE A x 01045 nACOE A 1x 08955 nAH2E A 1 2x 0791 yACH3OHE A 00583 yACOE A 05000 yAH2E A 04417 b For P 01 MPa A x1x E12x2 E A A0005 98 4E AA01 01E AA 2E A 0001 495 x is much smaller 00015 not good Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful Ionization Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1494 At 10 000 K the ionization reaction for Ar is Ar ArAE A eAE A with equilibrium constant of K 42 10A4E A What should the pressure be for a mole concentration of argon ions ArAE A of 10 From the reaction ionization we recognize that the concentration of electrons must equal that of argon ions so yAAr E A yAe E A 01 and yAArE A 1 yAAr E A yAe E A 08 Now K 42 10A4E A A yAr ye EyAr E A A P Po E AA 111E A A01 01 08E A A P 100E P 000042 A 08 01 01E A 100 336 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1495 Repeat the previous problem assuming the argon constitutes 1 of a gas mixture where we neglect any reactions of the other gases and find the pressure that will give a moleconcentration of ArAE A of 01 Previous problem At 10 000 K the ionization reaction for Ar is Ar ArAE A eAE A with equilibrium constant of K 42 10A4E A What should the pressure be for a mole concentration of argon ions ArAE A of 10 Ar ArAE A eAE A others Initial 001 0 0 099 Change x x x 0 Final 001x x x 099 Total 1 x From the reaction ionization we recognize that the concentration of electrons must equal that of argon ions so yAAr E A yAe E A 0001 x 1x x 0001001 yAArE A 001 x 1 x 000899 Now K 42 10A4E A A yAr ye EyAr E A A P Po E AA 111E A A0001 0001 000899E A A P 100E P 000042 8990 100 377 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1496 Operation of an MHD converter requires an electrically conducting gas It is proposed to use helium gas seeded with 10 mole percent cesium as shown in Fig P1496 The cesium is partly ionized Cs CsAE A e by heating the mixture to 1800 K 1 MPa in a nuclear reactor to provide free electrons No helium is ionized in this process so that the mixture entering the converter consists of He Cs CsAE A and e Determine the mole fraction of electrons in the mixture at 1800 K where ln K 1402 for the cesium ionization reaction described Reaction Cs CsAE A e Also He ln K 1402 initial 001 0 0 099 K 40633 change x x x 0 Equil 001x x x 099 total 1 x K A ye yCs EyCs E A A P P0 E A A x 001 xE A A x 1 xE A A P P0 E A or A x 001 xE A A x 1 xE A 40633 101 040633 Quadratic equation x 0009767 yAeE A A x 1 xE A 000967 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1497 One kilomole of argon gas at room temperature is heated to 20 000 K 100 kPa Assume that the plasma in this condition consists of an equilibrium mixture of Ar ArAE A ArAE A and e according to the simultaneous reactions 1 Ar ArAE A eAE A 2 ArAE A ArAE A eAE The ionization equilibrium constants for these reactions at 20 000 K have been calculated from spectroscopic data as ln KA1E A 311 and ln KA2E A 492 Determine the equilibrium composition of the plasma 1 Ar ArAE A eAE A 2 ArAE A ArAE A eAE ch a a a ch b b b Equil Comp nAArE A 1a nAArE A ab nAArE A b nAeE A ab nATOTE A 1ab KA1E A A yArye EyAr E AA P P0 E A A a ba b E1 a1 a bE A 1 22421 KA2E A A yArye EyAr E AA P P0 E A A ba b Ea b1 a bE A 1 00073 By trial error a 0978 57 b 0014 13 nAArE A 002143 nAArE A 096444 nAArE A 001413 nAeE A 09927 yAArE A 00107 yAArE A 0484 yAArE A 00071 yAeE A 04982 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1498 At 10 000 K the two ionization reactions for N and Ar as 1 Ar ArAE A eAE A 2 N NAE A eAE have equilibrium constants of KA1E A 42 10A4E A and KA2E A 63 10A4E A respectively If we start out with 1 kmol Ar and 05 kmol NA2E A what is the equilibrium composition at a pressure of 10 kPa At 10 000 K we assume all the nitrogen is dissociated to N Assume we shift the argon ionization with a and the nitrogen ionization with b we get Ar ArAE A eAE A N NAE A Initial 1 0 0 1 0 Change a a a b b b Final 1a a a b 1b b Tot 2 a b KA1E A 42 10A4E A A yArye EyAr E AA P P0 E A A a a b E1 a2 a bE A A 10 100E A 1 KA2E A 63 10A4E A A yNye EyN E AA P P0 E A A b a b E1 b2 a bE A A 10 100E A 2 Divide the second equation with the first to get A b 1 bE A A1 a EaE A A K2 EK1 E A 15 Ab ab a abE A 15 b ab 15 a 15 ab b 15 a 05 ab a15 05 b a A b 15 05 bE A trial and error on equation 1 a 0059 and b 0086 nAArE A 0941 nAArE A 0059 nANE A 0914 nANE A 0086 nAeE A 0145 yAArE A 0439 yAArE A 0027 yANE A 0426 yANE A 004 yAeE A 0068 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 1499 Plot to scale the equilibrium composition of nitrogen at 10 kPa over the temperature range 5000 K to 15 000 K assuming that NA2E A N NAE A and e are present For the ionization reaction N NAE A e the ionization equilibrium constant K has been calculated from spectroscopic data as T K 10000 12 000 14 000 16 000 100K 00626 151 151 92 1 NA2E A 2N 2 N NAE A eAE change a 2a change b b b Equil Comp nAN2E A 1 a nANE A 2a b nANE A b nAeE A b EQ1 KA1E A A y 2 N EyN2 E A A P P0 E A A 2a b2 E1 a1 a bE A A P P0 E A EQ2 KA2E A A yNye EyN E AA P P0 E A A b2 E2a b1 a bE A A P P0 E A For T 10 000 K b 0 so neglect EQ2 KA1E A A 4a2 E1a2E A A 10 100E A To extrapolate KA1E A above 6000 K ln KA1E A 16845 A118 260 TE from values at 5000 K 6000 K TK KA1E a yANE yAN2E 5000 00011 00524 00996 09004 6000 00570 03532 05220 04780 7000 09519 08391 09125 00875 8000 7866 09755 09876 00124 10000 15126 09987 09993 00007 For T 10 000 K a 10 KA2E A A b2 E2b2bE AA 10 100E A A b2 E4b2E A 01 TK KA2E b yANE yANE 10 000 62610A4E 01577 08538 00731 12 000 15110A2E 07244 04862 02659 14 000 0151 15512 01264 04368 16 000 092 18994 00258 04871 Note that b 0 is not a very good approximation in the vicinity of 10 000 K In this region it would be better to solve the original set simultaneously for a b The answer would be approximately the same Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful Applications Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14100 Are the three reactions in the Zeldovich mechanism pressure sensitive if we look at equilibrium conditions No All three reactions have two moles on the left and right hand sides and the net power to the pressure correction term is zero Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14101 Assume air is at 3000 K 1 MPa Find the time constant for the NO formation Repeat for 2000 K 800 kPa From the rate in Eq1440 τNO 8 10A16E A T PoPA12E A expA58 300 TE A Case a τNO 8 10A16E A 3000 1001000A12E A expA58 300 3000E A 209 10A4E A s 0209 ms Case b τNO 8 10A16E A 2000 100800A12E A expA58 300 2000E A 258 s Notice the significant difference in time constants Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14102 Consider air at 2600 K 1 MPa Find the equilibrium concentration of NO neglecting dissociations of oxygen and nitrogen 1 kmol air 078 NA2E A 021 OA2E A 001 Ar at 2600 K 1 MPa NA2E A OA2E A 2 NO Ar nAN2E A 078 x 078 021 0 001 nAO2E A 021 x x x 2x nANOE A 2x nAarE A 001 078x 021x 2x 001 ntot 10 From A11 at 2600 K ln K 5316 K 00049124 K A 4x2 E078 x021 xE A A P P0 E AA 0E A A x2 E078 x021 xE A A00049124 4E A 0001228 and 0 x 021 Solve for x x 00136 yANOE A A2x 10E A 00272 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14103 Redo the previous Problem but include the dissociation of oxygen and nitrogen 1 kmol air 078 NA2E A 021 OA2E A 001 Ar at 2600 K 1 MPa 1 OA2E A 2 O lnK 752 K1 0000542 yA 2 OE AyO2 PPo 2 NA2E A 2 N lnK 28313 K2 5056 1013 yA 2 NE AyN2 PPo 3 NA2E A OA2E A 2 NO lnK 5316 K3 000491 yA 2 NOE AyN2 yO2 Call the shifts abc respectively so we get nO2 021 a c nO 2a nN2 078 b c nN 2b nNO 2c nAr 001 ntot 1 a b From which the mole fractions are formed and substituted into the three equilibrium equations The result is 0000542 01 yA 2 OE AyO2 4a2 1ab021ac 5056 1014 yA 2 NE AyN2 4b2 1ab079bc 000491 yA 2 NOE AyN2 yO2 4c2 079bc021ac which give 3 eqs for the unknowns abc Trial and error assume b a 0 solve for c from the last eq then for a from the first and finally given the ac solve for b from the second equation The order chosen according to expected magnitude K3 K1 K2 a 0001626 b 099 107 c 001355 nO2 01948 nO 000325 nN2 07665 nN 198 107 nNO 00271 nAr 001 yNO nNO ntot 00271 100165 002706 Indeed it is a very small effect to include the additional dissociations Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14104 Calculate the equilibrium constant for the first reaction in the Zeldovich mechanism at 2600 K 500 kPa Notice this is not listed in A11 The reaction and equilibrium constant are O NA2E A NO N ln K AG0 ER TE A GA0E A HA0E A TSA0E A HA0E A AhE ANOE A AhE ANE A AhE AOE A AhE AN2E A 80034 90291 47860 472680 48216 249170 77963 315 516 kJkmol SA0E A AsE ANOE A AsE ANE A AsE AOE A AsE AN2E 282822 198322 206714 261615 12815 kJkmolK GA0E A 315 516 2600 12815 282 197 kJkmol ln K AG0 ER TE A A 282 197 8314472 2600E A 13054 K 214 10A6E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14105 Find the equilibrium constant for the reaction 2NO OA2E A 2NOA2E A from the elementary reaction in Table A11 to answer these two questions Which of the nitrogen oxides NO or NOA2E A is the more stable at 25AoE AC 100 kPa At what T do we have an equal amount of each The elementary reactions are the last two ones in Table A11 a NA2E A OA2E A 2 NO b NA2E A 2OA2E A 2 NOA2E Actual reaction is c b a lnKAcE A lnKAbE A lnKAaE A At 25AoE AC approx 300 K Table A11 lnKAaE A 69868 lnKAbE A 41355 so now lnKAcE A 41355 69868 285 KAcE A 24 10 A12E meaning reaction is pushed completely to the right and NOA2E A is the stable compound Assume we start at room T with 1 kmol NOA2E A then NO OA2E A NOA2E A TOT start 0 0 1 change 2x x 2x Final 2x x 12x 1x Equal amount of each yNO A 2x 1 xE A yNOA2E A A1 2x 1 xE A x 025 KT A1 2x2 4x3 E A A 052 4 0253 E A 4 lnK 1386 We quickly see lnK at 500 K 30725 40449 9724 lnK at 1000 K 23039 18709 433 Linear interpolation T 500 0406 500 703 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14106 If air at 300 K is brought to 2600 K 1 MPa instantly find the formation rate of NO The formation rate of NO is given by Eqs143940 using the equilibrium concentration in Eq1441 If we assume NO formation takes place and that the oxygen and nitrogen have not dissociated their concentrations are we neglect the argon NA2E A OA2E A 2 NO yANOeE A 2x yAO2E A 021 x and yAN2E A 079 x At 2600 K from A11 ln KA4E A 5316 KA4E A 00049124 so then we have yANOeE A 2x KA4E A 021 x 079 x A12E A solve for x 00136 yANOeE A 2x 00272 From the rate in Eq1440 τNO 8 10A16E A T PoPA12E A expA58 300 TE A τNO 8 10A16E A 2600 1001000A12E A expA58 300 2600E A 00036 s The formation rate is A dyNO EdtE A yNOe τNO A 00272 00036 sE A 756 sA1E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14107 Estimate the concentration of oxygen atoms in air at 3000 K 100 kPa and 00001 kPa Compare this to the result in Fig 1411 For the dissociation of oxygen and nitrogen we get from A11 OA2E A ln K 4356 NA2E A ln K 22367 from which it is evident that we need only to consider oxygen OA2E A 2 O NA2E Ar nAN2E A 078 x 021 0 078 001 nAO2E A 021 x x 2x nANOE A 2x nAarE A 001 021x 2x 078 001 ntot 1 x KT y2 O yO2 A P Po 21 AE A 2x 1 x 2E A A 1 x 021 xE A P Po A 4x2 E1 xE A A 1 021 xE A P Po A 4x2 E1 xE A A 1 021 xE A Po P KT Po P exp4356 P 100 kPa RHS 001283 x 00247 yO A 2x 1 xE A 00482 P 00001 kPa RHS 12830 x 021 yO A 2x 1 xE A 0347 We notice from Fig 1411 that at 3000 K all the oxygen is dissociated so nO natoms in air 021 so nO 042 yO 042121 0347 since air is mainly diatomic so natoms in air 2 scale is confusing Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14108 At what temperature range does air become a plasma From Fig 1411 we note that air becomes predominantly ions and electrons a plasma at about 1012 000 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14109 In a test of a gasturbine combustor saturatedliquid methane at 115 K is to be burned with excess air to hold the adiabatic flame temperature to 1600 K It is assumed that the products consist of a mixture of COA2E A HA2E AO NA2E A OA2E A and NO in chemical equilibrium Determine the percent excess air used in the combustion and the percentage of NO in the products CHA4E A 2x OA2E A 752x NA2E A 1 COA2E A 2 HA2E AO 2x2 OA2E A 752x NA2E Then NA2E A OA2E A 2 NO Also COA2E A HA2E AO initial 752x 2x2 0 1 2 change a a 2a 0 0 final 752xa 2x2a 2a 1 2 nATOTE A 1 952x 1600 K in A11 ln K 1055 K 262810A5E 262810A5E A K A y 2 NO EyN2yO2 E AA P P0 E AA0E A A y 2 NO EyN2yO2 E A A 4a2 E752x a2x 2 aE From A9 and B7 HARE A 174 873 1604327476241 0 0 89 292 kJ Air assumed 25 AoE AC HAPE A 1393 522 67 569 2241 826 52 907 752xa41 904 2x2a44 267 2a90 291 43 319 792 325 403 652 x 181 049 a Assume a 0 then from HAPE A HARE A 0 x 17417 and substitute A a2 E13098 a1483 aE A A2628105 E4E A get a 00113 Use this a in the energy equation x A703 042 181 04900113 403 652E A 17366 A a2 E13059a14732aE A A2628105 E4E A a 00112 x 17366 excess air 737 NO A200112100 195217366E A 0128 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14110 Find the equilibrium constant for the reaction in Problem 1492 Reaction 2 CO 2 HA2E A CHA4E A COA2E Equilibrium constant Eq 1415 ln K GA0E A ARE AT For CHA4E A at 600 K formula in Table A6 ACE AP0E A 5222 At 900 K AhE A 0 CH4E A 74 873 5222900 2982 43 446 kJkmol AsE A 0 CH4E A 186251 5222 ln 900 2982 243936 kJkmol K The integratedequation values are 43 656 and 240259 HA 0 900 KE A 143 446 1393 522 28 030 2110 527 18 397 20 17 657 259 993 kJ SA 0 900 KE A 1243936 1263646 2231074 2163060 280687 kJK GA 0 900 KE A 259 993 900280687 7375 kJ ln K A 7375 83145 900E A 09856 K 2679 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14111 A space heating unit in Alaska uses propane combustion is the heat supply Liquid propane comes from an outside tank at 44C and the air supply is also taken in from the outside at 44C The airflow regulator is misadjusted such that only 90 of the theoretical air enters the combustion chamber resulting in incomplete combustion The products exit at 1000 K as a chemical equilibrium gas mixture including only CO2 CO H2O H2 and N2 Find the composition of the products Hint use the water gas reaction in Example 144 Propane Liquid T1 44oC 2292 K Air TA2E A 44oC 2292 K 90 Theoretical Air Products T3 1000 K COA2E A CO HA2E AO H2 N2 Theoretical Air C3H8 5OA2E A 188NA2E A 3COA2E A 4H2O 188NA2E 90 Theoretical Air C3H8 45OA2E A 1692NA2E A aCOA2E A bCO cH2O dH2 1692N2 Carbon a b 3 Hydrogen c d 4 Oxygen 2a b c 9 Where 2 a 3 Reaction CO HA2E AO CO2 H2 Initial b c a d Change x x x x Equil b x c x a x d x Chose an Initial guess such as a 2 b 1 c 4 d 0 Note A different initial choice of constants will produce a different value for x but will result in the same number of moles for each product nCO2 2 x nCO 1 x nH2O 4 x nH2 x nN2 1692 The reaction can be broken down into two known reactions to find K 1 2COA2E A 2CO O2 1000 K lnK1 47052 2 2HA2E AO 2H2 O2 1000 K lnK2 46321 For the overall reaction lnK lnKA2E A