Solucionario-Termodinamica-Cengel-Boles-7ma-Edicion

LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS httpsolucionariosdelibrosblogspotcom 11 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 1 INTRODUCTION AND BASIC CONCEPTS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 12 Thermodynamics 11C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy and thus the bicyclist picks up speed There is no creation of energy and thus no violation of the conservation of energy principle 12C A car going uphill without the engine running would increase the energy of the car and thus it would be a violation of the first law of thermodynamics Therefore this cannot happen Using a level meter a device with an air bubble between two marks of a horizontal water tube it can shown that the road that looks uphill to the eye is actually downhill 13C There is no truth to his claim It violates the second law of thermodynamics preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 13 Mass Force and Units 14C The pound mentioned here must be lbf since thrust is a force and the lbf is the force unit in the English system You should get into the habit of never writing the unit lb but always use either lbm or lbf as appropriate since the two units have different dimensions 15C In this unit the word light refers to the speed of light The lightyear unit is then the product of a velocity and time Hence this product forms a distance dimension and unit 16C There is no acceleration thus the net force is zero in both cases 17E The weight of a man on earth is given His weight on the moon is to be determined Analysis Applying Newtons second law to the weight force gives 210 5 lbm 1lbf 174 lbm fts 32 10 fts 32 210 lbf 2 2 g W m mg W Mass is invariant and the man will have the same mass on the moon Then his weight on the moon will be 358 lbf 2 2 174 lbm fts 32 1lbf 210 5 lbm 5 47 fts mg W 18 The interior dimensions of a room are given The mass and weight of the air in the room are to be determined Assumptions The density of air is constant throughout the room Properties The density of air is given to be ρ 116 kgm3 ROOM AIR 6X6X8 m3 Analysis The mass of the air in the room is kg 3341 8 m 116 kgm 6 6 3 3 ρV m Thus 3277 N 2 2 kg ms 1 1 N ms kg981 3341 mg W preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 14 19 The variation of gravitational acceleration above the sea level is given as a function of altitude The height at which the weight of a body will decrease by 05 is to be determined 0 z Analysis The weight of a body at the elevation z can be expressed as W mg m z 9 807 332 10 6 In our case 0 995 9 81 0 995 0 995 m mg W W s s Substituting 14770 m 14774 m 3 32 10 9 81 0 995 9 81 6 z z Sea level 110 The mass of an object is given Its weight is to be determined Analysis Applying Newtons second law the weight is determined to be 1920 N ms 69 200 kg 2 mg W 111E The constantpressure specific heat of air given in a specified unit is to be expressed in various units Analysis Applying Newtons second law the weight is determined in various units to be F Btulbm 0240 C kcalkg 0240 C Jg 1005 kJkg K 1005 C 41868 kJkg F 1Btulbm C kJkg 1005 41868 kJ 1kcal C kJkg 1005 1000 g 1kg 1kJ C 1000 J kJkg 1005 C 1kJkg 1kJkg K C kJkg 1005 p p p p c c c c preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15 112 A rock is thrown upward with a specified force The acceleration of the rock is to be determined Analysis The weight of the rock is 2937 N kg ms 1 1 N 3 kg979 ms 2 2 W mg Then the net force that acts on the rock is 170 6 N 2937 200 down up net F F F Stone From the Newtons second law the acceleration of the rock becomes 569 ms2 1 N 1 kg ms 3 kg 1706 N 2 m F a preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 16 113 Problem 112 is reconsidered The entire EES solution is to be printed out including the numerical results with proper units Analysis The problem is solved using EES and the solution is given below The weight of the rock is Wmg m3 kg g979 ms2 The force balance on the rock yields the net force acting on the rock as Fup200 N Fnet Fup Fdown FdownW The acceleration of the rock is determined from Newtons second law Fnetma To Run the program press F2 or select Solve from the Calculate menu SOLUTION a5688 ms2 Fdown2937 N Fnet1706 N Fup200 N g979 ms2 m3 kg W2937 N m kg a ms2 1 2 3 4 5 6 7 8 9 10 1902 9021 5688 4021 3021 2354 1878 1521 1243 1021 1 2 3 4 5 6 7 8 9 10 0 40 80 120 160 200 m kg a ms 2 preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 17 114 During an analysis a relation with inconsistent units is obtained A correction is to be found and the probable cause of the error is to be determined Analysis The two terms on the righthand side of the equation E 25 kJ 7 kJkg do not have the same units and therefore they cannot be added to obtain the total energy Multiplying the last term by mass will eliminate the kilograms in the denominator and the whole equation will become dimensionally homogeneous that is every term in the equation will have the same unit Discussion Obviously this error was caused by forgetting to multiply the last term by mass at an earlier stage 115 A resistance heater is used to heat water to desired temperature The amount of electric energy used in kWh and kJ are to be determined Analysis The resistance heater consumes electric energy at a rate of 4 kW or 4 kJs Then the total amount of electric energy used in 2 hours becomes Total energy Energy per unit timeTime interval 4 kW2 h 8 kWh Noting that 1 kWh 1 kJs3600 s 3600 kJ Total energy 8 kWh3600 kJkWh 28800 kJ Discussion Note kW is a unit for power whereas kWh is a unit for energy 116 A gas tank is being filled with gasoline at a specified flow rate Based on unit considerations alone a relation is to be obtained for the filling time Assumptions Gasoline is an incompressible substance and the flow rate is constant Analysis The filling time depends on the volume of the tank and the discharge rate of gasoline Also we know that the unit of time is seconds Therefore the independent quantities should be arranged such that we end up with the unit of seconds Putting the given information into perspective we have t s V L and V Ls It is obvious that the only way to end up with the unit s for time is to divide the tank volume by the discharge rate Therefore the desired relation is V t V Discussion Note that this approach may not work for cases that involve dimensionless and thus unitless quantities preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 18 117 A pool is to be filled with water using a hose Based on unit considerations a relation is to be obtained for the volume of the pool Assumptions Water is an incompressible substance and the average flow velocity is constant Analysis The pool volume depends on the filling time the crosssectional area which depends on hose diameter and flow velocity Also we know that the unit of volume is m3 Therefore the independent quantities should be arranged such that we end up with the unit of seconds Putting the given information into perspective we have V m3 is a function of t s D m and V ms It is obvious that the only way to end up with the unit m3 for volume is to multiply the quantities t and V with the square of D Therefore the desired relation is V CD2Vt where the constant of proportionality is obtained for a round hose namely C π4 so that V πD24Vt Discussion Note that the values of dimensionless constants of proportionality cannot be determined with this approach 118 It is to be shown that the power needed to accelerate a car is proportional to the mass and the square of the velocity of the car and inversely proportional to the time interval Assumptions The car is initially at rest Analysis The power needed for acceleration depends on the mass velocity change and time interval Also the unit of power W is watt W which is equivalent to W Js Nms kgms2ms kgm2s3 Therefore the independent quantities should be arranged such that we end up with the unit kgm2s3 for power Putting the given information into perspective we have W kgm2s3 is a function of m kg V ms and t s It is obvious that the only way to end up with the unit kgm2s3 for power is to multiply mass with the square of the velocity and divide by time Therefore the desired relation is 2 is proportional to W m V t or t CmV W 2 where C is the dimensionless constant of proportionality whose value is ½ in this case Discussion Note that this approach cannot determine the numerical value of the dimensionless numbers involved preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 19 Systems Properties State and Processes 119C This system is a region of space or open system in that mass such as air and food can cross its control boundary The system can also interact with the surroundings by exchanging heat and work across its control boundary By tracking these interactions we can determine the energy conversion characteristics of this system 120C The system is taken as the air contained in the pistoncylinder device This system is a closed or fixed mass system since no mass enters or leaves it 121C Any portion of the atmosphere which contains the ozone layer will work as an open system to study this problem Once a portion of the atmosphere is selected we must solve the practical problem of determining the interactions that occur at the control surfaces which surround the systems control volume 122C Intensive properties do not depend on the size extent of the system but extensive properties do 123C If we were to divide the system into smaller portions the weight of each portion would also be smaller Hence the weight is an extensive property 124C If we were to divide this system in half both the volume and the number of moles contained in each half would be onehalf that of the original system The molar specific volume of the original system is N v V and the molar specific volume of one of the smaller systems is N N V V v 2 2 which is the same as that of the original system The molar specific volume is then an intensive property 125C For a system to be in thermodynamic equilibrium the temperature has to be the same throughout but the pressure does not However there should be no unbalanced pressure forces present The increasing pressure with depth in a fluid for example should be balanced by increasing weight 126C A process during which a system remains almost in equilibrium at all times is called a quasiequilibrium process Many engineering processes can be approximated as being quasiequilibrium The work output of a device is maximum and the work input to a device is minimum when quasiequilibrium processes are used instead of nonquasiequilibrium processes preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 110 127C A process during which the temperature remains constant is called isothermal a process during which the pressure remains constant is called isobaric and a process during which the volume remains constant is called isochoric 128C The state of a simple compressible system is completely specified by two independent intensive properties 129C The pressure and temperature of the water are normally used to describe the state Chemical composition surface tension coefficient and other properties may be required in some cases As the water cools its pressure remains fixed This cooling process is then an isobaric process 1 30C When analyzing the acceleration of gases as they flow through a nozzle the proper choice for the system is the volume within the nozzle bounded by the entire inner surface of the nozzle and the inlet and outlet crosssections This is a control volume since mass crosses the boundary 131C A process is said to be steadyflow if it involves no changes with time anywhere within the system or at the system boundaries preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 111 132 The variation of density of atmospheric air with elevation is given in tabular form A relation for the variation of density with elevation is to be obtained the density at 7 km elevation is to be calculated and the mass of the atmosphere using the correlation is to be estimated Assumptions 1 Atmospheric air behaves as an ideal gas 2 The earth is perfectly sphere with a radius of 6377 km and the thickness of the atmosphere is 25 km Properties The density data are given in tabular form as 0 5 10 15 20 25 0 02 04 06 08 1 12 14 z km ρ kgm 3 r km z km ρ kgm3 6377 0 1225 6378 1 1112 6379 2 1007 6380 3 09093 6381 4 08194 6382 5 07364 6383 6 06601 6385 8 05258 6387 10 04135 6392 15 01948 6397 20 008891 6402 25 004008 Analysis Using EES 1 Define a trivial function rho az in equation window 2 select new parametric table from Tables and type the data in a twocolumn table 3 select Plot and plot the data and 4 select plot and click on curve fit to get curve fit window Then specify 2nd order polynomial and enteredit equation The results are ρz a bz cz2 120252 0101674z 00022375z2 for the unit of kgm3 or ρz 120252 0101674z 00022375z2109 for the unit of kgkm3 where z is the vertical distance from the earth surface at sea level At z 7 km the equation would give ρ 060 kgm3 b The mass of atmosphere can be evaluated by integration to be 5 4 2 3 2 2 2 4 2 4 4 5 4 0 3 2 0 0 2 0 0 2 0 2 0 2 0 2 0 2 0 2 0 ch h cr b h cr br a h br a r h ar dz z r z r cz bz a dz z r cz bz a dV m h z h z V π π π ρ where r0 6377 km is the radius of the earth h 25 km is the thickness of the atmosphere and a 120252 b 0101674 and c 00022375 are the constants in the density function Substituting and multiplying by the factor 109 for the density unity kgkm3 the mass of the atmosphere is determined to be m 50921018 kg Discussion Performing the analysis with excel would yield exactly the same results EES Solution for final result a12025166 b010167 c00022375 r6377 h25 m4piar2hr2abrh22a2brcr2h33b2crh44ch551E9 preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 112 Temperature 133C The zeroth law of thermodynamics states that two bodies are in thermal equilibrium if both have the same temperature reading even if they are not in contact 134C They are Celsius C and kelvin K in the SI and fahrenheit F and rankine R in the English system 135C Probably but not necessarily The operation of these two thermometers is based on the thermal expansion of a fluid If the thermal expansion coefficients of both fluids vary linearly with temperature then both fluids will expand at the same rate with temperature and both thermometers will always give identical readings Otherwise the two readings may deviate 136 A temperature is given in C It is to be expressed in K Analysis The Kelvin scale is related to Celsius scale by TK TC 273 Thus TK 37C 273 310 K 137E The temperature of air given in C unit is to be converted to F and R unit Analysis Using the conversion relations between the various temperature scales 762 R 302 F 460 302 460 F R 32 150 81 32 C 81 F T T T T 138 A temperature change is given in C It is to be expressed in K Analysis This problem deals with temperature changes which are identical in Kelvin and Celsius scales Thus TK TC 45 K preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 113 139E The flash point temperature of engine oil given in F unit is to be converted to K and R units Analysis Using the conversion relations between the various temperature scales K 457 R 823 81 823 18 R K 460 363 460 F R T T T T 140E The temperature of ambient air given in C unit is to be converted to F K and R units Analysis Using the conversion relations between the various temperature scales 41967 R 23315 K C 40 45967 40 27315 40 32 40 81 C 40 T T T 141E The change in water temperature given in F unit is to be converted to C K and R units Analysis Using the conversion relations between the various temperature scales 10 R 56 K C 56 10 F 81 10 81 10 T T T 142E A temperature range given in F unit is to be converted to C unit and the temperature difference in F is to be expressed in K C and R Analysis The lower and upper limits of comfort range in C are 183C 81 32 65 81 32 F C T T 239C 81 32 75 81 32 F C T T A temperature change of 10F in various units are K 56 C 56 R 10 C K 81 10 18 F C F R T T T T T T preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 114 Pressure Manometer and Barometer 143C The pressure relative to the atmospheric pressure is called the gage pressure and the pressure relative to an absolute vacuum is called absolute pressure 144C The blood vessels are more restricted when the arm is parallel to the body than when the arm is perpendicular to the body For a constant volume of blood to be discharged by the heart the blood pressure must increase to overcome the increased resistance to flow 145C No the absolute pressure in a liquid of constant density does not double when the depth is doubled It is the gage pressure that doubles when the depth is doubled 146C If the lengths of the sides of the tiny cube suspended in water by a string are very small the magnitudes of the pressures on all sides of the cube will be the same 147C Pascals principle states that the pressure applied to a confined fluid increases the pressure throughout by the same amount This is a consequence of the pressure in a fluid remaining constant in the horizontal direction An example of Pascals principle is the operation of the hydraulic car jack 148E The pressure given in psia unit is to be converted to kPa Analysis Using the psia to kPa units conversion factor 1034 kPa 1psia 6 89 5 kPa 150 psia P 149 The pressure in a tank is given The tanks pressure in various units are to be determined Analysis Using appropriate conversion factors we obtain a 1500 kNm2 1kPa 1kNm 1500 kPa 2 P b 1500000 kgm s2 1kN 1000 kg ms 1kPa 1kNm 1500 kPa 2 2 P c 1500000000 kgkm s2 1km 1000 m 1kN 1000 kg ms 1kPa 1kNm 1500 kPa 2 2 P preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 115 150E The pressure in a tank in SI unit is given The tanks pressure in various English units are to be determined Analysis Using appropriate conversion factors we obtain a 31330 lbfft2 1kPa 1500 kPa 20886 lbfft 2 P b 2176 psia 2 2 2 2 lbfin 1 psia 1 in 144 1ft 1kPa 1500 kPa 20886 lbfft P 151E The pressure given in mm Hg unit is to be converted to psia Analysis Using the mm Hg to kPa and kPa to psia units conversion factors 290 psia 6 895 kPa 1psia 1mm Hg 1500 mm Hg 01333 kPa P 152 The pressure given in mm Hg unit is to be converted to kPa Analysis Using the mm Hg to kPa units conversion factor 1666 kPa 1mm Hg 1250 mm Hg 01333 kPa P preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 116 153 The pressure in a pressurized water tank is measured by a multifluid manometer The gage pressure of air in the tank is to be determined Assumptions The air pressure in the tank is uniform ie its variation with elevation is negligible due to its low density and thus we can determine the pressure at the airwater interface Properties The densities of mercury water and oil are given to be 13600 1000 and 850 kgm3 respectively Analysis Starting with the pressure at point 1 at the airwater interface and moving along the tube by adding as we go down or subtracting as we go up th e gh ρ terms until we reach point 2 and setting the result equal to Patm since the tube is open to the atmosphere gives Patm gh gh gh P 3 mercury 2 oil 1 water 1 ρ ρ ρ PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Solving for P1 3 mercury 2 oil 1 water atm 1 gh gh gh P P ρ ρ ρ or 2 oil water 1 mercury 3 atm 1 h h h g P P ρ ρ ρ Noting that P1gage P1 Patm and substituting 569 kPa 2 2 3 3 3 2 gage 1 Nm 1000 kPa 1 kg ms 1 1N m 30 850 kgm m 20 1000 kgm 981 ms 13600 kgm 0 46 m P Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly 154 The barometric reading at a location is given in height of mercury column The atmospheric pressure is to be determined Properties The density of mercury is given to be 13600 kgm3 Analysis The atmospheric pressure is determined directly from 1001 kPa 2 2 2 3 atm Nm 1000 kPa 1 kg ms 1 1N kgm 9 81 ms 0 750 m 13600 gh P ρ preparation If you are a student using this Manual you are using it without permission 117 155 The gage pressure in a liquid at a certain depth is given The gage pressure in the same liquid at a different depth is to be determined Assumptions The variation of the density of the liquid with depth is negligible Analysis The gage pressure at two different depths of a liquid can be expressed as 1 1 gh P ρ and 2 2 gh P ρ h2 2 h1 1 Taking their ratio 1 2 1 2 1 2 h h gh gh P P ρ ρ Solving for P2 and substituting gives 126 kPa 3 m 42 kPa 9 m 1 1 2 2 h P h P Discussion Note that the gage pressure in a given fluid is proportional to depth 156 The absolute pressure in water at a specified depth is given The local atmospheric pressure and the absolute pressure at the same depth in a different liquid are to be determined Assumptions The liquid and water are incompressible Properties The specific gravity of the fluid is given to be SG 085 We take the density of water to be 1000 kgm3 Then density of the liquid is obtained by multiplying its specific gravity by the density of water 3 3 850 kgm 085100 0 kgm SG 2 ρH O ρ Analysis a Knowing the absolute pressure the atmospheric pressure can be determined from Patm h P 960 kPa 2 2 3 atm Nm 1000 1 kPa 1000 kgm 981 ms 5 m kPa 145 gh P P ρ b The absolute pressure at a depth of 5 m in the other liquid is 1377 kPa 2 2 3 atm Nm 1000 1 kPa 850 kgm 981 ms 5 m kPa 960 gh P P ρ Discussion Note that at a given depth the pressure in the lighter fluid is lower as expected preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 118 157E It is to be shown that 1 kgfcm2 14223 psi Analysis Noting that 1 kgf 980665 N 1 N 022481 lbf and 1 in 254 cm we have 2 20463 lbf 1 N 022481 lbf 980665 N 980665 N 1 kgf and 14223 psi 2 2 2 2 2 14223 lbfin 1 in 254 cm 2 20463 lbfcm 2 20463 lbfcm 1 kgfcm 158E The pressure in chamber 3 of the twopiston cylinder shown in the figure is to be determined Analysis The area upon which pressure 1 acts is 2 2 2 1 1 7 069 in 4 3in 4 π π D A F1 F3 F2 and the area upon which pressure 2 acts is 2 2 2 2 2 1 767 in 4 in 51 4 π π D A The area upon which pressure 3 acts is given by 2 2 1 3 5 302 in 1 767 7 069 A A A The force produced by pressure 1 on the piston is then 1060 lbf 7 069 in 1psia 1lbfin 150 psia 2 2 1 1 1 P A F while that produced by pressure 2 is 441 8 lbf 250 psia 1 767 in 2 2 2 1 P A F According to the vertical force balance on the piston free body diagram 618 3 lbf 441 8 1060 2 1 3 F F F Pressure 3 is then 117 psia 2 3 3 3 302 in 5 618 3 lbf A F P preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 119 159 The pressure in chamber 1 of the twopiston cylinder shown in the figure is to be determined Analysis Summing the forces acting on the piston in the vertical direction gives 1 1 2 1 3 2 2 1 3 2 P A A A P A P F F F F1 F3 F2 which when solved for P1 gives 1 2 3 1 2 2 1 1 A A P A P A P since the areas of the piston faces are given by the above equation becomes D2 4 A π 908 kPa 2 2 2 1 2 3 2 1 2 2 1 10 4 700 kPa 1 10 2000 kPa 4 1 D D P D D P P 160 The mass of a woman is given The minimum imprint area per shoe needed to enable her to walk on the snow without sinking is to be determined Assumptions 1 The weight of the person is distributed uniformly on the imprint area of the shoes 2 One foot carries the entire weight of a person during walking and the shoe is sized for walking conditions rather than standing 3 The weight of the shoes is negligible Analysis The mass of the woman is given to be 70 kg For a pressure of 05 kPa on the snow the imprint area of one shoe must be 137 m2 2 2 2 Nm 1000 kPa 1 kg ms 1 1N 05 kPa kg981 ms 70 P mg P W A Discussion This is a very large area for a shoe and such shoes would be impractical to use Therefore some sinking of the snow should be allowed to have shoes of reasonable size preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 120 161 The vacuum pressure reading of a tank is given The absolute pressure in the tank is to be determined Properties The density of mercury is given to be ρ 13590 kgm3 Analysis The atmospheric or barometric pressure can be expressed as 30 kPa Pabs 100 0 kPa Nm 1000 kPa 1 kg ms 1 1 N 13590 kgm 9807 ms 0750 m 2 2 2 3 atm gh P ρ Patm 750 mmHg Then the absolute pressure in the tank becomes 700 kPa 30 1000 vac atm abs P P P 162E The vacuum pressure given in kPa unit is to be converted to various units Analysis Using the definition of vacuum pressure 18 kPa 80 98 applicable for pressures below atmospheri c pressure not vac atm abs gage P P P P Then using the conversion factors 18 kNm2 1 kPa 1kNm kPa 18 2 Pabs lbfin2 261 6895 kPa 1lbfin kPa 18 2 Pabs 261 psi 6895 kPa 1psi 18 kPa Pabs 135 mm Hg 01333 kPa 1mm Hg 18 kPa Pabs preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 121 163 A mountain hiker records the barometric reading before and after a hiking trip The vertical distance climbed is to be determined 630 mbar h Assumptions The variation of air density and the gravitational acceleration with altitude is negligible Properties The density of air is given to be ρ 120 kgm3 Analysis Taking an air column between the top and the bottom of the mountain and writing a force balance per unit base area we obtain 740 mbar 0630 bar 0740 Nm 100000 bar 1 kg ms 1 1 N kgm 981 ms 120 2 2 2 3 top bottom air top bottom air h P P gh P P A W ρ It yields h 934 m which is also the distance climbed 164 A barometer is used to measure the height of a building by recording reading at the bottom and at the top of the building The height of the building is to be determined Assumptions The variation of air density with altitude is negligible Properties The density of air is given to be ρ 118 kgm3 The density of mercury is 13600 kgm3 675 mmHg Analysis Atmospheric pressures at the top and at the bottom of the building are h 695 mmHg 9272 kPa Nm 1000 kPa 1 kg ms 1 1N kgm 981 ms 0695 m 13600 9006 kPa Nm 1000 kPa 1 kg ms 1 1 N kgm 981 ms 0675 m 13600 2 2 2 3 bottom bottom 2 2 2 3 top top gh P ρgh P ρ Taking an air column between the top and the bottom of the building and writing a force balance per unit base area we obtain 9006 kPa 9272 Nm 1000 kPa 1 kg ms 1 1 N kgm 981 ms 118 2 2 2 3 top bottom air top bottom air h P P gh P P A W ρ It yields h 231 m which is also the height of the building preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 122 165 Problem 164 is reconsidered The entire EES solution is to be printed out including the numerical results with proper units Analysis The problem is solved using EES and the solution is given below Pbottom695 mmHg Ptop675 mmHg g981 ms2 local acceleration of gravity at sea level rho118 kgm3 DELTAPabsPbottomPtopCONVERTmmHg kPa kPa Delta P reading from the barometers converted from mmHg to kPa DELTAPh rhoghConvertPa kPa Delta P due to the air fluid column height h between the top and bottom of the building DELTAPabsDELTAPh SOLUTION DELTAPabs2666 kPa DELTAPh2666 kPa g981 ms2 h2303 m Pbottom695 mmHg Ptop675 mmHg rho118 kgm3 166 A man is standing in water vertically while being completely submerged The difference between the pressures acting on the head and on the toes is to be determined Assumptions Water is an incompressible substance and thus the density does not change with depth htoe hhead Properties We take the density of water to be ρ 1000 kgm3 Analysis The pressures at the head and toes of the person can be expressed as head atm head gh P P ρ and toe atm toe gh P P ρ where h is the vertical distance of the location in water from the free surface The pressure difference between the toes and the head is determined by subtracting the first relation above from the second head toe head toe head toe h g h gh gh P P ρ ρ ρ Substituting 172 kPa 2 2 2 3 head toe Nm 1000 kPa 1 kg ms 1 1N 1000 kgm 981 ms 175 m 0 P P Discussion This problem can also be solved by noting that the atmospheric pressure 1 atm 101325 kPa is equivalent to 103m of water height and finding the pressure that corresponds to a water height of 175 m preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 123 167 A gas contained in a vertical pistoncylinder device is pressurized by a spring and by the weight of the piston The pressure of the gas is to be determined Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield W mg P Patm Fspring spring atm F A W P PA Thus 147 kPa 2 2 4 2 spring atm Nm 1000 kPa 1 m 10 35 15 0 N 32 kg981 ms kPa 95 A F mg P P 168 Problem 167 is reconsidered The effect of the spring force in the range of 0 to 500 N on the pressure inside the cylinder is to be investigated The pressure against the spring force is to be plotted and results are to be discussed Analysis The problem is solved using EES and the solution is given below g981 ms2 Patm 95 kPa mpiston32 kg Fspring150 N A35CONVERTcm2 m2 Wpistonmpistong FatmPatmACONVERTkPa Nm2 From the free body diagram of the piston the balancing vertical forces yield Fgas FatmFspringWpiston PgasFgasACONVERTNm2 kPa Fspring N Pgas kPa 0 50 100 150 200 250 300 350 400 450 500 104 1183 1325 1468 1611 1754 1897 204 2183 2325 2468 0 100 200 300 400 500 100 120 140 160 180 200 220 240 260 Fspring N Pgas kPa preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 124 169 Both a gage and a manometer are attached to a gas to measure its pressure For a specified reading of gage pressure the difference between the fluid levels of the two arms of the manometer is to be determined for mercury and water Properties The densities of water and mercury are given to be ρwater 1000 kgm3 and be ρHg 13600 kgm3 Analysis The gage pressure is related to the vertical distance h between the two fluid levels by g P h gh P ρ ρ gage gage a For mercury 0 60 m 1 kN 1000 kgm s 1 kPa kNm 1 kgm 981 ms 13600 80 kPa 2 2 2 3 gage g P h Hg ρ b For water 816 m 1 kN 1000 kgm s 1 kPa kNm 1 kgm 981 ms 1000 80 kPa 2 2 2 3 O H gage 2 g P h ρ preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 125 170 Problem 169 is reconsidered The effect of the manometer fluid density in the range of 800 to 13000 kgm3 on the differential fluid height of the manometer is to be investigated Differential fluid height against the density is to be plotted and the results are to be discussed Analysis The problem is solved using EES and the solution is given below Lets modify this problem to also calculate the absolute pressure in the tank by supplying the atmospheric pressure Use the relationship between the pressure gage reading and the manometer fluid column height Function fluiddensityFluid This function is needed since ifthenelse logic can only be used in functions or procedures The underscore displays whatever follows as subscripts in the Formatted Equations Window If fluidMercury then fluiddensity13600 else fluiddensity1000 end Input from the diagram window If the diagram window is hidden then all of the input must come from the equations window Also note that brackets can also denote comments but these comments do not appear in the formatted equations window FluidMercury Patm 101325 kPa DELTAP80 kPa Note how DELTAP is displayed on the Formatted Equations Window g9807 ms2 local acceleration of gravity at sea level rhoFluiddensityFluid Get the fluid density either Hg or H2O from the function To plot fluid height against density place around the above equation Then set up the parametric table and solve DELTAP RHOgh1000 Instead of dividiing by 1000 PakPa we could have multiplied by the EES function CONVERTPakPa hmmhconvertm mm The fluid height in mm is found using the builtin CONVERT function Pabs Patm DELTAP To make the graph hide the diagram window and remove the brackets from Fluid and from Patm Select New Parametric Table from the Tables menu Choose Pabs DELTAP and h to be in the table Choose Alter Values from the Tables menu Set values of h to range from 0 to 1 in steps of 02 Choose Solve Table or press F3 from the Calculate menu Choose New Plot Window from the Plot menu Choose to plot Pabs vs h and then choose Overlay Plot from the Plot menu and plot DELTAP on the same scale 0 2000 4000 6000 8000 10000 12000 14000 0 2200 4400 6600 8800 11000 ρ kgm3 hmm mm Manometer Fluid Height vs Manometer Fluid Density ρ kgm3 hmm mm 800 2156 3511 4867 6222 7578 8933 10289 11644 13000 10197 3784 2323 1676 1311 1076 9131 7928 7005 6275 preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 126 171 The air pressure in a tank is measured by an oil manometer For a given oillevel difference between the two columns the absolute pressure in the tank is to be determined Patm 98 kPa AIR 036 m Properties The density of oil is given to be ρ 850 kgm3 Analysis The absolute pressure in the tank is determined from 1010 kPa 2 2 3 atm Nm 1000 1kPa 850 kgm 981ms 036 m kPa 98 gh P P ρ 172 The air pressure in a duct is measured by a mercury manometer For a given mercurylevel difference between the two columns the absolute pressure in the duct is to be determined AIR P 15 mm Properties The density of mercury is given to be ρ 13600 kgm3 Analysis a The pressure in the duct is above atmospheric pressure since the fluid column on the duct side is at a lower level b The absolute pressure in the duct is determined from 102 kPa 2 2 2 3 atm Nm 1000 kPa 1 kg ms 1 1 N 13600 kgm 981 ms 0015 m kPa 100 gh P P ρ 173 The air pressure in a duct is measured by a mercury manometer For a given mercurylevel difference between the two columns the absolute pressure in the duct is to be determined 45 mm AIR P Properties The density of mercury is given to be ρ 13600 kgm3 Analysis a The pressure in the duct is above atmospheric pressure since the fluid column on the duct side is at a lower level b The absolute pressure in the duct is determined from 106 kPa 2 2 2 3 atm Nm 1000 kPa 1 kg ms 1 1 N 13600 kgm 981 ms 0045 m kPa 100 gh P P ρ preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 127 174E The systolic and diastolic pressures of a healthy person are given in mmHg These pressures are to be expressed in kPa psi and meter water column Assumptions Both mercury and water are incompressible substances Properties We take the densities of water and mercury to be 1000 kgm3 and 13600 kgm3 respectively Analysis Using the relation gh P ρ for gage pressure the high and low pressures are expressed as kPa 107 kPa 160 Nm 1000 kPa 1 kg ms 1 1N kgm 981 ms 008 m 13600 Nm 1000 kPa 1 kg ms 1 1N kgm 981 ms 012 m 13600 2 2 2 3 low low 2 2 2 3 high high gh P gh P ρ ρ Noting that 1 psi 6895 kPa 232 psi 6895kPa 1 psi 0 Pa 16 high P and 155 psi 6895kPa 1 psi 107 Pa low P For a given pressure the relation gh P ρ can be expressed for mercury and water as water ghwater P ρ and mercuryghmercury P ρ Setting these two relations equal to each other and solving for water height gives h mercury water mercury water mercury mercury water water h h gh gh P ρ ρ ρ ρ Therefore m 109 m 163 0 08 m kgm 1000 600 kgm 13 0 12 m kgm 1000 600 kgm 13 3 3 mercury low water mercury low water 3 3 mercury high water mercury high water h h h h ρ ρ ρ ρ Discussion Note that measuring blood pressure with a water monometer would involve differential fluid heights higher than the person and thus it is impractical This problem shows why mercury is a suitable fluid for blood pressure measurement devices preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 128 175 A vertical tube open to the atmosphere is connected to the vein in the arm of a person The height that the blood will rise in the tube is to be determined Assumptions 1 The density of blood is constant 2 The gage pressure of blood is 120 mmHg Properties The density of blood is given to be ρ 1050 kgm3 Blood h Analysis For a given gage pressure the relation gh P ρ can be expressed for mercury and blood as blood ghblood P ρ and mercuryghmercury P ρ Setting these two relations equal to each other we get mercury mercury blood blood gh gh P ρ ρ Solving for blood height and substituting gives 155 m 0 12 m kgm 1050 600 kgm 13 3 3 mercury blood mercury blood h h ρ ρ Discussion Note that the blood can rise about one and a half meters in a tube connected to the vein This explains why IV tubes must be placed high to force a fluid into the vein of a patient 176 A diver is moving at a specified depth from the water surface The pressure exerted on the surface of the diver by water is to be determined Assumptions The variation of the density of water with depth is negligible Properties The specific gravity of seawater is given to be SG 103 We take the density of water to be 1000 kgm3 Analysis The density of the seawater is obtained by multiplying its specific gravity by the density of water which is taken to be 1000 kgm3 Patm Sea h P 3 3 1030 kgm 103100 0 kgm SG 2 ρH O ρ The pressure exerted on a diver at 30 m below the free surface of the sea is the absolute pressure at that location 404 kPa 2 2 3 atm Nm 1000 1 kPa 1030 kgm 9807 ms 30 m kPa 101 gh P P ρ preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 129 177 Water is poured into the Utube from one arm and oil from the other arm The water column height in one arm and the ratio of the heights of the two fluids in the other arm are given The height of each fluid in that arm is to be determined Assumptions Both water and oil are incompressible substances Water oil hw1 hw2 ha Properties The density of oil is given to be ρ 790 kgm3 We take the density of water to be ρ 1000 kgm3 Analysis The height of water column in the left arm of the monometer is given to be hw1 070 m We let the height of water and oil in the right arm to be hw2 and ha respectively Then ha 4hw2 Noting that both arms are open to the atmosphere the pressure at the bottom of the Utube can be expressed as w1 w atm bottom gh P P ρ and a a w2 w atm bottom gh gh P P ρ ρ Setting them equal to each other and simplifying a a w2 w1 a a w2 w w1 w a a w2 w w1 w h h h h h h gh gh gh ρw ρ ρ ρ ρ ρ ρ ρ Noting that ha 4hw2 the water and oil column heights in the second arm are determined to be 0168 m 2 2 2 7901000 4 07 m w w w h h h 0673 m a a h h 7901000 0 168 m 07 m Discussion Note that the fluid height in the arm that contains oil is higher This is expected since oil is lighter than water preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 130 178 Fresh and seawater flowing in parallel horizontal pipelines are connected to each other by a double Utube manometer The pressure difference between the two pipelines is to be determined Assumptions 1 All the liquids are incompressible 2 The effect of air column on pressure is negligible PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The densities of seawater and mercury are given to be ρsea 1035 kgm3 and ρHg 13600 kgm3 We take the density of water to be ρ w 1000 kgm3 Analysis Starting with the pressure in the fresh water pipe point 1 and moving along the tube by adding as we go down or subtracting as we go up the gh ρ terms until we reach the sea water pipe point 2 and setting the result equal to P2 gives 2 sea sea air air Hg Hg w 1 P gh gh gh gh P w ρ ρ ρ ρ Rearranging and neglecting the effect of air column on pressure sea sea w Hg Hg sea sea Hg Hg w 2 1 h h h g gh gh gh P P w w ρ ρ ρ ρ ρ ρ Fresh Water hw Air hsea hair hHg Mercury Sea Water Substituting 339 kPa 2 2 3 3 3 2 2 1 39 kNm 3 kg ms 1000 1kN m 40 1035 kgm m 60 1000 kgm m 10 981 ms 13600 kgm P P Therefore the pressure in the fresh water pipe is 339 kPa higher than the pressure in the sea water pipe Discussion A 070m high air column with a density of 12 kgm3 corresponds to a pressure difference of 0008 kPa Therefore its effect on the pressure difference between the two pipes is negligible preparation If you are a student using this Manual you are using it without permission 131 179 Fresh and seawater flowing in parallel horizontal pipelines are connected to each other by a double Utube manometer The pressure difference between the two pipelines is to be determined Assumptions All the liquids are incompressible Fresh Water Mercury hw Oil hsea hoil hHg Sea Water Properties The densities of seawater and mercury are given to be ρsea 1035 kgm3 and ρHg 13600 kgm3 We take the density of water to be ρ w 1000 kgm3 The specific gravity of oil is given to be 072 and thus its density is 720 kgm3 Analysis Starting with the pressure in the fresh water pipe point 1 and moving along the tube by adding as we go down or subtracting as we go up the gh ρ terms until we reach the sea water pipe point 2 and setting the result equal to P2 gives 2 sea sea oil oil Hg Hg w 1 P gh gh gh gh P w ρ ρ ρ ρ Rearranging sea sea w oil oil Hg Hg sea sea oil oil Hg Hg w 2 1 h h h h g gh gh gh gh P P w w ρ ρ ρ ρ ρ ρ ρ ρ Substituting 834 kPa 2 2 3 3 3 3 2 2 1 34 kNm 8 kg ms 1000 1kN m 40 1035 kgm m 60 m 1000 kgm 70 720 kgm m 10 981 ms 13600 kgm P P Therefore the pressure in the fresh water pipe is 834 kPa higher than the pressure in the sea water pipe 180 The pressure indicated by a manometer is to be determined hA hB Properties The specific weights of fluid A and fluid B are given to be 10 kNm3 and 8 kNm3 respectively Analysis The absolute pressure P1 is determined from 1027 kPa 8 kNm 015 m kNm 005 m 10 1mm Hg 758 mm Hg 01333 kPa 3 3 atm atm 1 B B A A B A h h P gh gh P P γ γ ρ ρ Note that 1 kPa 1 kNm2 preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 132 181 The pressure indicated by a manometer is to be determined hB hA 100 kNm3 Properties The specific weights of fluid A and fluid B are given to be 100 kNm3 and 8 kNm3 respectively Analysis The absolute pressure P1 is determined from 962 kPa 8 kNm 015 m 100 kNm 005 m kPa 90 3 3 atm atm 1 B B A A B A h h P gh gh P P γ γ ρ ρ Note that 1 kPa 1 kNm2 182 The pressure indicated by a manometer is to be determined hA 12 kNm3 hB Properties The specific weights of fluid A and fluid B are given to be 10 kNm3 and 12 kNm3 respectively Analysis The absolute pressure P1 is determined from 983 kPa 12 kNm 015 m kNm 005 m 10 1mm Hg 720 mm Hg 01333 kPa 3 3 atm atm 1 B B A A B A h h P gh gh P P γ γ ρ ρ Note that 1 kPa 1 kNm2 preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 133 183 The gage pressure of air in a pressurized water tank is measured simultaneously by both a pressure gage and a manometer The differential height h of the mercury column is to be determined Assumptions The air pressure in the tank is uniform ie its variation with elevation is negligible due to its low density and thus the pressure at the airwater interface is the same as the indicated gage pressure Properties We take the density of water to be ρw 1000 kgm3 The specific gravities of oil and mercury are given to be 072 and 136 respectively Analysis Starting with the pressure of air in the tank point 1 and moving along the tube by adding as we go down or subtracting as we go u p the gh ρ terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere and setting the result equal to Patm gives atm w P gh gh gh P oil oil Hg Hg w 1 ρ ρ ρ Rearranging ghw gh gh P P w Hg Hg oil oil atm 1 ρ ρ ρ or hw h h g P Hg Hg oil oil w 1 gage SG SG ρ Substituting m 30 0 72 075 m 136 kPa m 1 kg ms 1000 kgm 981 ms 1000 80 kPa Hg 2 2 2 3 h Solving for hHg gives hHg 0582 m Therefore the differential height of the mercury column must be 582 cm Discussion Double instrumentation like this allows one to verify the measurement of one of the instruments by the measurement of another instrument preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 134 184 The gage pressure of air in a pressurized water tank is measured simultaneously by both a pressure gage and a manometer The differential height h of the mercury column is to be determined Assumptions The air pressure in the tank is uniform ie its variation with elevation is negligible due to its low density and thus the pressure at the airwater interface is the same as the indicated gage pressure Properties We take the density of water to be ρ w 1000 kgm3 The specific gravities of oil and mercury are given to be 072 and 136 respectively Analysis Starting with the pressure of air in the tank point 1 and moving along the tube by adding as we go down or subtracting as we go up the gh ρ terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere and setting the result equal to Patm gives PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course atm w P gh gh gh P oil oil Hg Hg w 1 ρ ρ ρ Rearranging ghw gh gh P P w Hg Hg oil oil atm 1 ρ ρ ρ or hw h h g P Hg Hg oil oil w 1 gage SG SG ρ Substituting m 30 0 72 075 m 136 kPa m 1 kg ms 1000 kgm 981 ms 1000 40 kPa Hg 2 2 2 3 h 40 kPa AIR Water hoil hw hHg Solving for hHg gives hHg 0282 m Therefore the differential height of the mercury column must be 282 cm Discussion Double instrumentation like this allows one to verify the measurement of one of the instruments by the measurement of another instrument 185 The top part of a water tank is divided into two compartments and a fluid with an unknown density is poured into one side The levels of the water and the liquid are measured The density of the fluid is to be determined Assumptions 1 Both water and the added liquid are incompressible substances 2 The added liquid does not mix with water Water Fluid hw hf Properties We take the density of water to be ρ 1000 kgm3 Analysis Both fluids are open to the atmosphere Noting that the pressure of both water and the added fluid is the same at the contact surface the pressure at this surface can be expressed as w w atm f f atm contact gh P gh P P ρ ρ Simplifying and solving for ρf gives 846 kgm3 65 cm 1000 kgm 55 cm 3 w w f f w f w f h h gh gh ρ ρ ρ ρ Discussion Note that the added fluid is lighter than water as expected a heavier fluid would sink in water preparation If you are a student using this Manual you are using it without permission 135 186 The fluid levels in a multifluid Utube manometer change as a result of a pressure drop in the trapped air space For a given pressure drop and brine level change the area ratio is to be determined Assumptions 1 All the liquids are incompressible 2 Pressure in the brine pipe remains constant 3 The variation of pressure in the trapped air space is negligible A Air B Brine pipe Water Mercury SG1356 SG11 Area A2 hb 5 mm Properties The specific gravities are given to be 1356 for mercury and 11 for brine We take the standard density of water to be ρw 1000 kgm3 Area A1 Analysis It is clear from the problem statement and the figure that the brine pressure is much higher than the air pressure and when the air pressure drops by 07 kPa the pressure difference between the brine and the air space increases also by the same amount Starting with the air pressure point A and moving along the tube by adding as we go down or subtracting as we go up the gh ρ terms until we reach the brine pipe point B and setting the result equal to PB before and after the pressure change of air give Before B w A P gh gh gh P br1 br Hg1 Hg w 1 ρ ρ ρ After B w A P gh gh gh P br2 br Hg2 Hg w 2 ρ ρ ρ Subtracting 0 br br Hg Hg 1 2 g h g h P P A A ρ ρ 0 br br Hg Hg 2 1 h SG h SG g P P w A A ρ 1 where and are the changes in the differential mercury and brine column heights respectively due to the drop in air pressure Both of these are positive quantities since as the mercurybrine interface drops the differential fluid heights for both mercury and brine increase Noting also that the volume of mercury is constant we have hHg hbr Hgright 2 Hgleft 1 h A h A and 2 2 1 2 700 kgm s 700 Nm 70 kPa A A P P hbr 0 005 m A 1 A 1 2 br 1 2 br br Hgleft Hgright Hg A h A h h h h h Substituting 11 0005 m 1356 00051 1000 kgm 981 ms 700 kgm s 1 2 2 3 2 A A It gives A2A1 0134 preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 136 187 A multifluid container is connected to a Utube For the given specific gravities and fluid column heights the gage pressure at A and the height of a mercury column that would create the same pressure at A are to be determined Assumptions 1 All the liquids are incompressible 2 The multi fluid container is open to the atmosphere PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The specific gravities are given to be 126 for glycerin and 090 for oil We take the standard density of water to be ρw 1000 kgm3 and the specific gravity of mercury to be 136 Analysis Starting with the atmospheric pressure on the top surface of the container and moving along the tube by adding as we go down or subtracting as we go up the gh ρ terms until we reach point A and setting the result equal to PA give A gly gly w oil oil atm P gh gh gh P w ρ ρ ρ Rearranging and using the definition of specific gravity gly gly oil oil atm A SG SG SG gh gh gh P P w w w w w ρ ρ ρ A 90 cm 70 cm 30 cm 15 cm 20 cm Water Oil SG090 Glycerin SG126 or SG SG SG gly gly oil oil Agage h h h g P w w w ρ Substituting 0471 kPa 2 2 3 2 gage A 471 kNm 0 kg ms 1000 1kN 1 26 0 70 m m 30 1 981 ms 1000 kgm 0 90 0 70 m P The equivalent mercury column height is 0 00353 m 0353 cm 1kN kg ms 1000 kgm 1000 kgm 981 ms 13600 kNm 0471 2 2 3 3 2 Hg Agage Hg g P h ρ Discussion Note that the high density of mercury makes it a very suitable fluid for measuring high pressures in manometers preparation If you are a student using this Manual you are using it without permission 137 Solving Engineering Problems and EES 188C Despite the convenience and capability the engineering software packages offer they are still just tools and they will not replace the traditional engineering courses They will simply cause a shift in emphasis in the course material from mathematics to physics They are of great value in engineering practice however as engineers today rely on software packages for solving large and complex problems in a short time and perform optimization studies efficiently 189 Determine a positive real root of the following equation using EES 2x3 10x05 3x 3 Solution by EES Software Copy the following line and paste on a blank EES screen to verify solution 2x310x053x 3 Answer x 2063 using an initial guess of x2 190 Solve the following system of 2 equations with 2 unknowns using EES x3 y2 775 3xy y 35 Solution by EES Software Copy the following lines and paste on a blank EES screen to verify solution x3y2775 3xyy35 Answer x2 y05 191 Solve the following system of 3 equations with 3 unknowns using EES 2x y z 7 3x2 2y z 3 xy 2z 4 Solution by EES Software Copy the following lines and paste on a blank EES screen to verify solution 2xyz7 3x22yz3 xy2z4 Answer x1609 y09872 z2794 preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 138 192 Solve the following system of 3 equations with 3 unknowns using EES x2y z 1 x 3y05 xz 2 x y z 2 Solution by EES Software Copy the following lines and paste on a blank EES screen to verify solution x2yz1 x3y05xz2 xyz2 Answer x1 y1 z0 preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 139 193E Specific heat of water is to be expressed at various units using unit conversion capability of EES Analysis The problem is solved using EES and the solution is given below EQUATION WINDOW GIVEN Cp418 kJkgC ANALYSIS Cp1CpConvertkJkgC kJkgK Cp2CpConvertkJkgC BtulbmF Cp3CpConvertkJkgC BtulbmR Cp4CpConvertkJkgC kCalkgC FORMATTED EQUATIONS WINDOW GIVEN Cp 418 kJkgC ANALYSIS Cp1 Cp 1 kJkgK kJkgC Cp2 Cp 0238846 BtulbmF kJkgC Cp3 Cp 0238846 BtulbmR kJkgC Cp4 Cp 0238846 kCalkgC kJkgC SOLUTION Cp418 kJkgC Cp1418 kJkgK Cp209984 BtulbmF Cp309984 BtulbmR Cp409984 kCalkgC preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 140 Review Problems 194 The weight of a lunar exploration module on the moon is to be determined Analysis Applying Newtons second law the weight of the module on the moon can be determined from 469 N 1 64 ms ms 89 2800 N 2 2 moon earth earth moon moon g g W mg W 195 The deflection of the spring of the twopiston cylinder with a spring shown in the figure is to be determined Analysis Summing the forces acting on the piston in the vertical direction gives PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1 1 2 1 3 2 2 1 3 2 P A A A P P A kx F F F Fs which when solved for the deflection of the spring and substituting D2 4 gives A π 172 cm m 00172 0 03 1000 0 08 0 03 10000 0 08 800 5000 4 4 2 2 2 2 2 2 2 1 3 2 2 2 2 1 1 π π D P D P D k P D x F1 Fs F3 F2 We expressed the spring constant k in kNm the pressures in kPa ie kNm2 and the diameters in m units 196 An airplane is flying over a city The local atmospheric pressure in that city is to be determined Assumptions The gravitational acceleration does not change with altitude Properties The densities of air and mercury are given to be 115 kgm3 and 13600 kgm3 Analysis The local atmospheric pressure is determined from 127 kPa 126 5 kNm kg ms 1000 1kN 115 kgm 981 ms 9000 m 25 kPa 2 2 2 3 plane atm gh P P ρ The atmospheric pressure may be expressed in mmHg as 948 mmHg 1m 1000 mm 1kPa Pa 1000 ms kgm 981 13600 5 kPa 126 2 3 atm Hg g P h ρ preparation If you are a student using this Manual you are using it without permission 141 197 The gravitational acceleration changes with altitude Accounting for this variation the weights of a body at different locations are to be determined Analysis The weight of an 80kg man at various locations is obtained by substituting the altitude z values in m into the relation 2 2 6 kg ms 1 1N ms 332 10 80kg9807 z mg W Sea level z 0 m W 809807332x1060 809807 7846 N Denver z 1610 m W 809807332x1061610 809802 7842 N Mt Ev z 8848 m W 809807332x1068848 809778 7822 N 198 A man is considering buying a 12oz steak for 315 or a 300g steak for 295 The steak that is a better buy is to be determined Assumptions The steaks are of identical quality Analysis To make a comparison possible we need to express the cost of each steak on a common basis Let us choose 1 kg as the basis for comparison Using proper conversion factors the unit cost of each steak is determined to be 12 ounce steak 926kg 045359 kg 1lbm 1lbm 12 oz 16 oz Unit Cost 315 300 gram steak 983kg 1kg 1000 g 300 g Unit Cost 295 Therefore the steak at the traditional market is a better buy preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 142 199E The mass of a substance is given Its weight is to be determined in various units Analysis Applying Newtons second law the weight is determined in various units to be 981 N 2 2 kg ms 1 1 N 1 kg981 ms mg W 000981 kN 2 2 kg ms 1000 1 kN 1 kg981 ms mg W 1kg ms2 ms kg981 1 2 mg W 1kgf 981 N kgf 1 kg ms 1 1 N kg981 ms 1 2 2 mg W 71 lbm fts2 322 fts 1 kg kg 2205 lbm 1 2 mg W lbf 221 2 2 lbm fts 322 1 lbf 322 fts 1 kg 1 kg 2205 lbm mg W 1100E The efficiency of a refrigerator increases by 3 per C rise in the minimum temperature This increase is to be expressed per F K and R rise in the minimum temperature Analysis The magnitudes of 1 K and 1C are identical so are the magnitudes of 1 R and 1F Also a change of 1 K or 1C in temperature corresponds to a change of 18 R or 18F Therefore the increase in efficiency is a 3 for each K rise in temperature and b c 318 167 for each R or F rise in temperature 1101E The boiling temperature of water decreases by 3C for each 1000 m rise in altitude This decrease in temperature is to be expressed in F K and R Analysis The magnitudes of 1 K and 1C are identical so are the magnitudes of 1 R and 1F Also a change of 1 K or 1C in temperature corresponds to a change of 18 R or 18F Therefore the decrease in the boiling temperature is a 3 K for each 1000 m rise in altitude and b c 318 54F 54 R for each 1000 m rise in altitude preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 143 1102E Hyperthermia of 5C is considered fatal This fatal level temperature change of body temperature is to be expressed in F K and R Analysis The magnitudes of 1 K and 1C are identical so are the magnitudes of 1 R and 1F Also a change of 1 K or 1C in temperature corresponds to a change of 18 R or 18F Therefore the fatal level of hypothermia is a 5 K b 518 9F c 518 9 R 1103E A house is losing heat at a rate of 2700 kJh per C temperature difference between the indoor and the outdoor temperatures The rate of heat loss is to be expressed per F K and R of temperature difference between the indoor and the outdoor temperatures Analysis The magnitudes of 1 K and 1C are identical so are the magnitudes of 1 R and 1F Also a change of 1 K or 1C in temperature corresponds to a change of 18 R or 18F Therefore the rate of heat loss from the house is a 2700 kJh per K difference in temperature and b c 270018 1500 kJh per R or F rise in temperature 1104 The average temperature of the atmosphere is expressed as Tatm 28815 65z where z is altitude in km The temperature outside an airplane cruising at 12000 m is to be determined Analysis Using the relation given the average temperature of the atmosphere at an altitude of 12000 m is determined to be Tatm 28815 65z 28815 6512 21015 K 63C Discussion This is the average temperature The actual temperature at different times can be different preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 144 1105 A new Smith absolute temperature scale is proposed and a value of 1000 S is assigned to the boiling point of water The ice point on this scale and its relation to the Kelvin scale are to be determined Analysis All linear absolute temperature scales read zero at absolute zero pressure and are constant multiples of each other For example TR 18 TK That is multiplying a temperature value in K by 18 will give the same temperature in R 0 S K 37315 1000 The proposed temperature scale is an acceptable absolute temperature scale since it differs from the other absolute temperature scales by a constant only The boiling temperature of water in the Kelvin and the Smith scales are 31515 K and 1000 K respectively Therefore these two temperature scales are related to each other by K 37315 1000 2 6799T K T T S The ice point of water on the Smith scale is TSice 26799 TKice 2679927315 7320 S 1106E An expression for the equivalent wind chill temperature is given in English units It is to be converted to SI units Analysis The required conversion relations are 1 mph 1609 kmh and TF 18TC 32 The first thought that comes to mind is to replace TF in the equation by its equivalent 18TC 32 and V in mph by 1609 kmh which is the regular way of converting units However the equation we have is not a regular dimensionally homogeneous equation and thus the regular rules do not apply The V in the equation is a constant whose value is equal to the numerical value of the velocity in mph Therefore if V is given in kmh we should divide it by 1609 to convert it to the desired unit of mph That is T T V equiv ambient F F 914 914 0 475 0 0203 1609 0 304 V 1609 or T T V equiv ambient F F 914 914 0 475 0 0126 0 240 V where V is in kmh Now the problem reduces to converting a temperature in F to a temperature in C using the proper convection relation 1 8 32 914 914 18 32 0 475 0 0126 0 240 T T V V equiv ambient C C which simplifies to T T V V equiv ambient C 330 330 0 475 0 0126 0 240 where the ambient air temperature is in C preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 145 1107E Problem 1106E is reconsidered The equivalent windchill temperatures in F as a function of wind velocity in the range of 4 mph to 40 mph for the ambient temperatures of 20 40 and 60F are to be plotted and the results are to be discussed Analysis The problem is solved using EES and the solution is given below Tambient20 V20 Tequiv914914Tambient0475 00203V 0304sqrtV 0 5 10 15 20 25 30 35 40 20 10 0 10 20 30 40 50 60 V mph Tequiv F Tamb 20F Tamb 40F Tamb 60F V mph Tequiv F 4 8 12 16 20 24 28 32 36 40 5994 5459 5107 485 4654 4502 4382 4288 4216 4161 The table is for Tambient60F 1108 One section of the duct of an airconditioning system is laid underwater The upward force the water will exert on the duct is to be determined Assumptions 1 The diameter given is the outer diameter of the duct or the thickness of the duct material is negligible 2 The weight of the duct and the air in is negligible Properties The density of air is given to be ρ 130 kgm3 We take the density of water to be 1000 kgm3 D 15 cm L 20 m Analysis Noting that the weight of the duct and the air in it is negligible the net upward force acting on the duct is the buoyancy force exerted by water The volume of the underground section of the duct is FB 0 15 m 420 m 0353 m 4 3 2 2 π π L D AL V Then the buoyancy force becomes 346 kN 2 3 2 3 000 kg ms 1 1kN 1000 kgm 981 ms 0353 m gV FB ρ Discussion The upward force exerted by water on the duct is 346 kN which is equivalent to the weight of a mass of 353 kg Therefore this force must be treated seriously preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 146 1109 A helium balloon tied to the ground carries 2 people The acceleration of the balloon when it is first released is to be determined Assumptions The weight of the cage and the ropes of the balloon is negligible Properties The density of air is given to be ρ 116 kgm3 The density of helium gas is 17th of this Analysis The buoyancy force acting on the balloon is D 12 m m 170 kg 10296 N kg ms 1 1 N kgm 981ms 9048 m 116 904 8 m 4π6 m 3 r 3 4π 2 3 2 3 balloon air 3 3 3 V V g FB balloon ρ The total mass is 319 9 kg 2 85 9 149 149 9 kg 9048 m 7 kgm 116 people He total 3 3 He He m m m m V ρ The total weight is 3138 N kg ms 1 1 N ms 3199 kg981 2 2 total g m W Thus the net force acting on the balloon is 7157 N 3138 10296 net W F F B Then the acceleration becomes 224 ms2 1 N 1 kg ms 3199 kg 7157 N 2 total net m F a preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 147 1110 Problem 1109 is reconsidered The effect of the number of people carried in the balloon on acceleration is to be investigated Acceleration is to be plotted against the number of people and the results are to be discussed Analysis The problem is solved using EES and the solution is given below Given D12 m Nperson2 mperson85 kg rhoair116 kgm3 rhoHerhoair7 Analysis g981 ms2 VballonpiD36 FBrhoairgVballon mHerhoHeVballon mpeopleNpersonmperson mtotalmHempeople Wmtotalg FnetFBW aFnetmtotal Nperson a ms2 1 2 3 4 5 6 7 8 9 10 34 2236 1561 112 8096 579 401 2595 1443 04865 1 2 3 4 5 6 7 8 9 10 0 5 10 15 20 25 30 35 Nperson a ms 2 preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 148 1111 A balloon is filled with helium gas The maximum amount of load the balloon can carry is to be determined Assumptions The weight of the cage and the ropes of the balloon is negligible D 12 m Properties The density of air is given to be ρ 116 kgm3 The density of helium gas is 17th of this Analysis The buoyancy force acting on the balloon is 10296 N kg ms 1 1 N kgm 981ms 9048 m 116 904 8 m 4π6 m 3 r 3 4π 2 3 2 3 balloon air 3 3 3 V V g FB balloon ρ The mass of helium is 149 9 kg 9048 m 7 kgm 116 3 3 He He V ρ m In the limiting case the net force acting on the balloon will be zero That is the buoyancy force and the weight will balance each other 1050 kg ms 981 N 10296 2 total g F m F mg W B B Thus 900 kg 149 9 1050 He total people m m m 1112 A 10m high cylindrical container is filled with equal volumes of water and oil The pressure difference between the top and the bottom of the container is to be determined PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The density of water is given to be ρ 1000 kgm3 The specific gravity of oil is given to be 085 Analysis The density of the oil is obtained by multiplying its specific gravity by the density of water 3 3 H O 850 kgm 085100 0 kgm SG 2 ρ ρ The pressure difference between the top and the bottom of the cylinder is the sum of the pressure differences across the two fluids Nm 1000 1 kPa 1000 kgm 981 ms 5 m kgm 981 ms 5 m 850 2 2 3 2 3 water oil water oil total 907 kPa gh gh P P P ρ ρ Water Oil SG 085 h 10 m preparation If you are a student using this Manual you are using it without permission 149 1113 The pressure of a gas contained in a vertical pistoncylinder device is measured to be 180 kPa The mass of the piston is to be determined Assumptions There is no friction between the piston and the cylinder P Patm W mg Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield 1kPa m 1000 kgm s 180 100 kPa25 10 981 ms 2 2 4 2 atm atm m A P P mg A P PA W It yields m 204 kg 1114 The gage pressure in a pressure cooker is maintained constant at 100 kPa by a petcock The mass of the petcock is to be determined Assumptions There is no blockage of the pressure release valve P Patm W mg Analysis Atmospheric pressure is acting on all surfaces of the petcock which balances itself out Therefore it can be disregarded in calculations if we use the gage pressure as the cooker pressure A force balance on the petcock ΣFy 0 yields 00408 kg 1 kPa kgm s 1000 ms 981 m 100 kPa4 10 2 2 2 6 gage gage g A P m A P W 1115 A glass tube open to the atmosphere is attached to a water pipe and the pressure at the bottom of the tube is measured It is to be determined how high the water will rise in the tube Properties The density of water is given to be ρ 1000 kgm3 Water Patm 99 kPa h Analysis The pressure at the bottom of the tube can be expressed as tube atm gh P P ρ Solving for h 214 m 1 kPa 1000 Nm 1 N kg ms 1 kgm 981 ms 1000 99 kPa 120 2 2 2 3 atm g P P h ρ preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 150 1116 The air pressure in a duct is measured by an inclined manometer For a given vertical level difference the gage pressure in the duct and the length of the differential fluid column are to be determined Assumptions The manometer fluid is an incompressible substance 12 cm 45 Properties The density of the liquid is given to be ρ 081 kgL 810 kgm3 Analysis The gage pressure in the duct is determined from 954 Pa 2 2 2 3 atm abs gage Nm 1 Pa 1 kg ms 1 1 N kgm 981 ms 012 m 810 gh P P P ρ The length of the differential fluid column is 170 cm 12 cm sin 45 sin θ h L Discussion Note that the length of the differential fluid column is extended considerably by inclining the manometer arm for better readability 1117E Equal volumes of water and oil are poured into a Utube from different arms and the oil side is pressurized until the contact surface of the two fluids moves to the bottom and the liquid levels in both arms become the same The excess pressure applied on the oil side is to be determined Assumptions 1 Both water and oil are incompressible substances 2 Oil does not mix with water 3 The crosssectional area of the Utube is constant Properties The density of oil is given to be ρoil 493 lbmft3 We take the density of water to be ρw 624 lbmft3 Analysis Noting that the pressure of both the water and the oil is the same at the contact surface the pressure at this surface can be expressed as w w atm a a blow contact gh P gh P P ρ ρ Noting that ha hw and rearranging 0227 psi 2 2 2 2 3 atm blow blow gage in 144 ft 1 lbm fts 322 1 lbf 32 2 fts 3012 ft 493 lbmft 624 gh P P P oil w ρ ρ Discussion When the person stops blowing the oil will rise and some water will flow into the right arm It can be shown that when the curvature effects of the tube are disregarded the differential height of water will be 237 in to balance 30in of oil preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 151 1118 It is given that an IV fluid and the blood pressures balance each other when the bottle is at a certain height and a certain gage pressure at the arm level is needed for sufficient flow rate The gage pressure of the blood and elevation of the bottle required to maintain flow at the desired rate are to be determined Assumptions 1 The IV fluid is incompressible 2 The IV bottle is open to the atmosphere 80 cm Properties The density of the IV fluid is given to be ρ 1020 kgm3 Analysis a Noting that the IV fluid and the blood pressures balance each other when the bottle is 08 m above the arm level the gage pressure of the blood in the arm is simply equal to the gage pressure of the IV fluid at a depth of 08 m 80 kPa 2 2 2 3 armbottle atm abs arm gage kNm 1 kPa 1 000 kg ms 1 1 kN kgm 981 ms 08 m 1020 gh P P P ρ b To provide a gage pressure of 15 kPa at the arm level the height of the bottle from the arm level is again determined from armbottle gage arm gh P ρ to be 15 m 1 kPa 1 kNm 1 kN 000 kg ms 1 kgm 981 ms 1020 15 kPa 2 2 2 3 gage arm armbottle g P h ρ Discussion Note that the height of the reservoir can be used to control flow rates in gravity driven flows When there is flow the pressure drop in the tube due to friction should also be considered This will result in raising the bottle a little higher to overcome pressure drop preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 152 1119E A water pipe is connected to a doubleU manometer whose free arm is open to the atmosphere The absolute pressure at the center of the pipe is to be determined Assumptions 1 All the liquids are incompressible 2 The solubility of the liquids in each other is negligible Properties The specific gravities of mercury and oil are given to be 136 and 080 respectively We take the density of water to be ρw 624 lbmft3 Analysis Starting with the pressure at the center of the water pipe and moving along the tube by adding as we go down or subtracting as we go up the gh ρ terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere and setting the result equal to Patm gives Patm gh gh gh gh P oil oil Hg Hg oil oil water water water pipe ρ ρ ρ ρ Solving for Pwater pipe oil oil Hg Hg oil oil water water water pipe h SG h SG h SG g h P P atm ρ Substituting 223 psia 2 2 2 2 3 pipe water in 144 ft 1 lbm fts 322 1 lbf 4012 ft 80 13 6 1512 ft 80 6012 ft 32 2 fts 3512 ft 624lbmft 142psia P Therefore the absolute pressure in the water pipe is 223 psia Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly 1120 The average atmospheric pressure is given as where z is the altitude in km The atmospheric pressures at various locations are to be determined 5 256 atm 0 02256 101325 1 z P Analysis The atmospheric pressures at various locations are obtained by substituting the altitude z values in km into the relation P z atm 101325 1 0 02256 5 256 Atlanta z 0306 km Patm 1013251 00225603065256 977 kPa Denver z 1610 km Patm 1013251 00225616105256 834 kPa M City z 2309 km Patm 1013251 00225623095256 765 kPa Mt Ev z 8848 km Patm 1013251 00225688485256 314 kPa preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 153 1121 The temperature of the atmosphere varies with altitude z as z T T β 0 while the gravitational acceleration varies by Relations for the variation of pressure in atmosphere are to be obtained a by ignoring and b by considering the variation of g with altitude 2 0 6 370320 1 z g g z Assumptions The air in the troposphere behaves as an ideal gas Analysis a Pressure change across a differential fluid layer of thickness dz in the vertical z direction is gdz dP ρ From the ideal gas relation the air density can be expressed as 0 z R T P RT P β ρ Then z gdz R T P dP 0 β Separating variables and integrating from z 0 where P P0 to z z where P P 0 0 0 z R T gdz P dP z P P β Performing the integrations 0 0 0 ln ln T z T R g P P β β Rearranging the desired relation for atmospheric pressure for the case of constant g becomes R g T z P P β β 0 0 1 b When the variation of g with altitude is considered the procedure remains the same but the expressions become more complicated dz z g z R T P dP 2 0 0 6 370320 1 β Separating variables and integrating from z 0 where P P0 to z z where P P 2 0 0 0 6 370320 1 0 z z T R g dz P dP z P P β Performing the integrations z P P z T kz kT kz kT R g P 0 0 2 0 0 0 ln 1 1 1 1 1 1 ln 0 β β β β where R 287 JkgK 287 m2s2K is the gas constant of air After some manipulations we obtain 0 0 0 0 0 1 1 ln 1 1 1 1 1 exp z T kz kT kz kT R g P P β β β where T0 28815 K β 00065 Km g0 9807 ms2 k 16370320 m1 and z is the elevation in m Discussion When performing the integration in part b the following expression from integral tables is used together with a transformation of variable z T x β 0 x bx a a bx a a bx a x dx 1 ln 1 2 2 Also for z 11000 m for example the relations in a and b give 2262 and 2269 kPa respectively preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 154 1122 The variation of pressure with density in a thick gas layer is given A relation is to be obtained for pressure as a function of elevation z Assumptions The property relation C n is valid over the entire region considered P ρ Analysis The pressure change across a differential fluid layer of thickness dz in the vertical z direction is given as gdz dP ρ Also the relation can be expressed as and thus C n P ρ n n P P C 0 0 ρ ρ n P P 1 0 0 ρ ρ Substituting dz P P g dP 1 n 0 0 ρ Separating variables and integrating from z 0 where C n to z z where P P P P 0 0 ρ z P P n g dz dP P P 0 0 1 0 0 ρ Performing the integrations gz n P P P P P n 0 1 1 0 0 0 1 1 ρ 0 0 1 0 1 1 P gz n n P P n n ρ Solving for P 1 0 0 0 1 1 n n P gz n n P P ρ which is the desired relation Discussion The final result could be expressed in various forms The form given is very convenient for calculations as it facilitates unit cancellations and reduces the chance of error preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 155 1123 A pressure transducers is used to measure pressure by generating analogue signals and it is to be calibrated by measuring both the pressure and the electric current simultaneously for various settings and the results are tabulated A calibration curve in the form of P aI b is to be obtained and the pressure corresponding to a signal of 10 mA is to be calculated Assumptions Mercury is an incompressible liquid Properties The specific gravity of mercury is given to be 1356 and thus its density is 13560 kgm3 Analysis For a given differential height the pressure can be calculated from g h P ρ For h 280 mm 00280 m for example 3 75 kPa kNm 1 kPa 1 kg ms 1000 1 kN 13561000 kgm 981 ms 00280 m 2 2 2 3 P Repeating the calculations and tabulating we have hmm 280 1815 2978 4131 7659 1027 1149 1362 1458 1536 PkPa 373 2414 3961 5495 1019 1366 1528 1812 1939 2043 I mA 421 578 697 815 1176 1443 1568 1786 1884 1964 A plot of P versus I is given below It is clear that the pressure varies linearly with the current and using EES the best curve fit is obtained to be PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course P 1300I 5100 kPa for 1964 4 21 I For I 10 mA for example we would get P 790 kPa 4 6 8 10 12 14 16 18 20 0 45 90 135 180 225 I mA P kPa Discussion Note that the calibration relation is valid in the specified range of currents or pressures preparation If you are a student using this Manual you are using it without permission 156 1124 The flow of air through a wind turbine is considered Based on unit considerations a proportionality relation is to be obtained for the mass flow rate of air through the blades Assumptions Wind approaches the turbine blades with a uniform velocity Analysis The mass flow rate depends on the air density average wind velocity and the crosssectional area which depends on hose diameter Also the unit of mass flow rate m is kgs Therefore the independent quantities should be arranged such that we end up with the proper unit Putting the given information into perspective we have m kgs is a function of ρ kgm3 D m and V ms It is obvious that the only way to end up with the unit kgs for mass flow rate is to multiply the quantities ρ and V with the square of D Therefore the desired proportionality relation is 2 m is proportional to ρD V or C D V m 2 ρ where the constant of proportionality is C π4 so that V D m 4 2 ρ π Discussion Note that the dimensionless constants of proportionality cannot be determined with this approach 1125 A relation for the air drag exerted on a car is to be obtained in terms of on the drag coefficient the air density the car velocity and the frontal area of the car Analysis The drag force depends on a dimensionless drag coefficient the air density the car velocity and the frontal area Also the unit of force F is newton N which is equivalent to kgms2 Therefore the independent quantities should be arranged such that we end up with the unit kgms2 for the drag force Putting the given information into perspective we have FD kgms2 CDrag Afront m2 ρ kgm3 and V ms It is obvious that the only way to end up with the unit kgms2 for drag force is to multiply mass with the square of the velocity and the fontal area with the drag coefficient serving as the constant of proportionality Therefore the desired relation is 2 front Drag V A C FD ρ Discussion Note that this approach is not sensitive to dimensionless quantities and thus a strong reasoning is required preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 157 Fundamentals of Engineering FE Exam Problems 1126 Consider a fish swimming 5 m below the free surface of water The increase in the pressure exerted on the fish when it dives to a depth of 25 m below the free surface is a 196 Pa b 5400 Pa c 30000 Pa d 196000 Pa e 294000 Pa Answer d 196000 Pa Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values rho1000 kgm3 g981 ms2 z15 m z225 m DELTAPrhogz2z1 Pa Some Wrong Solutions with Common Mistakes W1Prhogz2z11000 dividing by 1000 W2Prhogz1z2 adding depts instead of subtracting W3Prhoz1z2 not using g W4Prhog0z2 ignoring z1 1127 The atmospheric pressures at the top and the bottom of a building are read by a barometer to be 960 and 980 kPa If the density of air is 10 kgm3 the height of the building is a 17 m b 20 m c 170 m d 204 m e 252 m Answer d 204 m Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values rho10 kgm3 g981 ms2 P196 kPa P298 kPa DELTAPP2P1 kPa DELTAPrhogh1000 kPa Some Wrong Solutions with Common Mistakes DELTAPrhoW1h1000 not using g DELTAPgW2h1000 not using rho P2rhogW3h1000 ignoring P1 P1rhogW4h1000 ignoring P2 preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 158 1128 An apple loses 45 kJ of heat as it cools per C drop in its temperature The amount of heat loss from the apple per F drop in its temperature is a 125 kJ b 250 kJ c 50 kJ d 81 kJ e 41 kJ Answer b 250 kJ Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values QperC45 kJ QperFQperC18 kJ Some Wrong Solutions with Common Mistakes W1QQperC18 multiplying instead of dividing W2QQperC setting them equal to each other 1129 Consider a 2m deep swimming pool The pressure difference between the top and bottom of the pool is a 120 kPa b 196 kPa c 381 kPa d 508 kPa e 200 kPa Answer b 196 kPa Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values rho1000 kgm3 g981 ms2 z10 m z22 m DELTAPrhogz2z11000 kPa Some Wrong Solutions with Common Mistakes W1Prhoz1z21000 not using g W2Prhogz2z12000 taking half of z W3Prhogz2z1 not dividing by 1000 preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1130 At sea level the weight of 1 kg mass in SI units is 981 N The weight of 1 lbm mass in English units is a 1 lbf b 981 lbf c 322 lbf d 01 lbf e 0031 lbf Answer a 1 lbf Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values m1 lbm g322 fts2 Wmg322 lbf Some Wrong Solutions with Common Mistakes gSI981 ms2 W1W mgSI Using wrong conversion W2W mg Using wrong conversion W3W mgSI Using wrong conversion W4W mg Using wrong conversion 1131 During a heating process the temperature of an object rises by 10C This temperature rise is equivalent to a temperature rise of a 10F b 42F c 18 K d 18 R e 283 K Answer d 18 R Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TinC10 C TinRTinC18 R Some Wrong Solutions with Common Mistakes W1TinFTinC F setting C and F equal to each other W2TinFTinC1832 F converting to F W3TinK18TinC K wrong conversion from C to K W4TinKTinC273 K converting to K 1132 1133 Design and Essay Problems 21 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 2 ENERGY ENERGY TRANSFER AND GENERAL ENERGY ANALYSIS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 22 Forms of Energy 21C The macroscopic forms of energy are those a system possesses as a whole with respect to some outside reference frame The microscopic forms of energy on the other hand are those related to the molecular structure of a system and the degree of the molecular activity and are independent of outside reference frames 22C The sum of all forms of the energy a system possesses is called total energy In the absence of magnetic electrical and surface tension effects the total energy of a system consists of the kinetic potential and internal energies 23C Thermal energy is the sensible and latent forms of internal energy and it is referred to as heat in daily life 24C The mechanical energy is the form of energy that can be converted to mechanical work completely and directly by a mechanical device such as a propeller It differs from thermal energy in that thermal energy cannot be converted to work directly and completely The forms of mechanical energy of a fluid stream are kinetic potential and flow energies 25C Hydrogen is also a fuel since it can be burned but it is not an energy source since there are no hydrogen reserves in the world Hydrogen can be obtained from water by using another energy source such as solar or nuclear energy and then the hydrogen obtained can used as a fuel to power cars or generators Therefore it is more proper to view hydrogen is an energy carrier than an energy source 26E The total kinetic energy of an object is given is to be determined Analysis The total kinetic energy of the object is given by 300 Btu 2 2 2 2 037 ft s 25 1Btulbm 2 15 lbm 100 fts 2 KE m V 27 The total kinetic energy of an object is given is to be determined Analysis The total kinetic energy of the object is given by 200 kJ 2 2 2 2 m s 1000 1kJkg 2 100 kg 20 ms 2 KE m V PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 23 28E The specific potential energy of an object is to be determined Analysis In the English unit system the specific potential energy in Btu is given by 0128 Btulbm 2 2 2 037 ft s 25 1Btulbm 32 1 fts 100 ft pe gz 29E The total potential energy of an object is to be determined Analysis Substituting the given data into the potential energy expression gives 257 Btu 2 2 2 037 ft s 25 1Btulbm 200 lbm32 2 fts 10 ft PE mgz 210 The total potential energy of an object that is below a reference level is to be determined Analysis Substituting the given data into the potential energy expression gives 38 kJ 2 2 2 m s 1000 1kJkg ms 20 m 59 20 kg PE mgz 211 A person with his suitcase goes up to the 10th floor in an elevator The part of the energy of the elevator stored in the suitcase is to be determined Assumptions 1 The vibrational effects in the elevator are negligible Analysis The energy stored in the suitcase is stored in the form of potential energy which is mgz Therefore 103 kJ 2 2 2 suitcase m s 1000 1 kJkg ms 35 m 30 kg981 mg z PE E Therefore the suitcase on 10th floor has 103 kJ more energy compared to an identical suitcase on the lobby level Discussion Noting that 1 kWh 3600 kJ the energy transferred to the suitcase is 1033600 00029 kWh which is very small PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 24 212 A hydraulic turbinegenerator is to generate electricity from the water of a large reservoir The power generation potential is to be determined Assumptions 1 The elevation of the reservoir remains constant 2 The mechanical energy of water at the turbine exit is negligible Analysis The total mechanical energy water in a reservoir possesses is equivalent to the potential energy of water at the free surface and it can be converted to work entirely Therefore the power potential of water is its potential energy which is gz per unit mass and gz for a given mass flow rate m 1 574 kJkg m s 1000 1kJkg ms 160 m 981 2 2 2 mech gz pe e Generator 160 m Turbine Then the power generation potential becomes 5509 kW 1kJs 1kW 3500 kgs157 4 kJkg mech mech max me E W Therefore the reservoir has the potential to generate 1766 kW of power Discussion This problem can also be solved by considering a point at the turbine inlet and using flow energy instead of potential energy It would give the same result since the flow energy at the turbine inlet is equal to the potential energy at the free surface of the reservoir 213 Wind is blowing steadily at a certain velocity The mechanical energy of air per unit mass and the power generation potential are to be determined Assumptions The wind is blowing steadily at a constant uniform velocity Properties The density of air is given to be ρ 125 kgm3 Analysis Kinetic energy is the only form of mechanical energy the wind possesses and it can be converted to work entirely Therefore the power potential of the wind is its kinetic energy which is V22 per unit mass and for a given mass flow rate V 2 2 m 0 050 kJkg m s 1000 1 kJkg 2 10 ms 2 2 2 2 2 mech V ke e 35340 kgs 4 60 m 1 25 kgm 10 ms 4 2 3 2 π π ρ ρ D V VA m Wind turbine 60 m 10 ms Wind 1770 kW 35340 kgs0050 kJkg mech mech max me E W Therefore 1770 kW of actual power can be generated by this wind turbine at the stated conditions Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity and thus the power generation will change strongly with the wind conditions PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 25 214 A water jet strikes the buckets located on the perimeter of a wheel at a specified velocity and flow rate The power generation potential of this system is to be determined Assumptions Water jet flows steadily at the specified speed and flow rate Analysis Kinetic energy is the only form of harvestable mechanical energy the water jet possesses and it can be converted to work entirely Therefore the power potential of the water jet is its kinetic energy which is V22 per unit mass and V 2 2 for a given mass flow rate m Vj Nozzle Shaft kJkg 81 1000 m s 1kJkg 2 60 ms 2 2 2 2 2 mech V ke e 216 kW 1kJs 1kW 120 kgs18 kJkg mech mech max me E W Therefore 216 kW of power can be generated by this water jet at the stated conditions Discussion An actual hydroelectric turbine such as the Pelton wheel can convert over 90 of this potential to actual electric power PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 