lnK12 03655 K 14412 K yCO2yH2 yCOyH2O A P Po 1111 AE yCO2yH2 yCOyH2O 14412 A 2 xx E1 44 xE A x 06462 nCO2 26462 nCO 03538 nN2 1692 nH2O 33538 nH2 06462 yCO2 0111 yCO 0015 yN2 0707 yH2O 0140 yH2 0027 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14112 Derive the vant Hoff equation given in problem 1456 using Eqs1412 and 1415 Note the dAgE AT at constant P for each component can be expressed using the relations in Eqs 1218 and 1219 Eq 1412 GA0E A vACE A AgE A 0 CE A vADE A AgE A 0 DE A vAAE A AgE A 0 AE A vABE A AgE A 0 BE Eq 1415 ln K GA0E A ARE AT Eq 1419 GA0E A Η T SA0E Ad lnK dTE A A d dTE A EAGA0 A EAR ATE A A 1 RTE A EAdGA0 A EdTE A EAGA0 A EAR ATA2 AE A EA 1 AR ATA2 AE A GA0E A T EAdGA0 A EdTE A EA 1 AR ATA2 AE A GA0E A T SA0E A used Eq1219 EAdAg A EdTE A AsE EA 1 AR ATA2 AE A ΗA0E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14113 Find the equilibrium constant for reaction in Eq 1433 at 600 K see Problem 1488 Reaction CO 3 HA2E A CHA4E A HA2E AO Equilibrium constant Eq 1415 ln K GA0E A ARE AT For CHA4E A at 600 K formula in Table A6 ACE AP0E A 5222 A5 ACE AP0E A 3616 So use the average for the interval 298 600 K ACE AP0E A 4419 kJkmolK At 600 K AhE A 0 CH4E A 74 873 4419600 29815 61 534 kJkmol AsE A 0 CH4E A 186251 4419 ln 600 29815 217154 kJkmol K HA 0 600 KE A 161 534 1241 826 10 499 110 527 8 942 30 8 799 217 673 kJ SA 0 600 KE A 1217154 1213051 218321 3151078 24135 kJK GA 0 600 KE A 217 673 60024135 72 863 kJ ln K A 72 863 83145 600E A 146 K 22 10A6E Comment If ACE AP0E A at 298 K for CHA4E A was used then K 183 10A6E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14114 Combustion of stoichiometric benzene CA6E AHA6E A and air at 80 kPa with a slight heat loss gives a flame temperature of 2400 K Consider the dissociation of COA2E A to CO and OA2E A as the only equilibrium process possible Find the fraction of the COA2E A that is dissociated Combustion CA5E AHA12E A 75 OA2E A 75 376 NA2E A 6 COA2E A 3 HA2E AO 282 NA2E 2 COA2E A 2 CO 1 OA2E also HA2E AO NA2E Initial 6 0 0 3 282 Change 2z 2z z 0 0 Equil 62z 2z z 3 282 At 2400 K ln K 7715 K 4461 10A4E A and we have ntot 372 z Substitute into equilibrium equation K A 4z2 z E6 2z2372 zE A A 80 100E AA 1E A A z3 E3 z2372 zE A 00005576 Solve with limit 0 z 3 gives z 0507 so then nACO2 E A 6 2z 4986 Fraction dissociated 6 49866 0169 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14115 One kilomole of liquid oxygen OA2E A at 93 K and x kmol of gaseous hydrogen HA2E A at 25C are fed to a combustion chamber x is greater than 2 such that there is excess hydrogen for the combustion process There is a heat loss from the chamber of 1000 kJ per kmol of reactants Products exit the chamber at chemical equilibrium at 3800 K 400 kPa and are assumed to include only HA2E AO HA2E A and O a Determine the equilibrium composition of the products and also x the amount of H2 entering the combustion chamber b Should another substances have been included in part a as being present in the products Justify your answer x HA2E A 1 OA2E A 2 HA2E AO x 2 HA2E A 1 1 HA2E AO 1 HA2E A 1 O shift a a a and a 0 Equil 2a x2a a a 2 and ntot x a 2 2 HA2E AO 2 HA2E A 1 OA2E A ln K2 1906 3 1 OA2E A 2 O ln K3 0017 ln K1 05 ln K2 ln K3 09615 K1 03823 Equil A K1 PPo1E A A x2aa E2axaE A A03823 4E A 095575 Energy Eq Q HR HP Q 1 x1000 kJ Table A9 AhE A IGE A 5980 kJkmol or 092232932982 6054 kJkmol Fig D2 Tr 931546 0601 AhE Af 516ARE A1546 6633 HR x0 0 10 AhE A IGE A AhE Af 15980 6633 12613 kJ HP 2a241 826 171 981 x2a0 119077 a249170 73424 119077 x 511516 a 377844 Q HR 1000 1000 x 12613 Rearrange eq to x 42599 a 303331 Substitute it into the equilibrium eq A 103331 52599 a a E2a30333132599 aE A 0095575 Solve a 0198 LHS 009547 x 21898 yH2O A2a xaE A 0755 yH2 Ax2a xaE A 0162 yO A a xaE A 0083 Other substances and reactions 2 H2O H2 2 OH ln K 0984 H2 2 H ln K 0201 O2 2 O ln K 0017 All are significant as Ks are of order 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14116 Dry air is heated from 25C to 4000 K in a 100kPa constantpressure process List the possible reactions that may take place and determine the equilibrium composition Find the required heat transfer Air assumed to be 21 oxygen and 79 nitrogen by volume From the elementary reactions we have at 4000 K A11 1 OA2E A 2 O K1 2221 yA 2 OE AyO2 2 NA2E A 2 N K2 3141 106 yA 2 NE AyN2 3 NA2E A OA2E A 2 NO K3 008955 yA 2 NOE AyN2 yO2 Call the shifts abc respectively so we get nO2 021 a c nO 2a nN2 079 b c nN 2b nNO 2c ntot 1 a b From which the mole fractions are formed and substituted into the three equilibrium equations The result is K1 2221 yA 2 OE AyO2 4a2 1ab021ac K2 3141 106 yA 2 NE AyN2 4b2 1ab079bc K3 008955 yA 2 NOE AyN2 yO2 4c2 079bc021ac which give 3 eqs for the unknowns abc Trial and error assume b c 0 solve for a from K1 then for c from K3 and finally given the ac solve for b from K2 The order chosen according to expected magnitude K1 K3 K2 a 015 b 0000832 c 00244 nO2 00356 nO 03 nN2 0765 nN 000167 nNO 0049 Q Hex Hin nO2AhE AO2 nN2AhE AN2 nOAhE AfO AhE AO nNAhE AfN AhE AN nNOAhE AfNO AhE ANO 0 00356 138705 0765 130027 03249170 77675 000167472680 77532 004990291 132671 214 306 kJkmol air If no reaction Q nO2AhE AO2 nN2AhE AN2 131 849 kJkmol air Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful 14117 Saturated liquid butane note use generalized charts enters an insulated constant pressure combustion chamber at 25C and x times theoretical oxygen gas enters at the same pressure and temperature The combustion products exit at 3400 K Assuming that the products are a chemical equilibrium gas mixture that includes CO what is x Butane T1 25oC sat liq x1 0 Tc 4252 K Pc 38 MPa Tr1 07 Figs D1and D2 Pr1 010 P1 Pr1Pc 380 kPa Fig D2 A h E1 h 1 E AfE A 485 RTAcE Oxygen T2 25oC X theoretical air Products T3 3400 K C4H10 65X OA2E A 4 COA2E A 5 HA2E AO 65X1 OA2E 2COA2E A 2CO OA2E Initial 4 0 65X1 Change 2a 2a a Equil 42a 2a 65X1 a ntot 25 a 65X nACO2E A 4 2a nACOE A 2a nAO2E A 65X1 a nAH2OE A 5 yACOE A A 2a 25 a 65XE A yACO2E A A 4 2a 25 a 65XE A yAO2E A A65x 1 a E25 a 65XE The equilibrium constant is K y2 COyO2 y 2 CO2 A P1 Po 212 AEE A a 2a 2E A A 65X 65 a 65X 25 a E A P1 Po T3 3400 K Table A11 lnK 0346 K 14134 14134 A a 2a 2E A A 65X 65 a 65X 25 a E A 376 Equation 1 Need a second equation Energy eq Qcv HR HP Wcv Qcv 0 Wcv 0 HR Aho EfE A AhE AAC4H10E A 126 200 17 146 143 346 kJ Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States Copyright Act without the permission of the copyright owner is unlawful Products 3400 K HP nAho EfE A AhE AACO2E A nAho EfE A AhE AACOE A nAho EfE A AhE AAO2E A nAho EfE A AhE AAH2OE 4 2a393 522 177 836 2a110 527 108 440 65X 1 a0 114101 5241 826 149 073 463 765 kJkmol HP HR 1924820 541299a 7416565 X Equation 2 Two equations and two unknowns solve for X and a a 087 X 196 ENGLISH UNIT PROBLEMS UPDATED August 2013 SOLUTION MANUAL CHAPTER 14 English Units Fundamentals of Thermodynamics Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CHAPTER 14 CONTENT CHAPTER 14 SUBSECTION PROB NO Equilibrium 118 Chemical equilibrium Equilibrium Constant 119136 Simultaneous Reactions 137141 Review problems 142144 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Equilibrium Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14118E Carbon dioxide at 2200 lbfin2 is injected into the top of a 3mi deep well in connection with an enhanced oil recovery process The fluid column standing in the well is at a uniform temperature of 100 F What is the pressure at the bottom of the well assuming ideal gas behavior Z 1 Z 2 CO 2 cb Z1 Z2 3 miles 15 840 ft P1 2200 lbfin2 T 100 F const Equilibrium and ideal gas beahvior wREV 0 g PE RT ln P2P1 gZ2 Z1 0 ln P2 P1 gZ2 Z1 RT 322 fts2 15 840 ft 32174 lbmfts2lbf 351 ftlbflbmR 5597 R 08063 P2 2200 lbfin2 exp08063 4927 lbfin2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Chemical equilibrium Equilibrium Constant Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14119E Calculate the equilibrium constant for the reaction O2 2O at temperatures of 537 R and 10 000 R Find the change in Gibbs function at the two Ts from Table F6 537 R H0 2h0 O 1h0 O2 2 107 124 0 214 248 Btulbmol S0 2s0 O 1s0 O2 2 38442 48973 27911 Btulbmol R G0 H0 TS0 214 248 537 27911 199 260 Btulbmol ln K G0 R T 199 260 198589537 18585 10 000 R H0 2h0 O 1h0 O2 2 107 124 47 897 87 997 222 045 Btulbmol S0 2s0 O 1s0 O2 2 53210 74034 32386 Btulbmol R G0 H0 TS0 222 045 10 000 32386 101 815 Btulbmol ln K G0 R T 101 815 198589 10 000 5127 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14120E Consider the dissociation of oxygen O2 2O starting with 1 lbmol oxygen at 77 F and heating it at constant pressure 1 atm At which temperature will we reach a concentration of monatomic oxygen of 10 Look at initially 1 lbmol Oxygen and shift reaction with x O2 2 O Species O2 O Initial 1 0 Change x 2x Equil 1x 2x ntot 1 x 2x 1 x yO 2x 1 x 01 x 012 01 00526 yO2 09 K yO 2 y02 P Po 21 012 09 101325 0011258 ln K 4487 Recall Po 100 kPa for Table A11 PPo 101325 100 101325 Now look in Table A11 T 2982 K 5368 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14121E Redo problem 14120E for a total pressure of 04 atm Look at initially 1 lbmol Oxygen and shift reaction with x O2 2 O Species O2 O Initial 1 0 Change x 2x Equil 1x 2x ntot 1 x 2x 1 x yO 2x 1 x 01 x 012 01 00526 yO2 09 K yO 2 y02 P Po 21 012 09 101325 04 0004503 ln K 5403 Recall Po 100 kPa for Table A11 PPo 101325 100 101325 Now look in Table A11 T 28576 K 5143 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14122E Redo problem 14120E but start with 1 lbmol oxygen and 1 lbmol helium at 77 F 1 atm Look at initially 1 mol Oxygen and shift reaction with x O2 2 O Species O2 O He Initial 1 0 1 Change x 2x Equil 1x 2x 1 ntot 1 x 2x 1 2 x yO 2x 2 x 01 x 022 01 010526 yO2 1 x2 x 0425 K yO 2 y02 P Po 21 012 0425 101325 002384 ln K 3736 Recall Po 100 kPa for Table A11 PPo 101325 100 101325 Now look in Table A11 T 3096 K 5573 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14123E Pure oxygen is heated from 77 F to 5300 F in a steady flow process at a constant pressure of 30 lbfin2 Find the exit composition and the heat transfer The only reaction will be the dissociation of the oxygen O2 2O K5300 F K3200 K 0046467 Look at initially 1 mol Oxygen and shift the above reaction with x nO2 1 x nO 2x ntot 1 x yi nintot K yO 2 y02 P Po21 4x2 1 x2 1 x 1 x 2 1 x2 8x2 x2 K8 1 K8 x 007599 y02 0859 y0 0141 q n02exh 02ex n0exh Oex h 02in 1 xy02h 02 y0h O 0 h 02 45 581 h O 107124 26125 133 249 q 10760859 45581 0141 133249 62 345 Btulbmol O 2 q q32 1948 Btulbm 1424 if no dissociation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14124E Find the equilibrium constant for CO2 CO ½ O2 at 3960 R using Table A11 The elementary reaction in A11 is 2CO2 2CO O 2 so the wanted reaction is 12 times that and T 3960 R 2200 K so K K12 A11 exp10232 exp5116 0006 or ln K 05 ln KA11 05 10232 5116 K exp511 0006 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14125E Assume we have air at 4 atm 3600 R as 21 O2 and 79 N2 and we can neglect dissociations of O2 and N2 What is the equilibrium mole fraction of NO Find the enthalpy difference in the gases due to the formation of NO Air initially 079 N2 021 O2 Final mix 079 x N2 021 x O2 2x NO N2 O2 2 NO nN2 079 x x x 2x nO2 021 x nNO 0 2x n 10 From Table A11 T 3600 R 2000 K ln K 7825 K y 2 NO yN2yO2 P P00 4x2 079 x 021 x exp7825 00003996 So x2 0000099904 079 x 021 x x 000402 yNO 2x 000804 ΔH 2x hNO xhN2 xhO2 all at 3600 R 000402 2 38 818 24 875 24 135 25441 Btulbmol 3128 Btulbmol air Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14126E Air assumed to be 79 nitrogen and 21 oxygen is heated in a steady flow process at a constant pressure of 147 lbfin2 and some NO is formed At what temperature will the mole fraction of NO be 0001 079N2 021O2 heated at 147 lbfin2 forms NO At exit yNO 0001 N2 O2 2 NO nN2 079 x x x 2x nO2 021 x nNO 0 2x n 10 yNO 0001 2x 10 x 00005 nN2 07895 nO2 02095 K y 2 NO yN2yO2 P P00 106 0789502095 6046106 or ln K 12016 From Table A11 T 1444 K 2600 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14127E The combustion products from burning pentane C5H12 with pure oxygen in a stoichiometric ratio exists at 4400 R Consider the dissociation of only CO2 and find the equilibrium mole fraction of CO C5H12 8 O2 5 CO2 6 H2O At 4400 R 2 CO2 2 CO 1 O 2 ln K 7226 Initial 5 0 0 K 7272x104 Change 2z 2z z Equil 52z 2z z Assuming P Po 01 MPa K yCO 2 yO2 yCO2 2 P P0 2z 5 2z2 z 5 z 1 7272 104 Trial error on z z 02673 nCO2 44654 nCO 05346 nO2 02673 yCO 01015 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14128E Pure oxygen is heated from 77 F 147 lbfin2 to 5300 F in a constant volume container Find the final pressure composition and the heat transfer As oxygen is heated it dissociates O2 2O ln Keq 3069 C V Heater U2 U1 1Q2 H2 H1 P2v P1v Per mole O2 1q 2 h 2 h 1 R T1 n2n1T2 Shift x in reaction final composition 1 xO2 2xO n1 1 n2 1 x 2x 1 x yO2 1 x1 x yO 2x1 x Ideal gas and V2 V1 P2 P1n2T2n1T1 P2Po 1 xT2T 1 Substitute the molefractions and the pressure into the equilibrium equation Keq e3069 yO 2 y02 P2 Po 2x 1 x2 1 x 1 x 1 x 1 T2 T1 4x2 1 x T1 T2e3069 000433 x 00324 nO22 09676 nO2 00648 n2 10324 P2 Po 1 xT2T1 147 1 00324 5760 460 77 1628 psia 1q 2 0967645 581 00648107 124 26 125 0 19858953667 10324 5760 41 996 Btulbmol O2 yO2 09676 10324 0937 yO 0064810324 00628 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14129E Assume the equilibrium mole fractions of oxygen and nitrogen are close to those in air find the equilibrium mole fraction for NO at 5400 R 75 psia disregarding dissocoiations Assume the simple reaction to make NO as N2 O2 2 NO Ar nN2 078 x 078 021 0 001 nO2 021 x x x 2x nNO 2x nar 001 078x 021x 2x 001 ntot 10 From A11 at 5400 R 3000 K ln K 4205 K 0014921 K 4x2 078 x021 x P P0 0 x2 078 x021 x 0014921 4 000373 and 0 x 021 Solve for x x 00230 yNO 2x 10 0046 Comment Since the reaction is pressure insensitive P does not matter Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14130E Acetylene gas C2H2 is burned with stoichiometric air in a torch The reactants are supplied at the reference conditions P0 T0 The products come out from the flame at 5000 R after a small heat loss by radiation Consider the dissociation of CO2 into CO and O2 and no others Find the equilibrium composition of the products Are there any other reactions that should be considered Combustion C2H2 25 O2 376N2 2 CO2 1 H2O 94 N2 At 5000 R 2778 K 2 CO2 2 CO 1 O2 H2O N 2 ln K 39804 Initial 2 0 0 1 94 K 001868 Change 2z 2z z 0 0 Equil 22z 2z z 1 94 Assuming P Po 01 MPa and ntot 2 z 1 94 124 z K yCO 2 yO2 yCO2 2 P P0 2z 2 2z 2 z 124 z 1 001868 Solve z3 001868 1 z2124 z z 0428 yCO 2z 124 z 00667 yCO2 2 2z 124 z 0089 yO2 z 124 z 00334 yH2O 1 124 z 00779 yN2 94 124 z 07328 Looking in A11 there will be no dissociation of nitrogen and only a small dissociation of water and oxygen which could be included Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14131E Use the information in problem 14142E to estimate the enthalpy of reaction Ho at 1260 R using the vant Hoff equation see problem 1456 with finite differences for the derivatives d ln K Ho RT2 dT Po solve for H H EA AdlnK dTE A AR TA 2 EA AlnK TE R T 2 198589 1260A 2E A A03362 4607 1440 1080E A 37 403 Btulb mol Remark compare this to F6 values F4 F11 H HACE A 2HAH2 E A HACH4 E A 0146 12 1260537 2 5044 0538 16043 1260537 32190 37 304 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14132E A gas mixture of 1 lbmol carbon monoxide 1 lbmol nitrogen and 1 lbmol oxygen at 77 F 20 lbfin2 is heated in a constant pressure steady flow process The exit mixture can be assumed to be in chemical equilibrium with CO2 CO O2 and N2 present The mole fraction of CO2 at this point is 0176 Calculate the heat transfer for the process initial mix 1 CO 1 OA2E A 1 NA2E Constant pressure reactor Q Equil mix COA2E A CO OA2E A NA2E yACO2E A 0176 P 20 lbfinA2E reaction 2 COA2E 2 CO OA2E also NA2E initial 0 1 1 1 change 2x 2x x 0 equil 2x 12x 1x 1 yACO2E A 0176 A 2x 3xE A x 0242 65 A nCO2 04853 EnCO 05147 nO2 0757 35 nN2 1 E A A yCO2 0176 EyCO 0186 67 yO2 0274 67 E K A y 2 COyO2 Ey 2 CO2 E AA P P0 E AA 1E A A0186 6720274 67 E01762 E AA 20 14504E A 0426 07 Since Table A11 corresponds to a pressure PA0E A of 100 kPa which is 14504 lbfinA2E A Then from A11 TAPRODE A 3200 K 5760 R HARE A 47 518 Btu HAP E A 04853169 184 71 075 0514747 518 43 406 0757 350 45 581 10 43 050 27 842 Btu QACVE A HAP E A HARE A 27 842 47 518 75 360 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14133E Acetylene gas at 77 F is burned with 140 theoretical air which enters the burner at 77 F 147 lbfin2 80 relative humidity The combustion products form a mixture of CO2 H2O N2 O2 and NO in chemical equilibrium at 3500 F 147 lbfin2 This mixture is then cooled to 1340 F very rapidly so that the composition does not change Determine the mole fraction of NO in the products and the heat transfer for the overall process CA2E AHA2E A 35 OA2E A 1316 NA2E A water 2 COA2E A 1 HA2E AO 1 OA2E A 1316 NA2E A water Water PAVE A 08046 0368 lbfinA2E nAVE A nAAE AA PV EPA E A 351316A 0368 14332E A 0428 So total HA2E AO in products is 1 nAVE A 1428 a reaction NA2E A OA2E A 2 NO change x x 2x at 3500 F 3960 R 2200 K from A11 K 0001 074 Equilibrium products nACO2E A 2 nAH2OE A 1428 nAO2 E A 1 x nAN2 E A 1316 x nANOE A 0 2x nATOTE A 17588 K A 2x2 E1 x1316 xE A 0001 074 By trial and error x 00576 yANOE A A2 00576 17588E A 0006 55 b Final products same composition at 1340 F 1800 R HARE A 197 476 0428103 966 52 979 Btu HAP E A 2169 18414 358 1428103 96611 178 0942409761 13102409227 0115238 8189557 306 486 Btu QACVE A HAP E A HARE A 359 465 Btu Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14134E A special coal burner uses a stoichiometric mixture of coal and an oxygen argon mixture 11 mole ratio with the reactants supplied at Po To Consider the dissociation of CO2 into CO and O2 and no others Find the equilibrium composition of the products for T 8600 R Is the final temperature including dissociations higher or lower than 8600 R Combustion C OA2E A Ar COA2E A Ar Reaction 2 COA2E A 2 CO OA2E A COA2E A CO OA2E A Ar Start 1 0 0 1 Change 2x 2x x 0 Final 12x 2x x 1 tot 2 x At 8600 R K exp57694 320345 K yACOE A2E A yAO2E A yACO2E A2E A 1 A 2x 2 xE AA 2E A x 2 xE A A1 2x 2 xE AA 2E A Reduce and solve xA3E A exp576944 2 x 1 2xA2E A xA3E A 8009 2 x 1 2xA2E A x 048793 nACOE A 09759 nAO2E A 04879 nACO2E A 00241 nAArE A 1 Now we can do the energy equation HAP 8600 RE A 00241 113 961 169 184 1 49536 8600 4597 09759 68 977 47 518 04879 73 648 95 868 btulbmol Since HAP 8600 RE A 0 too high then T 8600 R We should have had HAPE A 0 if we match the energy equation to find the T we have to redo the equilibrium equation as K is different This is a larger trial and error problem to find the actual T and we even neglected other possible reactions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14135E An important step in the manufacture of chemical fertilizer is the production of ammonia according to the reaction N2 3H2 2NH3 Show that the equilibrium constant is K 6826 at 300 F 1NA2E A 3HA2E A 2NHA3E A at 300 F 760 R from Table F4 for NH3 M 17031 CAP0E A 0509 BtulbmR Assume ideal gas and constant specific heat for ammonia we could be more accurate if we used Table F8 realizing properties depend also on P We must use the reference h s values from Table F11 AhE A o NH3 300 FE A 19 656 050917031 30077 17723 Btulbmol AsE A o NH3 300 FE A 45969 050917031 ln A760 537E A 48980 Use Table F6 for nitrogen and hydrogen gases HA o 300 FE A 217723 10 1557 30 1552 41 659 Btulbmol SA 0 300 FE A 24898 148164 33360 510 BtuRlbmol GA 0 300 FE A 41 659 760510 2899 Btulbmol ln K A 2899 198589760E A 19208 K 6826 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14136E Consider the previous reaction in equilibrium at 300 F 750 psia For an initial composition of 25 nitrogen 75 hydrogen on a mole basis calculate the equilibrium composition 1NA2E A 3HA2E A 2NHA3E A at 300 F Beginning 1 3 0 Change x 3x 2x Final 1x 33x 2x Total n 4 2x nANH3E A 2x nAN2E A 1x nAH2E A 33x The reaction constant is K 6826 so the reaction constant equation is K A y 2 NH3 EyN2y 3 H2 E AA P P0 E AA 2E A A2x2222x2 E331x4 E AA P P0 E AA 2E or A x 1xE AA 2E AA2x 1xE AA 2E A A27 16E A 6826 A750 147E AA 2E A 29985 or A x 1xE AA2x 1xE A 17316 n y Trial Error NHA3E 1848 08591 x 09242 NA2E 00758 00352 HA2E 02273 01057 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Simultaneous Reactions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14137E Ethane is burned with 150 theoretical air in a gas turbine combustor The products exiting consist of a mixture of CO2 H2O O2 N2 and NO in chemical equilibrium at 2800 F 150 lbfin2 Determine the mole fraction of NO in the products Is it reasonable to ignore CO in the products Combustion CA2E AHA6E A 525 OA2E A 1974 NA2E A 2 COA2E A 3 HA2E AO 175 OA2E A 1974 NA2E a Products at 2800 F 150 lbfinA2E A Equilibrium mixture COA2E A HA2E AO OA2E A NA2E A NO NA2E OA2E 2 NO initial 1974 175 0 change x x 2x equil 1974x 175x 2x Equil comp nACO2E A 2 nAH2OE A 3 nAO2 E A 175x nAN2 E A 1974x nANO E A 2x K 128310A4E A A y 2 NO EyN2yO2 E AA P P0 E AA 0E A A 4x2 E1974x175xE Solving x 0032 95 yANOE A A20032 95 2649E A 0002 49 b 2 COA2E 2 CO OA2E initial 2 0 0 change 2a 2a 2x equil 22a 2a 2x K 525910A8E A A y 2 COyO2 Ey 2 CO2 E AA P P0 E AA 1E A A 2a 22aE AA 2E AA175xa 2649aE AA 150 14504E A Since Table A11 corresponds to a pressure PAoE A of 100 kPa which is 14504 lbfinA2E A This equation should be solved simultaneously with the equation solved in part a modified to include the unknown a Since x was found to be small and also a will be very small the two are practically independent Therefore use the value x 0032 95 in the equation above and solve for a A a 1aE AA 2E AA1750032 95a 2649aE A A14504 150E A525910A8E Solving a 0000 28 or yACOE A 2110A5E A negligible for most applications Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14138E One pound mole of water vapor at 147 lbfin2 720 R is heated to 5400 R in a constant pressure steady flow process Determine the final composition assuming that HA2E AO HA2E A H OA2E A and OH are present at equilibrium Reactions 1 2 HA2E AO 2 HA2E A OA2E A 2 2 HA2E AO HA2E A 2 OH change 2a 2a a change 2b b 2b 3 HA2E A 2 H change c 2c At equilibrium 5400 R 147 lbfin2 nAH2OE A 12a2b nAOH E A 2b nAH2 E A 2abc nAH E A 2c nAO2 E A a nATOTE A 1abc A K1 EPP0E A A2062103 E103E A A 2abc 12a2bE AA 2E AA a 1abcE A A K2 EPP0E A A2893103 E103E A A 2abc 1abcE AA 2b 12a2bE AA 2E A K3 EPP0E A A2496102 E103E A A 2a2 E2abc1abcE These three equations must be solved simultaneously for a b c a 00622 b 00570 c 00327 and nAH2OE A 07616 yAH2OE A 06611 nAH2 E A 01487 yAH2 E A 01291 nAO2 E A 00622 yAO2 E A 00540 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14139E Methane is burned with theoretical oxygen in a steady flow process and the products exit the combustion chamber at 5300 F 100 lbfin2 Calculate the equilibrium composition at this state assuming that only COA2E A CO HA2E AO HA2E A OA2E A and OH are present Combustion CHA4E A 2 OA2E A COA2E A 2 HA2E AO Dissociation reactions At equilibrium 1 2 HA2E AO 2 HA2E A OA2E nAH2OE A 22a2b change 2a 2a a nAH2 E A 2ab 2 2 HA2E AO HA2E A 2 OH nAO2 E A ac change 2b b 2b nAOH E A 2b 3 2 COA2E A 2 CO OA2E nACO2E A 12c change 2c 2c c nACO E A 2c nATOTE A 3abc Products at 5300F 100 lbfinA2E KA1E A 0007 328 A 2ab 22a2bE AA 2E AA ac 3abcE AA 100 14504E A KA2E A 0012 265 A 2b 22a2bE AA 2E AA 2ab 3abcE AA 100 14504E A KA3E A 0426 135 A 2c 12cE AA 2E AA ac 3abcE AA 100 14504E A These 3 equations must be solved simultaneously for a b c If solving by hand divide the first equation by the second and solve for c fnab This reduces the solution to 2 equations in 2 unknowns Solving a 00245 b 01460 c 02365 Substance H2O H2 O2 OH CO2 CO n 1659 0195 0260 0292 0527 0473 y 04871 00573 00763 00857 01547 01389 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14140E One pound mole of air assumed to be 78 nitrogen 21 oxygen and 1 argon at room temperature is heated to 7200 R 30 lbfin2 Find the equilibrium composition at this state assuming that only N2 O2 NO O and Ar are present 1 lbmol air 078 NA2E A 021 OA2E A 001 Ar heated to 7200 R 30 lbfinA2E A 1 NA2E A OA2E A 2 NO 2 OA2E A 2 O change a a 2a change b 2b Equil nAN2E A 078a nAArE A 001 nANOE A 2a nAO2E A 021ab nAO E A 2b nA E A 1b KA1E A 00895 A 4a2 E078a021abE A A 30 14504E AA 0E KA2E A 2221 A 4b2 E1b021abE A A 30 14504E A Divide 1st eqn by 2nd and solve for a as functionb using X A K1 EK2 E A A P P0 E A 0083 35 Get a A Xb2 E21bE A1 A 1 40781b EXb2 EA 1 Also A b2 E1b021abE A A K2 E4PP0E A 0268 44 2 Assume b 01269 From 1 get a 00299 Then check a b in 2 OK Therefore Subst N2 O2 Ar O NO n 07501 00532 001 02538 00598 y 06656 00472 00089 02252 00531 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14141E Acetylene gas and x times theoretical air x 1 at room temperature and 75 lbfin2 are burned at constant pressure in an adiabatic steady flow process The flame temperature is 4600 R and the combustion products are assumed to consist of NA2E A OA2E A COA2E A HA2E AO CO and NO Determine the value of x Combustion CA2E AHA2E A 25x OA2E A 94x NA2E A 2 COA2E A HA2E AO 25x1OA2E A 94x NA2E Eq products 4600 R 75 lbfin2 NA2E A OA2E A COA2E A HA2E AO CO NO 2 Reactions 1 2 COA2E A 2 CO OA2E A 2 NA2E A OA2E A 2 NO change 2a 2a a change b b 2b Equil Comp nAN2 E A 94xb nACO2E A 22a nAO2 E A 25x25ab nAH2OE A 1 nACO E A 2a nANO E A 2b nATOTE A 119x05a At 4600 R from A11 KA1E A 235910A3E A KA2E A 424910A3E A K1 EPP0E A A2359103 E5103E A 462210A4E A A a 1aE AA 2E AA25x25ab 119x05aE A KA2E A 424910A3E A A 2b2 E94b25x25abE Also from the energy Eq HAP E A HARE A 0 where HARE A 197 477 0 0 97 497 Btu HAP E A 94xb032 817 25x25ab034 605 22a169 18453 885 1103 96643 899 2a47 51833 122 2b38 81831 161 Substituting 394 992 x 236 411 a 72 536 b 377 178 97477 which results in a set of 3 equations in the 3 unknowns xab Trial and error solution from the last eq and the ones for K1 and K2 The result is x 112 a 01182 b 005963 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14142E The equilibrium reaction with methane as CH4 C 2H2 has ln K 03362 at 1440 R and ln K 4607 at 1080 R By noting the relation of K to temperature show how you would interpolate lnK in 1T to find K at 1260 R and compare that to a linear interpolation ln K 03362 at 1440 R ln K 4607 at 1080 R lnKA1260E A lnKA1440E A EA A 1 1260 A A 1 E1440 A A 1 1080 A A 1 1440 AE A 4607 03362 03362 EA A1440 1260 A 1 EA1440 1080 A 1E A 42708 21665 Linear interpolation lnKA1260E A lnKA1080E A A1260 1080 1440 1080E A lnKA1440E A lnKA1080E A 4607 A1 2E A 03362 4607 24716 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14143E In a test of a gasturbine combustor saturatedliquid methane at 210 R is to be burned with excess air to hold the adiabatic flame temperature to 2880 R It is assumed that the products consist of a mixture of CO2 H2O N2 O2 and NO in chemical equilibrium Determine the percent excess air used in the combustion and the percentage of NO in the products CHA4E A 2x OA2E A 752x NA2E 1 COA2E A 2 HA2E AO 2x2 OA2E A 752x NA2E Then NA2E A OA2E A 2 NO Also COA2E A HA2E AO init 752x 2x2 0 1 2 ch a a 2a 0 0 equil 752xa 2x2a 2a 1 2 nATOTE A 1 952x 2880 R ln K 1055 K 262810A5E 262810A5E A K A y 2 NO EyN2yO2 E AA P P0 E AA0E A A y 2 NO EyN2yO2 E A A 4a2 E752xa2x2aE HARE A 132 190 18544300 0 0 assume 77 F 38 344 Btu HAP E A 1169 184 29 049 2103 966 22 746 752xa18 015 2x2a19 031 2a38 818 18 624 340 639 173 535 x 77 838 a Assume a 0 then from HAP E A HARE A 0 x 1742 Subst A a2 E131a1484aE A A2628105 E4E A get a 001125 Use this a in energy equation x A302 295 77 838001125 173 535E A 1737 A a2 E13062a1474aE A A2628105 E4E A a 00112 x 1737 excess air 737 NO A200112100 19521737E A 0128 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 14144E Dry air is heated from 77 F to 7200 R in a 147 lbfin2 constantpressure process List the possible reactions that may take place and determine the equilibrium composition Find the required heat transfer Air assumed to be 21 oxygen and 79 nitrogen by volume From the elementary reactions at 4000 K 7200 R A11 1 OA2E A 2 O K1 2221 yA 2 OE AyO2 2 NA2E A 2 N K2 3141 106 yA 2 NE AyN2 3 NA2E A OA2E A 2 NO K3 008955 yA 2 NOE AyN2 yO2 Call the shifts abc respectively so we get nO2 021ac nO 2a nN2 079bc nN 2b nNO 2c ntot 1ab From which the molefractions are formed and substituted into the three equilibrium equations The result is corrected for 1 atm 147 lbfin2 101325 kPa versus the tables 100 kPa K1 21511 yA 2 OE AyO2 4a21ab021ac K2 3042106 yA 2 NE AyN2 4b21ab079bc K3 008955 yA 2 NOE AyN2 yO2 4c2079bc021ac which give 3 eqs for the unknowns abc Trial and error assume b c 0 solve for a from K1 then for c from K3 and finally given the ac solve for b from K2 The order chosen according to expected magnitude K1 K3 K2 a 015 b 0000832 c 00244 nO2 00356 nO 03 nN2 0765 nN 000167 nNO 0049 Q Hex Hin nO2AhE AO2 nN2AhE AN2 nOAhE AfO AhE AO nNAhE AfN AhE AN nNOAhE AfNO AhE ANO 0 00356 59 632 0765 55 902 03107 124 33 394 000167203 216 33 333 004938 818 57 038 92 135 Btulbmol air If no reaction Q nO2AhE AO2 nN2AhE AN2 56 685 Btulbmol air Updated June 2013 8e SOLUTION MANUAL CHAPTER 15 Fundamentals of Thermodynamics Borgnakke Sonntag Borgnakke Sonntag Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Fundamentals of Thermodynamics 8th Edition Borgnakke and Sonntag CONTENT CHAPTER 15 SUBSECTION PROB NO inText concept questions aj Study guide problems 113 Stagnation Properties 1423 Momentum Equation and Forces 2431 Adiabatic 1D Flow and Velocity of Sound 3239 Reversible Flow Through a Nozzle 4060 Normal Shocks 6170 Nozzles Diffusers and Orifices 7181 Review problems 8284 Problems solved with the