26 215 Two sites with specified wind data are being considered for wind power generation The site better suited for wind power generation is to be determined Assumptions 1The wind is blowing steadily at specified velocity during specified times 2 The wind power generation is negligible during other times Properties We take the density of air to be ρ 125 kgm3 it does not affect the final answer Analysis Kinetic energy is the only form of mechanical energy the wind possesses and it can be converted to work entirely Therefore the power potential of the wind is its kinetic energy which is V22 per unit mass and for a given mass flow rate Considering a unit flow area A 1 m V 2 2 m 2 the maximum wind power and power generation becomes 0 0245 kJkg m s 1000 1 kJkg 2 7 ms 2 2 2 2 2 1 1 mech1 V ke e 0 050 kJkg m s 1000 1 kJkg 2 10 ms 2 2 2 2 2 2 2 mech 2 V ke e Wind turbine V ms Wind 02144 kW 25 kgm 7 ms1 m 00245 kJkg 1 2 3 1 1 1 mech1 mech1 max1 V Ake m e E W ρ 0625 kW 25 kgm 10 ms1 m 0050 kJkg 1 2 3 2 2 mech 2 2 mech 2 max 2 V Ake m e E W ρ since 1 kW 1 kJs Then the maximum electric power generations per year become per m flow area 0 2144 kW3000 hyr 2 1 max1 max1 643 kWhyr t W E per m flow area 0 625 kW2000 hyr 2 2 max 2 max 2 1250 kWhyr t W E Therefore second site is a better one for wind generation Discussion Note the power generation of a wind turbine is proportional to the cube of the wind velocity and thus the average wind velocity is the primary consideration in wind power generation decisions PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 27 216 A river flowing steadily at a specified flow rate is considered for hydroelectric power generation by collecting the water in a dam For a specified water height the power generation potential is to be determined Assumptions 1 The elevation given is the elevation of the free surface of the river 2 The mechanical energy of water at the turbine exit is negligible Properties We take the density of water to be ρ 1000 kgm3 River 80 m Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free surface of the dam relative to free surface of discharge water and it can be converted to work entirely Therefore the power potential of water is its potential energy which is gz per unit mass and for a given mass flow rate gz m 0 7848 kJkg m s 1000 1kJkg ms 80 m 981 2 2 2 mech gz pe e The mass flow rate is 175000 kgs 1000 kgm 175 m s 3 3 V ρ m Then the power generation potential becomes 137 MW 1000 kJs 1MW 175000 kgs07848 kJkg mech mech max me E W Therefore 137 MW of power can be generated from this river if its power potential can be recovered completely Discussion Note that the power output of an actual turbine will be less than 137 MW because of losses and inefficiencies PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 28 217 A river is flowing at a specified velocity flow rate and elevation The total mechanical energy of the river water per unit mass and the power generation potential of the entire river are to be determined Assumptions 1 The elevation given is the elevation of the free surface of the river 2 The velocity given is the average velocity 3 The mechanical energy of water at the turbine exit is negligible Properties We take the density of water to be ρ 1000 kgm3 Analysis Noting that the sum of the flow energy and the potential energy is constant for a given fluid body we can take the elevation of the entire river water to be the elevation of the free surface and ignore the flow energy Then the total mechanical energy of the river water per unit mass becomes 0887 kJkg 2 2 2 2 2 mech m s 1000 1kJkg 2 3ms 981 ms 90 m 2 V gh ke pe e 3 ms 90 m River The power generation potential of the river water is obtained by multiplying the total mechanical energy by the mass flow rate 500000 kgs 1000 kgm 500 m s 3 3 V ρ m 444 MW 444000 kW 500000 kgs0887 kJkg mech mech max me E W Therefore 444 MW of power can be generated from this river as it discharges into the lake if its power potential can be recovered completely Discussion Note that the kinetic energy of water is negligible compared to the potential energy and it can be ignored in the analysis Also the power output of an actual turbine will be less than 444 MW because of losses and inefficiencies PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 29 Energy Transfer by Heat and Work 218C Energy can cross the boundaries of a closed system in two forms heat and work 219C The form of energy that crosses the boundary of a closed system because of a temperature difference is heat all other forms are work 220C An adiabatic process is a process during which there is no heat transfer A system that does not exchange any heat with its surroundings is an adiabatic system 221C Point functions depend on the state only whereas the path functions depend on the path followed during a process Properties of substances are point functions heat and work are path functions 222C a The cars radiator transfers heat from the hot engine cooling fluid to the cooler air No work interaction occurs in the radiator b The hot engine transfers heat to cooling fluid and ambient air while delivering work to the transmission c The warm tires transfer heat to the cooler air and to some degree to the cooler road while no work is produced No work is produced since there is no motion of the forces acting at the interface between the tire and road d There is minor amount of heat transfer between the tires and road Presuming that the tires are hotter than the road the heat transfer is from the tires to the road There is no work exchange associated with the road since it cannot move e Heat is being added to the atmospheric air by the hotter components of the car Work is being done on the air as it passes over and through the car 223C When the length of the spring is changed by applying a force to it the interaction is a work interaction since it involves a force acting through a displacement A heat interaction is required to change the temperature and hence length of the spring PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 210 224C a From the perspective of the contents heat must be removed in order to reduce and maintain the contents temperature Heat is also being added to the contents from the room air since the room air is hotter than the contents b Considering the system formed by the refrigerator box when the doors are closed there are three interactions electrical work and two heat transfers There is a transfer of heat from the room air to the refrigerator through its walls There is also a transfer of heat from the hot portions of the refrigerator ie back of the compressor where condenser is placed system to the room air Finally electrical work is being added to the refrigerator through the refrigeration system c Heat is transferred through the walls of the room from the warm room air to the cold winter air Electrical work is being done on the room through the electrical wiring leading into the room 225C a As one types on the keyboard electrical signals are produced and transmitted to the processing unit Simultaneously the temperature of the electrical parts is increased slightly The work done on the keys when they are depressed is work done on the system ie keyboard The flow of electrical current with its voltage drop does work on the keyboard Since the temperature of the electrical parts of the keyboard is somewhat higher than that of the surrounding air there is a transfer of heat from the keyboard to the surrounding air b The monitor is powered by the electrical current supplied to it This current and voltage drop is work done on the system ie monitor The temperatures of the electrical parts of the monitor are higher than that of the surrounding air Hence there is a heat transfer to the surroundings c The processing unit is like the monitor in that electrical work is done on it while it transfers heat to the surroundings d The entire unit then has electrical work done on it and mechanical work done on it to depress the keys It also transfers heat from all its electrical parts to the surroundings 226 The power produced by an electrical motor is to be expressed in different units Analysis Using appropriate conversion factors we obtain a 5 N ms J 1 1N m 1 W 1Js 5 W W b 5 kg m2s3 1N 1kg ms J 1 1N m 1 W 1Js 5 W 2 W 227E The power produced by a model aircraft engine is to be expressed in different units Analysis Using appropriate conversion factors we obtain a 738 lbf fts 1Btus 778169 lbf fts 1055056 W 1Btus 10 W W b 00134 hp 745 7 W 1hp 10 W W PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 211 Mechanical Forms of Work 228C The work done is the same but the power is different 229 A car is accelerated from rest to 100 kmh The work needed to achieve this is to be determined Analysis The work needed to accelerate a body the change in kinetic energy of the body 309 kJ 2 2 2 2 1 22 1000 kg m s 1 kJ 0 3600 s 100000 m 2 800 kg 1 2 1 V m V Wa 230E A construction crane lifting a concrete beam is considered The amount of work is to be determined considering a the beam and b the crane as the system Analysis a The work is done on the beam and it is determined from Btu 185 lbf ft 144000 778169 lbf ft 1Btu 144000 lbf ft 24 ft 174 lbm fts 32 1lbf 3000 lbm32174 fts 2 2 2 mg z W 24 ft b Since the crane must produce the same amount of work as is required to lift the beam the work done by the crane is 185 Btu 144000 lbf ft W 231E A man is pushing a cart with its contents up a ramp that is inclined at an angle of 10 from the horizontal The work needed to move along this ramp is to be determined considering a the man and b the cart and its contents as the system Analysis a Considering the man as the system letting l be the displacement along the ramp and letting θ be the inclination angle of the ramp Btu 6248 lbf ft 4862 778169 lbf ft 1Btu 4862 lbf ft 100 180 lbf100 ftsin10 Fl sinθ W This is work that the man must do to raise the weight of the cart and contents plus his own weight a distance of lsinθ b Applying the same logic to the cart and its contents gives Btu 2231 lbf ft 1736 778169 lbf ft 1Btu 1736 lbf ft 100 lbf100 ftsin10 Fl sinθ W PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 212 232E The work required to compress a spring is to be determined Analysis Since there is no preload F kx Substituting this into the work expression gives Btu 00107 lbf ft 833 778169 lbf ft 1Btu 8 33 lbf ft 12 in 1ft 0 1 in 2 lbfin 200 2 2 2 2 1 2 2 2 1 2 1 2 1 x k x xdx k kxdx Fds W F x 233E The work required to expand a soap bubble is to be determined Analysis The surface tension work is determined from Btu 10 211 6 7782 lbf ft 1Btu 0 00164 lbf ft 00164 lbf ft 0 12 ft 50 2 12 ft 0 005 lbfft4 2 2 2 1 2 1 π σ σ A A dA W s 234E The work required to stretch a steel rod in a specieid length is to be determined Assumptions The Youngs modulus does not change as the rod is stretched Analysis The original volume of the rod is 3 2 2 0 2 356 in 12 in 4 in 50 4 π π L D V The work required to stretch the rod 0125 in is Btu 10 411 4 9338 lbf in 1Btu 2 835 lbf in 835 lbf in 2 0 0 12512 in 2 2 356 in 30000 lbfin 2 2 2 2 3 2 1 2 2 0 ε E ε W V PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 213 235E The work required to compress a spring is to be determined Analysis The force at any point during the deflection of the spring is given by F F0 kx where F0 is the initial force and x is the deflection as measured from the point where the initial force occurred From the perspective of the spring this force acts in the direction opposite to that in which the spring is deflected Then 00214 Btu 12 in 1ft 778169 lbf ft 1Btu 200 lbf in lbf in 200 0 in 1 2 200 lbfin 100 lbf 1 0in 2 2 2 2 2 1 2 2 1 2 0 2 1 0 2 1 x k x x x F kx dx F Fds W F x 236 The work required to compress a spring is to be determined Analysis Since there is no preload F kx Substituting this into the work expression gives 0135 kJ 1kN m 1kJ 0 135 kN m kN m 0135 0 0 03 m 2 kNm 300 2 2 2 2 1 2 2 2 1 2 1 2 1 x k x xdx k kxdx Fds W F x PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 214 237 A ski lift is operating steadily at 10 kmh The power required to operate and also to accelerate this ski lift from rest to the operating speed are to be determined Assumptions 1 Air drag and friction are negligible 2 The average mass of each loaded chair is 250 kg 3 The mass of chairs is small relative to the mass of people and thus the contribution of returning empty chairs to the motion is disregarded this provides a safety factor Analysis The lift is 1000 m long and the chairs are spaced 20 m apart Thus at any given time there are 100020 50 chairs being lifted Considering that the mass of each chair is 250 kg the load of the lift at any given time is Load 50 chairs250 kgchair 12500 kg Neglecting the work done on the system by the returning empty chairs the work needed to raise this mass by 200 m is 24525 kJ kg m s 1000 1 kJ ms 200 m 12500 kg981 2 2 2 1 2 z mg z Wg At 10 kmh it will take t distance velocity 1 km 10 km h 01 h 360 s to do this work Thus the power needed is 681 kW s 360 kJ 24525 t W W g g The velocity of the lift during steady operation and the acceleration during start up are 2778 ms 36 kmh 1ms 10 kmh V 0 556 ms2 s 5 2 778 ms 0 t V a During acceleration the power needed is 96 kW 5 s m s 1000 1 kJkg 0 2 12500 kg 2 778 ms 1 2 1 2 2 2 2 1 22 t V m V Wa Assuming the power applied is constant the acceleration will also be constant and the vertical distance traveled during acceleration will be 139 m 2 0556 ms 5 s 02 1 1000 m 200 m 2 1 sin 2 1 2 2 2 2 at at h α and 34 1 kW 5 s kg m s 1000 1 kJkg ms 139 m 12500 kg981 2 2 2 1 2 t z mg z Wg Thus 437 kW 34 1 69 total g a W W W PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 215 238 A car is to climb a hill in 12 s The power needed is to be determined for three different cases Assumptions Air drag friction and rolling resistance are negligible Analysis The total power required for each case is the sum of the rates of changes in potential and kinetic energies That is g a W W W total a since the velocity is constant Also the vertical rise is h 100 msin 30 50 m Thus Wa 0 470 kW 12 s kg m s 1000 1 kJ 1150 kg981 ms 50 m 2 2 2 1 2 t z mg z Wg and 470 kW 47 0 0 total g a W W W b The power needed to accelerate is 43 1 kW 12 s kg m s 1000 1 kJ 0 2 1150 kg 30 ms 1 2 1 2 2 2 2 1 22 t V m V Wa and 901 kW 43 1 47 0 total g a W W W c The power needed to decelerate is 57 5 kW 12 s kg m s 1000 1 kJ 35 ms 5 ms 2 1150 kg 1 2 1 2 2 2 2 2 1 22 t V m V Wa and 105 kW breaking power 47 1 57 5 total g a W W W 239 A damaged car is being towed by a truck The extra power needed is to be determined for three different cases Assumptions Air drag friction and rolling resistance are negligible Analysis The total power required for each case is the sum of the rates of changes in potential and kinetic energies That is g a W W W total a Zero b Thus Wa 0 817 kW 05 m s 1000 kJkg 1 s 3600 kg981ms 50000 m 1200 sin 30 2 2 2 1 2 total o mgV mgV t z mg t z mg z W W z g c Thus Wg 0 313 kW 12 s 1000 m s 1 kJkg 0 3600 s 90000 m 2 1200 kg 1 2 1 2 2 2 2 1 2 2 total t V m V W W a PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 216 The First Law of Thermodynamics 240C No This is the case for adiabatic systems only 241C Energy can be transferred to or from a control volume as heat various forms of work and by mass transport 242C Warmer Because energy is added to the room air in the form of electrical work 243E The high rolling resistance tires of a car are replaced by low rolling resistance ones For a specified unit fuel cost the money saved by switching to low resistance tires is to be determined Assumptions 1The low rolling resistance tires deliver 2 mpg over all velocities 2 The car is driven 15000 miles per year Analysis The annual amount of fuel consumed by this car on high and lowrolling resistance tires are 428 6 galyear 35 milesgal 15000 milesyear Miles per gallon Miles driven per year Annual Fuel Consumption High 405 4 galyear 37 milesgal 15000 milesyear Miles per gallon Miles driven per year Annual Fuel Consumption Low Then the fuel and money saved per year become 232 galyear 4054 galyear 6 galyear 428 Annual Fuel Consumption Annual Fuel Consumption Fuel Savings Low High 510year 23 2 galyear220gal Fuel savingsUnit cost of fuel Cost savings Discussion A typical tire lasts about 3 years and thus the low rolling resistance tires have the potential to save about 150 to the car owner over the life of the tires which is comparable to the installation cost of the tires 244 The specific energy change of a system which is accelerated is to be determined Analysis Since the only property that changes for this system is the velocity only the kinetic energy will change The change in the specific energy is 045 kJkg 2 2 2 2 2 1 22 m s 1000 1kJkg 2 0 ms 30 ms 2 ke V V PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 217 245 The specific energy change of a system which is raised is to be determined Analysis Since the only property that changes for this system is the elevation only the potential energy will change The change in the specific energy is then 098 kJkg 2 2 2 1 2 m s 1000 1kJkg 0 m ms 100 89 pe z g z 246E A water pump increases water pressure The power input is to be determined Analysis The power input is determined from 50 psia 126 hp 07068 Btus hp 1 psia ft 5404 1Btu ft s50 10psia 21 3 3 1 2 P P W V Water 10 psia The water temperature at the inlet does not have any significant effect on the required power 247 A classroom is to be airconditioned using window airconditioning units The cooling load is due to people lights and heat transfer through the walls and the windows The number of 5kW window air conditioning units required is to be determined Assumptions There are no heat dissipating equipment such as computers TVs or ranges in the room Analysis The total cooling load of the room is determined from Q Q Q Q cooling lights people heat gain 15000 kJh Qcool Room 40 people 10 bulbs where Q Q Q lights people heat gain 10 100 W 1 kW 40 360 kJ h 4 kW 15000 kJ h 417 kW Substituting Qcooling 917 kW 1 4 417 Thus the number of airconditioning units required is 2 units kWunit 183 5 917 kW PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 218 248 The lighting energy consumption of a storage room is to be reduced by installing motion sensors The amount of energy and money that will be saved as well as the simple payback period are to be determined Assumptions The electrical energy consumed by the ballasts is negligible Analysis The plant operates 12 hours a day and thus currently the lights are on for the entire 12 hour period The motion sensors installed will keep the lights on for 3 hours and off for the remaining 9 hours every day This corresponds to a total of 9365 3285 off hours per year Disregarding the ballast factor the annual energy and cost savings become Energy Savings Number of lampsLamp wattageReduction of annual operating hours 24 lamps60 Wlamp 3285 hoursyear 4730 kWhyear Cost Savings Energy SavingsUnit cost of energy 4730 kWhyear008kWh 378year The implementation cost of this measure is the sum of the purchase price of the sensor plus the labor Implementation Cost Material Labor 32 40 72 This gives a simple payback period of 23 months 378 year 72 Annual cost savings Implementation cost Simple payback period 019 year Therefore the motion sensor will pay for itself in about 2 months 249 The classrooms and faculty offices of a university campus are not occupied an average of 4 hours a day but the lights are kept on The amounts of electricity and money the campus will save per year if the lights are turned off during unoccupied periods are to be determined Analysis The total electric power consumed by the lights in the classrooms and faculty offices is 528 kW 264 264 264 kW consumed per lamp No of lamps 400 6 110 W 264000 Power 264 kW consumed per lamp No of lamps 200 12 110 W 264000 Power lighting offices lighting classroom total lighting offices lighting classroom lighting E E E E E Noting that the campus is open 240 days a year the total number of unoccupied work hours per year is Unoccupied hours 4 hoursday240 daysyear 960 hyr Then the amount of electrical energy consumed per year during unoccupied work period and its cost are 41564yr 506880 kWhyr0082kWh Energy savingsUnit cost of energy savings Cost 506880 kWh 528 kW960 hyr Unoccupied hours Energy savings lighting total E Discussion Note that simple conservation measures can result in significant energy and cost savings PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 250 A room contains a light bulb a TV set a refrigerator and an iron The rate of increase of the energy content of the room when all of these electric devices are on is to be determined Assumptions 1 The room is well sealed and heat loss from the room is negligible 2 All the appliances are kept on Analysis Taking the room as the system the rate form of the energy balance can be written as Ėin Ėout dEsystem dt dEroom dt Ėin since no energy is leaving the room in any form and thus Ėout 0 Also Ėin Ėlights ĖTV Ėrefrig Ėiron 100 110 200 1000 W 1410 W Substituting the rate of increase in the energy content of the room becomes dEroom dt Ėin 1410 W Discussion Note that some appliances such as refrigerators and irons operate intermittently switching on and off as controlled by a thermostat Therefore the rate of energy transfer to the room in general will be less 251 A fan is to accelerate quiescent air to a specified velocity at a specified flow rate The minimum power that must be supplied to the fan is to be determined Assumptions The fan operates steadily Properties The density of air is given to be ρ 118 kgm3 Analysis A fan transmits the mechanical energy of the shaft shaft power to mechanical energy of air kinetic energy For a control volume that encloses the fan the energy balance can be written as Ėin Ėout dEsystem dt steady 0 Ėin Ėout Wshin ṁair keout ṁair Vout2 2 where ṁair ρ V 118 kgm39 m3s 1062 kgs Substituting the minimum power input required is determined to be Wshin ṁair Vout2 2 1062 kgs 8 ms2 2 1 Jkg 1 m2s2 340 Js 340 W Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another and it does not allow any energy to be created or destroyed during a process In reality the power required will be considerably higher because of the losses associated with the conversion of mechanical shaft energy to kinetic energy of air 252E A fan accelerates air to a specified velocity in a square duct The minimum electric power that must be supplied to the fan motor is to be determined Assumptions 1 The fan operates steadily 2 There are no conversion losses Properties The density of air is given to be ρ 0075 lbmft3 Analysis A fan motor converts electrical energy to mechanical shaft energy and the fan transmits the mechanical energy of the shaft shaft power to mechanical energy of air kinetic energy For a control volume that encloses the fanmotor unit the energy balance can be written as Ėin Ėout dEsystem dt steady 0 Ėin Ėout Welectin ṁair keout ṁair Vout2 2 where ṁair ρV A 0075 lbmft33 3 ft222 fts 1485 lbms Substituting the minimum power input required is determined to be Win ṁair Vout2 2 1485 lbms 22 fts2 2 1 Btulbm 25037 ft2s2 01435 Btus 151 W since 1 Btu 1055 kJ and 1 kJs 1000 W Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another and it does not allow any energy to be created or destroyed during a process In reality the power required will be considerably higher because of the losses associated with the conversion of electricaltomechanical shaft and mechanical shafttokinetic energy of air 253 A gasoline pump raises the pressure to a specified value while consuming electric power at a specified rate The maximum volume flow rate of gasoline is to be determined Assumptions 1 The gasoline pump operates steadily 2 The changes in kinetic and potential energies across the pump are negligible Analysis For a control volume that encloses the pumpmotor unit the energy balance can be written as Eindot Eoutdot dEsystem dttheta steady 0 Eindot Eoutdot Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Windot mdotPV1 mdotPV2 Windot mdotP2 P1 v Vdot ΔP since mdot Vdot v and the changes in kinetic and potential energies of gasoline are negligible Solving for volume flow rate and substituting the maximum flow rate is determined to be Vdotmax Windot ΔP 38 kJs 7 kPa 1 kPa m3 1 kJ 0543 m3s Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another and it does not allow any energy to be created or destroyed during a process In reality the volume flow rate will be less because of the losses associated with the conversion of electricaltomechanical shaft and mechanical shafttoflow energy 222 254 An inclined escalator is to move a certain number of people upstairs at a constant velocity The minimum power required to drive this escalator is to be determined Assumptions 1 Air drag and friction are negligible 2 The average mass of each person is 75 kg 3 The escalator operates steadily with no acceleration or breaking 4 The mass of escalator itself is negligible Analysis At design conditions the total mass moved by the escalator at any given time is Mass 30 persons75 kgperson 2250 kg The vertical component of escalator velocity is mssin45 80 sin 45 vert V V Under stated assumptions the power supplied is used to increase the potential energy of people Taking the people on elevator as the closed system the energy balance in the rate form can be written as 0 potential etc energies of change in internal kinetic Rate system by heat work and mass of net energy transfer Rate out in 14243 4243 1 dt dE E E t E dt dE E sys sys in vert in mgV t mg z t PE W That is under stated assumptions the power input to the escalator must be equal to the rate of increase of the potential energy of people Substituting the required power input becomes 125 kW 12 5 kJs 1000 m s 1 kJkg 2250 kg981 ms 08 mssin45 2 2 2 vert in mgV W When the escalator velocity is doubled to V 16 ms the power needed to drive the escalator becomes 250 kW 25 0 kJs 1000 m s 1 kJkg 2250 kg981 ms 16 mssin45 2 2 2 vert in mgV W Discussion Note that the power needed to drive an escalator is proportional to the escalator velocity PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 223 255 An automobile moving at a given velocity is considered The power required to move the car and the area of the effective flow channel behind the car are to be determined Analysis The absolute pressure of the air is 9331 kPa 1mm Hg 700 mm Hg 01333 kPa P and the specific volume of the air is 0 9012 m kg 9331 kPa 0 287 kPa m kg K293 K 3 3 P RT v The mass flow rate through the control volume is 8322 kgs m kg 09012 3 m 9036 ms 3 2 1 1 v A V m The power requirement is 442 kW 2 2 2 2 2 2 12 m s 1000 1kJkg 2 ms 63 82 ms 63 8322 kgs 90 2 V m V W The outlet area is 329 m2 8236 ms 8322 kgs 0 9012 m kg 3 2 2 2 2 V m A A V m v v PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 224 Energy Conversion Efficiencies 256C Mechanical efficiency is defined as the ratio of the mechanical energy output to the mechanical energy input A mechanical efficiency of 100 for a hydraulic turbine means that the entire mechanical energy of the fluid is converted to mechanical shaft work 257C The combined pumpmotor efficiency of a pumpmotor system is defined as the ratio of the increase in the mechanical energy of the fluid to the electrical power consumption of the motor in elect pump in elect fluid mech in elect mechin mechout motor pump pumpmotor W W W E W E E η η η The combined pumpmotor efficiency cannot be greater than either of the pump or motor efficiency since both pump and motor efficiencies are less than 1 and the product of two numbers that are less than one is less than either of the numbers 258C The turbine efficiency generator efficiency and combined turbinegenerator efficiency are defined as follows Mechanical energy extracted from the fluid energy output Mechanical fluid mech shaftout turbine E W η in shaft electout generator Mechanical power input power output Electrical W W η fluid mech out elect mechout in mech electout turbine generator turbinegen E W E E W η η η 259C No the combined pumpmotor efficiency cannot be greater that either of the pump efficiency of the motor efficiency This is because motor pump pumpmotor η η η and both ηpump and ηmotor are less than one and a number gets smaller when multiplied by a number smaller than one PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 225 260 A hooded electric open burner and a gas burner are considered The amount of the electrical energy used directly for cooking and the cost of energy per utilized kWh are to be determined Analysis The efficiency of the electric heater is given to be 73 percent Therefore a burner that consumes 3kW of electrical energy will supply 73 38 electric gas η η Efficiency 24 kW073 175 kW Energy input utilized Q of useful energy The unit cost of utilized energy is inversely proportional to the efficiency and is determined from 0137kWh 073 0 10 kWh Efficiency Cost of energy input Cost of utilized energy Noting that the efficiency of a gas burner is 38 percent the energy input to a gas burner that supplies utilized energy at the same rate 175 kW is 15700 Btuh 038 1 75 kW Efficiency utilized input gas kW 461 Q Q since 1 kW 3412 Btuh Therefore a gas burner should have a rating of at least 15700 Btuh to perform as well as the electric unit Noting that 1 therm 293 kWh the unit cost of utilized energy in the case of gas burner is determined the same way to be 0108kWh 038 1 20 29 3 kWh Efficiency Cost of energy input Cost of utilized energy 261 A worn out standard motor is replaced by a high efficiency one The reduction in the internal heat gain due to the higher efficiency under full load conditions is to be determined Assumptions 1 The motor and the equipment driven by the motor are in the same room 2 The motor operates at full load so that fload 1 Analysis The heat generated by a motor is due to its inefficiency and the difference between the heat generated by two motors that deliver the same shaft power is simply the difference between the electric power drawn by the motors 75 746 W0954 58648 W 75 746 W091 61484 W motor shaft electric efficient in motor shaft electric standard in η η W W W W Then the reduction in heat generation becomes 58648 2836 W 61484 in electric efficient in electric standard reduction W W Q PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 226 262 An electric car is powered by an electric motor mounted in the engine compartment The rate of heat supply by the motor to the engine compartment at full load conditions is to be determined Assumptions The motor operates at full load so that the load factor is 1 Analysis The heat generated by a motor is due to its inefficiency and is equal to the difference between the electrical energy it consumes and the shaft power it delivers 8 90 hp 664 kW 90 90 98 90 hp091 9890 hp shaft out in electric generation motor shaft electric in W W Q W W η since 1 hp 0746 kW Discussion Note that the electrical energy not converted to mechanical power is converted to heat 263 A worn out standard motor is to be replaced by a high efficiency one The amount of electrical energy and money savings as a result of installing the high efficiency motor instead of the standard one as well as the simple payback period are to be determined Assumptions The load factor of the motor remains constant at 075 Analysis The electric power drawn by each motor and their difference can be expressed as 1 Power ratingLoad factor1 savings Power Power ratingLoad factor Power ratingLoad factor efficient standard electric in efficient in standard electric efficient efficient shaft in efficient electric standard standard shaft in standard electric η η η η η η W W W W W W where ηstandard is the efficiency of the standard motor and ηefficient is the efficiency of the comparable high efficiency motor Then the annual energy and cost savings associated with the installation of the high efficiency motor are determined to be Energy Savings Power savingsOperating Hours 4 95 0 91 new old η η Power RatingOperating HoursLoad Factor1ηstandard 1ηefficient 75 hp0746 kWhp4368 hoursyear0751091 10954 9290 kWhyear Cost Savings Energy savingsUnit cost of energy 9290 kWhyear008kWh 743year The implementation cost of this measure consists of the excess cost the high efficiency motor over the standard one That is Implementation Cost Cost differential 5520 5449 71 This gives a simple payback period of months or 11 743 year 71 Annual cost savings Implementation cost Simple payback period 0096 year Therefore the highefficiency motor will pay for its cost differential in about one month PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 227 264E The combustion efficiency of a furnace is raised from 07 to 08 by tuning it up The annual energy and cost savings as a result of tuning up the boiler are to be determined Assumptions The boiler operates at full load while operating Boiler 70 55106 Analysis The heat output of boiler is related to the fuel energy input to the boiler by Boiler output Boiler inputCombustion efficiency or furnace in out Q η Q The current rate of heat input to the boiler is given to be 10 Btuh 55 6 in current Q Then the rate of useful heat output of the boiler becomes 385 10 Btuh 10 Btuh07 55 6 6 furnace current in out Q η Q The boiler must supply useful heat at the same rate after the tune up Therefore the rate of heat input to the boiler after the tune up and the rate of energy savings become 0 69 10 Btuh 481 10 10 55 481 10 Btuh 385 10 Btuh08 6 6 6 in new in current saved in 6 6 furnace new out new in Q Q Q Q Q η Then the annual energy and cost savings associated with tuning up the boiler become Energy Savings in saved Operation hours Q 069106 Btuh4200 hyear 289109 Btuyr Cost Savings Energy SavingsUnit cost of energy 289109 Btuyr435106 Btu 12600year Discussion Notice that tuning up the boiler will save 12600 a year which is a significant amount The implementation cost of this measure is negligible if the adjustment can be made by inhouse personnel Otherwise it is worthwhile to have an authorized representative of the boiler manufacturer to service the boiler twice a year PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 228 265E Problem 264E is reconsidered The effects of the unit cost of energy and combustion efficiency on the annual energy used and the cost savings as the efficiency varies from 07 to 09 and the unit cost varies from 4 to 6 per million Btu are the investigated The annual energy saved and the cost savings are to be plotted against the efficiency for unit costs of 4 5 and 6 per million Btu Analysis The problem is solved using EES and the solution is given below Given Qdotincurrent55E6 Btuh etafurnacecurrent07 etafurnacenew08 Hours4200 hyear UnitCost435E6 Btu Analysis QdotoutQdotincurrentetafurnacecurrent QdotinnewQdotoutetafurnacenew QdotinsavedQdotincurrentQdotinnew EnergysavingsQdotinsavedHours CostSavingsEnergySavingsUnitCost 068 072 076 08 084 088 092 0x100 109 2x109 3x109 4x109 5x109 6x109 ηfurnacenew Energysavings Btuyear ηfurnacenew EnergySaving s Btuyear CostSaving s year 07 072 074 076 078 08 082 084 086 088 09 000E00 642E08 125E09 182E09 237E09 289E09 338E09 385E09 430E09 473E09 513E09 0 3208 6243 9118 11846 14437 16902 19250 21488 23625 25667 068 072 076 08 084 088 092 0 5000 10000 15000 20000 25000 30000 ηfurnacenew CostSavings year 5x106 Btu 4x106 Btu 6x106 Btu Table values are for UnitCost 5E5 Btu PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 229 266 Several people are working out in an exercise room The rate of heat gain from people and the equipment is to be determined Assumptions The average rate of heat dissipated by people in an exercise room is 525 W Analysis The 8 weight lifting machines do not have any motors and thus they do not contribute to the internal heat gain directly The usage factors of the motors of the treadmills are taken to be unity since they are used constantly during peak periods Noting that 1 hp 746 W the total heat generated by the motors is 746 W 070 10077 6782 W 52 4 No of motors motor usage load motor motors η f f W Q The heat gain from 14 people is Q Q people person No of people W W 14 525 7350 Then the total rate of heat gain of the exercise room during peak period becomes 14132 W 7350 6782 people motors total Q Q Q 267 A room is cooled by circulating chilled water through a heat exchanger and the air is circulated through the heat exchanger by a fan The contribution of the fanmotor assembly to the cooling load of the room is to be determined Assumptions The fan motor operates at full load so that fload 1 Analysis The entire electrical energy consumed by the motor including the shaft power delivered to the fan is eventually dissipated as heat Therefore the contribution of the fanmotor assembly to the cooling load of the room is equal to the electrical energy it consumes 0 25 hp054 0463 hp 345 W motor shaft in electric internal generation η W W Q since 1 hp 746 W PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 230 268 A hydraulic turbinegenerator is to generate electricity from the water of a lake The overall efficiency the turbine efficiency and the shaft power are to be determined Assumptions 1 The elevation of the lake and that of the discharge site remains constant 2 Irreversible losses in the pipes are negligible Properties The density of water can be taken to be ρ 1000 kgm3 The gravitational acceleration is g 981 ms2 Analysis a We take the bottom of the lake as the reference level for convenience Then kinetic and potential energies of water are zero and the mechanical energy of water consists of pressure energy only which is 0 491 kJkg m s 1000 1kJkg ms 50 m 981 2 2 2 mechout in mech gh P e e ρ Then the rate at which mechanical energy of fluid supplied to the turbine and the overall efficiency become 2455 kW 5000 kgs0491 kJkg mechin mechin mechfluid e m e E 0760 2455 kW 1862 kW fluid mech electout turbinegen overall E W η η b Knowing the overall and generator efficiencies the mechanical efficiency of the turbine is determined from 0800 0 95 0 76 generator turbinegen turbine generator turbine turbinegen η η η η η η c The shaft power output is determined from the definition of mechanical efficiency 1960 kW 1964 kW 0 8002455 kW mechfluid turbine shaftout E W η Therefore the lake supplies 2455 kW of mechanical energy to the turbine which converts 1964 kW of it to shaft work that drives the generator which generates 1862 kW of electric power PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 231 269 Wind is blowing steadily at a certain velocity The mechanical energy of air per unit mass the power generation potential and the actual electric power generation are to be determined Assumptions 1 The wind is blowing steadily at a constant uniform velocity 2 The efficiency of the wind turbine is independent of the wind speed Wind turbine 80 m 7 ms Wind Properties The density of air is given to be ρ 125 kgm3 Analysis Kinetic energy is the only form of mechanical energy the wind possesses and it can be converted to work entirely Therefore the power potential of the wind is its kinetic energy which is V22 per unit mass and for a given mass flow rate V 2 2 m 0 0245 kJkg m s 1000 1kJkg 2 7 ms 2 2 2 2 2 mech V ke e 43982 kgs 4 80 m 1 25 kgm 7 ms 4 2 3 2 π π ρ ρ D V VA m 1078 kW 43982 kgs00245 kJkg mech mech max me E W The actual electric power generation is determined by multiplying the power generation potential by the efficiency 323 kW 0 301078 kW max wind turbine elect W W η Therefore 323 kW of actual power can be generated by this wind turbine at the stated conditions Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity and thus the power generation will change strongly with the wind conditions PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 232 270 Problem 269 is reconsidered The effect of wind velocity and the blade span diameter on wind power generation as the velocity varies from 5 ms to 20 ms in increments of 5 ms and the diameter varies from 20 m to 120 m in increments of 20 m is to be investigated Analysis The problem is solved using EES and the solution is given below Given V7 ms D80 m etaoverall030 rho125 kgm3 Analysis g981 ms2 ApiD24 mdotrhoAV WdotmaxmdotV22Convertm2s2 kJkg WdotelectetaoverallWdotmax 4 6 8 10 12 14 16 18 20 0 2000 4000 6000 8000 10000 12000 14000 16000 V ms Welect kW D20 m D40 m D60 m D100 m D80 m D120 m PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 233 271 Water is pumped from a lake to a storage tank at a specified rate The overall efficiency of the pumpmotor unit and the pressure difference between the inlet and the exit of the pump are to be determined Assumptions 1 The elevations of the tank and the lake remain constant 2 Frictional losses in the pipes are negligible 3 The changes in kinetic energy are negligible 4 The elevation difference across the pump is negligible Properties We take the density of water to be ρ 1000 kgm3 2 Analysis a We take the free surface of the lake to be point 1 and the free surfaces of the storage tank to be point 2 We also take the lake surface as the reference level z1 0 and thus the potential energy at points 1 and 2 are pe1 0 and pe2 gz2 The flow energy at both points is zero since both 1 and 2 are open to the atmosphere P1 P2 Patm Further the kinetic energy at both points is zero ke1 ke2 0 since the water at both locations is essentially stationary The mass flow rate of water and its potential energy at point 2 are 1 Storage tank Pump 20 m 70 kgs 1000 kgm 0 070 m s 3 3 V ρ m 0 196 kJkg m s 1000 1 kJkg ms 20 m 981 2 2 2 2 2 pe gz Then the rate of increase of the mechanical energy of water becomes 137 kW 70 kgs019 6 kJkg 0 2 2 mechin mechout mechfluid mpe m pe e m e E The overall efficiency of the combined pumpmotor unit is determined from its definition 0672 or 672 204 kW 7 kW 13 in elect mechfluid pumpmotor W E η b Now we consider the pump The change in the mechanical energy of water as it flows through the pump consists of the change in the flow energy only since the elevation difference across the pump and the change in the kinetic energy are negligible Also this change must be equal to the useful mechanical energy supplied by the pump which is 137 kW P P m P e m e E V ρ 1 2 mechin mechout mechfluid Solving for P and substituting 196 kPa 1 kJ kPa m 1 m s 0070 137 kJs 3 3 mechfluid V E P Therefore the pump must boost the pressure of water by 196 kPa in order to raise its elevation by 20 m Discussion Note that only twothirds of the electric energy consumed by the pumpmotor is converted to the mechanical energy of water the remaining onethird is wasted because of the inefficiencies of the pump and the motor PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 234 272 A large wind turbine is installed at a location where the wind is blowing steadily at a certain velocity The electric power generation the daily electricity production and the monetary value of this electricity are to be determined Assumptions 1 The wind is blowing steadily at a constant uniform velocity 2 The efficiency of the wind turbine is independent of the wind speed Properties The density of air is given to be ρ 125 kgm3 Analysis Kinetic energy is the only form of mechanical energy the wind possesses and it can be converted to work entirely Therefore the power potential of the wind is its kinetic energy which is V22 per unit mass and for a given mass flow rate V 2 2 m 0 032 kJkg m s 1000 1 kJkg 2 8 ms 2 2 2 2 2 mech V ke e 78540 kgs 4 100 m 1 25 kgm 8 ms 4 2 3 2 π π ρ ρ D V VA m 100 m Wind turbine 8 ms Wind 2513 kW 78540 kgs0032 kJkg mech mech max me E W The actual electric power generation is determined from 8042 kW 0 322513 kW max wind turbine elect W W η Then the amount of electricity generated per day and its monetary value become Amount of electricity Wind powerOperating hours8042 kW24 h 19300 kWh Revenues Amount of electricityUnit price 19300 kWh006kWh 1158 per day Discussion Note that a single wind turbine can generate several thousand dollars worth of electricity every day at a reasonable cost which explains the overwhelming popularity of wind turbines in recent years 273E A water pump raises the pressure of water by a specified amount at a specified flow rate while consuming a known amount of electric power The mechanical efficiency of the pump is to be determined PUMP Pump inlet 6 hp P 12 psi Assumptions 1 The pump operates steadily 2 The changes in velocity and elevation across the pump are negligible 3 Water is incompressible Analysis To determine the mechanical efficiency of the pump we need to know the increase in the mechanical energy of the fluid as it flows through the pump which is 333 Btus 471 hp psi ft 5404 1Btu 15 ft s12 psi 3 3 1 2 1 2 1 2 mechin mechout fluid mech P P P m P P m P e m e E V v v v since 1 hp 07068 Btus and there is no change in kinetic and potential energies of the fluid Then the mechanical efficiency of the pump becomes v V V ρ m 0786 or 786 6 hp 71 hp 4 shaft pump mechfluid pump W E η Discussion The overall efficiency of this pump will be lower than 838 because of the inefficiency of the electric motor that drives the pump PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 235 274 Water is pumped from a lower reservoir to a higher reservoir at a specified rate For a specified shaft power input the power that is converted to thermal energy is to be determined Assumptions 1 The pump operates steadily 2 The elevations of the reservoirs remain constant 3 The changes in kinetic energy are negligible Reservoir 45 m Pump Reservoir 2 1 Properties We take the density of water to be ρ 1000 kgm3 Analysis The elevation of water and thus its potential energy changes during pumping but it experiences no changes in its velocity and pressure Therefore the change in the total mechanical energy of water is equal to the change in its potential energy which is gz per unit mass and gz for a given mass flow rate That is m 13 2 kW 1000 N ms kW 1 kg ms 1 1N 1000 kgm 003 m s981 ms 45 m 2 2 3 3 mech mech g z mg z m pe m e E V ρ Then the mechanical power lost because of frictional effects becomes 68 kW 13 2 kW 20 mech pump in frict E W W Discussion The 68 kW of power is used to overcome the friction in the piping system The effect of frictional losses in a pump is always to convert mechanical energy to an equivalent amount of thermal energy which results in a slight rise in fluid temperature Note that this pumping process could be accomplished by a 132 kW pump rather than 20 kW if there were no frictional losses in the system In this ideal case the pump would function as a turbine when the water is allowed to flow from the upper reservoir to the lower reservoir and extract 132 kW of power from the water 275 The mass flow rate of water through the hydraulic turbines of a dam is to be determined Analysis The mass flow rate is determined from 49500 kgs 2 2 2 1 2 1 2 m s 1000 1kJkg 0 m ms 206 89 100000 kJs z g z W m z mg z W PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 236 276 A pump is pumping oil at a specified rate The pressure rise of oil in the pump is measured and the motor efficiency is specified The mechanical efficiency of the pump is to be determined Assumptions 1 The flow is steady and incompressible 2 The elevation difference across the pump is negligible Properties The density of oil is given to be ρ 860 kgm3 Analysis Then the total mechanical energy of a fluid is the sum of the potential flow and kinetic energies and is expressed per unit mass as To determine the mechanical efficiency of the pump we need to know the increase in the mechanical energy of the fluid as it flows through the pump which is 2 2 mech V Pv gh e 2 2 2 2 1 2 2 1 2 2 1 1 2 2 2 mechin mechout mechfluid V V P P V Pv V Pv m e m e E ρ V since and there is no change in the potential energy of the fluid Also V v V ρ m 19 9 ms m 4 008 m s 10 4 2 3 2 1 1 1 π πD A V V V 8 84 ms m 4 012 m s 10 4 2 3 2 2 2 2 π πD A V V V Substituting the useful pumping power is determined to be 26 3 kW 1 kN ms kW 1 kg ms 1000 1 kN 2 199 ms 860 kgm 8 84 ms m s 400 kNm 01 2 2 2 3 2 3 mechfluid u pump E W PUMP Pump inlet 1 2 Motor 35 kW Then the shaft power and the mechanical efficiency of the pump become 31 5 kW 0 9035 kW electric motor pumpshaft W W η 836 0 836 315 kW 3 kW 26 shaft pump pump u pump W W η Discussion The overall efficiency of this pumpmotor unit is the product of the mechanical and motor efficiencies which is 090836 075 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 237 277E Water is pumped from a lake to a nearby pool by a pump with specified power and efficiency The mechanical power used to overcome frictional effects is to be determined Assumptions 1 The flow is steady and incompressible 2 The elevation difference between the lake and the free surface of the pool is constant 3 The average flow velocity is constant since pipe diameter is constant Properties We take the density of water to be ρ 624 lbmft3 Analysis The useful mechanical pumping power delivered to water is 16 hp 0 8020 hp pump pump pumpu W W η Pool Lake Pump 80 ft 2 1 The elevation of water and thus its potential energy changes during pumping but it experiences no changes in its velocity and pressure Therefore the change in the total mechanical energy of water is equal to the change in its potential energy which is gz per unit mass and gz for a given mass flow rate That is m g z mg z m pe m e E V ρ mech mech Substituting the rate of change of mechanical energy of water becomes 1363 hp 550 lbf fts hp 1 lbm fts 322 1lbf 624 lbmft 15 ft s322 fts 80 ft 2 2 3 3 mech E Then the mechanical power lost in piping because of frictional effects becomes 237 hp 1363 hp 16 mech pump u frict E W W Discussion Note that the pump must supply to the water an additional useful mechanical power of 237 hp to overcome the frictional losses in pipes PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 238 278 A wind turbine produces 180 kW of power The average velocity of the air and the conversion efficiency of the turbine are to be determined Assumptions The wind turbine operates steadily Properties The density of air is given to be 131 kgm3 Analysis a The blade diameter and the blade span area are 42 m 88 60 s 1min 15 Lmin 36 kmh 1ms 250 kmh tip π πn V D 2 2 2 6140 m 4 8842 m 4 π πD A Then the average velocity of air through the wind turbine becomes 523 ms kgm 6140 m 31 1 000 kgs 42 2 3 A m V ρ b The kinetic energy of the air flowing through the turbine is 574 3 kW 2 42000 kgs52 3ms 1 V 2 1 KE 2 2 m Then the conversion efficiency of the turbine becomes 0313 313 574 3 kW 180 kW E K W η Discussion Note that about onethird of the kinetic energy of the wind is converted to power by the wind turbine which is typical of actual turbines PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 239 Energy and Environment 279C Energy conversion pollutes the soil the water and the air and the environmental pollution is a serious threat to vegetation wild life and human health The emissions emitted during the combustion of fossil fuels are responsible for smog acid rain and global warming and climate change The primary chemicals that pollute the air are hydrocarbons HC also referred to as volatile organic compounds VOC nitrogen oxides NOx and carbon monoxide CO The primary source of these pollutants is the motor vehicles 280C Smog is the brown haze that builds up in a large stagnant air mass and hangs over populated areas on calm hot summer days Smog is made up mostly of groundlevel ozone O3 but it also contains numerous other chemicals including carbon monoxide CO particulate matter such as soot and dust volatile organic compounds VOC such as benzene butane and other hydrocarbons Groundlevel ozone is formed when hydrocarbons and nitrogen oxides react in the presence of sunlight in hot calm days Ozone irritates eyes and damage the air sacs in the lungs where oxygen and carbon dioxide are exchanged causing eventual hardening of this soft and spongy tissue It also causes shortness of breath wheezing fatigue headaches nausea and aggravate respiratory problems such as asthma 281C Fossil fuels include small amounts of sulfur The sulfur in the fuel reacts with oxygen to form sulfur dioxide SO2 which is an air pollutant The sulfur oxides and nitric oxides react with water vapor and other chemicals high in the atmosphere in the presence of sunlight to form sulfuric and nitric acids The acids formed usually dissolve in the suspended water droplets in clouds or fog These acidladen droplets are washed from the air on to the soil by rain or snow This is known as acid rain It is called rain since it comes down with rain droplets As a result of acid rain many lakes and rivers in industrial areas have become too acidic for fish to grow Forests in those areas also experience a slow death due to absorbing the acids through their leaves needles and roots Even marble structures deteriorate due to acid rain 282C Carbon monoxide which is a colorless odorless poisonous gas that deprives the bodys organs from getting enough oxygen by binding with the red blood cells that would otherwise carry oxygen At low levels carbon monoxide decreases the amount of oxygen supplied to the brain and other organs and muscles slows body reactions and reflexes and impairs judgment It poses a serious threat to people with heart disease because of the fragile condition of the circulatory system and to fetuses because of the oxygen needs of the developing brain At high levels it can be fatal as evidenced by numerous deaths caused by cars that are warmed up in closed garages or by exhaust gases leaking into the cars 283C Carbon dioxide CO2 water vapor and trace amounts of some other gases such as methane and nitrogen oxides act like a blanket and keep the earth warm at night by blocking the heat radiated from the earth This is known as the greenhouse effect The greenhouse effect makes life on earth possible by keeping the earth warm But excessive amounts of these gases disturb the delicate balance by trapping too much energy which causes the average temperature of the earth to rise and the climate at some localities to change These undesirable consequences of the greenhouse effect are referred to as global warming or global climate change The greenhouse effect can be reduced by reducing the net production of CO2 by consuming less energy for example by buying energy efficient cars and appliances and planting trees PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 240 284E A person trades in his Ford Taurus for a Ford Explorer The extra amount of CO2 emitted by the Explorer within 5 years is to be determined Assumptions The Explorer is assumed to use 940 gallons of gasoline a year compared to 715 gallons for Taurus Analysis The extra amount of gasoline the Explorer will use within 5 years is Extra Gasoline Extra per yearNo of years 940 715 galyr5 yr 1125 gal Extra CO2 produced Extra gallons of gasoline usedCO2 emission per gallon 1125 gal197 lbmgal 22163 lbm CO2 Discussion Note that the car we choose to drive has a significant effect on the amount of greenhouse gases produced 285 A power plant that burns natural gas produces 059 kg of carbon dioxide CO2 per kWh The amount of CO2 production that is due to the refrigerators in a city is to be determined Assumptions The city uses electricity produced by a natural gas power plant Properties 059 kg of CO2 is produced per kWh of electricity generated given Analysis Noting that there are 300000 households in the city and each household consumes 700 kWh of electricity for refrigeration the total amount of CO2 produced is 123000 CO tonyear 2 23 10 CO kgyear 1 household700 kWhyear household059 kgkWh 300000 Amount of electricity consumedAmount of CO per kWh of CO produced Amount 2 8 2 2 Therefore the refrigerators in this city are responsible for the production of 123000 tons of CO2 286 A power plant that burns coal produces 11 kg of carbon dioxide CO2 per kWh The amount of CO2 production that is due to the refrigerators in a city is to be determined Assumptions The city uses electricity produced by a coal power plant Properties 11 kg of CO2 is produced per kWh of electricity generated given Analysis Noting that there are 300000 households in the city and each household consumes 700 kWh of electricity for refrigeration the total amount of CO2 produced is 231000 CO tonyear 2 31 10 CO kgyear 2 household700 kWhhousehold11 kgkWh 300000 Amount of electricity consumedAmount of CO per kWh of CO produced Amount 2 8 2 2 Therefore the refrigerators in this city are responsible for the production of 231000 tons of CO2 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 241 287E A household uses fuel oil for heating and electricity for other energy needs Now the household reduces its energy use by 20 The reduction in the CO2 production this household is responsible for is to be determined Properties The amount of CO2 produced is 154 lbm per kWh and 264 lbm per gallon of fuel oil given Analysis Noting that this household consumes 11000 kWh of electricity and 1500 gallons of fuel oil per year the amount of CO2 production this household is responsible for is 540 CO lbmyear 56 1500 galyr264 lbmgal kWhyr154 lbmkWh 11000 Amount of fuel oil consumedAmount of CO per gallon Amount of electricity consumedAmount of CO per kWh of CO produced Amount 2 2 2 2 Then reducing the electricity and fuel oil usage by 15 will reduce the annual amount of CO2 production by this household by CO lbmyear 8481 2 0 1556540 CO kgyear 0 15Current amount of CO production in CO produced Reduction 2 2 2 Therefore any measure that saves energy also reduces the amount of pollution emitted to the environment 288 A household has 2 cars a natural gas furnace for heating and uses electricity for other energy needs The annual amount of NOx emission to the atmosphere this household is responsible for is to be determined Properties The amount of NOx produced is 71 g per kWh 43 g per therm of natural gas and 11 kg per car given Analysis Noting that this household has 2 cars consumes 1200 therms of natural gas and 9000 kWh of electricity per year the amount of NOx production this household is responsible for is 9106 NO kgyear x 1200 thermsyr00043 kgtherm 9000 kWhyr00071 kgkWh kgcar cars11 2 Amount of gas consumedAmount of NO per gallon Amount of electricit y consumedAmount of NO per kWh No of carsAmount of NO produced per car of NO produced Amount x x x x Discussion Any measure that saves energy will also reduce the amount of pollution emitted to the atmosphere PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 242 Special Topic Mechanisms of Heat Transfer 289C The three mechanisms of heat transfer are conduction convection and radiation 290C Diamond has a higher thermal conductivity than silver and thus diamond is a better conductor of heat 291C No It is purely by radiation 292C In forced convection the fluid is forced to move by external means such as a fan pump or the wind The fluid motion in natural convection is due to buoyancy effects only 293C A blackbody is an idealized body that emits the maximum amount of radiation at a given temperature and that absorbs all the radiation incident on it Real bodies emit and absorb less radiation than a blackbody at the same temperature 294C Emissivity is the ratio of the radiation emitted by a surface to the radiation emitted by a blackbody at the same temperature Absorptivity is the fraction of radiation incident on a surface that is absorbed by the surface The Kirchhoffs law of radiation states that the emissivity and the absorptivity of a surface are equal at the same temperature and wavelength 295 The inner and outer surfaces of a brick wall are maintained at specified temperatures The rate of heat transfer through the wall is to be determined Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values 2 Thermal properties of the wall are constant Properties The thermal conductivity of the wall is given to be k 069 WmC Analysis Under steady conditions the rate of heat transfer through the wall is 1035 W 03 m 5 C 6 m 20 C5 069 Wm 2 cond L T kA Q PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 243 296 The inner and outer surfaces of a window glass are maintained at specified temperatures The amount of heat transferred through the glass in 5 h is to be determined Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values 2 Thermal properties of the glass are constant Properties The thermal conductivity of the glass is given to be k 078 WmC 6C Glass Analysis Under steady conditions the rate of heat transfer through the glass by conduction is 5616 W 0005 m 6 C 2 m 15 C2 078 Wm 2 cond L T kA Q 15C Then the amount of heat transferred over a period of 10 h becomes 202200 kJ 3600s 5616 kJs10 Qcond t Q 05 cm If the thickness of the glass is doubled to 1 cm then the amount of heat transferred will go down by half to 101100 kJ PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 244 297 Reconsider Prob 296 Using EES or other software investigate the effect of glass thickness on heat loss for the specified glass surface temperatures Let the glass thickness vary from 02 cm to 2 cm Plot the heat loss versus the glass thickness and discuss the results Analysis The problem is solved using EES and the solution is given below FUNCTION klookupmaterial If materialGlass then klookup078 If materialBrick then klookup072 If materialFiber Glass then klookup0043 If materialAir then klookup0026 If materialWoodoak then klookup017 END L2 m W2 m materialGlass Tin15 C Tout6 C k078 WmC t10 hr thickness05 cm ALW QdotlossAkTinToutthicknessconvertcmm QlosstotalQdotlosstconverthrsconvertJkJ Thicknes s cm Qlosstotal kJ 02 04 06 08 1 12 14 16 18 2 505440 252720 168480 126360 101088 84240 72206 63180 56160 50544 02 04 06 08 1 12 14 16 18 2 0 100000 200000 300000 400000 500000 600000 thickness cm Qlosstotal kJ PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 245 298 Heat is transferred steadily to boiling water in the pan through its bottom The inner surface temperature of the bottom of the pan is given The temperature of the outer surface is to be determined Assumptions 1 Steady operating conditions exist since the surface temperatures of the pan remain constant at the specified values 2 Thermal properties of the aluminum pan are constant Properties The thermal conductivity of the aluminum is given to be k 237 WmC Analysis The heat transfer surface area is 105C 500 W 04 cm A π r² π01 m² 00314 m² Under steady conditions the rate of heat transfer through the bottom of the pan by conduction is Q kA T L kA T T L 2 1 Substituting 500 W 237 W m C00314 m 105 C 0004 m 2 2 o o T which gives T2 1053C 299 The inner and outer glasses of a double pane window with a 1cm air space are at specified temperatures The rate of heat transfer through the window is to be determined Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values 2 Heat transfer through the window is onedimensional 3 Thermal properties of the air are constant 4 The air trapped between the two glasses is still and thus heat transfer is by conduction only Air 18C 1cm Q 6C Properties The thermal conductivity of air at room temperature is k 0026 WmC Table 23 Analysis Under steady conditions the rate of heat transfer through the window by conduction is 0125 kW 125 W 001 m 6 C 2 m 18 0026 Wm C2 2 cond o o L T kA Q PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 246 2100 Two surfaces of a flat plate are maintained at specified temperatures and the rate of heat transfer through the plate is measured The thermal conductivity of the plate material is to be determined Assumptions 1 Steady operating conditions exist since the surface temperatures of the plate remain constant at the specified values 2 Heat transfer through the plate is onedimensional 3 Thermal properties of the plate are constant 500 Wm2 Plate 2 cm 0C Analysis The thermal conductivity is determined directly from the steady onedimensional heat conduction relation to be 100C Wm C 01 100 0 C 500 Wm 0 02 m 2 2 1 2 1 T T Q A L k L T kAT Q 2101 A person is standing in a room at a specified temperature The rate of heat transfer between a person and the surrounding air by convection is to be determined Ts 34C Q Assumptions 1 Steady operating conditions exist 2 Heat transfer by radiation is not considered 3 The environment is at a uniform temperature Analysis The heat transfer surface area of the person is A πDL π03 m170 m 160 m² Under steady conditions the rate of heat transfer by convection is 336 W 20 C C160 m 34 15 Wm 2 2 conv hA T Q 2102 A spherical ball whose surface is maintained at a temperature of 110C is suspended in the middle of a room at 20C The total rate of heat transfer from the ball is to be determined Assumptions 1 Steady operating conditions exist since the ball surface and the surrounding air and surfaces remain at constant temperatures 2 The thermal properties of the ball and the convection heat transfer coefficient are constant and uniform Air 20C 110C D 9 cm Properties The emissivity of the ball surface is given to be ε 08 Q Analysis The heat transfer surface area is A πD² π 009 m2 002545 m2 Under steady conditions the rates of convection and radiation heat transfer are 1633 W 293 K K 383 K Wm 0800254 5 m 567 10 3435 W 20 C C002545 m 110 Wm 15 4 4 4 2 8 2 4 4 rad 2 2 conv o s T A T Q hA T Q εσ o o Therefore 507 W 1633 3435 rad conv total Q Q Q PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 247 2103 Reconsider Prob 2102 Using EES or other software investigate the effect of the convection heat transfer coefficient and surface emissivity on the heat transfer rate from the ball Let the heat transfer coefficient vary from 5 Wm2C to 30 Wm2C Plot the rate of heat transfer against the convection heat transfer coefficient for the surface emissivities of 01 05 08 and 1 and discuss the results Analysis The problem is solved using EES and the solution is given below Given D009 m TsConvertTempCK110 TfConvertTempCK20 h15 Wm2C epsilon08 Properties sigma567E8 Wm2K4 Analysis ApiD2 QdotconvhATsTf QdotradepsilonsigmaATs4Tf4 QdottotalQdotconvQdotrad h Wm2C Qtotal W 5 75 10 125 15 175 20 225 25 275 30 278 3353 3925 4498 507 5643 6216 6788 7361 7933 8506 5 10 15 20 25 30 30 40 50 60 70 80 90 h Wm2C Qtotal W ε01 ε05 ε08 ε1 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 248 2104 Hot air is blown over a flat surface at a specified temperature The rate of heat transfer from the air to the plate is to be determined Assumptions 1 Steady operating conditions exist 2 Heat transfer by radiation is not considered 3 The convection heat transfer coefficient is constant and uniform over the surface 30C 80C Air Analysis Under steady conditions the rate of heat transfer by convection is 22 kW 22000 W 30 C 4 m 80 C2 55 Wm 2 2 conv o hA T Q 2105 A 1000W iron is left on the iron board with its base exposed to the air at 20C The temperature of the base of the iron is to be determined in steady operation Air 20C Iron 1000 W Assumptions 1 Steady operating conditions exist 2 The thermal properties of the iron base and the convection heat transfer coefficient are constant and uniform 3 The temperature of the surrounding surfaces is the same as the temperature of the surrounding air Properties The emissivity of the base surface is given to be ε 06 Analysis At steady conditions the 1000 W of energy supplied to the iron will be dissipated to the surroundings by convection and radiation heat transfer Therefore Q Q Q total conv rad 1000 W where 293 K W 07 293 K K002 m 35 Wm 2 2 conv s s T T hA T Q and Q A T T T s o s rad 2 8 2 4 s 4 4 8 4 06002 m 567 10 Wm K T 293 K 006804 10 293 K W εσ 4 4 4 Substituting 293 K 0 06804 10 293 K 70 1000 W 4 4 8 s s T T Solving by trial and error gives 674 C 947 K sT Discussion We note that the iron will dissipate all the energy it receives by convection and radiation when its surface temperature reaches 947 K PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 249 2106 The backside of the thin metal plate is insulated and the front side is exposed to solar radiation The surface temperature of the plate is to be determined when it stabilizes Assumptions 1 Steady operating conditions exist 2 Heat transfer through the insulated side of the plate is negligible 3 The heat transfer coefficient is constant and uniform over the plate 4 Heat loss by radiation is negligible Properties The solar absorptivity of the plate is given to be α 08 α 08 25C 450 Wm2 Analysis When the heat loss from the plate by convection equals the solar radiation absorbed the surface temperature of the plate can be determined from 25 C 50 Wm 450 Wm 80 2 2 solar conv absorbed solar s o s A T A T hA T Q Q Q α Canceling the surface area A and solving for Ts gives s 322C T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 250 2107 Reconsider Prob 2106 Using EES or other software investigate the effect of the convection heat transfer coefficient on the surface temperature of the plate Let the heat transfer coefficient vary from 10 Wm2C to 90 Wm2C Plot the surface temperature against the convection heat transfer coefficient and discuss the results Analysis The problem is solved using EES and the solution is given below Given alpha08 qdotsolar450 Wm2 Tf25 C h50 Wm2C Analysis qdotsolarabsorbedalphaqdotsolar qdotconvhTsTf qdotsolarabsorbedqdotconv h Wm2C Ts C 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 61 49 43 394 37 3529 34 33 322 3155 31 3054 3014 298 295 2924 29 10 20 30 40 50 60 70 80 90 25 30 35 40 45 50 55 60 65 h Wm2C Ts C 2108 A hot water pipe at 80C is losing heat to the surrounding air at 5C by natural convection with a heat transfer coefficient of 25 W m2C The rate of heat loss from the pipe by convection is to be determined Assumptions 1 Steady operating conditions exist 2 Heat transfer by radiation is not considered 3 The convection heat transfer coefficient is constant and uniform over the surface L 10 m D 5 cm 80C Analysis The heat transfer surface area is A πDL 314x005 m10 m 1571 m² Q Under steady conditions the rate of heat transfer by convection is Air 5C 295 kW 2945 W 5 C m 80 C1571 25 Wm 2 2 conv hA T Q PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 251 2109 A spacecraft in space absorbs solar radiation while losing heat to deep space by thermal radiation The surface temperature of the spacecraft is to be determined when steady conditions are reached Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values 2 Thermal properties of the spacecraft are constant Properties The outer surface of a spacecraft has an emissivity of 06 and an absorptivity of 02 α 02 ε 06 1000 Wm2 Analysis When the heat loss from the outer surface of the spacecraft by radiation equals the solar radiation absorbed the surface temperature can be determined from 0 K K Wm 10 567 06 1000 Wm 20 4 4 s 4 2 8 2 4 space 4 solar rad absorbed solar T A A T A T Q Q Q s εσ α Canceling the surface area A and solving for Ts gives s 2769 K T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 252 2110 Reconsider Prob 2109 Using EES or other software investigate the effect of the surface emissivity and absorptivity of the spacecraft on the equilibrium surface temperature Plot the surface temperature against emissivity for solar absorptivities of 01 05 08 and 1 and discuss the results Analysis The problem is solved using EES and the solution is given below Given epsilon02 alpha06 qdotsolar1000 Wm2 Tf0 K space temperature Properties sigma567E8 Wm2K4 Analysis qdotsolarabsorbedalphaqdotsolar qdotradepsilonsigmaTs4Tf4 qdotsolarabsorbedqdotrad ε Ts K 01 02 03 04 05 06 07 08 09 1 648 5449 4924 4582 4334 4141 3984 3853 3741 3644 01 02 03 04 05 06 07 08 09 1 250 300 350 400 450 500 550 600 650 ε Ts K α01 α05 α08 α1 Table for ε 1 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 253 2111 A hollow spherical iron container is filled with iced water at 0C The rate of heat loss from the sphere and the rate at which ice melts in the container are to be determined Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values 2 Heat transfer through the shell is onedimensional 3 Thermal properties of the iron shell are constant 4 The inner surface of the shell is at the same temperature as the iced water 0C Properties The thermal conductivity of iron is k 802 WmC Table 23 The heat of fusion of water is at 1 atm is 3337 kJkg Iced water 0C 5C 04 cm Analysis This spherical shell can be approximated as a plate of thickness 04 cm and surface area A πD² 31402 m² 0126 m² Then the rate of heat transfer through the shell by conduction is 12632 W 0004 m 0 C 802 Wm C0126 m 5 2 cond o L T kA Q Considering that it takes 3337 kJ of energy to melt 1 kg of ice at 0C the rate at which ice melts in the container can be determined from 0038 kgs 3337 kJkg 12632 kJs ice hif Q m Discussion We should point out that this result is slightly in error for approximating a curved wall as a plain wall The error in this case is very small because of the large diameter to thickness ratio For better accuracy we could use the inner surface area D 192 cm or the mean surface area D 196 cm in the calculations PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 254 Review Problems 2112 A classroom has a specified number of students instructors and fluorescent light bulbs The rate of internal heat generation in this classroom is to be determined Assumptions 1 There is a mix of men women and children in the classroom 2 The amount of light and thus energy leaving the room through the windows is negligible Properties The average rate of heat generation from people seated in a roomoffice is given to be 100 W Analysis The amount of heat dissipated by the lamps is equal to the amount of electrical energy consumed by the lamps including the 10 additional electricity consumed by the ballasts Therefore 5600 W 100 W 56 No of people 40 W1118 792 W No of lamps Energy consumed per lamp person people lighting Q Q Q Then the total rate of heat gain or the internal heat load of the classroom from the lights and people become 6392 W 5600 792 people lighting total Q Q Q 2113 A decision is to be made between a cheaper but inefficient natural gas heater and an expensive but efficient natural gas heater for a house Assumptions The two heaters are comparable in all aspects other than the initial cost and efficiency Analysis Other things being equal the logical choice is the heater that will cost less during its lifetime The total cost of a system during its lifetime the initial operation maintenance etc can be determined by performing a life cycle cost analysis A simpler alternative is to determine the simple payback period The annual heating cost is given to be 1200 Noting that the existing heater is 55 efficient only 55 of that energy and thus money is delivered to the house and the rest is wasted due to the inefficiency of the heater Therefore the monetary value of the heating load of the house is Gas Heater η1 82 η2 95 Cost of useful heat 55Current annual heating cost 0551200yr660yr This is how much it would cost to heat this house with a heater that is 100 efficient For heaters that are less efficient the annual heating cost is determined by dividing 660 by the efficiency 82 heater Annual cost of heating Cost of useful heatEfficiency 660yr082 805yr 95 heater Annual cost of heating Cost of useful heatEfficiency 660yr095 695yr Annual cost savings with the efficient heater 805 695 110 Excess initial cost of the efficient heater 2700 1600 1100 The simple payback period becomes 10 years 110 yr 1100 Annaul cost savings Excess initial cost Simple payback period Therefore the more efficient heater will pay for the 1100 cost differential in this case in 10 years which is more than the 8year limit Therefore the purchase of the cheaper and less efficient heater is a better buy in this case PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 255 2114 A wind turbine is rotating at 20 rpm under steady winds of 30 kmh The power produced the tip speed of the blade and the revenue generated by the wind turbine per year are to be determined Assumptions 1 Steady operating conditions exist 2 The wind turbine operates continuously during the entire year at the specified conditions Properties The density of air is given to be ρ 120 kgm3 Analysis a The blade span area and the mass flow rate of air through the turbine are kgm 5027 m 8333 ms 50270 kgs 21 8 333 ms 3600 s 1 h 1 km 30 kmh 1000 m 5027 m 80 m 4 4 2 3 2 2 2 AV m V D A ρ π π Noting that the kinetic energy of a unit mass is V22 and the wind turbine captures 35 of this energy the power generated by this wind turbine becomes 6109 kW 2 2 2 2 m s 1000 1 kJkg 2 50270 kgs 8 333 ms 0 35 1 2 1 mV W η b Noting that the tip of blade travels a distance of πD per revolution the tip velocity of the turbine blade for an rpm of becomes n 302 kmh 5027 mmin 838 ms 80 m20 min tip π π n D V c The amount of electricity produced and the revenue generated per year are 321100year 5351 10 kWhyear006kWh Electricity producedUnit price generated Revenue 5351 10 kWhyear 610 9 kW365 24 hyear y produced Electricit 6 6 t W PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 256 2115 A wind turbine is rotating at 20 rpm under steady winds of 20 kmh The power produced the tip speed of the blade and the revenue generated by the wind turbine per year are to be determined Assumptions 1 Steady operating conditions exist 2 The wind turbine operates continuously during the entire year at the specified conditions Properties The density of air is given to be ρ 120 kgm3 Analysis a The blade span area and the mass flow rate of air through the turbine are kgm 5027 m 5556 ms 33510 kgs 21 5 556 ms 3600s h 1 1km 20 kmh 1000 m 5027 m 80 m 4 4 2 3 2 2 2 AV m V D A ρ π π Noting that the kinetic energy of a unit mass is V22 and the wind turbine captures 35 of this energy the power generated by this wind turbine becomes 1810 kW 2 2 2 2 m s 1000 1kJkg 2 33510 kgs 6 944 ms 0 35 1 2 1 mV W η b Noting that the tip of blade travels a distance of πD per revolution the tip velocity of the turbine blade for an rpm of becomes n 302 kmh 5027 mmin 838 ms 80 m20 min tip π π n D V c The amount of electricity produced and the revenue generated per year are 95130year 1585535 kWhyear006kWh Electricity producedUnit price Revenue generated 1585535 kWhyear 24 hyear 181 0 kW365 Electricity produced t W PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 257 2116E The energy contents unit costs and typical conversion efficiencies of various energy sources for use in water heaters are given The lowest cost energy source is to be determined Assumptions The differences in installation costs of different water heaters are not considered Properties The energy contents unit costs and typical conversion efficiencies of different systems are given in the problem statement Analysis The unit cost of each Btu of useful energy supplied to the water heater by each system can be determined from Unit cost of useful energy Unit cost of energy supplied Conversion efficiency Substituting Natural gas heater Btu 10 21 3 1025 Btu 1ft 055 0012ft Unit cost of useful energy 6 3 3 Heating by oil heater Btu 10 15 1 138700 Btu 1gal 055 115gal Unit cost of useful energy 6 Electric heater Btu 10 27 4 3412 Btu 1kWh 090 0084kWh Unit cost of useful energy 6 Therefore the lowest cost energy source for hot water heaters in this case is oil 2117 A home owner is considering three different heating systems for heating his house The system with the lowest energy cost is to be determined Assumptions The differences in installation costs of different heating systems are not considered Properties The energy contents unit costs and typical conversion efficiencies of different systems are given in the problem statement Analysis The unit cost of each Btu of useful energy supplied to the house by each system can be determined from Unit cost of useful energy Unit cost of energy supplied Conversion efficiency Substituting Natural gas heater kJ 10 13 5 105500 kJ 1 therm 087 124therm Unit cost of useful energy 6 Heating oil heater kJ 10 10 4 138500 kJ 1gal 087 125gal Unit cost of useful energy 6 Electric heater kJ 10 25 0 3600 kJ 1kWh 10 009kWh Unit cost of useful energy 6 Therefore the system with the lowest energy cost for heating the house is the heating oil heater PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 258 2118 The heating and cooling costs of a poorly insulated house can be reduced by up to 30 percent by adding adequate insulation The time it will take for the added insulation to pay for itself from the energy it saves is to be determined Assumptions It is given that the annual energy usage of a house is 1200 a year and 46 of it is used for heating and cooling The cost of added insulation is given to be 200 Analysis The amount of money that would be saved per year is determined directly from House Heat loss Money saved 1200 year046030 166 yr Then the simple payback period becomes 166yr 12 yr 200 Money saved Cost Payback period Therefore the proposed measure will pay for itself in less than one and a half year 2119 Caulking and weatherstripping doors and windows to reduce air leaks can reduce the energy use of a house by up to 10 percent The time it will take for the caulking and weatherstripping to pay for itself from the energy it saves is to be determined Assumptions It is given that the annual energy usage of a house is 1100 a year and the cost of caulking and weather stripping a house is 60 Analysis The amount of money that would be saved per year is determined directly from Money saved 1100 year010 110 yr Then the simple payback period becomes 110yr 0546 yr 60 Money saved Cost Payback period Therefore the proposed measure will pay for itself in less than half a year 2120E The energy stored in the spring of a railroad car is to be expressed in different units Analysis Using appropriate conversion factors we obtain a 160870 lbm ft2s2 1lbf 5000 lbf ft 32174 lbm fts 2 W b 1665 lbf yd 1ft 0 33303 yd 5000 lbf ft W c 74785 lbm mile2h2 2 2 2 h 1 3600s 5280 ft 1mile 1lbf 5000 lbf ft 32174 lbm fts W PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 259 2121E The work required to compress a gas in a gas spring is to be determined Assumptions All forces except that generated by the gas spring will be neglected Analysis When the expression given in the problem statement is substituted into the work integral relation and advantage is taken of the fact that the force and displacement vectors are collinear the result is 00228 Btu 778169 lbf ft 1Btu 1774 lbf ft 1774 lbf ft 12 in 1ft 1 in 4 in 41 1 lbf in 200 1 Constant Constant 04 04 14 1 1 1 2 2 1 2 1 k k k x x k dx x Fds W x F 2122E A man pushes a block along a horizontal plane The work required to move the block is to be determined considering a the man and b the block as the system Analysis The work applied to the block to overcome the friction is found by using the work integral 257 Btu 778169 lbf ft 1Btu 2000 lbf ft 2000 lbf ft 20 100 lbf100 ft 2 1 1 2 2 1 x fW x Fds W fW x fW W The man must then produce the amount of work 257 Btu W PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 260 2123 A diesel engine burning light diesel fuel that contains sulfur is considered The rate of sulfur that ends up in the exhaust and the rate of sulfurous acid given off to the environment are to be determined Assumptions 1 All of the sulfur in the fuel ends up in the exhaust 2 For one kmol of sulfur in the exhaust one kmol of sulfurous acid is added to the environment Properties The molar mass of sulfur is 32 kgkmol Analysis The mass flow rates of fuel and the sulfur in the exhaust are 18 kg airkg fuel 1867 kg fuelh 336 kg airh AF air fuel m m 0014 kgh 1867 kgh 750 10 10 750 6 fuel 6 Sulfur m m The rate of sulfurous acid given off to the environment is 0036 kgh 0014 kgh 32 3 16 32 2 1 Sulfur Sulfur H2SO3 H2SO3 m M M m Discussion This problem shows why the sulfur percentage in diesel fuel must be below certain value to satisfy regulations 2124 Lead is a very toxic engine emission Leaded gasoline contains lead that ends up in the exhaust The amount of lead put out to the atmosphere per year for a given city is to be determined Assumptions 35 of lead is exhausted to the environment Analysis The gasoline consumption and the lead emission are 335 kgyear 6 375 10 Lyear015 10 kgL035 GaolineConsumption Emission Lead 6 375 10 Lyear 5000 cars15000 kmcar year85 L100 km Consumption Gasoline 3 6 lead lead 6 f m Discussion Note that a huge amount of lead emission is avoided by the use of unleaded gasoline 2125E The power required to pump a specified rate of water to a specified elevation is to be determined Properties The density of water is taken to be 624 lbmft3 Table A3E Analysis The required power is determined from 113 kW 73756 lbf fts 1kW 8342 lbf fts lbf fts 8342 lbm fts 32174 1lbf 15850 galmin 32174 fts 300 ft 35315 ft s 62 4 lbmft 200 galmin 2 2 3 3 1 2 1 2 z g z z mg z W V ρ PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 261 2126 The power that could be produced by a water wheel is to be determined Properties The density of water is taken to be 1000 m3kg Table A3 Analysis The power production is determined from 0732 kW 2 2 2 3 3 1 2 1 2 m s 1000 1kJkg 1000 kgm 032060 m s 9 81 ms 14 m z g z z mg z W V ρ 2127 The flow of air through a flow channel is considered The diameter of the wind