Pr vr functions 44 73 English unit problems 85103 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful InText Concept Questions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 15a Is stagnation temperature always higher than free stream temperature Why Yes Since kinetic energy can only be positive we have h0 h1 V1 22 h1 If it is a gas with constant heat capacity we get T0 T1 V1 22Cp Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 15b By looking at Eq 1525 rank the speed of sound for a solid a liquid and a gas Speed of sound P ρ s c2 For a solid and liquid phase the density varies only slightly with temperature and constant s is also nearly constant T We thus expect the derivative to be very high that is we need very large changes in P to give small changes in density A gas is highly compressible so the formula reduces to Eq1528 which gives modest values for the speed of sound 15c Does speed of sound in an ideal gas depend on pressure What about a real gas No For an ideal gas the speed of sound is given by Eq1528 c kRT and is only a function of temperature T For a real gas we do not recover the simple expression above and there is a dependency on P particularly in the dense gas region above the critical point 15d Can a convergent adiabatic nozzle produce a supersonic flow No From Eq1533 and a nozzle so dP 0 it is required to have dA 0 to reach M 1 A convergent nozzle will have M 1 at the exit which is the smallest area For lower back pressures there may be a shock standing outside the exit plane Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 15e To maximize the mass flow rate of air through a given nozzle which properties should I try to change and in which direction higher or lower The mass flow rate is given by Eq1541 and if we have M 1 at the throat then Eq1542 gives the maximum mass flow rate possible Max flow for Higher upstream stagnation pressure Lower upstream stagnation temperature 15f How do the stagnation temperature and pressure change in an isentropic flow The stagnation temperature and stagnation pressure are constant Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 15g Which of the cases in Fig 1517 ah have entropy generation and which do not a There is no flow so sgen 0 b Subsonic flow reversible so sgen 0 c Limit for subsonic flow reversible so sgen 0 d The only supersonic reversible flow solution so sgen 0 e Supersonic reversible in nozzle sgen 0 irreversible outside f Supersonic reversible in nozzle sgen 0 compression outside g Shock stands at exit plane sgen 0 across shock h Shock is located inside nozzle sgen 0 across shock Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 15h How does the stagnation temperature and pressure change in an adiabatic nozzle flow with an efficiency of less than 100 The stagnation temperature stays constant energy eq The stagnation pressure drops s is generated less kinetic energy Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 15i Table A13 has a column for PoyPox why is there not one for ToyTox The stagnation pressure drops across the shock irreversible flow whereas the stagnation temperature is constant energy equation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 15j How high can a gas velocity Mach number be and still treat it as incompressible flow within 2 error The relative error in the P versus kinetic energy Eq1566 becomes e EA 0283 1 4 002 M V co V co 2 4 002 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful ConceptStudy Guide Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 151 Which temperature does a thermometer or thermocouple measure Would you ever need to make a correction to that Since the probe with the thermocouple in its tip is stationary relative to the moving fluid it will measure something close to the stagnation temperature If that is high relative to the free stream temperature there will be significant heat transfer convection and radiation from the probe and it will measure a little less For very high accuracy temperature measurements you must make some corrections for these effects Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 152 The jet engine thrust is found from the overall momentum equation Where is the actual force acting it is not a longrange force in the flow The compressor is generating the high pressure flow so the blades pushes hard on the flow and thus a force acts in the forward direction on the shaft holding the rotating blades The high pressure in the chamber with combustion also has a net force in the forward direction as the flow leaves in the backwards direction so less wall area there and higher velocity The pressure drop in the turbine means its blades pushes in the other direction but as the turbine exit pressure is higher than the ambient pressure the axial force is less than that of the compressor In front of the nozzle section there is a higher upstream pressure and a lower downstream pressure which is the pressure difference that generates the large exit velocity This higher pressure plus the pressure difference across the turbine equals the pressure in the combustion chamber which is the compressor exit pressure If the nozzle was not there this pressure would be lower and the pressure on all the surfaces including the compressor blades would provide less thrust High P Low P cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 153 Most compressors have a small diffuser at the exit to reduce the high gas velocity near the rotating blades and increase the pressure in the exit flow What does this do to the stagnation pressure For a reversible flow ideal case the stagnation pressure is constant However the reason it is done is to raise the pressure in a near reversible flow diffuser rather than let the flow reduce the peak velocities in a less reversible fashion which would lower the stagnation pressure 154 A diffuser is a divergent nozzle used to reduce a flow velocity Is there a limit for the Mach number for it to work like this Yes the flow must be subsonic If the flow was supersonic then increasing the flow area would increase the velocity Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 155 Sketch the variation in V T P ρ and M for a subsonic flow into a convergent nozzle with M 1 at the exit plane V M c M A kRTEA 2CpTo T Since we do not know the area versus length we plot it versus mach number M T P and ρ relative to the stagnation state is listed in Table A12 and given in eqs153436 A small spread sheet M step 01 did the calculations Only the first part 0 M 1 is the answer to this question The curves are plotted as the variables T To ρ ρo P Po V 2Cp To and for k 14 0 01 02 03 04 05 06 07 08 09 1 0 02 04 06 08 1 12 14 16 18 2 Mach number V P ρ T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 156 Sketch the variation in V T P ρ and M for a sonic M 1 flow into a divergent nozzle with M 2 at the exit plane V M c M kRT 2CpTo T Since we do not know the area versus length we plot it versus mach number M T P and ρ relative to the stagnation state is listed in Table A12 and given in eqs153436 Only the last part 1 M 2 is the answer to this question The curves are plotted as the variables T To ρ ρo P Po V 2Cp To and for k 14 0 01 02 03 04 05 06 07 08 09 1 0 02 04 06 08 1 12 14 16 18 2 Mach number V P ρ T Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 157 Can any low enough backup pressure generate an isentropic supersonic flow No Only one back pressure corresponds to a supersonic flow which is the exit pressure at state d in Figure 1513 However a pressure lower than that can give an isentropic flow in the nozzle case e with a drop in pressure outside the nozzle This is irreversible leading to an increase in s and therefore not isentropic 158 Is there any benefit to operate a nozzle choked Yes Since the mass flow rate is constant max value between points c and d in Fig 1512 a small variation in the back pressure will not have any influence The nozzle then provides a constant mass flow rate free of surges up or down which is very useful for flow calibrations or other measurements where a constant mass flow rate is essential Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 159 Can a shock be located upstream from the throat No The flow adjust so M 1 at the throat Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1510 The high velocity exit flow in Example 157 is at 183 K Can that flow be used to cool a room Being that cold it sounds like it could However when the flow enters a room it eventually would have to slow down and then it has the stagnation temperature If you let the flow run over a surface there will be a boundary layer with zero velocity at the surface and again there the temperature is close to the stagnation temperature Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1511 A convergentdivergent nozzle is presented for an application that requires a supersonic exit flow What features of the nozzle do you look at first You look at the cross section area change through the nozzle At the throat M 1 so in the divergent section the velocity increases and the ratio AA determines how the flow changes The exit area can then tell you what the exit mach number will be and if you can have a reversible flow or not Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1512 To increase the flow through a choked nozzle the flow can be heatedcooled or compressedexpanded four processes before or after the nozzle Explain which of these eight possibilities will help and which will not The mass flow rate through a choked nozzle is given by Eq1542 Since k and R are constant it varies with the upstream stagnation properties Po and To After nozzle Any downstream changes have no effects Before nozzle Upstream changes in Po and To has an influence a Heat This lowers mass flow rate To increases b Cool This raises mass flow rate To decreases c Compress Raises Po and To opposite effects Isentropic Po new Po rp and To new To rp k1 k Po new To new rp k1 2k Po To Po To So the mass flow rate increases d Expand Lowers Po and To opposite effects Assume isentropic then mass flow rate decreases Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1513 Suppose a convergentdivergent nozzle is operated as case h in Fig 1517 What kind of nozzle could have the same exit pressure but with a reversible flow A convergent nozzle having subsonic flow everywhere assuming the pressure ratio is higher than the critical Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Stagnation Properties Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1514 A stationary thermometer measures 80oC in an air flow that has a velocity of 200 ms What is the actual flow temperature We assume that the thermometer measures the stagnation temperature that is the probe bulb thermistor or thermocouple junction sits stationary h0 h1 V1 22 Τ1 Τ0 V1 22Cp Τ1 80oC 2000 1004 2002 JkgK ms2 60oC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1515 Steam leaves a nozzle with a pressure of 500 kPa a temperature of 350C and a velocity of 250 ms What is the isentropic stagnation pressure and temperature Stagnation enthalpy from energy equation and values from steam tables B13 h0 h1 V1 22 31677 kJkg 2502 2000 ms2 JkJ 31984 kJkg s0 s1 76329 kJkg K It can be linearly interpolated from the printed tables Computer software ho so To 365C Po 556 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1516 Steam at 1600 kPa 300oC flows so it has a stagnation total pressure of 1800 kPa Find the velocity and the stagnation temperature The stagnation state has the same entropy as the inlet state so 1 1600 kPa 300oC has h 303483 kJkg s 68844 kJkgK Stagnation 1800 kPa s0 68844 kJkgK has Table B13 T01 3165oC h01 3066155 kJkg V1 22 h01 h1 3066155 303483 31325 kJkg V1 2 ho1 h1 2 31325 1000 Jkg 2503 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1517 An object from space enters the earths upper atmosphere at 5 kPa 100 K with a relative velocity of 2500 ms or more Estimate the objects surface temperature ho1 h1 V1 22 250022000 3125 kJkg ho1 h1 2000 100 3125 3225 kJkg T 2767 K The value for h1 from ideal gas table A7 was estimated since the lowest T in the table is 200 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1518 The products of combustion of a jet engine leave the engine with a velocity relative to the plane of 500 ms a temperature of 525C and a pressure of 75 kPa Assuming that k 132 Cp 115 kJkg K for the products determine the stagnation pressure and temperature of the products relative to the airplane Energy Eq ho1 h1 V1 22 50022000 125 kJkg To1 T1 ho1 h1Cp 125115 1087 K To1 525 27315 1087 9068 K Isentropic process relates to the stagnation pressure Po1 P1To1T1kk1 759068798154125 127 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1519 Steam is flowing to a nozzle with a pressure of 400 kPa The stagnation pressure and temperature are measured to be 600 kPa and 350oC respectively What are the flow velocity and temperature Stagnation state Table B13 ho1 316566 kJkg so1 75463 kJkg K State 1 400 kPa s1 so1 75463 kJkg K T1 250 300 250 75463 73788 75661 73788 2947oC h1 296416 75463 73788 75661 73788 306675 296416 30559 kJkg Energy equation gives V1 22 ho1 h1 316566 30559 10976 kJkg V1 2 h o1 h 1 2 10976 1000 Jkg 4685 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1520 A meteorite melts and burn up at temperatures of 3000 K If it hits air at 5 kPa 50 K how high a velocity should it have to experience such a temperature Assume we have a stagnation T 3000 K h1 V1 22 hstagn Use table A7 hstagn 352536 kJkg h1 50 kJkg V1 22 352536 50 34754 kJkg remember convert to Jkg m2s2 V1 2 34754 1000 Jkg 2636 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1521 Air leaves a compressor in a pipe with a stagnation temperature and pressure of 150C 300 kPa and a velocity of 125 ms The pipe has a crosssectional area of 002 m2 Determine the static temperature and pressure and the mass flow rate ho1 h1 V1 22 12522000 78125 kJkg To1 T1 ho1 h1Cp 781251004 78 K T1 To1 T 150 78 1422 C 4154 K P1 Po1T1To1kk1 300 kPa 41544231535 281 kPa m ρAV AV v P1AV1 RT1 2812002125 02874154 59 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1522 I drive down the highway at 110 kmh on a day with 25C 1013 kPa I put my hand cross sectional area 001 m2 flat out the window What is the force on my hand and what temperature do I feel The air stagnates on the hand surface h1 V1 22 hstagn Use constant heat capacity Tstagn T1 E2 Cp E 25 EA05 110A2 A 10003600A2 A E1004E A 25465C V1 2 Assume a reversible adiabatic compression Pstagn P1 TstagnT1Akk1E A 1013 kPa 29861529815A35E 10185 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1523 A stagnation pressure of 110 kPa is measured for an airflow where the pressure is 100 kPa and 20C in the approach flow What is the incomming velocity Assume a reversible adiabatic compression To1 T1 Po1P1Ak1kE A 29315 K A110 100E AA02857E A 30124 K VA1 2 AEE2 ho1 h1 Cp To1 T1 1004 30124 29315 81246 kJkg V1 A 2 81246 1000EA 1275 ms 1 2 cb To the left a Pitot tube blue inner tube measures stagnation pressure and yellow outer tube with holes in it measures static pressure To the right is a stagnation point on a wall relative to the free stream flow at state 1 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1524 A 4cm inner diameter pipe has an inlet flow of 10 kgs water at 20AoE AC 200 kPa After a 90 degree bend as shown in Fig P1524 the exit flow is at 20AoE AC 190 kPa Neglect gravitational effects and find the anchoring forces FAxE A and FAyE A D 004 m Α Aπ 4E A DA2E A 0001257 mA2E A Vavg A m EρAE A A10 0001002 0001257E A 7971 ms Now we can do the x and y direction momentum equations for steady flow and the same magnitude of the velocity but different directions Xdir 0 AmE A Vavg 1 Fx AmE A 0 P1 Po A Ydir 0 AmE A 0 Fy AmE A Vavg 2 P2 Po A Fx AmE A Vavg 1 P1 Po A 10 797 100 0001257 1000 205 N Fy AmE A Vavg 2 P2 Po A 10 797 90 0001257 1000 193 N x y F Fx y Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1525 A jet engine receives a flow of 150 ms air at 75 kPa 5C across an area of 06 mA2E A with an exit flow at 450 ms 75 kPa 800 K Find the mass flow rate and thrust AmE A ρAV ideal gas ρ PRT AmE A PRTAV A 75 0287 27815E A 06 150 09395 06 150 84555 kgs Fnet AmE A Vex Vin 84555 450 150 25 367 N Inlet High P Low P exit cb Fnet The shaft must have axial load bearings to transmit thrust to aircraft Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1526 How large a force must be applied to a squirt gun to have 01 kgs water flow out at 20 ms What pressure inside the chamber is needed F Ad mV dtE A AmE A V 01 20 kg msA2E A 2 N Eq1521 vP 05 VA2E P 05 VA2E Av 05 20A2E A 0001 200 000 Pa 200 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1527 A jet engine at takeoff has air at 20AoE AC 100 kPa coming at 25 ms through the 10 m diameter inlet The exit flow is at 1200 K 100 kPa through the exit nozzle of 04 m diameter Neglect the fuel flow rate and find the net force thrust on the engine Α1 Aπ 4E A DA2E A 07854 mA2E A Α2 Aπ 4E A DA2E A 01257 mA2E v1 ART PE A A0287 29315 100E A 08409 mA3E Akg v2 3444 mA3E Akg AmE A AVv Α1V1v1 A07854 25 08409E A 480 kgs V2 m v2 A2 A480 3444 01257E A 1315 ms Now we can do the x direction momentum equation for steady flow and the same mass flow rate in and out Xdir 0 AmE A V1 Fx P1 Po A1 AmE A V2 P2 Po A2 Fx AmE A V1 P1 Po A1 AmE A V2 P2 Po A2 AmE A V2 V1 0 0 48 1315 25 61 920 N Inlet High P Low P exit cb Fnet Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1528 A water turbine using nozzles is located at the bottom of Hoover Dam 175 m below the surface of Lake Mead The water enters the nozzles at a stagnation pressure corresponding to the column of water above it minus 20 due to losses The temperature is 15C and the water leaves at standard atmospheric pressure If the flow through the nozzle is reversible and adiabatic determine the velocity and kinetic energy per kilogram of water leaving the nozzle The static pressure at the 175 m depth is P ρgZ AgZ vE A A 9807 175 0001001 1000E A 17145 kPa Pac 08 P 13716 kPa Use Bernoullis equation vP VAex 2 AE E2 Vex A 2vPEA Vex A 2 0001001 1000 13716EA 624 ms VAex 2 AE E2 vP 1373 kJkg Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1529 A water cannon sprays 1 kgs liquid water at a velocity of 100 ms horizontally out from a nozzle It is driven by a pump that receives the water from a tank at 15C 100 kPa Neglect elevation differences and the kinetic energy of the water flow in the pump and hose to the nozzle Find the nozzle exit area the required pressure out of the pump and the horizontal force needed to hold the cannon AmE A ρAV AVv A AmE AvV 1 A0001001 100E A 10 10A5E A mA2E A AW E Ap AmE Awp AmE AvPex Pin AmE AVAex 2 AE E2 Pex Pin VAex 2 AE E2v 100 100A2E A2 1000 0001 150 kPa F AmE AVex 1 100 100 N Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1530 An irrigation pump takes water from a lake and discharges it through a nozzle as shown in Fig P1530 At the pump exit the pressure is 900 kPa and the temperature is 20C The nozzle is located 15 m above the pump and the atmospheric pressure is 100 kPa Assuming reversible flow through the system determine the velocity of the water leaving the nozzle Assume we can neglect kinetic energy in the pipe in and out of the pump Incompressible flow so Bernoullis equation applies V1 V2 V3 0 Bernoulli vP3 P2 VA3 2 AEE VA2 2 AE2 gZ3 Z2 0 P3 P2 gZ3 Z2 v 900 A 9807 15 1000 0001002E A 753 kPa Bernoulli VA4 2 AEE2 vP3 P4 V4 2vP3 P4 A 2 0001002 653 1000EA 362 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1531 A water tower on a farm holds 1 m3 liquid water at 20C 100 kPa in a tank on top of a 5 m tall tower A pipe leads to the ground level with a tap that can open a 15 cm diameter hole Neglect friction and pipe losses and estimate the time it will take to empty the tank for water Incompressible flow so we can use Bernoulli Equation PAeE A Pi Vi 0 Ze 0 Zi H VAe 2 AE2 gZi VAeE A A 2gZEA A 2 9807 5EA 99 ms AmE A ρAVAeE A AVAeE Av mt m Vv A πDA2E A4 π 0015A2E A 4 177 10A4E A mA2E A t mvAVAeE A VAVAeE t EA 1 177 10A4 A 99E A 5716 sec 953 min Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1532 Find the speed of sound for air at 100 kPa at the two temperatures 0C and 30C Repeat the answer for carbon dioxide and argon gases From eq 1528 we have c0 A kRTEA A 14 0287 27315 1000EA 331 ms c30 A 14 0287 30315 1000EA 349 ms For Carbon Dioxide R 01889 kJkg K k 1289 c0 A 1289 01889 27315 1000EA 2579 ms c30 A 1289 01889 30315 1000EA 2717 ms For Argon R 02081 kJkg K k 1667 c0 A 1667 02081 27315 1000EA 3078 ms c30 A 1667 02081 30315 1000EA 3243 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1533 Find the expression for the anchoring force RAxE A for an incompressible flow like in Figure 156 Show that it can be written as RAxE A A Vi Ve EVi Ve E A PAiE A PAoE AAAiE A PAeE A PAoE AAAeE A Apply the Xdir momentum equation for a steady flow 0 RAxE A PAiE A PAoE AAAiE A PAeE A PAoE AAAeE A AmE AVi AmE AVAeE Bernoulli equation for the flow is 05 V 2 e V 2 i v PAeE A PAiE A 0 VAeE A VAiE A A 2v Pi Pe EVi Ve E Continuity equation gives AmE A AAiE AVAiE A v AAeE AVAeE A v Solve for RAxE A from the momentum equation RAxE A AmE A VAeE A VAiE A PAeE A PAoE AAAeE A PAiE A PAoE AAAiE A AiVi EvE A A 2v Pi Pe EVi Ve E A PAeE A PAoE AAAeE A PAiE A PAoE AAAiE Multiply in and use continuity equation for second term RAxE A A 2 Vi Ve E A PAiE AAAiE AVAiE A PAeE AAAeE AVAeE A PAeE A PAoE AAAeE A PAiE A PAoE AAAiE A 2 Vi Ve E A PAiE AAAiE AVAiE A PAeE AAAeE AVAeE A A1 2E APAeE A PAoE AAAeE AVAeE A A1 2E APAiE A PAoE AAAiE AVAiE A1 2E APAeE A PAoE AAAeE AVAiE A A1 2E APAiE A PAoE AAAiE AVAeE A Now put the first four terms together RAxE A A 2 Vi Ve E A A1 2E APAiE A PAoE AAAiE AVAiE A A1 2E APAeE A PAoE AAAeE AVAeE A1 2E APAeE A PAoE AAAeE AVAiE A A1 2E APAiE A PAoE AAAiE AVAeE A A 2 Vi Ve E A A1 2E APAiE A PAoE AAAiE AVAiE A VAeE A A1 2E APAeE A PAoE AAAeE AVAiE A VAeE A A Vi Ve EVi Ve E A PAiE A PAoE AAAiE A PAeE A PAoE AAAeE A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1534 Estimate the speed of sound for steam directly from Eq 1525 and the steam tables for a state of 6 MPa 400C Use table values at 5 and 7 MPa at the same entropy as the wanted state Eq 1525 is then done by finite difference Find also the answer for the speed of sound assuming steam is an ideal gas Eq1525 cA2E A AP ρE As AP ρE As State 6 MPa 400C s 65407 kJkg K 7 MPa s v 004205 mA3E Akg ρ 1v 23777 kgmA3E 5 MPa s v 005467 mA3E Akg ρ 1v 182909 kgmA3E cA2E A A 7000 5000 23777 182909E A 36456 1000 c 6038 ms From Table A8 Cp A133856 12353 50E A 20652 kJkg K Cv Cp R 20652 04615 16037 kJkg K k CpCv 1288 R 04615 kJkg K from A5 Now do the speed of sound from Eq1528 c A kRTEA A 1288 04615 67315 1000EA 6326 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1535 Use the CATT3 software to solve the previous problem From Eq 1525 cA2E A AP ρE AAsE A vA2E AAP vE AAsE Superheated vapor water at 400AoE AC 6000 kPa CATT3 v 004739 mA3E Akg s 6541 kJkg K At P 6200 kPa s 6541 kJkg K T 4051AoE AC v 00462 mA3E Akg At P 5800 kPa s 6541 kJkg K T 3948AoE AC v 004866 mA3E Akg cA2E A 004739A2E A A 62 58 00462 004866E A AMJ kgE A 036517 10A6E A mA2E AsA2E c 604 ms From Table A8 Cp A133856 12353 50E A 20652 kJkg K Cv Cp R 20652 04615 16037 kJkg K k CpCv 1288 R 04615 kJkg K from A5 Now do the speed of sound from Eq1528 c A kRTEA A 1288 04615 67315 1000EA 6326 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1536 If the sound of thunder is heard 5 seconds after the lightning is seen and the weather is 20C How far away is the lightning taking place The sound travels with the speed of sound in air ideal gas Use the formula in Eq1528 L c t A kRTEA t A 14 0287 29315 1000EA 5 1716 m For every 3 seconds after the lightning the sound travels about 1 km Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1537 Find the speed of sound for carbon dioxide at 2500 kPa 60AoE AC using either the tables or the CATT3 software same procedure as in Problem 1534 and compare that with Eq1528 From Eq 1525 cA2E A AP ρE AAsE A vA2E AAP vE AAsE Superheated carbon dioxide at 60AoE AC 2500 kPa CATT3 v 002291 mA3E Akg s 1521 kJkg K At P 2600 kPa s 1521 kJkg K T 6302AoE AC v 002221 mA3E Akg At P 2400 kPa s 1521 kJkg K T 569AoE AC v 002365 mA3E Akg cA2E A 002291A2E A A 26 24 002221 002365E A AMJ kgE A 78983 10A4E A mA2E AsA2E c 270 ms From Table A5 k CpCv 1289 R 01889 kJkg K Now do the speed of sound from Eq1528 c A kRTEA A 1289 01889 33315 1000EA 2848 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1538 A jet flies at an altitude of 12 km where the air is at 40AoE AC 45 kPa with a velocity of 1000 kmh Find the Mach number and the stagnation temperature on the nose From Table A5 k CpCv 14 R 0287 kJkg K Now do the speed of sound from Eq1528 c A kRTEA A 14 0287 23315 1000EA 306 ms V 1000 kmh 1000 kmh 1000 mkm 3600 sh 278 ms M V c 278306 0908 h0 h1 VA1 2 AE2 T0 T1 VA1 2 AE 2Cp 40 A 2782 E2 1004E A 15AoE AC Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1539 The speed of sound in liquid water at 25AoE AC is about 1500 ms Find the stagnation pressure and temperature for a M 01 flow at 25AoE AC 100 kPa Is it possible to get a significant mach number flow of liquid water V M c 01 1500 150 ms h0 h1 VA1 2 AE2 Bernoulli Eq P VA1 2 AE2v A 1502 E2 0001E A 1125 10A6E A Pa 1125 MPa P0 P1 P 100 11 250 11 350 kPa T0 T1 VA1 2 AE 2Cp 25 A 1502 E2 4180E A 277AoE AC Remark Notice the very high pressure To get a higher velocity you need a higher pressure to accelerate the fluid that is not feasible for any large flow rate Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1540 Steam flowing at 15 ms 1800 kPa 300AoE AC expands to 1600 kPa in a converging nozzle Find the exit velocity and area ratio Ae Ai Solve the problem with the steam tables Inlet state vi 014021 mA3E Akg hi 302921 kJkg si 68226 kJkgK Exit state Pese si ve 015371 mA3E Akg he 3000995 kJkg Energy Eq VAi 2 AE 2 hi VAe 2 AEE 2 he VAe 2 AEE VAi 2 AEE 2hi he Ve A 15 15 2000302921 3000995EA 238 ms Same mass flow rate so AeAi veviViVe A015371 014021E A A 15 238E A 006909 If we solved as ideal gas with constant specific heat we get k 1327 Te Ti PePiA k1kE A 57315 16001800A 02464E A 55676 K Ve A Vi 2 AE 2CpTi TeE A 15 15 2 187257315 55676E 248 ms AeAi veviViVe PiPeA 1kE AViVe A 1800 1600 07536E A A 15 248E A 00661 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1541 A convergent nozzle has a minimum area of 01 mA2E A and receives air at 175 kPa 1000 K flowing with 100 ms What is the back pressure that will produce the maximum flow rate and find that flow rate APA AEA Po E A 2 k1E AA k k1 E A 0528 Critical Pressure Ratio Find Po h0 h1 VA1 2 AEE2 104622 100A2E A2000 105122 kJkg T0 Ti 44 10044 K from table A7 P0 Pi T0TiAkk1E A 175 100441000A35E A 17771 kPa The mass flow rate comes from the throat properties PAE A 0528 Po 0528 17771 9383 kPa TAE A 08333 To 83697 K ρAE A EA PA A ERTA AE A A 9383 0287 83697E A 03906 kgmA3E V c EA kRTA AEA A 14 1000 0287 83697EA 5799 ms AmE A ρAV 03906 01 5799 2265 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1542 A convergentdivergent nozzle has a throat area of 100 mm2 and an exit area of 175 mm2 The inlet flow is helium at a stagnation pressure of 1 MPa stagnation temperature of 375 K What is the back pressure that will give sonic condition at the throat but subsonic everywhere else For this flow we have helium with kHe 1667 so we cannot use the tables for air We need the solution to the curve labeled c in Fig 1513 For critical flow at the throat we have from Table 151 last column PAE A 04867 Po 4867 kPa Now we need to find the conditions where the area ratio is AEAAE A 175100 175 that is solve for M in Eq 1543 given the area ratio This is nonlinear so we have to iterate on it Here k12k1 2 so look also at Fig 1510 for the general shape M 04 AAAE A 104 0751 0333304A2E A A2E A 15602 M 03 AAAE A 103 0751 0333303A2E A A2E A 19892 M 035 AAAE A 1035 0751 03333035A2E A A2E A 17410 M 034 AAAE A 1034 0751 03333034A2E A A2E A 17844 Now do a linear interpolation for the rest to get ME 0348 Eq 1535 PEPo 1 033330348A2E A A25E A 09058 PE 09058 1000 906 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1543 To what pressure should the steam in problem 1540 expand to reach Mach one Use constant specific heats to solve Find stagnation properties T0 T1 VA1 2 AE2Cp 57315 K 15A2E A msA2E A2 1872 JkgK 57321 K P0 P1 T0T1Akk1E A 1800 kPa 5732157315A 4058E A 1800765 kPa From Eq1535 we get k 1327 P P0 1 Ak 1 2E A M A2E A A kk1E A 1800765 kPa 1 01635A 4058E A 974 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1544 A jet plane travels through the air with a speed of 1000 kmh at an altitude of 6 km where the pressure is 40 kPa and the temperature is 12C Consider the inlet diffuser of the engine where air leaves with a velocity of 100 ms Determine the pressure and temperature leaving the diffuser and the ratio of inlet to exit area of the diffuser assuming the flow to be reversible and adiabatic V 1000 kmh 2778 ms v1 RTP 0287 2611540 1874 mA3E Akg h1 26148 kJkg ho1 26148 2778A2E A2000 30007 kJkg To1 2997 K Po1 P1 To1T1Akk1E A 40 299726115A35E A 64766 kPa h2 30007 100A2E A2000 29507 kJkg T2 2947 K P2 Po1 T2To1Akk1E A 64766 29472997A35E A 61 kPa v2 RT2P2 0287 kJkgK 2947 K61 kPa 1386 mA3E Akg A1A2 v1v2V2V1 187413861002778 0487 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1545 Air flows into a convergentdivergent nozzle with an exit area of 159 times the throat area of 0005 mA2E A The inlet stagnation state is 1 MPa 600 K Find the backpressure that will cause subsonic flow throughout the entire nozzle with M 1 at the throat What is the mass flow rate This corresponds to case c and is a reversible flow AEAAE A 159 Look at top in Table A12 M 1 ME 04 and PEPo 08956 PE 08956 Po 08956 1000 kPa 896 kPa To find the mass