channel downstream from the rotor and the power produced by the windmill are to be determined Analysis The specific volume of the air is 0 8409 m kg 100 kPa 0 287 kPa m kg K293 K 3 3 P RT v The diameter of the wind channel downstream from the rotor is 738 m 9 ms 7 m 10 ms 4 4 2 1 1 2 2 2 2 1 2 1 2 2 1 1 V V D D V D V D A V A V π π The mass flow rate through the wind mill is 457 7 kgs m kg 408409 7 m 10 ms 3 2 1 1 π v A V m The power produced is then 435 kW 2 2 2 2 2 2 12 m s 1000 1kJkg 2 9 ms 457 7 kgs 10 ms 2 V m V W PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 262 2128 The available head flow rate and efficiency of a hydroelectric turbine are given The electric power output is to be determined Assumptions 1 The flow is steady 2 Water levels at the reservoir and the discharge site remain constant 3 Frictional losses in piping are negligible Properties We take the density of water to be ρ 1000 kgm3 1 kgL Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free surface of the dam relative to free surface of discharge water and it can be converted to work entirely Therefore the power potential of water is its potential energy which is gz per unit mass and gz for a given mass flow rate m 0 8829 kJkg m s 1000 1kJkg ms 90 m 981 2 2 2 mech gz pe e The mass flow rate is 1 90 m 2 ηoverall 84 Generator Turbin 65000 kgs 1000 kgm 65 m s 3 3 V ρ m Then the maximum and actual electric power generation become 5739 MW 1000 kJs 1MW 65000 kgs08829 kJkg mech mech max me E W 482 MW 0 845739 MW max overall electric W W η Discussion Note that the power generation would increase by more than 1 MW for each percentage point improvement in the efficiency of the turbinegenerator unit PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 263 2129 An entrepreneur is to build a large reservoir above the lake level and pump water from the lake to the reservoir at night using cheap power and let the water flow from the reservoir back to the lake during the day producing power The potential revenue this system can generate per year is to be determined Assumptions 1 The flow in each direction is steady and incompressible 2 The elevation difference between the lake and the reservoir can be taken to be constant and the elevation change of reservoir during charging and discharging is disregarded 3 Frictional losses in piping are negligible 4 The system operates every day of the year for 10 hours in each mode Properties We take the density of water to be ρ 1000 kgm3 Reservoir 2 1 40 m Pump turbine Lake Analysis The total mechanical energy of water in an upper reservoir relative to water in a lower reservoir is equivalent to the potential energy of water at the free surface of this reservoir relative to free surface of the lower reservoir Therefore the power potential of water is its potential energy which is gz per unit mass and for a given mass flow rate This also represents the minimum power required to pump water from the lower reservoir to the higher reservoir gz m 8 kW 784 1000 N ms 1kW 1kg ms 1N 1000 kgm 2 m s981 ms 40 m 2 2 3 3 mech mech ideal min pump turbine max g z mg z m pe m e E W W W V ρ The actual pump and turbine electric powers are 1046 kW 075 8 kW 784 motor pump ideal pump elect η W W 588 6 kW 0 75784 8 kW ideal turbinegen turbine W W η Then the power consumption cost of the pump the revenue generated by the turbine and the net income revenue minus cost per year become 114500year 1046 kW365 10 hyear003kWh Unit price Cost pump elect t W 171900year 588 6 kW365 10 hyear008kWh Unit price Reveue turbine t W Net income Revenue Cost 171900 114500 57400year Discussion It appears that this pumpturbine system has a potential to generate net revenues of about 57000 per year A decision on such a system will depend on the initial cost of the system its life the operating and maintenance costs the interest rate and the length of the contract period among other things PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 264 Fundamentals of Engineering FE Exam Problems 2130 A 2kW electric resistance heater in a room is turned on and kept on for 50 min The amount of energy transferred to the room by the heater is a 2 kJ b 100 kJ c 3000 kJ d 6000 kJ e 12000 kJ Answer d 6000 kJ Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values We 2 kJs time5060 s WetotalWetime kJ Some Wrong Solutions with Common Mistakes W1EtotalWetime60 using minutes instead of s W2EtotalWe ignoring time 2131 In a hot summer day the air in a wellsealed room is circulated by a 050hp shaft fan driven by a 65 efficient motor Note that the motor delivers 050 hp of net shaft power to the fan The rate of energy supply from the fanmotor assembly to the room is a 0769 kJs b 0325 kJs c 0574 kJs d 0373 kJs e 0242 kJs Answer c 0574 kJs Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Eff065 Wfan05007457 kW EWfanEff kJs Some Wrong Solutions with Common Mistakes W1EWfanEff Multiplying by efficiency W2EWfan Ignoring efficiency W3EWfanEff07457 Using hp instead of kW PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 265 2132 A fan is to accelerate quiescent air to a velocity to 12 ms at a rate of 3 m3min If the density of air is 115 kgm3 the minimum power that must be supplied to the fan is a 248 W b 72 W c 497 W d 216 W e 162 W Answer a 248 W Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values rho115 V12 Vdot3 m3s mdotrhoVdot kgs WemdotV22 Some Wrong Solutions with Common Mistakes W1WeVdotV22 Using volume flow rate W2WemdotV2 forgetting the 2 W3WeV22 not using mass flow rate 2133 A 900kg car cruising at a constant speed of 60 kmh is to accelerate to 100 kmh in 4 s The additional power needed to achieve this acceleration is a 56 kW b 222 kW c 25 kW d 62 kW e 90 kW Answer a 56 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values m900 kg V160 kmh V2100 kmh Dt4 s WamV2362V13622000Dt kW Some Wrong Solutions with Common Mistakes W1WaV2362V13622Dt Not using mass W2WamV22V122000Dt Not using conversion factor W3WamV2362V13622000 Not using time interval W4WamV236V1361000Dt Using velocities PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 266 2134 The elevator of a large building is to raise a net mass of 400 kg at a constant speed of 12 ms using an electric motor Minimum power rating of the motor should be a 0 kW b 48 kW c 47 kW d 12 kW e 36 kW Answer c 47 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values m400 kg V12 ms g981 ms2 WgmgV1000 kW Some Wrong Solutions with Common Mistakes W1WgmV Not using g W2WgmgV22000 Using kinetic energy W3WgmgV Using wrong relation 2135 Electric power is to be generated in a hydroelectric power plant that receives water at a rate of 70 m3s from an elevation of 65 m using a turbinegenerator with an efficiency of 85 percent When frictional losses in piping are disregarded the electric power output of this plant is a 39 MW b 38 MW c 45 MW d 53 MW e 65 MW Answer b 38 MW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Vdot70 m3s z65 m g981 ms2 Eff085 rho1000 kgm3 WerhoVdotgzEff106 MW Some Wrong Solutions with Common Mistakes W1WerhoVdotzEff106 Not using g W2WerhoVdotgzEff106 Dividing by efficiency W3WerhoVdotgz106 Not using efficiency PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 267 2136 A 75 hp shaft compressor in a facility that operates at full load for 2500 hours a year is powered by an electric motor that has an efficiency of 93 percent If the unit cost of electricity is 006kWh the annual electricity cost of this compressor is a 7802 b 9021 c 12100 d 8389 e 10460 Answer b 9021 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Wcomp75 hp Hours2500 hyear Eff093 price006 kWh WeWcomp07457HoursEff CostWeprice Some Wrong Solutions with Common Mistakes W1cost Wcomp07457HourspriceEff multiplying by efficiency W2cost WcompHourspriceEff not using conversion W3cost WcompHourspriceEff multiplying by efficiency and not using conversion W4cost Wcomp07457Hoursprice Not using efficiency 2137 Consider a refrigerator that consumes 320 W of electric power when it is running If the refrigerator runs only one quarter of the time and the unit cost of electricity is 009kWh the electricity cost of this refrigerator per month 30 days is a 356 b 518 c 854 d 928 e 2074 Answer b 518 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values We0320 kW Hours0252430 hyear price009 kWh CostWehoursprice Some Wrong Solutions with Common Mistakes W1cost We2430price running continuously PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 268 2138 A 2kW pump is used to pump kerosene ρ 0820 kgL from a tank on the ground to a tank at a higher elevation Both tanks are open to the atmosphere and the elevation difference between the free surfaces of the tanks is 30 m The maximum volume flow rate of kerosene is a 83 Ls b 72 Ls c 68 Ls d 121 Ls e 178 Ls Answer a 83 Ls Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values W2 kW rho0820 kgL z30 m g981 ms2 WrhoVdotgz1000 Some Wrong Solutions with Common Mistakes WW1Vdotgz1000 Not using density 2139 A glycerin pump is powered by a 5kW electric motor The pressure differential between the outlet and the inlet of the pump at full load is measured to be 211 kPa If the flow rate through the pump is 18 Ls and the changes in elevation and the flow velocity across the pump are negligible the overall efficiency of the pump is a 69 b 72 c 76 d 79 e 82 Answer c 76 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values We5 kW Vdot 0018 m3s DP211 kPa EmechVdotDP EmechEffWe PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 269 The following problems are based on the optional special topic of heat transfer 2140 A 10cm high and 20cm wide circuit board houses on its surface 100 closely spaced chips each generating heat at a rate of 008 W and transferring it by convection to the surrounding air at 25C Heat transfer from the back surface of the board is negligible If the convection heat transfer coefficient on the surface of the board is 10 Wm2C and radiation heat transfer is negligible the average surface temperature of the chips is a 26C b 45C c 15C d 80C e 65C Answer e 65C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values A010020 m2 Q 100008 W Tair25 C h10 Wm2C Q hATsTair W Some Wrong Solutions with Common Mistakes Q hW1TsTair Not using area Q h2AW2TsTair Using both sides of surfaces Q hAW3TsTair Adding temperatures instead of subtracting Q100 hAW4TsTair Considering 1 chip only 2141 A 50cmlong 02cmdiameter electric resistance wire submerged in water is used to determine the boiling heat transfer coefficient in water at 1 atm experimentally The surface temperature of the wire is measured to be 130C when a wattmeter indicates the electric power consumption to be 41 kW Then the heat transfer coefficient is a 43500 Wm2C b 137 Wm2C c 68330 Wm2C d 10038 Wm2C e 37540 Wm2C Answer a 43500 Wm2C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values L05 m D0002 m ApiDL m2 We41 kW Ts130 C Tf100 C Boiling temperature of water at 1 atm We hATsTf W Some Wrong Solutions with Common Mistakes We W1hTsTf Not using area We W2hLpiD24TsTf Using volume instead of area We W3hATs Using Ts instead of temp difference PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 270 2142 A 3m2 hot black surface at 80C is losing heat to the surrounding air at 25C by convection with a convection heat transfer coefficient of 12 Wm2C and by radiation to the surrounding surfaces at 15C The total rate of heat loss from the surface is a 1987 W b 2239 W c 2348 W d 3451 W e 3811 W Answer d 3451 W Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values sigma567E8 Wm2K4 eps1 A3 m2 hconv12 Wm2C Ts80 C Tf25 C Tsurr15 C QconvhconvATsTf W QradepssigmaATs2734Tsurr2734 W QtotalQconvQrad W Some Wrong Solutions with Common Mistakes W1QlQconv Ignoring radiation W2QQrad ignoring convection W3QQconvepssigmaATs4Tsurr4 Using C in radiation calculations W4QQtotalA not using area 2143 Heat is transferred steadily through a 02m thick 8 m by 4 m wall at a rate of 24 kW The inner and outer surface temperatures of the wall are measured to be 15C to 5C The average thermal conductivity of the wall is a 0002 WmC b 075 WmC c 10 WmC d 15 WmC e 30 WmC Answer d 15 WmC Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values A84 m2 L02 m T115 C T25 C Q2400 W QkAT1T2L W Some Wrong Solutions with Common Mistakes QW1kT1T2L Not using area QW2k2AT1T2L Using areas of both surfaces QW3kAT1T2L Adding temperatures instead of subtracting QW4kALT1T2 Multiplying by thickness instead of dividing by it PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2144 The roof of an electrically heated house is 7 m long 10 m wide and 025 m thick It is made of a flat layer of concrete whose thermal conductivity is 092 WmC During a certain winter night the temperatures of the inner and outer surfaces of the roof are measured to be 15C and 4C respectively The average rate of heat loss through the roof that night was a 41 W b 177 W c 4894 W d 5567 W e 2834 W Answer e 2834 W Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values A710 m2 L025 m k092 WmC T115 C T24 C QcondkAT1T2L W Some Wrong Solutions with Common Mistakes W1QkT1T2L Not using area W2Qk2AT1T2L Using areas of both surfaces W3QkAT1T2L Adding temperatures instead of subtracting W4QkALT1T2 Multiplying by thickness instead of dividing by it 31 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 3 PROPERTIES OF PURE SUBSTANCES PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 32 Pure Substances Phase Change Processes Property Diagrams 31C A liquid that is about to vaporize is saturated liquid otherwise it is compressed liquid 32C A vapor that is about to condense is saturated vapor otherwise it is superheated vapor 33C No 34C The temperature will also increase since the boiling or saturation temperature of a pure substance depends on pressure 35C Because one cannot be varied while holding the other constant In other words when one changes so does the other one 36C At critical point the saturated liquid and the saturated vapor states are identical At triple point the three phases of a pure substance coexist in equilibrium 37C Yes 38C Case c when the pan is covered with a heavy lid Because the heavier the lid the greater the pressure in the pan and thus the greater the cooking temperature 39C At supercritical pressures there is no distinct phase change process The liquid uniformly and gradually expands into a vapor At subcritical pressures there is always a distinct surface between the phases PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 33 Property Tables 310C A perfectly fitting pot and its lid often stick after cooking as a result of the vacuum created inside as the temperature and thus the corresponding saturation pressure inside the pan drops An easy way of removing the lid is to reheat the food When the temperature rises to boiling level the pressure rises to atmospheric value and thus the lid will come right off 311C The molar mass of gasoline C8H18 is 114 kgkmol which is much larger than the molar mass of air that is 29 kgkmol Therefore the gasoline vapor will settle down instead of rising even if it is at a much higher temperature than the surrounding air As a result the warm mixture of air and gasoline on top of an open gasoline will most likely settle down instead of rising in a cooler environment 312C Yes Otherwise we can create energy by alternately vaporizing and condensing a substance 313C No Because in the thermodynamic analysis we deal with the changes in properties and the changes are independent of the selected reference state 314C The term hfg represents the amount of energy needed to vaporize a unit mass of saturated liquid at a specified temperature or pressure It can be determined from hfg hg hf 315C Yes It decreases with increasing pressure and becomes zero at the critical pressure 316C Yes the higher the temperature the lower the hfg value 317C Quality is the fraction of vapor in a saturated liquidvapor mixture It has no meaning in the superheated vapor region 318C Completely vaporizing 1 kg of saturated liquid at 1 atm pressure since the higher the pressure the lower the hfg 319C No Quality is a mass ratio and it is not identical to the volume ratio PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 34 320C The compressed liquid can be approximated as a saturated liquid at the given temperature Thus T f T P v v 321C Ice can be made by evacuating the air in a water tank During evacuation vapor is also thrown out and thus the vapor pressure in the tank drops causing a difference between the vapor pressures at the water surface and in the tank This pressure difference is the driving force of vaporization and forces the liquid to evaporate But the liquid must absorb the heat of vaporization before it can vaporize and it absorbs it from the liquid and the air in the neighborhood causing the temperature in the tank to drop The process continues until water starts freezing The process can be made more efficient by insulating the tank well so that the entire heat of vaporization comes essentially from the water 322 Complete the following table for H2 O T C P kPa v m3 kg Phase description 50 1235 772 Saturated mixture 1436 400 04624 Saturated vapor 250 500 04744 Superheated vapor 110 350 0001051 Compressed liquid PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 35 323 Problem 322 is reconsidered The missing properties of water are to be determined using EES and the solution is to be repeated for refrigerant134a refrigerant22 and ammonia Analysis The problem is solved using EES and the solution is given below Given T150 C v1772 m3kg P2400 kPa x21 T3250 C P3500 kPa T4110 C P4350 kPa Analysis Fluidsteamiapws Change the Fluid to R134a R22 and Ammonia and solve P1pressureFluid TT1 vv1 x1qualityFluid TT1 vv1 T2temperatureFluid PP2 xx2 v2volumeFluid PP2 xx2 v3volumeFluid PP3 TT3 x3qualityFluid PP3 TT3 v4volumeFluid PP4 TT4 x4qualityFluid PP4 TT4 x 100 for superheated vapor and x 100 for compressed liquid SOLUTION for water T C P kPa x v kgm3 5000 1235 0641 9 772 1436 1 40000 1 04624 2500 0 50000 100 04744 1100 0 35000 100 000105 1 SOLUTION for R134a T C P kPa x v kgm3 5000 341 100 772 891 40000 1 00512 2500 0 50000 1100 0 35000 100 008666 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 36 SOLUTION for R22 T C P kPa X v kgm3 5000 402 100 772 656 40000 1 005817 2500 0 50000 100 009959 1100 35000 100 0103 0 SOLUTION for Ammonia kgm3 T C P kPa X v 5000 2040 100 772 189 40000 4 1 0309 2500 0 50000 100 05076 1100 0 35000 100 05269 00 10 20 30 40 50 60 70 80 90 100 0 100 200 300 400 500 600 700 s kJkgK T C 8600 kPa 2600 kPa 500 kPa 45 kPa Steam 104 103 102 101 100 101 102 103 0 100 200 300 400 500 600 700 v m3kg T C 8600 kPa 2600 kPa 500 kPa 45 kPa Steam PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 37 103 102 101 100 101 102 100 101 102 103 104 105 v m kg 3 P kPa 250 C 170 C 110 C 75 C Steam 0 500 1000 1500 2000 2500 3000 100 101 102 103 104 105 h kJkg P kPa 250 C 170 C 110 C 75 C Steam 00 10 20 30 40 50 60 70 80 90 100 0 500 1000 1500 2000 2500 3000 3500 4000 s kJkgK h kJkg 8600 kPa 2600 kPa 500 kPa 45 kPa Steam PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 38 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 24E Complete the following table for H2 O Phase description 3 T F P psia u Btu lbm 300 6703 782 Saturated mixture 26722 40 23602 Saturated liquid 500 120 11744 Superheated vapor 400 400 37384 Compressed liquid 325E Problem 324E is reconsidered The missing properties of water are to be determined using EES and the lution is to be repeated for refrigerant134a refrigerant22 and ammonia ven below 1qualityFluid TT1 uu1 2temperatureFluid PP2 xx2 so Analysis The problem is solved using EES and the solution is gi Given T1300 F u1782 Btulbm P240 psia x20 T3500 F P3120 psia T4400 F P4420 psia Analysis Fluidsteamiapws P1pressureFluid TT1 uu1 x T u2intenergyFluid PP2 xx2 u3intenergyFluid PP3 TT3 x3qualityFluid PP3 TT3 u4intenergyFluid PP4 TT4 x4qualityFluid PP4 TT4 x 100 for superheated vapor and x 100 for compressed liquid Solution for steam T ºF P psia x u Btulbm 300 67028 06173 782 2672 40 0 236 500 120 100 1174 400 400 100 3738 preparation If you are a student using this Manual you are using it without permission 39 326 Complete the following table for H2 O PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course P k h kJ kg x P T C Pa hase description 12021 200 20458 07 S aturated mixture 140 3615 1800 0565 S 3 aturated mixture 17766 950 75274 00 Saturated liquid 80 500 33537 Compressed liquid 3500 800 31622 Superheated vapor P kPa v m3 kg Phase description 327 Complete the following table for Refrigerant134a T C 12 320 0000750 Compressed liquid 30 77064 00065 Saturated mixture 1873 550 003741 Saturated vapor 60 600 004139 Superheated vapor 328 Complete the following table for water h kJkg Condition description and quality if applicable P kPa T oC v m3kg 200 1202 08858 27063 x 1 Saturated vapor 2703 130 19593 x 0650 Twophase mixture 2018 400 15358 32770 Superheated vapor 80 0 0001004 12574 ressed liquid 30 Comp 450 14790 Insufficient information Approximated as sat rated liquid at ven temperat oC u the gi ure of 30 preparation If you are a student using this Manual you are using it without permission 310 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course gerant134a 329E Complete the following table for Refri T F P psia h Btu lbm x Phase description 6589 80 78 0566 Saturated mixture 15 29759 6992 06 Saturated mixture 10 70 1535 Compressed liquid 160 180 12946 Superheated vapor 110 16116 11723 10 Saturated vapor 330 A pistoncylinder device contains R134a at a specified state Heat is transferred to R134a The final pressure the olume change of the cylinder and the enthalpy change are to be determined a The fin re is equal to the in re which is d v Analysis al pressu itial pressu etermined from 904 kP 2 1kN a 2 2 2 1 2 1000 m 4 81 ms 88 kPa 4 m g P P p specific vol nd enthalpy of R e initial state of C and at the final state of 904 kPa C are from EES v1 02302 m h1 24776 kJkg v 2 02544 m3kg h2 26816 kJkg he initial and the final volumes and the volume change are m3 0 1957 0 2162 m 0 0 1957 m 0 85 kg02302 m kg 3 2 2 3 3 v V m c The total enthalpy change is determined from kgms atm P 025 π kg9 12 πD b The and 15 ume a 134a at th 904 kPa and 10 3kg R134a 085 kg 10C Q T 0 85 kg02544 m kg 3 1 1 v V m 00205 2162 V 174 kJkg k 6816 0 1 m H 331 e temp ure of R is to Ana e is higher than erheated vapor state From R134a tables 04619 ft lbm 3 v 2 h h 24776 kJ g 85 kg2 E Th erat 134a at a specified state be determined lysis Since the specified specific volum vg for 120 psia this is a sup Table A 13E 120 psia 140F T P preparation If you are a student using this Manual you are using it without permission 311 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course e initial temperature and the final pressure are to be determined nalysis This is a constant volume process The specific volume is 332 A rigid container that is filled with water is cooled Th A 0 150 m kg 3 1kg 0 150 m 3 1 m v v initial stat rheated va e temperature is determined Table A m kg a 1 3 1 1 C T P v is a const e cooling process v V m stant The ixture and thus the pressure is the saturation pressure at e final temperature Table A 4 0 150 m kg 40 C sat 40 C 2 3 1 2 2 7385 kPa P P T v v 333 initial pressure are to be determined Analysis 2 V The e is supe por Th to be 6 0 150 2 MP 395 This ant volum con final state is saturated m th A rigid container that is filled with R134a is heated The final temperature and This is a constant volume process The specific volume is 0 1348 m kg kg 348 m 3 e pressure is th ined by le A 13 200 kPa P 10 m 1 3 2 1 V v v The initial state is determined to be a mixture and thus th e saturation pressure at the given temperature Table A 11 sat 40 C 1 5125 kPa P P The final state is superheated vapor and the temperature is determ interpolation to be Tab 0 1348 m kg 2 3 2 2 663C T v R134a 40C 10 kg 1348 m3 P v 2 1 v 2 1 Q H2O Pa kg 2 M 1 150 L P preparation If you are a student using this Manual you are using it without permission 312 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course ssure and the total internal energy at the final state are Analysis 334E Left chamber of a partitioned system contains water at a specified state while the right chamber is evacuated The partition is now ruptured and heat is transferred to the water The pre to be determined The final specific volume is Water 500 2 lbm psia 15 ft3 Evacuated 15 ft3 ft lbm 51 lbm 2 3 2 2 m v 3ft 3 V t this s ixture and the pressure is the saturation pressure A m pecific volume and the final temperature the state is a saturated Table A 4E sat 300 F 2 6703 psia P P The quality and internal energy at the final state are 46038 Btulbm 0 229983025 51 269 0 2299 0 01745 ft lbm 6 4663 0 01745 ft lbm 51 2 2 3 3 2 2 fg f fg f x u u u x v v v The total internal energy is then 9208 Btu 2 lbm46038 Btulbm 2 2 mu U 335 ed state is to be determined Analysis The enthalpy of R134a at a specifi The specific volume is 0 03 m kg 9 m 3 3 v V m kg 300 Inspection of Table A11 indicates that this is a mixture of liquid and vapor Usi and the enthalpy are determined to be ng the properties at 10C line the quality kJkg 6008 fg f 336 The specific volume of R134a at a specified state is to be determined Analysis Since the given temperature is higher than the saturation temperature for 200 kPa this is a superheated vapor state The specific volume is then Table A 13 C 25 200 kPa 011646 m kg 3 v T P 18002 0 600819073 43 65 0 0 0007930 m kg 0 049403 0 0007930 m kg 0 03 3 3 fg f xh h h x v v v preparation If you are a student using this Manual you are using it without permission 313 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course le A11E the initial specific volume is 3 5463 ft lbm 0 01143 80 4 4300 3 and the initial volume will be e will be l and final conditions is 337E A springloaded pistoncylinder device is filled with R134a The water now undergoes a process until its volume increases by 40 The final temperature and the enthalpy are to be determined Analysis From Tab 0 0 01143 1 1 fg f x v v v P v 2 1 3 3 1 1 07093 ft lbm3546 3ft lbm 20 v V m With a 40 increase in the volume the final volum 3 3 1 2 09930 ft 41 0 7093 ft 41 V V The distance that the piston moves between the initia 0 3612 ft ft 1 0 7093ft 0 9930 4 4 2 3 2 π πD A x p V V As a result of the compression of the spring the pressure difference between the initial and final states is 142 psia 1 42 lbfin in 37 lbfin 0 3612 4 4 2 2 2 π πD k x A k x A F P p p 9 87 psia Table A 11E sat 30 F 1 P P 1129 psia 12 in 12 The initial pressure is The final pressure is then 1 42 9 87 1 2 P P P and the final specific volume is lbm 09930 ft 3 3 2 V v enthalpy are Table A13E accurately 4 965 ft lbm 20 2 m At this final state the temperature and from EES 4 965 ft lbm 29 psia 11 1 1 3 2 2 1199 Btulbm 815 F h T P v Note that it is very difficult to get the temperature and enthalpy from preparation If you are a student using this Manual you are using it without permission 314 338E A pistoncylinder device that is filled with water is cooled The final pressure and volume of the water are to be determined Analysis The initial specific volume is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2 4264 ft lbm 1lbm m 2 4264 ft 3 3 1 1 v V ined to be superheated th th pressure is determined to be A 6E 2 1 3 1 250 psia comp ice that is filled with R134a is heated The final volume is to be determined nalysis e is P 2 1 H2O 600F 1 lbm 24264 ft3 This is a constantpressure process The initial state is determ vapor and us e Table 600 F 1 P P T 2 4264 ft lbm v The saturation temperature at 250 psia is 4001F Since the final temperature is less than this temperature the final state is compressed liquid Using the in ressible liquid approximation 0 01663 ft lbm Table A 4E 3 200 F 2 v f v v The final volume is then 001663 ft3 1 lbm 0 01663 ft lbm 3 2 2 v V m 339 A pistoncylinder dev A This is a constant pressure process The initial specific volum R134a 264C 10 kg 1595 m3 P v 2 1 0 1595 m kg kg 10 1 595 m 3 V 3 1 v m The initial state is determined to be a mixture and thus the pressure is the 0 30138 m kg Table A 13 100 C 2 2 v T The final volume is then 30138 m3 10 kg 0 30138 m kg 3 2 2 v V m saturation pressure at the given temperature 100 kPa Table A 12 sat 264 C 1 P P The final state is superheated vapor and the specific volume is 100 kPa 3 2 P preparation If you are a student using this Manual you are using it without permission 315 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course specific volume is 340E The total internal energy and enthalpy of water in a container are to be determined Analysis The 2 ft lbm 1lbm 2 ft 3 3 m V v At this specific volume and the given pressure the state is a saturated mixture Water 100 psia 2 ft3 The quality internal energy and enthalpy at this state are Table A5E 697 7 Btulbm 29851 0 449088899 660 7 Btulbm 0 449080729 19 298 0 4490 0 01774 ft lbm 4 4327 0 01774 ft lbm 2 3 3 v v fg f fg f xh h h xu u u x v 6607 Btu 1 lbm697 7 Btulbm bm mh H The volume of a container that contains water at a specified state is to be determined nalysis The specific volume is determined from steam tables by interpolation to be The volum m3 fg f The total internal energy and enthalpy are then 1 lbm660 7 Btul mu U 6977 Btu 341 A 2 4062 m kg Table A 6 100 kPa 3 v P Water 3 kg 100 kPa 250C 250 C T e of the container is then 722 3kg 2 4062 m kg 3 v V m preparation If you are a student using this Manual you are using it without permission 316 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course y are to be determined at the Analysis e process The specific volume is 342 A rigid container that is filled with R134a is heated The temperature and total enthalp initial and final states This is a constant volum R134a 300 kPa 10 kg 14 L 0 0014 m kg 0 014 m 3 3 2 1 V v v the saturation tem interpolation 10 kg m The initial state is determined to be a mixture and thus the temperature is perature at the given pressure From Table A12 by 061C sat 300 kPa 1 T T and P v 2 1 52 kJkg 0 009321 kg 1 1 3 3 fg f fg x h h h v he total enthalpy is then 5452 kJ 10 kg5452 kJkg 1 1 mh H he final state is also saturated mixture Repeating the calculations at this state 54 0 00932119813 67 52 0 0007736 m kg 0 067978 0 0007736 m 0 0014 1 1 f x v v T T 2155C sat 600 kPa 2 T T 8464 kJkg 0173318090 0 0 01733 0 0008199 m k 0 034295 0 0008199 m kg 0 0014 3 2 2 fg f x v v v 3 g 8151 2 2 fg f x h h h 8464 kJ 10 kg8464 kJkg 2 2 mh H preparation If you are a student using this Manual you are using it without permission 317 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course e are to be determined 343 A pistoncylinder device that is filled with R134a is cooled at constant pressure The final temperature and the chang of total internal energy Analysis The initial specific volume is 0 12322 m kg 100 kg 12322 m 3 3 1 m V v The initial state is superheated and the internal energy at this state is 263 08 kJkg Table A 13 final specific volume is 0 12322 m kg 200 kPa 1 3 1 1 u P v The 0 06161 m kg 2 0 12322 m 2 3 3 1 2 kg v v This is a constant pressure process The final state is determined to be saturated xture whose temperature is mi Table A 12 sat 200 kPa 2 1009C T T The internal energy at the final state is Table A12 15261 kJkg 0 614018621 28 38 0 6140 0 0007533 m kg 0 099867 0007533 m kg 0 06161 2 2 3 2 2 fg f fg f x u u u x v v v ence the change in the internal energy is 11047 kJkg 26308 15261 u u u 10 R134a 0 20 kPa 0 kg 12322 m3 2 1 P v 3 0 H 1 2 preparation If you are a student using this Manual you are using it without permission 318 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Tv diagram and the change in internal energy is to be determined agram The internal A 6 1 u T P 2 v x P A 5 100 kPa 3 3 3 u P nergy is 345E ature changes with the weather conditions The change rcury in atmospheric pressure is to be determined at 200 and 212F are 11538 and 14709 psia respectively Table A4E One of m 3387 kPa 0491 psia inner cover page nalysis A change of 02 in of mercury in atmospheric pressure corresponds to 344 A pistoncylinder device fitted with stops contains water at a specified state Now the water is cooled until a final pressure The process is to be indicated on the Analysis The process is shown on Tv di energy at the initial state is 2728 9 kJkg Table 250 C 300 kPa 1 1 State 2 is saturated vapor at the initial pressure Then Table A 5 0 6058 m kg 1sat vapor 300 kPa 3 2 2 Process 23 is a constantvolume process Thus 1163 3 kJkg Table 0 6058 m kg 3 v The overall change in internal e 1566 kJkg 1163 3 2728 9 3 1 u u u The local atmospheric pressure and thus the boiling temper in the boiling temperature corresponding to a change of 02 in of me Properties The saturation pressures of water in ercury is equivalent to 1 inHg A 00982 psia 1inHg inHg 0491 psia 20 P At about boiling temperature the change in boiling temperature per 1 psia change in pressure is determined using data at 200 and 212F to be 3 783 Fpsia 11538 psia 14709 200 F 212 P T Then the change in saturation boiling temperature corresponding to a change of 0147 psia becomes 037F 3 783 Fpsia00982 psia 3 783 Fpsia boiling P T which is very small Therefore the effect of variation of atmospheric pressure on the boiling temperature is negligible P 02 inHg T v 2 1 3 300 kPa 100 kPa 250C Wa 300 250C ter kPa Q preparation If you are a student using this Manual you are using it without permission 319 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course n ospheric pressure is 1 atm 101325 kPa 2 all and thus its effect No air has leaked into the pan during cooling roperties The saturation pressure of water at 20C is 23392 kPa Table A4 F on the lid after cooling at the panlid interface e vertical direction to be or 346 A person cooks a meal in a pot that is covered with a wellfitting lid and leaves the food to cool to the room temperature It is to be determined if the lid will open or the pan will move up together with the lid when the perso attempts to open the pan by lifting the lid up Assumptions 1 The local atm The weight of the lid is sm on the boiling pressure and temperature is negligible 3 P Analysis Noting that the weight of the lid is negligible the reaction force can be determined from a force balance on the lid in th PA F PatmA 1 Nm since 1 Pa 6997 m Pa 2339 2 Pa 101325 4 m 30 4 2 π P P D P A P F atm atm 2 2 2 π ms 785 N 8 kg981 2 W mg hich is much less than the reaction force of 6997 N at the panlid interface Therefore the pan will move up together with pts to open the pan by lifting the lid up In fact it looks like the lid will not open even if the ass of the pan and its contents is several hundred kg water level drops by 10 cm in 45 min roperti nd thus at a saturation temperature of Tsat 100 22565 kJkg and vf 0001043 m3kg Table A4 P Patm 1 atm 2 92 kPa 33 N 6997 The weight of the pan and its contents is w the lid when the person attem m 347 Water is boiled at 1 atm pressure in a pan placed on an electric burner The during boiling The rate of heat transfer to the water is to be determined P es The properties of water at 1 atm a C are hfg Analysis The rate of evaporation of water is H2O 1 atm 0 001742 kgs 60 s 45 evap t m 704 kg 4 4 704 kg 0001043 evap evap m m f f v v hen the 025 m 4010 m 4 2 2 evap L D π π V T rate of heat transfer to water becomes 393 kW 0 001742 kgs22565 kJkg evap h fg m Q preparation If you are a student using this Manual you are using it without permission 320 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course The are T 933C h 22739 kJkg and v 0001038 m3kg Table A5 348 Water is boiled at a location where the atmospheric pressure is 795 kPa in a pan placed on an electric burner water level drops by 10 cm in 45 min during boiling The rate of heat transfer to the water is to be determined Properties The properties of water at 795 kPa sat fg f Analysis The rate of evaporation of water is 0 001751 kgs 727 kg 4 0001038 evap m m f f v v 4 727 kg 025 m 4010 m 4 2 2 evap evap L D m π π V 60 s 45 n the rate of heat transfer to water becomes 398 kW at a rate of 130 kgh The rate of heat ansfer from the steam to the cooling water is to be determined lysis Noting that 24060 kJ of heat is released as 1 kg of saturated apor at 40C condenses the rate of heat transfer from the steam to e cooling water in the tube is determined directly from evap hfg m Q evap t The 0 001751 kgs22739 kJkg evap h fg m Q 349 Saturated steam at Tsat 40C condenses on the outer surface of a cooling tube tr Assumptions 1 Steady operating conditions exist 2 The condensate leaves the condenser as a saturated liquid at 30C Properties The properties of water at the saturation temperature of 40C are hfg 24060 kJkg Table A4 D 3 cm L 35 m 40C H2O 795 kPa Ana v th 869 kW 312780 kJh 130 kgh24060 kJkg preparation If you are a student using this Manual you are using it without permission 321 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course oximately ρ 1000 kgm3 Analysis 9439 kPa sat98 C o P P Table A4 The pressure difference between the bottoms of two pans is 350 The boiling temperature of water in a 5cm deep pan is given The boiling temperature in a 40cm deep pan is to be determined Assumptions Both pans are full of water Properties The density of liquid water is appr The pressure at the bottom of the 5cm pan is the saturation pressure corresponding to the boiling temperature of 98C 40 cm cm 5 343 kPa 1 kPa gh P ρ hen the pressure at the bottom of the 40cm deep pan is P 9439 343 9782 kPa hen the boiling temperature becomes ed wi Analysis a force balance on the piston or 1000 kgm s 1000 kgm 9807 ms 035 m 2 2 3 T T 990C sat9782 kPa boiling T T Table A5 351 A vertical pistoncylinder device is filled with water and cover boiling temperature of water is to be determined th a 20kg piston that serves as the lid The The pressure in the cylinder is determined from PA PatmA W 11961 kPa 1000 kgm s 1 kPa 001 m 20 kg981 ms 100 kPa 2 2 2 atm A mg P P The boiling temperature is the saturation temperature corresponding to this pressure 1047C kPa Tsat11961 T Table A5 W mg Patm P preparation If you are a student using this Manual you are using it without permission 322 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course k vaporized is to be determined and the Tv diagram is to be drawn v V m constant and 352 A rigid tank that is filled with saturated liquidvapor mixture is heated The temperature at which the liquid in the tan is completely Analysis This is a constant volume process the specific volume is determined to be 012 m kg 15 kg 18 m 3 3 m V v W rized the tank will contain h saturated vapor only Thus 2 v g v e temperature at this point is the temperature that corresponds to this vg value 53 en the liquid is completely vapo 012 m kg 3 Th 2029C 012 m kg sat 3 v T T Table A4 g 3 A pistoncylinder device contains a saturated liquidvapor mixture of water at 800 kPa pressure The mixture is eated at constant pressure until the temperature rises to 200C The initial temperature the total mass of water the final olume are to be determined and the Pv diagram is to be drawn perature at the specified pressure 600 he total mass in this case can easily be determined by adding the ma h v Analysis a Initially two phases coexist in equilibrium thus we have a saturated liquidvapor mixture Then the temperature in the tank must be the saturation tem 1588C T T kPa sat b T ss of each phase 90C T 2 v 1 P v 2 1 H2O 7395 kg 2 852 4 543 g f t m m m 2852 kg m kg 03156 m 09 4 543 kg m kg 0001101 m 0005 3 3 3 3 g g g f f f m m v V v V At th eated vapor and its specific volume is 03521 m kg 200 C 600 kPa 3 2 2 2 v o T P Table A6 Then 2604 m3 m kg 7395 kg03521 3 2 2 v V mt c e final state water is superh preparation If you are a student using this Manual you are using it without permission 323 354 Problem 353 is reconsidered The effect of pressure on the total mass of water in the tank as the pressure ater is gainst pressure and results are Analysis is solved using EES and the solution is given below P1600 kPa f1 g109 m3 c ific volume m3kg pvsatf1 sat liq mass kg g1 mtotmf1mg1 1Vf1Vg1 specific volume1 m3 1temperatureSteamiapws PP1vspvol1C he final volume is calculated from the specific volume at the final T and P specific volume2 m3kg varies from 01 MPa to 1 MPa is to be investigated The total mass of w to be discussed to be plotted a The problem P2P1 T2200 C V V 0005 m3 spvsatf1volumeSteamiapws PP1x0 sat liq specifi spvsatg1volumeSteamiapwsPP1x1 sat vap spec mf1Vf1s volume m3kg m s kg Vg1spvsatg1 sat vap mas V spvol1V1mtot T T spvol2volumeSteamiapws PP2 TT2 V2mtotspvol2 P1 kPa mtot kg PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 100 200 300 5324 5731 6145 400 500 600 7395 700 800 900 1000 6561 6978 7812 823 8648 9066 100 200 300 400 500 600 700 800 900 1000 5 55 6 65 7 103 102 10 75 8 85 9 95 P1 kPa m tot kg 1 100 101 102 100 101 102 103 104 105 106 v m3kg SteamIAPWS P kPa 200C 1 2 P600 kPa preparation If you are a student using this Manual you are using it without permission 324 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 355E Superheated water vapor cools at constant volume until the temperature drops to 250F At the final state the 1700 ft3lbm and vg 13816 ft3lbm Thus at the k will contain saturated liquidvapor mixture since the final pressure must be the saturation pressure at pressure the quality and the enthalpy are to be determined Analysis This is a constant volume process v Vm constant and the initial specific volume is determined to be 30433 ft lbm 500 F 180 psia 3 1 1 1 v o T P Table A6E H2O 180 psia 500F At 250F vf 00 final state the tan vf v vg and the final temperature P P sat250 F 2984 psia o T b The quality at the final state is determined from v 2 1 0219 0 01700 3 0433 2 v f v x2 v 0 01700 13816 fg c The enthalpy at the final state is determined from 4260 Btulbm 0 219 94541 21863 fg f xh h h preparation If you are a student using this Manual you are using it without permission 325 356E Problem 355E is reconsidered The effect of initial pressure on the quality of water at the final state as the he nalysis The problem is solved using EES a iapwsTT1PP1 2v v2 P x pressure varies from 100 psi to 300 psi is to be investigated The quality is to be plotted against initial pressure and t results are to be discussed A nd the solution is given below T1500 F P1180 psia T2250 F v 1volumesteam v 1 P2pressuresteamiapwsTT2v h2enthalpysteamiapwsTT2vv2 x2qualitysteamiapwsTT2vv2 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1 psia 2 100 1222 1444 1667 1889 2111 2333 2556 2778 300 02761 02378 02084 01853 01665 0151 01379 01268 04037 03283 100 140 180 220 260 300 01 015 02 025 03 035 04 045 P1 psia x2 102 101 100 101 1200 102 103 104 0 200 400 600 800 1000 1400 v ft3lbm T F 1600 psia 780 psia 180 psia 2982 psia 005 01 02 05 12 13 14 15 Btulbm Steam R 1 2 preparation If you are a student using this Manual you are using it without permission 326 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 357 The properties of compressed liquid water at a specified state are to be determined using the compressed liquid tables 33502 kJkg 453 error 80 C 80 C 3 f f h h liquid table Table A7 33050 kJkg and also sing the saturated liquid approximation and the results are to be compared Analysis Compressed liquid can be approximated as saturated liquid at the given temperature Then from Table A4 by u T 80C 135 error 33497 kJkg 0001029 m kg 090 error 80 C f u u v v From compressed 35090 kJkg 80 C h T 000102 m kg MPa 20 3 u id approximation are listed above in parentheses P v The percent errors involved in the saturated liqu preparation If you are a student using this Manual you are using it without permission 327 358 Problem 357 is reconsidered Using EES the indicated properties of compressed liquid are to be determined nalysis s s Given T80 C Analysis luidsteamiapws Saturated liquid assumption fintenergyFluid TT x0 fenthalpyFluid TT x0 Compressed liquid treatment vvolumeFluid TT PP uintenergyFluid TT PP henthalpyFluid TT PP Percentage Errors errorv100vfvv erroru100ufuu errorh100hhfh SOLUTION errorh4527 erroru1351 errorv08987 Fluidsteamiapws h35090 kJkg hf33502 kJkg P20000 kPa T80 C u33050 kJkg uf33497 kJkg v000102 m3kg vf0001029 m3kg and they are to be compared to those obtained using the saturated liquid approximation A The problem i olved using EES and the solution is given below P20000 kPa F vfvolumeFluid TT x0 u h PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 328 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 359 Superheated steam in a pistoncylinder device is cooled at constant pressure until half of the mass condenses The final 17988C Tsat1 MPa T Table A5 the final state is specified to be x2 05 The specific tial and the final states are 025799 m kg 3 Table A6 kg 0 001127 0 19436 50 3 0128 2 m3 025799m kg 9775 3 tank is cooled until the vapor starts condensing The initial pressure in the tank is to be etermined is a constant volume process v V m constant and the ume is equal to the final specific volume that is 0 79270 m kg 3 Table A4 s condensing at 150C Then from 030 MPa 1 kg P tempe and the volume change are to be determined and the process should be shown on a Tv diagram Analysis b At the final state the cylinder contains saturated liquid rature vapor mixture and thus the final temperature must be the saturation temperature at the final pressure H2O 300C 1 MPa T v 2 1 H2O T1 250C P1 T v 2 1 25 15 C c The quality at olumes at the ini v 300 C MPa 1 1 1 v o T 10 P 10 MPa 2 2 2 f x P v v v 0 001127 50 2 fg x 009775 m Thus 08 kg00 1 2 v v V m 360 The water in a rigid d Analysis This initial specific vol 124 C 2 1 v g v v since the vapor start Table A6 1 250 T C 3 1 079270 m v preparation If you are a student using this Manual you are using it without permission 329 361 Heat is supplied to a pistoncylinder device that contains water at a specified state The volume of the tank the final temperature and pressure and the internal energy change of water are to be determined PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0001157 m3kg and u 85046 kJkg Table A4 e volume of the cylinder at the initial state is kg 0 0011 41 v V m Properties The saturated liquid properties of water at 200C are vf Analysis a The cylinder initially contains saturated liquid water Th f 3 3 0 001619 m 57 m kg 1 1 The volume at the final state is 0006476 m3 4 0 001619 V b The final state properties are 0 004626 m kg 14 kg 0006476 m 3 2 m V v 0 004626 m kg 2 2 3 2 P T kPa 21367 3713 C v 3 2201 5 kJkg 1 2 2 u x Table A4 or A5 or EES kJ 1892 c The total internal energy change is determined from kg22015 85046 kJkg 41 1 2 u m u U ror involved in using the enthalpy of water by the incompressible liquid approximation is to be determined A 7E 37651 B 400 F 1500 psia h T P 7504 Btulbm Table A 4E he error involved is 362E The er Analysis The state of water is compressed liquid From the steam tables Table tulbm Based upon the incompressible liquid approximation 3 F 400 1500 psia 400 F h f h T P T 039 100 37651 37651 37504 Percent Error which is quite acceptable in most engineering calculations Water 14 kg 200C sat liq Q preparation If you are a student using this Manual you are using it without permission 330 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course e 140 C h T liquid approximation Table A 4 kJkg 16 m kg 001080 C 0 3 140 C f P v v 363 The errors involved in using the specific volume and enthalpy of water by the incompressible liquid approximation ar to be determined Analysis The state of water is compressed liquid From the steam tables Table A 7 0 0010679 m kg 20 MPa 3 P v 07 kJkg 602 Based upon the incompressible 589 0 C 140 MPa 20 14 hf h T The errors involved are 214 113 60 60207 Error enthalpy Percent 100 0 0010679 0 001080 0 001067 Error specific volume Percent which are quite acceptable in most engineering calculations 64 A pistoncylinder device that is filled with R134a is heated The volume change is to be determined nalysis The initial specific volume is nd the i 0033608 m3 0 100 kg03360 8 m kg v V m 2 v P m3 The volum R134a 60 kPa 20C 100 g 9 100 2 07 16 589 3 A P v 2 1 0 33608 m kg Table A 13 C 20 3 1 1 1 v T 60 kPa P a nitial volume is 3 1 1 At the final state we have 050410 m kg Table A 13 C 100 60 kPa 3 2 2 T 3 2 2 0050410 0 100 kg05041 0 m kg v V m e change is then 00168 m3 0 033608 0 050410 1 2 V V V preparation If you are a student using this Manual you are using it without permission 331 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course perties of R134a at the given state are Table A13 120 C 3 365 A rigid vessel is filled with refrigerant134a The total volume and the total internal energy are to be determined Properties The pro m kg 0037625 32787 kJkg 800 kPa v T o Analysis The total volume and internal energy are determined from 3 2 kg32787 kJkg mu U 66 tal internal energy and the volume of the Analysis u P kJ 6557 00753 m 2 kg0037625 m kg 3 mv V 3 A rigid vessel contains R134a at specified temperature The pressure to liquid phase are to be determined a The specific volume of the refrigerant is 005 m kg 10 kg v m t 20C vf 00007362 m3kg and vg 014729 m3kg Table A11 Thus the tank ontains saturated liquidvapor mixture since vf v vg and the pressure must be the b are determ R134a 10 kg 20C R134a 2 kg 800 kPa 120C 05 m 3 3 V A c saturation pressure at the specified temperature 13282 kPa sat 20 C o P P The quality of the refrigerant134a and its total internal energy ined from 9042 kJ kg9042 kJkg 10 9042 kJkg 03361 19345 2539 03361 00007362 4729 00007362 005 mu U xu u u x fg f v f v ed from m3 t f m x m m v V 01 fg v c The mass of the liquid phase and its volume are determin 000489 kg00007362 m kg 6639 6639 kg 1 03361 10 1 3 f f f preparation If you are a student using this Manual you are using it without permission 332 367 The PessureEnthalpy diagram of R134a showing some constanttemperature and constantentropy lines are obtained using Property Plot feature of EES PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 100 0 100 200 300 400 500 101 102 103 104 105 R134a 70C Pa 40C P k 10C 10C h kJkg 30C 02 03 05 08 1 12 kJkgK 368 xture is heated until it reaches the critical state The mass of the quid water and the volume occupied by the liquid at the initial state are to be determined Analysis This is a constant volume process v V m constant to the critical state and thus the initial specific volume l specific volume of water he total mass is A rigid vessel that contains a saturated liquidvapor mi li will be equal to the final specific volume which is equal to the critica 0003106 m kg 3 2 1 vcr v v last row of Table A4 T 9660 kg m kg 0003106 m 03 3 3 v V m At 150C vf 0001091 m3kg and vg 039248 m3kg Table A 4 Then the quality of water at the initial state is 0005149 0001091 039248 0001091 0003106 1 1 fg f x v v v Then the mass of the liquid phase and its volume at the initial state are determined from m3 0105 kg 9610 kg0001091 m kg 9610 1 00051499660 1 3 1 f f f t f m x m m v V T v cp vcr H2O 150C preparation If you are a student using this Manual you are using it without permission 333 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course r molar mass M is the mass of one mole in grams or the mass of one kmol in ilograms These two are related to each other by m NM where N is the number of moles w pressure relative to its critical he specific gas constant that is different is the molar mass of the gas hane molar mass 16 kgkmol since or Methane on the other hand is 73 The specific volume of nitrogen at a specified state is to be determined ssumptions At specified conditions nitrogen behaves as an ideal gas roperties The gas constant of nitrogen is R 02968 kJkgK Table A1 Ideal Gas 369C Mass m is simply the amount of matte k 370C A gas can be treated as an ideal gas when it is at a high temperature or lo temperature and pressure 371C Ru is the universal gas constant that is the same for all gases whereas R is t for different gases These two are related to each other by R Ru M where M 372C Propane molar mass 441 kgkmol poses a greater fire danger than met propane is heavier than air molar mass 29 kgkmol and it will settle near the flo lighter than air and thus it will rise and leak out 3 A P Analysis According to the ideal gas equation of state 0495 m kg 3 300 kPa 273 K 02968 kPa m kg K227 3 P RT v 74E with oxygen is to be determined gas 53 psiaft lbmR Table A1E Analysis 3 The temperature in a container that is filled Assumptions At specified conditions oxygen behaves as an ideal Properties The gas constant of oxygen is R 033 3 The definition of the specific volume gives ft lbm 51 2 lbm m 3ft 3 3 v V Using th e ideal gas equation of state the temperature is 358 R psia ft lbm R 03353 psia15 ft lbm 80 3 3 R P T v preparation If you are a student using this Manual you are using it without permission 334 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course lume of a container that is filled with helium at a specified state is to be determined ssumptions At specified conditions helium behaves as an ideal gas 375 The vo A Properties The gas constant of helium is R 20769 kJkgK Table A1 Analysis According to the ideal gas equation of state 4154 m3 300 kPa 273 K 2 kg20769 kPa m kg K27 3 P mRT V roperties The universal gas constant is Ru 8314 kPam3kmolK The molar mass of helium is 40 kgkmol Table A1 nalysis The volume of the sphere is 376 A balloon is filled with helium gas The mole number and the mass of helium in the balloon are to be determined Assumptions At specified conditions helium behaves as an ideal gas P A He D 9 m 27C 200 kPa 3 3 3 381 7 m 45 m 3 4 3 4 π π r V Assuming ideal gas behavior the mole numbers of He is determined from 3061 kmol kPa m kmol K300 K 8314 kPa3817 m 200 3 3 R T P N u V Then the mass of He can be determined from 123 kg kmol40 kgkmol 3061 NM m preparation If you are a student using this Manual you are using it without permission 335 377 Problem 376 is to be reconsidered The effect of the balloon diameter on the mass of helium contained in the he diameter varies from 5 m to 15 m The ution is given below 9 m 27 C 200 kPa u8314 kJkmolK D m balloon is to be determined for the pressures of a 100 kPa and b 200 kPa as t mass of helium is to be plotted against the diameter for both cases Analysis The problem is solved using EES and the sol Given Data D T P R Solution PVNRuT273 V4piD233 mNMOLARMASSHelium m kg PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5 6111 8333 9444 1056 1167 1389 15 2101 3835 7222 6331 1278 3506 9725 1416 1976 2669 4502 5672 P200 kPa 5 7 9 11 500 13 15 0 100 200 300 400 600 D m m kg P200 kPa P100 kPa preparation If you are a student using this Manual you are using it without permission 336 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 378 Two rigid tanks connected by a valve to each other contain air at specified conditions The volume of the second tank A1 ideal gas the volume of the second tank and e mass of air in the first tank are determined to be and th l equilibrium pressure when the valve is opened are to be determined Assumptions At specified conditions air behaves as an ideal gas Properties The gas constant of air is R 0287 kPam3 e fina kgK Table Analysis Lets call the first and the second tanks A and B Treating air as an th 5846 kg kPa m kg K298 K 0287 kPa10 m 500 200 kPa kg0287 kPa m kg K308 K 5 3 3 1 3 1 1 1 A B P m RT m3 221 10846 kg 50 321 m 1 3 B A he al eq pressure b B V 1 A RT P m V Thus 5846 22 10 B A m m m V V V T n the fin uilibrium ecomes 2841 kPa m 321 kg0287 kPa m kg K293 K 0846 3 3 V mR 37 lastic ntains air at a specified state The volume is doubled at the same pressure The initial volume and the final temperature are to be determined ecified conditions air behaves as an ideal gas nalysis According to the ideal gas equation of state 1 T2 2 P 9E An e tank co Assumptions At sp A 590 F R 1050 4049 ft 3 2 2 1 2 1 2 3 460 R 65 2 460 R lbmol1073 psia ft lbmol R65 32 psia 32 T T T T nR T P u V V V V V Air V 1 m3 T 25C P 500 kPa Air m 5 kg T 35C P 200 kPa A B preparation If you are a student using this Manual you are using it without permission 337 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course used e 380 An ideal gas in a rigid tank is cooled to a final gage pressure The final temperature is to be determined Assumptions The gas is specified as an ideal gas so that ideal gas relation can be Analysis According to the ideal gas equation of state at constant volum 2 2 2 1 1 1 1 T P T P V V m m Ideal gas 1227C 200 kPa gage Patm 100 kPa 1 Since 2 1 V V hen Q T 477C 750 K kPa 200 273 K 50 100 kPa 1227 2 2 1 2 2 2 1 1 P T P T T P T P 381 On acuated The partition is removed and the as fills the entire tank The gas is also heated to a final pressure The final temperature is to be determined ssumptions The gas is specified as an ideal gas so that ideal gas relation can be used nalysis According to the ideal gas equation of state 100 e side of a twosided tank contains an ideal gas while the other side is ev g A A 1 1 1 2 1 2 3 2 V V V V P P Q Ideal gas 927C V1 Evacuated 2V1 Applying these 1 1 m m 3327C 3600 K 273 K 3 927 3 3 1 1 1 1 1 2 1 2 2 2 1 1 2 2 2 1 1 1 T T T T T T T P T P V V V V V V V V preparation If you are a student using this Manual you are using it without permission 338 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course termined Analysis erature remains constant the ideal gas equation gives 382 A pistoncylinder device containing argon undergoes an isothermal process The final pressure is to be de Assumptions At specified conditions argon behaves as an ideal gas Properties The gas constant of argon is R 02081 kJkgK Table A1 Since the temp 2 2 1 1 2 1 1 V V V V P P RT P RT P m for final pressure becomes 2 which w ed hen solv kPa 275 kPa 50 550 50 1 1 P P P P V V 83 An automobile tire is inflated with air The pressure rise of air in the tire when the tire is heated and the amount of air at must be bled off to reduce the temperature to the original value are to be determined ssumptions 1 At specified conditions air behaves as an ideal gas 2 The volume of the tire remains constant ing the volume of the tire to remain e can be determined fro 2 1 1 1 2 1 2 V V 3 th A Properties The gas constant of air is R 0287 kPam3kgK Table A1 Analysis Initially the absolute pressure in the tire is 310kPa 100 210 atm 1 P P P g Treating air as an ideal gas and assum constant the final pressure in the tir m 336 kPa 298 K 310 kPa 323 K 1 1 2 2 2 2 2 1 1 T P T P P P V V Thus th The am ue is 1 T T e pressure rise is P P P 2 1 336 310 26 kPa ount of air that needs to be bled off to restore pressure to its original val 00070 kg 00836 00906 kPa m kg K323 K 0287 2 1 3 2 2 m m m RT Argon 06 kg 005 m3 kPa 00836 kg kPa0025 m 310 00906 kg 0287 kPa m kg K298 K kPa0025 m 310 3 1 3 3 1 1 1 P m RT P m V V 550 Tire 25C preparation If you are a student using this Manual you are using it without permission 339 Compressibility Factor PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course it is from re the gas deviates from factor Z at the same reduced temperature and pressure 86C Reduced pressure is the pressure normalized with respect to the critical pressure and reduced temperature is the mperature normalized with respect to the critical temperature cr 2206 MPa 384C It represent the deviation from ideal gas behavior The further away 1 the mo ideal gas behavior 385C All gases have the same compressibility 3 te 387 The specific volume of steam is to be determined using the ideal gas relation the compressibility chart and the steam tables The errors involved in the first two approaches are also to be determined Properties The gas constant the critical pressure and the critical temperature of water are from Table A1 R 04615 kPam3kgK T 6471 K P cr Analysis a From the ideal gas equation of state m3 kg 70 error 04615 kPa m kg K62315 K 3 RT b From 001917 15000 kPa P v 6 the compressibility chart Fig A15 0 65 K 0453 2206 MPa MPa 10 Z T T P P P cr R R 85 error c From 104 6471 K 673 T cr Thus 001246 m kg 3 065001917 m kg 3 videal v Z the superheated steam table Table A6 001148 m kg 3 v 350 C 15 MPa T P H2O 15 MPa 350C preparation If you are a student using this Manual you are using it without permission 340 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 388 Problem 387 is reconsidered The problem is to be solved using the general compressibility factor feature of to st nalysis The problem is solved using EES and the solution is given below TCelsius273 K criticalTCRITSteamiapws criticalPCRITSteamiapws eT TcompTTidealgasT tablevolumeSteamiapwsPPtableTTtable EES data for steam as a real gas tablepressureSteamiapws TTtablevv iapwsPPtablevv COM lPco ompressibility factor leConvert TCelcius C EES or other software The specific volume of water for the three cases at 15 MPa over the temperature range of 350C 600C in 25C intervals is to be compared and the error involved in the ideal gas approximation is to be plotted again temperature A P15 MPaConvertMPakPa TCelsius 350 C T T P vVolm PtableP PcompPPidealgasP Ttabl v P TsattemperatureSteam MMMOLARMASSwater Ru8314 kJkmolK Universal gas constant RRuMM kJkgK Particular gas constant PidealgasvidealgasRTidealgas Ideal gas equation z PRESSTcompTcritica mpPcritical PcompvcompzRTcomp generalized C ErroridealgasAbsvtablevidealgasvtab ErrorcompAbsvtablevcompvtableConvert Errorcomp Errorideal gas 9447 6722 2725 04344 05995 1101 1337 1428 1437 1397 4353 3221 2523 2044 1692 1422 121 1039 8976 7802 350 375 400 25 50 475 500 525 550 575 600 4 4 1329 1245 300 350 400 450 500 550 600 0 10 20 30 40 50 60 70 TCelsius C Percent Error Ideal Gas Ideal Gas Compressibility Factor Compressibility Factor Steam at 15 MPa Specific Volume preparation If you are a student using this Manual you are using it without permission 341 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 389 The specific volume of R134a is to be determined using the ideal gas relation the compressibility chart and the R cr cr 134a t The errors involved in the first two approaches are also to be determined Properties The gas constant the critical pressure and the critical temperature of refrigerant134a are from Table A1 R 008149 kPam ables 3kgK T 3742 K P 4059 MPa Analysis a From the ideal gas equation of state kPa 008149 kPa m kg K343 K 3 133 error 003105 m kg 3 RT ty chart Fig A15 900 v P b From the compressibili 894 0 2 4059 MPa Z P 13error 76 m kg 3 3 3 022 09 MPa P PR 0917 343 K T TR cr 3742 K Tcr Thus 0027 0894003105 m kg videal v Z c From the superheated refrigerant table Table A1 0027413 m kg 3 70 C T 390 The sp olume of ste be determin ing the ideal gas relation the compressibility chart and the steam tables The ved in th o approaches are also to be determined Properties The gas constant the critical pressure and the critical temperature of water are from Table A1 R 04615 kPam3kgK Tcr 6471 K Pcr 2206 MPa Analysis a he ideal gas e ate 09 MPa v P ecific v am is to ed us errors invol e first tw From t quation of st 37 error 9533 m kg 3 a 3 K723 K 04615 kPa v From the compressibility chart Fig A15 00 m kg RT kP 500 3 P b 0 961 112 6471 K K 723 0159 2206 MPa MPa 35 Z T T T P P P cr R cr R Thus 04 error m kg 009161 3 0961009533 m kg 3 videal v Z c From the superheated steam table Table A6 009196 m kg 3 v 450 C 35 MPa T P H2O 35 MPa 450C R134a 09 MPa 70C preparation If you are a student using this Manual you are using it without permission 342 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course are from Table A1E nalysis art at th ig A15 391E Ethane in a rigid vessel is heated The final pressure is to be determined using the compressibility chart Properties The gas constant the critical pressure and the critical temperature of ethane R 03574 psiaft3lbmR Tcr 5498 R Pcr 708 psia A From the compressibility ch e initial state F 0 963 Z 0 1130 708 psia cr 1 P R psia 80 1019 5498 R 1 1 cr 1 1 P P T TR The speci e process Then 560 R T Ethane 80 psia 100F Q fic volume does not change during th 2 409 ft 80 psia m R560 R 096303574 psia ft 3 3 1 1 1 2 1 P Z RT v v At the final state lb lbm 01 8 68 R708 psia R5498 1819 5498 R R 1000 2 3 cr cr 2 cr 2 2 Z P RT T T T R R hus psia ft lbm 03574 24091ft lbm 3 2actual v v T 148 psia ft lbm 2409 74 psia ft lbm R1000 R 10035 3 3 2 2 2 v Z RT P 2 preparation If you are a student using this Manual you are using it without permission 343 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course e perature of ethane are from Table A1 3 Pcr 512 MPa Analysis d final states Fig A15 392 Ethylene is heated at constant pressure The specific volume change of ethylene is to be determined using th compressibility chart Properties The gas constant the critical pressure and the critical tem R 02964 kPam kgK Tcr 2824 K From the compressibility chart at the initial an 0 56 Z 0 977 512 MPa MPa 5 1038 2824 K K 293 1 cr 1 1 cr 1 1 P P P T T T R R 0 961 Z 977 0 1675 2824 KR K 473 1 1 2 cr 2 2 R R R P P T T T The specific volume change is 0 56293 K 0 961473 K 5000 kPa kPa m kg K 02964 3 1 1 2 2 Z T P Z T R v 00172 m kg 3 Ethylene 5 MPa 2 C 0 Q preparation If you are a student using this Manual you are using it without permission 344 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course the nd the steam tables 647 cr 22 393 Water vapor is heated at constant pressure The final temperature is to be determined using ideal gas equation compressibility charts a Properties The gas constant the critical pressure and the critical temperature of water are from Table A1 R 04615 kPam3kgK Tcr 1 K P 06 MPa Analysis a From the ideal gas equation 1246 K 273 K2 350 1 2 1 2 v T v T The pressure of the steam is 16529 kPa sat350 C 2 1 P P P b rom th 5 F e compressibility chart at the initial state Fig A1 0 75 0 593 Z a 0 963 6471 KR K 623 1 1 cr 1 1 1 1 R R R P P T T T v At the final state 0 88 1 50 2 0 75 2 0 749 2 1 2 1 2 Z P P R R R R v v Thus 0 749 MPa 2206 MP 16529 cr P 826 K 22060 kPa 1506471 K 088 kPa 16529 cr cr 2 2 2 2 2 2 2 P T Z P Z R P T v R v c From the superheated steam table 0 008806 m kg 1 350 C 3 1 1 1 v x T Table A4 750 K 477 C 0 01761 m kg 2 16529 kPa 2 3 1 2 2 T P v v from Table A6 or EES s Water 350C at vapor Q preparation If you are a student using this Manual you are using it without permission 345 394E Water vapor is heated at constant pressure The final temperature is to be determined using ideal gas equation the compressibility charts and the steam tables Properties The critical pressure and the critical temperature of water are from Table A1E R 05956 psiaft3lbmR Tcr 11648 R Pcr 3200 psia Analysis a From the ideal gas equation 1720 R 460 R2 400 1 2 1 2 v T v T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course b The properties of steam are Table A4E 3 1 2 sat400 F 2 1 v v P P P t the final state from the compressibility chart Fig A15 3 7278 ft lbm 2 1 8639 ft lbm 3 g400 F 1 v v 26 psia 247 A 0 985 1719 R3200 psia R11648 ft lbm 37278 0 0773 3200 psia psia 24726 2 3 actual 2 cr 2 2 Z P P PR v Thus 05956 psia ft lbm 3 cr cr 2 P RT R v 1571 R 098505956 psia ft lbm R 3 2 2 Z R T the superheated steam table 24726 psia37278 ft lbm 3 2 2 P v c From 1560 R 1100 F 3 7278 ft lbm 2 3 2 2 T v from Table A6E or EES 4 sa Water 00F t vapor Q 24726 psia P preparation If you are a student using this Manual you are using it without permission 346 395 Methane is heated at constant pressure The final temperature is to be determined using ideal gas equation and the compressibility charts PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course from Table A1 Properties The gas constant the critical pressure and the critical temperature of methane are R 05182 kPam3kgK Tcr 1911 K Pcr 464 MPa Analysis From the ideal gas equation 450 K 300 K 51 1 2 1 2 v T v T From the compressibility chart at the initial state Fig A15 0 80 0 88 1 1 v R Thus Z 1 724 464 MPa MPa 8 1 570 1911 K K 300 cr 1 1 cr 1 1 R R P P P T T T At the final state 0 975 21 51 0 80 51 1 724 2 1 2 1 2 Z P P R R R R v v 406 K 8000 kPa 12191 1K cr 2 2 2 2 2 T P P T v R v f these two results the accuracy of the second result is limited by the accuracy with which the charts may be read Accepting the error associated with reading charts the second temperature is the more accurate 396 The percent error involved in treating CO2 at a specified state as an ideal gas is to be determined Properties The critical pressure and the critical temperature of CO2 are from Table A1 739MPa 3042K and cr cr P T Analysis From the compressibility chart Fig A15 4640 kPa 0975 cr 2 2 P Z R Z O 0 69 0980 3042 K K 298 0677 739 MPa MPa 5 cr cr Z T T T P P P R R Then the error involved in treating CO2 as an ideal gas is 0449 or 449 069 1 1 1 1 Error ideal Z v v v CO2 5 MPa 25C a Methane 8 MP 300 K Q preparation If you are a student using this Manual you are using it without permission 347 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course t of ined From 397 CO2 gas flows through a pipe The volume flow rate and the density at the inlet and the volume flow rate at the exi the pipe are to be determ Properties The gas constant the critical pressure and the critical temperature of CO2 are Table A1 R 01889 kPam3kgK Tcr 3042 K Pcr 739 MPa Analysis 3 MPa 500 K CO 450 K 2 2 kgs a the ideal gas equation of state 006297 m kg 21 error 3000 kPa 2 kgs01889 kPa m kg K500 K 3 1 1 1 P mRT V 3 21 error 3176 kgm 3 ρ kPa m kg K500 K 01889 kPa 3000 3 1 1 1 RT P 005667 m kg 36 error 3 3000 kPa 2 kgs01889 kPa m kg K450 K 3 2 2 2 P mRT V b From the compressibility chart EES function for compressibility factor is used 0 9791 K 500 0407 739 MPa 1 1 Z T P P cr R 164 K 3042 MPa 3 1 1 T T P cr R 0 9656 450 K 2 2 Z T cr 0407 739 MPa 3 MPa 2 P P PR 148 3042 K 2 T T cr R Thus m3 kg 89 kPa m kg K500 K 3 1 RT1 m Z 3000 kPa 1 1 P V 097912 kgs018 006165 gm3 3244 k 1889 kPa m kg K500 K 097910 kPa 3000 3 1 1 1 1 Z RT P ρ 005472 m kg 3 096562 kgs01889 kPa m kg K450 K 3 2 2 2 Z mRT V 3000 kPa 2 P preparation If you are a student using this Manual you are using it without permission 348 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course roaches are also to be determined le A1 From the ideal gas equation of state 398 The specific volume of nitrogen gas is to be determined using the ideal gas relation and the compressibility chart The errors involved in these two app Properties The gas constant the critical pressure and the critical temperature of nitrogen are from Tab R 02968 kPam3kgK Tcr 1262 K Pcr 339 MPa Analysis a 86 4 error 10000 02968 kPa m kg K150 K 3 0004452m kg 3 P RT v From the comp ity chart Fig A15 kPa b ressibil N2 10 MPa 150 K 0 54 119 1262 K K 150 295 339 MPa P P cr R MPa 10 Z T T T P cr R Thus 07 error 0540004452 m kg 3 ideal 0002404m kg 3 v v Z preparation If you are a student using this Manual you are using it without permission 349 Other Equations of State PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course resents the olume ey ar m ritical isotherm has an inflection linder device The final volum onoxide is to be determined ubin equation of state roperties The gas constant and molar mass of C R 02968 kPam3kgK M 28011 kgkmol Analysis a From the ideal gas equation of state 399C The constant a represents the increase in pressure as a result of intermolecular forces the constant b rep v occupied by the molecules Th e determined fro the requirement that the c point at the critical point 3100 Carbon monoxide is heated in a pistoncy e of the carbon m using the ideal gas equation and the BenedictWebbR P O are Table A1 002294 m3 1000 kPa 0100 kg02968 kPa m kg K773 K 2 2 P mRT V 3 b Using the coefficients of Table 34 for carbon dioxide and the given data the BenedictWebbRubin equation of state for state 2 is 0 0060 0 0060 exp 1 773 1 054 10 0 000135 71 3 3 71 8 314 773 0 002632 1 773 8 673 10 8 314 773 135 9 0 05454 8 314773 1000 exp 1 1 2 2 2 3 5 6 3 2 2 5 2 2 2 2 2 3 6 3 2 2 2 2 0 0 2 0 2 2 2 v v v v v v v v v v v v v v γ γ α T c a a bR T T C A B R T R T P u u u The solution of this equation by an equation solver such as EES gives 6 460 m kmol 3 v 2 Then 002306 m3 0 100 kg 0 2306 m kg 0 2306 m kg 28011 kgkmol 460 m kmol 6 3 2 2 3 3 2 2 v V v v m M CO 1000 kPa 200C Q preparation If you are a student using this Manual you are using it without permission 350 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course igid container The final pressure of the methane is to be determined using the ideal gas quation and the BenedictWebbRubin equation of state 3101 Methane is heated in a r e Analysis a From the ideal gas equation of state 1565 kPa 80 kPa 573 K 1 2 1 2 T P T P Methane 80 kPa 20C Q K 293 The specific molar volume of the methane is b 3045 m kmol 80 kPa 8314 kPa m kmol K293 K 3 3 1 2 1 P RuT v v 1 Using the coefficients of Table 34 for methane and the given data the BenedictWebbRubin equation of state for state 2 gives 1565 kPa 0 0060 3045 exp 3045 0 0060 1 3045 573 2 578 10 3045 1 244 10 5 00 2 2 2 3 6 3 5 00 0 003380 3045 1 573 2 286 10 8 314 573 18791 0 04260 3045 8 314573 exp 5 4 2 2 6 2 2 2 2 3 6 v v v v v γ γ α T c a 1 1 3 2 2 2 0 0 2 0 2 2 a bR T C A B R T R T P u u u 2 2 v v T 045 8 314 573 3 preparation If you are a student using this Manual you are using it without permission 351 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course l gas are Table A1 ol Analysis 3102E Carbon monoxide is heated in a rigid container The final pressure of the CO is to be determined using the idea equation and the BenedictWebbRubin equation of state Properties The gas constant and molar mass of CO R 02968 kPam3kgK M 28011 kgkm a From the ideal gas equation of state CO Q 147 p 70F sia psia 530 R 1 1 2 T b The specific molar volume of the CO in SI units is 3495 14 7 psia 1260 R P T2 P 2420 m kmol 8314 kPa m kmol K294 K 3 3 1 RuT v v kPa 101 1 2 1 P Using th e 2 gives e coefficients of Table 34 for CO and the given data the BenedictWebbRubin equation of state for stat 2408 kPa he pressure in English unit is 0 0060 2420 exp 2420 0 0060 1 2420 700 1 054 10 2420 1 350 10 71 3 2420 3 71 8 314 700 0 002632 2420 1 700 8 673 10 8 314 700 13587 0 05454 2420 8 314700 exp 1 1 2 2 2 3 5 6 4 3 2 2 5 2 2 2 2 3 6 3 2 2 2 2 0 0 2 0 2 2 2 v v v v v v v γ γ α T c a a bR T T C A B R T R T P u u u T 3492 psia 68948 kPa 1psia 240 8 kPa P2 preparation If you are a student using this Manual you are using it without permission 352 3103E The temperature of R134a in a tank at a specified state is to be determined using the ideal gas relation the van der Waals equation and the refrigerant tables PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course e of R134a are Table A1E cr Properties The gas constant critical pressure and critical temperatur R 01052 psiaft3lbmR Tcr 6736 R P 588 7 psia Analysis a From the ideal gas equation of state 529 R psia ft lbm R 01052 psia03479 ft lbm 160 3 3 R P T v b The van der Waals constants for the refrigerant are determined from 00150 ft lbm 01052 psia ft lbm R6736 R 3 3 RTcr b 5887 psia 8 8 3591 ft 645887 psia 270105 2 psia ft lbm R 6736 R 64 27 2 6 2 2 3 2 2 cr cr cr P P R T a Then psialbm 600 R 00150 03479 03479 3591 01052 160 1 1 2 2 b a R P T v v c From the superheated refrigerant table Table A13E 620 R 03479 ft lbm 160 psia 3 160F T P v preparation If you are a student using this Manual you are using it without permission 353 3104 The pressure of nitrogen in a tank at a specified state is to be determined using the ideal gas relation and the roperti r mass T le A1 3 kgkmol Analysis BeattieBridgeman equation The error involved in each case is to be determined P es The gas constant and mola of nitrogen are ab N2 0041884 m3kg 150 K R 02968 kPam3kgK and M 2801 a From the ideal gas equation of state 63 error kPa m kg K150 K 02968 3 3 1063 kPa P RT b Th 0041884 m kg v e constants in the BeattieBridgeman equation are K kmol 42 10 m c 005076 11733 000691 005046 1 1 133193 11733 002617 1362315 1 1 3 3 4 b B B a A A o o v v since 11733 m kmol kgkmol0041884 m kg 28013 3 3 v v M Substitu ting e error negligibl 11733 133193 005076 11733 150 11733 42 10 1 11733 8314 1 2 3 4 2 2 3 2 10004 kPa v v v v A B T c R T P u 150 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 354 3105 Problem 3104 is reconsidered Using EES or other software the pressure results of the ideal gas and BeattieBridgeman equations with nitrogen data supplied by EES are to be compared The temperature is to be plotted rated liquid and saturated vapor lines of nitrogen lysis s given below barvM Conversion from m3kg to m3kmol The constants for the BeattieBridgeman equation of state are found in text 91 cc4201E4 nd T150 K v0041884 m3kg e 1000 kPa ASSNitrogen RRuM idealgasRTidealgasv Ideal gas equation PBBBeattBridgTBBvMRu BeattieBridgeman equation of state Function PBB kPa Ptable kPa Pidealgas kPa v m3kg TBB K Tideal gas K Ttable K versus specific volume for a pressure of 1000 kPa with respect to the satu over the range of 110 K T 150 K Ana The problem is solved using EES and the solution i Function BeattBridgTvMRu v Ao1362315 aa002617 Bo005046 bb0006 BBo1bbvbar AAo1aavbar The BeattieBridgeman equation of state is BeattBridgRuTvbar21ccvbarT3vbarBAvbar2 E Pexp r TtableT TBBTTidealgasT PtablePRESSURENitrogenTTtablevv EES data for nitrogen as a real gas TtabletemperatureNitrogen PPtablevv MMOLARM Ru8314 kJkmolK Universal gas constant Particular gas constant P 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 001 002 0025 003 0035 004 005 9123 9552 105 1168 1301 1444 1746 3369 6739 8423 1011 1179 1348 1685 1038 1038 1061 1172 1301 1443 1745 103 102 101 70 80 90 100 110 120 130 140 150 160 v m3kg T K 1000 kPa Nitrogen T vs v for P1000 kPa EES Table Value EES Table Value BeattieBridgeman BeattieBridgeman Ideal Gas Ideal Gas PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 355 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 3106 oxide is compressed in a pistoncylinder device in a polytropic process The final temperature is to be ined usi Carbon di determ ng the ideal gas and van der Waals equations Properties The gas constant molar mass critical pressure and critical temperature of carbon dioxide are Table A1 R 01889 kPam3kgK M 4401 kgkmol Tcr 3042 K Pcr 739 MPa Analysis a The specific volume at the initial state is 008935 m kg 0 kPa 01889 kPa m kg K473 K 3 3 1 RT 100 1 v P 1 According to process specification 0 03577 m kg 3000 kPa 0 08935 m kg 1000 kPa 3 3 21 1 1 n 1 P 2 1 2 P v v The final temperature is then 568 K kPa m kg K 01889 kPa003577 m kg 3000 3 3 2 2 R P v onstants for carbon dioxide are determined from T2 b The van der Waals c 00009720 m kg kPa 3 kPakg 01885 m 270188 9 kPa m kg K 3042 K 27 2 6 2 2 3 2 2 cr T R 647390 kPa 64 cr P a 8 7390 8 cr P 01889 kPa m kg K3042 K 3 cr RT b Applying the van der Waals equation to the initial state 0 18 0 0 5 1000 v v a this equ y trialerror or by EES giv kg 0 3 v1 ccording to process specification 89473 000972 188 0 2 v RT b 2 v P Solving ation b es m 08821 A m kg 0 03531 3000 kPa m kg 1000 kPa 0 08821 3 21 1 3 1 2 1 1 2 n P P v v Applying the van der Waals equation to the final state T RT b a P 0 1889 0 0009720 0 03531 03531 0 0 1885 3000 2 2 v v Solving for the final temperature gives T2 573 K CO2 1 MPa 200C preparation If you are a student using this Manual you are using it without permission 356 Special Topic Vapor Pressure and Phase Equilibrium PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course A glass of water is left in a room at the free surface of the water far from the perature roperti and 1706 kPa at 15 Analysis The vapor pressure at the water surface is the saturation pressure of water at the 3107 The vapor pressures and in the room glass are to be determined Assumptions The water in the glass is at a uniform tem P es The saturation pressure of water is 2339 kPa at 20C C Table A4 water temperature 1706 kPa sat15C sat water surface water P P P T v Noting that the air in the room is not saturated the vapor pressure in the room far from the glass is 0936 kPa 40 2 339 kPa T v P P P sat20C sat air air φ φ 3108 ed to be 52 kPa The validity of is claim is to be evaluated Properties The saturation pressure of water at 30C is 4247 kPa Table A4 Analysis The maximum vapor pressure in the air is the saturation pressure of water hich is less than the claimed value of 52 kPa Therefore the claim is false 109 tive humidity of air over a swimming pool are given The water temperature of the um conditions are established is to be determined Assumptions constant Properti Analysis kPa The vapor pressure in the air at the beach when the air temperature is 30C is claim th at the given temperature which is 4247 kPa sat30C sat max air P P P T v w 3 la The temperature and re swimming pool when phase equilibri The temperature and relative humidity of air over the pool remain es The saturation pressure of water at 20C is 2339 kPa Table A4 The vapor pressure of air over the swimming pool is 09357 40 2 339 kPa sat20 C sat air air P P P T v φ φ Phase equilibrium will be established when the vapor pressure at the water surface equals the vapor pressure of air far from the surface Therefore 09357 kPa air water surface v v P P a 60C iscussi e water temperature drops to 60C in an environment at 20C when phase equilibrium is tablished H2O 15C Patm 20C POOL 30C WATER and sat 0 9357 kP sat water v T T T P D on Note that th es preparation If you are a student using this Manual you are using it without permission 357 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course summer day yet no condensation occurs on the drink The laim that the temperature of the drink is below 10C is to be evaluated of air is 3110 A person buys a supposedly cold drink in a hot and humid c Properties The saturation pressure of water at 35C is 5629 kPa Table A4 Analysis The vapor pressure 3 940 kPa 70 5 629 kPa sat35 C sat air air P P P T v φ φ 287C P T That is the vapor in the air will condense at temperatures below 287C Noting that no condensat served on the 111E A thermos bottle halffilled with water is left open to air in a room at a specified temperature and pressure The mperature of water when phase equilibrium is established is to be determined ssure of water at 70F is 03633 psia Table A4E 35C 70 The saturation temperature corresponding to this pressure called the dewpoint temperature is T T sat3940 kPa sat sat v n is ob io can the claim that the drink is at 10C is false 3 te Assumptions The temperature and relative humidity of air over the bottle remain constant Properties The saturation pre Analysis The vapor pressure of air in the room is 01272 psia 0 35 0 3633 psia sat70 F sat air air P P P T v φ φ P e hase eq the vapor pressure at the water surface quals the vapor pressure of air far from the surface Therefore nd 411F sat 0 1272 psia sat water v T T T P blished different temperatur ities The room at con Properti and 317 kPa at 25C Table A4 kPa 127 uilibrium will be established when Thermos ttle F 35 bo 70 01272 psia air water surface v v P P a Discussion Note that the water temperature drops to 41F in an environment at 70F when phase equilibrium is esta 3112 Two rooms are identical except that they are maintained at es and relative humid th tains more moisture is to be determined es The saturation pressure of water is 2339 kPa at 20C Analysis The vapor pressures in the two rooms are 1 sat 1 1 P P T v Room 1 40 3 17 kPa φ1Psat25 C φ 129 kPa Room 2 2 sat 2 2 2 P P P T v 0 55 2 339 kPa sat20 C φ φ Therefore room 1 at 30C and 40 relative humidity contains more moisture preparation If you are a student using this Manual you are using it without permission 358 Review Problems PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m rmined nitrogen behaves as an ideal gas nalysis 3113 Nitrogen gas in a rigid tank is heated to a final gage pressure The final te perature is to be dete Assumptions At specified conditions A e According to the ideal gas equation of state at constant volum 602C 875 K 100 100 kPa 273 K 250 100 kPa 227 2 2 1 2 2 1 2 2 1 P T P T T T V low rates and the density of CO2 at the The vol ermined from ideal gas relation as 2 1 1 1 1 P P m m V Since 2 1 2 1 T P T P V V Nitrogen gas 227C 100 kPa Patm 100 kPa Q gage 3114 Carbon dioxide flows through a pipe at a given state The volume and mass f given state and the volume flow rate at the exit of the pipe are to be determined Analysis 3 MPa 500 K 04 kmols CO2 450 K a ume and mass flow rates may be det 05543 m s 3 00 K kmols8314 kPam 40 3 NR T1 u V kmolK5 3000 kPa 1 P 1760 kgs kPam kgK500 K 01889 kPa0554 3m 3000 3 3 1 1 1 1 s RT P m V he density is T 3176 kgm3 0 5543 m s kgs 1760 3 1 1 1 V m ρ b The volume flow rate at the exit is 04988 m s 3 3000 kPa 50 K 2 2 P u V kmols8314 kPam kmolK4 40 3 R T N preparation If you are a student using this Manual you are using it without permission 359 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course nditions before the heat addition process is specified The pressure after the heat addition process is be determined 3115 The cylinder co to Assumptions 1 The contents of cylinder are approximated by the air properties 2 Air is an ideal gas Analysis The final pressure may be determined from the ideal gas relation 3627 kPa 273 K 1600 T 1400 kPa 273 K 450 1 1 2 2 T P P 3116 The cylinder conditions before the heat addition process is specified The temperature after the heat addition process is to be determined ssumptions 1 The contents of cylinder is approximated by the air roperties 2 Air is an ideal gas lysis