flow rate we have the maximum flow rate from Eq 1542 AmE A AAE A Po kRTo Ak 1 2E A A k12k1E 0005 1000 1000 N A 140287 600 1000E A sm 12A 2408E 825 kgs Notice kPa must be converted to Pa and also units for RT kJkg 1000 msA2E A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1546 A nozzle is designed assuming reversible adiabatic flow with an exit Mach number of 28 while flowing air with a stagnation pressure and temperature of 2 MPa and 150C respectively The mass flow rate is 5 kgs and k may be assumed to be 140 and constant Determine the exit pressure temperature exit area and the throat area Since ME 28 we must have choked flow with MAE A 1 then the mass flow rate is maximum and follows Eq1542 AmE AAAE A Po kRTo Ak 1 2E A A k12k1E 2000 1000 NmA2E A A 140287 42315 1000EA sm 12A 2408E A 392972 kgsmA2E AAE A AmE A 392972 kgsmA2E A 5 392972 mA2E A 1272 10A3E A mA2E From Table A12 ME 28 TE To 038941 PE Po 0036848 AE AAE A AEAAE A 1272 10A3E A 35001 4452 10A3E A mA2E A TE To TE To 42315 Κ 038941 1648 K PE Po PE Po 2000 kPa 0036848 737 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1547 An air flow at 600 kPa 600 K M 03 flows into a convergentdivergent nozzle with M 1 at the throat Assume a reversible flow with an exit area twice the throat area and find the exit pressure and temperature for subsonic exit flow to exist To find these properties we need the stagnation properties from the inlet state From Table A12 Mi 03 PiPo 093947 TiTo 098232 Po 600 kPa 093947 63866 kPa To 600 K 098232 6108 K This flow is case c in Figure 1513 From Table A12 AEAAE A 2 PEPo 09360 TETo 098127 PE 09360 Po 0936 63866 kPa 5978 kPa TE 098127 To 098127 6108 K 5994 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1548 Air at 150 kPa 290 K expands to the atmosphere at 100 kPa through a convergent nozzle with exit area of 001 m2 Assume an ideal nozzle What is the percent error in mass flow rate if the flow is assumed incompressible Te Ti Pe PiA k1 k E A 25828 K VAe 2 AEE2 hi he Cp Ti Te 1004 290 25828 3183 kJkg Ve 2523 ms ve RTe Pe A0287 25828 100E A 07412 mA3E Akg AmE A AVe ve A001 2523 07413E A 34 kgs Incompressible Flow vi RTP 0287 290150 055487 m3kg VAe 2 AEE2 v P vi Pi Pe 055487 150 100 2774 kJkg Ve 235 ms AmE A AVe vi 001 235 055487 423 kgs m incompressible m compressible A423 34E A 125 about 25 overestimation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1549 Find the exit pressure and temperature for supersonic exit flow to exist in the nozzle flow of Problem 1547 We assume a reversible as the possibility which is case d in Figure 1513 To find these properties we need the stagnation properties from the inlet state From Table A12 Mi 03 PiPo 093947 TiTo 098232 Po 600 kPa 093947 63866 kPa To 600 K 098232 6108 K From Table A12 AEAAE A 2 PEPo 009352 TETo 050813 PE 009352 Po 009352 63866 597 kPa TE 050813 To 050813 6108 3104 K This is significant lower P and T but then we also have M 22 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1550 Air is expanded in a nozzle from a stagnation state of 2 MPa 600 K to a backpressure of 19 MPa If the exit crosssectional area is 0003 mA2E A find the mass flow rate This corresponds to case c and is a reversible flow PEPox 1920 095 Table A12 ME 0268 TE TToE To 09854 600 5912 K cE kRTE A 14 1000 0287 5912EA 4874 ms VE MEcE 0268 4874 1306 ms vE RTP 0287 kJkgK 5912 K1900 kPa 00893 mA3E Akg AmE A AEVEvE 0003 mA2E A 1306 ms00893 mA3E Akg 4387 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1551 A 1m3 insulated tank contains air at 1 MPa 560 K The tank is now discharged through a small convergent nozzle to the atmosphere at 100 kPa The nozzle has an exit area of 2 105 m2 a Find the initial mass flow rate out of the tank b Find the mass flow rate when half the mass has been discharged a The back pressure ratio PBPo1 1001000 01 PAE APocrit 05283 so the initial flow is choked with the maximum possible flow rate ME 1 PE 05283 1000 5283 kPa TE TAE A 08333 560 K 4667 K VE c A kRT EA A 14 1000 0287 4667EA 433 ms vE RTAE APE 0287 kJkgK 4667 K5283 kPa 02535 mA3E Akg AmE A1 AVEvE 2105 mA2E A 433 ms02535 mA3E Akg 00342 kgs b The initial mass is m1 P1VRT1 1000 10287 560 6222 kg with a mass at state 2 as m2 m12 3111 kg Assume an adiabatic reversible expansion of the mass that remains in the tank P2 P1v1v2AkE A 100 05A14E A 3789 kPa T2 T1v1v2Ak1E A 560 05A04E A 424 K The pressure ratio is still less than critical and the flow thus choked PBPo2 1003789 0264 PAE APocrit ME 1 PE 05283 3789 2002 kPa TE TAE A 08333 424 3537 K VE c A kRT EA A 14 1000 0287 3537EA 377 ms AmE A2 AVEPERTE A21053772002 E02873537E A 00149 kgs We could also have found the two mass flow rates from Eq 1542 without finding the exit plane conditions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1552 A convergentdivergent nozzle has a throat diameter of 005 m and an exit diameter of 01 m The inlet stagnation state is 500 kPa 500 K Find the back pressure that will lead to the maximum possible flow rate and the mass flow rate for three different gases as air hydrogen or carbon dioxide There is a maximum possible flow when M 1 at the throat TAE A A 2 k1E A To PAE A Po A 2 k1E AA k k1 E A ρAE A ρo A 2 k1E AA 1 k1 E A AmE A ρAE AAAE AV ρAE AAAE Ac PAE AAAE A kRT EA AAE A πDA2E A4 0001963 mA2E A k TAE A PAE A c ρAE A AmE A a 1400 4167 2641 4482 2209 1944 b 1409 4151 2634 17045 0154 0515 c 1289 4369 2739 3489 3318 2273 AEAAE A DEDAE AA2E A 4 There are 2 possible solutions corresponding to points c and d in Fig 1513 and Fig 1517 For these we have Subsonic solution Supersonic solution ME PEPo ME PEPo a 01466 0985 2940 00298 b 01464 0985 2956 00293 c 01483 0986 2757 00367 PB PE 0985 500 4925 kPa all cases point c a PB PE 00298 500 149 kPa point d b PB PE 00293 500 1465 kPa point d c PB PE 00367 500 1835 kPa point d Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1553 Air is expanded in a nozzle from a stagnation state of 2 MPa 600 K to a static pressure of 200 kPa The mass flow rate through the nozzle is 5 kgs Assume the flow is reversible and adiabatic and determine the throat and exit areas for the nozzle PAE A Po A 2 k1 E A k k1 E 2 05283 1056 MPa TAE A To A 2 k1E A 600 08333 500 K vAE A RTAE APAE A 0287 5001056 01359 mA3E Akg The critical speed of sound is cAE A A kRT EA A 14 1000 0287 500EA 4482 ms AAE A AmE AvAE AcAE A 5 013594482 000152 mA2E A P2Po 2002000 01 MA AE2 E 1701 V2cAE We used the column in Table A12 with mach no based on throat speed of sound V2 1701 4482 7624 ms T2 To T2To 600 05176 31056 K v2 RT2P2 0287 kJkgK 31056 K200 kPa 04456 mA3E Akg A2 AmE Av2V2 5 kgs 04456 mA3E Akg 7624 ms 000292 mA2E A Velocity Density Area Mach 20 MPa 02 MPa P Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1554 Air flows into a convergentdivergent nozzle with an exit area of 20 times the throat area of 0005 mA2E A The inlet stagnation state is 12 MPa 600 K Find the backpressure that will cause a reversible supersonic exit flow with M 1 at the throat What is the mass flow rate This flow is case d in Fig1517 the only reversible supersonic flow AEAAE A 2 see Table A12 M 1 ME 22 and PEPo 009399 PE 009399 1200 1128 kPa Using Eq1542 for the chocked flow rate AmE A AAE A Po kRTo Ak 1 2E A A k12k1E 0005 1200 1000 A 140287 600 1000EA 12A 2408E 99 kgs Notice kPa must be converted to Pa and also units for RT kJkg 1000 msA2E A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1555 What is the exit pressure that will allow a reversible subsonic exit flow in the previous problem This flow is case c in Fig1517 and c in Fig 1513 the only reversible subsonic flow with M 1 at the throat AEAAE A 2 see Table A12 M 1 PEPo 09360 TETo 098127 PE 09360 Po 0936 1200 1123 kPa TE 098127 To 098127 600 5888 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1556 A flow of helium flows at 500 kPa 500 K with 100 ms into a convergent divergent nozzle Find the throat pressure and temperature for reversible flow and M 1 at the throat We need to find the stagnation properties first k 1667 To T1 VA1 2 AE2Cp 500 100A2E A2 5193 500963 K Po P1 T0T1Akk1E A 500 500963500A 25E A 50241 kPa From the analysis we get Eqs153738 PAE A Po A 2 k 1 kk1E A 50241 A 2 1667 1 25E A 2447 kPa TAE A To A 2 k 1E A 500963 A 2 1667 1E A 3757 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1557 Assume the same tank and conditions as in Problem 1551 After some flow out the nozzle flow changes to become subsonic Find the mass in the tank and the mass flow rate out at that instant The initial mass is m1 P1VRT1 1000 10287 560 6222 kg The flow changes to subsonic when the pressure ratio reaches critical PBPo3 05283 Po3 1893 kPa v1v3 Po3P1A1kE A 18931000A07143E A 03046 m3 m1v1v3 6222 03046 1895 kg At this point the tank temperatures is assuming isentropic expansion T3 T1v1v3Ak1E A 560 03046A04E A 348 K PE PB 100 kPa ME 1 TE 08333 348 290 K VE kRTE 3414 ms AmE A3 AVEPERTE A21053414100 E0287290E A 00082 kgs We could have used Eq1542 with Po Po3 1893 kPa To T3 348 K P AIR e cb g Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1558 A given convergent nozzle operates so it is choked with stagnation inlet flow properties of 400 kPa 400 K To increase the flow a reversible adiabatic compressor is added before the nozzle to increase the stagnation flow pressure to 500 kPa What happens to the flow rate Since the nozzle is choked the mass flow rate is given by Eq1542 The compressor changes the stagnation pressure and temperature Isentropic Po new Po rp and To new To A rp k1 k AE E Po new To new A rp k1 2k AE E Po To so the mass flow rate is multiplied with the factor A rp k1 2k AE E A 500 400 24 28 E A 121 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1559 A 1m3 uninsulated tank contains air at 1 MPa 560 K The tank is now discharged through a small convergent nozzle to the atmosphere at 100 kPa while heat transfer from some source keeps the air temperature in the tank at 560 K The nozzle has an exit area of 2 105 m2 a Find the initial mass flow rate out of the tank b Find the mass flow rate when half the mass has been discharged a The back pressure ratio PBPo1 1001000 01 PAE APocrit 05283 so the initial flow is choked with the maximum possible flow rate given in Eq1542 AmE A AAE A Po kRTo Ak 1 2E A A k12k1E 2 105 1 106 A 140287 560 1000EA 12A 2408E 0034 kgs b The initial mass is m1 P1VRT1 1000 10287 560 6222 kg with a mass at state 2 as m2 m12 3111 kg P2 P12 500 kPa T2 T1 PBP2 100500 02 PAE APocrit The flow is choked and the mass flow rate is similar to the previously stated one but with half the stagnation pressure same stagnation temperature AmE A2 AAE A Po kRTo Ak 1 2E A A k12k1E A AmE A1 2 0017 kgs P AIR e cb Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1560 Assume the same tank and conditions as in Problem 1559 After some flow out the nozzle flow changes to become subsonic Find the mass in the tank and the mass flow rate out at that instant The initial mass is m1 P1VRT1 1000 10287 560 6222 kg Flow changes to subsonic when the pressure ratio reaches critical PBPo3 05283 P3 Po3 PB05283 10005283 1893 kPa m3 m1P3P1 1178 kg T3 T1 TE TAE A 08333 560 4667 K PE PB 100 kPa VE c A kRT EA A 14 1000 0287 4667EA 433 ms AmE A3 AVEPERTE A2105433100 E02874667E A 000646 kgs Flow is critical so we could have used Eq1542 with Po 1893 kPa To 560 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Normal Shocks Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1561 The products of combustion use air enter a convergent nozzle of a jet engine at a total pressure of 125 kPa and a total temperature of 650C The atmospheric pressure is 45 kPa and the flow is adiabatic with a rate of 25 kgs Determine the exit area of the nozzle The critical pressure Table 151 Pcrit P2 125 05283 66 kPa Pamb The flow is then choked with M 1 at the exit and a pressure drop outside of the nozzle T2 92315 08333 7693 K V2 c2 A 14 1000 0287 7693EA 556 ms v2 RT2P2 0287 769366 33453 mA3E Akg A2 AmE Av2 V2 25 33453556 01504 mA2E A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1562 Redo the previous problem for a mixture with k 13 and molecular mass of 31 The critical pressure Table 151 Pcrit P2 125 05457 682 kPa Pamb The flow is then choked T2 92315 08696 8028 K The gas constant is R 831451 31 02682 kJkgK V2 c2 A 13 1000 02682 8028EA 529 ms v2 RT2P2 02682 8028682 3157 mA3E Akg A2 AmE Av2 V2 25 3157529 0149 mA2E A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1563 At what Mach number will the normal shock occur in the nozzle of Problem 1552 flowing with air if the back pressure is halfway between the pressures at c and d in Fig 1517 First find the two pressures that will give exit at c and d See solution to 1552 a AEAAE A DEDAE AA2E A 4 For case c PE 4925 kPa For case d PE 149 kPa Actual case PE 4925 1492 2537 kPa Assume Mx 24 My 05231 PoyPox 054015 AxAA AEx E 24031 AxAA AEy E 1298 Work backwards to get PE AEAA AEy E AEAA AEx E AxAA AEy E AxAA AEx E 4 129824031 21605 ME 02807 PEPoy 094675 PE PEPoy PoyPox Pox 094675 054015 500 2557 kPa Repeat if Mx 25 PE 2338 kPa Interpolate to match the desired pressure Mx 241 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1564 Consider the nozzle of Problem 1553 and determine what back pressure will cause a normal shock to stand in the exit plane of the nozzle This is case g in Fig 1517 What is the mass flow rate under these conditions We assume reversible flow up to the shock Table A12 PEPo 2002000 01 ME 21591 Mx Shock functions Table A13 My 05529 PyPx 5275 PB Py 5275 Px 5275 200 1055 kPa AmE A 5 kgs same as in Problem 1553 since M 1 at throat Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1565 A normal shock in air has upstream total pressure of 500 kPa stagnation temperature of 500 K and Mx 14 Find the downstream stagnation pressure From the normal shock relations in Section 158 found in Table A13 we get Mx 14 Po yPo x 095819 Po y 0 95819 Po x 0 95819 500 4791 kPa Remark The stagnation temperature would be unchanged energy equation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1566 How much entropy per kg flow is generated in the shock in Example 159 The change in entropy is sgen sy sx Cp ln Ty Tx R ln Py Px 1004 ln 132 0287 ln 24583 027874 025815 00206 kJkg K Notice that this function could have been tabulated also as sgenR A k k 1E A ln Ty Tx ln Py Px Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1567 Consider the diffuser of a supersonic aircraft flying at M 14 at such an altitude that the temperature is 20C and the atmospheric pressure is 50 kPa Consider two possible ways in which the diffuser might operate and for each case calculate the throat area required for a flow of 50 kgs a The diffuser operates as reversible adiabatic with subsonic exit velocity b A normal shock stands at the entrance to the diffuser Except for the normal shock the flow is reversible and adiabatic and the exit velocity is subsonic This is shown in Fig P1567 a Assume a convergentdivergent diffuser with M 1 at the throat