The ratio of the initial to the final mass is A p Ana 23 22 1 22 22 1 AF AF 2 1 m m The final temperature may be determined from ideal gas relation 1817 K 950 K cm 75 150 cm 23 22 3 3 2 m m T 3117 al temperature are to be determined Analysis the in initial pressure is then the saturation pressure Table A 11 0 sat 40 C 1 3 1 5125 kPa P P v This is a constant volume cooling process v V m constant The final temperature is then 2 2 P v 1 1 2 V 2 1 T V A rigid container that is filled with R13a is heated The initial pressure and the fin The initial specific volume is 0090 m3kg Using this with itial temperature reveals that the initial state is a mixture The 090 m kg 40 C 1 T state is superheated vapor and the final Table A 13 090 m kg 0 280 kPa 2 3 1 50C T v Combustion chamber 950 K 75 cm3 R134a 40C 1 kg 0090 m3 P v 1 2 Combustion chamber 14 MPa 450C preparation If you are a student using this Manual you are using it without permission 360 3118E A pistoncylinder device that is filled with water is cooled The final pressure and volume of the water are to be determined PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis The initial specific volume is 2 649 ft lbm lbm 1 3 1 1 v m 649 ft 3 his is a co superheated vapor and thus the pressure is determined to be Table A 6E 649 ft lbm 2 400 F 2 1 3 1 1 180 psia P P T v he saturation temperature at 180 psia is 3731F Since the final mperature is less than this temperature the final state is compressed ng the incompressible liquid approximation 0 01613 ft lbm Table A 4E F 1613 ft3 3 3119 n the figure is to be determined ssumptions At specified conditions helium behaves as an i gas ies The gas constant of helium is R 20769 kJkgK Table A1 lysis Since the water vapor in chamber 2 is condensing the pressure in is chamber is the saturation pressure 1555 kPa sat 200 C 2 P P Table A4 P v 2 1 H2O 400F 1 lbm 2649 ft3 P2A2 2 V T ed to be nstantpressure process The initial state is determin T te liquid Usi 3 100 2 v f v The final volume is then 00 1 lbm 0 01613 ft lbm 2 2 v V m The volume of chamber 1 of the twopiston cylinder shown i A deal Propert Ana th Summing the forces acting on the piston in the vertical direction gives 248 8 kPa 10 1555 kPa 4 2 2 1 2 2 1 2 2 1 D D P A P A P P1A1 According to the ideal gas equation of state 395 m 273 K kg20769 kPa m kg K200 1 3 1 mRT V 3 2488 kPa 1P preparation If you are a student using this Manual you are using it without permission 361 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course At specified conditions air behaves as an ideal gas aft3lbmR Table A1E Analysis ng the pressure in this chamber is pressure 186 0 psia P P Table A11E ion gives 3120E The volume of chamber 1 of the twopiston cylinder shown in the figure is to be determined Assumptions Properties The gas constant of air is R 03704 psi Since R134a in chamber 2 is condensi ration the satu sat 120 F 2 Summing the forces acting on the piston in the vertical direct F2 1 1 2 1 3 2 2 1 3 2 P A A A P A P F F F F1 F3 which when solved for P1 gives 2 3 2 2 1 1 A P P A P 1 1 A A D2 4 A π the above since the areas of the piston faces are given by equation becomes 9933 psia 3 2 30 psia 1 3 186 0 psia 2 1 2 2 1 3 1 2 1 D D P D D P P 2 2 2 2 According to the ideal gas equation of state 10 8 ft3 9933 psia 460 R lbm0370 4 psia ft lbm R120 05 1 1 P mRT V 3 preparation If you are a student using this Manual you are using it without permission 362 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course rmined 3121E The difference in the volume of chamber 1 for two cases of pressure in chamber 3 is to be dete Assumptions At specified conditions air behaves as an ideal gas Properties The gas constant of air is R 03704 psiaft3lbmR Table A1 Analysis Since R134a in chamber 2 is condensing the pressure in this chamber is the saturation pressure 0 psia Table A11E 186 sat 120 F 2 P P F1 F3 F2 Summing the forces acting on the piston in the vertical direction gives 1 1 2 1 3 2 2 1 3 2 P A A A P A P F F F which when solved for P1 gives 1 2 3 1 2 2 1 1 A A P A P A P since the areas of the piston faces are given by the above equation becomes D2 4 A π psia 116 3 2 60 kPa 1 3 186 0 psia 2 1 2 2 2 1 2 3 2 1 2 2 1 D D P D D P P According to the ideal gas equation of state 3 3 1 1 0926 ft 116 psia 460 R 05 lbm0370 4 psia ft lbm R120 P mRT V For a chamber 3 pressure of 30 psia the volume of chamber 1 was determined to be 108 ft3 Then the change in the volume of chamber 1 is 0154 ft3 0 926 1 08 1 2 V V V preparation If you are a student using this Manual you are using it without permission 363 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course nd the of ethane are from Table A1 lysis 3122 Ethane is heated at constant pressure The final temperature is to be determined using ideal gas equation a compressibility charts Properties The gas constant the critical pressure and the critical temperature 3 R 02765 kPam kgK Tcr 3055 K Pcr 448 MPa Ana From the ideal gas equation 5968 K 373 K 61 1 2 1 2 T v T v rom th itial state Fig A15 Ethane 10 MPa 100C F e compressibility chart at the in Q 0 35 0 61 Z 232 448 MPa MPa 10 221 K 373 1 1 cr 1 1 R R P P P T v 2 232 Z R R v v Thus 2 1 3055 K cr 1 1 R T T At the final state P 2 P 1 R R 0 83 0 56 61 0 35 61 2 1 2 460 K 4480 kPa 0563055 K 083 kPa 10000 cr cr 2 2 2 2 2 2 2 P T Z P Z R P T v R v Of these two results the accuracy of the second result is limited by the accuracy with which the charts may be read Accepting the error associated with reading charts the second temperature is the more accurate nd pressure of nitrogen drop to new values The amount of nitrogen that has escaped is to be determined roperti 3kgK Table A1 nalysis Treating N2 as an ideal gas the initial and the final masses in the tank are determined to be 3123 A large tank contains nitrogen at a specified temperature and pressure Now some nitrogen is allowed to escape and the temperature a P es The gas constant for nitrogen is 02968 kPam A 92 0 kg kPa m kg K293 K 02968 kPa20 m 400 1366 kg kPa m kg K296 K 02968 kPa20 m 600 3 3 2 2 2 3 3 1 1 1 RT P m RT P m V V Thus the amount of N2 that escaped is 446 kg 92 0 1366 2 1 m m m N2 600 kPa 23C 20 m3 preparation If you are a student using this Manual you are using it without permission 364 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course olume it is to be determined if the final phase is a liquid or a vapor qual to the itial sp The criti 3kg Thus if the final specific olume is smaller than this value the water will exist as a liquid otherwise as a v 3124 The rigid tank contains saturated liquidvapor mixture of water The mixture is heated until it exists in a single phase For a given tank v Analysis This is a constant volume process v V m constant and thus the final specific volume will be e in ecific volume 1 2 v v cal specific volume of water is 0003106 m v apor cr 0002 m3kg 2 kg 0004 m 4 v V v V m L 3 Thus liquid 02 m kg 2 kg 04 m 400 cr 3 3 v V v V m L Thus vapor states are connected to each other Now a valve is opened and l equilibrium with the surroundings The final pressure in the tanks determined roperti Analysis ideal gas the total volume and the total mass of H2O V 4 L m 2 kg T 50C H2 V 05 m3 T20C 3125 Two rigid tanks that contain hydrogen at two different the two gases are allowed to mix while achieving therma is to be P es The gas constant for hydrogen is 4124 kPam3kgK Table A1 Lets call the first and the second tanks A and B Treating H2 as an H2 are A B 10 m 05 05 3 B A V V V H2 V 05 m3 T50C 02218 kg 00563 01655 00563 kg 4124 kPa m kg K323 K 3 1 B A B B m m m RT m kPa05 m 150 01655 kg kPa m kg K293 K 4124 kPa05 m 400 3 1 3 3 1 1 A A P RT P m V V P400 kPa P150 kPa Then the final pressure can be determined from 264 kPa 2 m 10 02218 kg4124 kPa m kg K288 K V mRT P 3 3 preparation If you are a student using this Manual you are using it without permission 365 3126 Problem 3125 is reconsidered The effect of the surroundings temperature on the final equilibrium pressure in d nalysis em is solved using EES and the solution is given below VA A20 C A400 kPa VB05 m3 TB B150 kPa T215 C olution RuMOLARMASSH2 u8314 kJkmolK totalVAVB RT2273 2 k T the tanks is to be investigated The final pressure in the tanks is to be plotted versus the surroundings temperature and the results are to be discusse A The probl Given Data 05 m3 T P 50 C P S R R V mtotalmAmB PAVAmARTA273 PBVBmBRTB273 P2Vtotalmtotal P PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Pa 2 C 10 5 0 5 10 15 20 25 30 2406 2452 2497 2543 2589 2635 268 2726 10 5 0 5 10 15 20 25 2772 30 250 240 260 270 280 T2 C P2 kPa preparation If you are a student using this Manual you are using it without permission 366 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 3127 essure in an automobile tire increases during a trip while its volume remains constant The percent increase in lute heric pressure is 90 kPa P P 200 90 290 220 90 310 gage1 atm ge2 atm kPa kPa n ideal gas and the volume is constant the ratio of absolute r and before the trip are The pr the abso temperature of the air in the tire is to be determined Assumptions 1 The volume of the tire remains constant 2 Air is an ideal gas Properties The local atmosp Analysis The absolute pressures in the tire before and after the trip are P 1 P P 2 P ga Noting that air is a temperatures afte 290 kPa 1069 310 kPa 1 2 1 2 2 2 2 P P T T T V he absolute temperature of air in the tire will increase by 69 during this trip n a tank at a specified state is to be determined using the ideal gas relation the roperties The gas constant the critical pressure and the critical temperature of water are from Table A1 04615 kPa m kg K 3 R Ana From t eal gas equation of state H2O 002 m3kg 400C TIRE 200 kPa 0035 m3 1 T 1 1 P P V Therefore t 3128 The temperature of steam i generalized chart and the steam tables P 2206 MPa 6471 K cr cr P T lysis a he id 15529 kPa 002 m kg kPa m kg K673 K 615 3 3 v RT b e compressibility chart Fig A15a 04 P From th 0 57 148 kPa m kg K6471 K 04615 002 m kg2206 0 kPa 1 040 1 K K 3 3 3 cr cr actual cr R R P P RT T T v v 12574 kPa 22060 0 57 cr P P P R c From the superheated steam table 12515 kPa P T 002 m kg 400 C 3 v from EES 647 67 R T Thus preparation If you are a student using this Manual you are using it without permission 367 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course d The volume of the tank is to be determined 3129 One section of a tank is filled with saturated liquid R134a while the other side is evacuated The partition is removed and the temperature and pressure in the tank are measure Analysis The mass of the refrigerant contained in the tank is 3358 kg 00008934 m kg 3 v1 m 003 m 3 1 V since m3 191 kg00568 0 m kg 3358 3 2 2 tank v V V m 00008934 m kg 3 14 MPa 1 v f v Evacuated R134a P12 MPa V 003 m3 At the final state Table A13 005680 m kg C 30 400 kPa 3 2 2 2 v T P Thus preparation If you are a student using this Manual you are using it without permission 368 3130 Problem 3129 is reconsidered The effect of the initial pressure of refrigerant134 on the volume of the tank a to 15 MPa The volume of the tank is to be plotted versus the in ssed nalysis The problem is solved using EES and t e solution is given below x100 Vol Soluti R134aPP1xx1 1 v2vol Vol P1 kPa Vol2 m3 m kg is to be investigated as the initial pressure varies from 05 MP itial pressure and the results are to be discu A h Given Data 1003 m3 P11200 kPa T230 C P2400 kPa on v1volume Vol1mv umeR134aPP2TT2 2mv2 500 600 700 800 900 1000 1100 1200 1300 1400 1500 2114 2078 2045 2015 1986 1958 1932 1907 1883 1859 1836 3723 3659 3601 3547 3496 3448 3402 3358 3315 3273 3232 500 700 900 1100 1300 1500 18 185 19 195 2 205 21 215 P1 kPa Vol2 m 3 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 369 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 3131 ank contains 5 L of liquid propane at the ambient temperature Now a leak develops at the top of the tank 421C ρ 581 kgm3 and hfg 4278 kJkg Table A3 1 atm is simply the saturation pressure at that temperature sat atm 1 421 C iquid propane is 2905 kg kgm 0005 m 581 3 3 ρV he amount of heat absorbed is simply the total heat of vaporization 278 kJ kg A propane t and propane starts to leak out The temperature of propane when the pressure drops to 1 atm and the amount of heat transferred to the tank by the time the entire propane in the tank is vaporized are to be determined Properties The properties of propane at 1 atm are Tsat Analysis The temperature of propane when the pressure drops to T T The initial mass of l m T Q mh fg absorbed 2905 kg4 1243 kJ 132 An isobutane tank contains 5 L of l e top of the tank and isobut s t k out The temperature of isobutane when the pressure drops to 1 atm and the amount of hea erred e tan he time the entire isobutane in the tank is vaporized are to be determined Pro The propertie butane at 1 atm are Tsat 117C ρ 5938 kg m3 and hfg 3671 kJkg Table A3 Ana tu butane when the pressure drops to 1 atm is simply the saturation pressure at that tem re C The initial m u ane is 2969kg 0005 m 93 3 3 he amount of heat absorbed is simply the total heat of vaporization Q mh fg absorbed 2969 kg3671 kJ kg 3 iquid isobutane at the ambient temperature Now a leak develops at th ane start o lea t transf to th k by t perties s of iso lysis The tempera re of iso peratu T T 117 sat atm 1 ass of liq id isobut 8 kgm ρV 5 m T 1090 kJ 3133 A tank contains helium at a specified state Heat is transferred to helium until it reaches a specified temperature The final gage pressure of the helium is to be determined Assumptions 1 Helium is an ideal gas Properties The local atmospheric pressure is given to be 100 kPa Analysis Noting that the specific volume of helium in the tank remains constant from ideal gas relation we have 3438 kPa 273K 77 273K 110 100 kPa 300 1 2 1 2 T P T P Then the gage pressure becomes 244 kPa 100 343 8 atm 2 gage2 P P P Isobutane 5 L 20C Leak Propane 5 L 20C Leak Helium 77ºC 110 kPa gage Q preparation If you are a student using this Manual you are using it without permission 370 3134 The first eight virial coefficients of a BenedictWebbRubin gas are to be obtained Analysis The BenedictWebbRubin equation of state is given by 1 2 3 6 3 2 2 0 0 0 v v v v v v γ α T c a bR T T C A B R T R T P u u u exp 1 2 2 v γ a Expanding the last term in a series gives 3 1 6 3 4 2 v v γ γ γ n of state and es 2 1 1 exp 2 2 v v γ Substituting this into the BenedictWebbRubin equatio rearranging the first terms giv PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2 9 2 2 7 2 1 2 1 1 T T c a c a bR T T C A R TB T R v v γ γ γ γ α γ The viri 1 c 6 2 5 3 2 0 0 0 T P u u u v v v v v al equation of state is 9 8 7 6 5 4 3 2 v v v v v v v v v h T g T f T e T d T c T b T a T R T P u omparing the BenedictWebbRubin equation of state to the virial equation of state the virial coefficients are C 2 0 0 0 0 T c a bR T T b T C A R TB T a u u 2 1 T c d T γ 2 2 1 2 1 c h T γ γ 2 1 T g T c γ γ 135 The table is completed as follows P kPa T oC v m3kg u kJkg Condition description and quality if applicable T f a e T α 0 T 3 300 250 07921 27289 Superheated vapor 300 13352 03058 15600 x 0504 Twophase mixture 10142 100 Insuffic ient information 3000 180 0001127 76192 Compressed liquid Approximated as saturated liquid at the given temperature of 180oC preparation If you are a student using this Manual you are using it without permission 371 3136 The table is completed as follows PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course P lity if kPa T oC v m3kg u kJkg Condition description and qua applicable 200 08858 25291 Saturated vapor 1202 23223 125 05010 18310 x 0650 Twophase mixture 7829 400 00352 29672 Superheated vapor 1000 30 0001004 12573 Compressed liquid 12090 105 Insufficient information Approximated as saturated liquid at the given temperature of 30oC preparation If you are a student using this Manual you are using it without permission 372 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course ed in a tank It is now cooled The process will be indicated on the Pv and T v iagrams Ana ro at the fina e kg 0 25 300 1 1 1 v T P 11135 C Table A 5 m 964 150 2 3 1 2 T v Using Property Pl ure of EES and by addi e poi ng diagrams 3137 Water at a specified state is contain d lysis The p perties initial and l states ar Table A 6 7964 m 3 C kPa 0 kg 70 kPa 2 v P ot feat ng stat nts we obtain followi 104 103 102 101 100 101 102 100 101 102 103 104 105 106 v m3kg P kPa 250C 1114C SteamIAPWS 1 2 150 kPa 300 kPa 103 102 101 100 101 102 0 100 200 300 400 500 600 700 v m3kg T C 300 kPa 150 kPa SteamIAPWS 1 2 250C 11135C preparation If you are a student using this Manual you are using it without permission 373 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course he process will be indicated on the Pv and T v diagrams A 5 C Table A 6 517 8 0 6058 m kg 2 3 3 T v lowing diagrams 3138 Water at a specified state is contained in a pistoncylinder device fitted with stops Water is now heated until a final pressure T Analysis The properties at the three states are C Table A 5 133 5 m kg 50 300 kPa 1 3 1 1 T P v C Table 133 5 0 6058 m kg sat vap 1 300 kPa 2 3 2 2 2 T x P v 600 kPa 2 P Using Property Plot feature of EES and by adding state points we obtain fol 104 103 102 101 100 101 102 100 101 102 103 104 105 106 v m kg 3 P kPa 5178C 1588C 1335C SteamIAPWS 1 2 3 05 300 kPa 600 kPa 103 102 101 100 101 102 0 100 200 300 400 500 600 700 v m3kg T C 600 kPa 300 kPa SteamIAPWS 1 2 3 05 5178C 1588C 1335C Water 300 kPa 05 m3kg Q preparation If you are a student using this Manual you are using it without permission 374 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course an equation gas give 3139E Argon contained in a pistoncylinder device at a given state undergoes a polytropic process The final temperature is to be determined using the ideal gas relation and the BeattieBridgem Analysis a The polytropic relations for an ideal 986 R 61 60 1 1 2 1 2 1000 psia 460 R 2000 psia 300 n n P P T T b Th e constants in the BeattieBridgeman equation are expressed as K kmol 599 10 m 3 3 4 c 0 003931 1 1 002328 1307802 1 1 b B B a A A o o v v v v Substituting these coefficients into the BeattieBridgeman equation and using data in SI units P 1000 psia 6895 kPa T760 R 4222 K Ru 8314 kJkmolK 2 3 2 1 v v v v A B T c R T P u and solving using an equation solver such as EES gives 8 201 ft lbmol 0 5120 m kmol 3 3 v From the polytropic equation 0 3319 m kmol 2 05120 m kmol 1 3 61 1 3 1 2 1 1 2 n P P v v Substituting this value into the BeattieBridgeman equation and using da 8314 kJkmolK ta in SI units P 2000 psia 13790 kPa and Ru 2 3 2 1 v v v v A B T c R T P u and solving using an equation solver such as EES gives 300F Argon 1000 psia 2 K 958 R 532 2 T preparation If you are a student using this Manual you are using it without permission 375 3140E The specific volume of nitrogen at a given state is to be determined using the ideal gas relation the Benedict WebbRubin equation and the compressibility factor Properties The properties of nitrogen are Table A1E PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Pcr 492 psia nalysis a From the ideal gas equation of state R 03830 psiaft3lbmR M 28013 lbmlbmol Tcr 2271 R A Nitrogen 00 psia 100F 4 03447 ft lbm 3 03830 psia ft lbm R360 R 3 RT v 400 psia P b Using the coefficients of Table 34 for nitrogen and the given data in SI units BenedictWebbRubin equation of state is the 0 0053 0 0053 exp 1 200 10 200 exp 1 1 2 2 2 3 4 6 4 3 2 2 2 2 2 3 6 3 2 0 0 0 v v v v v v v v v v v v v γ γ α T c a a bR T T C A B R T R T P u u u as EES gives 7 379 1 272 10 54 2 2 54 8 314 200 0 002328 1 8 164 10 8 314 200 10673 0 04074 8 314200 2758 5 v2 The solution of this equation by an equation solver such 0 5666 m kmol 3 v Then 03240 ft lbm 3 m kg 1 1602 ft lbm 28013 kgkmol 5666 m kmol 0 3 3 3 M v v c From the compressibility chart Fig A15 0 94 psia 400 1 585 2271 R R 360 cr P T T TR hus 03240 ft lbm 3 0 9403447 ft lbm 3 videal v Z Z 0 813 492 psia cr P PR T preparation If you are a student using this Manual you are using it without permission 376 Fundamentals of Engineering FE Exam Problems PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course and 40C Now a valve is opened and half of mass of the gas is llowed tu 00C d 20C e 172C e foll ank EES s can be obtained easily by modifying numerical values m12 P14 P222 tm T140273 K m205m1 kg P1P2m1T1m2T2 T2CT2273 C 1P2 T2 Using C instead of K Disregarding the decrease in mass m1T1m1W3T2 Disregarding the decrease in mass and not converting to deg C 3142 o be 190 kPa gage before a trip and 215 kPa gage after the trip at location where the atmospheric pressure is 95 kPa If the temperature of air in the tire before the trip is 25C the air mperature after the trip is a 511C b 642C c 272C d 283C e 250C a 511C copyingandpasting the following lines on a blank EES reen Similar problems and their solutions can be obtained easily by modifying numerical values When R V and m are constant P1P2T1T2 Patm95 P1190Patm kPa P2215Patm kPa T125273 K P1P2T1T2 T2CT2273 C Some Wrong Solutions with Common Mistakes P1P2T1273W1T2 Using C instead of K P1PatmP2PatmT1W2T2273 Using gage pressure instead of absolute pressure P1PatmP2PatmT1273W3T2 Making both of the mistakes above W4T2T1273 Assuming the temperature to remain constant 3141 A rigid tank contains 2 kg of an ideal gas at 4 atm a to escape If the final pressure in the tank is 22 atm the final tempera re in the tank is a 71C b 44C c 1 Answer a 71C Solution Solved by EES Software Solutions can be verified by copyingandpasting th screen Similar problems and their solution owing lines on a bl When Rconstant and V constant P1P2m1T1m2T2 kg atm a Some Wrong Solutions with Common Mistakes P 1 m1T1273m2W P1P2m1T1m1W2T2273 P1P2 W4T2T12732 Taking T2 to be half of T1 since half of the mass is discharged The pressure of an automobile tire is measured t a te Answer Solution Solved by EES Software Solutions can be verified by sc preparation If you are a student using this Manual you are using it without permission 377 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course r mixture of water at 200 kPa If 25 of the mass is quid and the 75 of the mass is vapor the total mass in the tank is EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES milar problems and their solutions can be obtained easily by modifying numerical values 075 vf g lutions with Common Mistakes amIAPWSx0PP1 RT273 Treating steam as ideal gas s and using degC 144 Water is boiled at 1 atm pressure in a coffee maker equipped with an immersiontype electric heating element The ffee maker initially contains 1 kg of water Once boiling started it is observed that half of the water in the coffee maker vaporated in 10 minutes If the heat loss from the coffee maker is negligible the power rating of the heating element is ES Software Solutions can be verified by copyingandpasting the following lines on a blank EES ems and their solutions can be obtained easily by modifying numerical values kPa hfg kJ mIAPWS x0PP mIAPWS x1PP tions with Common Mistakes stead of seconds for time 3143 A 300m3 rigid tank is filled with saturated liquidvapo li a 451 kg b 556 kg c 300 kg d 331 kg e 195 kg Answer a 451 kg Solution Solved by screen Si Vtank300 m3 P1200 kPa x vfVOLUMESteamIAPWS x0PP1 vgVOLUMESteamIAPWS x1PP1 vvfxvg mVtankv k Some Wrong So R04615 kJkgK TTEMPERATURESte P1VtankW1m P1VtankW2mRT Treating steam as ideal ga W3mVtank Taking the density to be 1 kgm3 3 co e a 38 kW b 22 kW c 19 kW d 16 kW e 08 kW Answer c 19 kW Solution Solved by E screen Similar probl m11 kg P101325 time1060 s mevap05m1 Powertimemevap hfENTHALPYStea hgENTHALPYStea hfghghf Some Wrong Solu W1Powertimemevaphg Using hg W2Powertime60mevaphg Using minutes in W3Power2Power Assuming all the water evaporates preparation If you are a student using this Manual you are using it without permission 378 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 000 kPa e 1618 kPa 618 kPa olution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES tank1 m3 tions with Common Mistakes 1PVtankmRT273 Treating steam as ideal gas and using degC 146 Water is boiling at 1 atm pressure in a stainless steel pan on an electric range It is observed that 2 kg of liquid ater evaporates in 30 minutes The rate of heat transfer to the water is b 232 kW c 297 kW d 047 kW e 312 kW nswer a 251 kW ar probl ution ined easi ng nume kg 101325 kPa ENTHALPYSteamIAPWS x0PP gENTHALPYSteamIAPWS x1PP f olutions with Common Mistakes phg Using hg sing minutes instead of seconds for time 3145 A 1m3 rigid tank contains 10 kg of water in any phase or phases at 160C The pressure in the tank is a 738 kPa b 618 kPa c 370 kPa d 2 Answer b S screen Similar problems and their solutions can be obtained easily by modifying numerical values V m10 kg vVtankm T160 C PPRESSURESteamIAPWSvvTT Some Wrong Solu R04615 kJkgK W W2PVtankmRT Treating steam as ideal gas 3 w a 251 kW A Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Simil ems and their sol s can be obta ly by modifyi rical values mevap2 P time3060 s Qtimemevaphfg kJ hf h hfghgh Some Wrong S W1Qtimemeva W2Qtime60mevaphg U W3Qtimemevaphf Using hf preparation If you are a student using this Manual you are using it without permission 379 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course nsfer to min c 418 kJmin d 535 kJmin e 2257 kJmin nswer b 451 kJmin es on a blank EES reen Similar problems and their solutions can be obtained easily by modifying numerical values kg Pa in fENTHALPYSteamIAPWS x0PP 1Qtimemevaphg Using hg Using seconds instead of minutes for time 3Qtimemevaphf Using hf 3m3 ri ains ste nd 500 c 26 kg d 35 kg e 52 kg nswer d 35 kg olution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES ems and their solutions can be obtained easily by modifying numerical values SteamIAPWSTT1PP1 ideal gas and using degC 3147 Water is boiled in a pan on a stove at sea level During 10 min of boiling its is observed that 200 g of water has evaporated Then the rate of heat tra the water is a 084 kJmin b 451 kJ A Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lin sc mevap02 P101325 k time10 m Qtimemevaphfg kJ h hgENTHALPYSteamIAPWS x1PP hfghghf Some Wrong Solutions with Common Mistakes W W2Qtime60mevaphg W 3148 A rigid gid vessel cont am at 4 MPa a C The mass of the steam is a 3 kg b 9 kg A S screen Similar probl V3 m3 mVv1 m3kg P14000 kPa T1500 C v1VOLUME Some Wrong Solutions with Common Mistakes R04615 kJkgK P1VW1mRT1273 Treating steam as ideal gas P1VW2mRT1 Treating steam as preparation If you are a student using this Manual you are using it without permission 3149 Consider a sealed can that is filled with refrigerant134a The contents of the can are at the room temperature of 25C Now a leak developes and the pressure in the can drops to the local atmospheric pressure of 90 kPa The temperature of the refrigerant in the can is expected to drop to rounded to the nearest integer a 0C b 29C c 16C d 5C e 25C Answer b 29C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values T125 C P290 kPa T2TEMPERATURER134ax0PP2 Some Wrong Solutions with Common Mistakes W1T2T1 Assuming temperature remains constant 41 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 42 Moving Boundary Work 41C Yes 42C The area under the process curve and thus the boundary work done is greater in the constant pressure case PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course J 43 1 kPa m 1 kN m m 1 kN m 1 k 3 2 3 44 Helium is compressed in a pistoncylinder device The initial and final temperatures of helium and the work required to compress it are to be determined Assumptions The process is quasiequilibrium Properties The gas constant of helium is R 20769 kJkgK Table A1 Analysis The initial specific volume is 7 m kg 1kg 7 m 3 3 1 1 m V v 2 1 7 3 V m3 P kPa Using the ideal gas equation 200 5051 K 2 0769 kJkg K 150 kPa7 m kg 3 1 1 1 R P T v Since the pressure stays constant 2165 K 505 1 K m 7 m 3 3 3 1 1 2 2 T T V V and the work integral expression gives 600 kJ kPa m 1 1 kJ 7 m 150 kPa3 3 3 1 2 2 1 out V V V P Pd Wb That is W b in 600 kJ preparation If you are a student using this Manual you are using it without permission 43 45E The boundary work done during the process shown in the figure is to be determined Assumptions The process is quasiequilibrium Analysis The work done is equal to the the sumof the areas under the process lines 12 and 23 P psia 1 3 2 514 Btu 3 3 3 3 2 3 2 1 2 2 1 out psia ft 5404 1Btu 33 ft 00 psia2 3 psia ft 5404 1Btu 331 ft 2 15psia 300 2 Area V V V V P P P Wb 300 15 1 2 33 V ft3 The negative sign shows that the work is done on the system 46 The work done during the isothermal process shown in the figure is to be determined Assumptions The process is quasiequilibrium Analysis From the ideal gas equation v P RT For an isothermal process 06 m kg 200 kPa 02 m kg 600 kPa 3 3 1 2 2 1 P P v v Substituting ideal gas equation and this result into the boundary work integral produces 3955 kJ 3 3 3 3 1 2 1 1 2 1 2 1 out kPa m 1 kJ 1 m 06 3 kg200 kPa06 m ln 02 m ln v v v v v v mP d mRT Pd Wb The negative sign shows that the work is done on the system PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 44 47 A pistoncylinder device contains nitrogen gas at a specified state The boundary work is to be determined for the polytropic expansion of nitrogen Properties The gas constant for nitrogen is 02968 kJkgK Table A2 N2 130 kPa 120C Analysis The mass and volume of nitrogen at the initial state are 0 07802 kg 273 K 0 2968 kJkgK120 130 kPa007 m 3 1 1 1 RT P m V 3 3 2 2 2 0 08637 m 100 kPa 273 K 0 07802 kg02968 kPam kgK100 P mRT V The polytropic index is determined from 1 249 100 kPa008637 m 130 kPa007 m 3 3 2 2 1 1 n P P n n n n V V The boundary work is determined from 186 kJ 1 249 1 130 kPa007 m 100 kPa008637 m 1 3 3 1 1 2 2 n P P Wb V V 48 A pistoncylinder device with a set of stops contains steam at a specified state Now the steam is cooled The compression work for two cases and the final temperature are to be determined Analysis a The specific volumes for the initial and final states are Table A6 Steam 03 kg 1 MPa 400C Q 0 23275 m kg C 250 1MPa 0 30661 m kg C 400 1MPa 3 2 2 2 3 1 1 1 v v T P T P Noting that pressure is constant during the process the boundary work is determined from 2216 kJ 0 23275m kg kg1000 kPa 0 30661 30 3 2 1 v mP v Wb b The volume of the cylinder at the final state is 60 of initial volume Then the boundary work becomes 3679 kJ 0 30661m kg 0 60 kg1000 kPa 0 30661 30 0 60 3 1 1 v mP v Wb The temperature at the final state is 1518C Table A5 2 3 2 2 0 30661 m kg 0 60 MPa 50 T P v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 45 49 A pistoncylinder device contains nitrogen gas at a specified state The final temperature and the boundary work are to be determined for the isentropic expansion of nitrogen N2 130 kPa 180C Properties The properties of nitrogen are R 02968 kJkgK k 1395 Tables A2a A2b Analysis The mass and the final volume of nitrogen are 0 06768 kg 273 K 0 2968 kJkgK180 130 kPa007 m 3 1 1 1 RT P m V 3 2 1 395 2 1 395 3 2 2 1 1 0 09914 m 80 kPa 130 kPa007 m V V V V k k P P The final temperature and the boundary work are determined as 395 K 0 06768 kg02968 kPam kgK kPa009914 m 80 3 3 2 2 2 mR P T V 296 kJ 1 395 1 130 kPa007 m 80 kPa009914 m 1 3 3 1 1 2 2 k P P Wb V V 410 Saturated water vapor in a cylinder is heated at constant pressure until its temperature rises to a specified value The boundary work done during this process is to be determined Assumptions The process is quasiequilibrium Properties Noting that the pressure remains constant during this process the specific volumes at the initial and the final states are Table A4 through A6 1 2 P kPa 071643 m kg C 200 00 kPa 3 060582 m kg vapor Sat kPa 300 3 2 2 2 3 300 kPa 1 1 v v v T P P g 300 Analysis The boundary work is determined from its definition to be V 1659 kJ 3 3 1 2 1 2 2 1 out kPa m 1 1 kJ 060582 m kg kg300 kPa071643 5 v v V V V mP P Pd Wb Discussion The positive sign indicates that work is done by the system work output PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 46 411 Refrigerant134a in a cylinder is heated at constant pressure until its temperature rises to a specified value The boundary work done during this process is to be determined Assumptions The process is quasiequilibrium Properties Noting that the pressure remains constant during this process the specific volumes at the initial and the final states are Table A11 through A13 0052427 m kg 0 C 7 00 kPa 5 00008059 m kg liquid Sat 00 kPa 5 3 2 2 2 3 500 kPa 1 1 v v v T P P f P kPa 1 2 v 500 Analysis The boundary work is determined from its definition to be 6204 kg m kg 00008059 m 005 3 3 1 1 v V m and 1600 kJ 3 3 1 2 1 2 2 1 out kPa m 1 1 kJ 00008059m kg kg500 kPa0052427 6204 v v V V V mP P Pd Wb Discussion The positive sign indicates that work is done by the system work output PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 47 412 Problem 411 is reconsidered The effect of pressure on the work done as the pressure varies from 400 kPa to 1200 kPa is to be investigated The work done is to be plotted versus the pressure Analysis The problem is solved using EES and the solution is given below Knowns Vol1L200 L x10 saturated liquid state P900 kPa T270 C Solution Vol1Vol1LconvertLm3 The work is the boundary work done by the R134a during the constant pressure process WboundaryPVol2Vol1 The mass is Vol1mv1 v1volumeR134aPPxx1 Vol2mv2 v2volumeR134aPPTT2 Plot information v1v1 v2v2 P1P P2P T1temperatureR134aPPxx1 T2T2 P kPa Wboundary kJ 200 400 500 600 700 800 900 1000 1100 1200 1801 1661 1601 1546 1493 1442 1393 1344 1297 1250 200 300 400 500 600 700 800 900 1000 1100 1200 1200 1300 1400 1500 1600 1700 1800 P kPa Wboundary kJ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 48 104 103 102 101 101 102 103 104 v m3kg P kPa R134a 104 103 102 101 100 50 0 50 100 150 200 250 v m3kg T C 500 kPa R134a PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 49 413 Water is expanded isothermally in a closed system The work produced is to be determined Assumptions The process is quasiequilibrium Analysis From water table 1555 1 2 8816 1 V m3 P kPa 10200 m kg 0 0 001157 0 80 0 12721 001157 0 001157 m kg 0 9 kPa 1554 3 2 3 f 200 C 1 sat 200 C 2 1 fg f x P P P v v v v v The definition of specific volume gives 3 3 3 3 1 2 1 2 8816 m m kg 0001157 1 m 010200 m kg v V v V The work done during the process is determined from 10 kJ 1355 5 3 3 1 2 2 1 out kPa m 1 1kJ 15549 kPa8816 1m V V V P Pd Wb 414 Air in a cylinder is compressed at constant temperature until its pressure rises to a specified value The boundary work done during this process is to be determined Assumptions 1 The process is quasiequilibrium 2 Air is an ideal gas Properties The gas constant of air is R 0287 kJkgK Table A1 P 2 1 T 12C Analysis The boundary work is determined from its definition to be 272 kJ 600 kPa kg0287 kJkg K285 Kln 150 kPa 24 ln ln 2 1 2 1 1 2 1 1 ou P P mRT P Pd W t b V V V V V Discussion The negative sign indicates that work is done on the system work input 415 Several sets of pressure and volume data are taken as a gas expands The boundary work done during this process is to be determined using the experimental data Assumptions The process is quasiequilibrium Analysis Plotting the given data on a PV diagram on a graph paper and evaluating the area under the process curve the work done is determined to be 025 kJ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 410 416 A gas in a cylinder expands polytropically to a specified volume The boundary work done during this process is to be determined Assumptions The process is quasiequilibrium Analysis The boundary work for this polytropic process can be determined directly from 2033 kPa m 02 kPa 003 m 350 15 3 3 2 1 1 2 n P P V V P kPa 2 1 PV V m3 15 and 129 kJ 3 3 2 1 1 1 2 2 out kPa m 1 1 kJ 15 1 003 kPa m 350 02 2033 1 n P P Pd Wb V V V 02 00 Discussion The positive sign indicates that work is done by the system work output PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 411 417 Problem 416 is reconsidered The process described in the problem is to be plotted on a PV diagram and the effect of the polytropic exponent n on the boundary work as the polytropic exponent varies from 11 to 16 is to be plotted Analysis The problem is solved using EES and the solution is given below Function BoundWorkP1V1P2V2n This function returns the Boundary Work for the polytropic process This function is required since the expression for boundary work depens on whether n1 or n1 If n1 then BoundWorkP2V2P1V11nUse Equation 322 when n1 else BoundWork P1V1lnV2V1 Use Equation 320 when n1 endif end Inputs from the diagram window n15 P1 350 kPa V1 003 m3 V2 02 m3 GasAIR System The gas enclosed in the pistoncylinder device Process Polytropic expansion or compression PVn C P2V2nP1V1n n 13 Polytropic exponent Input Data Wb BoundWorkP1V1P2V2nkJ If we modify this problem and specify the mass then we can calculate the final temperature of the fluid for compression or expansion m1 m2 Conservation of mass for the closed system Lets solve the problem for m1 005 kg m1 005 kg Find the temperatures from the pressure and specific volume T1temperaturegasPP1vV1m1 T2temperaturegasPP2vV2m2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 412 002 004 006 008 01 012 014 016 018 02 0 20 40 60 80 100 120 140 160 V m3 P kPa 11 12 13 14 15 16 11 12 13 14 15 16 17 18 19 n Wb kJ n Wb kJ 11 1156 1211 1267 1322 1378 1433 1489 1544 16 1814 1725 1641 1563 149 1422 1358 1298 1242 1189 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 413 418 Nitrogen gas in a cylinder is compressed polytropically until the temperature rises to a specified value The boundary work done during this process is to be determined Assumptions 1 The process is quasiequilibrium 2 Nitrogen is an ideal gas Properties The gas constant for nitrogen is R 02968 kJkgK Table A2a Analysis The boundary work for this polytropic process can be determined from P 2 PV n C 1 890 kJ 1 14 300K 2 kg02968 kJkg K360 1 1 1 2 2 1 1 1 2 2 out n T mR T n P P Pd Wb V V V V Discussion The negative sign indicates that work is done on the system work input 419 A gas whose equation of state is R T P u 10 v 2 v expands in a cylinder isothermally to a specified volume The unit of the quantity 10 and the boundary work done during this process are to be determined Assumptions The process is quasiequilibrium P Analysis a The term 2 10 v must have pressure units since it is added to P T 350 K Thus the quantity 10 must have the unit kPam6kmol2 b The boundary work for this process can be determined from 2 2 2 2 10 10 10 V V V V v v N T NR N N R T R T P u u u V 4 2 and 403 kJ 3 3 3 2 2 6 3 3 1 2 2 1 2 2 1 2 2 2 1 out kPa m 1 kJ 1 m 2 1 m 4 1 10 kPa m kmol 05kmol m 2 kmol831 4 kJkmol K350 Kln 4 m 02 1 1 10 ln 10 V V V V V V V V N NR T d N NR T Pd W u u b Discussion The positive sign indicates that work is done by the system work output PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 414 420 Problem 419 is reconsidered Using the integration feature the work done is to be calculated and compared and the process is to be plotted on a PV diagram Analysis The problem is solved using EES and the solution is given below Input Data N02 kmol v1bar2N m3kmol v2bar4N m3kmol T350 K Ru8314 kJkmolK The quation of state is vbarP10vbar2RuT P is in kPa using the EES integral function the boundary work WbEES is WbEESNintegralPvbar v1bar v2bar001 We can show that Wbhand integeral of Pdvbar is one should solve for PFvbar and do the integral by hand for practice Wbhand NRuTlnv2barv1bar 101v2bar1v1bar To plot P vs vbar define Pplot fvbarplot T as vbarplotPplot10vbarplot2RuT PPplot and vbarvbarplot just to generate the parametric table for plotting purposes To plot P vs vbar for a new temperature or vbarplot range remove the and from the above equation and reset the vbarplot values in the Parametric Table Then press F3 or select Solve Table from the Calculate menu Next select New Plot Window under the Plot menu to plot the new data Pplot vplot 2909 2618 238 2182 2014 187 1746 1637 154 1455 10 1111 1222 1333 1444 1556 1667 1778 1889 20 9 11 13 15 17 19 21 0 40 80 120 160 200 240 280 320 vplot m3kmol Pplot kPa Area Wboundary T 350 K 1 2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 415 421 CO2 gas in a cylinder is compressed until the volume drops to a specified value The pressure changes during the process with volume as V 2 The boundary work done during this process is to be determined a P Assumptions The process is quasiequilibrium P 2 1 P aV2 V m3 Analysis The boundary work done during this process is determined from 533 kJ 3 3 3 6 1 2 2 1 2 2 1 out kPa m 1 kJ 1 m 03 1 m 01 1 kPa m 8 1 1 V V V V V a d a Pd Wb 03 01 Discussion The negative sign indicates that work is done on the system work input 422E A gas in a cylinder is heated and is allowed to expand to a specified pressure in a process during which the pressure changes linearly with volume The boundary work done during this process is to be determined Assumptions The process is quasiequilibrium Analysis a The pressure of the gas changes linearly with volume and thus the process curve on a PV diagram will be a straight line The boundary work during this process is simply the area under the process curve which is a trapezoidal Thus At state 1 20 psia 5 psiaft 7 ft 15 psia 3 3 1 1 b b b a P V P psia 1 2 P aV b V ft3 100 At state 2 15 3 2 2 3 2 2 ft 24 20 psia 5 psiaft psia 100 V V V b a P 7 and 181 Btu 3 3 1 2 2 1 out psia ft 54039 1 Btu 7ft 24 2 100 15psia 2 Area V V P P Wb Discussion The positive sign indicates that work is done by the system work output PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 416 423 A pistoncylinder device contains nitrogen gas at a specified state The boundary work is to be determined for the isothermal expansion of nitrogen Properties The properties of nitrogen are R 02968 kJkgK k 14 Table A2a Analysis We first determine initial and final volumes from ideal gas relation and find the boundary work using the relation for isothermal expansion of an ideal gas N2 130 kPa 180C 3 1 1 0 2586 m 130 kPa 273 K 0 25 kg 0 2968 kJkgK180 P mRT V 3 2 2 0 4202 m 80 kPa 273 K 0 25 kg 0 2968 kJkgK180 P mRT V 163 kJ 3 3 3 1 2 1 1 2586 m 0 0 4202 m 130 kPa0258 6 m ln ln V V PV Wb 424 A pistoncylinder device contains air gas at a specified state The air undergoes a cycle with three processes The boundary work for each process and the net work of the cycle are to be determined Properties The properties of air are R 0287 kJkgK k 14 Table A2a Air 2 MPa 350C Analysis For the isothermal expansion process 3 1 1 0 01341 m 2000 kPa 273 K 0 15 kg 0 287 kJkgK350 P mRT V 3 2 2 0 05364 m 500 kPa 273 K 0 15 kg 0 287 kJkgK350 P mRT V 3718 kJ 3 3 3 1 2 1 1 1 2 01341 m 0 0 05364 m 2000 kPa001341 m ln ln V V PV Wb For the polytropic compression process 3 3 21 3 21 3 3 3 2 2 0 01690 m 2000 kPa 500 kPa005364 m V V V V n n P P 3486 kJ 21 1 500 kPa005364 m 2000 kPa001690 m 1 3 3 2 2 3 3 3 2 n P P Wb V V For the constant pressure compression process 697 kJ 3 3 1 3 3 1 2000 kPa001341 001690m V P V Wb The net work for the cycle is the sum of the works for each process 465 kJ 6 97 3486 3718 1 3 2 3 2 1 net b b b W W W W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 417 425 A saturated water mixture contained in a springloaded pistoncylinder device is heated until the pressure and temperature rises to specified values The work done during this process is to be determined Assumptions The process is quasiequilibrium Analysis The initial state is saturated mixture at 90C The pressure and the specific volume at this state are Table A4 P 2 1 v 23686 m kg 0 0 001036 0 10 2 3593 001036 0 183 kPa 70 3 1 1 fg f x P v v v 800 kPa The final specific volume at 800 kPa and 250C is Table A6 0 29321 m kg 3 2 v Since this is a linear process the work done is equal to the area under the process line 12 2452 kJ 3 3 1 2 2 1 out kPa m 1 1kJ 0 23686m 1 kg029321 2 800kPa 70183 2 Area v P m v P Wb 426 A saturated water mixture contained in a springloaded pistoncylinder device is cooled until it is saturated liquid at a specified temperature The work done during this process is to be determined Assumptions The process is quasiequilibrium Analysis The initial state is saturated mixture at 1 MPa The specific volume at this state is Table A5 1 2 1 MPa v P 059097 m kg 0 0 001127 0 30 0 19436 001127 0 3 1 fg f xv v v The final state is saturated liquid at 100C Table A4 001043 m kg 0 42 kPa 101 3 2 2 f P v v Since this is a linear process the work done is equal to the area under the process line 12 480 kJ 3 3 1 2 2 1 out kPa m 1 1kJ 0 059097m 15 kg0001043 2 10142kPa 1000 2 Area v P m v P Wb The negative sign shows that the work is done on the system in the amount of 480 kJ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 418 427 An ideal gas undergoes two processes in a pistoncylinder device The process is to be sketched on a PV diagram an expression for the ratio of the compression to expansion work is to be obtained and this ratio is to be calculated for given values of n and r Assumptions The process is quasiequilibrium PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis a The processes on a PV diagram is as follows b The ratio of the compressiontoexpansion work is called the backwork ratio BWR Process 12 2 1 1 2 PdV Wb The process is PVn constant n P V constant and the integration results in P const PVn const 3 1 2 V P n T mR T n P P Wb 1 1 1 2 1 1 2 2 1 2 V V where the ideal gas law has been used However the compression work is 1 1 2 b12 comp n T mR T W W Process 23 3 2 2 3 PdV Wb The process is P constant and the integration gives 2 3 2 3 V V P Wb where P P2 P3 Using the ideal gas law the expansion work is 2 3 b23 exp T mR T W W The backwork ratio is defined as 1 1 1 1 1 1 1 1 1 1 1 2 3 2 1 2 3 2 1 2 2 2 3 1 2 2 3 1 2 exp comp T T T T n T T T T T T n T T T T n T mR T n T T mR W W BWR Since process 12 is polytropic the temperaturevolume relation for the ideal gas is n n n r r T T 1 1 1 1 2 2 1 1 V V where r is the compression ratio V1 V 2 Since process 23 is constant pressure the combined ideal gas law gives then and 2 1 2 3 2 3 2 3 2 2 2 3 3 3 r T T P P T P T P V V V V V V The backwork ratio becomes 1 1 1 1 1 r r n BWR n c For n 14 and r 6 the value of the BWR is 0256 1 6 6 1 1 41 1 41 1 BWR preparation If you are a student using this Manual you are using it without permission 419 Closed System Energy Analysis 428E The table is to be completed using conservation of energy principle for a closed system Analysis The energy balance for a closed system can be expressed as 1 2 1 2 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in e m e E E W Q E E E 43 42 1 4243 1 Application of this equation gives the following completed table Qin Btu Wout Btu E1 Btu E2 Btu m lbm e2 e1 Btulbm 350 510 1020 860 3 533 350 130 550 770 5 440 560 260 600 900 2 150 500 0 1400 900 7 714 650 50 1000 400 3 200 429E A pistoncylinder device involves expansion work and work input by a stirring device The net change of internal energy is to be determined Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved 3 The thermal energy stored in the cylinder itself is negligible Analysis This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as since KE PE 0 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in U W W E E E 43 42 1 4243 1 where Win 1028 Btu and 1928 Btu 77817 lbf ft 1Btu 15000 lbf ft 15000 lbf ft out W Substituting 900 Btu 1028 1928 out in W W U PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 420 430E The heat transfer during a process that a closed system undergoes without any internal energy change is to be determined Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero 2 The compression or expansion process is quasiequilibrium Analysis The energy balance for this stationary closed system can be expressed as out in out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in since KE PE 0 0 W Q U W Q E E E 43 42 1 4243 1 Then 1414 Btu 77817 lbf ft 1Btu 10 lbf ft 11 6 Qin 431 Motor oil is contained in a rigid container that is equipped with a stirring device The rate of specific energy increase is to be determined Analysis This is a closed system since no mass enters or leaves The energy balance for closed system can be expressed as E W Q E E E 43 42 1 4243 1 shin in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in Then 25 W 52 51 1 shin in W Q E Dividing this by the mass in the system gives 167 Jkg s 15 kg 25 Js m E e PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 421 432 An insulated rigid tank is initially filled with a saturated liquidvapor mixture of water An electric heater in the tank is turned on and the entire liquid in the tank is vaporized The length of time the heater was kept on is to be determined and the process is to be shown on a Pv diagram Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero 2 The device is well insulated and thus heat transfer is negligible 3 The energy stored in the resistance wires and the heat transferred to the tank itself is negligible Analysis We take the contents of the tank as the system This is a closed system since no mass enters or leaves Noting that the volume of the system is constant and thus there is no boundary work the energy balance for this stationary closed system can be expressed as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course V KE PE 0 since 1 2 1 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u m u t I Q u m u U W E E E e 43 42 1 4243 1 The properties of water are Tables A4 through A6 25697 kJkg vapor sat m kg 029065 98003 kJkg 20523 025 46697 029065 m kg 0001053 11594 025 0001053 2052 3 kJkg 46697 11594 m kg 0 001053 25 0 kPa 150 029065 m kg 2 3 1 2 1 1 3 1 1 3 1 1 3 g fg f fg f fg f g f u u x u u u x u u x P v v v v v v v We H2O V const T 1 2 v Substituting 602 min s 33613 1 kJs 98003kJkg 1000 VA 2 kg25697 110 V 8 A t t preparation If you are a student using this Manual you are using it without permission 422 433 Problem 432 is reconsidered The effect of the initial mass of water on the length of time required to completely vaporize the liquid as the initial mass varies from 1 kg to 10 kg is to be investigated The vaporization time is to be plotted against the initial mass Analysis The problem is solved using EES and the solution is given below PROCEDURE P2X2v1P2x2 FluidSteamIAPWS If v1 VCRITFluid then P2pressureFluidvv1x1 x21 else P2pressureFluidvv1x0 x20 EndIf End Knowns m2 kg P1150 kPa y075 moisture Volts110 V I8 amp Solution Conservation of Energy for the closed tank EdotinEdotoutDELTAEdot EdotinWdotele kW WdoteleVoltsICONVERTJskW kW Edotout0 kW DELTAEdotmu2u1DELTAts kW DELTAtminDELTAtsconvertsmin min The quality at state 1 is FluidSteamIAPWS x11y u1INTENERGYFluidPP1 xx1 kJkg v1volumeFluidPP1 xx1 m3kg T1temperatureFluidPP1 xx1 C Check to see if state 2 is on the saturated liquid line or saturated vapor line Call P2X2v1P2x2 u2INTENERGYFluidPP2 xx2 kJkg v2volumeFluidPP2 xx2 m3kg T2temperatureFluidPP2 xx2 C PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 423 103 102 101 100 101 102 0 100 200 300 400 500 600 700 v m3kg T C 150 kPa 655 kPa Steam IAPWS 1 2 tmin min m kg 3011 6021 9032 1204 1505 1806 2107 2409 271 3011 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 0 50 100 150 200 250 300 350 m kg tmin min PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 424 434 Saturated water vapor is isothermally condensed to a saturated liquid in a pistoncylinder device The heat transfer and the work done are to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or expansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as Water 200C sat vapor since KE PE 0 1 2 in out 1 2 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u m u W Q u m u U Q W E E E b b 43 42 1 4243 1 Heat The properties at the initial and final states are Table A4 85046 kJkg 0 001157 m kg 0 0 C 20 9 kPa 1554 2594 2 kJkg 0 12721 m kg 1 0 C 20 2 3 2 2 2 2 1 1 3 1 1 1 f f g g u u x T P P u u x T v v v v T 2 v 1 The work done during this process is 196 0 kJkg kPa m 1 1 kJ 012721 m kg 15549 kPa0001157 3 3 1 2 2 1 out v v V P Pd wb That is w b in 1960 kJkg Substituting the energy balance equation we get 1940 kJkg 1743 7 196 0 in 1 2 in out fg b b u w u u w q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 425 435 Water contained in a rigid vessel is heated The heat transfer is to be determined Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved 3 The thermal energy stored in the vessel itself is negligible Analysis We take water as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course since KE PE 0 1 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u m u U Q E E E 43 42 1 4243 1 The properties at the initial and final states are Table A4 1643 5 kJkg 0 52501927 4 66 631 0 5250 0 39248 0 001091 2066 0 001091 0 2066 m kg 0 0 C 15 67576 kJkg 0 1232087 0 41906 0 2066 m kg 0 001043 0 123 1 6720 0 001043 123 0 C 100 2 2 2 2 3 1 2 2 1 3 1 1 1 fg f fg f fg f fg f x u u u x T xu u u x x T v v v v v v v v v T 2 1 Water 10 L 100C x 0123 Q The mass in the system is 0 04841 kg 2066 m kg 0 m 0100 3 3 1 1 v V m Substituting 469 kJ 67576 kJkg 0 04841 kg16435 1 2 in u m u Q preparation If you are a student using this Manual you are using it without permission 426 436 Saturated vapor water is cooled at constant temperature and pressure to a saturated liquid The heat rejected is to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or expansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as Water 300 kPa sat vap 2 1 out 2 1 out 1 2 out 1 2 ou out 1 2 ou out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in since KE PE 0 h h q h m h Q h m h Q u m u W Q u m u U W Q E E E t b t b 43 42 1 4243 1 Q T 1 2 v since U Wb H during a constant pressure quasiequilibrium process Since water changes from saturated liquid to saturated vapor we have 21635 kJkg 300 kPa out fg f g h h h q Table A5 Note that the temperature also remains constant during the process and it is the saturation temperature at 300 kPa which is 1335C PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 427 437 Saturated vapor water is cooled at constant pressure to a saturated liquid The heat transferred and the work done are to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or expansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as 2 1 out 1 2 out 1 2 ou out 1 2 ou out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in since KE PE 0 h h q h h q u u w q u u u w q E E E t b t b 43 42 1 4243 1 Water 40 kPa sat vap Q since u wb h during a constant pressure quasiequilibrium process Since water changes from saturated liquid to saturated vapor we have P 1 2 v 23184 kJkg 40 kPa out fg f g h h h q Table A5 The specific volumes at the initial and final states are 001026 m kg 0 993 m kg 3 3 40 kPa 2 3 40 kPa 1 f g v v v v Then the work done is determined from 1597 kJkg 3 3 1 2 2 1 out kPa m 1 1kJ 3 9933m 40 kPa0001026 v v V P Pd wb PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 428 438 A cylinder is initially filled with saturated liquid water at a specified pressure The water is heated electrically as it is stirred by a paddlewheel at constant pressure The voltage of the current source is to be determined and the process is to be shown on a Pv diagram Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 The cylinder is well insulated and thus heat transfer is negligible 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or expansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as KE PE 0 since 1 2 in pw 1 2 pwin in e bout pwin in e potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in h m h W t I h m h W W Q U W W W E E E V 43 42 1 4243 1 H2O P const Wpw We since U Wb H during a constant pressure quasiequilibrium process The properties of water are Tables A4 through A6 4731 kg m kg 0001057 m 0005 15936 kJkg 2213 1 50 48701 50 kPa 175 m kg 0001057 kJkg 48701 liquid sat kPa 175 3 3 1 1 2 2 2 2 3 175 kPa 1 175 kPa 1 1 v V v v m x h h h x P h h P fg f f f P 2 1 Substituting v 2239 V 1 kJs 1000 VA 8 A45 60 s kJ 4835 kJ 4835 48701kJkg 4731 kg15936 kJ 400 V V V t I t I PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 429 439 A cylinder equipped with an external spring is initially filled with steam at a specified state Heat is transferred to the steam and both the temperature and pressure rise The final temperature the boundary work done by the steam and the amount of heat transfer are to be determined and the process is to be shown on a Pv diagram Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 The thermal energy stored in the cylinder itself is negligible 3 The compression or expansion process is quasiequilibrium 4 The spring is a linear spring Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves Noting that the spring is not part of the system it is external the energy balance for this stationary closed system can be expressed as out 1 2 in 1 2 out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in since KE PE 0 b b in W u m u Q u m u U W Q E E E 43 42 1 4243 1 H2O 200 kPa 200C Q The properties of steam are Tables A4 through A6 33120 kJkg m kg 16207 kPa 250 16207 m kg 03702 kg m 06 03702 kg m kg 108049 m 04 26546 kJkg 108049 m kg C 200 kPa 200 2 2 3 2 2 3 3 2 2 3 3 1 1 1 3 1 1 1 u T P m m u T P C 606 v V v v V v P 1 2 v b The pressure of the gas changes linearly with volume and thus the process curve on a PV diagram will be a straight line The boundary work during this process is simply the area under the process curve which is a trapezoidal Thus 45 kJ 3 3 1 2 2 1 kPa m 1 1 kJ 04m 06 2 250kPa 200 2 V P V P Area Wb c From the energy balance we have Qin 03702 kg33120 26546kJkg 45 kJ 288 kJ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 430 440 Problem 439 is reconsidered The effect of the initial temperature of steam on the final temperature the work done and the total heat transfer as the initial temperature varies from 150C to 250C is to be investigated The final results are to be plotted against the initial temperature Analysis The problem is solved using EES and the solution is given below The process is given by P2P1kxAA and as the spring moves x amount the volume changes by V2V1 P2P1SpringconstV2 V1 P2 is a linear function of V2 where Springconst kA the actual spring constant divided by the piston face area Conservation of mass for the closed system is m2m1 The conservation of energy for the closed system is Ein Eout DeltaE neglect DeltaKE and DeltaPE for the system Qin Wout m1u2u1 DELTAUm1u2u1 Input Data P1200 kPa V104 m3 T1200 C P2250 kPa V206 m3 FluidSteamIAPWS m1V1spvol1 spvol1volumeFluidTT1 PP1 u1intenergyFluid TT1 PP1 spvol2V2m2 The final temperature is T2temperatureFluidPP2vspvol2 u2intenergyFluid PP2 TT2 Wnetother 0 WoutWnetother Wb Wb integral of P2dV2 for 05V206 and is given by WbP1V2V1Springconst2V2V12 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 431 103 102 101 100 101 102 100 101 102 103 104 105 106 v m3kg P kPa 200C 606C Steam IAPWS 1 2 Area Wb 150 170 190 210 230 250 500 545 590 635 680 725 0 20 40 60 80 100 T1 C T2 C Wout kJ T1 C Qin kJ T2 C Wout kJ 150 160 170 180 190 200 210 220 230 240 250 2815 2828 2842 2855 2869 2883 2898 2912 2927 2942 2957 5082 5279 5475 567 5864 6058 625 6443 6634 6826 7017 45 45 45 45 45 45 45 45 45 45 45 150 170 190 210 230 250 280 282 284 286 288 290 292 294 296 T1 C Qin kJ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 432 441 A room is heated by an electrical radiator containing heating oil Heat is lost from the room The time period during which the heater is on is to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 141C and 377 MPa 2 The kinetic and potential energy changes are negligible 0 pe ke 3 Constant specific heats at room temperature can be used for air This assumption results in negligible error in heating and airconditioning applications 4 The local atmospheric pressure is 100 kPa 5 The room is airtight so that no air leaks in and out during the process Properties The gas constant of air is R 0287 kPam3kgK Table A1 Also cv 0718 kJkgK for air at room temperature Table A2 Oil properties are given to be ρ 950 kgm3 and cp 22 kJkgC Analysis We take the air in the room and the oil in the radiator to be the system This is a closed system since no mass crosses the system boundary The energy balance for this stationary constantvolume closed system can be expressed as 10C Q Radiator Room 0 PE since KE oil 1 2 air 1 2 oil air out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc T T mc U U t Q W E E E p v 43 42 1 4243 1 The mass of air and oil are 2850 kg kgm 0030 m 950 6232 kg 273 K kPa m kg K10 0287 kPa50 m 100 3 3 oil oil oil 3 3 1 air air V V ρ m RT P m Substituting 340 min 2038 s t t C50 10 C 2850 kg22 kJkg C20 10 C 6232 kg0718 kJkg 0 35 kJs 81 Discussion In practice the pressure in the room will remain constant during this process rather than the volume and some air will leak out as the air expands As a result the air in the room will undergo a constant pressure expansion process Therefore it is more proper to be conservative and to using H instead of use U in heating and airconditioning applications PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 433 442 A saturated water mixture contained in a springloaded pistoncylinder device is heated until the pressure and volume rise to specified values The heat transfer and the work done are to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or expansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as 225 kPa 1 2 v P since KE PE 0 1 2 ou in 1 2 ou in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u m u W Q u m u U W Q E E E t b t b 43 42 1 4243 1 75 kPa The initial state is saturated mixture at 75 kPa The specific volume and internal energy at this state are Table A5 55330 kJkg 0 082111 8 36 384 0 1783 m kg 0 001037 0 08 2 2172 001037 0 1 3 1 fg f fg f xu u u xv v v The mass of water is 1122 kg 1783 m kg 0 m 2 3 3 1 1 v V m The final specific volume is 0 4458 m kg 1122 kg 5 m 3 3 2 2 m V v The final state is now fixed The internal energy at this specific volume and 225 kPa pressure is Table A6 4 kJkg 2 1650 u Since this is a linear process the work done is equal to the area under the process line 12 450 kJ 3 3 1 2 2 1 out kPa m 1 1kJ 2m 5 2 225kPa 75 2 Area V V P P Wb Substituting into energy balance equation gives 12750 kJ 55330 kJkg 1122 kg16504 450 kJ 1 2 out in u m u W Q b PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 434 443 R134a contained in a springloaded pistoncylinder device is cooled until the temperature and volume drop to specified values The heat transfer and the work done are to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or expansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as since KE PE 0 1 2 in out 1 2 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u m u W Q u m u U Q W E E E b b 43 42 1 4243 1 2 1 600 kPa v P The initial state properties are Table A13 96 kJkg 357 0 055522 m kg C 15 kPa 600 1 3 1 1 1 u T P v The mass of refrigerant is 5 4033 kg 055522 m kg 0 m 03 3 3 1 1 v V m The final specific volume is 0 018507 m kg 5 4033 kg m 01 3 3 2 2 m V v The final state at this specific volume and at 30C is a saturated mixture The properties at this state are Table A11 43 kPa 84 2844 kJkg 0 07902420052 59 12 0 079024 0 0007203 0 22580 0 0007203 018507 0 2 2 2 2 2 P x u u u x fg f f g f v v v v Since this is a linear process the work done is equal to the area under the process line 12 6844 kJ 3 3 2 1 2 1 in kPa m 1 1kJ 10 m 03 2 8443kPa 600 2 Area V V P P Wb Substituting into energy balance equation gives 1849 kJ 35796 kJkg 5 4033 kg2844 6844 kJ 1 2 in out u m u W Q b PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 435 444E Saturated R134a vapor is condensed at constant pressure to a saturated liquid in a pistoncylinder device The heat transfer and the work done are to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or expansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as since KE PE 0 1 2 in out 1 2 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u m u W Q u m u U Q W E E E b b 43 42 1 4243 1 R134a 100F Q The properties at the initial and final states are Table A11E 768 Btulbm 44 0 01386 ft lbm 0 0 F 10 45 Btulbm 107 0 34045 ft lbm 1 0 F 10 2 3 2 2 2 1 3 1 1 1 f f g g u u x T u u x T v v v v T 1 2 v Also from Table A11E 080 Btulbm 71 683 Btulbm 62 13893 psia 2 1 fg fg h u P P The work done during this process is 8 396 Btulbm 5404 psia ft 1Btu ft lbm 386 034045 psia001 13893 3 3 1 2 2 1 out v v v P Pd wb That is w b in 8396 Btulbm Substituting into energy balance equation gives 71080 Btulbm 62683 8 396 in 1 2 in out fg b b u w u u w q Discussion The heat transfer may also be determined from 71080 Btulbm out 1 2 out hfg q h h q since U Wb H during a constant pressure quasiequilibrium process PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 436 445 Saturated R134a liquid is contained in an insulated pistoncylinder device Electrical work is supplied to R134a The time required for the refrigerant to turn into saturated vapor and the final temperature are to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or expansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as fg e fg e b e b e mh t W mh h m h H H W u m u W W u m u U W W E E E in 1 2 1 2 in 1 2 out in 1 2 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in since KE PE 0 43 42 1 4243 1 T 2 1 v since U Wb H during a constant pressure quasiequilibrium process The electrical power and the enthalpy of vaporization of R134a are 34 kJkg Table A 11 202 20 W 10 V2 A 5 C in fg e h I W V Substituting 225 h 8093s kJkg 2 kg20234 0 020 kJs t t The temperature remains constant during this phase change process 5C 1 2 T T PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 437 446 Two tanks initially separated by a partition contain steam at different states Now the partition is removed and they are allowed to mix until equilibrium is established The temperature and quality of the steam at the final state and the amount of heat lost from the tanks are to be determined PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions Analysis a We take the contents of both tanks as the system This is a closed system since no mass enters or leaves Noting that the volume of the system is constant and thus there is no boundary work the energy balance for this stationary closed system can be expressed as KE PE 0 since 1 2 1 2 out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in W u m u u m u U U Q E E E B A B A 43 42 1 4243 1 TANK A 2 kg 1 MPa 300C Q TANK B 3 kg 150C x05 The properties of steam in both tanks at the initial state are Tables A4 through A6 2793 7 kJkg 0 25799 m kg C 300 kPa 1000 1 3 1 1 1 A A A A u T P v 1595 4 kJkg 050 1927 4 66 631 019679 m kg 0001091 039248 050 0001091 1927 4 kJkg 63166 039248 m kg 001091 0 50 0 C 150 1 1 3 1 1 3 1 1 fg f B fg f B fg f g f B x u u u x u u x T v v v v v The total volume and total mass of the system are 5 kg 2 3 1 106 m 3 kg 0 19679 m kg 2 kg 0 25799 m kg 3 3 3 1 1 B A B B A A B A m m m m m v v V V V Now the specific volume at the final state may be determined 0 22127 m kg 5 kg 1 106 m 3 3 2 m V v which fixes the final state and we can determine other properties 1282 8 kJkg 03641 1982 1 56111 0 001073 0 60582 0 001073 22127 0 22127 m kg 0 00 kPa 3 2 2 2 2 sat 300 kPa 2 3 2 2 fg f f g f x u u u x T T P 03641 C 1335 v v v v v b Substituting 3959 kJ 1595 4 kJkg 3 kg1282 8 2793 7 kJkg 2 kg1282 8 1 2 1 2 out B A B A u m u u m u U U Q or Qout 3959 kJ preparation If you are a student using this Manual you are using it without permission 438 Specific Heats u and h of Ideal Gases 447C It can be either The difference in temperature in both the K and C scales is the same 448C It can be used for any kind of process of an ideal gas 449C It can be used for any kind of process of an ideal gas 450C Very close but no Because the heat transfer during this process is Q mcpT and cp varies with temperature 451C The energy required is mcpT which will be the same in both cases This is because the cp of an ideal gas does not vary with pressure 452C The energy required is mcpT which will be the same in both cases This is because the cp of an ideal gas does not vary with volume 453C Modeling both gases as ideal gases with constant specific heats we have T c h T c u p v Since both gases undergo the same temperature change the gas with the greatest cv will experience the largest change in internal energy Similarly the gas with the largest cp will have the greatest enthalpy change Inspection of Table A2a indicates that air will experience the greatest change in both cases 454 The desired result is obtained by multiplying the first relation by the molar mass M MR Mc Mc p v or u p R c c v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 439 455 The enthalpy change of oxygen is to be determined for two cases of specified temperature changes Assumptions At specified conditions oxygen behaves as an ideal gas Properties The constantpressure specific heat of oxygen at room temperature is cp 0918 kJkgK Table A2a Analysis Using the specific heat at constant pressure 459 kJkg 150K 0918 kJkg K200 T c h p If we use the same room temperature specific heat value the enthalpy change will be the same for the second case However if we consider the variation of specific heat with temperature and use the specific heat values from Table A2b we have cp 0956 kJkgK at 175C 450 K and cp 0918 kJkgK at 25C 300 K Then 478 kJkg 150K 0956 kJkg K200 1 1 T c h p 459 kJkg 0K 0918 kJkg K50 1 2 T c h p The two results differ from each other by about 4 The pressure has no influence on the enthalpy of an ideal gas 456E Air is compressed isothermally in a compressor The change in the specific volume of air is to be determined Assumptions At specified conditions air behaves as an ideal gas Properties The gas constant of air is R 03704 psiaft3lbmR Table A1E Analysis At the compressor inlet the specific volume is P 1 2 v 9816 ft lbm 20 psia 460 R 03704 psia ft lbm R70 3 3 1 1 P RT v Similarly at the compressor exit 1 309 ft lbm 150 psia 460 R 03704 psia ft lbm R70 3 3 2 2 P RT v The change in the specific volume caused by the compressor is ft lbm 851 3 9 816 1 309 1 2 v v v 457 The total internal energy changes for neon and argon are to be determined for a given temperature change Assumptions At specified conditions neon and argon behave as an ideal gas Properties The constantvolume specific heats of neon and argon are 06179 kJkgK and 03122 kJkgK respectively Table A2a Analysis The total internal energy changes are 1977 kJ 20K 2 kg06179 kJkg K180 neon T mc U v 999 kJ 20K 2 kg03122 kJkg K180 argon T mc U v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 440 458 The enthalpy changes for neon and argon are to be determined for a given temperature change Assumptions At specified conditions neon and argon behave as an ideal gas Properties The constantpressure specific heats of argon and neon are 05203 kJkgK and 10299 kJkgK respectively Table A2a Analysis The enthalpy changes are 390 kJkg 05203 kJkg K25 100K argon T c h p 772 kJkg 10299 kJkg K25 100K neon T c h p 459E The enthalpy change of oxygen gas during a heating process is to be determined using an empirical specific heat relation constant specific heat at average temperature and constant specific heat at room temperature Analysis a Using the empirical relation for c p T from Table A2Ec 3 2 dT cT bT a cp where a 6085 b 02017102 c 005275105 and d 005372109 Then 170 Btulbm 31999 lbmlbmol Btulbmol 54423 Btulbmol 54423 800 1500 0 05372 10 800 1500 0 05275 10 800 1500 0 2017 10 800 0851500 6 4 4 9 4 1 3 3 5 3 1 2 2 2 2 1 4 1 4 2 4 1 3 1 3 2 3 1 2 1 2 2 2 1 1 2 2 1 3 2 2 1 M h h T d T T c T T b T T T a dT dT cT bT a T dT c h p b Using the constant cp value from Table A2Eb at the average temperature of 1150 R 169 Btulbm 800 R 0242 Btulbm R1500 Btulbm R 0242 1 2 avg 1150 R avg T T c h c c p p p c Using the constant cp value from Table A2Ea at room temperature 153 Btulbm 800R 0219 Btulbm R1500 Btulbm R 0219 1 2 avg 537 R avg T T c h c c p p p PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 441 460 The internal energy change of hydrogen gas during a heating process is to be determined using an empirical specific heat relation constant specific heat at average temperature and constant specific heat at room temperature Analysis a Using the empirical relation for c p T from Table A2c and relating it to cv T 3 2 dT cT bT R a R c T c u u p v where a 2911 b 01916102 c 04003105 and d 08704109 Then 6194 kJkg 2016 kgkmol 12487 kJkmol kJkmol 12487 200 800 0 8704 10 200 800 0 4003 10 200 800 0 1961 10 200 8 314800 2911 4 4 9 4 1 3 3 5 3 1 2 2 2 2 1 4 1 4 2 4 1 3 1 3 2 3 1 2 1 2 2 2 1 1 2 2 1 3 2 2 1 M u u T d T T c T T b T T T R a dT dT cT bT R a c T dT u u u v b Using a constant cp value from Table A2b at the average temperature of 500 K 6233 kJkg 200K 10389 kJkg K800 kJkg K 10389 1 2 avg 500 K avg T T c u c c v v v c Using a constant cp value from Table A2a at room temperature 6110 kJkg 200K 10183 kJkg K800 kJkg K 10183 1 2 avg 300 K avg T T c u c c v v v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 442 461E A springloaded pistoncylinder device is filled with air The air is now cooled until its volume decreases by 50 The changes in the internal energy and enthalpy of the air are to be determined Assumptions At specified conditions air behaves as an ideal gas Properties The gas constant of air is R 03704 psiaft3lbmR Table A1E The specific heats of air at room temperature are cv 0171 BtulbmR and cp 0240 BtulbmR Table A2Ea Analysis The mass of the air in this system is P 1 2 0 7336 lbm 460 R psia ft lbm R460 03704 psia1 ft 250 3 3 1 1 1 RT P m V The final specific volume is then 0 6816 ft lbm 07336 lbm ft 051 3 3 2 2 m V v v As the volume of the air decreased the length of the spring will increase by 1100 in 0 9167 ft 10 12 ft ft 50 4 4 2 3 2 π πD A x p V V The final pressure is then 3 psia 249 249 3 lbfin 10 in 4 5 lbfin1100 in psia 250 4 2 2 2 1 1 1 1 2 π πD k x P A k x P A F P P P P p p Employing the ideal gas equation of state the final temperature will be 458 7 R lbm0370 4 psia ft lbm R 07336 psia05 ft 2493 3 3 2 2 2 mR P T V Using the specific heats Btulbm 1107 Btulbm 789 920R Btulbm R4587 0240 920R Btulbm R4587 0171 T c h T c u p v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 443 Closed System Energy Analysis Ideal Gases 462C No it isnt This is because the first law relation Q W U reduces to W 0 in this case since the system is adiabatic Q 0 and U 0 for the isothermal processes of ideal gases Therefore this adiabatic system cannot receive any net work at constant temperature 463 Oxygen is heated to experience a specified temperature change The heat transfer is to be determined for two cases Assumptions 1 Oxygen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 1548 K and 508 MPa 2 The kinetic and potential energy changes are negligible 0 pe ke 3 Constant specific heats can be used for oxygen Properties The specific heats of oxygen at the average temperature of 20120270C343 K are cp 0927 kJkgK and cv 0667 kJkgK Table A2b Analysis We take the oxygen as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for a constantvolume process can be expressed as 1 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc U Q E E E v 43 42 1 4243 1 O2 T1 20C T2 120C Q The energy balance during a constantpressure process such as in a piston cylinder device can be expressed as 1 2 in out in out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc H Q U W Q U W Q E E E p b b 43 42 1 4243 1 O2 T1 20C T2 120C Q since U Wb H during a constant pressure quasiequilibrium process Substituting for both cases 667 kJ 20K 1 kg 0 667 kJkg K120 1 2 const in T T mc Q v V 927 kJ 20K 1 kg 0 927 kJkg K120 1 2 const in T T mc Q p P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 444 464E The air in a rigid tank is heated until its pressure doubles The volume of the tank and the amount of heat transfer are to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 221F and 547 psia 2 The kinetic and potential energy changes are negligible pe ke 0 3 Constant specific heats at room temperature can be used for air This assumption results in negligible error in heating and airconditioning applications Properties The gas constant of air is R 03704 psiaft3lbmR Table A1E Air 20 lbm 50 psia 80F Analysis a The volume of the tank can be determined from the ideal gas relation 800 ft3 50 psia 20 lbm0370 4 psia ft lbm R540 R 3 1 1 P mRT V Q b We take the air in the tank as our system The energy balance for this stationary closed system can be expressed as 1 2 1 2 in in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc u m u Q U Q E E E v 43 42 1 4243 1 The final temperature of air is 1080 R 540 R 2 1 1 2 2 2 2 1 1 P T P T T P T P V V The internal energies are Table A17E u u u u 1 2 540 R 1080 R 9204 Btu lbm 18693 Btu lbm Substituting Qin 20 lbm18693 9204Btulbm 1898 Btu Alternative solutions The specific heat of air at the average temperature of Tavg 54010802 810 R 350F is from Table A2Eb cvavg 0175 BtulbmR Substituting Qin 20 lbm 0175 BtulbmR1080 540 R 1890 Btu Discussion Both approaches resulted in almost the same solution in this case PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 445 465E Heat is transferred to air contained in a rifid container The internal energy change is to be determined Assumptions 1 Air is an ideal gas since it is probably at a high temperature and low pressure relative to its critical point values of 2385 R and 547 psia 2 The kinetic and potential energy changes are negligible 0 pe ke 3 Constant specific heats at room temperature can be used for air Analysis We take air as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as Air since KE PE 0 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u q E E E 43 42 1 4243 1 Q Substituting 50 Btulbm qin u 466E Paddle Wheel work is applied to nitrogen in a rigid container The final temperature is to be determined Assumptions 1 Nitrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 1262 K 2271 R and 339 MPa 492 psia 2 The kinetic and potential energy changes are negligible 0 pe ke 3 Constant specific heats at room temperature can be used for nitrogen Properties For nitrogen cv 0177 BtulbmR at room temperature and R Table A1E and A 2Ea 03830 psia ft lbm R 3 Analysis We take the nitrogen as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as since KE PE 0 1 2 in pw potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc U W E E E v 43 42 1 4243 1 Nitrogen 20 psia 100F Wpw The mass of nitrogen is 0 09325 lbm psia ft lbm R560 R 03830 psia1 ft 20 3 3 1 1 RT P m V Substituting 489 F 949 R 2 2 1 2 in pw 560R 0 09325 lbm 0 177 Btulbm R 77817 lbf ft 1Btu 5000 lbf ft T T T T mc W v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 446 467 Nitrogen in a pistoncylinder device undergoes an isobaric process The final pressure and the heat transfer are to be determined Assumptions 1 Nitrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 1262 K 2271 R and 339 MPa 492 psia 2 The kinetic and potential energy changes are negligible 0 pe ke 3 Constant specific heats at room temperature can be used for nitrogen Properties For nitrogen cp 1039 kJkgK at room temperature Table A2a Analysis Since this is an isobaric pressure the pressure remains constant during the process and thus 1MPa 1 2 P P We take the nitrogen as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as Nitrogen 1 MPa 427C Q since KE PE 0 2 1 out 1 2 out 1 2 ou out 1 2 ou out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc q h h q u u w q u u u w q E E E p t b t b 43 42 1 4243 1 since u wb h during a constant pressure isobaric quasiequilibrium process Substituting 416 kJkg 27 F F427 1 039 kJkg 2 1 out T T c q p PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 447 468 A resistance heater is to raise the air temperature in the room from 5 to 25C within 11 min The required power rating of the resistance heater is to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 141C and 377 MPa 2 The kinetic and potential energy changes are negligible ke pe 0 3 Constant specific heats at room temperature can be used for air This assumption results in negligible error in heating and airconditioning applications 4 Heat losses from the room are negligible 5 The room is airtight so that no air leaks in and out during the process Properties The gas constant of air is R 0287 kPam3kgK Table A1 Also cv 0718 kJkgK for air at room temperature Table A2 Analysis We take the air in the room to be the system This is a closed system since no mass crosses the system boundary The energy balance for this stationary constantvolume closed system can be expressed as 0 PE KE since 1 2 avg in e potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in Q T T mc U W E E E v 43 42 1 4243 1 456 m3 5C AIR We or 1 2 avg in T T mc t We v The mass of air is 150 6 kg kPa m kg K278 K 0287 kPa120 m 100 120 m 5 6 4 3 3 1 1 3 RT P m V V Substituting the power rating of the heater becomes 328 kW 11 60 s 5 C C25 1506 kg0718 kJkg in o e W Discussion In practice the pressure in the room will remain constant during this process rather than the volume and some air will leak out as the air expands As a result the air in the room will undergo a constant pressure expansion process Therefore it is more proper to be conservative and to use H instead of using U in heating and airconditioning applications PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 448 469 A student living in a room turns her 150W fan on in the morning The temperature in the room when she comes back 10 h later is to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 141C and 377 MPa 2 The kinetic and potential energy changes are negligible ke pe 0 3 Constant specific heats at room temperature can be used for air This assumption results in negligible error in heating and airconditioning applications 4 All the doors and windows are tightly closed and heat transfer through the walls and the windows is disregarded Properties The gas constant of air is R 0287 kPam3kgK Table A1 Also cv 0718 kJkgK for air at room temperature Table A2 Analysis We take the room as the system This is a closed system since the doors and the windows are said to be tightly closed and thus no mass crosses the system boundary during the process The energy balance for this system can be expressed as 1 2 1 2 potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer T T mc u m u W U W E E E in e in e out in v 43 42 1 4243 1 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course The mass of air is 1742 kg 0287 kPa m kg K288 K kPa144 m 100 144 m 6 6 4 3 3 1 1 3 RT P m V V ROOM 4 m 6 m 6 m Fan The electrical work done by the fan is W W t e e 015 kJ s10 3600 s 5400 kJ Substituting and using the cv value at room temperature 5400 kJ 1742 kg0718 kJkgCT2 15C T2 582C Discussion Note that a fan actually causes the internal temperature of a confined space to rise In fact a 100W fan supplies a room with as much energy as a 100W resistance heater preparation If you are a student using this Manual you are using it without permission 449 470E One part of an insulated rigid tank contains air while the other side is evacuated The internal energy change of the air and the final air pressure are to be determined when the partition is removed Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 2215F and 547 psia 2 The kinetic and potential energy changes are negligible 0 pe ke 3 Constant specific heats at room temperature can be used for air This assumption results in negligible error in heating and airconditioning applications 3 The tank is insulated and thus heat transfer is negligible Vacuum 15 ft3 AIR 15 ft3 100 psia 100F Analysis We take the entire tank as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as 1 2 potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer T T mc U E E E out in v 0 43 42 1 4243 1 Since the internal energy does not change the temperature of the air will also not change Applying the ideal gas equation gives 50 psia 2 100 psia 2 2 1 2 2 1 2 1 1 2 2 2 1 1 P P P P P P V V V V V V PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 450 471 Air in a closed system undergoes an isothermal process The initial volume the work done and the heat transfer are to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 1325 K and 377 MPa 2 The kinetic and potential energy changes are negligible 0 pe ke 3 Constant specific heats can be used for air Properties The gas constant of air is R 0287 kJkgK Table A1 Analysis We take the air as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as out in 2 1 out in 1 2 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in since 0 b b b W Q T T W Q T T mc U W Q E E E v 43 42 1 4243 1 Air 600 kPa 200C Q The initial volume is 04525 m3 600 kPa 2 kg0287 kPa m kg K473 K 3 1 1 1 P mRT V Using the boundary work relation for the isothermal process of an ideal gas gives 5471 kJ 80 kPa 2 kg0287 kPa m kg K473 Kln 600 kPa ln ln 3 2 1 2 1 1 2 2 1 out P P mRT mRT d mRT m Pd Wb v v v v v From energy balance equation 5471 kJ out in Wb Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 451 472 Argon in a pistoncylinder device undergoes an isothermal process The mass of argon and the work done are to be determined Assumptions 1 Argon is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 151 K and 486 MPa 2 The kinetic and potential energy changes are negligible 0 pe ke Properties The gas constant of argon is R 02081kJkgK Table A1 Analysis We take argon as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as out in 2 1 out in 1 2 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in since 0 b b b W Q T T W Q T T mc U W Q E E E v 43 42 1 4243 1 Argon 200 kPa 100C Q Thus 1500 kJ in out Q Wb Using the boundary work relation for the isothermal process of an ideal gas gives 2 1 2 1 1 2 2 1 out ln ln P P mRT mRT d mRT m Pd Wb v v v v v Solving for the mass of the system 1394 kg 50 kPa kPa m kg K373 Kln 200 kPa 02081 kJ 1500 ln 3 2 1 out P P RT W m b PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 452 473 Argon is compressed in a polytropic process The work done and the heat transfer are to be determined Assumptions 1 Argon is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 151 K and 486 MPa 2 The kinetic and potential energy changes are negligible 0 pe ke Properties The properties of argon are R 02081kJkgK and cv 03122 kJkgK Table A2a Analysis We take argon as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as Argon 120 kPa 10C Pv n constant Q 1 2 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc U W Q E E E b v 43 42 1 4243 1 Using the boundary work relation for the polytropic process of an ideal gas gives 1095 kJkg 1 120 800 112 0 2081 kJkg K283 K 1 1 21 20 1 1 2 1 out n n b P P n RT w Thus b in 1095 kJkg w The temperature at the final state is 3882 K 120 kPa 283 K 800 kPa 21 20 1 1 2 1 2 n n P P T T From the energy balance equation 76 6 kJkg 283K 0 3122 kJkg K3882 109 5 kJkg 1 2 out in T T c w q b v Thus out 766 kJkg q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 453 474 Carbon dioxide contained in a springloaded pistoncylinder device is heated The work done and the heat transfer are to be determined Assumptions 1 CO2 is an ideal gas since it is at a high temperature relative to its critical temperature of 3042 K 2 The kinetic and potential energy changes are negligible 0 pe ke Properties The properties of CO2 are R 01889 kJkgK and cv 0657 kJkgK Table A2a P kPa 2 1 Analysis We take CO2 as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as 1000 1 2 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc U W Q E E E b v 43 42 1 4243 1 100 V m3 The initial and final specific volumes are 3 3 1 1 1 0 5629 m 100 kPa 1 kg01889 kPa m kg K298 K P mRT V 3 3 2 2 2 0 1082 m 1000 kPa 1 kg01889 kPa m kg K573 K P mRT V Pressure changes linearly with volume and the work done is equal to the area under the process line 12 250 1 kJ kPa m 1 1 kJ 0 5629m 01082 2 1000kPa 100 2 Area 3 3 1 2 2 1 out V V P P Wb Thus b in 2501 kJ W Using the energy balance equation 69 4 kJ 25K 1 kg 0 657 kJkg K300 250 1 kJ 1 2 out in T T mc W Q b v Thus out 694 kJ Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 454 475 A pistoncylinder device contains air A paddle wheel supplies a given amount of work to the air The heat transfer is to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 1325 K and 377 MPa 2 The kinetic and potential energy changes are negligible 0 pe ke 3 Constant specific heats can be used for air Analysis We take the air as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as pwin out in 2 1 in out in pw 1 2 in out in pw potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in since 0 W W Q T T Q W W T T mc U Q W W E E E b b b v 43 42 1 4243 1 Wpw Air 500 kPa 27C Q Using the boundary work relation on a unit mass basis for the isothermal process of an ideal gas gives 94 6 kJkg 0 287 kJkg K300 Kln3 ln 3 ln 1 2 out RT RT wb v v Substituting into the energy balance equation expressed on a unit mass basis gives 446 kJkg 50 94 6 pwin out in w w q b Discussion Note that the energy content of the system remains constant in this case and thus the total energy transfer output via boundary work must equal the total energy input via shaft work and heat transfer PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 455 476 A cylinder is initially filled with air at a specified state Air is heated electrically at constant pressure and some heat is lost in the process The amount of electrical energy supplied is to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 Air is an ideal gas with variable specific heats 3 The thermal energy stored in the cylinder itself and the resistance wires is negligible 4 The compression or expansion process is quasiequilibrium AIR P const We Properties The initial and final enthalpies of air are Table A17 h h h h 1 2 298 K 350 K 29818 kJ kg 35049 kJ kg Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this closed system can be expressed as Q out 1 2 ein bout out in e potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer Q h m h W U W Q W E E E out in 43 42 1 4243 1 since U Wb H during a constant pressure quasiequilibrium process Substituting Wein 15 kg35049 29818kJkg 60 kJ 845 kJ or 0235 kWh 3600 kJ 1 kWh 845kJ We in Alternative solution The specific heat of air at the average temperature of Tavg 25 772 51C 324 K is from Table A2b cpavg 10065 kJkgC Substituting 845 kJ 60 kJ 25 C 15 kg10065 kJkg C 77 out 1 2 ein Q T T mc W p or 0235 kWh 3600 kJ 1 kWh 845 kJ ein W Discussion Note that for small temperature differences both approaches give the same result PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 456 477 A cylinder initially contains nitrogen gas at a specified state The gas is compressed polytropically until the volume is reduced by onehalf The work done and the heat transfer are to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 The N2 is an ideal gas with constant specific heats 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or expansion process is quasiequilibrium Properties The gas constant of N2 is R 02968 kPam3kgK Table A1 The cv value of N2 at the anticipated average temperature of 350 K is 0744 kJkgK Table A2b Analysis We take the contents of the cylinder as the system This is a closed system since no mass crosses the system boundary The energy balance for this closed system can be expressed as 1 2 out in b 1 2 out in b potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T Q W u m u U Q W E E E v 43 42 1 4243 1 N2 100 kPa 17C PV13 C Q The final pressure and temperature of nitrogen are 357 0 K 05 290 K 100 kPa kPa 2462 2462 kPa 100 kPa 2 1 1 2 1 2 2 2 2 2 1 1 1 13 1 31 2 1 2 31 1 1 31 2 2 T P P T T P T P P P P P V V V V V V V V Then the boundary work for this polytropic process can be determined from 995 kJ 1 13 290K 15 kg02968 kJkg K3570 1 1 1 2 1 1 2 2 2 1 bin n T mR T n P P Pd W V V V Substituting into the energy balance gives 247 kJ 290K 15 kg0744 kJkgK3570 5 kJ 99 1 2 bin out T T mc W Q v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 457 478 Problem 477 is reconsidered The process is to be plotted on a PV diagram and the effect of the polytropic exponent n on the boundary work and heat transfer as the polytropic exponent varies from 11 to 16 is to be investigated The boundary work and the heat transfer are to be plotted versus the polytropic exponent Analysis The problem is solved using EES and the solution is given below Procedure WorkP2V2P1V1nWin If n1 then WinP1V1lnV2V1 Else WinP2V2P1V11n endif End Input Data Vratio05 V2V1 Vratio n13 Polytropic exponent P1 100 kPa T1 17273 K m15 kg MMmolarmassnitrogen Ru8314 kJkmolK RRuMM V1mRT1P1 Process equations V2VratioV1 P2V2T2P1V1T1The combined ideal gas law for states 1 and 2 plus the polytropic process relation give P2 and T2 P2V2nP1V1n Conservation of Energy for the closed system Ein Eout DeltaE we neglect Delta KE and Delta PE for the system the nitrogen Qout Winmu2u1 u1intenergyN2 TT1 internal energy for nitrogen as an ideal gas kJkg u2intenergyN2 TT2 Call WorkP2V2P1V1nWin The following is required for the Pv plots PplotspvplotTplotP1V1mT1 The combined ideal gas law for states 1 and 2 plus the polytropic process relation give P2 and T2 PplotspvplotnP1V1mn spVplotRTplotPplotm3 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 458 0 02 04 06 08 1 0 200 400 600 800 1000 1200 1400 1600 1800 0 500 1000 1500 2000 2500 3000 3500 4000 4500 spvplot m3kg Pplot kPa n10 Pplot n13 n2 Pressure vs specific volume as function of polytropic exponent n Qout kJ Win kJ 11 1156 1211 1267 1322 1378 1433 1489 1544 16 6941 5759 4529 3249 1917 5299 9141 2418 3987 5623 9266 9449 9637 9829 1003 1023 1044 1065 1087 111 11 12 13 14 15 16 60 40 20 0 20 40 60 80 n Qout 11 12 13 14 15 16 925 965 1005 1045 1085 1125 n Win PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 459 479 It is observed that the air temperature in a room heated by electric baseboard heaters remains constant even though the heater operates continuously when the heat losses from the room amount to 6500 kJh The power rating of the heater is to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 141C and 377 MPa 2 The kinetic and potential energy changes are negligible ke pe 0 3 The temperature of the room is said to remain constant during this process Analysis We take the room as the system This is a closed system since no mass crosses the boundary of the system The energy balance for this system reduces to out ein out in e potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 0 Q W U Q W E E E 43 42 1 4243 1 ROOM We Tair const Q since U mcvT 0 for isothermal processes of ideal gases Thus 181 kW 3600 kJh 1 kW 6500 kJh out ein Q W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 460 480 A cylinder equipped with a set of stops on the top is initially filled with air at a specified state Heat is transferred to the air until the piston hits the stops and then the pressure doubles The work done by the air and the amount of heat transfer are to be determined and the process is to be shown on a Pv diagram Assumptions 1 Air is an ideal gas with variable specific heats 2 The kinetic and potential energy changes are negligible 3 The thermal energy stored in the cylinder itself is negligible ke pe 0 Properties The gas constant of air is R 0287 kPam3kgK Table A1 AIR 200 kPa Analysis We take the air in the cylinder to be the system This is a closed system since no mass crosses the boundary of the system The energy balance for this closed system can be expressed as bout 1 3 in 1 3 bout in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in W u m u Q u m u U W Q E E E 43 42 1 4243 1 Q The initial and the final volumes and the final temperature of air are determined from P 3 1 2 1200 K 300 K 2 200 kPa kPa 400 258 m 1 29 2 2 129 m 200 kPa kg0287 kPa m kg K300 K 3 1 1 3 1 3 3 3 3 3 1 1 1 3 1 3 3 3 1 1 1 T P P T T P T P P mRT V V V V V V V v No work is done during process 23 since V2 V3 The pressure remains constant during process 12 and the work done during this process is 258 kJ kPa m 1 1 kJ 129 m kPa258 200 3 3 1 2 2 2 1 V V V P Pd Wb The initial and final internal energies of air are Table A17 kJkg kJkg 93333 21407 1200 K 3 300 K 1 u u u u Substituting Qin 3 kg93333 21407kJkg 258 kJ 2416 kJ Alternative solution The specific heat of air at the average temperature of Tavg 300 12002 750 K is from Table A 2b cvavg 0800 kJkgK Substituting 3 kg0800 kJkgK1200 300 K 258 kJ 2418 kJ bout 1 3 bout 1 3 in W T mc T W u m u Q v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 461 481 Air at a specified state contained in a pistoncylinder device with a set of stops is heated until a final temperature The amount of heat transfer is to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature relative to its critical temperature of 3042 K 2 The kinetic and potential energy changes are negligible 0 pe ke Properties The properties of air are R 0287 kJkgK and cv 0718 kJkgK Table A2a Analysis We take air as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as Air 100 kPa 27C 04 m3 Q 1 2 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc U W Q E E E b v 43 42 1 4243 1 The volume will be constant until the pressure is 300 kPa 900 K 100 kPa 300 K 300 kPa 1 2 1 2 P T P T P kPa 2 3 1 The mass of the air is 300 0 4646 kg kPa m kg K300 K 0287 kPa04 m 100 3 3 1 1 1 RT P m V 100 The boundary work done during process 23 is V m3 4 0 kJ 0 4646 kg0287 kPa m kg K1200 900K 3 2 3 2 3 2 out T mR T P Wb V V Substituting these values into energy balance equation 340 kJ 300K 0 4646 kg 0 718 kJkg K1200 40 kJ 1 3 out in T T mc W Q b v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 462 482 Air at a specified state contained in a pistoncylinder device undergoes an isothermal and constant volume process until a final temperature The process is to be sketched on the PV diagram and the amount of heat transfer is to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature relative to its critical temperature of 3042 K 2 The kinetic and potential energy changes are negligible 0 pe ke Properties The properties of air are R 0287 kJkgK and cv 0718 kJkgK Table A2a Analysis a The processes 12 isothermal and 23 constantvolume are sketched on the PV diagram as shown Air 100 kPa 27C 04 m3 b We take air as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system fort he process 13 can be expressed as 1 3 in 2 out 1 potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc U Q W E E E b v 43 42 1 4243 1 Q P kPa The mass of the air is 1 2 3 1 394 kg kPa m kg K1200 K 0287 kPa08 m 600 3 3 1 1 1 RT P m V 600 300 The work during process 12 is determined from boundary work relation for an isothermal process to be 332 8 kJ 300 kPa 394 kg0287 kPa m kg K1200 Kln 600 kPa 1 ln ln 3 2 1 1 1 2 1 2 1 P P mRT mRT W b out V V V m3 since 2 1 1 2 P P V V for an isothermal process Substituting these values into energy balance equation 568 kJ 1200K 1 394 kg 0 718 kJkg K300 8 kJ 332 1 3 out12 in T T mc W Q b v Thus Qout 568 kJ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 463 483 Argon at a specified state contained in a pistoncylinder device with a set of stops is compressed at constant temperature until a final volume The process is to be sketched on the PV diagram and the amount of heat transfer is to be determined Assumptions 1 Argon is an ideal gas 2 The kinetic and potential energy changes are negligible 0 pe ke Properties The properties of argon are R 02081 kJkgK and cv 03122 kJkgK Table A2a Analysis a The processes 12 isothermal and 23 isobaric are sketched on the PV diagram as shown Argon 100 kPa 927C 04 m3 b We take argon as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system fort he process 13 can be expressed as 1 3 out23 in 2 in 1 potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc U W Q W E E E b b v 43 42 1 4243 1 Q The mass of the air is P kPa 1 2 3 0 6407 kg kPa m kg K300 K 02081 kPa04 m 100 3 3 1 1 1 RT P m V 300 The pressure at state 2 is 12 isothermal process 100 200 kPa m 02 100 kPa 04 m 3 3 2 1 1 2 V P V P V m3 The temperature at state 3 is 23 isobaric process 900 K m 02 300 K 06 m 3 3 2 3 2 3 V T V T The compression work during process 12 is determined from boundary work relation for an isothermal process to be 27 7 kJ m 04 0 6407 kg02081 kPa m kg K300 Kln 02 m ln 3 3 3 1 2 1 2 1 V V mRT W b out Thus 7 kJ 27 2 1 b in W The expansion work during process 23 is determined from boundary work relation for a constantpressure process to be 80 0 kJ 300K kPa m kg K900 0 6407 kg02081 3 2 3 2 3 2 out23 T mR T P Wb V V Substituting these values into energy balance equation 1722 kJ 300K 0 6407 kg 0 3122 kJkg K900 27 7 kJ kJ 80 1 3 2 in 1 out23 in T T mc W W Q b b v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 464 484 An ideal gas at a specified state contained in a pistoncylinder device undergoes an isothermal process during which heat is lost from the gas The final pressure and volume are to be determined Assumptions 1 The kinetic and potential energy changes are negligible 0 pe ke Ideal gas 100 kPa 06 m3 Properties The properties of air are R 0287 kJkgK and cv 0718 kJkgK Table A2a Analysis We take the ideal gas as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system fort he process 13 can be expressed as Q out out 1 2 1 3 out out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 0 since b b W Q T T T T mc U W Q E E E v 43 42 1 4243 1 V T constant 2 1 P The boundary work for an isothermal process is determined from 1 2 1 1 1 2 2 1 2 1 out ln ln V V V V V V V P mRT mRT dV Pd Wb Thus 0 3679 1 exp 100 kPa06 m 60 kJ exp exp exp 3 1 1 out 1 1 1 2 V V V V P Q P W out b and 0221 m3 0 367906 m 0 3679 3 1 2 V V The final pressure is found from the combined ideal gas law 272 kPa 03679 1 100 kPa 2 1 1 2 2 2 2 1 1 1 V V V V P P T P T P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 465 Closed System Energy Analysis Solids and Liquids 485 An iron block is heated The internal energy and enthalpy changes are to be determined for a given temperature change Assumptions Iron is an incompressible substance with a constant specific heat Properties The specific heat of iron is 045 kJkgK Table A3b Analysis The internal energy and enthalpy changes are equal for a solid Then 225 kJ 25K 1 kg045 kJkg K75 mc T U H 486E Liquid water experiences a process from one state to another The internal energy and enthalpy changes are to be determined under different assumptions Analysis a Using the properties from compressed liquid tables 30 Btulbm Table A 7E 73 6736 Btulbm F 100 psia 2000 1821 Btulbm 404 psia ft 5 1Btu 0 17812 psia 0 01602 ft lbm50 1807 Btulbm 07 Btulbm Table A 4E 18 2 2 2 2 3 3 sat 50 F 1 50 F 1 h u T P P P v h h u u T f f f Btulbm 5508 Btulbm 4929 30 1821 73 36 1807 67 1 2 1 2 h h h u u u b Using incompressible substance approximation and property tables Table A4E 6803 Btulbm 6803 Btulbm 1807 Btulbm 07 Btulbm 18 100 F 2 100 F 2 50 F 1 50 F 1 f f f f h h u u h h u u Btulbm 4996 Btulbm 4996 03 1807 68 03 1807 68 1 2 1 2 h h h u u u c Using specific heats and taking the specific heat of water to be 100 BtulbmR Table A3Ea 50 Btulbm 50R 100 Btulbm R100 c T u h PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 466 487E A person shakes a canned of drink in a iced water to cool it The mass of the ice that will melt by the time the canned drink is cooled to a specified temperature is to be determined Assumptions 1 The thermal properties of the drink are constant and are taken to be the same as those of water 2 The effect of agitation on the amount of ice melting is negligible 3 The thermal energy capacity of the can itself is negligible and thus it does not need to be considered in the analysis Properties The density and specific heat of water at the average temperature of 75452 60F are ρ 623 lbmft3 and cp 10 BtulbmF Table A3E The heat of fusion of water is 1435 Btulbm Analysis We take a canned drink as the system The energy balance for this closed system can be expressed as 2 1 out 1 2 canned drink out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T Q u m u U Q E E E 43 42 1 4243 1 Cola 75F Noting that 1 gal 128 oz and 1 ft3 748 gal 9575 oz the total amount of heat transfer from a ball is 23 4 Btucan 45 F Btulbm F75 01 0 781 lbmcan 0 781 lbmcan 128 fluid oz 1gal 748 gal 1ft 62 3 lbmft 12 ozcan 2 1 out 3 3 T mc T Q m ρV Noting that the heat of fusion of water is 1435 Btulbm the amount of ice that will melt to cool the drink is per can of drink 1435 Btulbm 4 Btucan 23 out ice 0163 lbm hif Q m since heat transfer to the ice must be equal to heat transfer from the can Discussion The actual amount of ice melted will be greater since agitation will also cause some ice to melt PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 467 488 An iron whose base plate is made of an aluminum alloy is turned on The minimum time for the plate to reach a specified temperature is to be determined Assumptions 1 It is given that 85 percent of the heat generated in the resistance wires is transferred to the plate 2 The thermal properties of the plate are constant 3 Heat loss from the plate during heating is disregarded since the minimum heating time is to be determined 4 There are no changes in kinetic and potential energies 5 The plate is at a uniform temperature at the end of the process Properties The density and specific heat of the aluminum alloy plate are given to be ρ 2770 kgm3 and cp 875 kJkgC Analysis The mass of the irons base plate is 0 4155 kg 2770 kgm 0 005 m 0 03 m 2 3 LA m ρ ρV Air 22C Noting that only 85 percent of the heat generated is transferred to the plate the rate of heat transfer to the irons base plate is IRON 1000 W 900 W 0 90 1000 W in Q We take plate to be the system The energy balance for this closed system can be expressed as 1 2 in 1 2 plate in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T t Q u m u U Q E E E 43 42 1 4243 1 Solving for t and substituting 900 Js 22 C 0 4155 kg875 Jkg C200 in plate 719 s Q mc T t which is the time required for the plate temperature to reach the specified temperature PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 468 489 Stainless steel ball bearings leaving the oven at a specified uniform temperature at a specified rate are exposed to air and are cooled before they are dropped into the water for quenching The rate of heat transfer from the ball bearing to the air is to be determined Assumptions 1 The thermal properties of the bearing balls are constant 2 The kinetic and potential energy changes of the balls are negligible 3 The balls are at a uniform temperature at the end of the process Properties The density and specific heat of the ball bearings are given to be ρ 8085 kgm3 and cp 0480 kJkgC Analysis We take a single bearing ball as the system The energy balance for this closed system can be expressed as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2 1 out 1 2 ball out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T Q u m u U Q E E E 43 42 1 4243 1 Steel balls 900C Water 25C Furnace The total amount of heat transfer from a ball is 0 1756 kJball 850 C 0 007315 kg 0 480 kJkg C900 0 007315 kg 6 0 012 m 8085 kgm 6 2 1 out 3 3 3 T mc T Q D m π ρ π ρV Then the rate of heat transfer from the balls to the air becomes 234 kW 1405 kJmin 800 ballsmin 0 1756 kJball out per ball ball total Q n Q Therefore heat is lost to the air at a rate of 234 kW 490 Carbon steel balls are to be annealed at a rate of 2500h by heating them first and then allowing them to cool slowly in ambient air at a specified rate The total rate of heat transfer from the balls to the ambient air is to be determined Assumptions 1 The thermal properties of the balls are constant 2 There are no changes in kinetic and potential energies 3 The balls are at a uniform temperature at the end of the process Properties The density and specific heat of the balls are given to be ρ 7833 kgm3 and cp 0465 kJkgC Analysis We take a single ball as the system The energy balance for this closed system can be expressed as 2 1 out 1 2 ball out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T Q u m u U Q E E E 43 42 1 4243 1 Steel balls 900C Air 35C Furnace b The amount of heat transfer from a single ball is 100 C 0781 kJ per ball 0 0021 kg 0 465 kJkg C900 0 00210 kg 6 0 008 m 7833 kgm 6 2 1 out 3 3 3 T T mc Q D m p π ρ π ρV Then the total rate of heat transfer from the balls to the ambient air becomes 542 W 1953 kJh 0 781 kJball 2500 ballsh out ball out Q n Q preparation If you are a student using this Manual you are using it without permission 469 491 An egg is dropped into boiling water The amount of heat transfer to the egg by the time it is cooked is to be determined Assumptions 1 The egg is spherical in shape with a radius of r0 275 cm 2 The thermal properties of the egg are constant 3 Energy absorption or release associated with any chemical andor phase changes within the egg is negligible 4 There are no changes in kinetic and potential energies Properties The density and specific heat of the egg are given to be ρ 1020 kgm3 and cp 332 kJkgC Analysis We take the egg as the system This is a closes system since no mass enters or leaves the egg The energy balance for this closed system can be expressed as 1 2 1 2 egg in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T u m u U Q E E E 43 42 1 4243 1 Boiling Water Egg 8C Then the mass of the egg and the amount of heat transfer become 212 kJ 8 C 0 0889 kg 3 32 kJkg C80 0 0889 kg 6 0 055 m 1020 kgm 6 1 2 in 3 3 3 T T mc Q D m p π ρ π ρV 492E Large brass plates are heated in an oven at a rate of 300min The rate of heat transfer to the plates in the oven is to be determined Assumptions 1 The thermal properties of the plates are constant 2 The changes in kinetic and potential energies are negligible Properties The density and specific heat of the brass are given to be ρ 5325 lbmft3 and cp 0091 BtulbmF Analysis We take the plate to be the system The energy balance for this closed system can be expressed as Plates 75F 1 2 1 2 plate in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T u m u U Q E E E 43 42 1 4243 1 The mass of each plate and the amount of heat transfer to each plate is 213 lbm 12 ft 2 ft2 ft 532 5 lbmft 21 3 LA m ρ ρV 17930 Btuplate 75 F 213 lbmplate 0 091 Btulbm F1000 1 2 in T mc T Q Then the total rate of heat transfer to the plates becomes 5379000 Btumin 89650 Btus 17930 Btuplate 300 platesmin in per plate plate total Q n Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 470 493 Long cylindrical steel rods are heattreated in an oven The rate of heat transfer to the rods in the oven is to be determined Assumptions 1 The thermal properties of the rods are constant 2 The changes in kinetic and potential energies are negligible Properties The density and specific heat of the steel rods are given to be ρ 7833 kgm3 and cp 0465 kJkgC Analysis Noting that the rods enter the oven at a velocity of 2 mmin and exit at the same velocity we can say that a 3m long section of the rod is heated in the oven in 1 min Then the mass of the rod heated in 1 minute is 7875 kg 4 7833 kgm 2 m 0 08 m 4 2 3 2 π π ρ ρ ρ D L LA m V We take the 2m section of the rod in the oven as the system The energy balance for this closed system can be expressed as Steel rod 30C Oven 900C 1 2 1 2 rod in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T u m u U Q E E E 43 42 1 4243 1 Substituting 24530 kJ 30 C 7875 kg 0 465 kJkg C700 1 2 in T mc T Q Noting that this much heat is transferred in 1 min the rate of heat transfer to the rod becomes 24530 kJ1 min 24530 kJmin 409 kW in in t Q Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 471 494 An electronic device is on for 5 minutes and off for several hours The temperature of the device at the end of the 5 min operating period is to be determined for the cases of operation with and without a heat sink Assumptions 1 The device and the heat sink are isothermal 2 The thermal properties of the device and of the sink are constant 3 Heat loss from the device during on time is disregarded since the highest possible temperature is to be determined Properties The specific heat of the device is given to be cp 850 JkgC The specific heat of aluminum at room temperature of 300 K is 902 JkgC Table A3 Analysis We take the device to be the system Noting that electrical energy is supplied the energy balance for this closed system can be expressed as Electronic device 25C 1 2 in e 1 2 device in e potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T t W u m u U W E E E 43 42 1 4243 1 Substituting the temperature of the device at the end of the process is determined to be 466C 2 2 25 C 25 Js5 60 s 0 020 kg850 Jkg C T T without the heat sink Case 2 When a heat sink is attached the energy balance can be expressed as 1 heat sink 2 1 device 2 in e heat sink device in e T mc T T mc T t W U U W Substituting the temperature of the deviceheat sink combination is determined to be with heat sink 25 C 0 500 kg902 Jkg C 25 C 25 Js5 60 s 0 020 kg850 Jkg C 2 2 2 C 410 T T T Discussion These are the maximum temperatures In reality the temperatures will be lower because of the heat losses to the surroundings PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 472 495 Problem 494 is reconsidered The effect of the mass of the heat sink on the maximum device temperature as the mass of heat sink varies from 0 kg to 1 kg is to be investigated The maximum temperature is to be plotted against the mass of heat sink Analysis The problem is solved using EES and the solution is given below Knowns T1 is the maximum temperature of the device Qdotout 25 W mdevice20 g Cpdevice850 JkgC A5 cm2 DELTAt5 min Tamb25 C msink05 kg Cpal taken from Table A3b at 300K Cpal0902 kJkgC T2Tamb Solution The device without the heat sink is considered to be a closed system Conservation of Energy for the closed system Edotin Edotout DELTAEdot we neglect DELTA KE and DELTA PE for the system the device Edotin Edotout DELTAEdot Edotin 0 Edotout Qdotout Use the solid material approximation to find the energy change of the device DELTAEdot mdeviceconvertgkgCpdeviceT2T1deviceDELTAtconvertmins The device with the heat sink is considered to be a closed system Conservation of Energy for the closed system Edotin Edotout DELTAEdot we neglect DELTA KE and DELTA PE for the device with the heat sink Edotin Edotout DELTAEdotcombined Use the solid material approximation to find the energy change of the device DELTAEdotcombined mdeviceconvertgkgCpdeviceT2T1devicesinkmsinkCpalT2 T1devicesinkconvertkJJDELTAtconvertmins msink kg T1devicesink C 0 01 02 03 04 05 06 07 08 09 1 4662 9496 6299 5108 4485 4103 3844 3657 3515 3405 3316 0 02 04 06 08 1 0 100 200 300 400 500 msink kg T1devicesink C PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 473 496 The face of a person is slapped For the specified temperature rise of the affected part the impact velocity of the hand is to be determined Assumptions 1 The hand is brought to a complete stop after the impact 2 The face takes the blow without significant movement 3 No heat is transferred from the affected area to the surroundings and thus the process is adiabatic 4 No work is done on or by the system 5 The potential energy change is zero PE 0 and E U KE Analysis We analyze this incident in a professional manner without involving any emotions First we identify the system draw a sketch of it and state our observations about the specifics of the problem We take the hand and the affected portion of the face as the system This is a closed system since it involves a fixed amount of mass no mass transfer We observe that the kinetic energy of the hand decreases during the process as evidenced by a decrease in velocity from initial value to zero while the internal energy of the affected area increases as evidenced by an increase in the temperature There seems to be no significant energy transfer between the system and its surroundings during this process Under the stated assumptions and observations the energy balance on the system can be expressed as hand 2 tissue affected hand tissue affected potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 2 0 0 KE 0 V m T mc U E E E 43 42 1 4243 1 That is the decrease in the kinetic energy of the hand must be equal to the increase in the internal energy of the affected area Solving for the velocity and substituting the given quantities the impact velocity of the hand is determined to be or 149 kmh 1kJkg 1000 m s kg 21 C 81 C kJkg 83 0 15 kg 2 2 2 2 hand hand affected tissue hand 414 ms V m mc T V Discussion Reconstruction of events such as this by making appropriate assumptions are commonly used in forensic engineering PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 474 Special Topic Biological Systems 497C Metabolism refers to the chemical activity in the cells associated with the burning of foods The basal metabolic rate is the metabolism rate of a resting person which is 84 W for an average man 498C The food we eat is not entirely metabolized in the human body The fraction of metabolizable energy contents are 955 for carbohydrates 775 for proteins and 977 for fats Therefore the metabolizable energy content of a food is not the same as the energy released when it is burned in a bomb calorimeter 499C Yes Each body rejects the heat generated during metabolism and thus serves as a heat source For an average adult male it ranges from 84 W at rest to over 1000 W during heavy physical activity Classrooms are designed for a large number of occupants and thus the total heat dissipated by the occupants must be considered in the design of heating and cooling systems of classrooms 4100C 1 kg of natural fat contains almost 8 times the metabolizable energy of 1 kg of natural carbohydrates Therefore a person who fills his stomach with carbohydrates will satisfy his hunger without consuming too many calories 4101 Six people are fast dancing in a room and there is a resistance heater in another identical room The room that will heat up faster is to be determined Assumptions 1 The rooms are identical in every other aspect 2 Half of the heat dissipated by people is in sensible form 3 The people are of average size Properties An average fast dancing person dissipates 600 Calh of energy sensible and latent Table 42 Analysis Three couples will dissipate E 6 persons600 Calhperson41868 kJCal 15072 kJh 4190 W of energy About half of this is sensible heat Therefore the room with the people dancing will warm up much faster than the room with a 2kW resistance heater PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 475 4102 Two men are identical except one jogs for 30 min while the other watches TV The weight difference between these two people in one month is to be determined Assumptions The two people have identical metabolism rates and are identical in every other aspect Properties An average 68kg person consumes 540 Calh while jogging and 72 Calh while watching TV Table 42 Analysis An 80kg person who jogs 05 h a day will have jogged a total of 15 h a month and will consume 34578 kJ 68 kg 80 kg 1 Cal 72 Calh15 h 41868 kJ 540 consumed E more calories than the person watching TV The metabolizable energy content of 1 kg of fat is 33100 kJ Therefore the weight difference between these two people in 1month will be 1045 kg 33100 kJkg 34578 kJ content of fat Energy consumed fat E m 4103 A bicycling woman is to meet her entire energy needs by eating 30g candy bars The number of candy bars she needs to eat to bicycle for 1h is to be determined Assumptions The woman meets her entire calorie needs from candy bars while bicycling Properties An average 68kg person consumes 639 Calh while bicycling and the energy content of a 20g candy bar is 105 Cal Tables 41 and 42 Analysis Noting that a 20g candy bar contains 105 Calories of metabolizable energy a 30g candy bar will contain 1575 Cal 20 g 30 g 105 Cal candy E of energy If this woman is to meet her entire energy needs by eating 30g candy bars she will need to eat 4candybarsh 1575Cal 639Calh Ncandy PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 476 4104 A 75kg man eats 1L of ice cream The length of time this man needs to jog to burn off these calories is to be determined Assumptions The man meets his entire calorie needs from the ice cream while jogging Properties An average 68kg person consumes 540 Calh while jogging and the energy content of a 100ml of ice cream is 110 Cal Tables 41 and 42 Analysis The rate of energy consumption of a 55kg person while jogging is 596 Calh 68 kg 75 kg 540 Calh consumed E Noting that a 100ml serving of ice cream has 110 Cal of metabolizable energy a 1liter box of ice cream will have 1100 Calories Therefore it will take 185 h 596 Calh 1100 Cal t of jogging to burn off the calories from the ice cream 4105 A man with 20kg of body fat goes on a hunger strike The number of days this man can survive on the body fat alone is to be determined Assumptions 1 The person is an average male who remains in resting position at all times 2 The man meets his entire calorie needs from the body fat alone Properties The metabolizable energy content of fat is 33100 Calkg An average resting person burns calories at a rate of 72 Calh Table 42 Analysis The metabolizable energy content of 20 kg of body fat is 662000 kJ 33100 kJkg 20 kg fat E The person will consume 7235 kJday 1 Cal 41868 kJ 72 Calh 24 h consumed E Therefore this person can survive t 662000 kJ 7235 kJ day 915 days on his body fat alone This result is not surprising since people are known to survive over 100 days without any food intake PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 477 4106 Two 50kg women are identical except one eats her baked potato with 4 teaspoons of butter while the other eats hers plain every evening The weight difference between these two woman in one year is to be determined Assumptions 1 These two people have identical metabolism rates and are identical in every other aspect 2 All the calories from the butter are converted to body fat Properties The metabolizable energy content of 1 kg of body fat is 33100 kJ The metabolizable energy content of 1 teaspoon of butter is 35 Calories Table 41 Analysis A person who eats 4 teaspoons of butter a day will consume 51100 Calyear 1 year 365 days 4 teaspoonsday 35 Calteaspoon consumed E Therefore the woman who eats her potato with butter will gain 65 kg 1 Cal 41868 kJ 33100 kJkg 51100 Cal fat m of additional body fat that year 4107 A woman switches from 1L of regular cola a day to diet cola and 2 slices of apple pie It is to be determined if she is now consuming more or less calories Properties The metabolizable energy contents are 300 Cal for a slice of apple pie 87 Cal for a 200ml regular cola and 0 for the diet drink Table 43 Analysis The energy contents of 2 slices of apple pie and 1L of cola are 435 Cal 87 Cal 5 600 Cal 300 Cal 2 cola pie E E Therefore the woman is now consuming more calories 4108 A person drinks a 12oz beer and then exercises on a treadmill The time it will take to burn the calories from a 12 oz can of regular and light beer are to be determined Assumptions The drinks are completely metabolized by the body Properties The metabolizable energy contents of regular and light beer are 150 and 100 Cal respectively Exercising on a treadmill burns calories at an average rate of 700 Calh given Analysis The exercising time it will take to burn off beer calories is determined directly from a Regular beer 129 min 0214 h 700 Calh 150 Cal tregular beer b Light beer 86 min 0143h 700 Calh 100 Cal tlight beer PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 478 4109 A man and a woman have lunch at Burger King and then shovel snow The shoveling time it will take to burn off the lunch calories is to be determined for both Assumptions The food intake during lunch is completely metabolized by the body Properties The metabolizable energy contents of different foods are as given in the problem statement Shoveling snow burns calories at a rate of 360 Calh for the woman and 480 Calh for the man given Analysis The total calories consumed during lunch and the time it will take to burn them are determined for both the man and woman as follows Man Lunch calories 720400225 1345 Cal Shoveling time tshoveling man 1345 Cal 480 Cal h 280 h Woman Lunch calories 3304000 730 Cal Shoveling time tshoveling woman 730 Cal 360 Cal h 203 h 4110 Two friends have identical metabolic rates and lead identical lives except they have different lunches The weight difference between these two friends in a year is to be determined Assumptions 1 The diet and exercise habits of the people remain the same other than the lunch menus 2 All the excess calories from the lunch are converted to body fat Properties The metabolizable energy content of body fat is 33100 Calkg text The metabolizable energy contents of different foods are given in problem statement Analysis The person who has the double whopper sandwich consumes 1600 800 800 Cal more every day The difference in calories consumed per year becomes Calorie consumption difference 800 Calday365 daysyear 292000 Calyear Therefore assuming all the excess calories to be converted to body fat the weight difference between the two persons after 1 year will be 369 kgyr 1 Cal 41868 kJ 33100 kJkg 292000 Calyr Enegy content of fat Calorie intake difference Weight difference fat intake e E or about 80 pounds of body fat per year PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 479 4111E A person eats dinner at a fastfood restaurant The time it will take for this person to burn off the dinner calories by climbing stairs is to be determined Assumptions The food intake from dinner is completely metabolized by the body Properties The metabolizable energy contents are 270 Cal for regular roast beef 410 Cal for big roast beef and 150 Cal for the drink Climbing stairs burns calories at a rate of 400 Calh given Analysis The total calories consumed during dinner and the time it will take to burn them by climbing stairs are determined to be Dinner calories 270410150 830 Cal Stair climbing time t 830 Cal 400 Cal h 208 h 4112 Three people have different lunches The person who consumed the most calories from lunch is to be determined Properties The metabolizable energy contents of different foods are 530 Cal for the Big Mac 640 Cal for the whopper 350 Cal for french fries and 5 Cal for each olive given Analysis The total calories consumed by each person during lunch are Person 1 Lunch calories 530 Cal Person 2 Lunch calories 640 Cal Person 3 Lunch calories 350550 600 Cal Therefore the person with the Whopper will consume the most calories 4113 A 100kg man decides to lose 10 kg by exercising without reducing his calorie intake The number of days it will take for this man to lose 10 kg is to be determined Assumptions 1 The diet and exercise habits of the person remain the same other than the new daily exercise program 2 The entire calorie deficiency is met by burning body fat Properties The metabolizable energy content of body fat is 33100 Calkg text Analysis The energy consumed by an average 68kg adult during fastswimming fast dancing jogging biking and relaxing are 860 600 540 639 and 72 Calh respectively Table 42 The daily energy consumption of this 100kg man is 6316 Cal 68 kg 72 Calh 23 h 100 kg 639 Calh 1h 540 600 860 Therefore this person burns 63164000 2316 more Calories than he takes in which corresponds to 0293 kg 1 Cal 41868 kJ 33100 kJkg 2316 Cal fat m of body fat per day Thus it will take only 341 days 0293 kg 10 kg t for this man to lose 10 kg PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 480 4114 The effect of supersizing in fast food restaurants on monthly weight gain is to be determined for a given case Properties The metabolizable energy content of 1 kg of body fat is 33100 kJ Analysis The increase in the body fat mass due to extra 1000 calories is 379 kgmonth 3 0 daysmonth 1Cal 4 1868 kJ 33100 kJkg 1000 Calday mfat 4115E The range of healthy weight for adults is usually expressed in terms of the body mass index BMI in SI units as BMI kg m2 W H 2 This formula is to be converted to English units such that the weight is in pounds and the height in inches Analysis Noting that 1 kg 22 lbm and 1 m 3937 in the weight in lbm must be divided by 22 to convert it to kg and the height in inches must be divided by 3937 to convert it to m before inserting them into the formula Therefore BMI kg m lbm in lbm in 2 2 W H W H W H 2 2 2 2 2 2 39 37 705 2 Every person can calculate their own BMI using either SI or English units and determine if it is in the healthy range PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 481 4116 A person changes hisher diet to lose weight The time it will take for the body mass index BMI of the person to drop from 30 to 20 is to be determined Assumptions The deficit in the calori intake is made up by burning body fat Properties The metabolizable energy contents are 350 Cal for a slice of pizza and 87 Cal for a 200ml regular cola The metabolizable energy content of 1 kg of body fat is 33100 kJ Analysis The lunch calories before the diet is 1224 Cal 2 87 Cal 3 35 0 Cal 2 3 coke pizza old e e E The lunch calories after the diet is 787 Cal 1 87 Cal 2 35 0 Cal 1 2 coke pizza old e e E The calorie reduction is 437 Cal 787 1224 reduction E The corresponding reduction in the body fat mass is 0 05528 kg 1Cal 4 1868 kJ 33100 kJkg 437 Cal fat m The weight of the person before and after the diet is 51 2 kg m 61 20 BMI 76 8 kg m 61 30 BMI 2 2pizza 2 2 2 2pizza 1 1 h W h W Then it will take 463 days 005528 kgday 51 2 kg 76 8 Time fat 2 1 m W W for the BMI of this person to drop to 20 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 482 Review Problems 4117 During a phase change a constant pressure process becomes a constant temperature process Hence infinite 0 finite T p T h c and the specific heat at constant pressure has no meaning The specific heat at constant volume does have a meaning since finite finite finite v v T u c 4118 Nitrogen is heated to experience a specified temperature change The heat transfer is to be determined for two cases Assumptions 1 Nitrogen is an ideal gas since it is at a high temperature and probably low pressure relative to its critical point values of 1262 K and 339 MPa 2 The kinetic and potential energy changes are negligible 0 pe ke 3 Constant specific heats can be used for nitrogen Properties The specific heats of nitrogen at the average temperature of 202502135C408 K are cp 1045 kJkgK and cv 0748 kJkgK Table A2b Analysis We take the nitrogen as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for a constantvolume process can be expressed as 1 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc U Q E E E v 43 42 1 4243 1 N2 T1 20C T2 250C Q The energy balance during a constantpressure process such as in a piston cylinder device can be expressed as 1 2 in out in out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc H Q U W Q U W Q E E E p b b 43 42 1 4243 1 N2 T1 20C T2 250C Q since U Wb H during a constant pressure quasiequilibrium process Substituting for both cases 1720kJ 20K 10 kg 0 748 kJkg K250 1 2 const in T T mc Q v V 2404 kJ 20K 10 kg 1 045 kJkg K250 1 2 const in T T mc Q p P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 483 4119 A classroom has 40 students each dissipating 84 W of sensible heat It is to be determined if it is necessary to turn the heater on in the room to avoid cooling of the room Properties Each person is said to be losing sensible heat to the room air at a rate of 84 W Analysis We take the room is losing heat to the outdoors at a rate of 333 kW 3600 s 1 h 12000 kJh loss Q The rate of sensible heat gain from the students is 336 kW 3360 W 84 Wstudent 40 students gain Q which is greater than the rate of heat loss from the room Therefore it is not necessary to turn the heater 4120E An insulated rigid vessel contains air A paddle wheel supplies work to the air The work supplied and final temperature are to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 2385 R and 547 psia 2 The kinetic and potential energy changes are negligible 0 pe ke 3 Constant specific heats can be used for air Properties The specific heats of air at room temperature are cp 0240 BtulbmR and cv 0171 BtulbmR Table A2Ea Analysis We take the air as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as 1 2 in pw potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc U W E E E v 43 42 1 4243 1 Wpw Air 2 lbm 30 psia 60F As the specific volume remains constant during this process the ideal gas equation gives 2333F 693 3 R 30 psia 520 R 40 psia 1 2 1 2 P T P T Substituting 593 Btu 520R 2 lbm 0 171 Btulbm R6933 1 2 pwin T T mc W v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 484 4121 Air at a given state is expanded polytropically in a pistoncylinder device to a specified pressure The final temperature is to be determined Air 2 MPa 300C Analysis The polytropic relations for an ideal gas give 77 C 196 K 51 50 1 1 2 1 2 2000 kPa 80 kPa 273 K 300 n n P P T T 4122 Nitrogen in a rigid vessel is heated The work done and heat transfer are to be determined Assumptions 1 Nitrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 1262 K and 339 MPa 2 The kinetic and potential energy changes are negligible 0 pe ke 3 Constant specific heats at room temperature can be used for nitrogen Properties For nitrogen cv 0743 kJkgK at room temperature Table A2a Analysis We take the nitrogen as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as Nitrogen 100 kPa 25C 1 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc U Q E E E v 43 42 1 4243 1 Q There is no work done since the vessel is rigid 0 kJkg w Since the specific volume is constant during the process the final temperature is determined from ideal gas equation to be 894 K 100 psia 298 K 300 kPa 1 2 1 2 P T P T Substituting 4428 kJkg 298K 0 743 kJkg K894 2 1 in T T c q v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 485 4123 A wellinsulated rigid vessel contains saturated liquid water Electrical work is done on water The final temperature is to be determined Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero 2 The thermal energy stored in the tank itself is negligible Analysis We take the contents of the tank as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as 1 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u m u U W E E E e 43 42 1 4243 1 P 2 1 v The amount of electrical work added during 30 minutes period is 900 kJ 900000 J 1VA 50 V10 A30 60 s 1 W ein I t W V The properties at the initial state are Table A4 001008 m kg 0 53 kJkg 167 3 40 C 1 40 C 1 f u f u v v Substituting 46753 kJkg 3 kg 900 kJ 16753 kJkg in 1 2 1 2 in m W u u u m u W e e The final state is compressed liquid Noting that the specific volume is constant during the process the final temperature is determined using EES to be from EES 001008 m kg 0 53 kJkg 467 2 3 1 2 2 1189C T u v v Discussion We cannot find this temperature directly from steam tables at the end of the book Approximating the final compressed liquid state as saturated liquid at the given internal energy the final temperature is determined from Table A4 to be 1115 C 46753 kJkg sat 2 u T T Note that this approximation resulted in an answer which is not very close to the exact answer determined using EES PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 486 4124 The boundary work expression during a polytropic process of an ideal gas is to be derived Assumptions The process is quasiequilibrium Analysis For a polytropic process Constant 2 2 1 1 n n P P v v Substituting this into the boundary work expression gives 1 1 1 1 1 1 1 1 2 1 1 1 1 2 1 1 1 1 1 1 2 2 1 1 1 1 2 1 1 2 1 1 1 2 1 out n n n n n n n n n n n b n RT n P n P n P d P Pd w v v v v v v v v v v v v v v v v v v When the specific volume ratio is eliminated with the expression for the constant 1 1 1 1 2 1 out n n b P P n RT w where 1 n PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 487 4125 A cylinder equipped with an external spring is initially filled with air at a specified state Heat is transferred to the air and both the temperature and pressure rise The total boundary work done by the air and the amount of work done against the spring are to be determined and the process is to be shown on a Pv diagram Assumptions 1 The process is quasiequilibrium 2 The spring is a linear spring Analysis a The pressure of the gas changes linearly with volume during this process and thus the process curve on a PV diagram will be a straight line Then the boundary work during this process is simply the area under the process curve which is a trapezoidal Thus 135 kJ 3 3 1 2 2 1 out b kPa m 1 1 kJ 015m 045 2 800kPa 00 1 2 Area V P V P W Air 100 kPa 015 m3 Q b If there were no spring we would have a constant pressure process at P 100 kPa The work done during this process is 2 100 1 P kPa kJ 30 kPa m 1 1 kJ 015 m kg 100 kPa 045 3 3 1 2 2 1 outno spring b V V V P Pd W 800 Thus V m3 015 045 105 kJ 30 135 bno spring spring W W W b PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 488 4126 A cylinder equipped with a set of stops for the piston is initially filled with saturated liquidvapor mixture of water a specified pressure Heat is transferred to the water until the volume increases by 20 The initial and final temperature the mass of the liquid when the piston starts moving and the work done during the process are to be determined and the process is to be shown on a Pv diagram Assumptions The process is quasiequilibrium Analysis a Initially the system is a saturated mixture at 125 kPa pressure and thus the initial temperature is 1060C sat125 kPa 1 T T H2O 5 kg The total initial volume is 3 1 4127 m 1 3750 3 0 001048 2 g g f f m m v v V Then the total and specific volumes at the final state are 09905 m kg 5 kg m 4953 4953 m 12 4127 21 3 3 3 3 3 1 3 m V v V V P Thus 2 1 3 3736C 3 3 3 3 9905 m kg 0 300 kPa T v P b When the piston first starts moving P2 300 kPa and V2 V1 4127 m3 The specific volume at this state is v 08254 m kg 5 kg 4127 m 3 3 2 2 m V v which is greater than vg 060582 m3kg at 300 kPa Thus no liquid is left in the cylinder when the piston starts moving c No work is done during process 12 since V1 V2 The pressure remains constant during process 23 and the work done during this process is 2476 kJ 3 3 2 3 2 3 2 kPa m 1 1 kJ 4 127 m 300 kPa 4953 V V V P Pd Wb PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 489 4127E A spherical balloon is initially filled with air at a specified state The pressure inside is proportional to the square of the diameter Heat is transferred to the air until the volume doubles The work done is to be determined Assumptions 1 Air is an ideal gas 2 The process is quasiequilibrium Properties The gas constant of air is R 006855 BtulbmR Table A1E Analysis The dependence of pressure on volume can be expressed as 2 3 2 2 1 3 3 6 6 6 1 π π π V V V k kD P D P D D AIR 7 lbm 30 psia 600 R or 2 3 2 2 2 3 1 1 2 3 6 V V P P k π Also 1 587 2 2 3 3 2 1 2 1 2 V V P P and 1905 R 600 R 2 1587 1 1 1 2 2 2 2 2 2 1 1 1 T P P T T P T P V V V V Thus 376 Btu 600R 5 7 lbm006855 Btulbm R1905 3 5 3 5 3 6 5 3 6 1 2 1 1 2 2 5 3 1 5 3 2 2 3 2 1 2 3 2 1 T T mR P P k d k Pd Wb V V V V V V V π π PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 490 4128E Problem 4127E is reconsidered Using the integration feature the work done is to be determined and compared to the hand calculated result Analysis The problem is solved using EES and the solution is given below N2 m7 lbm P130 psia T1600 R V22V1 RRumolarmassair Ru1545 ftlbflbmolR P1Convertpsialbfft2V1mRT1 V14piD1233 CP1D1N D1D23V1V2 P2CD2N P2Convertpsialbfft2V2mRT2 PCDNConvertpsialbfft2 V4piD233 WboundaryEESintegralPVV1V2convertftlbfBtu WboundaryHANDpiC2N3D2N3D1N3convertftlbfBtuconvertft2in2 0 05 1 15 2 25 3 275 310 345 380 415 450 N WboundaryEES Btu N Wboundary Btu 0 03333 06667 1 1333 1667 2 2333 2667 3 2879 3006 314 3281 3431 3589 3756 3933 412 4318 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 491 4129 A cylinder is initially filled with saturated R134a vapor at a specified pressure The refrigerant is heated both electrically and by heat transfer at constant pressure for 6 min The electric current is to be determined and the process is to be shown on a Tv diagram Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are negligible 2 The thermal energy stored in the cylinder itself and the wires is negligible 3 The compression or expansion process is quasi equilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as R134a P const We V KE PE 0 since 1 2 in 1 2 ein in ein in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in h m h I t Q h m h W Q Q U W W Q E E E out b 43 42 1 4243 1 since U Wb H during a constant pressure quasiequilibrium process The properties of R134a are Tables A11 through A13 T 1 2 31451 kJkg C 70 kPa 240 24728 kJkg vapor sat kPa 240 2 1 2 240kPa 1 1 h T P h h P g v Substituting 128 A I I 1 kJs 24728 kJkg 1000 VA 12 kg 31451 60 s 6 110 V 300000 VAs PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 492 4130 Saturated water vapor contained in a springloaded pistoncylinder device is condensed until it is saturated liquid at a specified temperature The heat transfer is to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or expansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as since KE PE 0 1 2 out 1 2 in out 1 2 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u u w q u m u W Q u m u U Q W E E E in b b b 43 42 1 4243 1 P 1 2 v 1235 kPa 1555 kPa The properties at the initial and final states are Table A4 2 kJkg 2594 1272 m kg 0 kPa 1555 200 C 1 3 200 C 1 sat 200 C 1 g g u u P P v v 33 kJkg 209 001012 m kg 0 35 kPa 12 50 C 2 3 50 C 2 sat 50 C 2 f f u u P P v v Since this is a linear process the work done is equal to the area under the process line 12 989 kJkg kPa m 1 1kJ 0 1272 m kg 0001012 2 1235kPa 1555 2 Area 3 3 1 2 2 1 out v v P P wb Thus 989 kJkg in b w Substituting into energy balance equation gives 2484 kJkg 2594 2 kJkg 20933 98 9 kJkg 1 2 out u u w q b in PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 493 4131 A cylinder is initially filled with helium gas at a specified state Helium is compressed polytropically to a specified temperature and pressure The heat transfer during the process is to be determined Assumptions 1 Helium is an ideal gas with constant specific heats 2 The cylinder is stationary and thus the kinetic and potential energy changes are negligible 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or expansion process is quasiequilibrium Properties The gas constant of helium is R 20769 kPam3kgK Table A1 Also cv 31156 kJkgK Table A2 Analysis The mass of helium and the exponent n are determined to be 1 536 0 264 50 150 400 0264 m 05 m 400 kPa 150 kPa 293 K K 413 0123 kg kPa m kg K 293 K 20769 kPa 05 m 150 2 1 1 2 1 1 2 2 3 3 1 2 1 1 2 2 2 2 2 1 1 1 3 3 1 1 1 n P P P P T P T P RT P RT P RT P m n n n n V V V V V V V V V He PV n C Q Then the boundary work for this polytropic process can be determined from 572 kJ 1 1536 293K 0123 kg20769 kJkg K413 1 1 1 2 1 1 2 2 2 1 in b n T mR T n P P Pd W V V V We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves Taking the direction of heat transfer to be to the cylinder the energy balance for this stationary closed system can be expressed as bin 1 2 bin 1 2 in 1 2 bin in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in W T T mc W u m u Q u m u U W Q E E E v 43 42 1 4243 1 Substituting Qin 0123 kg31156 kJkgK413 293K 572 kJ 112 kJ The negative sign indicates that heat is lost from the system PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 494 4132 Nitrogen gas is expanded in a polytropic process The work done and the heat transfer are to be determined Assumptions 1 Nitrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 1262 K and 339 MPa 2 The kinetic and potential energy changes are negligible 0 pe ke 3 Constant specific heats can be used Properties The properties of nitrogen are R 02968 kJkgK and cv 0743 kJkgK Table A2a Analysis We take nitrogen in the cylinder as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as N2 2 MPa 1200 K Q 1 2 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc U W Q E E E b v 43 42 1 4243 1 Using the boundary work relation for the polytropic process of an ideal gas gives kJkg 4041 1 2000 120 1145 0 2968 kJkg K1200 K 1 1 0 45 1 45 1 1 2 1 out n n b P P n RT w The temperature at the final state is 587 3 K 2000 kPa 200 kPa 1200 K 0 45 1 45 1 1 2 1 2 n n P P T T Substituting into the energy balance equation 511 kJkg 1200K 0 743 kJkg K5873 404 1 kJkg 1 2 out in T T c w q b v The negative sign indicates that heat is lost from the device That is out 511 kJkg q 4133 A cylinder and a rigid tank initially contain the same amount of an ideal gas at the same state The temperature of both systems is to be raised by the same amount The amount of extra heat that must be transferred to the cylinder is to be determined Analysis In the absence of any work interactions other than the boundary work the H and U represent the heat transfer for ideal gases for constant pressure and constant volume processes respectively Thus the extra heat that must be supplied to the air maintained at constant pressure is mR T T c m c T mc T mc U H Q p p in extra v v IDEAL GAS P const Q where IDEAL GAS V const Q R R M u 8314 kJ kmol K 25 kg kmol 03326 kJ kg K Substituting Qin extra 12 kg03326 kJkgK15 K 599 kJ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 495 4134 The heating of a passive solar house at night is to be assisted by solar heated water The length of time that the electric heating system would run that night with or without solar heating are to be determined Assumptions 1 Water is an incompressible substance with constant specific heats 2 The energy stored in the glass containers themselves is negligible relative to the energy stored in water 3 The house is maintained at 22C at all times Properties The density and specific heat of water at room temperature are ρ 1 kgL and c 418 kJkgC Table A3 Analysis a The total mass of water is 80C water 50000 kJh 22C 1000 kg 1 kgL 50 20 L ρV mw Taking the contents of the house including the water as our system the energy balance relation can be written as 1 water 2 water air water out in e potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T U U U U Q W E E E 43 42 1 4243 1 or water 1 2 out ein T mc T Q t W Substituting 15 kJst 50000 kJh10 h 1000 kg418 kJkgC22 80C It gives t 17170 s 477 h b If the house incorporated no solar heating the energy balance relation above would simplify further to 0 out ein Q t W Substituting 15 kJst 50000 kJh10 h 0 It gives t 33333 s 926 h PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 496 4135 An electric resistance heater is immersed in water The time it will take for the electric heater to raise the water temperature to a specified temperature is to be determined Assumptions 1 Water is an incompressible substance with constant specific heats 2 The energy stored in the container itself and the heater is negligible 3 Heat loss from the container is negligible Properties The specific heat of water at room temperature is c 418 kJkgC Table A3 Analysis Taking the water in the container as the system the energy balance can be expressed as 1 water 2 in e water in e potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T t W U W E E E 43 42 1 4243 1 Resistance Heater Water Substituting 1800 Jst 40 kg4180 JkgC80 20C Solving for t gives t 5573 s 929 min 155 h 4136 One ton of liquid water at 50C is brought into a room The final equilibrium temperature in the room is to be determined Assumptions 1 The room is well insulated and well sealed 2 The thermal properties of water and air are constant Properties The gas constant of air is R 0287 kPam3kgK Table A1 The specific heat of water at room temperature is c 418 kJkgC Table A3 Analysis The volume and the mass of the air in the room are ROOM 15C 95 kPa 4 m 5 m 6 m Heat Water 50C V 4 5 6 120 m³ 1379 kg kPa m kg K 288 K 02870 kPa 120 m 95 3 3 1 1 1 air RT P m V Taking the contents of the room including the water as our system the energy balance can be written as air water potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 0 U U U E E E 43 42 1 4243 1 or 0 air 1 2 water 1 2 T mc T T mc T v Substituting 0 15 C C 1379 kg0718 kJkg 50 C C 1000 kg418 kJkg f f T T It gives Tf 492C where Tf is the final equilibrium temperature in the room PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4137 A room is to be heated by 1 ton of hot water contained in a tank placed in the room The minimum initial temperature of the water is to be determined if it to meet the heating requirements of this room for a 24h period Assumptions 1 Water is an incompressible substance with constant specific heats 2 Air is an ideal gas with constant specific heats 3 The energy stored in the container itself is negligible relative to the energy stored in water 4 The room is maintained at 20C at all times 5 The hot water is to meet the heating requirements of this room for a 24h period Properties The specific heat of water at room temperature is c 418 kJkgC Table A3 Analysis Heat loss from the room during a 24h period is Qloss 8000 kJh24 h 192000 kJ Taking the contents of the room including the water as our system the energy balance can be written as Ėin Ėout ΔĖsystem Net energy transfer by heat work and mass Change in internal kinetic potential etc energies Qout ΔU ΔUwater ΔUair 0 or Qout mcT2 T1water Substituting 192000 kJ 1000 kg418 kJkgC20 T1 It gives T1 659C where T1 is the temperature of the water when it is first brought into the room 498 4138 A sample of a food is burned in a bomb calorimeter and the water temperature rises by 32C when equilibrium is established The energy content of the food is to be determined Assumptions 1 Water is an incompressible substance with constant specific heats 2 Air is an ideal gas with constant specific heats 3 The energy stored in the reaction chamber is negligible relative to the energy stored in water 4 The energy supplied by the mixer is negligible Properties The specific heat of water at room temperature is c 418 kJkgC Table A3 The constant volume specific heat of air at room temperature is cv 0718 kJkgC Table A2 Analysis The chemical energy released during the combustion of the sample is transferred to the water as heat Therefore disregarding the change in the sensible energy of the reaction chamber the energy content of the food is simply the heat transferred to the water Taking the water as our system the energy balance can be written as in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in U Q E E E 43 42 1 4243 1 T 32C Water Reaction chamber Food or 1 water 2 water T mc T U Qin Substituting Qin 3 kg418 kJkgC32C 4013 kJ for a 2g sample Then the energy content of the food per unit mass is 20060 kJkg 1 kg 1000 g 2 g 4013 kJ To make a rough estimate of the error involved in neglecting the thermal energy stored in the reaction chamber we treat the entire mass within the chamber as air and determine the change in sensible internal energy 023 kJ 0102 kg 0718 kJkg C 32 C chamber 1 2 chamber o o T T mc U v which is less than 1 of the internal energy change of water Therefore it is reasonable to disregard the change in the sensible energy content of the reaction chamber in the analysis PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 499 4139 A man drinks one liter of cold water at 3C in an effort to cool down The drop in the average body temperature of the person under the influence of this cold water is to be determined Assumptions 1 Thermal properties of the body and water are constant 2 The effect of metabolic heat generation and the heat loss from the body during that time period are negligible Properties The density of water is very nearly 1 kgL and the specific heat of water at room temperature is c 418 kJkgC Table A3 The average specific heat of human body is given to be 36 kJkgC Analysis The mass of the water is 1 kg 1 L 1 kgL ρV mw We take the man and the water as our system and disregard any heat and mass transfer and chemical reactions Of course these assumptions may be acceptable only for very short time periods such as the time it takes to drink the water Then the energy balance can be written as water body potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 0 U U U E E E 43 42 1 4243 1 or 0 water 1 2 body 1 2 T mc T T mc T Substituting 0 3 C 1 kg418 kJkg C 39 C 68 kg36 kJkg C o o o o f f T T It gives Tf 384C Then T39 384 06C Therefore the average body temperature of this person should drop about half a degree celsius PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4100 4140 A 03L glass of water at 20C is to be cooled with ice to 5C The amount of ice or cold water that needs to be added to the water is to be determined Assumptions 1 Thermal properties of the ice and water are constant 2 Heat transfer to the glass is negligible 3 There is no stirring by hand or a mechanical device it will add energy Properties The density of water is 1 kgL and the specific heat of water at room temperature is c 418 kJkgC Table A3 The specific heat of ice at about 0C is c 211 kJkgC Table A3 The melting temperature and the heat of fusion of ice at 1 atm are 0C and 3337 kJkg Analysis a The mass of the water is Ice cubes 0C Water 20C 03 L 03 kg L 30 1 kgL V ρ mw We take the ice and the water as our system and disregard any heat and mass transfer This is a reasonable assumption since the time period of the process is very short Then the energy balance can be written as water ice potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 0 0 U U U E E E 43 42 1 4243 1 0 0 C 0 C water 1 2 liquid ice 2 1 solid T mc T mc T mh T mc if o o Noting that T1 ice 0C and T2 5C and substituting gives m0 3337 kJkg 418 kJkgC50C 03 kg418 kJkgC520C 0 m 00546 kg 546 g b When T1 ice 20C instead of 0C substituting gives m211 kJkgC020C 3337 kJkg 418 kJkgC50C 03 kg418 kJkgC520C 0 m 00487 kg 487 g Cooling with cold water can be handled the same way All we need to do is replace the terms for ice by a term for cold water at 0C 0 0 water 1 2 coldwater 1 2 water coldwater T mc T T T mc U U Substituting mcold water 418 kJkgC5 0C 03 kg418 kJkgC520C 0 It gives m 09 kg 900 g Discussion Note that this is about 16 times the amount of ice needed and it explains why we use ice instead of water to cool drinks Also the temperature of ice does not seem to make a significant difference PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4101 4141 Problem 4140 is reconsidered The effect of the initial temperature of the ice on the final mass of ice required as the ice temperature varies from 26C to 0C is to be investigated The mass of ice is to be plotted against the initial temperature of ice Analysis The problem is solved using EES and the solution is given below Knowns rhowater 1 kgL V 03 L T1ice 0 C T1 20 C T2 5 C Cice 211 kJkgC Cwater 418 kJkgC hif 3337 kJkg T1ColdWater 0 C mwater rhowaterV kg The mass of the water The system is the water plus the ice Assume a short time period and neglect any heat and mass transfer The energy balance becomes Ein Eout DELTAEsys kJ Ein 0 kJ Eout 0kJ DELTAEsys DELTAUwaterDELTAUicekJ DELTAUwater mwaterCwaterT2 T1kJ DELTAUice DELTAUsolidiceDELTAUmeltedicekJ DELTAUsolidice miceCice0T1ice micehifkJ DELTAUmeltedicemiceCwaterT2 0kJ micegramsmiceconvertkggg Cooling with Cold Water DELTAEsys DELTAUwaterDELTAUColdWaterkJ DELTAUwater mwaterCwaterT2ColdWater T1kJ DELTAUColdWater mColdWaterCwaterT2ColdWater T1ColdWaterkJ mColdWatergramsmColdWaterconvertkggg T1ice C micegrams g 26 24 22 20 18 16 14 12 10 8 6 4 2 0 4594 4642 4691 474 4791 4843 4897 4951 5007 5064 5122 5181 5242 5305 24 20 16 12 8 4 0 45 46 47 48 49 50 51 52 53 54 T1ice C micegrams g PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4102 4142 A rigid tank filled with air is connected to a cylinder with zero clearance The valve is opened and air is allowed to flow into the cylinder The temperature is maintained at 30C at all times The amount of heat transfer with the surroundings is to be determined Assumptions 1 Air is an ideal gas 2 The kinetic and potential energy changes are negligible ke pe 0 3 There are no work interactions involved other than the boundary work Properties The gas constant of air is R 0287 kPam3kgK Table A1 Analysis We take the entire air in the tank and the cylinder to be the system This is a closed system since no mass crosses the boundary of the system The energy balance for this closed system can be expressed as bout in 1 2 bout in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 0 W Q u m u U W Q E E E 43 42 1 4243 1 Air T 30C Q since u uT for ideal gases and thus u2 u1 when T1 T2 The initial volume of air is 3 3 1 1 2 2 1 2 2 2 2 1 1 1 080 m 1 04 m 200 kPa 400 kPa V V V V T T P P T P T P The pressure at the piston face always remains constant at 200 kPa Thus the boundary work done during this process is 80 kJ kPa m 1 1 kJ 04m 200 kPa08 3 3 1 2 2 2 1 bout V V V P Pd W Therefore the heat transfer is determined from the energy balance to be 80 kJ in bout Q W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4103 4143 A wellinsulated room is heated by a steam radiator and the warm air is distributed by a fan The average temperature in the room after 30 min is to be determined Assumptions 1 Air is an ideal gas with constant specific heats at room temperature 2 The kinetic and potential energy changes are negligible 3 The air pressure in the room remains constant and thus the air expands as it is heated and some warm air escapes Properties The gas constant of air is R 0287 kPam3kgK Table A1 Also cp 1005 kJkgK for air at room temperature Table A2 Analysis We first take the radiator as the system This is a closed system since no mass enters or leaves The energy balance for this closed system can be expressed as KE PE 0 since 2 1 out 1 2 out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u m u Q W u m u U Q E E E 43 42 1 4243 1 Steam radiator 10C 4 m 4 m 5 m Using data from the steam tables Tables A4 through A6 some properties are determined to be 20882 kJkg 40 417 16941 m kg 0 001043 kPa 100 26546 kJkg 108049 m kg C 200 kPa 200 3 1 2 2 1 3 1 1 1 fg f g f u u P u T P v v v v v 00139 kg m kg 108049 m 0015 17487 kJkg 06376 20882 41740 0 6376 0 001043 1 6941 0 001043 08049 1 3 3 1 1 2 2 2 2 v V v v v m x u u u x fg f fg f Substituting Qout 00139 kg 26546 17487kJkg 1258 kJ The volume and the mass of the air in the room are V 445 80 m3 and 985 kg kPa m kg K 283 K 02870 kPa 80 m 100 3 3 1 1 1 air RT P m V The amount of fan work done in 30 min is 216kJ 0 120 kJs30 60 s fanin fanin t W W We now take the air in the room as the system The energy balance for this closed system is expressed as 1 2 fanin in bout fanin in system out in T T mc H W Q U W W Q E E E p since the boundary work and U combine into H for a constant pressure expansion or compression process It can also be expressed as 1 2 avg fanin in T T mc t W Q p Substituting 1258 kJ 216 kJ 985 kg1005 kJkgCT2 10C which yields T2 123C Therefore the air temperature in the room rises from 10C to 123C in 30 min PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4104 4144 An insulated cylinder is divided into two parts One side of the cylinder contains N2 gas and the other side contains He gas at different states The final equilibrium temperature in the cylinder when thermal equilibrium is established is to be determined for the cases of the piston being fixed and moving freely Assumptions 1 Both N2 and He are ideal gases with constant specific heats 2 The energy stored in the container itself is negligible 3 The cylinder is wellinsulated and thus heat transfer is negligible Properties The gas constants and the constant volume specific heats are R 02968 kPam3kgK is cv 0743 kJkgC for N2 and R 20769 kPam3kgK is cv 31156 kJkgC for He Tables A1 and A2 Analysis The mass of each gas in the cylinder is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 07691 kg kPa m kg K 313 K 20769 1 m kPa 500 4 287 kg kPa m kg K 393 K 02968 1 m kPa 500 3 3 He 1 1 1 He 3 3 N 1 1 1 N 2 2 RT P m RT P m V V Taking the entire contents of the cylinder as our system the 1st law relation can be written as He 1 m3 500 kPa 40C N2 1 m3 500 kPa 120C He 1 2 N 1 2 He N potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 0 0 2 2 T mc T T T mc U U U E E E v v 43 42 1 4243 1 Substituting 0 40 C C 07691 kg 31156 kJkg 120 C C 4287 kg 0743 kJkg f f T T It gives Tf 857C where Tf is the final equilibrium temperature in the cylinder The answer would be the same if the piston were not free to move since it would effect only pressure and not the specific heats Discussion Using the relation PV NRuT it can be shown that the total number of moles in the cylinder is 0153 0192 0345 kmol and the final pressure is 515 kPa preparation If you are a student using this Manual you are using it without permission 4105 4145 An insulated cylinder is divided into two parts One side of the cylinder contains N2 gas and the other side contains He gas at different states The final equilibrium temperature in the cylinder when thermal equilibrium is established is to be determined for the cases of the piston being fixed and moving freely Assumptions 1 Both N2 and He are ideal gases with constant specific heats 2 The energy stored in the container itself except the piston is negligible 3 The cylinder is wellinsulated and thus heat transfer is negligible 4 Initially the piston is at the average temperature of the two gases Properties The gas constants and the constant volume specific heats are R 02968 kPam3kgK is cv 0743 kJkgC for N2 and R 20769 kPam3kgK is cv 31156 kJkgC for He Tables A1 and A2 The specific heat of copper piston is c 0386 kJkgC Table A3 Analysis The mass of each gas in the cylinder is He 1 m3 500 kPa 40C N2 1 m3 500 kPa 120C 07691 kg kPa m kg K 313 K 20769 1 m kPa 500 4 287 kg kPa m kg K 393 K 02968 1 m kPa 500 3 3 He 1 1 1 He 3 3 N 1 1 1 N 2 2 RT P m RT P m V V Copper Taking the entire contents of the cylinder as our system the 1st law relation can be written as 0 0 Cu 1 2 He 1 2 N 1 2 Cu He N potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 2 2 T mc T T mc T T T mc U U U U E E E v v 43 42 1 4243 1 where T1 Cu 120 40 2 80C Substituting 0 80 C kg 0386 kJkg C 8 40 C 07691 kg 31156 kJkg C 120 C kg 0743 kJkg C 4287 f f f T T T o o o It gives Tf 837C where Tf is the final equilibrium temperature in the cylinder The answer would be the same if the piston were not free to move since it would effect only pressure and not the specific heats PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4106 4146 Problem 4145 is reconsidered The effect of the mass of the copper piston on the final equilibrium temperature as the mass of piston varies from 1 kg to 10 kg is to be investigated The final temperature is to be plotted against the mass of piston Analysis The problem is solved using EES and the solution is given below Knowns Ru8314 kJkmolK VN211 m3 CvN20743 kJkgK From Table A2a at 27C RN202968 kJkgK From Table A2a TN21120 C PN21500 kPa VHe11 m3 CvHe31156 kJkgK From Table A2a at 27C THe140 C PHe1500 kPa RHe20769 kJkgK From Table A2a mPist8 kg CvPist0386 kJkgK Use Cp for Copper from Table A3b at 27C Solution mass calculations PN21VN21mN2RN2TN21273 PHe1VHe1mHeRHeTHe1273 The entire cylinder is considered to be a closed system neglecting the piston Conservation of Energy for the closed system Ein Eout DELTAEnegPist we neglect DELTA KE and DELTA PE for the cylinder Ein Eout DELTAEneglPist Ein 0 kJ Eout 0 kJ At the final equilibrium state N2 and He will have a common temperature DELTAEneglPist mN2CvN2T2neglPistTN21mHeCvHeT2neglPistTHe1 The entire cylinder is considered to be a closed system including the piston Conservation of Energy for the closed system Ein Eout DELTAEwithPist we neglect DELTA KE and DELTA PE for the cylinder Ein Eout DELTAEwithPist At the final equilibrium state N2 and He will have a common temperature DELTAEwithPist mN2CvN2T2withPistTN21mHeCvHeT2withPist THe1mPistCvPistT2withPistTPist1 TPist1TN21THe12 Total volume of gases VtotalVN21VHe1 Final pressure at equilibrium Neglecting effect of piston P2 is P2neglPistVtotalNtotalRuT2neglPist273 Including effect of piston P2 is NtotalmN2molarmassnitrogenmHemolarmassHelium P2withPistVtotalNtotalRuT2withPist273 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4107 mPist kg T2neglPist C T2withPist C 1 2 3 4 5 6 7 8 9 10 8565 8565 8565 8565 8565 8565 8565 8565 8565 8565 8529 8496 8468 8443 842 8399 8381 8364 8348 8334 1 2 3 4 5 6 7 8 9 10 83 835 84 845 85 855 86 mPist kg T2 C Without piston With piston PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4108 4147 A pistoncylinder device initially contains saturated liquid water An electric resistor placed in the tank is turned on until the tank contains saturated water vapor The volume of the tank the final temperature and the power rating of the resistor are to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions Properties The initial properties of steam are Table A4 9 kPa 1554 85226 kJkg 001157 m kg 0 0 C 200 1 1 3 1 1 1 P h x T v Analysis a We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves Noting that the volume of the system is constant and thus there is no boundary work the energy balance for this stationary closed system can be expressed as KE PE 0 since 1 2 bout in e 1 2 bout in e potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in h m h H U W W Q u m u U W W E E E 43 42 1 4243 1 We Water 14 kg 200C sat liq since H U W bout for a constantpressure process The initial and final volumes are 0006476 m3 0001619 m 4 0 001619 m kg0001157 m kg 41 3 2 3 3 1 1 V v V m b Now the final state can be fixed by calculating specific volume 0 004626 m kg 14 kg 0006476 m 3 3 2 2 m V v The final state is saturated mixture and both pressure and temperature remain constant during the process Other properties are Table A4 or EES 02752 0 90565 kJkg 004626 m kg 0 9 kPa 1554 2 2 1 2 3 2 1 2 x h T T P P C 200 v c Substituting 7475 kJ 85226kJkg kg90565 41 ein W Finally the power rating of the resistor is 00623 kW 20 60 s 7475 kJ ein ein t W W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4109 4148 A pistoncylinder device contains an ideal gas An external shaft connected to the piston exerts a force For an isothermal process of the ideal gas the amount of heat transfer the final pressure and the distance that the piston is displaced are to be determined PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Assumptions 1 The kinetic and potential energy changes are negligible 2 The friction between the piston and the cylinder is negligible ke pe 0 Analysis a We take the ideal gas in the cylinder to be the system This is a closed system since no mass crosses the system boundary The energy balance for this stationary closed system can be expressed as out in b 1 2 1 ideal gas 2 ideal gas out in b potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 0 and 0 since Q W PE KE T T T T mc U Q W E E E v 43 42 1 4243 1 W GAS 1 bar 24C Q Thus the amount of heat transfer is equal to the boundary work input 01 kJ bin out W Q b The relation for the isothermal work of an ideal gas may be used to determine the final volume in the cylinder But we first calculate initial volume 3 2 1 2 1 0 002262 m m 20 4 012 m 4 π π L D V Then 3 2 3 2 3 1 2 1 1 in b 0 001454 m 0002262 m 100 kPa0002262 m ln kJ 10 ln V V V V PV W The final pressure can be determined from ideal gas relation applied for an isothermal process 1556 kPa 2 3 2 3 2 2 1 1 m 0001454 100 kPa0002262 m P P P P V V c The final position of the piston and the distance that the piston is displaced are cm 71 0 07146 m 0 1285 20 0 0 1285 m 4 012 m 0 001454 m 4 2 1 2 2 2 3 2 2 2 L L L L L L D π π V preparation If you are a student using this Manual you are using it without permission 4110 4149 A pistoncylinder device with a set of stops contains superheated steam Heat is lost from the steam The pressure and quality if mixture the boundary work and the heat transfer until the piston first hits the stops and the total heat transfer are to be determined Assumptions 1 The kinetic and potential energy changes are negligible 0 pe ke 2 The friction between the piston and the cylinder is negligible Steam 035 kg 35 MPa Q Analysis a We take the steam in the cylinder to be the system This is a closed system since no mass crosses the system boundary The energy balance for this stationary closed system can be expressed as 0 PE since KE out in b potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in U Q W E E E 43 42 1 4243 1 Denoting when piston first hits the stops as state 2 and the final state as 3 the energy balance relations may be written as u u 1 3 out13 in b 1 2 out12 bin m u Q W m u Q W The properties of steam at various states are Tables A4 through A6 250 C 47 56 242 56 C 242 sat 1 1 sat35 MPa T T T T 9 kJkg 2623 0 05875 m kg C 250 MPa 53 1 3 1 1 1 u T P v 4 kJkg 1045 0 001235 m kg 0 MPa 53 2 3 2 2 1 2 u x P P v 85155 kJkg C 200 001235 m kg 0 3 3 3 3 3 2 3 u P x T 1555 kPa 000062 v v b Noting that the pressure is constant until the piston hits the stops during which the boundary work is done it can be determined from its definition as 7045 kJ 3 2 1 1 bin 0001235m 0 35 kg3500 kPa005875 v mP v W c Substituting into energy balance relations 6229 kJ 2623 9 kJkg 0 35 kg1045 4 7045 kJ Qout12 d 6908 kJ 2623 9 kJkg 0 35 kg85155 7045 kJ Qout13 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4111 4150 An insulated rigid tank is divided into two compartments each compartment containing the same ideal gas at different states The two gases are allowed to mix The simplest expression for the mixture temperature in a specified format is to be obtained Analysis We take the both compartments together as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as 2 2 1 1 3 2 1 2 3 2 1 3 1 potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 0 KE PE 0 since 0 m T m T m T m T m c T T m c T W Q U E E E v v 43 42 1 4243 1 m2 T2 m1 T1 and 2 1 3 m m m Solving for final temperature we find 2 3 2 1 3 1 3 m T m m T m T 4151 A relation for the explosive energy of a fluid is given A relation is to be obtained for the explosive energy of an ideal gas and the value for air at a specified state is to be evaluated Properties The specific heat ratio for air at room temperature is k 14 Analysis The explosive energy per unit volume is given as e u u v explosion 1 2 1 For an ideal gas u1 u2 cvT1 T2 1 1 1 P RT R c c p v v and thus 1 1 1 1 k c c c c c R c p p v v v v Substituting 1 2 1 1 1 2 1 explosion 1 1 T T k P P RT T T c e v which is the desired result Using the relation above the total explosive energy of 20 m³ of air at 5 MPa and 100C when the surroundings are at 20C is determined to be 53619 kJ kPa m 1 1kJ 373 K 293 K 1 1 41 5000 kPa 20 m 1 1 3 3 1 2 1 1 explosion explosion T T k P e E V V PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4112 4152 Using the relation for explosive energy given in the previous problem the explosive energy of steam and its TNT equivalent at a specified state are to be determined Assumptions Steam condenses and becomes a liquid at room temperature after the explosion Properties The properties of steam at the initial and the final states are Table A4 through A6 10483 kJkg liquid Comp C 25 30470 kJkg m kg 0032811 C 500 MPa 10 25 C 2 2 1 3 1 1 1 o uf u T u T P v STEAM 10 MPa 500C 25C Analysis The mass of the steam is 6096 kg m kg 0032811 m 20 3 3 1 v V m Then the total explosive energy of the steam is determined from 1793436 kJ 10483 kJkg 6096 kg 30470 2 1 explosive u m u E which is equivalent to 3250 kJkg of TNT 5518 kg of TNT 1793436 kJ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4113 4153 Carbon dioxide is compressed polytropically in a pistoncylinder device The final temperature is to be determined treating the carbon dioxide as an ideal gas and a van der Waals gas Assumptions The process is quasiequilibrium Properties The gas constant of carbon dioxide is R 01889 kJkgK Table A1 Analysis a The initial specific volume is CO2 1 MPa 200C PV 15 const 0 08935 m kg 1000 kPa 01889 kJkg K473 K 3 1 1 1 P RT v From polytropic process expression 004295 m kg 3000 008935 m kg 1000 3 51 1 3 1 2 1 1 2 n P P v v The final temperature is then 6821 K 0 1889 kJkg K 3000 kPa004295 m kg 3 2 2 2 R P T v b The van der Waals equation of state for carbon dioxide is R T P u 0 0428 8 365 2 v v When this is applied to the initial state the result is 8 314473 0 0428 365 8 1000 1 12 v v whose solution by iteration or by EES is 3 882 m kmol 3 v1 The final molar specific volume is then 1 866 m kmol 3000 3882 m kmol 1000 3 51 1 3 1 2 1 1 2 n P P v v Substitution of the final molar specific volume into the van der Waals equation of state produces 6809 K 0 0428 1 866 866 1 365 8 8 314 3000 1 0 0428 365 8 1 2 2 2 v v P R T u PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4114 4154 Carbon dioxide contained in a springloaded pistoncylinder device is compressed in a polytropic process The final temperature is to be determined using the compressibility factor Properties The gas constant the critical pressure and the critical temperature of CO2 are from Table A1 R 01889 kPam3kgK Tcr 3042 K Pcr 739 MPa Analysis From the compressibility chart at the initial state Fig A15 or EES CO2 1 MPa 200C 0 991 Z 0 135 739 MPa MPa 1 155 3042 K K 473 1 cr 1 1 cr 1 1 P P P T T T R R The specific volume does not change during the process Then 0 08855 m kg 1000 kPa 099101889 kPa m kg K473 K 3 3 1 1 1 2 1 P Z RT v v At the final state 01 11 4 kPa m kg K3042 K7390 kPa 01889 m kg 008855 0 406 739 MPa MPa 3 2 3 3 cr cr 2actual 2 cr 2 2 Z P RT P P P R R v v Thus 1406 K 89 kPa m kg K 10018 kPa008855 m kg 3000 3 3 2 2 2 2 Z R P T v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4115 4155E Two adiabatic tanks containing water at different states are connected by a valve The valve is now opened allowing the water vapor from the highpressure tank to move to the lowpressure tank until the pressure in the two becomes equal The final pressure and the final mass in each tank are to be determined Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero 2 There are no heat or work interactions involved Analysis We take both tanks as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as B B A A B B A A u m u m u m u m U U U U U E E E 2 2 2 2 1 1 1 1 2 1 1 2 potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 0 0 43 42 1 4243 1 Water 10 ft3 15 psia x 075 Water 10 ft3 450 psia x 010 where the highpressure and lowpressure tanks are denoted by A and B respectively The specific volume in each tank is 50402 Btulbm 0 1068352 43567 0 12084 ft lbm 0 01955 0 10 1 0324 0 01955 10 0 psia 450 1 3 1 1 1 fg f A fg f A A A xu u u x x P v v v 85355 Btulbm 0 7589652 18116 19727 ft lbm 0 01672 0 7526297 0 01672 75 0 psia 15 1 3 1 1 1 fg f B fg f B B B xu u u x x P v v v The total mass contained in the tanks is 8326 lbm 0 5069 8275 727 ft lbm 19 ft 10 12084 ft lbm 0 ft 10 3 3 3 3 1 1 B B A A m v V v V Similarly the initial total internal energy contained in both tanks is 42155 kJ 0 506985355 827550402 1 1 1 1 1 B B A A u m u m U The internal energy and the specific volume are 506 3 Btulbm 8326 lbm 42155 Btu 1 2 1 m U u u 0 2402 ft lbm 8326 lbm 20 ft 3 3 2 1 m V v v Now the final state is fixed The pressure in this state may be obtained by iteration in water tables Table A5E We used EES to get the exact result 2 313 psia P Since the specific volume is now the same in both tanks and both tanks have the same volume the mass is equally divided between the tanks at the end of this process 4163 lbm 2 8326 lbm 2 2 2 m m m B A PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4116 Fundamentals of Engineering FE Exam Problems 4156 A frictionless pistoncylinder device and a rigid tank contain 3 kmol of an ideal gas at the same temperature pressure and volume Now heat is transferred and the temperature of both systems is raised by 10C The amount of extra heat that must be supplied to the gas in the cylinder that is maintained at constant pressure is a 0 kJ b 27 kJ c 83 kJ d 249 kJ e 300 kJ Answer d 249 kJ Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Note that CpCvR and thus QdiffmRdTNRudT N3 kmol Ru8314 kJkmolK Tchange10 QdiffNRuTchange Some Wrong Solutions with Common Mistakes W1Qdiff0 Assuming they are the same W2QdiffRuTchange Not using mole numbers W3QdiffRuTchangeN Dividing by N instead of multiplying W4QdiffNRairTchange Rair0287 using R instead of Ru 4157 The specific heat of a material is given in a strange unit to be C 360 kJkgF The specific heat of this material in the SI units of kJkgC is a 200 kJkgC b 320 kJkgC c 360 kJkgC d 480 kJkgC e 648 kJkgC Answer e 648 kJkgC Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values C360 kJkgF CSIC18 kJkgC Some Wrong Solutions with Common Mistakes W1CC Assuming they are the same W2CC18 Dividing by 18 instead of multiplying PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4117 4158 A 3m3 rigid tank contains nitrogen gas at 500 kPa and 300 K Now heat is transferred to the nitrogen in the tank and the pressure of nitrogen rises to 800 kPa The work done during this process is a 500 kJ b 1500 kJ c 0 kJ d 900 kJ e 2400 kJ Answer b 0 kJ Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values V3 m3 P1500 kPa T1300 K P2800 kPa W0 since constant volume Some Wrong Solutions with Common Mistakes R0297 W1WVP2P1 Using WVDELTAP W2WVP1 W3WVP2 W4WRT1lnP1P2 4159 A 05m3 cylinder contains nitrogen gas at 600 kPa and 300 K Now the gas is compressed isothermally to a volume of 01 m3 The work done on the gas during this compression process is a 720 kJ b 483 kJ c 240 kJ d 175 kJ e 143 kJ Answer b 483 kJ Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values R831428 V105 m3 V201 m3 P1600 kPa T1300 K P1V1mRT1 WmRT1 lnV2V1 constant temperature Some Wrong Solutions with Common Mistakes W1WRT1 lnV2V1 Forgetting m W2WP1V1V2 Using VDeltaP P1V1T1P2V2T1 W3WV1V2P1P22 Using PaveDelta V W4WP1V1P2V2 Using WP1V1P2V2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4118 4160 A wellsealed room contains 60 kg of air at 200 kPa and 25C Now solar energy enters the room at an average rate of 08 kJs while a 120W fan is turned on to circulate the air in the room If heat transfer through the walls is negligible the air temperature in the room in 30 min will be a 256C b 498C c 534C d 525C e 634C Answer e 634C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values R0287 kJkgK Cv0718 kJkgK m60 kg P1200 kPa T125 C Qsol08 kJs time3060 s Wfan012 kJs Applying energy balance EinEoutdEsystem gives timeWfanQsolmCvT2T1 Some Wrong Solutions with Common Mistakes Cp1005 kJkgK timeWfanQsolmCpW1T2T1 Using Cp instead of Cv timeWfanQsolmCvW2T2T1 Subtracting Wfan instead of adding timeQsolmCvW3T2T1 Ignoring Wfan timeWfanQsol60mCvW4T2T1 Using min for time instead of s 4161 A 2kW baseboard electric resistance heater in a vacant room is turned on and kept on for 15 min The mass of the air in the room is 75 kg and the room is tightly sealed so that no air can leak in or out The temperature rise of air at the end of 15 min is a 85C b 124C c 240C d 334C e 548C Answer d 334C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values R0287 kJkgK Cv0718 kJkgK m75 kg time1560 s We2 kJs Applying energy balance EinEoutdEsystem gives timeWemCvDELTAT kJ Some Wrong Solutions with Common Mistakes Cp1005 kJkgK timeWemCpW1DELTAT Using Cp instead of Cv timeWe60mCvW2DELTAT Using min for time instead of s PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4119 4162 A room contains 75 kg of air at 100 kPa and 15C The room has a 250W refrigerator the refrigerator consumes 250 W of electricity when running a 120W TV a 18kW electric resistance heater and a 50W fan During a cold winter day it is observed that the refrigerator the TV the fan and the electric resistance heater are running continuously but the air temperature in the room remains constant The rate of heat loss from the room that day is a 5832 kJh b 6192 kJh c 7560 kJh d 7632 kJh e 7992 kJh Answer e 7992 kJh Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values R0287 kJkgK Cv0718 kJkgK m75 kg P1100 kPa T115 C time3060 s Wref0250 kJs WTV0120 kJs Wheater18 kJs Wfan005 kJs Applying energy balance EinEoutdEsystem gives EoutEin since Tconstant and dE0 EgainWrefWTVWheaterWfan QlossEgain3600 kJh Some Wrong Solutions with Common Mistakes Egain1WrefWTVWheaterWfan Subtracting Wrefrig instead of adding W1QlossEgain13600 kJh Egain2WrefWTVWheaterWfan Subtracting Wfan instead of adding W2QlossEgain23600 kJh Egain3WrefWTVWheaterWfan Subtracting Wrefrig and Wfan instead of adding W3QlossEgain33600 kJh Egain4WrefWheaterWfan Ignoring the TV W4QlossEgain43600 kJh PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4120 4163 A pistoncylinder device contains 5 kg of air at 400 kPa and 30C During a quasiequilibrium isothermal expansion process 15 kJ of boundary work is done by the system and 3 kJ of paddlewheel work is done on the system The heat transfer during this process is a 12 kJ b 18 kJ c 24 kJ d 35 kJ e 60 kJ Answer a 12 kJ Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values R0287 kJkgK Cv0718 kJkgK m5 kg P1400 kPa T30 C Woutb15 kJ Winpw3 kJ Noting that Tconstant and thus dEsystem0 applying energy balance EinEoutdEsystem gives QinWinpwWoutb0 Some Wrong Solutions with Common Mistakes W1QinQinCv Dividing by Cv W2QinWinpwWoutb Adding both quantities W3QinWinpw Setting it equal to paddlewheel work W4QinWoutb Setting it equal to boundaru work PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4121 4164 A container equipped with a resistance heater and a mixer is initially filled with 36 kg of saturated water vapor at 120C Now the heater and the mixer are turned on the steam is compressed and there is heat loss to the surrounding air At the end of the process the temperature and pressure of steam in the container are measured to be 300C and 05 MPa The net energy transfer to the steam during this process is a 274 kJ b 914 kJ c 1213 kJ d 988 kJ e 1291 kJ Answer d 988 kJ Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values m36 kg T1120 C x11 saturated vapor P2500 kPa T2300 C u1INTENERGYSteamIAPWSTT1xx1 u2INTENERGYSteamIAPWSTT2PP2 Noting that Eout0 and dUsystemmu2u1 applying energy balance EinEoutdEsystem gives Eout0 Einmu2u1 Some Wrong Solutions with Common Mistakes Cpsteam18723 kJkgK Cvsteam14108 kJkgK W1EinmCpSteamT2T1 Assuming ideal gas and using Cp W2EinmCvsteamT2T1 Assuming ideal gas and using Cv W3Einu2u1 Not using mass h1ENTHALPYSteamIAPWSTT1xx1 h2ENTHALPYSteamIAPWSTT2PP2 W4Einmh2h1 Using enthalpy PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4122 4165 A 6pack canned drink is to be cooled from 18C to 3C The mass of each canned drink is 0355 kg The drinks can be treated as water and the energy stored in the aluminum can itself is negligible The amount of heat transfer from the 6 canned drinks is a 22 kJ b 32 kJ c 134 kJ d 187 kJ e 223 kJ Answer c 134 kJ Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values C418 kJkgK m60355 kg T118 C T23 C DELTATT2T1 C Applying energy balance EinEoutdEsystem and noting that dUsystemmCDELTAT gives QoutmCDELTAT kJ Some Wrong Solutions with Common Mistakes W1QoutmCDELTAT6 Using one can only W2QoutmCT1T2 Adding temperatures instead of subtracting W3Qoutm10DELTAT Using specific heat of air or forgetting specific heat 4166 A glass of water with a mass of 045 kg at 20C is to be cooled to 0C by dropping ice cubes at 0C into it The latent heat of fusion of ice is 334 kJkg and the specific heat of water is 418 kJkgC The amount of ice that needs to be added is a 56 g b 113 g c 124 g d 224 g e 450 g Answer b 113 g Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values C418 kJkgK hmelting334 kJkgK mw045 kg T120 C T20 C DELTATT2T1 C Noting that there is no energy transfer with the surroundings and the latent heat of melting of ice is transferred form the water and applying energy balance EinEoutdEsystem to icewater gives dEicedEw0 dEicemicehmelting dEwmwCDELTAT kJ Some Wrong Solutions with Common Mistakes W1micehmeltingT1T2mwCDELTAT0 Multiplying hlatent by temperature difference W2micemw taking mass of water to be equal to the mass of ice PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4123 4167 A 2kW electric resistance heater submerged in 5kg water is turned on and kept on for 10 min During the process 300 kJ of heat is lost from the water The temperature rise of water is a 04C b 431C c 574C d 718C e 1800C Answer b 431C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values C418 kJkgK m5 kg Qloss300 kJ time1060 s We2 kJs Applying energy balance EinEoutdEsystem gives timeWeQloss dUsystem dUsystemmCDELTAT kJ Some Wrong Solutions with Common Mistakes timeWe mCW1T Ignoring heat loss timeWeQloss mCW2T Adding heat loss instead of subtracting timeWeQloss m10W3T Using specific heat of air or not using specific heat 4168 15 kg of liquid water initially at 12C is to be heated to 95C in a teapot equipped with a 800 W electric heating element inside The specific heat of water can be taken to be 418 kJkgC and the heat loss from the water during heating can be neglected The time it takes to heat the water to the desired temperature is a 59 min b 73 min c 108 min d 140 min e 170 min Answer c 108 min Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values C418 kJkgK m15 kg T112 C T295 C Qloss0 kJ We08 kJs Applying energy balance EinEoutdEsystem gives time60WeQloss dUsystem time in minutes dUsystemmCT2T1 kJ Some Wrong Solutions with Common Mistakes W1time60WeQloss mCT2T1 Adding temperatures instead of subtracting W2time60WeQloss CT2T1 Not using mass PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4124 4169 An ordinary egg with a mass of 01 kg and a specific heat of 332 kJkgC is dropped into boiling water at 95C If the initial temperature of the egg is 5C the maximum amount of heat transfer to the egg is a 12 kJ b 30 kJ c 24 kJ d 18 kJ e infinity Answer b 30 kJ Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values C332 kJkgK m01 kg T15 C T295 C Applying energy balance EinEoutdEsystem gives Ein dUsystem dUsystemmCT2T1 kJ Some Wrong Solutions with Common Mistakes W1Ein mCT2 Using T2 only W2EinmENTHALPYSteamIAPWSTT2x1ENTHALPYSteamIAPWSTT2x0 Using hfg 4170 An apple with an average mass of 018 kg and average specific heat of 365 kJkgC is cooled from 22C to 5C The amount of heat transferred from the apple is a 085 kJ b 621 kJ c 177 kJ d 112 kJ e 71 kJ Answer d 112 kJ Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values C365 kJkgK m018 kg T122 C T25 C Applying energy balance EinEoutdEsystem gives Qout dUsystem dUsystemmCT2T1 kJ Some Wrong Solutions with Common Mistakes W1Qout CT2T1 Not using mass W2Qout mCT2T1 adding temperatures PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4125 4171 The specific heat at constant pressure for an ideal gas is given by cp 0927104T kJkg K where T is in kelvin The change in the enthalpy for this ideal gas undergoing a process in which the temperature changes from 27 to 47C is most nearly a 197 kJkg b 220 kJkg c 255 kJkg d 297 kJkg e 321 kJkg Answer a 197 kJkg Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values T127273 K T247273 K Performing the necessary integration we obtain DELTAh09T2T127E42T22T12 4172 The specific heat at constant volume for an ideal gas is given by cv 0727104T kJkg K where T is in kelvin The change in the enthalpy for this ideal gas undergoing a process in which the temperature changes from 27 to 127C is most nearly a 70 kJkg b 721 kJkg c 795 kJkg d 821 kJkg e 840 kJkg Answer c 795 kJkg Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values T127273 K T2127273 K Performing the necessary integration we obtain DELTAh07T2T127E42T22T12 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4126 4173 An ideal gas has a gas constant R 03 kJkgK and a constantvolume specific heat cv 07 kJkgK If the gas has a temperature change of 100C choose the correct answer for each of the following 1 The change in enthalpy is in kJkg a 30 b 70 c 100 d insufficient information to determine Answer c 100 2 The change in internal energy is in kJkg a 30 b 70 c 100 d insufficient information to determine Answer b 70 3 The work done is in kJkg a 30 b 70 c 100 d insufficient information to determine Answer d insufficient information to determine 4 The heat transfer is in kJkg a 30 b 70 c 100 d insufficient information to determine Answer d insufficient information to determine 5 The change in the pressurevolume product is in kJkg a 30 b 70 c 100 d insufficient information to determine Answer a 30 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values R03 kJkgK cv07 kJkgK DELTAT100 K I cpRcv DELTAhcpDELTAT II DELTAucvDELTAT V PVRDELTAT PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4174 An ideal gas undergoes a constant temperature isothermal process in a closed system The heat transfer and work are respectively a 0 cvΔT b cvΔT 0 c cpΔT RΔT d R lnT2T1 R lnT2T1 Answer d R lnT2T1 R lnT2T1 4175 An ideal gas under goes a constant volume isochoric process in a closed system The heat transfer and work are respectively a 0 cvΔT b cvΔT 0 c cpΔT RΔT d R lnT2T1 R lnT2T1 Answer b cvΔT 0 4176 An ideal gas under goes a constant pressure isobaric process in a closed system The heat transfer and work are respectively a 0 cvΔT b cvΔT 0 c cpΔT RΔT d R lnT2T1 R lnT2T1 Answer c cpΔT RΔT 4177 An ideal gas under goes a constant entropy isentropic process in a closed system The heat transfer and work are respectively a 0 cvΔT b cvΔT 0 c cpΔT RΔT d R lnT2T1 R lnT2T1 Answer a 0 cvΔT 4178 4183 Design and Essay Problems 4182 A claim that fruits and vegetables are cooled by 6C for each percentage point of weight loss as moisture during vacuum cooling to be evaluated Analysis Assuming the fruits and vegetables are cooled from 30C and 0C the average heat of vaporization can be taken to be 2466 kJkg which is the value at 15C and the specific heat of products can be taken to be 4 kJkgC Then the vaporization of 001 kg water will lower the temperature of 1 kg of produce by 24664 6C Therefore the vacuum cooled products will lose 1 percent moisture for each 6C drop in temperature Thus the claim is reasonable 51 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 52 Conservation of Mass 51C Mass flow rate is the amount of mass flowing through a crosssection per unit time whereas the volume flow rate is the amount of volume flowing through a crosssection per unit time 52C Flow through a control volume is steady when it involves no changes with time at any specified position 53C The amount of mass or energy entering a control volume does not have to be equal to the amount of mass or energy leaving during an unsteadyflow process 54C No a flow with the same volume flow rate at the inlet and the exit is not necessarily steady unless the density is constant To be steady the mass flow rate through the device must remain constant 55E A pneumatic accumulator arranged to maintain a constant pressure as air enters or leaves is considered The amount of air added is to be determined Assumptions 1 Air is an ideal gas Properties The gas constant of air is R 03704 psiaft3lbmR Table A1E Analysis At the beginning of the filling the mass of the air in the container is 0 200 lbm 460 R 0 3704 psia ft lbm R80 200 psia02 ft 3 3 1 1 1 1 RT P m V During the process both pressure and temperature remain constant while volume increases by 5 times Thus 1 00 lbm 5 0 200 5 1 2 2 2 2 m RT P m V The amount of air added to the container is then 08 lbm 0 200 1 00 1 2 m m m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 53 56E Helium at a specified state is compressed to another specified state The mass flow rate and the inlet area are to be determined Assumptions Flow through the compressor is steady PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The gas cosntant of helium is R 26809 psiaft3lbmR Table A1E Analysis The mass flow rate is determined from 007038 lbms psia ft lbm R1060 R 26809 0 01 ft 100 fts200 psia 3 2 2 2 2 2 2 2 2 RT A V P A V m v The inlet area is determined from 01333 ft2 50 fts15 psia 007038 lbms26809 psia ft lbm R530 R 3 1 1 1 1 1 1 V P mRT V m A v 15 psia 70F 50 fts 200 psia 600F 001 ft2 Compressor 57 Air is accelerated in a nozzle The mass flow rate and the exit area of the nozzle are to be determined Assumptions Flow through the nozzle is steady Properties The density of air is given to be 221 kgm3 at the inlet and 0762 kgm3 at the exit Analysis a The mass flow rate of air is determined from the inlet conditions to be 0796 kgs 2 21 kgm 0009 m 4 0 ms 2 3 1 1 1 V A m ρ V1 40 ms A1 90 cm2 V2 180 ms AIR b There is only one inlet and one exit and thus m m m 1 2 Then the exit area of the nozzle is determined to be 58 cm2 2 3 2 2 2 2 2 2 0 0058 m kgm 180 ms 0762 0796 kgs V m A A V m ρ ρ preparation If you are a student using this Manual you are using it without permission 54 58 Water flows through the tubes of a boiler The velocity and volume flow rate of the water at the inlet are to be determined Assumptions Flow through the boiler is steady Properties The specific volumes of water at the inlet and exit are Tables A6 and A7 0 001017 m kg C 65 7 MPa 3 1 1 1 v T P 6 MPa 450C 80 ms 7 MPa 65C Steam 0 05217 m kg C 450 6 MPa 3 2 2 2 v T P Analysis The crosssectional area of the tube is 2 2 2 0 01327 m 4 0 13 m 4 π πD Ac The mass flow rate through the tube is same at the inlet and exit It may be determined from exit data to be 2035 kgs m kg 005217 0 01327 m 80 ms 3 2 2 2 v A V m c The water velocity at the inlet is then 1560 ms 2 3 1 1 0 01327 m 2035 kgs0001017 m kg Ac m V v The volumetric flow rate at the inlet is 00207 m s 3 0 01327 m 1560 ms 2 1 1 AcV V 59 Air is expanded and is accelerated as it is heated by a hair dryer of constant diameter The percent increase in the velocity of air as it flows through the drier is to be determined Assumptions Flow through the nozzle is steady PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The density of air is given to be 120 kgm3 at the inlet and 105 kgm3 at the exit Analysis There is only one inlet and one exit and thus Then m m m 1 2 1263 or and increase of kgm 095 kgm 120 3 3 2 1 1 2 2 2 1 1 2 1 263 ρ ρ ρ ρ V V AV AV m m V2 V1 Therefore the air velocity increases 263 as it flows through the hair drier preparation If you are a student using this Manual you are using it without permission 55 510 A rigid tank initially contains air at atmospheric conditions The tank is connected to a supply line and air is allowed to enter the tank until the density rises to a specified level The mass of air that entered the tank is to be determined Properties The density of air is given to be 118 kgm3 at the beginning and 720 kgm3 at the end V1 1 m3 ρ1 118 kgm3 Analysis We take the tank as the system which is a control volume since mass crosses the boundary The mass balance for this system can be expressed as Mass balance V V 1 2 1 2 system ρ ρ m m m m m m i out in Substituting 602 kg 720118 kgm 1m 3 3 1 2 ρ V ρ mi Therefore 602 kg of mass entered the tank 511 A cyclone separator is used to remove fine solid particles that are suspended in a gas stream The mass flow rates at the two outlets and the amount of fly ash collected per year are to be determined Assumptions Flow through the separator is steady Analysis Since the ash particles cannot be converted into the gas and viceversa the mass flow rate of ash into the control volume must equal that going out and the mass flow rate of flue gas into the control volume must equal that going out Hence the mass flow rate of ash leaving is kgs 001 0 00110 kgs in ash ash m y m The mass flow rate of flue gas leaving the separator is then 999 kgs 0 01 10 ash in flue gas m m m The amount of fly ash collected per year is 315400 kgyear 0 01 kgs365 24 3600 syear ash ash t m m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 56 512 Air flows through an aircraft engine The volume flow rate at the inlet and the mass flow rate at the exit are to be determined Assumptions 1 Air is an ideal gas 2 The flow is steady Properties The gas constant of air is R 0287 kPam3kgK Table A1 Analysis The inlet volume flow rate is 180 m s 3 1 m 180 ms 2 1 1 1 A V V The specific volume at the inlet is 0 8409 m kg 100 kPa 273 K 0 287 kPa m kg K20 3 3 1 1 1 P RT v Since the flow is steady the mass flow rate remains constant during the flow Then 2141 kgs 8409 m kg 0 m s 180 3 3 1 1 v V m 513 A spherical hotair balloon is considered The time it takes to inflate the balloon is to be determined Assumptions 1 Air is an ideal gas Properties The gas constant of air is R 0287 kPam3kgK Table A1 Analysis The specific volume of air entering the balloon is 0 7008 m kg 120 kPa 273 K 0 287 kPa m kg K20 3 3 P RT v The mass flow rate at this entrance is 3 362 kgs m kg 07008 3 ms 4 m 01 4 3 2 2 π π v v V D A V m c The initial mass of the air in the balloon is 9339 kg m kg 607008 5 m 6 3 3 3 π π v v V D m i i Similarly the final mass of air in the balloon is 2522 kg 607008 m kg 15 m 6 3 3 3 π π v v V D m f f The time it takes to inflate the balloon is determined from 120 min 722 s 3 362 kgs 9339 kg 2522 m m m t i f PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 57 514 A water pump increases water pressure The diameters of the inlet and exit openings are given The velocity of the water at the inlet and outlet are to be determined Assumptions 1 Flow through the pump is steady 2 The specific volume remains constant Properties The inlet state of water is compressed liquid We approximate it as a saturated liquid at the given temperature Then at 15C and 40C we have Table A4 0 001001 m kg 0 15 C 3 1 v x T 700 kPa 0 001008 m kg 0 40 C 3 1 v x T Water 70 kPa 15C Analysis The velocity of the water at the inlet is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 637 ms 2 2 1 1 1 π 0 01 m πD A V 3 1 1 kgs0001001 m kg 50 4 4m m v v ass flow rate and the specific volume remains constant the velocity at the pump exit is Since the m 283 ms 2 1 2 1 2 637 ms 0015 m D V V A V 2 2 1 1 001 m D A sing the ecific volume at 40C the water velocity at the inlet becomes U sp 642 ms 2 3 2 1 1 1 1 1 0 01 m 1008 m kg kg 50 4 4m m v v s000 π πD A V which is a 08 increase in velocity preparation If you are a student using this Manual you are using it without permission 58 515 Refrigerant134a flows through a pipe Heat is supplied to R134a The volume flow rates of air at the inlet and exit the mass flow rate and the velocity at the exit are to be determined Properties The specific volumes of R134a at the inlet and exit are Table A13 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course v T P 0 1142 m kg C 20 200 kPa 3 1 1 1 v T P 1 0 1374 m kg C 40 180 kPa 3 2 1 Analysis a b The volume flow rate at the inlet and the mass flow rate are kgs 2696 03079 m s 3 5 ms 4 0 28 m 1142 m kg 0 1 4 1 1 5 ms 4 0 28 m 4 2 3 1 2 1 1 1 2 1 2 1 1 π π π π D V A V m D V V A c c v v V c Noting that mass flow rate is constant the volume flow rate and the velocity at the exit of the pipe are determined from ms 602 03705 m s 3 4 0 28 m s 3705 m 0 2 696 kgs01374 m kg 2 3 2 2 3 2 2 π Ac V m V v V te of air that needs Assumptions Infiltration of air into the smoking lounge is negligible Properties The minimum fresh air requirements for a smoking lounge is given to be Analysis The required minimum flow rate of air that needs to be supplied to the from V r Vair per person No of persons The volume flow rate of fresh air can be expressed as 516 A smoking lounge that can accommodate 15 smokers is considered The required minimum flow ra to be supplied to the lounge and the diameter of the duct are to be determined 30 Ls per person lounge is determined directly 045 m s 3 30 Ls person15 persons 450 Ls ai 4 D2 V VA π V Solving for the diameter D and substituting 0268 m 8 ms 4 0 45 m s 4 3 π πV D V Therefore the diameter of the fresh air duct should be at least 268 cm if the velocity of air is not to exceed 8 ms R134a 200 kPa 20C 5 ms Q 180 kPa 40C Smoking Lounge 15 smokers preparation If you are a student using this Manual you are using it without permission 59 517 The minimum fresh air requirements of a residential building is specified to be 035 air changes per hour The size of the fan that needs to be installed and the diameter of the duct are to be determined Analysis The volume of the building and the required minimum volume flow rate of fresh air are PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 210000 Lh h 600 m m200 m 03 3 3 2 om V The volume flow rate of fresh ir can be expressed as Solving for the diameter D and substituting 210 m 600 m 0 35h ACH 3 Vroom V 3500 Lmin ro a 4 D2 V VA π V 0136 m 4 ms 4210 3600 m s 4 3 π πV D V Therefore the diameter of the fresh air duct should be at least 136 cm if the velocity of air is not to exceed 4 ms 035 ACH House 200 m2 preparation If you are a student using this Manual you are using it without permission 510 Flow Work and Energy Transfer by Mass 518C Energy can be transferred to or from a control volume as heat various forms of work and by mass 519C Flow energy or flow work is the energy needed to push a fluid into or out of a control volume Fluids at rest do not possess any flow energy 520C Flowing fluids possess flow energy in addition to the forms of energy a fluid at rest possesses The total energy of a fluid at rest consists of internal kinetic and potential energies The total energy of a flowing fluid consists of internal kinetic potential and flow energies 521E A water pump increases water pressure The flow work required by the pump is to be determined Assumptions 1 Flow through the pump is steady 2 The state of water at the pump inlet is saturated liquid 3 The specific volume remains constant Properties The specific volume of saturated liquid water at 10 psia is 0 01659 ft lbm 3 10 psia v f v Table A5E PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Then the flow work relation gives 01228 Btulbm 3 3 1 2 1 1 2 2 flow psia ft 5404 1Btu 0 01659 ft lbm50 10psia P P P P w v v v ed by the compressor is to be determined Properties The gas constant of air is R 0287 kPam3kgK Table A1 nalysis Combining the flow work expression with the ideal gas equation of state gives 50 psia Water 10 psia 522 An air compressor compresses air The flow work requir Assumptions 1 Flow through the compressor is steady 2 Air is an ideal gas Compressor 1 MPa 400C 120 kPa 20C A 109 kJkg 20K 0 287 kJkg K400 1 2 1 1 2 2 flow T T R P P w v v preparation If you are a student using this Manual you are using it without permission 511 523E Steam is leaving a pressure cooker at a specified pressure The velocity flow rate the total and flow energies and the rate of energy transfer by mass are to be determined Assumptions 1 The flow is steady and the initial startup period is disregarded 2 The kinetic and potential energies are negligible and thus they are not considered 3 Saturation conditions exist within the cooker at all times so that steam leaves the cooker as a saturated vapor at 20 psia Properties The properties of saturated liquid water and water vapor at 20 psia are vf 001683 ft3lbm vg 20093 ft3lbm ug 10818 Btulbm and hg 11562 Btulbm Table A5E Analysis a Saturation conditions exist in a pressure cooker at all times after the steady operating conditions are established Therefore the liquid has the properties of saturated liquid and the exiting steam has the properties of saturated vapor at the operating pressure The amount of liquid that has evaporated the mass flow rate of the exiting steam and the exit velocity are PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 341 fts 2 2 1ft 015 in c c g A A ρ b Noting that h u P 2 3 3 3 liquid 10 lbms20093 ft lbm 144 in 1765 0 1059 lbmmin 45 min 766 lbm 4 0 13368 ft gal 06 m g m V t m v V lbms 10 1765 3 v and that the kinetic and potential energies are disregarded the flow and total energies of the h pe ke h θ 1081 8 w leaving the cooker by mass is simply the product of the mass flow rate and the total energy f the exiting steam per unit mass Discussion The numerical value of the energy leaving the cooker with steam alone does not mean much since this value depends on the reference point selected for enthalpy it could even be negative The significant quantity is the difference between the enthalpies of the exiting vapor and the liquid inside which is hfg since it relates directly to the amount of energy supplied to the cooker H2O Sat vapor P 20 psia Q 3 4 766 lbm 1gal 001683 ft lbm f m m v exiting steam are u h P e 1156 2 v Btulbm 11562 Btulbm 744 flo Note that the kinetic energy in this case is ke V22 341 fts2 2 581 ft2s2 00232 Btulbm which is very small compared to enthalpy c The rate at which energy is o 204 Btus lbms11562 Btulbm 765 10 1 3 mass mθ E preparation If you are a student using this Manual you are using it without permission 512 524 Air flows steadily in a pipe at a specified state The diameter of the pipe the rate of flow energy and the rate of energy transport by mass are to be determined Also the error involved in the determination of energy transport by mass is to be determined PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The properties of air are R 0287 kJkgK and cp 1008 kJkgK at 350 K from Table A2b Analysis a The diameter is determined as follows 0 3349 m kg 300 kPa 273 K 0 287 kJkgK77 3 P RT v 300 kPa 77C Air 25 ms 18 kgmin 2 3 0 004018 m 25 ms 18 60 kgs03349 m kg V m A v 00715 m π π D 40004018 m 4 2 A The rate of flow energy is determined from Pa0334 9 m kg 3 c The rate of energy transport by mass is b 1860 kgs300k flow mPv W 3014 kW 10594 kW 1000 2 2 2 2 mass s m 1kJkg 2 25 ms 1 273 K kgs 1008 kJkgK77 1860 2 1 V m c T ke m h E p d If we neglect kinetic energy in the calculation of energy transport by mass mass p Therefore the error involved if neglect the kinetic energy is only 009 273 K 1860 kgs100 5 kJkgK77 mc T mh E kW 10584 preparation If you are a student using this Manual you are using it without permission 513 Steady Flow Energy Balance Nozzles and Diffusers 525C No 526C It is mostly converted to internal energy as shown by a rise in the fluid temperature 527C The kinetic energy of a fluid increases at the expense of the internal energy as evidenced by a decrease in the fluid temperature 528C Heat transfer to the fluid as it flows through a nozzle is desirable since it will probably increase the kinetic energy of the fluid Heat transfer from the fluid will decrease the exit velocity PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 529 Air is decelerated in a diffuser from 230 ms to 30 ms The exit temperature of air and the exit area of the diffuser are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Air is an ideal gas with variable specific heats 3 Potential energy changes are negligible 4 The device is adiabatic and thus heat transfer is negligible 5 There are no work interactions Properties The gas constant of air is 0287 kPam3kgK Table A1 The enthalpy of air at the inlet temperature of 400 K is h1 40098 kJkg Table A17 Analysis a There is only one inlet and one exit and thus ṁ1 ṁ2 ṁ We take diffuser as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout ΔĖsystem steady 0 Rate of net energy transfer Rate of change in internal kinetic by heat work and mass potential etc energies Ėin Ėout ṁh1 V12 2 ṁh2 V22 2 since Q W Δpe 0 0 h2 h1 V22 V12 2 or h2 h1 V22 V12 2 40098 kJkg 30 ms2 230 ms2 2 1 kJkg 1000 m2s2 42698 kJkg From Table A17 T2 4256 K b The specific volume of air at the diffuser exit is ν2 RT2P2 0287 kPam3kgK 4256 K100 kPa 1221 m3kg From conservation of mass ṁ 1ν2 A2 V2 A2 ṁ ν2 V2 60003600 kgs1221 m3kg 30 ms 00678 m2 530 Air is accelerated in a nozzle from 45 ms to 180 ms The mass flow rate the exit temperature and the exit area of the nozzle are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Air is an ideal gas with constant specific heats 3 Potential energy changes are negligible 4 The device is adiabatic and thus heat transfer is negligible 5 There are no work interactions Properties The gas constant of air is 0287 kPam3kgK Table A1 The specific heat of air at the anticipated average temperature of 450 K is cp 102 kJkgC Table A2 Analysis a There is only one inlet and one exit and thus mdot1 mdot2 mdot Using the ideal gas relation the specific volume and the mass flow rate of air are determined to be ν1 RT1 P1 0287 kPa m3kg K473 K 300 kPa 04525 m3kg mdot 1ν1 A1V1 104525 m3kg00110 m245 ms 1094 kgs b We take nozzle as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Edotin Edotout ΔEdotsystem 0 steady 0 Edotin Edotout mdoth1 V12 2 mdoth2 V22 2 since Qdot Wdot Δpe 0 0 h2 h1 V22 V12 2 0 cpave T2 T1 V22 V12 2 Substituting 0 102 kJkg KT2 200 C 180 ms2 45 ms2 2 1 kJkg 1000 m2s2 It yields T2 1852C c The specific volume of air at the nozzle exit is ν2 RT2 P2 0287 kPa m3kg K1852 273 K 100 kPa 1315 m3kg mdot 1ν2 A2V2 1094 kgs 11315 m3kg A2 180 ms A2 000799 m2 799 cm2 516 531 Problem 530 is reconsidered The effect of the inlet area on the mass flow rate exit velocity and the exit area as the inlet area varies from 50 cm2 to 150 cm2 is to be investigated and the final results are to be plotted against the inlet area Analysis The problem is solved using EES and the solution is given below Function HCalWorkFluid Tx Px Function to calculate the enthalpy of an ideal gas or real gas If Air WorkFluid then HCalENTHALPYAirTTx Ideal gas equ else HCalENTHALPYWorkFluidTTx PPxReal gas equ endif end HCal System control volume for the nozzle Property relation Air is an ideal gas Process Steady state steady flow adiabatic no work Knowns obtain from the input diagram WorkFluid Air T1 200 C P1 300 kPa Vel1 45 ms P2 100 kPa Vel2 180 ms A1110 cm2 Am1A1convertcm2m2 Property Data since the Enthalpy function has different parameters for ideal gas and real fluids a function was used to determine h h1HCalWorkFluidT1P1 h2HCalWorkFluidT2P2 The Volume function has the same form for an ideal gas as for a real fluid v1volumeworkFluidTT1pP1 v2volumeWorkFluidTT2pP2 Conservation of mass mdot1 mdot2 Mass flow rate mdot1Am1Vel1v1 mdot2 Am2Vel2v2 Conservation of Energy SSSF energy balance h1Vel1221000 h2Vel2221000 Definition AratioA1A2 A2Am2convertm2cm2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 517 A1 cm2 A2 cm2 m1 kgs T2 C 50 60 70 80 90 100 110 120 130 140 150 3632 4359 5085 5812 6538 7265 7991 8718 9444 1017 109 0497 05964 06958 07952 08946 09941 1093 1193 1292 1392 1491 1852 1852 1852 1852 1852 1852 1852 1852 1852 1852 1852 50 70 90 110 130 150 30 40 50 60 70 80 90 100 110 04 06 08 1 12 14 16 A1 cm2 A2 cm 2 m1 kgs PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 532E Air is accelerated in an adiabatic nozzle The velocity at the exit is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Air is an ideal gas with constant specific heats 3 Potential energy changes are negligible 4 There are no work interactions 5 The nozzle is adiabatic Properties The specific heat of air at the average temperature of 70064526725F is cp 0253 BtulbmR Table A2Eb Analysis There is only one inlet and one exit and thus mdot1 mdot2 mdot We take nozzle as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Edotin Edotout ΔEdotsystem 0 steady 0 Edotin Edotout mdoth1 V12 2 mdoth2 V22 2 h1 V12 2 h2 V22 2 Solving for exit velocity V2 V12 2h1 h2 05 V12 2cp T1 T2 05 80 fts2 20253 Btulbm R700 645R 25037 ft2s2 1 Btulbm 05 8386 fts 533 Air is decelerated in an adiabatic diffuser The velocity at the exit is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Air is an ideal gas with constant specific heats 3 Potential energy changes are negligible 4 There are no work interactions 5 The diffuser is adiabatic Properties The specific heat of air at the average temperature of 2090255C 328 K is cp 1007 kJkgK Table A2b Analysis There is only one inlet and one exit and thus mdot1 mdot2 mdot We take diffuser as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Edotin Edotout ΔEdotsystem 0 steady 0 Edotin Edotout mdoth1 V12 2 mdoth2 V22 2 h1 V12 2 h2 V22 2 Solving for exit velocity V2 V12 2h1 h2 05 V12 2cp T1 T2 05 500 ms2 21007 kJkg K20 90K 1000 m2s2 1 kJkg 05 3302 ms 534 Heat is lost from the steam flowing in a nozzle The velocity and the volume flow rate at the nozzle exit are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Potential energy change is negligible 3 There are no work interactions Analysis We take the steam as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Energy balance Edotin Edotout ΔEdotsystem 0 steady 0 Edotin Edotout mdoth1 V12 2 mdoth2 V22 2 Qdotout since Wdot Δpe 0 or h1 V12 2 h2 V22 2 Qdotout mdot The properties of steam at the inlet and exit are Table A6 P1 800 kPa ν1 038429 m3kg h1 32677 kJkg T1 400C P2 200 kPa ν2 131623 m3kg h2 30721 kJkg T1 300C The mass flow rate of the steam is mdot 1 ν1 A1V1 1 038429 m3s008 m210 ms 2082 kgs Substituting 32677 kJkg 10 ms2 2 1 kJkg 1000 m2s2 30721 kJkg V22 2 1 kJkg 1000 m2s2 25 kJs 2082 kgs V2 606 ms The volume flow rate at the exit of the nozzle is Vdot2 mdot ν2 2082 kgs131623 m3kg 274 m3s 535 Steam is accelerated in a nozzle from a velocity of 40 ms to 300 ms The exit temperature and the ratio of the inlettoexit area of the nozzle are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Potential energy changes are negligible 3 There are no work interactions 4 The device is adiabatic and thus heat transfer is negligible Properties From the steam tables Table A6 P1 3 MPa v1 009938 m3kg T1 400C h1 32317 kJkg Analysis a There is only one inlet and one exit and thus m1 m2 m We take nozzle as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout ΔĖsystem 0 Ėin Ėout mh1 V12 2 mh2 V22 2 since Q W Δpe 0 0 h2 h1 V22 V12 2 or h2 h1 V22 V12 2 32317 kJkg 300 ms2 40 ms2 2 1 kJkg 1000 m2s2 31875 kJkg Thus P2 25 MPa T2 3766C h2 31875 kJkg v2 011533 m3kg b The ratio of the inlet to exit area is determined from the conservation of mass relation 1v2 A2 V2 1v1 A1 V1 A1 A2 v1 v2 V2 V1 009938 m3kg300 ms 011533 m3kg40 ms 646 536E Air is decelerated in a diffuser from 600 fts to a low velocity The exit temperature and the exit velocity of air are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Air is an ideal gas with variable specific heats 3 Potential energy changes are negligible 4 The device is adiabatic and thus heat transfer is negligible 5 There are no work interactions Properties The enthalpy of air at the inlet temperature of 50F is h1 12188 Btulbm Table A17E Analysis a There is only one inlet and one exit and thus m1 m2 m We take diffuser as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout ΔĖsystem 0 Ėin Ėout mh1 V12 2 mh2 V22 2 since Q W Δpe 0 0 h2 h1 V22 V12 2 or h2 h1 V22 V12 2 12188 Btulbm 0 600 fts22 1 Btulbm 25037 ft2s2 12907 Btulbm From Table A17E T2 540 R b The exit velocity of air is determined from the conservation of mass relation 1v2 A2 V2 1v1 A1 V1 1 RT2 P2 A2 V2 1 RT1 P1 A1 V1 Thus V2 A1 T2 P1 A2 T1 P2 V1 14 540 R13 psia 510 R145 psia 600 fts 142 fts 537 CO2 gas is accelerated in a nozzle to 450 ms The inlet velocity and the exit temperature are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 CO2 is an ideal gas with variable specific heats 3 Potential energy changes are negligible 4 The device is adiabatic and thus heat transfer is negligible 5 There are no work interactions Properties The gas constant and molar mass of CO2 are 01889 kPam3kgK and 44 kgkmol Table A1 The enthalpy of CO2 at 500C is h1 30797 kJkmol Table A20 Analysis a There is only one inlet and one exit and thus m1 m2 m Using the ideal gas relation the specific volume is determined to be ν1 RT1 P1 01889 kPam3kgK773 K 1000 kPa 0146 m3kg Thus m 1ν1 A1 V1 V1 m ν1 A1 60003600 kgs0146 m3kg40104 m2 608 ms b We take nozzle as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout ΔĖsystem 0 Ėin Ėout mh1 V12 2 mh2 V22 2 since Q W Δpe 0 0 h2 h1 V22 V12 2 Substituting h2 h1 V22 V12 2 M 30797 kJkmol 450 ms2 608 ms22 1 kJkg 1000 m2s244 kgkmol 26423 kJkmol Then the exit temperature of CO2 from Table A20 is obtained to be T2 6858 K 538 R134a is accelerated in a nozzle from a velocity of 20 ms The exit velocity of the refrigerant and the ratio of the inlettoexit area of the nozzle are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Potential energy changes are negligible 3 There are no work interactions 4 The device is adiabatic and thus heat transfer is negligible Properties From the refrigerant tables Table A13 P1700 kPa T1120C v10043358 m3kg h135890 kJkg and P2400 kPa T230C v20056796 m3kg h227507 kJkg Analysis a There is only one inlet and one exit and thus ṁ1ṁ2ṁ We take nozzle as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout ΔĖsystem 0 Ėin Ėout ṁh1 V12 2 ṁh2 V222 since Q W Δpe 0 0 h2 h1 V22 V122 Substituting 0 27507 35890 kJkg V22 20 ms22 1 kJkg 1000 m2s2 It yields V2 4099 ms b The ratio of the inlet to exit area is determined from the conservation of mass relation 1v2 A2 V2 1v1 A1 V1 A1A2 v1v2 V2V1 0043358 m3kg4099 ms 0056796 m3kg20 ms 1565 539 Nitrogen is decelerated in a diffuser from 275 ms to a lower velocity The exit velocity of nitrogen and the ratio of the inlettoexit area are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Nitrogen is an ideal gas with variable specific heats 3 Potential energy changes are negligible 4 The device is adiabatic and thus heat transfer is negligible 5 There are no work interactions Properties The molar mass of nitrogen is M 28 kgkmol Table A1 The enthalpies are Table A18 T17C280 K h18141 kJkmol T227C300 K h28723 kJkmol Analysis a There is only one inlet and one exit and thus ṁ1ṁ2ṁ We take diffuser as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout ΔĖsystem 0 Ėin Ėout ṁh1 V12 2 ṁh2 V222 since Q W Δpe 0 0 h2 h1 V22 V122 h2M h1M V22 V122 Substituting 0 8723 8141 kJkmol 28 kgkmol V22 275 ms22 1 kJkg 1000 m2s2 It yields V2 185 ms b The ratio of the inlet to exit area is determined from the conservation of mass relation 1v2 A2 V2 1v1 A1 V1 A1A2 v1v2 V2V1 RT1P1RT2P2 V2V1 or A1A2 T1P1T2P2 V2V1 280 K60 kPa185 ms 300 K85 kPa200 ms 0887 525 540 Problem 539 is reconsidered The effect of the inlet velocity on the exit velocity and the ratio of the inletto exit area as the inlet velocity varies from 210 ms to 350 ms is to be investigated The final results are to be plotted against nalysis The problem is solved using EES and the solution is given below ENTHALPYWorkFluidTTx PPxReal gas equ teady state steady flow adiabatic no work s a ers tion was used to determine h for an ideal gas as for a real fluid lv and conservation of mass the area ratio ARatio 1Vel1221000 h2Vel2221000 the inlet velocity A Function HCalWorkFluid Tx Function to calculate the enth Px alpy of an ideal gas or real gas If N2 WorkFluid then HCalENTHALPYWorkFluidTTx Ideal gas equ else HCal endif end HCal System control volume for the nozzle Property relation Nitrogen is an ideal gas Process S Knowns WorkFluid N2 T1 7 C P1 60 kPa m Vel1 275 P2 85 kP T2 27 C Property Data since the Enthalpy function has different paramet for ideal gas and real fluids a func h1HCalWorkFluidT1P1 h2HCalWorkFluidT2P2 The Volume function has the same form v1volumeworkFluidTT1pP1 v2volumeWorkFluidTT2pP2 From the definition of mass flow rate mdot AVe A1A2 is ARatioVel1v1 Vel2v2 Conservation of Energy SSSF energy balance h PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 526 Vel1 ms Vel2 ms ARatio 210 224 238 252 266 280 294 308 322 336 350 5001 9261 1227 148 1708 1918 2117 2308 2492 267 2844 03149 05467 06815 07766 08488 09059 09523 09908 1023 1051 1075 200 220 240 260 280 300 320 340 360 50 100 150 200 250 300 Vel1 ms Vel2 ms 200 220 240 260 280 300 320 340 360 03 04 05 06 07 08 09 1 11 Vel1 ms ARatio PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 541 R134a is decelerated in a diffuser from a velocity of 120 ms The exit velocity of R134a and the mass flow rate of the R134a are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Potential energy changes are negligible 3 There are no work interactions Properties From the R134a tables Tables A11 through A13 P1800 kPa sat vapor v10025621 m3kg h126729 kJkg P2900 kPa T240C v20023375 m3kg h227417 kJkg Analysis a There is only one inlet and one exit and thus ṁ1ṁ2ṁ Then the exit velocity of R134a is determined from the steadyflow mass balance to be 1v2 A2 V2 1v1 A1 V1 V2 v2v1 A1A2 V1 0023375 m3kg 18 0025621 m3kg120 ms 608 ms b We take diffuser as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout ΔĖsystem 0 Ėin Ėout Qin ṁh1 V122 ṁh2 V222 since W Δpe 0 Qin ṁh2 h1 V22 V122 Substituting the mass flow rate of the refrigerant is determined to be 2 kJs ṁ 27417 26729 kJkg 608 ms2 120 ms22 1 kJkg 1000 m2s2 It yields ṁ 1308 kgs 542 Steam is accelerated in a nozzle from a velocity of 60 ms The mass flow rate the exit velocity and the exit area of the nozzle are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Potential energy changes are negligible 3 There are no work interactions Properties From the steam tables Table A6 P1 4 MPa v1 007343 m3kg T1 400C h1 32145 kJkg and P2 2 MPa v2 012551 m3kg T2 300C h2 30242 kJkg Analysis a There is only one inlet and one exit and thus m1 m2 m The mass flow rate of steam is m 1v1 V1 A1 1007343 m3kg 60 ms50x104 m2 4085 kgs b We take nozzle as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem0 steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout mh1 V122 Qout mh2 V222 since W Δpe 0 Qout m h2 h1 V22 V122 Substituting the exit velocity of the steam is determined to be 75 kJs 4085 kgs 30242 32145 V22 60 ms22 1 kJkg1000 m2s2 It yields V2 5895 ms c The exit area of the nozzle is determined from m 1v2 V2 A2 A2 m v2 V2 4085 kgs012551 m3kg5895 ms 870 x 104 m2 Turbines and Compressors 543C Yes 544C The volume flow rate at the compressor inlet will be greater than that at the compressor exit 545C Yes Because energy in the form of shaft work is being added to the air 546C No 547 R134a at a given state is compressed to a specified state The mass flow rate and the power input are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible Analysis We take the compressor as the system which is a control volume since mass crosses the boundary Noting that one fluid stream enters and leaves the compressor the energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem0 steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout Win m h1 m h2 since Δke Δpe 0 Win m h2 h1 From R134a tables Tables A11 A12 A13 P1 100 kPa h1 23633 kJkg T1 24C v1 01947 m3kg P2 800 kPa T2 60C h2 29681 kJkg The mass flow rate is m V1 v1 135 60 m3s 01947 m3kg 01155 kgs Substituting Win m h2 h1 01155 kgs29681 23633 kJkg 699 kW 548 Saturated R134a vapor is compressed to a specified state The power input is given The exit temperature is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Heat transfer with the surroundings is negligible Analysis We take the compressor as the system which is a control volume since mass crosses the boundary Noting that one fluid stream enters and leaves the compressor the energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem0 steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout Win m h1 m h2 since Δke Δpe 0 Win m h2 h1 From R134a tables Table A12 P1 180 kPa h1 24286 kJkg x1 0 v1 01104 m3kg The mass flow rate is m V1 v1 035 60 m3s 01104 m3kg 005283 kgs Substituting for the exit enthalpy Win m h2 h1 235 kW 005283 kgsh2 24286 kJkg h2 28734 kJkg From Table A13 P2 700 kPa h2 28734 kJkg T2 488C 549 Steam expands in a turbine The change in kinetic energy the power output and the turbine inlet area are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Potential energy changes are negligible 3 The device is adiabatic and thus heat transfer is negligible Properties From the steam tables Tables A4 through 6 P1 6 MPa T1 400C v10047420 m3kg h131783 kJkg and P2 40 kPa x2 092 h2 hf x2hfg 31762 092 23921 23185 kJkg Analysis a The change in kinetic energy is determined from Δke V22 V122 50 ms2 80 ms22 1 kJkg 1000 m2s2 195 kJkg b There is only one inlet and one exit and thus m1 m2 m We take the turbine as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout ΔĖsystem π0 steady0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ėin Ėout mh1 V12 2 Ẇout mh2 V22 2 since Q Δpe 0 Ẇout mh2 h1 V22 V122 Then the power output of the turbine is determined by substitution to be Ẇout 20 kgs23185 31783 195kJkg 14590 kW 146 MW c The inlet area of the turbine is determined from the mass flow rate relation m 1v1 A1 V1 A1 mv1V1 20 kgs0047420 m3kg80 ms 00119 m2 532 550 Problem 549 is reconsidered The effect of the turbine exit pressure on the power output of the turbine as the utput is to be plotted against the exit pressure nalysis The problem is solved using EES and the solution is given below PWS 2 ss nvertm2s2 ELTAkeVel222Convertm2s2 kJkgVel122Convertm2s2 kJkg exit pressure varies from 10 kPa to 200 kPa is to be investigated The power o A Knowns T1 450 C P1 6000 kPa Vel1 80 ms P2 40 kPa X2092 Vel2 50 ms 40 50 60 70 80 90 100 110 120 130 mdot112 kgs FluidSteamIA Property Data h1enthalpyFluidTT1PP1 h2enthalpyFluidPP2xx2 T2 T2temperatureFluidPP2x 1volumeFluidTT1pP1 x v v2volumeFluidPP2xx2 Conservation of ma mdot1 mdot2 Mass flow rate mdot1A1Vel1v1 0 40 80 120 160 200 P2 kPa mdot2 A2Vel2v2 Conservation of Energy Steady Flow energy balance mdot1h1Vel122Convertm2s2 kJkg mdot2h2Vel222Co kJkgWdotturbconvertMWkJs D P2 kPa W turb MW T2 C 10 3111 5222 7333 9444 1156 1367 1578 1789 200 9297 1202 1095 1039 101 9909 976 9638 9535 9446 9367 4581 6993 824 9116 9802 1037 1086 1129 1167 92 94 96 98 10 102 104 106 108 11 Wturb Mw 0 40 80 120 160 200 P2 kPa PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 551 Steam expands in a turbine The mass flow rate of steam for a power output of 5 MW is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 The device is adiabatic and thus heat transfer is negligible Properties From the steam tables Tables A4 through 6 P1 10 MPa T1 500C h1 33751 kJkg P2 10 kPa x2 090 h2 hf x2 hfg19181 090 23921 23447 kJkg Analysis There is only one inlet and one exit and thus m1 m2 m We take the turbine as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout ΔĖsystem π0 steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ėin Ėout mh1 Ẇout mh2 since Q Δke Δpe 0 Ẇout mh2 h1 Substituting the required mass flow rate of the steam is determined to be 5000 kJs m23447 33751 kJkg m 4852 kgs 552E Steam expands in a turbine The rate of heat loss from the steam for a power output of 4 MW is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible Properties From the steam tables Tables A4E through 6E P1 1000 psia T1 900F h1 14486 Btulbm P2 5 psia satvapor h211307 Btulbm Analysis There is only one inlet and one exit and thus m1 m2 m We take the turbine as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout ΔĖsystem π0 steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ėin Ėout mh1 Qout Ẇout mh2 since Δke Δpe 0 Qout mh2 h1 Ẇout Substituting Qout 450003600 lbms11307 14486Btulbm 4000 kJs 1 Btu 1055 kJ 1820 Btus 553 Air is compressed at a rate of 10 Ls by a compressor The work required per unit mass and the power required are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The constant pressure specific heat of air at the average temperature of 203002160C433 K is cp 1018 kJkgK Table A2b The gas constant of air is R 0287 kPam3kgK Table A1 Analysis a There is only one inlet and one exit and thus m1 m2 m We take the compressor as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout Δ Ėsystem 70 steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ėin Ėout Win mh1 mh2 since Δke Δpe 0 Win mh2 h1 mcpT2 T1 Thus win cpT2 T1 1018 kJkg K300 20K 2850 kJkg b The specific volume of air at the inlet and the mass flow rate are v1 RT1P1 0287 kPa m3 kg K20 273 K120 kPa 07008 m3 kg m V1v1 0010 m3s 07008 m3kg 001427 kgs Then the power input is determined from the energy balance equation to be Win mcpT2 T1 001427 kgs1018 kJkg K300 20K 4068 kW 554 Argon gas expands in a turbine The exit temperature of the argon for a power output of 190 kW is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Potential energy changes are negligible 3 The device is adiabatic and thus heat transfer is negligible 4 Argon is an ideal gas with constant specific heats Properties The gas constant of Ar is R 02081 kPam3kgK The constant pressure specific heat of Ar is cp 05203 kJkgC Table A2a Analysis There is only one inlet and one exit and thus m1 m2 m The inlet specific volume of argon and its mass flow rate are v1 RT1P1 02081 kPa m3 kg K723 K1600 kPa 009404 m3 kg Thus m 1v1 A1V1 1009404 m3kg 0006 m255 ms 3509 kgs We take the turbine as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout Δ Ėsystem 70 steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ėin Ėout mh1 V122 Wout mh2 V222 since Q Δpe 0 Wout mh2 h1 V22 V122 Substituting 190 kJs 3509 kgs05203 kJkg CT2 450C 150 ms2 55 ms22 1 kJkg 1000 m2s2 It yields T2 327C 555 Helium is compressed by a compressor For a mass flow rate of 90 kgmin the power input required is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Helium is an ideal gas with constant specific heats Properties The constant pressure specific heat of helium is cp 51926 kJkgK Table A2a Analysis There is only one inlet and one exit and thus m1 m2 m We take the compressor as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout Δ Ėsystem 70 steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ėin Ėout Win mh1 Qout mh2 since Δke Δpe 0 Win Qout mh2 h1 mcpT2 T1 Thus Win Qout mcpT2 T1 9060 kgs20 kJkg 9060 kgs51926 kJkg K430 310K 965 kW 556 CO2 is compressed by a compressor The volume flow rate of CO2 at the compressor inlet and the power input to the compressor are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Helium is an ideal gas with variable specific heats 4 The device is adiabatic and thus heat transfer is negligible Properties The gas constant of CO2 is R 01889 kPam3kgK and its molar mass is M 44 kgkmol Table A1 The inlet and exit enthalpies of CO2 are Table A20 T1 300 K h1 9431 kJkmol T2 450 K h2 15483 kJkmol Analysis a There is only one inlet and one exit and thus m1 m2 m The inlet specific volume of air and its volume flow rate are v1 RT1 P1 01889 kPam3kgK300 K 100 kPa 05667 m3kg V mv1 05 kgs05667 m3kg 0283 m3s b We take the compressor as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔĖsystem steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout Win mh1 mh2 since Q Δke Δpe 0 Win mh2 h1 mh2 h1 M Substituting Win 05 kgs15483 9431 kJkmol 44 kgkmol 688 kW 557 Air is expanded in an adiabatic turbine The mass flow rate of the air and the power produced are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 The turbine is wellinsulated and thus there is no heat transfer 3 Air is an ideal gas with constant specific heats Properties The constant pressure specific heat of air at the average temperature of 5001272314C587 K is cp 1048 kJkgK Table A2b The gas constant of air is R 0287 kPam3kgK Table A1 Analysis a There is only one inlet and one exit and thus m1 m2 m We take the turbine as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔĖsystem steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout mh1 V122 mh2 V222 Wout Wout mh1 h2 V12 V222 mcp T1 T2 V12 V222 The specific volume of air at the inlet and the mass flow rate are v1 RT1 P1 0287 kPam3kgK500 273 K 1300 kPa 01707 m3kg m A1V1 v1 02 m240 ms 01707 m3kg 4688 kgs Similarly at the outlet v2 RT2 P2 0287 kPam3kgK127 273 K 100 kPa 1148 m3kg V2 mv2 A2 4688 kgs1148 m3kg 1 m2 5382 ms b Substituting into the energy balance equation gives Wout mcp T1 T2 V12 V222 4688 kgs1048 kJkgK500 127K 40 ms2 5382 ms2 2 1 kJkg 1000 m2s2 18300 kW 558E Air is expanded in an adiabatic turbine The mass flow rate of the air and the power produced are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 The turbine is wellinsulated and thus there is no heat transfer 3 Air is an ideal gas with constant specific heats Properties The constant pressure specific heat of air at the average temperature of 8002502525F is cp 02485 BtulbmR Table A2Eb The gas constant of air is R 03704 psiaft3lbmR Table A1E Analysis There is only one inlet and one exit and thus m1 m2 m We take the turbine as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔĖsystem steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout mh1 V122 mh2 V222 Wout Wout mh1 h2 V12 V222 mcp T1 T2 V12 V222 The specific volume of air at the exit and the mass flow rate are v2 RT2 P2 03704 psiaft3lbmR250 460 R 60 psia 4383 ft3lbm m V2 v2 50 ft3s 4383 ft3lbm 1141 kgs V2 mv2 A2 1141 lbms4383 ft3lbm 12 ft2 4168 fts Similarly at the inlet v1 RT1 P1 03704 psiaft3lbmR800 460 R 500 psia 09334 ft3lbm V1 mv1 A1 1141 lbms09334 ft3lbm 06 ft2 1775 fts Substituting into the energy balance equation gives Wout mcp T1 T2 V12 V222 1141 lbms 02485 BtulbmR800 250R 1775 fts2 4168 ms22 1 Btulbm 25037 ft2s2 1559 Btus 1645 kW 559 Steam expands in a twostage adiabatic turbine from a specified state to another state Some steam is extracted at the end of the first stage The power output of the turbine is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 The turbine is adiabatic and thus heat transfer is negligible Properties From the steam tables Tables A5 and A6 P1 5 MPa T1 600C h1 36669 kJkg P2 05 MPa x2 1 h2 27481 kJkg P3 10 kPa x2 085 h3 hf xhfg 19181 08523921 22251 kJkg Analysis We take the entire turbine including the connection part between the two stages as the system which is a control volume since mass crosses the boundary Noting that one fluid stream enters the turbine and two fluid streams leave the energy balance for this steadyflow system can be expressed in the rate form as E in E out ΔE system 0 Rate of net energy transfer Rate of change in internal kinetic by heat work and mass potential etc energies E in E out m1 h1 m2 h2 m3 h3 W out W out m1 h1 01 h2 09 h3 Substituting the power output of the turbine is W out m1 h1 01 h2 09 h3 20 kgs36669 01 27481 09 22251 kJkg 27790 kW 560 Steam is expanded in a turbine The power output is given The rate of heat transfer is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible Properties From the steam tables Table A4 A5 A6 P1 6 MPa T1 600C h1 36588 kJkg P2 05 MPa T2 200C h2 28558 kJkg Analysis We take the turbine as the system which is a control volume since mass crosses the boundary Noting that one fluid stream enters and leaves the compressor the energy balance for this steadyflow system can be expressed in the rate form as E in E out ΔE system 0 Rate of net energy transfer Rate of change in internal kinetic by heat work and mass potential etc energies E in E out m h1 V1² 2 m h2 V2² 2 W out Q out since Δpe 0 Q out W out m h1 h2 V1² V2² 2 Substituting Q out W out m h1 h2 V1² V2² 2 20000 kW 26 kgs 36588 28558 kJkg 0 180 ms² 2 1 kJkg 1000 m²s² 455 kW 561 Helium at a specified state is compressed to another specified state The power input is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Helium is an ideal gas Properties The properties of helium are cp 51926 kJkgK and R 20769 kPam³kgK Table A2a Analysis a There is only one inlet and one exit and thus m1 m2 m We take the compressor as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as E in E out ΔE system 0 Rate of net energy transfer Rate of change in internal kinetic by heat work and mass potential etc energies E in E out W in m h1 m h2 since Δke Δpe 0 W in m h2 h1 m cp T2 T1 The mass flow rate is determined from m A1 V1 v1 A1 V1 P1 R T1 01 m² 15 ms 150 kPa 20769 kPam³kgK293 K 03697 kgs Substituting W in m cp T2 T1 03697 kgs51926 kJkgK200 20K 346 kW The flow power input is determined from W fw m P2 v2 P1 v1 m R T2 T1 03697 kgs20769 kJkgK200 20K 138 kW Throttling Valves 562C The temperature of a fluid can increase decrease or remain the same during a throttling process Therefore this claim is valid since no thermodynamic laws are violated 563C No Because air is an ideal gas and h hT for ideal gases Thus if h remains constant so does the temperature 564C If it remains in the liquid phase no But if some of the liquid vaporizes during throttling then yes 565C Yes 566 Refrigerant134a is throttled by a capillary tube The quality of the refrigerant at the exit is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Heat transfer to or from the fluid is negligible 4 There are no work interactions involved Analysis There is only one inlet and one exit and thus ṁ1 ṁ2 ṁ We take the throttling valve as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout ΔĖsystem 0 steady 0 Ėin Ėout ṁh1 ṁh2 h1 h2 since Q W Δke Δpe 0 The inlet enthalpy of R134a is from the refrigerant tables Table A11 T1 50 C sat liquid h1 hf 12349 kJkg The exit quality is T2 20 C h2 h1 x2 h2 hf hfg 12349 254921291 0460 567 Steam is throttled from a specified pressure to a specified state The quality at the inlet is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Heat transfer to or from the fluid is negligible 4 There are no work interactions involved Analysis There is only one inlet and one exit and thus ṁ1 ṁ2 ṁ We take the throttling valve as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout ΔĖsystem 0 steady 0 Ėin Ėout ṁh1 ṁh2 h1 h2 Since Q W Δke Δpe 0 The enthalpy of steam at the exit is Table A6 P2 100 kPa T2 120 C h2 27161 kJkg The quality of the steam at the inlet is Table A5 P1 2000 kPa h1 h2 27161 kJkg x1 h2 hf hfg 27161 9084718898 0957 568 Refrigerant134a is throttled by a valve The pressure and internal energy after expansion are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Heat transfer to or from the fluid is negligible 4 There are no work interactions involved Properties The inlet enthalpy of R134a is from the refrigerant tables Tables A11 through 13 P1 08 MPa T1 25 C h1 hf25 C 8641 kJkg Analysis There is only one inlet and one exit and thus ṁ1 ṁ2 ṁ We take the throttling valve as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout ΔĖsystem 0 steady 0 Ėin Ėout ṁh1 ṁh2 h1 h2 since Q W Δke Δpe 0 Then T2 20 C h2 h1 hf 2549 kJkg hg 23841 kJkg uf 2539 kJkg ug 21884 kJkg Obviously hf h2 hg thus the refrigerant exists as a saturated mixture at the exit state and thus P2 Psat 20 C 13282 kPa Also x2 h2 hf hfg 8641 254921291 02861 Thus u2 uf x2ufg 2539 02861 19345 8074 kJkg 569 Steam is throttled by a wellinsulated valve The temperature drop of the steam after the expansion is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Heat transfer to or from the fluid is negligible 4 There are no work interactions involved Properties The inlet enthalpy of steam is Tables A6 P1 8 MPa T1 350C h1 29881 kJkg Analysis There is only one inlet and one exit and thus m1 m2 m We take the throttling valve as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout ΔĖsystem steady 0 Ėin Ėout m h1 m h2 h1 h2 since Q W Δke Δpe 0 Then the exit temperature of steam becomes P2 2 MPa h2 h1 T2 285C 548 570 Problem 569 is reconsidered The effect of the exit pressure of steam on the exit temperature after throttling a the exit pressu s re varies from 6 MPa to 1 MPa is to be investigated The exit temperature of steam is to be plotted against nalysis The problem is solved using EES and the solution is given below teamiapws WorkingFluid can be changed to ammonia or other fluids kPa nergy balance 1Pin P2Pout h1hin h2hout use arrays to place points on property plot the exit pressure A WorkingFluidS Pin8000 kPa Tin350 C Pout2000 Analysis mdotinmdotout steadystate mass balance mdotin1 mass flow rate is arbitrary mdotinhinQdotWdotmdotouthout0 steadystate e Qdot0 assume the throttle to operate adiabatically Wdot0 throttles do not have any means of producing power hinenthalpyWorkingFluidTTinPPin property table lookup TouttemperatureWorkingFluidPPouthhout property table lookup xoutqualityWorkingFluidPPouthhout xout is the quality at the outlet P P out kPa T out C 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500 6000 3259 331 2705 2777 2846 2912 2976 3037 3095 3152 3207 1000 2000 3000 4000 5000 6000 270 280 290 300 310 320 330 340 Pout kPa Tout C 0 500 1000 1500 2000 2500 3000 3500 100 101 102 103 104 105 106 h kJkg P kPa 285C 350C Steam IAPWS 1 2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 571E Refrigerant134a is throttled by a valve The temperature and internal energy change are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Heat transfer to or from the fluid is negligible 4 There are no work interactions involved Analysis There is only one inlet and one exit and thus m1 m2 m We take the throttling valve as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout ΔĖsystem steady 0 Ėin Ėout m h1 m h2 h1 h2 since Q W Δke Δpe 0 The properties are Tables A11E through 13E P1 160 psia x1 0 h1 4852 Btulbm u1 4810 Btulbm T1 1095F P2 30 psia h2 h1 4852 Btulbm T2 154F u2 4541 Btulbm ΔT T2 T1 154 1095 941F Δu u2 u1 4541 4810 27 Btulbm That is the temperature drops by 941F and internal energy drops by 27 Btulbm Mixing Chambers and Heat Exchangers 572C Under the conditions of no heat and work interactions between the mixing chamber and the surrounding medium 573C Under the conditions of no heat and work interactions between the heat exchanger and the surrounding medium 574C Yes if the mixing chamber is losing heat to the surrounding medium 575 Hot and cold streams of a fluid are mixed in a mixing chamber Heat is lost from the chamber The energy carried from the mixing chamber is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work and heat interactions Analysis We take the mixing device as the system which is a control volume The energy balance for this steadyflow system can be expressed in the rate form as Ėin Ėout ΔĖsystem steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ėin Ėout m1 e1 m2 e2 m3 e3 Qout From a mass balance m3 m1 m2 5 15 20 kgs Substituting into the energy balance equation solving for the exit enthalpy gives m1 e1 m2 e2 Qout m3 e3 5 kgs150 kJkg 15 kgs50 kJkg 55 kW 20 kgs 747 kJkg 576 A hot water stream is mixed with a cold water stream For a specified mixture temperature the mass flow rate of cold water is to be determined Assumptions 1 Steady operating conditions exist 2 The mixing chamber is wellinsulated so that heat loss to the surroundings is negligible 3 Changes in the kinetic and potential energies of fluid streams are negligible 4 Fluid properties are constant 5 There are no work interactions Properties Noting that T Tsat 250 kPa 12741C the water in all three streams exists as a compressed liquid which can be approximated as a saturated liquid at the given temperature Thus h1 hf 80C 33502 kJkg h2 hf 20C 83915 kJkg h3 hf 42C 17590 kJkg Analysis We take the mixing chamber as the system which is a control volume The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance min mout Δmsystem 0 steady 0 m1 m2 m3 Energy balance Ėin Ėout ΔĖsystem 0 steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ėin Ėout m1h1 m2h2 m3h3 since Q W Δke Δpe 0 Combining the two relations and solving for m2 gives m1h1 m2h2 m1 m2h3 m2 h1 h3h3 h2 m1 Substituting the mass flow rate of cold water stream is determined to be m2 33502 17590 kJkg 17590 83915 kJkg 05 kgs 0865 kgs 577E Liquid water is heated in a chamber by mixing it with saturated water vapor If both streams enter at the same rate the temperature and quality if saturated of the exit stream is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 The device is adiabatic and thus heat transfer is negligible Properties From steam tables Tables A5E through A6E h1 hf 65F 3308 Btulbm h2 hg 20 psia 11562 Btulbm Analysis We take the mixing chamber as the system which is a control volume since mass crosses the boundary The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance min mout Δmsystem 0 steady 0 min mout m1 m2 m3 2m m1 m2 m Energy balance Ėin Ėout ΔĖsystem 0 steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ėin Ėout m1h1 m2h2 m3h3 since Q W Δke Δpe 0 Combining the two gives mh1 mh2 2mh3 or h3 h1 h2 2 Substituting h3 3308 115622 5946 Btulbm At 20 psia hf 19627 Btulbm and hg 11562 Btulbm Thus the exit stream is a saturated mixture since hf h3 hg Therefore T3 Tsat 20 psia 228F and x3 h3 hf hfg 5946 19627 11562 19627 0415 578 Two streams of refrigerant134a are mixed in a chamber If the cold stream enters at twice the rate of the hot stream the temperature and quality if saturated of the exit stream are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 The device is adiabatic and thus heat transfer is negligible Properties From R134a tables Tables A11 through A13 h1 hf 20C 7932 kJkg h2 h 1 MPa 80C 31425 kJkg Analysis We take the mixing chamber as the system which is a control volume since mass crosses the boundary The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance min mout Δmsystem 0 steady 0 min mout m1 m2 m3 3m2 since m1 2m2 Energy balance Ėin Ėout ΔĖsystem 0 steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ėin Ėout m1h1 m2h2 m3h3 since Q W Δke Δpe 0 Combining the two gives 2m2h1 m2h2 3m2h3 or h3 2h1 h2 3 Substituting h3 27932 314253 15763 kJkg At 1 MPa hf 10732 kJkg and hg 27099 kJkg Thus the exit stream is a saturated mixture since hf h3 hg Therefore T3 Tsat 1 MPa 3937C and x3 h3 hf hfg 15763 10732 27099 10732 0307 554 579 Problem 578 is reconsidered The effect of the mass flow rate of the cold stream of R134a on the temperature and the quality of the exit stream as the ratio of the mass flow rate of the cold stream to that of the hot stream varies from 1 against the coldtohot mass flow rate ratio nalysis The problem is solved using EES and the solution is given below dmdothot mdot1mdot2 mdot2 Pa 134a Sum of mdotinmdotout across the control surface ment mdot2h2 3QUALITYR134ahh3PP3 m to 4 is to be investigated The mixture temperature and quality are to be plotted A Input Data mfrac 2 mfrac mdotcol T120 C P11000 kPa T280 C P21000 kPa mdot1mfrac P31000 k mdot11 Conservation of mass for the R mdot1 mdot2 mdot3 Conservation of Energy for steadyflow neglect changes in KE and PE We assume no heat transfer and no work occur Edotin Edotout DELTAEdotcv DELTAEdotcv0 Steadyflow require Edotinmdot1h1 Edotoutmdot3h3 Property data are given by h1 enthalpyR134aTT1PP1 h2 enthalpyR134aTT2PP2 T3 temperatureR134aPP3hh3 x frac T 3 C x3 1 125 15 175 2 225 25 275 3 325 35 375 4 3937 01162 3937 3937 3937 3937 3937 3937 3937 3937 3937 3937 3937 3937 05467 0467 04032 0351 03075 02707 02392 02119 0188 01668 01481 01313 1 15 2 25 3 35 4 01 015 02 025 03 035 04 045 05 055 30 35 40 45 x3 T3 C T3 x3 mfrac PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 580E Steam is condensed by cooling water in a condenser The rate of heat transfer in the heat exchanger and the rate of condensation of steam are to be determined Assumptions 1 Steady operating conditions exist 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid 3 Changes in the kinetic and potential energies of fluid streams are negligible 4 Fluid properties are constant Properties The specific heat of water is 10 BtulbmF Table A3E The enthalpy of vaporization of water at 85F is 10452 Btulbm Table A4E Analysis We take the tubeside of the heat exchanger where cold water is flowing as the system which is a control volume The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem 0 steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout Qin m h1 m h2 since Δke Δpe 0 Qin m cp T2 T1 Then the rate of heat transfer to the cold water in this heat exchanger becomes Q m cp Tout Tinwater 138 lbms10 BtulbmF73F 60F 1794 Btus Noting that heat gain by the water is equal to the heat loss by the condensing steam the rate of condensation of the steam in the heat exchanger is determined from Q m hfgsteam msteam Q hfg 1794 Btus 10452 Btulbm 172 lbms 581 Steam is condensed by cooling water in the condenser of a power plant If the temperature rise of the cooling water is not to exceed 10C the minimum mass flow rate of the cooling water required is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid 5 Liquid water is an incompressible substance with constant specific heats at room temperature Properties The cooling water exists as compressed liquid at both states and its specific heat at room temperature is c 418 kJkgC Table A3 The enthalpies of the steam at the inlet and the exit states are Tables A5 and A6 P3 20 kPa x3 095 h3 hf x3 hfg 25142 095 23575 24911 kJkg P4 20 kPa sat liquid h4 hf20 kPa 25142 kJkg Analysis We take the heat exchanger as the system which is a control volume The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance for each fluid stream min mout Δmsystem 0 steady 0 min mout m1 m2 mw and m3 m4 ms Energy balance for the heat exchanger Ein Eout ΔEsystem 0 steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout m1 h1 m3 h3 m2 h2 m4 h4 since Q W Δke Δpe 0 Combining the two mw h2 h1 ms h3 h4 Solving for mw mw h3 h4 h2 h1 ms h3 h4cp T2 T1 ms Substituting mw 24911 25142 kJkg 418 kJkgC10C 200003600 kgs 2977 kgs 582 Ethylene glycol is cooled by water in a heat exchanger The rate of heat transfer in the heat exchanger and the mass flow rate of water are to be determined Assumptions 1 Steady operating conditions exist 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid 3 Changes in the kinetic and potential energies of fluid streams are negligible 4 Fluid properties are constant Properties The specific heats of water and ethylene glycol are given to be 418 and 256 kJkgC respectively Analysis a We take the ethylene glycol tubes as the system which is a control volume The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem 0 steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout m h1 Qout m h2 since Δke Δpe 0 Qout m cp T1 T2 Then the rate of heat transfer becomes Q m cp Tin Toutglycol 32 kgs256 kJkgC80C 40C 3277 kW b The rate of heat transfer from glycol must be equal to the rate of heat transfer to the water Then Q m cp Tout Tinwater mwater Q cp Tout Tin 3277 kJs 418 kJkgC70C 20C 157 kgs 558 583 Problem 582 is reconsidered The effect of the inlet temperature of cooling water on the mass flow rate of water as the inlet temperature varies from 10C to 40C at constant exit temperature is to be investigated The mass flow nalysis The problem is solved using EES and the solution is given below s eg ethylene glycol h mass steam r across the control surface g1 mdotegheg2 d approximation applied for water and ethylene glycol eg2 CpegTeg2 rate of water is to be plotted against the inlet temperature A Input Data Tw120 C Tw270 C w water mdoteg2 kg Teg180 C Teg240 C Cpw418 kJkgK Cpeg256 kJkgK Conservation of mass for the water mdotwinmdotwoutmdotw Conservation of mass for the ethylene glycol mdoteginmdotegoutmdoteg Conservation of Energy for steadyflow neglect changes in KE and PE in eac We assume no heat transfer and no work occu Edotin Edotout DELTAEdotcv DELTAEdotcv0 Steadyflow requirement Edotinmdotwhw1 mdotegheg1 Edotoutmdotwhw2 mdotegheg2 Qexchanged mdoteghe Property data are given by hw1 CpwTw1 liqui hw2 CpwTw2 heg1 CpegTeg1 h T w1 C m w kgs 10 15 20 25 30 35 40 2613 1307 1425 1568 1742 196 224 10 15 20 25 30 35 40 12 14 16 18 2 22 24 26 28 mw kgs Tw1 C PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 584 Oil is to be cooled by water in a thinwalled heat exchanger The rate of heat transfer in the heat exchanger and the exit temperature of water is to be determined Assumptions 1 Steady operating conditions exist 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid 3 Changes in the kinetic and potential energies of fluid streams are negligible 4 Fluid properties are constant Properties The specific heats of water and oil are given to be 418 and 220 kJkgC respectively Analysis We take the oil tubes as the system which is a control volume The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout mdot h1 Qdotout mdot h2 since Δke Δpe 0 Qdotout mdot cp T1 T2 Then the rate of heat transfer from the oil becomes Qdot mdot cp Tin Toutoil 2 kgs22 kJkgC150C 40C 484 kW Noting that the heat lost by the oil is gained by the water the outlet temperature of the water is determined from Qdot mdot cp Tout Tinwater Tout Tin Qdot mdotwater cp 22C 484 kJs 15 kgs418 kJkgC 992C 585 Cold water is heated by hot water in a heat exchanger The rate of heat transfer and the exit temperature of hot water are to be determined Assumptions 1 Steady operating conditions exist 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid 3 Changes in the kinetic and potential energies of fluid streams are negligible 4 Fluid properties are constant Properties The specific heats of cold and hot water are given to be 418 and 419 kJkgC respectively Analysis We take the cold water tubes as the system which is a control volume The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout Qdotin mdot h1 mdot h2 since Δke Δpe 0 Qdotin mdot cp T2 T1 Then the rate of heat transfer to the cold water in this heat exchanger becomes Qdot mdot cp Tout Tincold water 060 kgs418 kJkgC45C 15C 7524 kW Noting that heat gain by the cold water is equal to the heat loss by the hot water the outlet temperature of the hot water is determined to be Qdot mdot cp Tin Touthot water Tout Tin Qdot mdot cp 100C 7524 kW 3 kgs419 kJkgC 940C 586 Air is preheated by hot exhaust gases in a crossflow heat exchanger The rate of heat transfer and the outlet temperature of the air are to be determined Assumptions 1 Steady operating conditions exist 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid 3 Changes in the kinetic and potential energies of fluid streams are negligible 4 Fluid properties are constant Properties The specific heats of air and combustion gases are given to be 1005 and 110 kJkgC respectively Analysis We take the exhaust pipes as the system which is a control volume The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem steady 0 Ein Eout Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies mdot h1 Qdotout mdot h2 since Δke Δpe 0 Qdotout mdot cp T1 T2 Then the rate of heat transfer from the exhaust gases becomes Qdot mdot cp Tin Toutgas 095 kgs11 kJkgC160C 95C 6793 kW The mass flow rate of air is mdot PVdot RT 95 kPa06 m3s 0287 kPam3kgK x 293 K 06778 kgs Noting that heat loss by the exhaust gases is equal to the heat gain by the air the outlet temperature of the air becomes Qdot mdot cp Tcout Tcin Tcout Tcin Qdot mdot cp 20C 6793 kW 06778 kgs1005 kJkgC 120C 562 587E An adiabatic open feedwater heater mixes steam with feedwater The outlet mass flow rate and the outlet velocity are to be determined for two exit temperatures Assumptions Steady operating conditions exist Analysis From a mass balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10 psia 3 120 F 3 f P v v The exit velocity is then Steam 10 psia 200F 01 lbms 3 1 2 10 psia 120F Water 10 psia 100F 2 lbms 21 lbms 2 10 2 1 3 m m m The specific volume at the exit is Table A4E 0 01620 ft lbm 120 F 3 T 3 01733 fts 2 3 2 3 3 3 3 3 3 ft 50 lbms001620 ft lbm 12 4 4 π πD m A m V v v hen the temperature at the exit is 180F we have f T P v v W lbms 21 2 10 2 1 3 m m m 0 01651 ft lbm F 180 10 psia 3 180 F 3 3 3 01766 fts 2 3 4m m V v v 2 3 3 3 3 3 3 ft 50 lbms001651 ft lbm 12 4 π πD A he mass flow rate at the exit is same while the exit velocity slightly increases when the exit temperature is 180F instead of 120F T preparation If you are a student using this Manual you are using it without permission 588E Air is heated in a steam heating system For specified flow rates the volume flow rate of air at the inlet is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid 5 Air is an ideal gas with constant specific heats at room temperature Properties The gas constant of air is 03704 psiaft3lbmR Table A1E The constant pressure specific heat of air is cp 0240 BtulbmF Table A2E The enthalpies of steam at the inlet and the exit states are Tables A4E through A6E P3 30 psia T3 400F P4 25 psia T4 212F h3 12379 Btulbm h4 hf212F 18021 Btulbm Analysis We take the entire heat exchanger as the system which is a control volume The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance for each fluid stream ṁin ṁout Δṁsystem 0 steady 0 ṁin ṁout ṁ1 ṁ2 ṁa and ṁ3 ṁ4 ṁs Energy balance for the entire heat exchanger Ėin Ėout ΔĖsystem 0 steady 0 ṁ1h1 ṁ3h3 ṁ2h2 ṁ4h4 since Q W Δke Δpe 0 Combining the two ṁa h2 h1 ṁs h3 h4 Solving for ṁa ṁa h3 h4h2 h1 ṁs h3 h4cpT2 T1 ṁs Substituting ṁa 12379 18021Btulbm 0240 BtulbmF130 80F 15 lbmmin 1322 lbmmin 2204 lbms Also v1 RT1P1 03704 psiaft3lbmR540 R147 psia 1361 ft3lbm Then the volume flow rate of air at the inlet becomes V1 ṁa v1 2204 lbms1361 ft3lbm 300 ft3s 589 Two streams of cold and warm air are mixed in a chamber If the ratio of hot to