Remember that for the supersonic flow the convergent front will reduce the velocity Relate the inlet state to the stagnation state state table A12 with M 14 PPo 031424 Po 50 kPa 031424 15911 kPa TTo 071839 To 25315 K 071839 352385 K The maximum flow rate for chocked flow from Eq1542 AmE A AAE A Po kRTo Ak 1 2E A A k12k1E 15911 103 A 140287 25315 1000EA 12A 2408E A 34259 kgsmA2E AAE A AmE A 34259 kgsmA2E A 01459 mA2E A b Across the shock we have Mx 14 so from Table A13 My 07397 PoyPox 095819 The stagnation properties after the shock are Toy Tox 25315 071839 352385 K as before Poy Pox 095819 50 kPa 031424 095819 15246 kPa Only the stagnation pressure changed due to the shock so Eq1542 gives AmE A AAE A 095819 34259 kgsmA2E A 328266 kgsmA2E AAE A AmE A 328266 kgsmA2E A 01523 mA2E A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1568 A flow into a normal shock in air has a total pressure 400 kPa stagnation temperature of 600 K and Mx 12 Find the upstream temperature Tx the specific entropy generation in the shock and the downstream velocity From Table A12 Mx 12 has TxTo 07764 PxPo 041238 Tx 07764 To 07764 600 K 46584 K From Table A13 TyTx 1128 PyPx 15133 My 084217 The change in entropy is sgen sy sx Cp ln Ty Tx R ln Py Px 1004 ln 1128 0287 ln 15133 012093 011890 000203 kJkg K From the shock relations we had Ty 1128 Tx 1128 46584 K 52547 K Vy My cy My kRTy 084217 A 14 0287 52547 1000EA 084217 45949 387 ms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1569 Consider the nozzle in problem 1542 flowing helium What should the backpressure be for a normal shock to stand at the exit plane This is case g in Fig1517 What is the exit velocity after the shock Reversible flow up to the shock with M 1 at the throat Px o Po Tx o To AEAAE A 175 100 175 As we do not have a Table 12 made for helium k 53 1667 we need to use the formulas The area to M relation is given in Eq1543 so A AAE A 175 1Mx A 2 k 11 13M2 x EA A 2E A 3 M2 x A 2E A 16 Mx There are two solutions and we need the one with Mx 1 see fig 1510 we should expect a solution around Mx 2 With trial and error on 3 M2 x 16 175 Mx which gives Mx 22033 Now we can do the normal shock from Eq1553 Mx 22033 M2 y A 220332 3 E5 220332 1E A 03375 The temperature ratio from Eq 1549 and the Tx from Eq 1534 Ty Tx A 1 05k1 M2 Ex 1 05k1 M2 y E A Tx o 1 05k1 M2 y A 375 1 033303375E A 3371 K Vy My cy My kRTy A 03375EA A 1667 20771 3371 1000EA 058095 10804 6276 ms A M x M y A E Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1570 Find the specific entropy generation in the shock of the previous Problem Reversible flow up to the shock with M 1 at the throat Px o Po Tx o To AEAAE A 175 100 175 Table A12 ME Mx 2042 PxPo x 012 Tx To x 05454 Now we can do the normal shock from Table A13 Mx 2042 My 05704 PyPx 46984 TyTx 17219 The change in entropy is sgen sy sx Cp ln Ty Tx R ln Py Px 1004 ln 17219 0287 ln 46984 05456 044405 01015 kJkg K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Nozzles Diffusers and Orifices Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1571 Steam at 600 kPa 300C is fed to a set of convergent nozzles in a steam turbine The total nozzle exit area is 0005 m2 and they have a discharge coefficient of 094 The mass flow rate should be estimated from the measurement of the pressure drop across the nozzles which is measured to be 200 kPa Determine the mass flow rate Inlet B13 hi 30616 kJkg si 73724 kJkg K Exit Pe ses Pe Pi 200 400 kPa ses si 73724 kJkg K hes 2961 kJkg and ves 05932 mA3E Akg Ves A 2 100030616 2961EA 44855 ms AmE As AVesves 0005 4485505932 3781 kgs AmE Aa CD AmE As 094 3781 3554 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1572 Air enters a diffuser with a velocity of 200 ms a static pressure of 70 kPa and a temperature of 6C The velocity leaving the diffuser is 60 ms and the static pressure at the diffuser exit is 80 kPa Determine the static temperature at the diffuser exit and the diffuser efficiency Compare the stagnation pressures at the inlet and the exit Stagnation T at the inlet To1 T1 VA1 2 AE2Cp 26715 200A2E A2000 1004 2871 K Energy Eq gives the same stagnation T at exit To2 To1 T2 To2 VA2 2 AE2Cp 2871 60A2E A2000 1004 2853 K To1 T1 T1 Ak1 kE A Po1 P1 P1 Po1 P1 1825 Po1 883 kPa To2 T2 T2 Ak 1 kE A Po2 P2 P2 Po2 P2 177 Po2 818 kPa T ex s T1 Po2P1Ak1kE A 26715 10454 2793 K ηD T ex s T1 To1 T1 A2793 26715 2871 26715E A 0608 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1573 Repeat Problem 1544 assuming a diffuser efficiency of 80 V 1000 kmh 2778 ms v1 RTP 0287 2611540 1874 mA3E Akg h1 26148 kJkg ho1 26148 2778A2E A2000 30007 kJkg To1 2997 K Po1 P1 To1T1Akk1E A 40 299726115A35E A 64766 kPa Same as problem 1544 except ηD 080 We thus have from 1544 h3 h1 ho1 h1 h3 26148 30007 26148 08 h3 29235 kJkg T3 2919 K Po2 P3 P1 Τ3Τ1Akk1E A 40 291926115A35E A 5906 kPa To2 To1 2997 K h2 30007 100A2E A2000 29507 kJkg T2 2947 K P2 Po2 T2To1Akk1E A 5906 29472997A35E A 5568 kPa v2 RT2P2 0287 29475568 1519 mA3E Akg A1A2 v1v2V2V1 18741519 1002778 0444 h 0 1 0 2 3 1 2 s Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1574 A sharpedged orifice is used to measure the flow of air in a pipe The pipe diameter is 100 mm and the diameter of the orifice is 25 mm Upstream of the orifice the absolute pressure is 150 kPa and the temperature is 35C The pressure drop across the orifice is 15 kPa and the coefficient of discharge is 062 Determine the mass flow rate in the pipeline T Ti A k1 k E AP Pi 30815 A04 14E A A 15 150E A 88 K vi RTiPi 05896 mA3E Akg Pe 135 kPa Te 29935 K ve 06364 mA3E Akg AmE Ai AmE Ae Vi Ve DeDiA2E A vive 00579 hi he VAe 2 AE1 00579A2E A2 CpTi Te Ve s A 2 1000 1004 881 005792 EA 1331 ms AmE A CD AVv 062 π4 0025A2E A 1331 06364 006365 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1575 A critical nozzle is used for the accurate measurement of the flow rate of air Exhaust from a car engine is diluted with air so its temperature is 50C at a total pressure of 100 kPa It flows through the nozzle with throat area of 700 mm2 by suction from a blower Find the needed suction pressure that will lead to critical flow in the nozzle and the mass flow rate PAE A 05283 Po 5283 kPa TAE A 08333 To 2693 K vAE A RTAE APAE A 0287 26935283 1463 mA3E Akg cAE A A kRT EA A 14 1000 0287 2693EA 3289 ms AmE A AcAE AvAE A 700 10A6E A 32891463 0157 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1576 Air is expanded in a nozzle from 700 kPa 200C to 150 kPa in a nozzle having an efficiency of 90 The mass flow rate is 4 kgs Determine the exit area of the nozzle the exit velocity and the increase of entropy per kilogram of air Compare these results with those of a reversible adiabatic nozzle T2s T1P2P1Ak1kE A 4732 150700A0286E A 3046 K V2s A2E A 2 1000 10044732 3046 338 400 Jkg V2 A2E A 09 338 400 mA2E AsA2E A V2 552 ms h2 VA2 2 AEE2 h1 T2 T1 VA2 2 AEE2Cp T2 4732 552A2E A2 1000 1004 3214 K v2 RTP 0287 kJkg 3214 K150 kPa 06149 m3kg A2 AmE A v2V2 4 kgs 06149 m3kg 552 ms 000446 mA2E A 4460 mmA2E A s2 s1 10035 lnA 3214 4732 E A 0287 lnA 150 700 E A 00539 kJkg K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1577 Steam at a pressure of 1 MPa and temperature of 400C expands in a nozzle to a pressure of 200 kPa The nozzle efficiency is 90 and the mass flow rate is 10 kgs Determine the nozzle exit area and the exit velocity First do the ideal reversible adiabatic nozzle s2s s1 74651 kJkg K h1 32639 kJkg T2s 1904C h2s 2851 kJkg Now the actual nozzle can be calculated h1 h2ac ηDh1 h2s 0932639 2851 3716 kJkg h2ac 28923 kJkg T2 2109C v2 11062 mA3E Akg V2 A 200032639 28923EA 862 ms A2 AmE Av2V2 10 kgs 11062 mA3E Akg 862 ms 001283 mA2E A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1578 Steam at 800 kPa 350C flows through a convergentdivergent nozzle that has a throat area of 350 mm2 The pressure at the exit plane is 150 kPa and the exit velocity is 800 ms The flow from the nozzle entrance to the throat is reversible and adiabatic Determine the exit area of the nozzle the overall nozzle efficiency and the entropy generation in the process ho1 31617 kJkg so1 74089 kJkg K PAE APo1 2k1Akk1E A 054099 PAE A 4327 kPa At PAE AsAE A so1 hAE A 29993 kJkg vAE A 05687 mA3E Akg h VA2E A2 VAE A A 20003161729993EA 5699 ms AmE A AVAE AvAE A 350 10A6E A 569905687 03507 kgs he ho1 VAe 2 AE2 31617 800A2E A2 1000 28417 kJkg Exit Pe he ve 1395 mA3E Akg se 7576 kJkg K Ae AmE AveVe 03507 1395800 6115 10A4E A mA2E A sgen se so1 7576 74089 0167 kJkg K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1579 A convergent nozzle with exit diameter of 2 cm has an air inlet flow of 20C 101 kPa stagnation conditions The nozzle has an isentropic efficiency of 95 and the pressure drop is measured to 50 cm water column Find the mass flow rate assuming compressible adiabatic flow Repeat calculation for incompressible flow Convert P to kPa P 50 cm H2O 05 98064 4903 kPa T0 20C 29315 K P0 101 kPa Assume inlet Vi 0 Pe P0 P 101 4903 96097 kPa Te T0 Pe P0 A k1 k E A 29315 A96097 101E AA02857E A 28901 VAe 2 AEE2 hi he Cp Ti Te 1004 29315 28901 41545 kJkg 42545 Jkg Ve 9115 ms VAe ac 2 AE E2 η VAe s 2 AE E2 095 41545 394678 Ve ac 8885 ms Te ac Ti A VAe ac 2 AEAE E2 Cp E 29315 A39468 10035E A 2892 K ρe ac Pe RTp A 96097 0287 2892E A 1158 kgmA3E AmE A ρAV 1158 kgmA3E A Aπ 4E A 002A2E A mA2E A 8885 ms 00323 kgs For incompressible flow VAe 2 AEE2 Pi Pev Pi Pe RTi Pi 4903 kPa A0287 kJkgK 29315 K 101 kPaE A 4903 0833 kJkg 4084 kJkg 4084 Jkg VAe ac 2 AE E2 η VAe s 2 AE E2 095 4084 3880 Jkg Ve ac 8809 ms AmE A ρAV 10833 kgmA3E A Aπ 4E A 002A2E A mA2E A 8809 ms 00332 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1580 The coefficient of discharge of a sharpedged orifice is determined at one set of conditions by use of an accurately calibrated gasometer The orifice has a diameter of 20 mm and the pipe diameter is 50 mm The absolute upstream pressure is 200 kPa and the pressure drop across the orifice is 82 mm of mercury The temperature of the air entering the orifice is 25C and the mass flow rate measured with the gasometer is 24 kgmin What is the coefficient of discharge of the orifice at these conditions P 82 101325760 1093 kPa T Ti A k1 k E A PPi 29815 A04 14E A 1093200 466 vi RTiPi 04278 mA3E Akg ve RTePe 04455 mA3E Akg Vi VeAeviAive 01536 Ve VAe 2 AE VAi 2 AE2 VAe 2 AE1 01536A2E A2 hi he CpT Ve A 2 1000 1004 4661 015362EA 979 ms AmE As AeVeve Aπ 4E A 002A2E A 97904455 0069 kgs CD AmE AaAmE As 2460 0069 058 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1581 A convergent nozzle is used to measure the flow of air to an engine The atmosphere is at 100 kPa 25C The nozzle used has a minimum area of 2000 mm2 and the coefficient of discharge is 095 A pressure difference across the nozzle is measured to 25 kPa Find the mass flow rate assuming incompressible flow Also find the mass flow rate assuming compressible adiabatic flow Assume Vi 0 vi RTiPi 0287 29815100 08557 mA3E Akg Incompressible flow Ves A2E A2 hi hes viPi Pe 21393 kJkg Ves A 2 1000 21393EA 6541 ms AmE As AVesvi 2000 10A6E A 654108557 0153 kgs AmE Aa CD AmE As 01454 kgs Compressible flow Tes Ti PePiAk1kE A 29815975100A02857E A 296 K h CpT 10035 215 21575 Ves A2E A2 Ves A 2 1000 21575EA 6569 ms ves 0287 296975 08713 mA3E Akg AmE As AVesves 2000 10A6E A 656908713 01508 kgs AmE Aa CD AmE As 01433 kgs Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Review Problems Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1582 Atmospheric air is at 20C 100 kPa with zero velocity An adiabatic reversible compressor takes atmospheric air in through a pipe with crosssectional area of 01 m2 at a rate of 1 kgs It is compressed up to a measured stagnation pressure of 500 kPa and leaves through a pipe with crosssectional area of 001 m2 What are the required compressor work and the air velocity static pressure and temperature in the exit pipeline CV compressor out to standing air and exit to stagnation point AmE A ho1 AW E A c AmE Ah VA2E A2ex AmE Ahoex AmE Aso1 AmE Asoex Proex Pro1 PstexPo1 1028500100 514 Toex 463 K hoex 46538 kJkg ho1 20945 kJkg AW E Ac AmE Ahoex ho1 146538 20945 2559 kW Pex PoexTexToexAkk1E A Tex Toex VA 2 AEex E2Cp AmE A 1 kgs ρAVex PexAVexRTex Now select 1 unknown amongst Pex Tex Vex and write the continuity eq Am E A and solve the nonlinear equation Say use Tex then Vex 2CpToex Tex AmE A 1 kgs PoexTexToexAkk1E AA 2CpToex TexRTex solve for TexToex close to 1 Tex 4626 K Vex 283 ms Pex 4986 kPa Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1583 The nozzle in Problem 1546 will have a throat area of 0001272 mA2E A and an exit area 350 times as large Suppose the back pressure is raised to 14 MPa and that the flow remains isentropic except for a normal shock wave Verify that the shock mach number MAxE A is close to 2 and find the exit mach number the temperature and the mass flow rate through the nozzle Since a shock is present we must have choked flow with MAE A 1 then the mass flow rate is maximum and follows Eq1542 AmE AAAE A Po kRTo Ak 1 2E A A k12k1E 2000 kPa 1000A 140287 42315 1000EA ms 12A 2408E A 392972 kgsmA2E AmE A AAE A 392972 kgsmA2E A 5 kgs Assume Mx 2 then Table A13 My 057735 PoyPox 072087 AEAAx AEE 350 From A12 Mx 2 AxAAx AE 16875 upstream My 057735 AyAAy AEE 12225 downstream Now since Ay Ax then see Ex 159 page 734 AEAAy AEE AEAAx AEE AyAAy AEE AAx AEE Ax 35 1222516875 25356 From A12 with AAAE A then ME 0246 PEPoy 095728 PE 095728 072087 20 138 MPa OK close to the 14 MPa TE 09876 42315 418 K Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1584 At what Mach number will the normal shock occur in the nozzle of Problem 1553 if the back pressure is 14 MPa trial and error on Mx Relate the inlet and exit conditions to the shock conditions with reversible flow before and after the shock It becomes trial and error Assume Mx 18 My 06165 PoyPox 08127 AEAA AEx E A2AAE A 00024350001516 16062 AxAAE Ax 1439 AxAAE Ay 11694 AEAA AEy E AEAA AEx EAxAA AEy EAxAA AEx E A1606211694 E1439E A 13053 ME 05189 PEPoy 08323 PE PEPoyPoyPoxPox 08323 08127 2000 1353 kPa 14 MPa So select the mach number a little less Mx 17 My 064055 PoyPox 085573 AxAAE Ax 13376 AxAA AEy E 11446 AEAA AEy E AEAA AEx EAxAA AEy EAxAA AEx E A1606211446 E13376E A 13744 ME 0482 PEPoy 0853 PE PEPoyPoyPoxPox 0853 085573 2000 14599 kPa Now interpolate between the two Mx 1756 and we check My 06266 PoyPox 0832 AxAAE Ax 13926 AxAA AEy E 11586 AEAA AEy E 16062 1158613926 13363 ME 05 PEPoy 0843 PE 0843 0832 2000 14027 kPa OK Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Solution using the Pr or vr functions Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1544 A jet plane travels through the air with a speed of 1000 kmh at an altitude of 6 km where the pressure is 40 kPa and the temperature is 12C Consider the inlet diffuser of the engine where air leaves with a velocity of 100 ms Determine the pressure and temperature leaving the diffuser and the ratio of inlet to exit area of the diffuser assuming