cold air is 16 the mixture temperature and the rate of heat gain of the room are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 The device is adiabatic and thus heat transfer is negligible Properties The gas constant of air is R 0287 kPam3kgK The enthalpies of air are obtained from air table Table A17 as h1 h280 K 28013 kJkg h2 h307 K 30723 kJkg hroom h297 K 29718 kJkg Analysis a We take the mixing chamber as the system which is a control volume since mass crosses the boundary The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance ṁin ṁout Δṁsystem 0 steady 0 ṁin ṁout ṁ1 16ṁ1 ṁ3 26ṁ1 since ṁ2 16ṁ1 Energy balance Ėin Ėout ΔĖsystem 0 steady 0 ṁ1h1 ṁ2h2 ṁ3h3 since Q W Δke Δpe 0 Combining the two gives ṁ1h1 22ṁ1h2 32ṁ1h3 or h3 h1 22h232 Substituting h3 28013 22 3072332 29876 kJkg From air table at this enthalpy the mixture temperature is T3 Th 29876 kJkg 2986 K 256C b The mass flow rates are determined as follows v1 RT1P 0287 kPam3kgK7 273 K105 kPa 07654 m3kg ṁ1 V1v1 075 m3s 07654 m3kg 09799 kgs ṁ3 32ṁ1 3209799 kgs 3136 kgs The rate of heat gain of the room is determined from Qgain ṁ3hroom h3 3136 kgs29718 29876 kJkg 493 kW The negative sign indicates that the room actually loses heat at a rate of 493 kW 590 A heat exchanger that is not insulated is used to produce steam from the heat given up by the exhaust gases of an internal combustion engine The temperature of exhaust gases at the heat exchanger exit and the rate of heat transfer to the water are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 Exhaust gases are assumed to have air properties with constant specific heats Properties The constant pressure specific heat of the exhaust gases is taken to be cp 1045 kJkgC Table A2 The inlet and exit enthalpies of water are Tables A4 and A5 Twin 15C x 0 sat liq Pwout 2 MPa x 1 sat vap hwin 6298 kJkg hwout 27983 kJkg Analysis We take the entire heat exchanger as the system which is a control volume The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance for each fluid stream ṁin ṁout Δṁsystem 0 steady 0 ṁin ṁout Energy balance for the entire heat exchanger Ėin Ėout ΔĖsystem 0 steady 0 ṁexhhexhin ṁwhwin ṁexhhexhout ṁwhwout Qout since W Δke Δpe 0 or ṁexhc𝑝Texhin ṁwhwin ṁexhc𝑝Texhout ṁwhwout Qout Noting that the mass flow rate of exhaust gases is 15 times that of the water substituting gives 15ṁw 1045 kJkgC400C ṁw 6298 kJkg 15ṁw 1045 kJkgCTexhout ṁw 27983 kJkg Qout 1 The heat given up by the exhaust gases and heat picked up by the water are Qexh ṁexhc𝑝 Texhin Texhout 15ṁw 1045 kJkgC400 TexhoutC 2 Qw ṁw hwout hwin ṁw 27983 6298kJkg 3 The heat loss is Qout fheat loss Qexh 01 Qexh 4 The solution may be obtained by a trialerror approach Or solving the above equations simultaneously using EES software we obtain Texhout 2061C Qw 9726 kW ṁw 003556 kgs ṁexh 05333 kgs 591 A chilledwater heatexchange unit is designed to cool air by water The maximum water outlet temperature is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid 5 Air is an ideal gas with constant specific heats at room temperature Properties The gas constant of air is 0287 kPam3kgK Table A1 The constant pressure specific heat of air at room temperature is cp 1005 kJkgC Table A2a The specific heat of water is 418 kJkgK Table A3 Analysis The water temperature at the heat exchanger exit will be maximum when all the heat released by the air is picked up by the water First the inlet specific volume and the mass flow rate of air are v1 RT1P1 0287 kPa m3kg K303 K100 kPa 08696 m3kg ma V1v1 5 m3s08696 m3kg 5750 kgs We take the entire heat exchanger as the system which is a control volume The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance for each fluid stream min mout Δmsystem steady 0 min mout m1 m3 ma and m2 m4 mw Energy balance for the entire heat exchanger Ein Eout ΔEsystem steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout m1h1 m2h2 m3h3 m4h4 since Q W Δke Δpe 0 Combining the two ma h1 h2 mw h4 h2 ma c pa T1 T3 mw c pw T4 T2 Solving for the exit temperature of water T4 T2 ma c pa T1 T3mw c pw 8C 5750 kgs1005 kJkg C30 18C2 kgs418 kJkg C 163C PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 592 Refrigerant134a is condensed in a condenser by cooling water The rate of heat transfer to the water and the mass flow rate of water are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work and heat interactions between the condenser and the surroundings Analysis We take the condenser as the system which is a control volume The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout mRh1 mwh3 mRh2 mwh4 mR h1 h2 mw h4 h3 mw c p T4 T3 If we take the refrigerant as the system the energy balance can be written as Ein Eout ΔEsystem steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout mRh1 mRh2 Qout Qout mR h1 h2 a The properties of refrigerant at the inlet and exit states of the condenser are from Tables A11 through A13 P1 1200 kPa T1 85C h1 31673 kJkg P2 1200 kPa T2 Tsat1200 kPa ΔTsubcool 463 63 40C h2 hf40C 10826 kJkg The rate of heat rejected to the water is Qout mR h1 h2 0042 kgs31673 10826kJkg 876 kW 525 kJmin b The mass flow rate of water can be determined from the energy balance on the condenser Qout mw c p ΔT w 876 kW mw 418 kJkg C12C mw 0175 kgs 105 kgmin The specific heat of water is taken as 418 kJkgC Table A3 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 593 Refrigerant22 is evaporated in an evaporator by air The rate of heat transfer from the air and the temperature change of air are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work and heat interactions between the evaporator and the surroundings Analysis We take the condenser as the system which is a control volume The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout mRh1 mah3 mRh2 mah4 mR h2 h1 ma c p ΔTa If we take the refrigerant as the system the energy balance can be written as Ein Eout ΔEsystem steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout mRh1 Qin mRh2 Qin mR h2 h1 a The mass flow rate of the refrigerant is mR V1 v1 2253600 m3s 00253 m3kg 002472 kgs The rate of heat absorbed from the air is Qin mR h2 h1 002472 kgs3980 2202kJkg 439 kW b The temperature change of air can be determined from an energy balance on the evaporator QL mR h3 h2 ma c p Ta1 Ta2 439 kW 05 kgs1005 kJkg C ΔTa ΔTa 87C The specific heat of air is taken as 1005 kJkgC Table A2 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 594 Two mass streams of the same idela gas are mixed in a mixing chamber Heat is transferred to the chamber Three expressions as functions of other parameters are to be obtained Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions Analysis a We take the mixing device as the system which is a control volume The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout m1 h1 m2 h2 Qin m3 h3 From a mass balance m3 m1 m2 Since h cp T Then m1 cp T1 m2 cp T2 Qin m3 cp T3 T3 m1m3 T1 m2m3 T2 Qinm3 cp b Expression for volume flow rate V3 m3 v3 m3 RT3P3 V3 m3 P3 m1m3 T1 m2m3 T2 Qinm3 cp P3 P1 P2 P V3 m1 R T1 P1 m2 R T2 P2 R Qin P3 cp V3 V1 V2 R Qin P cp c If the process is adiabatic then V3 V1 V2 Pipe and duct Flow 595 Heat is supplied to the argon as it flows in a heater The exit temperature of argon and the volume flow rate at the exit are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible Properties The gas constant of argon is 02081 kPam3kgK The constant pressure specific heat of air at room temperature is cp 05203 kJkgC Table A2a Analysis a We take the pipeheater in which the argon is heated as the system which is a control volume The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout m h1 Qin m h2 Qin m h2 h1 Qin m cp T2 T1 100 kPa 300 K 624 kgs Argon 100 kPa Substituting and solving for the exit temperature T2 T1 Qin m cp 300 K 150 kW 624 kgs05203 kJkgK 3462 K 732C b The exit specific volume and the volume flow rate are v2 RT2 P2 02081 kPam3kgK3462 K 100 kPa 07204 m3kg V2 m v2 624 kgs08266 m3kg 450 m3s 596 Saturated liquid water is heated in a steam boiler at a specified rate The rate of heat transfer in the boiler is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions Analysis We take the pipe in which the water is heated as the system which is a control volume The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout m h1 Qin m h2 Qin m h2 h1 2 MPa sat liq 4 kgs Water 2 MPa 250C The enthalpies of water at the inlet and exit of the boiler are Table A5 A6 P1 2 MPa x0 h1 hf 2 MPa 90847 kJkg P2 2 MPa T2 250C h2 29033 kJkg Substituting Qin 4 kgs29033 90847kJkg 7980 kW 597E Saturated liquid water is heated in a steam boiler The heat transfer per unit mass is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions Analysis We take the pipe in which the water is heated as the system which is a control volume The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem steady 0 Rate of net energy transfer by heat work and mass Rate of change in internal kinetic potential etc energies Ein Eout m h1 Qin m h2 Qin m h2 h1 qin h2 h1 500 psia sat liq Water 500 psia 600F The enthalpies of water at the inlet and exit of the boiler are Table A5E A6E P1 500 psia x0 h1 hf 500 psia 44951 Btulbm P2 500 psia T2 600F h2 12986 Btulbm Substituting qin 12986 44951 8491 Btulbm 598 Air at a specified rate is heated by an electrical heater The current is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 The heat losses from the air is negligible Properties The gas constant of air is 0287 kPam3kgK Table A1 The constant pressure specific heat of air at room temperature is cp 1005 kJkgC Table A2a Analysis We take the pipe in which the air is heated as the system which is a control volume The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem 0 Wein 100 kPa 15C Air 100 kPa 03 m3s 30C Ein Eout mh1 Wein mh2 Wein mh2 h1 VI mcpT2 T1 The inlet specific volume and the mass flow rate of air are ν1 RT1 P1 0287 kPa m3kg K288 K 100 kPa 08266 m3kg ṁ V1 v1 03 m3s 08266 m3kg 03629 kgs Substituting into the energy balance equation and solving for the current gives I mcp T2 T1 V 03629 kgs1005 kJkg K30 15K 110 V 1000 VI 1 kJs 497 Amperes 599E The cooling fan of a computer draws air which is heated in the computer by absorbing the heat of PC circuits The electrical power dissipated by the circuits is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 All the heat dissipated by the circuits are picked up by the air drawn by the fan Properties The gas constant of air is 03704 psiaft3lbmR Table A1E The constant pressure specific heat of air at room temperature is cp 0240 BtulbmF Table A2Ea Analysis We take the pipe in which the air is heated as the system which is a control volume The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem 0 Wein 147 psia 70F Air 147 psia 05 ft3s 80F Ein Eout mh1 Wein mh2 Wein mh2 h1 Wein mcpT2 T1 The inlet specific volume and the mass flow rate of air are ν1 RT1 P1 03704 psia ft3 lbm R530 R 147 psia 1335 ft3lbm ṁ V1 v1 05 ft3s 1335 ft3lbm 003745 lbms Substituting Weout 003745 lbms0240 Btulbm R80 70Btulbm 1 kW 094782 Btus 00948 kW 5100 A desktop computer is to be cooled safely by a fan in hot environments and high elevations The air flow rate of the fan and the diameter of the casing are to be determined Assumptions 1 Steady operation under worst conditions is considered 2 Air is an ideal gas with constant specific heats 3 Kinetic and potential energy changes are negligible Properties The specific heat of air at the average temperature of Tavg 45602525 C 3255 K is cp 10065 kJkgC The gas constant for air is R 0287 kJkgK Table A2 Analysis The fan selected must be able to meet the cooling requirements of the computer at worst conditions Therefore we assume air to enter the computer at 6663 kPa and 45C and leave at 60C We take the air space in the computer as the system which is a control volume The energy balance for this steadyflow system can be expressed in the rate form as Ein Eout ΔEsystem 0 Qin mh1 mh2 since Δke Δpe 0 Qin mcpT2 T1 Then the required mass flow rate of air to absorb heat at a rate of 60 W is determined to be Q mcpTout Tin m Q cpTout Tin 60 W 10065 JkgC60 45C 000397 kgs 0238 kgmin The density of air entering the fan at the exit and its volume flow rate are ρ P RT 6663 kPa 0287 kPam³kgK60 273K 06972 kgm³ V ṁ ρ 0238 kgmin 06972 kgm³ 0341 m³min For an average exit velocity of 110 mmin the diameter of the casing of the fan is determined from V AcV πD²4 V D 4V πV 40341 m³min π110 mmin 0063 m 63 cm 582 5108 Problem 5107 is reconsidered The effect of the inner pipe diameter on the rate of heat loss as the pipe ss is to be plotted against the diameter nalysis The problem is solved using EES and the solution is given below 3 kJkgC water is a control volume The energy sed in the rate form as sumption outmdothout out cp T2 diameter varies from 15 cm to 75 cm is to be investigated The rate of heat lo A Knowns D 0025 m rho 965 kgm Vel 06 ms T1 90 C T2 88 C cp 421 Analysis The mass flow rate of e piD24 Ar a mdot rhoAreaVel We take the section of the pipe in the basement to be the system which is balance for this steadyflow system can be expres Edotin Edotout DELTAEdotsys DELTAEdotsys 0 Steadyflow as Edotin mdothin Edotout Qdot hin cp T1 h 001 002 003 004 005 006 007 008 0 5 10 15 20 25 30 D m Q out kW 0015 0025 0035 0045 0055 0065 0075 2872 1149 3191 6254 1034 1544 2157 Qout kW D m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 589 5117 Problem 5116 is reconsidered The effect of the exit velocity on the mass flow rate and the exit volume flow rate is to be investiagted Analysis The problem is solved using EES and the solution is given below Given T1300 K P100 kPa Vel10 ms Wdotein15 kW T280273 K Vel221 ms Properties cp1005 kJkgK R0287 kJkgK Analysis WdoteinmdotcpT2T1mdotVel22Vel12Convertm2s2kJkg energy balance on hair dryer v2RT2P Voldot2mdotv2 Vel2 ms m kgs Vol2 m3s 5 75 10 125 15 175 20 225 25 002815 002813 002811 002808 002804 0028 002795 00279 002783 002852 00285 002848 002845 002841 002837 002832 002826 00282 5 9 13 17 21 25 002815 00282 002825 00283 002835 00284 002845 00285 002855 5 9 13 17 21 25 00278 002785 00279 002795 0028 002805 00281 002815 Vel2 ms m kgs Vel2 ms Vol2 m 3s PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 590 Charging and Discharging Processes 5118 An insulated rigid tank is evacuated A valve is opened and air is allowed to fill the tank until mechanical equilibrium is established The final temperature in the tank is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Air is an ideal gas with constant specific heats 3 Kinetic and potential energies are negligible 4 There are no work interactions involved 5 The device is adiabatic and thus heat transfer is negligible Properties The specific heat ratio for air at room temperature is k 14 Table A2 initially evacuated Air Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance 0 since initial out 2 system out in m m m m m m m i Energy balance C by heat work and mass Net energy transfer 0 since initial out 2 2 etc energies potential in internal kinetic hange system out in pe E E W Q m u h m E E E i i 43 42 1 4243 1 w balances ke Combining the t o i i p i p i kT T c c T c T c T h u 2 2 2 v v Substituting 133C 406 K 290 K 14 T2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 591 5119 Helium flows from a supply line to an initially evacuated tank The flow work of the helium in the supply line and the final temperature of the helium in the tank are to be determined Properties The properties of helium are R 20769 kJkgK cp 51926 kJkgK cv 31156 kJkgK Table A2a Analysis The flow work is determined from its definition but we first determine the specific volume PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4 0811 m kg 200 kPa 273 K 2 0769 kJkgK120 3 line P RT v 8162 kJkg 200 kPa40811 m kg 3 flow Pv w Noting that the flow work in the supply line is converted to sensible internal energy in the tank the final helium temperature in the tank is determined as follows 6550 K tank tank tank tank line line line tank 3 1156 kJkgK 7 kJkg 2040 2040 7 kJkg 273 K 5 1926 kJkgK120 T T c T u c T h h u p v Helium Initially evacuated 200 kPa 120C Alternative Solution Noting the definition of specific heat ratio the final temperature in the tank can also be determined from 6551 K 273 K 1 667120 line tank which is practically the same result kT T preparation If you are a student using this Manual you are using it without permission 592 5120 An evacuated bottle is surrounded by atmospheric air A valve is opened and air is allowed to fill the bottle The amount of heat transfer through the wall of the bottle when thermal and mechanical equilibrium is established is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Air is an ideal gas with variable specific heats 3 Kinetic and potential energies are negligible 4 There are no work interactions involved 5 The direction of heat transfer is to the air in the bottle will be verified Properties The gas constant of air is 0287 kPam3kgK Table A1 Analysis We take the bottle as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance 100 kPa 27C 20 L Evacuated 0 since initial out 2 system out in m m m m m m m i Energy balance C by heat work and mass Net energy transfer 0 since initial out 2 2 in m h Q i i potential etc energies in internal kinetic hange system out in pe ke E E W u m E E E 43 42 1 4243 1 ing the two balances Combin ih m u Q 2 2 in where PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 21407 kJkg 300 K 2 2 u T Ti Substitutin 19 kJkg 300 002323 kg 0287 kPa m kg K300 K A17 Table 3 2 2 2 h RT m i g Qin 002323 kg21407 30019 kJkg 20 kJ or Qout 20 kJ Discussion The negative sign for heat transfer indicates that the assumed direction is wrong Therefore we reverse the direction 100 kPa0020 m 3 P V preparation If you are a student using this Manual you are using it without permission 593 5121 A rigid tank initially contains air at atmospheric conditions The tank is connected to a supply line and air is allowed to enter the tank until mechanical equilibrium is established The mass of air that entered and the amount of heat transfer are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Air is an ideal gas with variable specific heats 3 Kinetic and potential energies are negligible 4 There are no work interactions involved 5 The direction of heat transfer is to the tank will be verified Properties The gas constant of air is 0287 kPam3kgK Table A1 The properties of air are Table A17 25002 kJkg 350 K 21049 kJkg 295 K 29517 kJkg K 295 2 2 1 1 u T u T h T i i Analysis a We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m m m m m m i Pi 600 kPa Ti 22C V1 2 m3 P1 100 kPa T1 22C Q 1 2 system out in Energy balance 0 1 1 since 2 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in pe ke W m u m u m h Q E E E i i 43 42 1 4243 1 The initial and the final masses in the tank are kg 11946 0287 kPa m kg K350 K 3 2 2 m RT 600 kPa 2 m 2362 kg kg K295 K m 100 kPa 2 m 3 2 3 3 P P V V hen from e mass balance b The heat transfer during this process is determined from Discussion The negative sign for heat transfer indicates that the assumed direction is wrong Therefore we reversed the direction 0287 kPa 1 1 1 RT m T th m m m i 2 1 11946 2 362 9584 kg 339 kJ out 1 1 2 2 in kJ 339 2362 kg 21049 kJkg 11946 kg 25002 kJkg kg 29517 kJkg 9584 Q m u m u m h Q i i preparation If you are a student using this Manual you are using it without permission 594 5122 A rigid tank initially contains superheated steam A valve at the top of the tank is opened and vapor is allowed to escape at constant pressure until the temperature rises to 500C The amount of heat transfer is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process by using constant average properties for the steam leaving the tank 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The direction of heat transfer is to the tank will be verified Properties The properties of water are Tables A4 through A6 STEAM 2 MPa Q 34683 kJkg 31169 kJkg 017568 m kg C 500 MPa 2 30242 kJkg 27732 kJkg 012551 m kg C 300 MPa 2 2 2 3 2 2 2 1 1 3 1 1 1 h u T P h u T P v v Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m m m m m m e 2 1 system out in Energy balance 0 1 1 since 2 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in pe ke W m u m u m h Q E E E e e 43 42 1 4243 1 The state and thus the enthalpy of the steam leaving the tank is changing during this process But for simplicity we assume constant properties for the exiting steam at the average values Thus 32462 kJkg 2 2 2 1 he 34683 kJkg 30242 h h The initial and the final masses in the tank are kg 1138 017568 m kg 3 2 2 v m 02 1594 kg m kg 012551 m 02 3 2 3 3 1 1 1 V v V m m hen from e mass and energy balance relations T th 0456 kg 1 138 1 594 2 1 m m me 6068 kJ 1594 kg 27732 kJkg 1138 kg 31169 kJkg kg 32462 kJkg 0456 1 1 2 2 m u m u m h Q e e in preparation If you are a student using this Manual you are using it without permission 595 5123 A cylinder initially contains saturated liquidvapor mixture of water The cylinder is connected to a supply line and the steam is allowed to enter the cylinder until all the liquid is vaporized The final temperature in the cylinder and the mass of the steam that entered are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 The expansion process is quasiequilibrium 3 Kinetic and potential energies are negligible 3 There are no work interactions involved other than boundary work 4 The device is insulated and thus heat transfer is negligible PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 00 kPa 2 x h h h P Analysis a The cylinder contains saturated vapor at the final state at a pressure of 200 kPa thus the final temperature in be T Tsat 200 kPa 1202C which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as P 200 kPa m1 10 kg H2O Pi 05 MPa Ti 350C Properties The properties of steam are Tables A4 through A6 60 1 1 1 fg f x 1 18256 kJkg 22016 06 50471 31681 kJkg C 350 MPa 05 27063 kJkg sat vapor 200 kPa 2 i i g h T P 200 kPa 2 h h P i the cylinder must 2 b We take the cylinder as the system 1 2 system out in m m m m m m i Mass balance Energy balance pe ke Q m u m u W m h i i 43 42 1 since the boundary work and U combine into H for constant pressure expansion and compression processes Solving for m2 and substituting 0 1 1 since 2 2 out b potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in E E E 4243 1 Combining the two relations gives 1 1 2 2 1 2 bout 0 m u m u m h m W i or 1 1 2 2 1 2 0 m h m h m h m i 2907 kg 10 kg 27063 kJkg 31681 18256 kJkg 31681 1 2 1 2 m h h h h m i i Thus mi m2 m1 2907 10 1907 kg preparation If you are a student using this Manual you are using it without permission 596 5124E A scuba divers air tank is to be filled with air from a compressed air line The temperature and mass in the tank at the final state are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Air is an ideal gas with constant specific heats 3 Kinetic and potential energies are negligible 4 There are no work interactions involved 5 The tank is wellinsulated and thus there is no heat transfer Properties The gas constant of air is 03704 psiaft3lbmR Table A1E The specific heats of air at room temperature are cp 0240 BtulbmR and cv 0171 BtulbmR Table A2Ea Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m m m m m m i 1 2 system out in m c T m c T c T m v The initial and final masses are given by 20 psia 70F 2 ft3 120 psia 100F Air Energy balance 1 1 2 2 1 1 2 2 potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in m c T m c T c T m m u m u h m E E E i p i i i v v 43 42 1 4243 1 Combining the two balances m 1 1 2 2 1 2 i p v 0 2038 lbm 460 R 03704 psia ft lbm R70 2 0 psia 2 ft 3 3 1 1 1 RT P m V 2 2 3 3 2 2 m2 9 647 03704 psia ft lbm R 12 0 psia 2 ft T T RT P V ubstituting S 0 2038 0 171530 647 9 0 171 0 2038 0 24560 647 9 2 2 2 T T T The final mass is then whose solution is 2674 F 7274 R T2 0890 lbm 727 4 647 9 9 647 2 2 T m preparation If you are a student using this Manual you are using it without permission 597 5125 R134a from a tank is discharged to an airconditioning line in an isothermal process The final quality of the R134a in the tank and the total heat transfer are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the exit remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m m m m m m m m e e Energy balance m h m u m u Q m u u E E E 1 1 2 2 in 1 1 2 2 potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 43 42 1 4243 1 he initial state properties of R134a in the t 3 e v quid state and found the exiting enthalpy The final specific volume in the container is AC line Liquid R134a 5 kg 24C 2 1 1 2 system out in m e e m m h Q in e e Combining the two balances he m m m u m u Q 2 1 1 1 2 2 in T ank are 98 kJkg 84 8444 kJkg 0 24 C 1 1 1 h u x T Table A11 0008261 m kg 0 Note that we assumed that the refrigerant leaving the tank is at saturated li accordingly The volume of the tank is 3 3 1 1 m 0 004131 m kg 5 kg 0 0008261 v V m 001652 m kg 0 25 kg m 0 004131 3 3 2 2 m V v The final state is now fixed The properties at this state are Table A11 16473 kJkg 0 506115865 kJkg 8444 kJkg 0 0008261 0 031834 0 0008261 01652 0 01652 m kg 0 C 24 2 2 2 2 3 2 2 fg f fg f x u u u x T 05061 v v v v Substituting into the energy balance equation 2264 kJ 4 75 kg8498 kJkg 5 kg8444 kJkg 0 25 kg16473 kJkg 2 1 1 1 2 2 in he m m m u m u Q preparation If you are a student using this Manual you are using it without permission 598 5126E Oxygen is supplied to a medical facility from 10 compressed oxygen tanks in an isothermal process The mass of oxygen used and the total heat transfer to the tanks are to be determined Assumptions 1 This is an unsteady process but it can be analyzed as a uniformflow process 2 Oxygen is an ideal gas with constant specific heats 3 Kinetic and potential energies are negligible 4 There are no work interactions involved Properties The gas constant of oxygen is 03353 psiaft3lbmR Table A1E The specific heats of oxygen at room temperature are cp 0219 BtulbmR and cv 0157 BtulbmR Table A2Ea Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m m m m m m m m e Energy balance m u m u m h Q E E E 1 1 2 2 in 1 1 2 2 1 1 2 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in v v 43 42 1 4243 1 The initial and final masses and the mass used are Oxygen 1500 psia 80F 15 ft3 2 1 1 2 system out in me e p e e e e e m c T m c T m c T Q m h m u m u Q in Combining the two balances cpTe m m m c T m c T Q 2 1 1 1 2 2 in v v 124 3 lbm 460 R 03353 psia ft lbm R80 150 0 psia15 ft 3 3 1 1 1 RT P m V 2485 lbm 460 R 03353 psia ft lbm R80 30 0 psia15 ft 2 P V 3 3 2 m RT Substituting into the energy balance equation 2 9941 lbm 2485 124 3 2 1 m m me 3328 Btu 9941 0 219540 124 3 0 157540 2485 0 157540 1 1 2 2 in e e p m c T m c T m c T Q v v preparation If you are a student using this Manual you are using it without permission 599 5127E Steam is supplied from a line to a weighted pistoncylinder device The final temperature and quality if appropriate of the steam in the piston cylinder and the total work produced as the device is filled are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Kinetic and potential energies are negligible 3 The process is adiabatic Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively and also noting that the initial mass in the system is zero the mass and energy balances for this uniformflow system can be expressed as Mass balance 2 system out in m m m m m i Energy balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course out potential etc energies in internal kinetic ange system out in m u W h m E E E b i i 43 42 1 4243 1 ombining the two balances he boundary work is determined from Ch by heat work and mass Net energy transfer 2 2 2 2 out m u m h W i i b C 2 2 out u h m W i b T 2 2 1 1 2 2 1 2 out v v v V V Pm m P m P Wb Substituting the energy balance equation simplifies into The enthalpy of steam at the inlet is alue into the energy balance equation and using an iterative solution of this equation gives or better yet sing EES oftware v u The final mass is 2 2 2 2 2 u h P u h m Pm i i v v 2 1226 4 Btulbm Table A 6E F 450 300 psia i i i h T P Substituting this v u s 2 T F 4251 4575 ft lbm 2 5 Btulbm 1135 3 2 2 4 069 lbm 4575 ft lbm 2 ft 10 3 3 2 2 2 v V m and the work produced is Btu 3701 3 3 2 out psia ft 5404 1Btu 200 psia10 ft PV Wb preparation If you are a student using this Manual you are using it without permission 5100 5128E Oxygen is supplied from a line to a weighted pistoncylinder device The final temperature of the oxygen in the piston cylinder and the total work produced as the device is filled are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Kinetic and potential energies are negligible 3 The process is adiabatic 4 Oxygen is an ideal gas with constant specific heats Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively and also noting that the initial mass in the system is zero the mass and energy balances for this uniformflow system can be expressed as Mass balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m m m m Energy balance 2 2 out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in m u h m u W h m E E E i i b i i 43 42 1 4243 1 ombining the two balances he boundary work is determined from 2 system out in mi out m Wb 2 2 C 2 2 out u h m W i b T 2 2 1 1 2 2 1 2 out v v v V V Pm m P m P Wb S g ubstitutin the energy balance equation simplifies into c T c T RT u h P v i p i v perature 2 2 2 2 2 2 2 2 u h m Pm i v Solving for the final tem 450F i i p p i p p i T T c c T c R c T c T c T RT v v 2 2 2 The work produced is Btu 3701 3 3 2 out psia ft 5404 1Btu 200 psia10 ft PV Wb preparation If you are a student using this Manual you are using it without permission 5101 5129 A rigid tank initially contains saturated R134a vapor The tank is connected to a supply line and R134a is allowed to enter the tank The mass of the R134a that entered and the heat transfer are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The direction of heat transfer is to the tank will be verified Properties The properties of refrigerant are Tables A11 through A13 R134a 003 m3 14 MPa Sat vapor 16 MPa 36C R134a 10233 kJkg C 36 MPa 16 kJkg 13443 00009400 m kg liquid sat MPa 16 kJkg 25637 001411 m kg vapor sat MPa 14 36 C 16 MPa 2 3 16 MPa 2 2 14 MPa 1 3 14 MPa 1 1 f i i i f f g g h h T P u u P u u P v v v v Q Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m m m m m m i 1 2 system out in Energy balance 0 1 1 since 2 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in pe ke W m u m u m h Q E E E i i 43 42 1 4243 1 a The initial and the final masses in the tank are 3192 kg 00009400 m kg 3 2 2 m v m 003 2 127 kg kg m 003 3 2 3 3 V V Then from e mass balance c The heat transfer during this process is determined from the energy balance to be m 001411 1 1 1 v m th 2979 kg 2 127 3192 1 2 m m mi 697 kJ 2127 kg 25637 kJkg 3192 kg 13443 kJkg kg 10233 kJkg 2979 1 1 2 2 in m u m u m h Q i i preparation If you are a student using this Manual you are using it without permission 5102 5130 A rigid tank initially contains saturated liquid water A valve at the bottom of the tank is opened and half of the mass in liquid form is withdrawn from the tank The temperature in the tank is maintained constant The amount of heat transfer is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid leaving the device remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The direction of heat transfer is to the tank will be verified Properties The properties of water are Tables A4 through A6 85226 kJkg liquid sat C 200 kJkg 85046 m kg 0001157 liquid sat C 200 200 C 200 C 1 3 200 C 1 1 o o o o f e e f f h h T u u T v v Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m m m m m m e Mass balance in 2 1 system out Energy balance H2O Sat liquid T 200C V 03 m3 Q 0 1 1 since 2 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in pe ke W m u m u m h Q E E E e e 43 42 1 4243 1 The initial and the final masses in the tank are 1297 kg kg 2594 2594 kg 0001157m kg 1 1 1 m v m 03 2 1 1 2 1 2 3 3 m m V m m m Now we determine the final internal energy Then from the mass balance 1297 kg 129 7 259 4 2 1 e 86646 kJkg 0 009171 1743 7 85046 009171 0 C 200 0 009171 0 001157 0 12721 0 001157 002313 0 0002313 m kg 1297 kg m 03 2 2 2 2 2 2 3 3 2 2 fg f fg f x u u u x T x m v v v V v Then the heat transfer during this process is determined from the energy balance by substitution to be 2308 kJ 2594 kg 85046 kJkg 1297 kg 86646 kJkg 1297 kg 85226 kJkg Q preparation If you are a student using this Manual you are using it without permission 5103 5131 A rigid tank initially contains saturated liquidvapor mixture of refrigerant134a A valve at the bottom of the tank is opened and liquid is withdrawn from the tank at constant pressure until no liquid remains inside The amount of heat transfer is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid leaving the device remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The direction of heat transfer is to the tank will be verified Properties The properties of R134a are Tables A11 through A13 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0025621 m kg 00008458 m kg kPa 800 800 kPa 800 kPa 2 3 3 f e e g g f h h P u u P v v he tank as the system which is a control volume since mass crosses the boundary Noting that the icroscopi energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively essed as Mass balance R134a Sat vapor P 800 kPa V 012 m3 Q 1 0025621 m kg kPa 800 24679 kJkg 9479 kJkg 3 800 kPa 2 2 g g f P u u v v 9547 kJkg liquid sat kPa 800 24679 kJkg vapor sat Analysis We take t m c the mass and energy balances for this uniformflow system can be expr 2 1 system out in m m m m m m e Energy balance potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in pe ke W u m E E E 43 42 1 4243 1 he initial mass initial internal energy and final mass in the tank are 2 2 in m u m h Q e e 0 1 since 1 T 4 684 kg 0025621 m kg 2 m 01 4229 2 kJ 3513 24679 9479 3547 3898 kg 3513 3547 m kg 0025621 075 m 012 m kg 00008458 025 m 012 3 3 2 2 1 1 1 3 3 3 3 1 v V v V v V m m u m u m u U m m m g g f f g g f f g f Then from the mass and energy balances 3430 kg 4 684 3898 2 1 m m me 2012 kJ 4229 kJ 4684 kg 24679 kJkg 3430 kg 9547 kJkg Qin preparation If you are a student using this Manual you are using it without permission 5104 5132E A rigid tank initially contains saturated liquidvapor mixture of R134a A valve at the top of the tank is opened and vapor is allowed to escape at constant pressure until all the liquid in the tank disappears The amount of heat transfer is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid leaving the device remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved Properties The properties of R134a are Tables A11E through A13E PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 029316 ft lbm 001413 ft lbm psia 160 160 psia 160 psia 2 3 3 g e e g g f h h P u u P v v he tank as the system which is a control volume since mass crosses the boundary Noting that the icroscopi energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively essed as Mass balance R134a Sat vapor P 160 psia V 2 ft3 Q 1 029316 ft lbm psia 160 10850 Btulbm 4810 Btulbm 3 160 psia 2 2 g g f v P u u v 11718 Btulbm vapor sat psia 160 10850 Btulbm vapor sat Analysis We take t m c the mass and energy balances for this uniformflow system can be expr 2 1 system out in m m m m m m e Energy balance potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in pe ke W E E E 43 42 1 4243 1 he initial mass initial internal energy and final mass in the tank are 0 1 1 since 2 2 in m u m u m h Q e e T 6822 lbm ft lbm 029316 ft 2 1043 Btu 648 10850 4810 7077 1356 lbm 648 7 077 ft lbm 029316 08 ft 2 ft lbm 001413 02 ft 2 3 3 2 2 1 1 1 3 3 3 3 1 v V v V v V m m u m u m u U m m m g g f f g g f f g f Then from the mass and energy balances 6 736 lbm 6 822 1356 2 1 m m me 486 Btu 1043 Btu 6822 lbm 10850 Btulbm lbm 11718 Btulbm 6736 1 1 2 2 in m u m u m h Q e e preparation If you are a student using this Manual you are using it without permission 5105 5133 A rigid tank initially contains saturated R134a liquidvapor mixture The tank is connected to a supply line and R 134a is allowed to enter the tank The final temperature in the tank the mass of R134a that entered and the heat transfer are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The direction of heat transfer is to the tank will be verified Properties The properties of refrigerant are Tables A11 through A13 kJkg 33506 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 100 i C T MPa 10 kJkg 24448 002936 m kg vapor sat 00 kPa 7 16726 07 7057 70 700 kPa 2 3 700 kPa 2 2 1 1 1 i g g fg f P u u P x u u u x v v 18765 kJkg 003063 m kg 00008020 004342 07 00008020 14 C 3 1 1 i fg f h x T v v v the tank as the system which is a control volume nce mass crosses the boundary Noting that the microscopic energies f flowing and nonflowing fluids are represented by enthalpy h and xpressed as ass balance 1 1 MPa 100C 04 m3 R134a R134a Analysis We take si o internal energy u respectively the mass and energy balances for this uniformflow system can be e 1 2 system out in m m m m m m i M Energy balance Change by heat work and mass Net energy transfer 0 1 1 since 2 2 in etc energies potential in internal kinetic system out in pe ke W m u m u m h Q E E E i i 43 42 1 4243 1 perature is the saturation mperature at this pressure b The initial and the final masses in the tank are a The tank contains saturated vapor at the final state at 800 kPa and thus the final tem te 267C sat 700 kPa 2 T T 1362 kg m kg 002936 m 04 1306 kg m kg 003063 m 04 3 3 2 2 3 3 1 1 v V v V m m Then from the mass balance c The heat transfer during this process is determined from the energy balance to be 05653 kg 1306 1362 1 2 m m mi 691 kJ 1306 kg 18765 kJkg 1362 kg 24448 kJkg kg 33506 kJkg 05653 1 1 2 2 in m u m u m h Q i i preparation If you are a student using this Manual you are using it without permission 5106 5134 A hotair balloon is considered The final volume of the balloon and work produced by the air inside the balloon as it expands the balloon skin are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process 2 Air is an ideal gas with constant specific heats 3 Kinetic and potential energies are negligible 4 There is no heat transfer Properties The gas constant of air is 0287 kPam3kgK Table A1 Analysis The specific volume of the air at the entrance and exit and in the balloon is 0 8840 m kg 100 kPa 273 K 0287 kPa m kg K35 3 3 P RT v The mass flow rate at the entrance is then PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2 262 kgs m kg 80 v i m 840 2 ms m 1 3 2 AiVi hile that at the outlet is w 5656 kgs 0 0 8840 m kg 3 v m e ms 1 m 50 2 AeVe pplying a mass balance to the balloon 1 2 m m m m e i he volum in the balloon then changes by the amount and the final volume of the balloon is In order to push back the boundary of the balloon against the surrounding atmosphere the amount of work that must be done is A system out in m m m 1 2 t m m m m e i 203 6 kg 0 5656 kgs2 60 s 262 2 T e 3 3 1 2 180 m 203 6 kg08840 m kg v V m m 255 m3 180 75 1 2 V V V 18000 kJ 3 3 out kPa m 1 1kJ 100 kPa180 m P V Wb preparation If you are a student using this Manual you are using it without permission 5107 5135 An insulated rigid tank initially contains helium gas at high pressure A valve is opened and half of the mass of helium is allowed to escape The final temperature and pressure in the tank are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process by using constant average properties for the helium leaving the tank 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The tank is insulated and thus heat transfer is negligible 5 Helium is an ideal gas with constant specific heats Properties The specific heat ratio of helium is k 1667 Table A2 Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m m m m m m e Mass balance in 2 1 system out 1 2 1 2 1 2 1 2 given m m m m m e Energy balance potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 43 42 1 4243 1 0 1 1 s 2 2 m u m u m h e e ince pe ke Q W E E E d thus the enthalpy of helium leaving the tank is changing during this process But for simplicity we operties for the exiting steam at the average values and energy balances He 015 m3 3 MPa 130C Note that the state an assume constant pr 0 1 2 1 1 2 1 2 1 1 m h m u m u e Combining the mass 1 2 2 1 1 2 2 2 0 2 Dividing by m12 0 he c T c T T T c or u u p v v ividing by cv vc c k T T T k T p since 4 2 0 1 2 2 1 D 257 K 403 K 1667 2 1667 4 2 4 1 2 k T k T Solving for T2 The final pressure in the tank is 956 kPa 403 3000 kPa 257 2 1 1 1 1 2 2 2 2 2 1 1 2 1 m T P m T P m RT m RT P P V V preparation If you are a student using this Manual you are using it without permission 5108 5136E An insulated rigid tank equipped with an electric heater initially contains pressurized air A valve is opened and air is allowed to escape at constant temperature until the pressure inside drops to 30 psia The amount of electrical work transferred is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the exit temperature and enthalpy of air remains constant 2 Kinetic and potential energies are negligible 3 The tank is insulated and thus heat transfer is negligible 4 Air is an ideal gas with variable specific heats Properties The gas constant of air is R 03704 psiaft3lbmR Table A1E The properties of air are Table A17E PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 580 R 13866 Btu lbm e tank as the system which is a control volume since mass crosses the boundary Noting that the of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as m m m m m m e i i T h T u T u 580 R 9890 Btu lbm 580 R 9890 Btu lbm 1 1 2 2 Analysis We take th microscopic energies Mass balance AIR 60 ft3 75 psia 120F We 2 1 system out in The initial and the final masses of air in the tank are Energy balance 0 1 1 since 2 2 in e potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in pe ke Q m u m u m h W E E E e e 43 42 1 4243 1 838 lbm psia ft lbm R 580 R 03704 psia 60 ft 30 2095 lbm psia ft lbm R 580 R 03704 ft 3 3 2 2 2 3 3 1 RT P m RT m V Then from the mass and energy balances 75 psia 60 1 PV 1 m m m e 1 2 20 95 8 38 1257 lbm 500 Btu 2095 lbm 9890 Btulbm 838 lbm 9890 Btulbm lbm 13866 Btulbm 1257 1 1 2 2 ein m u m u m h W e e preparation If you are a student using this Manual you are using it without permission 5109 5137 A vertical cylinder initially contains air at room temperature Now a valve is opened and air is allowed to escape at constant pressure and temperature until the volume of the cylinder goes down by half The amount air that left the cylinder and the amount of heat transfer are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the exit temperature and enthalpy of air remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions other than boundary work 4 Air is an ideal gas with constant specific heats 5 The direction of heat transfer is to the cylinder will be verified Properties The gas constant of air is R 0287 kPam3kgK Table A1 Analysis a We take the cylinder as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m m m m m m e AIR 300 kPa 02 m3 20C 2 1 system out in Energy balance 0 1 1 since 2 2 bin in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in pe ke m u m u m h W Q E E E e e 43 42 1 4243 1 The initial and the final masses of air in the cylinder are 1 2 3 2 2 0357 kg 0287 kPa m kg K 293 K m RT m 1 3 2 2 3 3 1 kPa 01 m 300 0714 kg kg K 293 K m kPa 02 m 300 P P V V b This i hus the Wb and the U terms can be combined into H to yield 1 1 1 0287 kPa RT m Then from the mass balance m m m e 1 2 0 714 0 357 0357 kg s a constant pressure process and t Q m h m h m h e e 2 2 1 1 Noting that the temperature of the air remains constant during this process we have hi h1 h2 h lso A 1 2 1 2 m m me Thus 0 m h m m Q 1 1 2 1 1 2 1 preparation If you are a student using this Manual you are using it without permission 5110 5138 A balloon is initially filled with helium gas at atmospheric conditions The tank is connected to a supply line and helium is allowed to enter the balloon until the pressure rises from 100 to 125 kPa The final temperature in the balloon is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Helium is an ideal gas with constant specific heats 3 The expansion process is quasiequilibrium 4 Kinetic and potential energies are negligible 5 There are no work interactions involved other than boundary work 6 Heat transfer is negligible Properties The gas constant of helium is R 20769 kJkgK Table A1 The specific heats of helium are cp 51926 and cv 31156 kJkgK Table A2a Analysis We take the cylinder as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance 1 2 system out in m m m m m m i Energy balance 0 1 1 since 2 2 out b potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in pe ke Q m u m u W h m E E E i i 43 42 1 4243 1 kg 30093 K kPa m kg K 20769 kPa 50 m 125 50 m 100 kPa 40 m kPa 125 6 641 kg kPa m kg K 290 K 20769 kPa 40 m 100 2 2 3 3 2 2 2 2 3 3 1 1 2 2 1 1 3 3 1 1 1 1 T T RT m P P P P RT P m V V V V V 2 2 P V Then from the mass balance 6 641 kg 3 3009 2 1 2 T m m mi Noting that P varies linearly with V the boundary work done during this process is 1125 kJ 40 m 50 2 125 kPa 100 2 3 1 2 2 1 Wb V P V P sing specific heats the energy balance relation reduces to g U 1 1 2 2 bout m c T m c T m c T W i p i v v Substitutin 3 1156290 6 641 3 1156 3009 3 5 1926 298 6 641 3009 3 1125 2 2 2 T T T It yields T2 315 K PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5111 5139 The air in an insulated rigid compressedair tank is released until the pressure in the tank reduces to a specified value The final temperature of the air in the tank is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process 2 Air is an ideal gas with constant specific heats 3 Kinetic and potential energies are negligible 4 There are no work interactions involved 5 The tank is wellinsulated and thus there is no heat transfer Properties The gas constant of air is 0287 kPam3kgK Table A1 The specific heats of air at room temperature are cp 1005 kJkgK and cv 0718 kJkgK Table A2a Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m m m m m m m m e Energy balance m u m u h m E E E 1 1 2 2 1 1 2 2 1 1 2 2 potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 0 v v 43 42 1 4243 1 The initial and final masses are given by Air 4000 kPa 20C 05 m3 2 1 1 2 system out in me e p e e e e e m c T m c T c T m m h m u m u 0 Combining the two balances cpTe m m m c T m c T 0 2 1 1 1 2 2 v v 2378 kg 273 K 0287 kPa m kg K20 3 1 1 m RT m 50 400 0 kPa 3 1PV 2 2 3 3 2 3484 m 50 200 0 kPa P m V 2 2 0287 kPa m kg K T T RT he temperature of air leaving the tank changes from the initial temperature in the tank the final temperature during the discharging process We assume that the temperature of the air leaving the tank is the average of initial and final temperatures in the tank Substituting into the energy balance equation gives T to 0 c T m m m c T c T m 2 1 005 293 3484 2378 2378 0 718293 3484 0 718 0 2 2 2 2 2 1 1 1 2 2 T T T T e p v v whose solution by trialerror or by an equation solver such as EES is 32 C K 241 T2 preparation If you are a student using this Manual you are using it without permission 5112 5140 An insulated pistoncylinder device with a linear spring is applying force to the piston A valve at the bottom of the cylinder is opened and refrigerant is allowed to escape The amount of refrigerant that escapes and the final temperature of the refrigerant are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process assuming that the state of fluid leaving the device remains constant 2 Kinetic and potential energies are negligible Properties The initial properties of R134a are Tables A11 through A13 11 kJkg 354 32503 kJkg 02423 m kg 0 C 120 MPa 21 1 1 3 1 1 1 h u T P v Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m m m m m m e Mass balance in 2 1 system out Energy balance R 134a 08 m3 12 MPa 120C 0 1 1 since 2 2 in b potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in pe ke Q m u m u m h W E E E e e 43 42 1 4243 1 The initial mass and the relations for the final and exiting masses are 2 3 2 1 3302 m m e 2 3 2 2 2 3 m 05 m 05 m 08 v v V V m v m 3 1 1 1 3302 kg 002423 m kg v m Noting that the spring is linear the boundary work can be determined from 270 kJ 0805m 2 600 kPa 1200 2 3 2 1 2 1 bin V V P P W Substituting the energy balance 3302 kg32503 kJkg m 50 m 50 3302 270 2 2 3 2 3 u he v v Eq 1 where th age of initial and final enthalpies of the refrigerant in the cylinder That is e enthalpy of exiting fluid is assumed to be the aver 2 35411 kJkg 2 2 2 1 h h h he Final state properties of the refrigerant h2 u2 and v2 are all functions of final pressure known and temperature unknown The solution may be obtained by a trialerror approach by trying different final state temperatures until Eq 1 is satisfied Or solving the above equations simultaneously using an equation solver with builtin thermodynamic functions such as EES we obtain T2 968C me 2247 kg h2 33620 kJkg u2 30777 kJkg v2 004739 m3kg m2 1055 kg preparation If you are a student using this Manual you are using it without permission 5113 5141 Steam at a specified state is allowed to enter a pistoncylinder device in which steam undergoes a constant pressure expansion process The amount of mass that enters and the amount of heat transfer are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid entering the device remains constant 2 Kinetic and potential energies are negligible Properties The properties of steam at various states are Tables A4 through A6 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1 2 3 3 1 1 1 0 16667 m kg 06 kg m 01 P P m v V 1 2004 4 kJkg 16667 m kg 0 kPa 800 1 3 1 u P v 800 kPa 2 2 P v Analysis ake the tank as the system which is a control volume since mass crosses the boundary Noting that the es of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively gy balances for this uniformflow system can be expressed as Mass balance Q Steam 5 MPa 500C Steam 06 kg 01 m3 800 kPa 0 29321 m kg 3 2715 9 kJkg 250 C 2 2 u T 3434 7 kJkg 5 MPa i i h P iT 500C a We t microscopic energi the mass and ener 1 2 system out in m m m m m m i Energy balance e Change in internal kinetic by pe ke 43 42 1 ing that the pressure remains constant the boundary s determined from 1 2 bout V P V W heat work and mass Net energy transfer 4243 1 system out in E E E 1 1 sinc 2 2 bout in potential etc energies m u m u m h W Q i i 0 Not work i 3 80 kJ 01m 800 kPa2 01 The final mass and the mass that has entered are 0082 kg 60 682 0 0 682 kg m kg 029321 m 02 1 2 3 3 2 2 2 m m m m i v V b Finally substituting into energy balance equation 4479 kJ in in kg20044 kJkg 60 0 682 kg27159 kJkg 0 082 kg34347 kJkg kJ 80 Q Q preparation If you are a student using this Manual you are using it without permission 5114 5142 Steam is supplied from a line to a pistoncylinder device equipped with a spring The final temperature and quality if appropriate of the steam in the cylinder and the total work produced as the device is filled are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Kinetic and potential energies are negligible 3 The process is adiabatic Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively and also noting that the initial mass in the system is zero the mass and energy balances for this uniformflow system can be expressed as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course out potential etc energies in internal kinetic ange system out in m u W h m E E E b i i 43 42 1 4243 1 ances Because of the spring the relation between the pressure and volume is a linear relation According to the data in the roblem statement Mass balance 2 system out in m m m m m i Energy balance Ch by heat work and mass Net energy transfer 2 2 2 2 out m u m h W i i b Combining the two bal 2 2 out u h m W i b p V 5 P 300 2700 nal vapor volume is then The fi 3 2 2 222 m 300 1500 2700 5 V The work needed to compress the spring is 2000 kJ 2700 2700 2 V 2 222 300 2 222 270 300 2 5 300 5 2 2 2 2 out V V V V V d Pd Wb i i i h T P Substituting the information found into the energy balance equation gives 0 The enthalpy of steam at the inlet is 2796 0 kJkg Table A 6 C 200 kPa 1500 2 222 2796 0 2000 2 2 2 2 2 out 2 2 out u u h W u h m W i b i b v v V Using an iterative solution of this equation with steam tables gives or better yet using EES software 1458 m kg 0 8 kJkg 2664 3 2 2 2 v u T C 2332 preparation If you are a student using this Manual you are using it without permission 5115 5143 Air is supplied from a line to a pistoncylinder device equipped with a spring The final temperature of the steam in the cylinder and the total work produced as the device is filled are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Kinetic and potential energies are negligible 3 The process is adiabatic 4 Air is an ideal gas with constant specific heats Properties The gas constant of air is 0287 kPam3kgK Table A1 The specific heats of air at room temperature are cp 1005 kJkgK and cv 0718 kJkgK Table A2a Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively and also noting that the initial mass in the system is zero the mass and energy balances for this uniformflow system can be expressed as Mass balance 2 system out in m m m m m i Energy balance 2 2 out 2 2 out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in m u m h W m u W h m E E E i i b b i i 43 42 1 4243 1 Combining the two balances PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course g the relation between the pressure and volume is a linear relation According to the data in the problem statement 2 2 out u h m W i b Because of the sprin V 5 P 300 2700 The final a volume is then ir 3 2 3 148 m 300 2000 5 V 2700 work needed to compress the spring is The 3620 kJ 3 14 270 300 2 5 300 5 2 2 2 out V V V V V d Pd Wb 3 148 300 8 2700 2700 2 0 V2 und into the energy balance equation gives Substituting the information fo 0 718 1 005 600 0 287 3 148 2000 3620 2 2 2 2 2 2 out 2 2 out T T c T c T RT P W u h m W i p b i b v V The final temperature is then 4099 C 6829 K T2 preparation If you are a student using this Manual you are using it without permission 5118 5146 The rate of accumulation of water in a pool and the rate of discharge are given The rate supply of water to the pool is to be determined Assumptions 1 Water is supplied and discharged steadily 2 The rate of evaporation of water is negligible 3 No water is supplied or removed through other means Analysis The conservation of mass principle applied to the pool requires that the rate of increase in the amount of water in the pool be equal to the difference between the rate of supply of water and the rate of discharge That is e i e i e i dt dt since the density of water is constant and thus the conservation of mass dt d m dm m m m dm V V V pool pool pool is equivalent to conservation of volume The rate of discharge of water is 3 s ction of e pool times the rate at which the water level rises PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 01539 m s 007 m 44 ms 4 2 e 2 e π π V D AeV e V The rate of accumulation of water in the pool is equal to the cros se th 1 35 m min 00225 m s 9 m0025 mmin 6 m 3 pool V A dV 3 level crosssection dt supplied to the pool is determined to be Substituting the rate at which water is 00379 m s 3 0 01539 0 0225 d pool V V V e i dt herefore water is supplied at a rate of 00379 m3s 379 Ls xit is to be determined Properties The density of air is given to be 418 kgm3 at the inlet nalysis There is only one inlet and one exit and thus T 5147 Air is accelerated in a nozzle The density of air at the nozzle e Assumptions Flow through the nozzle is steady A m m m 1 2 Then 264 kgm3 120 ms 4 18 kgm 2 3 1 1 ρ ρ V A 380 ms 1 2 2 2 2 2 2 1 1 1 2 1 ρ ρ V A A V A V m m Discussion Note that the density of air decreases considerably despite a decrease in the crosssectional area of the nozzle AIR 1 2 preparation If you are a student using this Manual you are using it without permission 5119 5148E A heat exchanger that uses hot air to heat cold water is considered The total flow power and the flow works for both the air and water streams are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid 5 Air is an ideal gas with constant specific heats at room temperature Properties The gas constant of air is 03704 psiaft3lbmR 006855 BtulbmR Table A1E The specific volumes of water at the inlet and exit are Table A4E PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course psia 20 3 90 F 4 4 4 3 f T P P v v Analysis The specific volume of air at the inlet and the mass flow rate are 0 01610 ft lbm 0 F 9 psia 17 0 01602 ft lbm 50 F 50 F 3 3 f T v v 3 1222 ft lbm 460 R 0 3704 psia ft lbm R200 3 3 1 v RT 20 psia 1 1 P 0 1364 lbms ft lbm 1222 m V1 100 60 ft s 3 3 1 v ideal gas equation of state gives 2 1 4 3 Water AIR Combining the flow work expression with the 6855 Btulbm 200R 0 06855 Btulbm R100 1 2 1 1 2 2 flow T R T P P w v v The flow work of water is 000864 Btulbm 3 3 3 3 3 4 4 flow psia ft 5404 1Btu 20 psia 0 01602 ft lbm 17 psia 0 01610 ft lbm v v P P w The net flow power for the heat exchanger is 1329 hp 07068 Btus 1hp 9393 Btus 0 lbms000864 Btulbm 50 0 1364 lbms6855 Btulbm flow air flow air flow w m w m W preparation If you are a student using this Manual you are using it without permission 5124 5155 The mass flow rate of a compressed air line is divided into two equal streams by a Tfitting in the line The velocity of the air at the outlets and the rate of change of flow energy flow power across the Tfitting are to be determined Assumptions 1 Air is an ideal gas with constant specific heats 2 The flow is steady 3 Since the outlets are identical it is presumed that the flow divides evenly between the two Properties The gas constant of air is R 0287 kPam3kgK Table A1 14 MPa 36C 14 MPa 36C Analysis The specific volumes of air at the inlet and outlets are 0 05614 m kg 1600 kPa 273 K 0 287 kPa m kg K40 3 3 1 1 1 P RT v 16 MPa 40C 50 ms 0 06335 m kg 1400 kPa 2 3 2 P v v ing an even division of the inle 273 K 0 287 kPa m kg K36 3 3 2 RT t flow rate the mass balance can be ritten as Assum w 2821 ms 2 50 005614 0 06335 2 2 1 1 2 2 1 3 2 2 2 2 1 1 A V V A 1 V A A V V v v v v The mass flow rate at the inlet is 04372 kgs 005614 m kg 50 ms 4 0 025 m 4 3 2 1 1 2 1 1 1 πD AV 1 π v v V m 1 while that at the outlets is 02186 kgs 2 0 4372 kgs 2 1 3 2 m m m Substituting the above results into the flow power expression produces 0496 kW 04372 kgs1600 kPa005614 m kg 02186 kgs1400 kPa006335 m kg 2 2 3 3 1 1 1 2 2 2 flow v v m P m P W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5127 5158 Water is boiled at a specified temperature by hot gases flowing through a stainless steel pipe submerged in water The rate of evaporation of is to be determined Water 150C Heater Hot gases Assumptions 1 Steady operating conditions exist 2 Heat losses from the outer surfaces of the boiler are negligible Properties The enthalpy of vaporization of water at 150C is hfg 21138 kJkg Table A4 Analysis The rate of heat transfer to water is given to be 74 kJs Noting that the enthalpy of vaporization represents the amount of energy needed to vaporize a unit mass of a liquid at a specified temperature the rate of evaporation of water is determined to be 00350 kgs 21138 kJkg 74 kJs boiling evaporation fg h Q m 5159 Cold water enters a steam generator at 20C and leaves as saturated vapor at Tsat 200C The fraction of heat used to preheat the liquid water from 20C to saturation temperature of 200C is to be determined Assumptions 1 Steady operating conditions exist 2 Heat losses from the steam generator are negligible 3 The specific heat of water is constant at the average temperature Properties The heat of vaporization of water at 200C is hfg 19398 kJkg Table A4 and the specific heat of liquid water is c 418 kJkgC Table A3 Analysis The heat of vaporization of water represents the amount of heat needed to vaporize a unit mass of liquid at a specified temperature Using the average specific heat the amount of heat transfer needed to preheat a unit mass of water from 20C to 200C is Water 200C Heater Cold water 20C Steam 7524 kJkg 20 C C200 4 18 kJkg preheating c T q and herefore the fraction of heat used to preheat the water is 2692 2 kJkg 364 1 8 1939 preheating boiling total q q q T or 26922 752 4 Fraction t o preheat total preheating 280 02795 q q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5137 5171 A constantpressure R134a vapor separation unit separates the liquid and vapor portions of a saturated mixture into two separate outlet streams The flow power needed to operate this unit and the mass flow rate of the two outlet streams are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work and heat interactions Analysis The specific volume at the inlet is Table A12 0 03533 m kg 0 0007772 0 55 0 06360 0 0007772 55 0 320 kPa 3 1 1 1 1 f g f x x P v v v v The mass flow rate at the inlet is then PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 1698 kgs 003533 m kg 3 1 1 m v 0 006 m s 3 1 V For each kg of mixture processed 055 kg of vapor are rocessed Therefore 0 1698 45 0 0 1698 0 55 70 1 2 m m he flow p wer for this unit is R134a 320 kPa x 055 6 Ls Vapor separation unit Saturated liquid Saturated vapor p 0 45 1 2 1 3 m m m m kgs 007642 kgs 009340 T o 0 kW 01698 kgs320 kPa003533 m kg 007642 kgs320 kPa00007772 m kg 009340 kgs320 kPa006360 m kg 3 3 3 1 1 1 3 3 3 2 2 2 flow v v v m P m P m P W preparation If you are a student using this Manual you are using it without permission 5138 5172E A small positioning control rocket in a satellite is driven by a container filled with R134a at saturated liquid state The number of bursts this rocket experience before the quality in the container is 90 or more is to be determined Analysis The initial and final specific volumes are 001171 ft lbm Table A11E 0 10 F 3 1 1 1 v x T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10 F 3 T he initial and final masses in the container are 0 90 2 2 2 fg f x x v v v 2 4394 ft lbm 0 01171 0 90 2 7091 0 01171 2 T 170 8 lbm m kg 001171 ft 2 3 3 1 1 v V m 0 8199 lbm ft lbm 24394 ft 2 3 3 2 2 v V m Then m 170 0 lbm 0 8199 170 8 2 1 m m urst is ber of bursts that can be executed is then The amount of mass released during each control b 025 lbm 0 05 lbms5 s m t mb The num 680 bursts 025 lbmburst lbm 1700 b b m m N preparation If you are a student using this Manual you are using it without permission 5139 5173E The relationships between the mass flow rate and the time for the inflation and deflation of an air bag are given The volume of this bag as a function of time are to be plotted Assumptions Uniform flow exists at the inlet and outlet Properties The specific volume of air during inflation and deflation are given to be 15 and 13 ft3lbm respectively Analysis The volume of the airbag at any time is given by PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course out in d m dt m t v v V Applying at different time periods as given in problem statement give out flow time flow time in 10 ms 0 1000 ms s 1 10 ms 15 ft lbm 20 lbms 0 3 t tdt t t V 10 ms ft ms 0 0 015 0 2 3 2 t t t t V 12 ms 10 1000 ms s 1 15 ft lbm20 lbms 10 ms ms 10 3 t tdt t t V V 12 ms 10 ms 10 0 03 ft ms 10 ms 2 3 t t t V V 25 ms 12 12 ms 1000 ms s 1 30 12 ms 13 ft lbm 16 lbms 12 ms 0 03 ft ms 12 ms ms 12 3 2 3 t dt t t t t V V 25 ms 12 12 ms 0 011556 ft ms 12 ms 0 03 ft ms 12 ms ms 12 2 3 2 3 t dt t t t t V V 25 ms 12 ms 12 0 13867 144 ms 2 0 011556 12 ms 0 03 ft ms 12 ms 2 2 2 3 t t t t t V V 30 ms 25 ms 25 0 13867 625 ms 2 0 011556 25 ms 2 2 t t t t V V 50 ms 30 1000 ms s 1 13 ft lbm16 lbms 30 ms ms 12 3 t dt t t V V 50 ms 30 ms 30 0 208 ft ms 30 ms 3 t t t V V preparation If you are a student using this Manual you are using it without permission 5140 The results with some suitable time intervals are PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Time ms V ft3 0 0 2 006 4 024 6 054 8 096 10 150 12 210 15 295 20 413 25 502 27 470 30 413 40 205 46 080 4985 0 Alternative solution The net volume flow rate is obtained from out in v v V m m which is sketched on the figure below The volume of the airbag is given by Vdt V The results of a graphical interpretation of the volume is also given in the figure below Note that the evaluation of the bove integral is simply the area under the process curve a 0 10 20 30 40 50 60 400 300 200 100 0 100 200 300 400 time mlllisec V ft 3s 208 150 150 0 10 20 30 40 50 0 1 2 3 4 5 6 Time ms Volume ft 3 0 10 20 30 40 50 60 0 1 2 3 4 5 6 V ft 3 210 503 414 time millisec preparation If you are a student using this Manual you are using it without permission 5145 5178 The average air velocity in the circular duct of an airconditioning system is not to exceed 8 ms If the fan converts 80 percent of the electrical energy into kinetic energy the size of the fan motor needed and the diameter of the main duct are to be determined PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2 Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus m E CV CV and 0 0 The inlet velocity is negligible 0 1 V 3 There are no heat and work interactions other than the electrical power consumed by the fan motor 4 Air is an ideal gas with constant specific heats at room temperature m Properties The density of air is given to be ρ 120 kgm3 The constant pressure specific heat of air at room temperature is cp 1005 kJkgC Table A2 Analysis We take the fanmotor assembly as the system This is a control volume since mass crosses the system boundary during the process We note that there is only one inlet and one exit and thus m m 1 2 The change in the kinetic energy of air as it is accelerated from zero to 8 ms at a rate of 130 m3min is 8 ms 130 m3min 0 0832 kW m s 1000 1kJkg 2 0 kgs 8 ms 62 2 E K 20 kgm 130 m min 156 kgmin 26 kgs 1 2 2 2 2 1 2 2 3 3 V V m m V ρ It is stated that this represents 80 of the electrical energy consumed by the motor Then the total electrical power consumed by the motor is determined to be 0104 kW 70 motor motor W KE W 08 0 0832 kW 80 E K The diameter of the main duct is 0587 m 60 s 8 ms 4 π π π V D D V VA V Therefore the m min 1 4130 m min 4 3 2 V otor should have a rated power of at least 0104 kW and the diameter of the duct should be at least 587 m c preparation If you are a student using this Manual you are using it without permission 5149 5182 A submarine that has an airballast tank originally partially filled with air is considered Air is pumped into the ballast tank until it is entirely filled with air The final temperature and mass of the air in the ballast tank are to be determined Assumptions 1 Air is an ideal gas with constant specific heats 2 The process is adiabatic 3 There are no work interactions Properties The gas constant of air is R 0287 kPam3kgK Table A1 The specific heat ratio of air at room temperature is k 14 Table A2a The specific volume of water is taken 0001 m3kg Analysis The conservation of mass principle applied to the air gives min dt dma and as applied to the water becomes mout dt dmw The first law for the ballast tank produces a a w w dt dt w a h m h m d mu d mu 0 ombining is with the conservation of mass expressions rearranging and canceling the common dt term produces tegrating this result from the beginning to the end of the process gives C th w w a a w a h dm h dm d mu d mu In w w a a w a m m h m m h mu mu mu mu 1 2 1 2 1 2 1 2 Substituting the ideal gas equation of state and the specific heat models for the enthalpies and internal energies expands this to w w w p w w w c T m RT P RT P P P 2 1 2 V V V c T c T m RT c T c T RT 1 1 1 2 in 1 1 1 2 2 V v v When the common terms are cancelled this result becomes 3868 K 100 1 1 2 2 V V T 14293 700100 1 288 700 1 2 in 1 V V kT T he final mass from the ideal gas relation is T 9460 kg 0 287 kPa m kg K3868 K 1500 kPa700 m 3 3 2 2 2 RT P m V PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5150 5183 A submarine that has an airballast tank originally partially filled with air is considered Air is pumped into the ballast tank in an isothermal manner until it is entirely filled with air The final mass of the air in the ballast tank and the total heat transfer are to be determined Assumptions 1 Air is an ideal gas with constant specific heats 2 There are no work interactions Properties The gas constant of air is R 0287 kPam3kgK Table A1 The specific heats of air at room temperature are cp 1005 kJkgK and cv 0718 kJkgK Table A2a The specific volume of water is taken 0001 m3kg Analysis The initial air mass is 1814 kg 0 287 kPa m kg K28815 K 1500 kPa100 m 3 3 1 1 1 1 RT P m V and the initial water mass is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 000 kg 600 0 001 m kg 3 1 mw v 600 m 3 V1 nd the final mass of air in the tank is a 12697 kg 0 287 kPa m kg K28 3 2 2 RT m K 815 1500 kPa700 m 3 2 2 P V he first la when adapted to this system gives p w w w w a i i e e e e i i 1 2 1 2 in 1 1 2 in 1 1 2 2 v v ubstituting T w Qin 2 m c T m m h m u c T m m c T Q m h m h m u m u Q m u m u m h h m Noting that 6298 w w h u kJkg S 0 kJ 1 005 288 1814 12697 000 6298 600 600000 6298 0 718 288 1814 0 718 288 12697 Qin The process is adiabatic preparation If you are a student using this Manual you are using it without permission 5151 5184 A cylindrical tank is charged with nitrogen from a supply line The final mass of nitrogen in the tank and final temperature are to be determined for two cases Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Air is an ideal gas with constant specific heats 3 Kinetic and potential energies are negligible 4 There are no work interactions involved Properties The gas constant of nitrogen is 02968 kPam3kgK Table A1 The specific heats of nitrogen at room temperature are cp 1039 kJkgK and cv 0743 kJkgK Table A2a Analysis a We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance 1 2 system out in m m m m m m i 200 kPa 25C 01 m3 800 kPa 25C Air Energy balance es PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1 1 2 2 1 1 2 2 m m c T c T m i i potential etc energi in internal kinetic ange system by heat work and mass out in T c m u m u h m E E E i p i v v 43 42 1 Combining the two balances final masses are given by Ch Net energy transfer 4243 1 1 1 2 2 1 2 m c T m c T m c T m i p v v The initial and 0 2261 kg 02968 kPa m kg K298 K m 10 200 kPa 3 3 1 1 RT 1 m P V 2 2 3 3 2 2 2 RT m 5 269 kPa m kg K 02968 m 10 80 0 kPa T T P V Substituting 0 2261 0 743298 269 5 0 743 1 039298 0 2261 269 5 2 T 2 2 T T n is T en whose solutio 3790 K 2 The final mass is th 07112 kg 379 0 269 5 269 5 m When there is rapid heat transfer between the nitrogen and tank such that the cylinder and nitrogen remain in thermal equilibrium during the process the energy balance equation may be written as 2 2 T b 1 1 1 2 2 2 1 2 m c T c T m m c T c T m m c T m t t nit t t nit p i v v Substituting 50 0 43298 0 2261 0 743298 50 0 43 269 5 0 743 1 039298 0 2261 269 5 2 2 2 2 T T T T whose solution is The final mass is then T2 3007 K 08962 kg 300 7 269 5 300 7 269 5 m2 preparation If you are a student using this Manual you are using it without permission 5152 5185 The air in a tank is released until the pressure in the tank reduces to a specified value The mass withdrawn from the tank is to be determined for three methods of analysis Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process 2 Air is an ideal gas with constant specific heats 3 Kinetic and potential energies are negligible 4 There are no work or heat interactions involved Properties The gas constant of air is 0287 kPam3kgK Table A1 The specific heats of air at room temperature are cp 1005 kJkgK and cv 0718 kJkgK Also k 14 Table A2a Analysis a We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance 2 1 1 2 system out in m m m m m m m m m e e Energy balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m u m u h m E E E 1 1 2 2 potential etc energies in internal kinetic ange system out in 0 v v 43 42 1 ombining the two balances Ch by heat work and mass Net energy transfer 4243 1 e p e e e e e m c T m c T c T m m h m u m u 1 1 2 2 1 1 2 2 0 C cpTe m m m c T m c T 0 2 1 1 1 2 2 v v The initial and final masses are given by 9 354 kg 80 0 kPa 1m 3 1 m P V 273 K 0287 kPa m kg K25 3 1 RT 1 2 2 3 3 2 The temperature of air leaving th 2 2 6 522 kPa m kg K 0287 15 0 kPa 1m T T RT m e tank changes from the initial temperature in the tank to the final temperature during the ischargin rocess We assume that the temperature of the air leaving the tank is the average of initial and final bstituting into the energy balance equation gives P V d g p temperatures in the tank Su 1 005 298 522 6 9 354 9 354 0 718298 522 6 0 718 0 T2 T 2 0 2 2 2 1 1 1 2 2 T T c T m m m c T m c T e p v v T2 ubstituting the final mass is 2 u whose sol tion is 1910 K S 2 736 kg 191 522 6 2 m nd the mass withdrawn is in two parts first from 800 kPa to 400 kPa and from 400 kPa to 150 kPa the solution will be as rom 800 kPa to 400 kPa a e 6618 kg 2 736 9 354 2 1 m m m b Considering the process follows F 2 2 3 3 2 2 2 1394 0287 kPa m kg K 400 kPa 1m T T RT P V Air 800 kPa 25C 1 m3 m preparation If you are a student using this Manual you are using it without permission 5153 2 1 005 298 1394 9 354 0 718298 354 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9 1394 0 718 0 2 2 2 2 T T T T 1 K T2 245 5 687 kg 245 1 1394 2 m 5 687 9 354 m m m 3 667 kg 2 1 1e From 400 kPa to 150 kPa 2 1 005 245 1 522 6 5 687 5 687 0 718245 1 522 6 0 718 0 2 2 2 2 T T T T 5 K T2 186 2 803 kg 186 5 2 m 6 522 The total mass withdrawn is 2 884 3 667 2 1 me c The m alance may be written as 2 884 kg 2 803 5 687 2 1 2 m m me e e m m 6551 kg ass b me dt dm When this is combined with the ideal gas equation of state it becomes me dt d P T R V since the tank volume remains constant during the process An energy balance on the tank gives dt dT c P dP c c dt dT T P dt dP c dt dP c dt d P T c T R dt dP R c dt c T dm dt d mT c h m dt mu d p p p p p e e v v v v V V When this result is integrated it gives 184 7 K 800 kPa 298 K 150 kPa 41 40 1 1 2 1 2 k k P P T T The final mass is 2 830 kg 0287 kPa m kg K184 7 K 15 0 kPa 1m 3 3 2 2 2 RT P m V and the mass withdrawn is Discussion The result in first method is in error by 14 while that in the second method is in error by 04 6524 kg 2 830 9 354 2 1 m m me preparation If you are a student using this Manual you are using it without permission 5154 5186 A tank initially contains saturated mixture of R134a A valve is opened and R134a vapor only is allowed to escape slowly such that temperature remains constant The heat transfer necessary with the surroundings to maintain the temperature and pressure of the R134a constant is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the exit remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m m m m m m m m e e Energy balance m h m u m u Q m u u E E E 1 1 2 2 in 1 1 2 2 potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 43 42 1 4243 1 The specific volume at the initial state is 2 1 1 2 system out in R134a 04 kg 1 L 25C m e e m m h Q in e e Combining the two balances he m m m u m u Q 2 1 1 1 2 2 in 00025 m kg kg 40 1 1 m 0 001 m 3 3 V v The initial state properties of R134a in the tank are 9624 kJkg 0 0572615687 26 87 0 05726 0 0008313 0 029976 0 0008313 0025 0 0025 m kg 0 1 3 1 v v C 26 1 1 1 fg f fg f x u u u x T v v Table A11 t leaving the bottle is he specific volume at the final state is 1 The enthalpy of saturated vapor refrigeran 26468 kJkg 26 C g e h h T 001 m kg kg 10 3 2 2 m v 0 001 m 3 V he intern energy at the final state is T al 13661 kJkg 0 314615687 26 87 0 3146 0 0008313 0 029976 0 0008313 01 0 01 m kg 0 C 26 1 2 2 2 3 2 2 fg f fg f x u u u x T v v v v Table A11 Substituting into the energy balance equation 546 kJ kg26468 kJkg 10 40 kg9624 kJkg 40 kg13661 kJkg 10 2 1 1 1 2 2 in he m m m u m u Q preparation If you are a student using this Manual you are using it without permission 5156 5188 Problem 5187 is reconsidered The effects of turbine exit area and turbine exit pressure on the exit velocity and power output of the turbine as the exit pressure varies from 10 kPa to 50 kPa with the same quality and the exit area to varies from 1000 cm2 to 3000 cm2 is to be investigated The exit velocity and the power output are to be plotted against the exit pressure for the exit areas of 1000 2000 and 3000 cm2 Analysis The problem is solved using EES and the results are tabulated and plotted below FluidSteamIAPWS A1150 cm2 T1600 C P17000 kPa Vel1 60 ms A21400 cm2 P225 kPa qout 20 kJkg mdot A1Vel1v1convertcm2m2 v1volumeFluid TT1 PP1 specific volume of steam at state 1 Vel2mdotv2A2convertcm2m2 v2volumeFluid x095 PP2 specific volume of steam at state 2 T2temperatureFluid PP2 vv2 C not required but good to know conservation of Energy for steadyflow Eindot Eoutdot DeltaEdot For steadyflow DeltaEdot 0 DELTAEdot0 For the turbine as the control volume neglecting the PE of each flow steam EdotinEdotout h1enthalpyFluidTT1 PP1 Edotinmdoth1 Vel122Convertm2s2 kJkg h2enthalpyFluidx095 PP2 Edotoutmdoth2 Vel222Convertm2s2 kJkg mdot qout Wdotout PowerWdotout Qdotoutmdotqout P2 kPa Power kW Vel2 ms 10 1444 1889 2333 2778 3222 3667 4111 4556 50 22158 1895 6071 9998 12212 13573 14464 15075 15507 15821 2253 1595 1239 1017 8632 7511 6654 5978 543 4977 Table values are for A21000 cm2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5157 10 15 20 25 30 35 40 45 50 6000 8000 10000 12000 14000 16000 18000 P2 kPa Power kW A23000 cm2 A22000 cm2 A21000 cm2 10 15 20 25 30 35 40 45 50 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 P2 kPa Vel2 ms A23000 cm2 A22000 cm2 A21000 cm2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5160 5191 An insulated cylinder equipped with an external spring initially contains air The tank is connected to line and air is allowed to enter the cylinder until i a supply ts volume doubles The mass of the air that entered and the final near spring 5 The device is insulated specific heats of air at room temperature are cv ass and energy balances for this uniformflow sed as Mass balance m m m m i temperature in the cylinder are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 The expansion process is quasiequilibrium 3 Kinetic and potential energies are negligible 4 The spring is a li and thus heat transfer is negligible 6 Air is an ideal gas with constant specific heats Properties The gas constant of air is R 0287 kJkgK Table A1 The 0718 and cp 1005 kJkgK Table A2a Also u cvT and h cpT Analysis We take the cylinder as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the m Air P 150 kPa T1 22C V1 011 m3 Pi 07 MPa Ti 22C Fspring system can be expres out in m m 1 2 system nergy ba nce energies kinetic internal system Q u E 43 42 1 Combining the two relations E la potential etc Change in by heat work and mass Net energy transfer 1 2 2 bout m m u W m h i i 0 pe ke since 1 out in E E 4243 1 1 1 2 2 1 2 m u m u W m h m b out i o cv r m W m c T m b out i p v 1 1 2 2 1 2 T m c T The initial and the final masses in the tank are 2 2 3 2 2 2 2 3 kPa m kg K 0287 01949 kg K 295 K kg m 011 T T RT m 3 1 4599 600 kPa 022 m P V 3 1 1 1 kPa m 0287 150 kPa RT P m V Then from the mass balance becomes 0 1949 459 9 1 2 T m m mi 2 s a lin hus the boundary work for this process can be determined from The spring i ear spring and t 4125 kJ 011 m 022 2 Wb 600 kPa 150 2 3 1 2 2 1 V P V P Area Substituting into the energy balance the final temperature of air T2 is determined to be 0 718295 0 1949 0 718 459 9 1 005 295 0 1949 459 9 4125 2 2 2 T T T It yields T2 351 K Thus 1 309 kg 3514 4599 4599 2 2 T m and mi m2 m1 1309 01949 111 kg 5161 5192 R134a is allowed to leave a pistoncylinder device with a pair of stops The work done and the heat transfer are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid leaving the device is assumed to be constant 2 Kinetic and potential energies are negligible PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 500 kPa 3 2 P v he tank as the system which is a control volume since mass crosses the boundary Noting that the of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as R134a 2 kg 800 kPa 80C Q Properties The properties of R134a at various states are Tables A11 through A13 97 kJkg 316 29084 kJkg 032659 m kg 0 C 80 kPa 800 1 1 3 1 1 1 h u T P v 46 kJkg 263 24240 kJkg 20 C 2 2 2 2 h u T 042115 m kg 0 Analysis a We take t microscopic energies Mass balance 2 1 system out in m m m m m m e Energy balance system out in m u m u m h Q W E E E e e 43 42 1 4243 1 3 3 3 3 1 1 1 m v V e refrigerant withdrawn from the cylinder is assumed to be the average of initial and final enthalpies of the cylinder potential etc energies Change in internal kinetic by heat work and mass Net energy transfer 0 pe 1 1 since ke 2 2 out in b The volumes at the initial and final states and the mass that has left the cylinder are 006532 m kg0032659 m kg 2 1kg 1 2 004212 m 122 kg0042115 m kg 1 2 2 1 2 2 2 m m m m m v v V 2 1 e The enthalpy of th the refrigerant in 29021 kJkg 26346 1 231697 1 2 2 1 h h he Noting that the pressure remains constant after the piston starts moving the boundary work is determined from b Substituting 116 kJ 3 2 1 2 bin 0 04212m 500 kPa 0 06532 V P V W 607 kJ out out 2 kg29084 kJkg 1 kg24240 kJkg 1 kg29021 kJkg 6 kJ 11 Q Q preparation If you are a student using this Manual you are using it without permission 5162 5193 The pressures across a pump are measured The mechanical efficiency of the pump and the temperature rise of water are to be determined Assumptions 1 The flow is steady and incompressible 2 The pump is driven by an external motor so that the heat generated by the motor is dissipated to the atmosphere 3 The elevation difference between the inlet and outlet of the pump is negligible z1 z2 4 The inlet and outlet diameters are the same and thus the inlet and exit velocities are equal V1 V2 Properties We take the density of water to be 1 kgL 1000 kgm3 and its specific heat to be 418 kJkg C Table A3 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis a The mass flow rate of water through the pump is 18 kgs 1 kgL18 Ls V ρ m The motor draws 6 kW of power and is 95 percent efficient Thus the mechanical shaft power it delivers to the pump is kW 75 0 95 6 kW electric motor pumpshaft W W η To determine the mechanical efficiency of the pump we need to know the increase in the mechanical energy of the fluid as it flows through the pump which is PUMP Pump inlet Motor 6 kW 2 2 2 mechin mechout mechfluid 2 gz V m P E E E ρ 1 2 1 1 2 2 gz V P m ρ implifying it for this case and substituting the given values S kW 63 kPa m 1 kJ 1 kgm 1000 18 kgs 300 1 2 mechfluid ρ P m P E kPa 100 3 3 Then the mechanical efficiency of the pump becomes 632 kW 63 mechfluid W E pump η 0 632 57 kW pumpshaft b Of the 57kW mechanical power supplied by the pump only 36 kW is imparted to the fluid as mechanical energy The sts itself s a heatin effect in the fluid ss E W The temperature rise of water due to this mechanical inefficiency is determined from the thermal energy balance remaining 21 kW is converted to thermal energy due to frictional effects and this lost mechanical energy manife a g Emechlo kW 12 63 75 mechfluid shaft pump mc T u m u E 1 2 mechloss Solving for T 0028C 18 kgs418 kJkg K kW 12 mechloss c m E T Therefore the water will experience a temperature rise of 0028C which is very small as it flows through the pump iscussion In an actual application the temperature rise of water will probably be less since part of the heat generated will be transferred to the casing of the pump and from the casing to the surrounding air If the entire pump motor were submerged in water then the 21 kW dissipated to the air due to motor inefficiency would also be transferred to the surrounding water as heat This would cause the water temperature to rise more D preparation If you are a student using this Manual you are using it without permission 5164 5195 The turbocharger of an internal combustion engine consisting of a turbine a compressor and an aftercooler is considered The temperature of the air at the compressor outlet and the minimum flow rate of ambient air are to be determined Compressor Turbine Aftercooler Exhaust gases Air Cold air Assumptions 1 All processes are steady since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Air properties are used for exhaust gases 4 Air is an ideal gas with constant specific heats 5 The mechanical efficiency between the turbine and the compressor is 100 6 All devices are adiabatic 7 The local atmospheric pressure is 100 kPa Properties The constant pressure specific heats of exhaust gases warm air and cold ambient air are taken to be cp 1063 1008 and 1005 kJkgK respectively Table A2b Analysis a An energy balance on turbine gives 1 063 kW 350K 002 kgs106 3 kJkg K400 exh2 exh1 exh exh T T T c m W p This is also the power input to the compressor since the mechanical efficiency between the turbine and the compressor is assumed to be 100 An energy balance on the compressor gives the air temperature at the compressor outlet PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course T T m c W p b An energy balance on the aftercooler gives the mass flow rate of cold ambient air ca1 ca2 ca ca a3 a2 a a m T T m c T T m c p p he volum flow rate may be determined if we first calculate specific volume of cold ambient air at the inlet of aftercooler C 1086 50K 0018 kgs100 8 kJkg K 1 063 kW T T a2 a2 a1 a2 a a C 0 05161 kgs ca ca m 30 C C40 1005 kJkg 80 C C1086 kgs100 8 kJkg 0018 T e That is 0 8696 m kg 100 kPa 273 