the flow to be reversible and adiabatic V 1000 kmh 2778 ms v1 ART PE A A0287 26115 40E A 1874 mA3E Akg h1 26148 kJkg Pr1 06862 ho1 26148 2778A2E A2000 30007 kJkg To1 2997 K Pro1 11107 The ratio of the pressures equals the ratio of the Pr functions when s constant Po1 P Pro1 Pr1 40 1110706862 6474 kPa h2 30007 100A2E A2000 29507 T2 2947 K Pr2 10462 P2 6474 1046211107 61 kPa v2 RT2P2 0287 294761 1386 mA3E Akg A1A2 v1v2V2V1 187413861002778 0487 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1573 Repeat Problem 1544 assuming a diffuser efficiency of 80 V 1000 kmh 2778 ms v1 ART PE A A0287 26115 40E A 1874 mA3E Akg h1 26148 kJkg Pr1 06862 ho1 26148 2778A2E A2000 30007 kJkg To1 2997 K Pro1 11107 Same as problem 1544 except ηD 080 We thus have from 1544 h3 h1 ho1 h1 h3 26148 30007 26148 08 h3 29235 kJkg Pr3 10129 Po2 P3 40 1012906862 5904 kPa Pro2 Pro1 11107 h2 30007 100A2E A2000 29507 kJkg T2 2947 K Pr2 10462 P2 Po2 Pr2 Pro2 5904 1046211107 556 kPa v2 RT2P2 0287 2947556 1521 mA3E Akg A1A2 v1v2V2V1 18741521 1002778 0444 h 0 1 0 2 3 1 2 s ENGLISH UNIT PROBLEMS UPDATED JULY 2013 SOLUTION MANUAL CHAPTER 15 English Units Fundamentals of Thermodynamics Borgnakke Sonntag Fundamentals of Thermodynamics Borgnakke Sonntag 8e Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful CHAPTER 15 CONTENT CHAPTER 15 SUBSECTION PROB NO Stagnation properties 8587 Momentum Equation and Forces 8889 Velocity of Sound 90 Reversible Flow Through a Nozzle 9199 Normal Shocks 100 Nozzles Diffusers and Orifices 101103 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Stagnation properties Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1585E Steam leaves a nozzle with a velocity of 800 fts The stagnation pressure is 100 lbfin2 and the stagnation temperature is 500 F What is the static pressure and temperature h1 ho1 V1 22 12791 8002 2 32174 778 12663 Btu lbm Recall conversion 1 Btulbm 25 037 ft2s2 32174 7781693 Assume isentropic flow s1 s0 17085 Btulbm R h s Computer table P1 88 lbfin2 T 466 F Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1586E Air leaves the compressor of a jet engine at a temperature of 300 F a pressure of 45 lbfin2 and a velocity of 400 fts Determine the isentropic stagnation temperature and pressure ho1 h1 V1 22 40022 32174 778 32 Btulbm Recall conversion 1 Btulbm 25 037 ft2s2 32174 7781693 To1 T 1 ho1 h1Cp 32024 133 To1 T T 300 133 3133 F 773 R Po1 P1 To1T1 k k1 457737596735 4782 lbfin2 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1587E A meteorite melts and burn up at temperatures of 5400 R If it hits air at 075 lbfin2 90 R how high a velocity should it have to reach such temperature Assume we have a stagnation T 5400 R h1 V1 22 hstagn Extrapolating from table F5 hstagn 15156 h1 214 Btulbm V1 22 15156 214 14942 Btulbm V1 2 32174 778 14942 8649 fts Recall conversion 1 Btulbm 25 037 ft2s2 32174 7781693 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Momentum Equation and Forces Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1588E A jet engine receives a flow of 500 fts air at 10 lbfin2 40 F inlet area of 7 ft2 with an exit at 1500 fts 10 lbfin2 1400 R Find the mass flow rate and thrust m ρAV ideal gas ρ PRT m PRTA V 10 psi 144 in2ft2 5334 lbfftlbmR 4997 R 7 ft2 500 fts 1891 lbms Fnet m Vex Vin 32174 lbmfts2lbf 1891 lbms 1500 500 fts 5877 lbf Inlet High P Low P exit cb Fnet The shaft must have axial load bearings to transmit thrust to aircraft Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1589E A water turbine using nozzles is located at the bottom of Hoover Dam 575 ft below the surface of Lake Mead The water enters the nozzles at a stagnation pressure corresponding to the column of water above it minus 20 due to friction The temperature is 60 F and the water leaves at standard atmospheric pressure If the flow through the nozzle is reversible and adiabatic determine the velocity and kinetic energy per kilogram of water leaving the nozzle P ρ gZ gZv 32174 575 0016035 144 32174 249 lbfin2 Pac 08P 1992 lbfin2 and Bernoulli vP Vex 22 Vex 2v P 2g Z 2 32174 fts2 575 ft 1924 fts Vex 2 2 vP gZ 575778 0739 Btulbm Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Velocity of Sound Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1590E Find the speed of sound for air at 15 lbfin2 at the two temperatures of 32 F and 90 F Repeat the answer for carbon dioxide and argon gases We assume all gases can be treated as ideal gas behavior From eq 1528 we have c32 kRT 14 32174 lbmfts2lbf 5334 lbfftlbmR 4917 R 1087 fts c90 14 32174 5334 5497 1149 fts For Carbon Dioxide R 351 lbfftlbmR k 1289 c32 1289 32174 351 4917 846 fts c90 1289 32174 351 5497 8945 fts For Argon R 3868 lbfftlbmR k 1667 c32 1667 32174 3868 4917 1010 fts c90 1667 32174 3868 5497 1068 fts Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1591E A jet flies at an altitude of 40 000 ft where the air is at 40 F 65 psia with a velocity of 625 mih Find the Mach number and the stagnation temperature on the nose From Table F4 k CpCv 14 R 5334 ftlbflbmR Now do the speed of sound from Eq1528 c kRT 14 32174 lbmfts2lbf 5334 lbfftlbmR 4197 R 1004 fts V 625 mih 625 mih 146667 fts mih 917 fts M V c 9171004 0913 h0 h1 V1 22 T0 T1 V1 2 2Cp 40 9172 2 024 25037 2997 F Recall conversion 1 Btulbm 25 037 ft2s2 32174 7781693 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Flow Through Nozzles Shocks Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1592E Steam flowing at 50 fts 200 psia 600 F expands to 150 psia in a converging nozzle Find the exit velocity and area ratio Ae Ai Solve the problem with the steam tables Inlet state vi 3058 ft3lbm hi 132205 Btulbm si 16767 BtulbmR Exit state Pese si ve 38185 ft3lbm he 129069 Btulbm Energy Eq Vi 2 2 hi Ve 2 2 he Ve 2 Vi 2 2hi he Ve 50 50 2 25037 132205 129069 1254 fts Recall conversion 1 Btulbm 25 037 ft2s2 32174 7781693 Same mass flow rate so AeAi veviViVe 38185 3058 50 1254 00498 If we solved as ideal gas with constant specific heat we get k 1327 Te Ti PePi k1k 10597 150200 02464 9872 R Ve Vi 2 2CpTi Te 50 50 2 0447 2503710597 9872 12749 fts AeAi veviViVe PiPe 1kViVe 200 150 07536 50 12749 00487 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1593E A convergent nozzle has a minimum area of 1 ft2 and receives air at 25 lbfin2 1800 R flowing with 330 fts What is the back pressure that will produce the maximum flow rate and find that flow rate E A 2 k1E AA k k1 E A 0528 Critical Pressure Ratio P Po Find Po Cp 463445 44979450 0273 BtulbmR from table F5 h0 h1 VA1 2 AE2 T0 Ti VA2E A2Cp T0 1800 EA 330A2 A2 E25 037 0273E A 180797 TAE A 08333 To 15066 R Recall conversion 1 Btulbm 25 037 ftA2E AsA2E A 32174 7781693 P0 Pi T0TiAkk1E A 25 1807971800A35E A 2539 lbfinA2E PAE A 0528 Po 0528 2539 13406 lbfinA2E ρAE A EA PA A ERTA AE A A 13406 144 5334 15066E A 0024 lbmftA3E V c EA kRTA AEA A 14 5334 15066 32174EA 19026 fts AmE A ρAV 0024 1 19026 4566 lbms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1594E A jet plane travels through the air with a speed of 600 mih at an altitude of 20000 ft where the pressure is 575 lbfinA2E A and the temperature is 25 F Consider the diffuser of the engine where air leaves at with a velocity of 300 fts Determine the pressure and temperature leaving the diffuser and the ratio of inlet to exit area of the diffuser assuming the flow to be reversible and adiabatic V 600 mih 880 fts v1 5334 48467575 144 31223 ftA3E Albm h1 11591 Btulbm ho1 11591 880A2E A2 25 037 13138 Btulbm Recall conversion 1 Btulbm 25 037 ftA2E AsA2E A 32174 7781693 Table F5 To1 5492 R Po1 P1 To1T1Akk1E A 575 549248467A35E A 89 lbfinA2E h2 13138 300A2E A2 32174 778 12958 Btulbm T2 542 R P2 Po1 T2To1Akk1E A 89 5425492A35E A 85 lbfinA2E A v2 5334 54285 144 2362 ftA3E Albm A1A2 v1v2V2V1 312232362300880 045 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1595E An air flow at 90 psia 1100 R M 03 flows into a convergentdivergent nozzle with M 1 at the throat Assume a reversible flow with an exit area twice the throat area and find the exit pressure and temperature for subsonic exit flow to exist To find these properties we need the stagnation properties from the inlet state From Table A12 Mi 03 PiPo 093947 TiTo 098232 Po 90 093947 958 psia To 1100 098232 11198 R This flow is case c in Figure 1513 From Table A12 AEAAE A 2 PEPo 09360 TETo 098127 PE 09360 Po 0936 958 8967 psia TE 098127 To 098127 11198 1099 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1596E Air is expanded in a nozzle from 300 lbfin2 1100 R to 30 lbfinA2E A The mass flow rate through the nozzle is 10 lbms Assume the flow is reversible and adiabatic and determine the throat and exit areas for the nozzle Velocity Density Area Mach 300 psia 30 psia P PAE A Po A 2 k1 E A k k1 E 300 05283 1585 lbfinA2E A TAE A To 2k1 1100 08333 9166 R vAE A RTAE APAE A 5334 91661585 144 21421 ftA3E Albm The critical speed of sound is cAE A A kRT EA A 14 32174 5334 9166EA 1484 fts AAE A AmE AvAE AcAE A 10 214211484 00144 ftA2E A P2Po 30300 01 Table A11 MA AE2 E 1701 V2cAE A We used the column in Table A12 with mach no based on throat speed of sound V2 1701 1484 2524 fts T2 9166 05176 4744 R v2 RT2P2 5334 474430 144 58579 ftA3E Albm A2 AmE Av2V2 10 58579 2524 00232 ftA2E A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1597E A 50ft3 uninsulated tank contains air at 150 lbfinA2E A 1000 R The tank is now discharged through a small convergent nozzle to the atmosphere at 147 lbfinA2E A while heat transfer from some source keeps the air temperature in the tank at 1000 R The nozzle has an exit area of 2 104 ftA2E A a Find the initial mass flow rate out of the tank b Find the mass flow rate when half the mass has been discharged c Find the mass of air in the tank and the mass flow rate out of the tank when the nozzle flow changes to become subsonic P AIR e cb PBPo 147150 0098 PAE APocrit 05283 a The flow is choked max possible flow rate ME 1 PE 05283 150 79245 lbfinA2E A TE TAE A 08333 1000 8333 R VE c A kRT EA A 14 5334 8333 32174EA 1415 fts vE RTAE APE 5334 833379245 144 3895 ftA3E Albm Mass flow rate is AmE A1 AVEvE 2 10A4E A 14153895 00727 lbms b m1 P1VRT1 150 50 1445334 1000 20247 lbm m2 m12 10124 lbm P2 P12 75 lbfinA2E A T2 T1 PBP2 14775 0196 PAE APocrit The flow is choked and the velocity is the same as in a PE 05283 75 39623 lbfinA2E A ME 1 AmE A2 AVEPERTE A2 104 1415 39623 144 E5334 1000E A 00303 lbms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful c Flow changes to subsonic when the pressure ratio reaches critical PBPo 05283 P3 27825 lbfinA2E A m3 m1P3P1 3756 lbm T3 T1 VE 1415 fts AmE A3 AVEPERTE A2 104 1415 27825 144 E5334 1000E A 002125 lbms Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1598E A flow of helium flows at 75 psia 900 R with 330 fts into a convergentdivergent nozzle Find the throat pressure and temperature for reversible flow and M 1 at the throat We need to find the stagnation properties first k 1667 T0 T1 VA1 2 AE2Cp 900 330A2E A2 25037 124 901754 R Recall conversion 1 Btulbm 25 037 ftA2E AsA2E A 32174 7781693 P0 P1 T0T1Akk1E A 75 901754900A 25E A 75366 psia From the analysis we get Eqs153738 PAE A P0 A 2 k 1 kk1E A 75366 A 2 1667 1 25E A 367 psia TAE A T0 A 2 k 1E A 901754 A 2 1667 1E A 6762 R Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 1599E The products of combustion enter a nozzle of a jet engine at a total pressure of 18 lbfinA2E A and a total temperature of 1200 F The atmospheric pressure is 675 lbfinA2E A The nozzle is convergent and the mass flow rate is 50 lbms Assume the flow is adiabatic Determine the exit area of the nozzle Pcrit P2 18 05283 95 lbfinA2E A Pamb The flow is then choked T2 1660 08333 1382 R V2 c2 A kRTEA A 14 32174 5334 1382EA 1822 fts v2 5334 138295 144 539 ftA3E Albm A2 AmE A v2 V2 50 5391822 1479 ftA2E A Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 15100E A normal shock in air has upstream total pressure of 75 psia stagnation temperature of 900 R and Mx 14 Find the downstream stagnation pressure From the normal shock relations in Section 158 found in Table A13 we get Mx 14 Po yPo x 095819 Po y 095819 Po x 095819 75 7186 psi Remark The stagnation temperature would be unchanged energy equation Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Nozzles Diffusers and Orifices Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 15101E Air enters a diffuser with a velocity of 600 fts a static pressure of 10 lbfinA2E A and a temperature of 20 F The velocity leaving the diffuser is 200 fts and the static pressure at the diffuser exit is 117 lbfinA2E A Determine the static temperature at the diffuser exit and the diffuser efficiency Compare the stagnation pressures at the inlet and the exit To1 T1 A V1 2 AE 2gcCp E 480 A6002E A2 25 037 024 510 R Recall conversion 1 Btulbm 25 037 ftA2E AsA2E A 32174 7781693 To1 T1 T1 Ak1 kE A Po1 P1 P1 Po1 P1 21875 Po1 122 lbfinA2E A To2 To1 T2 To2 VA2 2 AE2Cp 510 200A2E A2 25 037 024 5067 R To1 T1 T1 Ak1 kE A Po1 P1 P1 Po1 P1 21875 Po1 122 lbfinA2E A To2 T2 T2 Ak1 kE A Po2 P2 P2 Po2 P2 0267 Po2 1197 lbfinA2E A Texs T1 Po2P1Ak1kE A 480 R 10528 5053 R ηD Texs T1 To1 T1 A5053 480 51 480E A 0844 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 15102E Repeat Problem 1594 assuming a diffuser efficiency of 80 From solution to 1594 h1 11591 Btulbm v1 31223 ftA3E Albm ho1 11591 880A2E A2 25 037 13138 Btulbm Recall conversion 1 Btulbm 25 037 ftA2E AsA2E A 32174 7781693 Table F5 To1 5492 R h 0 1 0 2 3 1 2 s ηD h3 h1ho1 h1 08 h3 12829 Btulbm T3 53629 R Po2 P3 P1 Τ3Τ1Akk1E A 575 5362948467A35E A 8194 lbfinA2E A To2 To1 5492 R h2 13138 300A2E A2 25 037 12958 Btulbm T2 542 R P2 Po2 T2To1Akk1E A 8194 5425492A35E A 7824 lbfinA2E A v2 A5334 542 7824 144E A 2566 ftA3E Albm A1A2 v1V2v2V1 31223 3002566 880 0415 Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 15103E A convergent nozzle with exit diameter of 1 in has an air inlet flow of 68 F 147 lbfinA2E A stagnation conditions The nozzle has an isentropic efficiency of 95 and the pressure drop is measured to 20 in water column Find the mass flow rate assuming compressible adiabatic flow Repeat calculation for incompressible flow Convert P to lbfinA2E P 20 in H2O 20 003613 07226 lbfinA2E T0 68 F 5277 R P0 147 lbfinA2E Assume inlet Vi 0 Pe P0 P 147 07226 13977 lbfinA2E Te T0 Pe P0 A k1 k E A 5277 A13977 147E AA02857E A 52015 R VAe 2 AE2 hi he Cp Ti Te 024 5277 52015 1812 Btulbm VAe ac 2 AE E2 η VAe 2 AE2 095 1812 17214 Btulbm Ve ac A 2 25 037 17214EA 2936 fts Recall conversion 1 Btulbm 25 037 ftA2E AsA2E A 32174 7781693 Te ac Ti A VAe ac 2 AEAE E2 Cp E 5277 A17214 024E A 52053 R ρe ac Pe RTe ac A 13977 144 5334 52053E A 007249 lbmftA3E AmE A ρAV 007249 Aπ 4E A A 1 12E AA2E A 2936 0116 lbms Incompressible ρi P0 RT0 A 147 144 5334 5277E A 00752 lbmftA3E VAe 2 AE2 vi Pi Pe P ρi A07226 144 00752 778E A 17785 Btulbm VAe ac 2 AE E2 η VAe 2 AE2 095 17785 16896 Btulbm Ve ac A 2 25 037 16896EA 29084 fts AmE A ρAV 00752 Aπ 4E A A 1 12E AA2E A 29084 0119 lbms