K 0 287 kJkg K30 3 ca P RT v 449 Ls 00449 m s 3 0 05161 kgs 0 8696 m kg 3 ca ca v V m preparation If you are a student using this Manual you are using it without permission 5165 5196 Heat is transferred to a pressure cooker at a specified rate for a specified time period The cooking temperature and the water remaining in the cooker are to be determined Assumptions 1 This process can be analyzed as a uniformflow process since the properties of the steam leaving the control volume remain constant during the entire cooking process 2 The kinetic and potential energies of the streams are negligible ke pe 0 3 The pressure cooker is stationary and thus its kinetic and potential energy changes are zero that is KE PE 0 and Esystem Usystem 4 The pressure and thus temperature in the pressure cooker remains constant 5 Steam leaves as a saturated vapor at the cooker pressure 6 There are no boundary electrical or shaft work interactions involved 7 Heat is transferred to the cooker at a constant rate Analysis We take the pressure cooker as the system This is a control volume since mass crosses the system boundary during the process We observe that this is an unsteadyflow process since changes occur within the control volume Also there is one exit and no inlets for mass flow a The absolute pressure within the cooker is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course s P P P ab 175 kPa 100 75 atm gage Since saturation conditions exist in the cooker at all times the cooking temperature must be the saturation temperature corresponding to this pressure From Table A5 it is 11604C sat 175 kPa 2 T T which is about 16C higher than the ordinary cooking temperature b Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance or 2 CV 1 1 CV 2 system out in m m m m m m m m m e e Energy balance potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in pe ke W m u m u m h Q E E E 43 42 1 4243 1 ombining the mass and energy balances gives The amount of heat transfer during this process is found from the pressure cooker as saturated vapor at 175 kPa at all mes Thus The initial internal energy is found after the quality is determined 0 1 1 since 2 2 in e e C 1 1 2 2 2 1 in m u m u h m m Q e 900 kJ 60 s kJs30 50 in in T Q Q Steam leaves ti 2 kJkg 2700 g 175 kPa he h 0 00499 0 001 1004 0 001 0006 0006 m kg 1kg m 0006 1 1 3 3 1 1 fg f x m v v v V v Thus 497 kJkg kJkg 0 004992037 7 48682 1 1 fg f x u u u and preparation If you are a student using this Manual you are using it without permission 5166 497 kJ 1 kg497 kJkg 1 1 1 m u U PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course he mass of the system at the final state is m2 V v2 Substituting this into the energy equation yields T 1 1 2 2 2 1 in m u u h m Q e v V v V There are two unknowns in this equation u2 and v2 Thus we need to relate them to a single unknown before we can etermine these unknowns Assuming there is still some liquid water left in the cooker at the final state ie saturation conditions exist v2 and u2 can be expressed as fg ecall tha uring a boiling process at constant pressure the properties of each phase remain constant only the amounts When these expressions are substituted into the above energy equation x2 becomes the only unknown and it is ined to be 3 3 v d kJkg 2037 7 82 486 0 001 m kg 1 004 001 0 2 3 2 2 2 x x u u u x x f fg f v v v 2 2 R t d change determ x2 0 009 Thus 0010 m kg 0 001 m kg 0 009 1 004 0 001 2 and 06 kg m kg 001 m 0006 3 3 2 2 v V m Therefore after 30 min there is 06 kg water liquid vapor left in the pressure cooker Discussion Note that almost half of the water in the pressure cooker has evaporated during cooking preparation If you are a student using this Manual you are using it without permission 5167 5197 A water tank open to the atmosphere is initially filled with water The tank discharges to the atmosphere through a long pipe connected to a valve The initial discharge velocity from the tank and the time required to empty the tank are to be determined Assumptions 1 The flow is incompressible 2 The draining pipe is horizontal 3 The tank is considered to be empty when the water level drops to the center of the valve Analysis a Substituting the known quantities the discharge velocity can be expressed as gz gz fL D gz V 0 1212 0 015100 m010 m 51 2 51 2 D D0 Then the initial discharge velocity becomes 154 ms ms 2 m 0 1212 9 81 0 1212 2 1 1 gz V z where z is the water height relative to the center of the orifice at th b The flow rate of water from the tank can be obtained by multiplying the discharge velocity by the pipe crosssectional area at time gz D V A 0 1212 4 2 pipe 2 π V Then the am er t ows through the pipe during a differential time interval dt is ount of wat hat fl gzdt D dt d 0 1212 4 π V V 1 2 which from conservation of mass mu e of water in the tank st be equal to the decrease in the volum D dz dz A d 4 0 tank π V 2 where dz is the change in the water level in the tank d 2 uring dt Note that dz is a negative quantity since the positive direction o is upwards Therefore we used dz to get a positive quantity for the amount of water discharged Setting Eqs 1 and 2 equal to each other and rearranging f z dz z g D D gz dz D dt dz D gzdt D 1 1212 0 2 2 0 2 0 2 π π D 2 0 1212 0 1212 4 4 2 0 2 The last relation can be integrated easily since the variables are separated Letting tf be the discharge time and integrating it from t 0 when z z1 to t tf when z 0 completely drained tank gives 2 1 2 1 0 2 0 0 1 2 2 0 2 z D D t f 1z 1 1 2 2 0 2 1 2 2 0 0 1212 0 1212 0 1212 z g D D g D t dz z g D dt f z z Simplifying and substituting the values given the draining time is determined to be t 721h 25940 s 1212 9 81 ms 0 m 2 m 10 210 m 0 1212 2 2 2 2 1 2 2 0 g z D D t f Discussion The draining time can be shortened considerably by installing a pump in the pipe PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5169 5201 An evacuated bottle is surrounded by atmospheric air A valve is opened and air is allowed to fill the bottle The amount of heat transfer through the wall of the bottle when thermal and mechanical equilibrium is established is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Air is an ideal gas 3 Kinetic and potential energies are negligible 4 There are no work interactions involved 5 The direction of heat transfer is to the air in the bottle will be verified Analysis We take the bottle as the system It is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance 0 since initial out 2 system out in m m m m m m m i P0 T0 V Evacuated Energy balance 0 pe ke since initial out 2 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in E E W m u m h Q E E E i i 43 42 1 4243 1 Combining the two balances 2 2 2 2 in p i i c T m c T h m u Q v but Ti T2 T0 and v R Substituting cp c V V v 0 0 0 0 0 2 0 2 in P RT RT P m RT c T m c Q p Therefor Qout P0V Heat is lost from the tank e PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5170 Fundamentals of Engineering FE Exam Problems 5202 Steam is accelerated by a nozzle steadily from a low velocity to a velocity of 280 ms at a rate of 25 kgs If the temperature and pressure of the steam at the nozzle exit are 400C and 2 MPa the exit area of the nozzle is a 84 cm2 b 107 cm2 c 135 cm2 d 196 cm2 e 230 cm2 Answer c 135 cm2 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Vel10 ms Vel2280 ms m25 kgs T2400 C P22000 kPa The rate form of energy balance is Edotin Edotout DELTAEdotcv v2VOLUMESteamIAPWSTT2PP2 m1v2A2Vel2 A2 in m2 Some Wrong Solutions with Common Mistakes R04615 kJkgK P2v2idealRT2273 m1v2idealW1A2Vel2 assuming ideal gas P1v2idealRT2 m1v2idealW2A2Vel2 assuming ideal gas and using C mW3A2Vel2 not using specific volume PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5171 5203 Steam enters a diffuser steadily at 05 MPa 300C and 122 ms at a rate of 35 kgs The inlet area of the diffuser is a 15 cm2 b 50 cm2 c 105 cm2 d 150 cm2 e 190 cm2 Answer d 150 cm2 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Vel1122 ms m35 kgs T1300 C P1500 kPa The rate form of energy balance is Edotin Edotout DELTAEdotcv v1VOLUMESteamIAPWSTT1PP1 m1v1AVel1 A in m2 Some Wrong Solutions with Common Mistakes R04615 kJkgK P1v1idealRT1273 m1v1idealW1AVel1 assuming ideal gas P1v2idealRT1 m1v2idealW2AVel1 assuming ideal gas and using C mW3AVel1 not using specific volume 5204 An adiabatic heat exchanger is used to heat cold water at 15C entering at a rate of 5 kgs by hot air at 90C entering also at rate of 5 kgs If the exit temperature of hot air is 20C the exit temperature of cold water is a 27C b 32C c 52C d 85C e 90C Answer b 32C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cw418 kJkgC Cpair1005 kJkgC Tw115 C mdotw5 kgs Tair190 C Tair220 C mdotair5 kgs The rate form of energy balance for a steadyflow system is Edotin Edotout mdotairCpairTair1Tair2mdotwCwTw2Tw1 Some Wrong Solutions with Common Mistakes Tair1Tair2W1Tw2Tw1 Equating temperature changes of fluids Cvair0718 kJkgK mdotairCvairTair1Tair2mdotwCwW2Tw2Tw1 Using Cv for air W3Tw2Tair1 Setting inlet temperature of hot fluid exit temperature of cold fluid W4Tw2Tair2 Setting exit temperature of hot fluid exit temperature of cold fluid PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5172 5205 A heat exchanger is used to heat cold water at 15C entering at a rate of 2 kgs by hot air at 85C entering at rate of 3 kgs The heat exchanger is not insulated and is loosing heat at a rate of 25 kJs If the exit temperature of hot air is 20C the exit temperature of cold water is a 28C b 35C c 38C d 41C e 80C Answer b 35C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cw418 kJkgC Cpair1005 kJkgC Tw115 C mdotw2 kgs Tair185 C Tair220 C mdotair3 kgs Qloss25 kJs The rate form of energy balance for a steadyflow system is Edotin Edotout mdotairCpairTair1Tair2mdotwCwTw2Tw1Qloss Some Wrong Solutions with Common Mistakes mdotairCpairTair1Tair2mdotwCwW1Tw2Tw1 Not considering Qloss mdotairCpairTair1Tair2mdotwCwW2Tw2Tw1Qloss Taking heat loss as heat gain Tair1Tair2W3Tw2Tw1 Equating temperature changes of fluids Cvair0718 kJkgK mdotairCvairTair1Tair2mdotwCwW4Tw2Tw1Qloss Using Cv for air 5206 An adiabatic heat exchanger is used to heat cold water at 15C entering at a rate of 5 kgs by hot water at 90C entering at rate of 4 kgs If the exit temperature of hot water is 50C the exit temperature of cold water is a 42C b 47C c 55C d 78C e 90C Answer b 47C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cw418 kJkgC Tcold115 C mdotcold5 kgs Thot190 C Thot250 C mdothot4 kgs Qloss0 kJs The rate form of energy balance for a steadyflow system is Edotin Edotout mdothotCwThot1Thot2mdotcoldCwTcold2Tcold1Qloss Some Wrong Solutions with Common Mistakes Thot1Thot2W1Tcold2Tcold1 Equating temperature changes of fluids W2Tcold290 Taking exit temp of cold fluidinlet temp of hot fluid PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5173 5207 In a shower cold water at 10C flowing at a rate of 5 kgmin is mixed with hot water at 60C flowing at a rate of 2 kgmin The exit temperature of the mixture will be a 243C b 350C c 400C d 443C e 552C Answer a 243C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cw418 kJkgC Tcold110 C mdotcold5 kgmin Thot160 C mdothot2 kgmin The rate form of energy balance for a steadyflow system is Edotin Edotout mdothotCwThot1mdotcoldCwTcold1mdothotmdotcoldCwTmix Some Wrong Solutions with Common Mistakes W1TmixTcold1Thot12 Taking the average temperature of inlet fluids 5208 In a heating system cold outdoor air at 7C flowing at a rate of 4 kgmin is mixed adiabatically with heated air at 70C flowing at a rate of 3 kgmin The exit temperature of the mixture is a 34C b 39C c 45C d 63C e 77C Answer a 34C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cair1005 kJkgC Tcold17 C mdotcold4 kgmin Thot170 C mdothot3 kgmin The rate form of energy balance for a steadyflow system is Edotin Edotout mdothotCairThot1mdotcoldCairTcold1mdothotmdotcoldCairTmix Some Wrong Solutions with Common Mistakes W1TmixTcold1Thot12 Taking the average temperature of inlet fluids PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5174 5209 Hot combustion gases assumed to have the properties of air at room temperature enter a gas turbine at 1 MPa and 1500 K at a rate of 01 kgs and exit at 02 MPa and 900 K If heat is lost from the turbine to the surroundings at a rate of 15 kJs the power output of the gas turbine is a 15 kW b 30 kW c 45 kW d 60 kW e 75 kW Answer c 45 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cpair1005 kJkgC T11500 K T2900 K mdot01 kgs Qdotloss15 kJs The rate form of energy balance for a steadyflow system is Edotin Edotout WdotoutQdotlossmdotCpairT1T2 Alternative Variable specific heats using EES data WdotoutvariableQdotlossmdotENTHALPYAirTT1ENTHALPYAirTT2 Some Wrong Solutions with Common Mistakes W1WoutmdotCpairT1T2 Disregarding heat loss W2WoutQdotlossmdotCpairT1T2 Assuming heat gain instead of loss 5210 Steam expands in a turbine from 4 MPa and 500C to 05 MPa and 250C at a rate of 1350 kgh Heat is lost from the turbine at a rate of 25 kJs during the process The power output of the turbine is a 157 kW b 207 kW c 182 kW d 287 kW e 246 kW Answer a 157 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values T1500 C P14000 kPa T2250 C P2500 kPa mdot13503600 kgs Qdotloss25 kJs h1ENTHALPYSteamIAPWSTT1PP1 h2ENTHALPYSteamIAPWSTT2PP2 The rate form of energy balance for a steadyflow system is Edotin Edotout WdotoutQdotlossmdoth1h2 Some Wrong Solutions with Common Mistakes W1Woutmdoth1h2 Disregarding heat loss W2WoutQdotlossmdoth1h2 Assuming heat gain instead of loss u1INTENERGYSteamIAPWSTT1PP1 u2INTENERGYSteamIAPWSTT2PP2 W3WoutQdotlossmdotu1u2 Using internal energy instead of enthalpy W4WoutQdotlossmdotu1u2 Using internal energy and wrong direction for heat PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5175 5211 Steam is compressed by an adiabatic compressor from 02 MPa and 150C to 08 MPa and 350C at a rate of 130 kgs The power input to the compressor is a 511 kW b 393 kW c 302 kW d 717 kW e 901 kW Answer a 511 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P1200 kPa T1150 C P2800 kPa T2350 C mdot130 kgs Qdotloss0 kJs h1ENTHALPYSteamIAPWSTT1PP1 h2ENTHALPYSteamIAPWSTT2PP2 The rate form of energy balance for a steadyflow system is Edotin Edotout WdotinQdotlossmdoth2h1 Some Wrong Solutions with Common Mistakes W1WinQdotlossh2h1mdot Dividing by mass flow rate instead of multiplying W2WinQdotlossh2h1 Not considering mass flow rate u1INTENERGYSteamIAPWSTT1PP1 u2INTENERGYSteamIAPWSTT2PP2 W3WinQdotlossmdotu2u1 Using internal energy instead of enthalpy W4WinQdotlossu2u1 Using internal energy and ignoring mass flow rate PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5176 5212 Refrigerant134a is compressed by a compressor from the saturated vapor state at 014 MPa to 09 MPa and 60C at a rate of 0108 kgs The refrigerant is cooled at a rate of 110 kJs during compression The power input to the compressor is a 494 kW b 604 kW c 714 kW d 750 kW e 813 kW Answer c 714 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P1140 kPa x11 P2900 kPa T260 C mdot0108 kgs Qdotloss110 kJs h1ENTHALPYR134axx1PP1 h2ENTHALPYR134aTT2PP2 The rate form of energy balance for a steadyflow system is Edotin Edotout WdotinQdotlossmdoth2h1 Some Wrong Solutions with Common Mistakes W1WinQdotlossmdoth2h1 Wrong direction for heat transfer W2Win mdoth2h1 Not considering heat loss u1INTENERGYR134axx1PP1 u2INTENERGYR134aTT2PP2 W3WinQdotlossmdotu2u1 Using internal energy instead of enthalpy W4WinQdotlossu2u1 Using internal energy and wrong direction for heat transfer PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5177 5213 Refrigerant134a expands in an adiabatic turbine from 12 MPa and 100C to 018 MPa and 50C at a rate of 125 kgs The power output of the turbine is a 463 kW b 664 kW c 727 kW d 892 kW e 1120 kW Answer a 463 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P11200 kPa T1100 C P2180 kPa T250 C mdot125 kgs Qdotloss0 kJs h1ENTHALPYR134aTT1PP1 h2ENTHALPYR134aTT2PP2 The rate form of energy balance for a steadyflow system is Edotin Edotout WdotoutQdotlossmdoth2h1 Some Wrong Solutions with Common Mistakes W1WoutQdotlossh2h1mdot Dividing by mass flow rate instead of multiplying W2WoutQdotlossh2h1 Not considering mass flow rate u1INTENERGYR134aTT1PP1 u2INTENERGYR134aTT2PP2 W3WoutQdotlossmdotu2u1 Using internal energy instead of enthalpy W4WoutQdotlossu2u1 Using internal energy and ignoring mass flow rate 5214 Refrigerant134a at 14 MPa and 90C is throttled to a pressure of 06 MPa The temperature of the refrigerant after throttling is a 22C b 56C c 82C d 80C e 900C Answer d 80C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P11400 kPa T190 C P2600 kPa h1ENTHALPYR134aTT1PP1 T2TEMPERATURER134ahh1PP2 Some Wrong Solutions with Common Mistakes W1T2T1 Assuming the temperature to remain constant W2T2TEMPERATURER134ax0PP2 Taking the temperature to be the saturation temperature at P2 u1INTENERGYR134aTT1PP1 W3T2TEMPERATURER134auu1PP2 Assuming uconstant v1VOLUMER134aTT1PP1 W4T2TEMPERATURER134avv1PP2 Assuming vconstant PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5178 5215 Air at 27C and 5 atm is throttled by a valve to 1 atm If the valve is adiabatic and the change in kinetic energy is negligible the exit temperature of air will be a 10C b 15C c 20C d 23C e 27C Answer e 27C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values The temperature of an ideal gas remains constant during throttling and thus T2T1 T127 C P15 atm P21 atm T2T1 C Some Wrong Solutions with Common Mistakes W1T2T1P1P2 Assuming vconstant and using C W2T2T1273P1P2273 Assuming vconstant and using K W3T2T1P2P1 Assuming vconstant and pressures backwards and using C W4T2T1273P2P1 Assuming vconstant and pressures backwards and using K 5216 Steam at 1 MPa and 300C is throttled adiabatically to a pressure of 04 MPa If the change in kinetic energy is negligible the specific volume of the steam after throttling will be a 0358 m3kg b 0233 m3kg c 0375 m3kg d 0646 m3kg e 0655 m3kg Answer d 0646 m3kg Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P11000 kPa T1300 C P2400 kPa h1ENTHALPYSteamIAPWSTT1PP1 v2VOLUMESteamIAPWShh1PP2 Some Wrong Solutions with Common Mistakes W1v2VOLUMESteamIAPWSTT1PP2 Assuming the volume to remain constant u1INTENERGYSteamTT1PP1 W2v2VOLUMESteamIAPWSuu1PP2 Assuming uconstant W3v2VOLUMESteamIAPWSTT1PP2 Assuming Tconstant PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 61 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 6 THE SECOND LAW OF THERMODYNAMICS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 62 The Second Law of Thermodynamics and Thermal Energy Reservoirs 61C Transferring 5 kWh of heat to an electric resistance wire and producing 5 kWh of electricity 62C An electric resistance heater which consumes 5 kWh of electricity and supplies 6 kWh of heat to a room 63C Transferring 5 kWh of heat to an electric resistance wire and producing 6 kWh of electricity 64C No Heat cannot flow from a lowtemperature medium to a higher temperature medium 65C A thermalenergy reservoir is a body that can supply or absorb finite quantities of heat isothermally Some examples are the oceans the lakes and the atmosphere 66C Yes Because the temperature of the oven remains constant no matter how much heat is transferred to the potatoes 67C The surrounding air in the room that houses the TV set PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 63 Heat Engines and Thermal Efficiency 68C No Such an engine violates the KelvinPlanck statement of the second law of thermodynamics 69C Heat engines are cyclic devices that receive heat from a source convert some of it to work and reject the rest to a sink 610C No Because 100 of the work can be converted to heat 611C It is expressed as No heat engine can exchange heat with a single reservoir and produce an equivalent amount of work 612C a No b Yes According to the second law no heat engine can have and efficiency of 100 613C No Such an engine violates the KelvinPlanck statement of the second law of thermodynamics 614C No The KelvinPlank limitation applies only to heat engines engines that receive heat and convert some of it to work 615C Method b With the heating element in the water heat losses to the surrounding air are minimized and thus the desired heating can be achieved with less electrical energy input 616E The rate of heat input and thermal efficiency of a heat engine are given The power output of the heat engine is to be determined Sink Source HE ηth 40 3104 Btuh net W Assumptions 1 The plant operates steadily 2 Heat losses from the working fluid at the pipes and other components are negligible Analysis Applying the definition of the thermal efficiency to the heat engine 472 hp 2544 5 Btuh 1hp 40 3 10 Btuh 4 th net QH W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 64 617 The power output and thermal efficiency of a heat engine are given The rate of heat input is to be determined Assumptions 1 The plant operates steadily 2 Heat losses from the working fluid at the pipes and other components are negligible Sink Source HE QH ηth 40 Analysis Applying the definition of the thermal efficiency to the heat engine 30 hp 559 kJs 1hp 0 7457 kJs 04 hp 30 th net η W QH 618 The power output and thermal efficiency of a power plant are given The rate of heat rejection is to be determined and the result is to be compared to the actual case in practice Assumptions 1 The plant operates steadily 2 Heat losses from the working fluid at the pipes and other components are negligible Analysis The rate of heat supply to the power plant is determined from the thermal efficiency relation 1500 MW 04 MW 600 th netout η W QH sink Furnace HE ηth 40 The rate of heat transfer to the river water is determined from the first law relation for a heat engine 600 MW 900 MW 600 1500 Wnetout Q Q H L In reality the amount of heat rejected to the river will be lower since part of the heat will be lost to the surrounding air from the working fluid as it passes through the pipes and other components 619 The work output and heat input of a heat engine are given The heat rejection is to be determined sink Furnace HE QL QH Assumptions 1 The plant operates steadily 2 Heat losses from the working fluid at the pipes and other components are negligible Analysis Applying the first law to the heat engine gives 450 kJ 250 kJ 700 kJ Wnet Q Q H L Wnet PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 65 620 The heat rejection and thermal efficiency of a heat engine are given The heat input to the engine is to be determined Assumptions 1 The plant operates steadily 2 Heat losses from the working fluid at the pipes and other components are negligible sink Furnace HE qL qH Analysis According to the definition of the thermal efficiency as applied to the heat engine H L H H q q q q w th th net η η wnet which when rearranged gives 1667 kJkg 40 1 1000 kJkg 1 th η L H q q 621 The power output and fuel consumption rate of a power plant are given The thermal efficiency is to be determined Assumptions The plant operates steadily Properties The heating value of coal is given to be 30000 kJkg sink HE 60 th coal Furnace 150 MW Analysis The rate of heat supply to this power plant is 500 MW 18 10 kJh 60000 kgh 30000 kJkg 9 HVcoal coal q m QH Then the thermal efficiency of the plant becomes 300 0300 500 MW 150 MW netout th H Q W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 66 622 The power output and fuel consumption rate of a car engine are given The thermal efficiency of the engine is to be determined Assumptions The car operates steadily Properties The heating value of the fuel is given to be 44000 kJkg Analysis The mass consumption rate of the fuel is sink HE Fuel 22 Lh Engine 55 kW 176 kgh 08 kgL22 Lh fuel fuel V ρ m The rate of heat supply to the car is 2151 kW 774400 kJh 176 kgh44000 kJkg HVcoal coal q m QH Then the thermal efficiency of the car becomes 256 0256 2151 kW 55 kW netout th H Q W η 623 The United States produces about 51 percent of its electricity from coal at a conversion efficiency of about 34 percent The amount of heat rejected by the coalfired power plants per year is to be determined Analysis Noting that the conversion efficiency is 34 the amount of heat rejected by the coal plants per year is kWh 10 3646 12 kWh 1 878 10 0 34 kWh 878 10 1 12 12 coal th coal out coal out coal in coal th W W Q W Q W Q W η η sink HE Coal Furnace 18781012 kWh ηth 34 out Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 67 624 The projected power needs of the United States is to be met by building inexpensive but inefficient coal plants or by building expensive but efficient IGCC plants The price of coal that will enable the IGCC plants to recover their cost difference from fuel savings in 5 years is to be determined Assumptions 1 Power is generated continuously by either plant at full capacity 2 The time value of money interest inflation etc is not considered Properties The heating value of the coal is given to be 28106 kJton Analysis For a power generation capacity of 150000 MW the construction costs of coal and IGCC plants and their difference are 9 9 9 9 IGCC 9 coal 30 10 195 10 225 10 on cost difference Constructi 150000000 kW1500kW 225 10 on cost Constructi 150000000 kW1300kW 195 10 on cost Constructi The amount of electricity produced by either plant in 5 years is kWh 150000000 kW5 365 24 h 6570 10 12 W t We The amount of fuel needed to generate a specified amount of power can be determined from Heating value Heating value or in fuel in in η η η e e e W Q m W Q Q W Then the amount of coal needed to generate this much electricity by each plant and their difference are 10 0 352 10 tons 1 760 10 112 2 10 tons 1 760 1kWh kJ 3600 0 4828 10 kJton kWh 6 570 10 Heating value 10 tons 2 112 1kWh kJ 3600 0 4028 10 kJton kWh 10 6 570 Heating value 9 9 9 coal IGCC plant coal coal plant coal 9 6 12 IGCC plant coal 9 6 12 coal plant coal m m m W m W m e e η η For to pay for the construction cost difference of 30 billion the price of coal should be mcoal 852ton 352 10 tons 0 30 10 Construction cost difference Unit cost of coal 9 9 mcoal Therefore the IGCC plant becomes attractive when the price of coal is above 852 per ton PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 68 625 Problem 624 is reconsidered The price of coal is to be investigated for varying simple payback periods plant construction costs and operating efficiency Analysis The problem is solved using EES and the solution is given below Given Wdot15E7 kW Costcoal1300 kW etacoal040 CostIGCC1500 kW etaIGCC048 HVcoal28000 kJkg PaybackYears5 yr Analysis timePaybackYearsConvertyr h ConstCostcoalWdotCostcoal ConstCostIGCCWdotCostIGCC ConstCostDifConstCostIGCCConstCostcoal WeWdottime mcoalcoalWeetacoalHVcoalConvertkWh kJ mcoalIGCCWeetaIGCCHVcoalConvertkWh kJ DELTAmcoalmcoalcoalmcoalIGCC UnitCostcoalConstCostDifDELTAmcoal1000 1 2 3 4 5 6 7 8 9 10 0 50 100 150 200 250 300 350 400 450 PaybackYears yr UnitCostcoal ton PaybackYears yr UnitCostcoal ton 1 2 3 4 5 6 7 8 9 10 4262 2131 1421 1065 8524 7103 6088 5327 4735 4262 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 69 ηcoal UnitCostcoal ton 03 031 032 033 034 035 036 037 038 039 04 041 042 043 044 045 2841 3109 3409 375 414 459 5114 5734 6478 7387 8524 9985 1193 1466 1875 2557 03 032 034 036 038 04 042 044 046 0 50 100 150 200 250 300 ηcoal UnitCostcoal ton 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 0 50 100 150 200 250 300 350 400 CostIGCC kW UnitCostcoal ton CostIGCC kW UnitCostcoal ton 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 0 4262 8524 1279 1705 2131 2557 2983 3409 3836 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 610 626 The projected power needs of the United States is to be met by building inexpensive but inefficient coal plants or by building expensive but efficient IGCC plants The price of coal that will enable the IGCC plants to recover their cost difference from fuel savings in 3 years is to be determined Assumptions 1 Power is generated continuously by either plant at full capacity 2 The time value of money interest inflation etc is not considered Properties The heating value of the coal is given to be 28106 kJton Analysis For a power generation capacity of 150000 MW the construction costs of coal and IGCC plants and their difference are 9 9 9 9 IGCC 9 coal 30 10 195 10 225 10 on cost difference Constructi 150000000 kW1500kW 225 10 on cost Constructi 150000000 kW1300kW 195 10 on cost Constructi The amount of electricity produced by either plant in 3 years is 150000000 kW3 365 24 h 3942 10 kWh 12 W t We The amount of fuel needed to generate a specified amount of power can be determined from Heating value Heating value or in fuel in in η η η e e e W Q m W Q Q W Then the amount of coal needed to generate this much electricity by each plant and their difference are 1 055 10 0 211 10 tons 10 267 1 1 055 10 tons 1kWh kJ 3600 0 4828 10 kJton kWh 3 942 10 Heating value 10 tons 1 267 1kWh kJ 3600 0 4028 10 kJton kWh 10 3 942 Heating value 9 9 9 coal IGCC plant coal coal plant coal 9 6 12 IGCC plant coal 9 6 12 coal plant coal m m m W m W m e e η η For to pay for the construction cost difference of 30 billion the price of coal should be mcoal 142ton 211 10 tons 0 30 10 Construction cost difference Unit cost of coal 9 9 mcoal Therefore the IGCC plant becomes attractive when the price of coal is above 142 per ton 627E The power output and thermal efficiency of a solar pond power plant are given The rate of solar energy collection is to be determined sink HE Solar pond Source 350 kW 4 Assumptions The plant operates steadily Analysis The rate of solar energy collection or the rate of heat supply to the power plant is determined from the thermal efficiency relation to be 10 Btuh 2986 7 1 h 3600 s 1055 kJ 1 Btu 004 kW 350 th netout η W QH PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 611 628 A coalburning power plant produces 300 MW of power The amount of coal consumed during a oneday period and the rate of air flowing through the furnace are to be determined Assumptions 1 The power plant operates steadily 2 The kinetic and potential energy changes are zero Properties The heating value of the coal is given to be 28000 kJkg Analysis a The rate and the amount of heat inputs to the power plant are 937 5 MW 0 32 MW 300 th netout in η W Q 10 MJ 18 3600 s 937 5 MJs24 7 in in t Q Q The amount and rate of coal consumed during this period are 3348 kgs 3600 s 24 893 10 kg 2 28 MJkg 10 MJ 18 6 coal coal 7 HV in coal t m m q Q m 10 kg 2893 6 b Noting that the airfuel ratio is 12 the rate of air flowing through the furnace is 4018 kgs 12 kg airkg fuel3348 kgs AF coal air m m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 612 Refrigerators and Heat Pumps 629C The difference between the two devices is one of purpose The purpose of a refrigerator is to remove heat from a cold medium whereas the purpose of a heat pump is to supply heat to a warm medium 630C The difference between the two devices is one of purpose The purpose of a refrigerator is to remove heat from a refrigerated space whereas the purpose of an airconditioner is remove heat from a living space 631C No Because the refrigerator consumes work to accomplish this task 632C No Because the heat pump consumes work to accomplish this task 633C The coefficient of performance of a refrigerator represents the amount of heat removed from the refrigerated space for each unit of work supplied It can be greater than unity 634C The coefficient of performance of a heat pump represents the amount of heat supplied to the heated space for each unit of work supplied It can be greater than unity 635C No The heat pump captures energy from a cold medium and carries it to a warm medium It does not create it 636C No The refrigerator captures energy from a cold medium and carries it to a warm medium It does not create it 637C No device can transfer heat from a cold medium to a warm medium without requiring a heat or work input from the surroundings 638C The violation of one statement leads to the violation of the other one as shown in Sec 64 and thus we conclude that the two statements are equivalent PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 613 639E The COP and the power input of a residential heat pump are given The rate of heating effect is to be determined Reservoir Reservoir HP H Q 5 hp COP 24 Assumptions The heat pump operates steadily Analysis Applying the definition of the heat pump coefficient of performance to this heat pump gives 30530 Btuh 1hp 42 5 hp 25445 Btuh COP HPWnetin QH 640 The cooling effect and the rate of heat rejection of an air conditioner are given The COP is to be determined Assumptions The air conditioner operates steadily Reservoir Reservoir AC netin W Q H L Q Analysis Applying the first law to the air conditioner gives kW 50 2 52 netin L H Q Q W Applying the definition of the coefficient of performance 4 05 kW 20 kW COP netin R W QL 641 The power input and the COP of a refrigerator are given The cooling effect of the refrigerator is to be determined Reservoir Reservoir R netin W QH COP14 L Q Assumptions The refrigerator operates steadily Analysis Rearranging the definition of the refrigerator coefficient of performance and applying the result to this refrigerator gives 42 kW 3 kW 41 COP RWnetin QL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 614 642 A refrigerator is used to keep a food department at a specified temperature The heat gain to the food department and the heat rejection in the condenser are given The power input and the COP are to be determined Assumptions The refrigerator operates steadily 12C 30C R H Q in W 4800 kJh 3300 kJh L Q Analysis The power input is determined from 0417 kW 3600 kJh 1kW 1500 kJh 1500 kJh 3300 4800 in L H Q Q W The COP is 22 1500 kJh 3300 kJh COP in W QL 643 The COP and the refrigeration rate of a refrigerator are given The power consumption and the rate of heat rejection are to be determined Assumptions The refrigerator operates steadily cool space Kitchen air R L Q COP12 Analysis a Using the definition of the coefficient of performance the power input to the refrigerator is determined to be 083 kW 5 0 kJmin 12 60 kJmin COPR netin QL W b The heat transfer rate to the kitchen air is determined from the energy balance 110 kJmin 50 60 Wnetin Q Q L H 644E The heat absorption the heat rejection and the power input of a commercial heat pump are given The COP of the heat pump is to be determined Reservoir Reservoir HP H Q 2 hp L Q Assumptions The heat pump operates steadily Analysis Applying the definition of the heat pump coefficient of performance to this heat pump gives 297 25445 Btuh 1hp 2 hp 15090 Btuh COP netin HP W QH PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 615 645 The cooling effect and the COP of a refrigerator are given The power input to the refrigerator is to be determined Reservoir Reservoir R Wnetin Q H COP16 L Q Assumptions The refrigerator operates steadily Analysis Rearranging the definition of the refrigerator coefficient of performance and applying the result to this refrigerator gives 434 kW 3600s h 1 160 25000 kJh COPR netin QL W 646 The COP and the power consumption of a refrigerator are given The time it will take to cool 5 watermelons is to be determined Assumptions 1 The refrigerator operates steadily 2 The heat gain of the refrigerator through its walls door etc is negligible 3 The watermelons are the only items in the refrigerator to be cooled Properties The specific heat of watermelons is given to be c 42 kJkgC Analysis The total amount of heat that needs to be removed from the watermelons is 2520 kJ 8 C C 20 10 kg 42 kJkg 5 watermelons o mc T QL cool space Kitchen air R 450 W COP 25 The rate at which this refrigerator removes heat is 1125 kW 25 045 kW COP netin R W QL That is this refrigerator can remove 1125 kJ of heat per second Thus the time required to remove 2520 kJ of heat is 373 min 2240 s 1125 kJs kJ 2520 L L Q Q t This answer is optimistic since the refrigerated space will gain some heat during this process from the surrounding air which will increase the work load Thus in reality it will take longer to cool the watermelons PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 616 647 An air conditioner with a known COP cools a house to desired temperature in 15 min The power consumption of the air conditioner is to be determined Assumptions 1 The air conditioner operates steadily 2 The house is wellsealed so that no air leaks in or out during cooling 3 Air is an ideal gas with constant specific heats at room temperature Properties The constant volume specific heat of air is given to be cv 072 kJkgC Analysis Since the house is wellsealed constant volume the total amount of heat that needs to be removed from the house is House 3520C H Q COP 28 Outside AC 8640 kJ 20 C C 35 800 kg 072 kJkg House T mc QL v This heat is removed in 30 minutes Thus the average rate of heat removal from the house is kW 84 60 s 30 8640 kJ t Q Q L L Using the definition of the coefficient of performance the power input to the air conditioner is determined to be kW 171 28 48 kW COPR netin QL W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 617 648 Problem 647 is reconsidered The rate of power drawn by the air conditioner required to cool the house as a function for air conditioner EER ratings in the range 5 to 15 is to be investigated Representative costs of air conditioning units in the EER rating range are to be included Analysis The problem is solved using EES and the results are tabulated and plotted below Since it is well sealed we treat the house as a closed system constant volume to determine the rate of heat transfer required to cool the house Apply the first law closed system on a rate basis to the house Input Data T135 C T220 C cv 072 kJkgC mhouse800 kg DELTAtime30 min EER5 COPEER3412 Assuming no work done on the house and no heat energy added to the house in the time period with no change in KE and PE the first law applied to the house is Edotin Edotout DELTAEdot Edotin 0 Edotout QdotL DELTAEdot mhouseDELTAuhouseDELTAtime DELTAuhouse cvT2T1 Using the definition of the coefficient of performance of the AC Wdotin QdotLCOP kJminconvertkJminkW kW QdotH WdotinconvertKWkJmin QdotL kJmin EER Win kW 5 6 7 8 9 10 11 12 13 14 15 3276 273 234 2047 182 1638 1489 1365 126 117 1092 5 7 9 11 13 15 1 15 2 25 3 35 EER Win kW PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 618 649 A refrigerator is used to cool bananas to a specified temperature The power input is given The rate of cooling and the COP are to be determined Assumptions The refrigerator operates steadily Properties The specific heat of banana is 335 kJkgC Analysis The rate of cooling is determined from 132 kJmin 13 C C24 215 60 kgmin 3 35 kJkg 2 1 T T mc Q p L The COP is 157 kW 41 132 60 kW COP Win QL 650 A refrigerator is used to cool water to a specified temperature The power input is given The flow rate of water and the COP of the refrigerator are to be determined Assumptions The refrigerator operates steadily Properties The specific heat of water is 418 kJkgC and its density is 1 kgL Analysis The rate of cooling is determined from 685 kW 2 65 kW 570 60 kW in W Q Q H L The mass flow rate of water is 0 09104 kgs 5 C C23 4 18 kJkg 6 85 kW 2 1 2 1 T T c Q m T T mc Q p L p L The volume flow rate is 546 Lmin 1min 60 s 1kgL 0 09104 kgs ρ m V The COP is 258 2 65 kW 6 85 kW COP in W QL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 619 651 The rate of heat loss the rate of internal heat gain and the COP of a heat pump are given The power input to the heat pump is to be determined Assumptions The heat pump operates steadily Outside House H Q HP COP 25 60000 kJh Analysis The heating load of this heat pump system is the difference between the heat lost to the outdoors and the heat generated in the house from the people lights and appliances QH 60 000 4 000 56 000 kJ h Using the definition of COP the power input to the heat pump is determined to be 622 kW 3600 kJh 1 kW 25 56000 kJh COPHP netin QH W 652E The COP and the refrigeration rate of an ice machine are given The power consumption is to be determined Assumptions The ice machine operates steadily QL Ice Machine ice 25F water 55F Outdoors R COP 24 Analysis The cooling load of this ice machine is 4732 Btuh 28 lbmh 169 Btulbm L L mq Q Using the definition of the coefficient of performance the power input to the ice machine system is determined to be 0775 hp 2545 Btuh 1 hp 24 4732 Btuh COPR netin QL W 653E An office that is being cooled adequately by a 12000 Btuh window airconditioner is converted to a computer room The number of additional airconditioners that need to be installed is to be determined Assumptions 1 The computers are operated by 7 adult men 2 The computers consume 40 percent of their rated power at any given time Properties The average rate of heat generation from a person seated in a roomoffice is 100 W given Analysis The amount of heat dissipated by the computers is equal to the amount of electrical energy they consume Therefore Outside AC Computer room 7000 Btuh 4060 W 13853 Btuh 700 3360 700 W 100 W 7 No of people Usage factor 84 kW04 336 kW Rated power people computers total person people computers Q Q Q Q Q Q since 1 W 3412 Btuh Then noting that each available air conditioner provides 7000 Btuh cooling the number of airconditioners needed becomes 2 Air conditioners 1 98 7000 Btuh 13853 Btuh Cooling capacity of AC Cooling load No of air conditioners PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 620 654 A decision is to be made between a cheaper but inefficient airconditioner and an expensive but efficient air conditioner for a building The better buy is to be determined Assumptions The two air conditioners are comparable in all aspects other than the initial cost and the efficiency Analysis The unit that will cost less during its lifetime is a better buy The total cost of a system during its lifetime the initial operation maintenance etc can be determined by performing a life cycle cost analysis A simpler alternative is to determine the simple payback period The energy and cost savings of the more efficient air conditioner in this case is 13500 kWhyear 05 120000 kWhyear132 1 1 COP Annual cooling load1 COP Annual energy usage of B Annual energy usage of A savings Energy B A Air Cond A COP 32 13500 kWhyear010kWh 1350year Energy savingsUnit cost of energy Cost savings The installation cost difference between the two airconditioners is Air Cond B COP 50 Cost difference Cost of B cost of A 7000 5500 1500 Therefore the more efficient airconditioner B will pay for the 1500 cost differential in this case in about 1 year Discussion A cost conscious consumer will have no difficulty in deciding that the more expensive but more efficient air conditioner B is clearly the better buy in this case since air conditioners last at least 15 years But the decision would not be so easy if the unit cost of electricity at that location was much less than 010kWh or if the annual airconditioning load of the house was much less than 120000 kWh 655 A house is heated by resistance heaters and the amount of electricity consumed during a winter month is given The amount of money that would be saved if this house were heated by a heat pump with a known COP is to be determined Assumptions The heat pump operates steadily Analysis The amount of heat the resistance heaters supply to the house is equal to he amount of electricity they consume Therefore to achieve the same heating effect the house must be supplied with 1200 kWh of energy A heat pump that supplied this much heat will consume electrical power in the amount of 500 kWh 24 1200 kWh COPHP netin QH W which represent a savings of 1200 500 700 kWh Thus the homeowner would have saved 700 kWh0085 kWh 5950 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 621 656 Refrigerant134a flows through the condenser of a residential heat pump unit For a given compressor power consumption the COP of the heat pump and the rate of heat absorbed from the outside air are to be determined Assumptions 1 The heat pump operates steadily 2 The kinetic and potential energy changes are zero PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The enthalpies of R134a at the condenser inlet and exit are 9547 kJkg 0 kPa 800 27122 kJkg C 35 kPa 800 2 2 2 1 1 1 h x P h T P Analysis a An energy balance on the condenser gives the heat rejected in the condenser 3 164 kW 9547 kJkg 0 018 kgs27122 2 1 h m h QH QH 800 kPa x0 Win Condenser Evaporator Compressor Expansion valve QL 800 kPa 35C The COP of the heat pump is 264 kW 21 3 164 kW COP Win QH b The rate of heat absorbed from the outside air 196 kW 21 3 164 Win Q Q H L 657 A commercial refrigerator with R134a as the working fluid is considered The evaporator inlet and exit states are specified The mass flow rate of the refrigerant and the rate of heat rejected are to be determined Assumptions 1 The refrigerator operates steadily 2 The kinetic and potential energy changes are zero QH 100 kPa 26C Condenser Evaporator Compressor Expansion valve 100 kPa x02 Win QL Properties The properties of R134a at the evaporator inlet and exit states are Tables A11 through A13 23474 kJkg C 26 kPa 100 6071 kJkg 20 kPa 100 2 2 2 1 1 1 h T P h x P Analysis a The refrigeration load is 0 72 kW 0 600 kW 21 COP in W QL The mass flow rate of the refrigerant is determined from 000414 kgs 6071 kJkg 23474 72 kW 0 1 2 h h Q m L R b The rate of heat rejected from the refrigerator is 132 kW 0 60 0 72 Win Q Q L H preparation If you are a student using this Manual you are using it without permission 622 PerpetualMotion Machines 658C This device creates energy and thus it is a PMM1 659C This device creates energy and thus it is a PMM1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 623 Reversible and Irreversible Processes 660C Adiabatic stirring processes are irreversible because the energy stored within the system can not be spontaneously released in a manor to cause the mass of the system to turn the paddle wheel in the opposite direction to do work on the surroundings 661C The chemical reactions of combustion processes of a natural gas and air mixture will generate carbon dioxide water and other compounds and will release heat energy to a lower temperature surroundings It is unlikely that the surroundings will return this energy to the reacting system and the products of combustion react spontaneously to reproduce the natural gas and air mixture 662C No Because it involves heat transfer through a finite temperature difference 663C This process is irreversible As the block slides down the plane two things happen a the potential energy of the block decreases and b the block and plane warm up because of the friction between them The potential energy that has been released can be stored in some form in the surroundings eg perhaps in a spring When we restore the system to its original condition we must a restore the potential energy by lifting the block back to its original elevation and b cool the block and plane back to their original temperatures The potential energy may be restored by returning the energy that was stored during the original process as the block decreased its elevation and released potential energy The portion of the surroundings in which this energy had been stored would then return to its original condition as the elevation of the block is restored to its original condition In order to cool the block and plane to their original temperatures we have to remove heat from the block and plane When this heat is transferred to the surroundings something in the surroundings has to change its state eg perhaps we warm up some water in the surroundings This change in the surroundings is permanent and cannot be undone Hence the original process is irreversible 664C Because reversible processes can be approached in reality and they form the limiting cases Work producing devices that operate on reversible processes deliver the most work and work consuming devices that operate on reversible processes consume the least work 665C When the compression process is nonquasi equilibrium the molecules before the piston face cannot escape fast enough forming a high pressure region in front of the piston It takes more work to move the piston against this high pressure region 666C When an expansion process is nonquasiequilibrium the molecules before the piston face cannot follow the piston fast enough forming a low pressure region behind the piston The lower pressure that pushes the piston produces less work PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 624 667C The irreversibilities that occur within the system boundaries are internal irreversibilities those which occur outside the system boundaries are external irreversibilities 668C A reversible expansion or compression process cannot involve unrestrained expansion or sudden compression and thus it is quasiequilibrium A quasiequilibrium expansion or compression process on the other hand may involve external irreversibilities such as heat transfer through a finite temperature difference and thus is not necessarily reversible PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 625 The Carnot Cycle and Carnots Principle 669C The four processes that make up the Carnot cycle are isothermal expansion reversible adiabatic expansion isothermal compression and reversible adiabatic compression 670C They are 1 the thermal efficiency of an irreversible heat engine is lower than the efficiency of a reversible heat engine operating between the same two reservoirs and 2 the thermal efficiency of all the reversible heat engines operating between the same two reservoirs are equal 671C False The second Carnot principle states that no heat engine cycle can have a higher thermal efficiency than the Carnot cycle operating between the same temperature limits 672C Yes The second Carnot principle states that all reversible heat engine cycles operating between the same temperature limits have the same thermal efficiency 673C a No b No They would violate the Carnot principle PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 626 Carnot Heat Engines 674C No 675C The one that has a source temperature of 600C This is true because the higher the temperature at which heat is supplied to the working fluid of a heat engine the higher the thermal efficiency 676E The source and sink temperatures of a heat engine are given The maximum work per unit heat input to the engine is to be determined 510 R 1260 R HE Wnet QL QH Assumptions The heat engine operates steadily Analysis The maximum work per unit of heat that the engine can remove from the source is the Carnot efficiency which is determined from 0595 1260 R 510 R 1 1 thmax net H L H T T Q W η 677 Two pairs of thermal energy reservoirs are to be compared from a workproduction perspective Assumptions The heat engine operates steadily Analysis For the maximum production of work a heat engine operating between the energy reservoirs would have to be completely reversible Then for the first pair of reservoirs HE TL TH Wnet QL QH 0519 675 K 325 K 1 1 thmax H L T T η For the second pair of reservoirs 0560 625 K 275 K 1 1 thmax H L T T η The second pair is then capable of producing more work for each unit of heat extracted from the hot reservoir PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 627 678 The source and sink temperatures of a heat engine and the rate of heat supply are given The maximum possible power output of this engine is to be determined Assumptions The heat engine operates steadily Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency which is determined from 25C 477C HE 65000 kJmin 0600 or 600 273 K 477 298 K 1 1 thC thmax H L T T η η Then the maximum power output of this heat engine is determined from the definition of thermal efficiency to be 653 kW 39000 kJmin 65000 kJmin 0600 th netout QH W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 628 679 Problem 678 is reconsidered The effects of the temperatures of the heat source and the heat sink on the power produced and the cycle thermal efficiency as the source temperature varies from 300C to 1000C and the sink temperature varies from 0C to 50C are to be studied The power produced and the cycle efficiency against the source temperature for sink temperatures of 0C 25C and 50C are to be plotted Analysis The problem is solved using EES and the results are tabulated and plotted below TH 477 C TL 25 C QdotH 65000 kJmin First Law applied to the heat engine QdotH QdotL Wdotnet 0 WdotnetKWWdotnetconvertkJminkW Cycle Thermal Efficiency Temperatures must be absolute etath 1 TL 273TH 273 Definition of cycle efficiency etathWdotnet QdotH 300 400 500 600 700 800 900 1000 450 500 550 600 650 700 750 800 850 TH C Wnet kW TL25C TL50C TL0C TH C WnetkW kW ηth 300 400 500 600 700 800 900 1000 5672 6439 7007 7446 7794 8077 8312 851 05236 05944 06468 06873 07194 07456 07673 07855 Values for TL 0C 300 400 500 600 700 800 900 1000 04 045 05 055 06 065 07 075 08 TH C ηth TL25C TL50C TL0C PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 629 680E The sink temperature of a Carnot heat engine the rate of heat rejection and the thermal efficiency are given The power output of the engine and the source temperature are to be determined Assumptions The Carnot heat engine operates steadily Analysis a The rate of heat input to this heat engine is determined from the definition of thermal efficiency 3200 Btumin 800 Btumin 1 0 75 1 th H H H L Q Q Q Q η Then the power output of this heat engine can be determined from 60F TH HE 800 Btumin 566 hp 2400 Btumin 3200 Btumin 075 th netout QH W η b For reversible cyclic devices we have L H L H T T Q Q rev Thus the temperature of the source TH must be 2080 R 520 R 800 Btumin Btumin 3200 rev L L H H T Q Q T 681E The claim of an inventor about the operation of a heat engine is to be evaluated Assumptions The heat engine operates steadily Analysis If this engine were completely reversible the thermal efficiency would be 15000 Btuh HE H Q p 5 h 1000 R 550 R 045 1000 R 550 R 1 1 thmax H L T T η When the first law is applied to the engine above 27720 Btuh 15000 Btuh 1hp 5 hp 25445 Btuh net L H Q W Q The actual thermal efficiency of the proposed heat engine is then 0 459 1hp 25445 Btuh 27720 Btuh 5 hp net th H Q W η Since the thermal efficiency of the proposed heat engine is greater than that of a completely reversible heat engine which uses the same isothermal energy reservoirs the inventors claim is invalid PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 630 682 The work output and thermal efficiency of a Carnot heat engine are given The heat supplied to the heat engine the heat rejected and the temperature of heat sink are to be determined Assumptions 1 The heat engine operates steadily 2 Heat losses from the working fluid at the pipes and other components are negligible sink 1200C HE QL QH ηth 40 Analysis Applying the definition of the thermal efficiency and an energy balance to the heat engine the unknown values are determined as follows 500 kJ 1250 kJ 04 kJ 500 th net η W QH 750 kJ 500 1250 Wnet Q Q H L 611C 883 8 K 273 K 1200 1 0 40 1 thmax L L H L T T T T η 683 The work output and heat rejection of a a Carnot heat engine are given The heat supplied to the heat engine and the source temperature are to be determined 27C TH HE QL QH 150 kJ Assumptions 1 The heat engine operates steadily 2 Heat losses from the working fluid at the pipes and other components are negligible Analysis Applying the definition of the thermal efficiency and an energy balance to the heat engine the unknown values are determined as follows 900 kJ 1050 kJ 900 150 Wnet Q Q L H 0 857 1050 kJ 900 kJ net th QH W η 1827C 2100 K 273 K 27 1 0 857 1 thmax H H H L T T T T η 684 The thermal efficiency and waste heat rejection of a Carnot heat engine are given The power output and the temperature of the source are to be determined Assumptions 1 The heat engine operates steadily 2 Heat losses from the working fluid at the pipes and other components are negligible 288 K Source HE 14 kW QL net W QH ηth 75 Analysis Applying the definition of the thermal efficiency and an energy balance to the heat engine the power output and the source temperature are determined as follows 42 kW 0 7556 kW 56 kW 14 kW 1 0 75 1 th net th H H H H L Q W Q Q Q Q η η 879C 1152 K 273 K 15 1 0 75 1 th H H H L T T T T η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 631 685 A geothermal power plant uses geothermal liquid water at 150ºC at a specified rate as the heat source The actual and maximum possible thermal efficiencies and the rate of heat rejected from this power plant are to be determined Assumptions 1 The power plant operates steadily 2 The kinetic and potential energy changes are zero 3 Steam properties are used for geothermal water Properties Using saturated liquid properties Table A4 10483 kJkg 0 C 25 37704 kJkg 0 C 90 63218 kJkg 0 C 150 sink sink sink source source source2 geo1 source source1 h x T h x T h x T Analysis a The rate of heat input to the plant is 53580 kW 37704 kJkg 210 kgs63218 geo2 geo1 geo in h h m Q The actual thermal efficiency is 01493 149 53580 kW kW 8000 in netout th Q W η b The maximum thermal efficiency is the thermal efficiency of a reversible heat engine operating between the source and sink temperatures 02955 296 273 K 150 273 K 25 1 1 thmax H L T T η c Finally the rate of heat rejection is 45580 kW 8000 53580 netout in out W Q Q 686 The claim that the efficiency of a completely reversible heat engine can be doubled by doubling the temperature of the energy source is to be evaluated Assumptions The heat engine operates steadily Analysis The upper limit for the thermal efficiency of any heat engine occurs when a completely reversible engine operates between the same energy reservoirs The thermal efficiency of this completely reversible engine is given by H L H H L T T T T T ηthrev 1 HE TL TH Wnet QL QH If we were to double the absolute temperature of the high temperature energy reservoir the new thermal efficiency would be H L H H L H H L T T T T T T T T 2 2 2 2 1 ηthrev The thermal efficiency is then not doubled as the temperature of the high temperature reservoir is doubled PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 632 Carnot Refrigerators and Heat Pumps 687C By increasing TL or by decreasing TH 688C The difference between the temperature limits is typically much higher for a refrigerator than it is for an air conditioner The smaller the difference between the temperature limits a refrigerator operates on the higher is the COP Therefore an airconditioner should have a higher COP 689C The deep freezer should have a lower COP since it operates at a much lower temperature and in a given environment the COP decreases with decreasing refrigeration temperature 690C No At best when everything is reversible the increase in the work produced will be equal to the work consumed by the refrigerator In reality the work consumed by the refrigerator will always be greater than the additional work produced resulting in a decrease in the thermal efficiency of the power plant 691C No At best when everything is reversible the increase in the work produced will be equal to the work consumed by the refrigerator In reality the work consumed by the refrigerator will always be greater than the additional work produced resulting in a decrease in the thermal efficiency of the power plant 692C Bad idea At best when everything is reversible the increase in the work produced will be equal to the work consumed by the heat pump In reality the work consumed by the heat pump will always be greater than the additional work produced resulting in a decrease in the thermal efficiency of the power plant 693 The minimum work per unit of heat transfer from the lowtemperature source for a refrigerator is to be determined Assumptions The refrigerator operates steadily R TH TL QL QH Wnetin Analysis Application of the first law gives 1 netin L H L L H L Q Q Q Q Q Q W For the minimum work input this refrigerator would be completely reversible and the thermodynamic definition of temperature would reduce the preceding expression to 0110 1 273 K 303 K 1 in net L H L T T Q W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 633 694 The validity of a claim by an inventor related to the operation of a heat pump is to be evaluated Assumptions The heat pump operates steadily Analysis Applying the definition of the heat pump coefficient of performance 273 K 293 K HP H Q QL 75 267 75 kW 200 kW COP netin HP W QH The maximum COP of a heat pump operating between the same temperature limits is 14 7 1 273 K293 K 1 1 1 COP HPmax TL TH Since the actual COP is less than the maximum COP the claim is valid 695 The power input and the COP of a Carnot heat pump are given The temperature of the lowtemperature reservoir and the heating load are to be determined TL 26C HP H Q L Q 425 kW Assumptions The heat pump operates steadily Analysis The temperature of the lowtemperature reservoir is 2646 K L L L H H T T T T T COP K 299 299 K 78 HPmax The heating load is 370 kW H H in H Q Q W Q COP 4 25 kW 78 HPmax 696 The refrigerated space and the environment temperatures for a refrigerator and the rate of heat removal from the refrigerated space are given The minimum power input required is to be determined Assumptions The refrigerator operates steadily Analysis The power input to a refrigerator will be a minimum when the refrigerator operates in a reversible manner The coefficient of performance of a reversible refrigerator depends on the temperature limits in the cycle only and is determined from 8C 25C R 300 kJmin 8 03 1 273 K 8 273 K 25 1 1 1 Rrev TH TL COP The power input to this refrigerator is determined from the definition of the coefficient of performance of a refrigerator 0623 kW 3736 kJmin 803 kJmin 300 Rmax netinmin COP Q W L PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 634 697 An inventor claims to have developed a refrigerator The inventor reports temperature and COP measurements The claim is to be evaluated 12C 25C R COP 65 Analysis The highest coefficient of performance a refrigerator can have when removing heat from a cool medium at 12C to a warmer medium at 25C is 71 1 273 K 12 273 K 25 1 1 1 COP COP Rrev Rmax TH TL The COP claimed by the inventor is 65 which is below this maximum value thus the claim is reasonable However it is not probable 698E An airconditioning system maintains a house at a specified temperature The rate of heat gain of the house and the rate of internal heat generation are given The maximum power input required is to be determined Assumptions The airconditioner operates steadily Analysis The power input to an airconditioning system will be a minimum when the airconditioner operates in a reversible manner The coefficient of performance of a reversible airconditioner or refrigerator depends on the temperature limits in the cycle only and is determined from 1767 1 460 R 460 R 70 100 1 1 1 COP Rrev L H T T 100F House 75F AC 800 kJmin The cooling load of this airconditioning system is the sum of the heat gain from the outside and the heat generated within the house 900 Btumin 100 800 L Q The power input to this refrigerator is determined from the definition of the coefficient of performance of a refrigerator 120 hp 5093 Btumin 1767 900 Btumin COP Rmax netinmin QL W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 635 699 A heat pump maintains a house at a specified temperature The rate of heat loss of the house and the power consumption of the heat pump are given It is to be determined if this heat pump can do the job Assumptions The heat pump operates steadily Analysis The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only and is determined from 1475 273 K 273 K 22 2 1 1 1 1 COP HPrev TL TH 5 kW House 22C HP 110000 kJh The required power input to this reversible heat pump is determined from the definition of the coefficient of performance to be 207 kW 3600 s 1 h 1475 110000 kJh COPHP netinmin QH W This heat pump is powerful enough since 5 kW 207 kW 6100E The power required by a reversible refrigerator with specified reservoir temperatures is to be determined Assumptions The refrigerator operates steadily 450 R 540 R R Wnetin 15000 Btuh Analysis The COP of this reversible refrigerator is 5 45 0 R 540 R 450 R COP Rmax L H L T T T Using this result in the coefficient of performance expression yields 0879 kW 341214 Btuh 1kW 5 15000 Btuh COP Rmax netin QL W 6101 The power input and heat rejection of a reversed Carnot cycle are given The cooling load and the source temperature are to be determined Assumptions The refrigerator operates steadily TL 300 K R 2000 kW H Q L Q 200 kW Analysis Applying the definition of the refrigerator coefficient of performance 1800 kW 200 2000 Wnetin Q Q H L Applying the definition of the heat pump coefficient of performance 9 200 kW 1800 kW COP netin R W QL The temperature of the heat source is determined from 3C 270 K 300 9 COP Rmax L L L L H L T T T T T T PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 636 6102 The power input and the cooling load of an air conditioner are given The rate of heat rejected in the condenser the COP and the rate of cooling for a reversible operation are to be determined Assumptions The air conditioner operates steadily 22C 33C AC H Q QL 18000 Btuh 34 kW Analysis a The rate of heat rejected is 31230 kJh 1kW kW 3600 kJh 43 1Btu 18000 Btuh 1055 kJ Win Q Q L H b The COP is 1552 kW 43 3412 Btuh 1kW 18000 Btuh COP in W QL c The rate of cooling if the air conditioner operated as a Carnot refrigerator for the same power input is 2682 22 K 33 295 K COPrev L H L T T T 311130 Btuh 1kW kW 3412 Btuh 43 2682 COP inmin rev max W QL 6103 The rate of heat removal and the source and sink temperatures are given for a Carnot refrigerator The COP of the refrigerator and the power input are to be determined Assumptions The refrigerator operates steadily 15C 36C R H Q L Q in W 16000 kJh Analysis The COP of the Carnot refrigerator is determined from 1371 36 15 K 288 K Rmax L H L T T T COP The power input is 0324 kW 1167 kJh 16000 kJh 1371 Rmax in in in L W W W Q COP The rate of heat rejected is 17167 kJh 1167 kJh 16000 kJh Wnetin Q Q L H PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 637 6104 A heat pump maintains a house at a specified temperature in winter The maximum COPs of the heat pump for different outdoor temperatures are to be determined Analysis The coefficient of performance of a heat pump will be a maximum when the heat pump operates in a reversible manner The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only and is determined for all three cases above to be 293 273K 273K 20 10 1 1 1 1 COP HPrev TL TH TL 20C HP 117 273K 273K 20 5 1 1 1 1 COP HPrev TL TH 586 273K 273K 20 30 1 1 1 1 COP HPrev TL TH 6105E A heat pump maintains a house at a specified temperature The rate of heat loss of the house is given The minimum power inputs required for different source temperatures are to be determined Assumptions The heat pump operates steadily Analysis a The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner If the outdoor air at 25F is used as the heat source the COP of the heat pump and the required power input are determined to be 1015 460 R 460 R 78 25 1 1 1 1 COP COP HPrev max HP TL TH 25F or 50F House 78F HP 55000 Btuh and 213 hp 2545 Btuh 1 hp 1015 55000 Btuh COP HPmax netinmin QH W b If the wellwater at 50F is used as the heat source the COP of the heat pump and the required power input are determined to be 192 460 R 460 R 78 50 1 1 1 1 COP COP HPrev HPmax TL TH and 113 hp 2545 Btuh 1 hp 192 55000 Btuh COP HPmax netinmin QH W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 638 6106 A Carnot heat pump consumes 66kW of power when operating and maintains a house at a specified temperature The average rate of heat loss of the house in a particular day is given The actual running time of the heat pump that day the heating cost and the cost if resistance heating is used instead are to be determined Analysis a The coefficient of performance of this Carnot heat pump depends on the temperature limits in the cycle only and is determined from 1296 273 K 273 K 25 2 1 1 1 1 COP HPrev TL TH 2C House 25C HP 66 kW 55000 kJh The amount of heat the house lost that day is 1320000 kJ 55000 kJh 24 h 1 day H H Q Q Then the required work input to this Carnot heat pump is determined from the definition of the coefficient of performance to be 101880 kJ 1296 1320000 kJ COPHP netin QH W Thus the length of time the heat pump ran that day is 429 h 15440 s 66 kJs kJ 101880 in net netin W W t b The total heating cost that day is 241 66 kW 429 h 0085 kWh price price Cost netin t W W c If resistance heating were used the entire heating load for that day would have to be met by electrical energy Therefore the heating system would consume 1320000 kJ of electricity that would cost 312 0085 kWh 3600 kJ 1 kWh 1320000kJ price New Cost QH PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 639 6107 A Carnot heat engine is used to drive a Carnot refrigerator The maximum rate of heat removal from the refrigerated space and the total rate of heat rejection to the ambient air are to be determined Assumptions The heat engine and the refrigerator operate steadily Analysis a The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency which is determined from 0744 1173 K 300 K 1 1 thC thmax H L T T η η 27C 900C HE 800 kJmin R 5C Then the maximum power output of this heat engine is determined from the definition of thermal efficiency to be 5952 kJmin 800 kJmin 0744 th netout QH W η which is also the power input to the refrigerator netin W The rate of heat removal from the refrigerated space will be a maximum if a Carnot refrigerator is used The COP of the Carnot refrigerator is 837 1 273 K 5 273 K 27 1 1 1 COP Rrev TH TL Then the rate of heat removal from the refrigerated space becomes 4982 kJmin 837 5952 kJmin COP netin Rrev R W QL b The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine and the heat discarded by the refrigerator LHE Q H R Q 55772 kJmin 5952 4982 2048 kJmin 5952 800 netin R R netout HE HE W Q Q W Q Q L H H L and 5782 kJmin 5577 2 204 8 R HE ambient H L Q Q Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 640 6108E A Carnot heat engine is used to drive a Carnot refrigerator The maximum rate of heat removal from the refrigerated space and the total rate of heat rejection to the ambient air are to be determined Assumptions The heat engine and the refrigerator operate steadily Analysis a The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency which is determined from 0 75 2160 R 540 R 1 1 thC thmax H L T T η η 80F 1700F HE 700 Btumin R 20F Then the maximum power output of this heat engine is determined from the definition of thermal efficiency to be 525 Btumin 700 Btumin 075 th netout QH W η which is also the power input to the refrigerator netin W The rate of heat removal from the refrigerated space will be a maximum if a Carnot refrigerator is used The COP of the Carnot refrigerator is 08 1 460 R 460 R 20 80 1 1 1 Rrev TH TL COP Then the rate of heat removal from the refrigerated space becomes 4200 Btumin 80 525 Btumin COP netin Rrev R W QL b The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine and the heat discarded by the refrigerator LHE Q H R Q 472 5 Btumin 525 4200 175 Btumin 525 700 netin R R netout HE HE W Q Q W Q Q L H H L and 4725 175 R HE ambient 4900 Btumin H L Q Q Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 641 6109 A heat pump that consumes 4kW of power when operating maintains a house at a specified temperature The house is losing heat in proportion to the temperature difference between the indoors and the outdoors The lowest outdoor temperature for which this heat pump can do the job is to be determined Assumptions The heat pump operates steadily Analysis Denoting the outdoor temperature by TL the heating load of this house can be expressed as K 1056 kWK 297 3800 kJh K 297 L L H T T Q The coefficient of performance of a Carnot heat pump depends on the temperature limits in the cycle only and can be expressed as TL House 24C HP 3800 kJhK 297 K 1 1 1 1 COPHP L H L T T T or as 4 kW 4 kW K 1056 kWK 297 COP netin HP L H T W Q Equating the two relations above and solving for TL we obtain TL 2635 K 95C PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 642 6110 An airconditioner with R134a as the working fluid is considered The compressor inlet and exit states are specified The actual and maximum COPs and the minimum volume flow rate of the refrigerant at the compressor inlet are to be determined Assumptions 1 The airconditioner operates steadily 2 The kinetic and potential energy changes are zero Properties The properties of R134a at the compressor inlet and exit states are Tables A11 through A13 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 30061 kJkg C 70 MPa 21 05120 m kg 0 55 kJkg 255 1 kPa 400 2 2 2 3 1 1 1 1 h T P h x P v Analysis a The mass flow rate of the refrigerant and the power consumption of the compressor are 0 03255 kgs 05120 m kg 0 60s 1min 1000 L 1m Lmin 100 3 3 1 1 v V R m QH 400 kPa sat vap Condenser Evaporator Compressor QL 12 MPa 70C Win Expansion valve 1 467 kW 25555 kJkg 0 03255 kgs30061 1 2 in h h m W R The heat gains to the room must be rejected by the airconditioner That is 5 067 kW kW 90 60s 1min 250 kJmin equipment heat Q Q QL Then the actual COP becomes 345 1 467 kW 5 067 kW COP in W QL b The COP of a reversible refrigerator operating between the same temperature limits is 2114 1 273 273 23 37 1 1 1 COPmax L H T T c The minimum power input to the compressor for the same refrigeration load would be 0 2396 kW 2114 5 067 kW COPmax inmin QL W The minimum mass flow rate is 0 005318 kgs 25555 kJkg 30061 2396 kW 0 1 2 inmin min h h W mR Finally the minimum volume flow rate at the compressor inlet is 163 Lmin 0 0002723 m s 0 005318 kgs 0 05120 m kg 3 3 1 min min1 v V m R preparation If you are a student using this Manual you are using it without permission 643 6111 The COP of a completely reversible refrigerator as a function of the temperature of the sink is to be calculated and plotted 250 K TH R H Q L Q netin W Assumptions The refrigerator operates steadily Analysis The coefficient of performance for this completely reversible refrigerator is given by 25 0 K 250 K COP Rmax H L H L T T T T Using EES we tabulate and plot the variation of COP with the sink temperature as follows 300 340 380 420 460 500 0 1 2 3 4 5 TH K COPRmax TH K COPRmax 300 5 320 3571 340 2778 360 2273 380 1923 400 1667 420 1471 440 1316 460 119 480 1087 500 1 6112 An expression for the COP of a completely reversible refrigerator in terms of the thermalenergy reservoir temperatures TL and TH is to be derived Assumptions The refrigerator operates steadily Analysis Application of the first law to the completely reversible refrigerator yields L H Q Q W netin This result may be used to reduce the coefficient of performance R TH TL QL QH Wnetin 1 1 COP netin Rrev L H L H L L Q Q Q Q Q W Q Since this refrigerator is completely reversible the thermodynamic definition of temperature tells us that L H L H T T Q Q When this is substituted into the COP expression the result is L H L L H T T T T T 1 1 COP Rrev PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 644 Special Topic Household Refrigerators 6113C The energy consumption of a household refrigerator can be reduced by practicing good conservation measures such as 1 opening the refrigerator door the fewest times possible and for the shortest duration possible 2 cooling the hot foods to room temperature first before putting them into the refrigerator 3 cleaning the condenser coils behind the refrigerator 4 checking the door gasket for air leaks 5 avoiding unnecessarily low temperature settings 6 avoiding excessive ice buildup on the interior surfaces of the evaporator 7 using the powersaver switch that controls the heating coils that prevent condensation on the outside surfaces in humid environments and 8 not blocking the air flow passages to and from the condenser coils of the refrigerator 6114C It is important to clean the condenser coils of a household refrigerator a few times a year since the dust that collects on them serves as insulation and slows down heat transfer Also it is important not to block air flow through the condenser coils since heat is rejected through them by natural convection and blocking the air flow will interfere with this heat rejection process A refrigerator cannot work unless it can reject the waste heat 6115C Todays refrigerators are much more efficient than those built in the past as a result of using smaller and higher efficiency motors and compressors better insulation materials larger coil surface areas and better door seals 6116C It is a bad idea to overdesign the refrigeration system of a supermarket so that the entire airconditioning needs of the store can be met by refrigerated air without installing any airconditioning system This is because the refrigerators cool the air to a much lower temperature than needed for air conditioning and thus their efficiency is much lower and their operating cost is much higher 6117C It is a bad idea to meet the entire refrigeratorfreezer requirements of a store by using a large freezer that supplies sufficient cold air at 20C instead of installing separate refrigerators and freezers This is because the freezers cool the air to a much lower temperature than needed for refrigeration and thus their efficiency is much lower and their operating cost is much higher PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 645 6118 A refrigerator consumes 300 W when running and 74 worth of electricity per year under normal use The fraction of the time the refrigerator will run in a year is to be determined Assumptions The electricity consumed by the light bulb is negligible Analysis The total amount of electricity the refrigerator uses a year is 1057 kWhyear 007kWh 74year Unit cost of energy Total cost of energy Total electric energy used total We The number of hours the refrigerator is on per year is 3524 hyear 03 kW 1057 kWhyear Total operating hours total e e W W t Noting that there are 365248760 hours in a year the fraction of the time the refrigerator is on during a year is determined to be 0402 8760 hyear 3524year Total hours per year Total operating hours Time fraction on Therefore the refrigerator remained on 402 of the time 6119 The light bulb of a refrigerator is to be replaced by a 25 energy efficient bulb that consumes less than half the electricity It is to be determined if the energy savings of the efficient light bulb justify its cost Assumptions The new light bulb remains on the same number of hours a year PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis The lighting energy saved a year by the energy efficient bulb is 1320 Wh 132 kWh 40 18W60 hyear Lighting power savedOperating hours Lighting energy saved This means 132 kWh less heat is supplied to the refrigerated space by the light bulb which must be removed from the refrigerated space This corresponds to a refrigeration savings of Refrigeration energy saved Lighting energy saved COP 132 kWh 13 kWh 102 Then the total electrical energy and money saved by the energy efficient light bulb become Total energy saved Lighting Refrigeration energy saved kWh year Money saved Total energy savedUnit cost of energy 234 kWh year008 kWh 1 32 102 2 34 019 year That is the light bulb will save only 19 cents a year in energy costs and it will take 25019 132 years for it to pay for itself from the energy it saves Therefore it is not justified in this case preparation If you are a student using this Manual you are using it without permission 646 6120 A person cooks three times a week and places the food into the refrigerator before cooling it first The amount of money this person will save a year by cooling the hot foods to room temperature before refrigerating them is to be determined Assumptions 1 The heat stored in the pan itself is negligible 2 The specific heat of the food is constant Properties The specific heat of food is c 390 kJkgC given Analysis The amount of hot food refrigerated per year is mfood 5 kgpan3 pansweek52 weeksyear 780 kgyear The amount of energy removed from food as it is unnecessarily cooled to room temperature in the refrigerator is 406year Energy savedUnit cost of energy 4056 kWhyear010kWh saved Money 4056 kWhyear 3600 kJ 1kWh 15 219024 kJyear COP Energy removed saved Energy 23 C 219024 kJyear 780 kgyear390 kJkg C95 removed Energy saved food out E c T m Q Therefore cooling the food to room temperature before putting it into the refrigerator will save about four dollars a year PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 647 6121 The door of a refrigerator is opened 8 times a day and half of the cool air inside is replaced by the warmer room air The cost of the energy wasted per year as a result of opening the refrigerator door is to be determined for the cases of moist and dry air in the room Assumptions 1 The room is maintained at 20C and 95 kPa at all times 2 Air is an ideal gas with constant specific heats at room temperature 3 The moisture is condensed at an average temperature of 4C 4 Half of the air volume in the refrigerator is replaced by the warmer kitchen air each time the door is opened Properties The gas constant of air is R 0287 kPam3kgK Table A1 The specific heat of air at room temperature is cp 1005 kJkgC Table A2a The heat of vaporization of water at 4C is hfg 2492 kJkg Table A4 Analysis The volume of the refrigerated air replaced each time the refrigerator is opened is 03 m3 half of the 06 m3 air volume in the refrigerator Then the total volume of refrigerated air replaced by room air per year is 876 m year 03 m 8day365 daysyear 3 3 air replaced V The density of air at the refrigerated space conditions of 95 kPa and 4C and the mass of air replaced per year are 1 195 kgm 0 287 kPam kgK4 273 K 95 kPa 3 3 o o o RT P ρ 1047 kgyear kgm 876 m year 1195 3 3 air air ρV m The amount of moisture condensed and removed by the refrigerator is 628 kgyear 1047 kg airyear0006 kgkg air air moisture removed per kg air moisture m m The sensible latent and total heat gains of the refrigerated space become 32486 kJyear 16836 15650 6 28 kgyear2492 kJkg 15650 kJyear 16836 kJyear 4 C 1047 kgyear1005 kJkg C20 gainlatent gainsensible total gain fg moisture latent gain refrig room air sensible gain Q Q Q h m Q T T c m Q p For a COP of 14 the amount of electrical energy the refrigerator will consume to remove this heat from the refrigerated space and its cost are 048year 645 kWhyear0075kWh Energy usedUnit cost of energy of energy used total Cost 6 45 kWhyear 3600 kJ 1kWh 14 32486 kJyear COP Electrical energy used total gaintotal Q If the room air is very dry and thus latent heat gain is negligible then the amount of electrical energy the refrigerator will consume to remove the sensible heat from the refrigerated space and its cost become 025year 334 kWhyear0075kWh Energy usedUnit cost of energy of energy used sensible Cost 3 34 kWhyear 3600 kJ 1kWh 14 16836 kJyear COP Electrical energy used sensible gainsensible Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 648 Review Problems 6122 The source and sink temperatures of a heat engine are given The maximum work per unit heat input to the engine is to be determined 290 K 1280 K HE Wnet QL QH Assumptions The heat engine operates steadily Analysis The maximum work per unit of heat that the engine can remove from the source is the Carnot efficiency which is determined from 0773 1280 K 290 K 1 1 thmax net H L H T T Q W η 6123 The work output and the source and sink temperatures of a Carnot heat engine are given The heat supplied to and rejected from the heat engine are to be determined Assumptions 1 The heat engine operates steadily 2 Heat losses from the working fluid at the pipes and other components are negligible 50C 1200C HE QL QH Analysis Applying the definition of the thermal efficiency and an energy balance to the heat engine the unknown parameters are determined as follows 500 kJ 273 K 0 781 1200 273 K 50 1 1 thmax H L T T η 640 kJ 0781 kJ 500 th net η W QH 140 kJ 500 640 Wnet Q Q H L 6124E The operating conditions of a heat pump are given The minimum temperature of the source that satisfies the second law of thermodynamics is to be determined Assumptions The heat pump operates steadily TL 530 R HP 18 kW H Q L Q Analysis Applying the first law to this heat pump gives 25860 Btuh 1kW kW 341214 Btuh 81 32000 Btuh netin W Q Q H L In the reversible case we have H L H L Q Q T T Then the minimum temperature may be determined to be 428 R 32000 Btuh 530 R 25860 Btuh H L H L Q Q T T PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 649 6125 A heat pump with a specified COP is to heat a house The rate of heat loss of the house and the power consumption of the heat pump are given The time it will take for the interior temperature to rise from 3C to 22C is to be determined Assumptions 1 Air is an ideal gas with constant specific heats at room temperature 2 The house is wellsealed so that no air leaks in or out 3 The COP of the heat pump remains constant during operation Properties The constant volume specific heat of air at room temperature is cv 0718 kJkgC Table A2 Analysis The house is losing heat at a rate of 22C 3C QH 40000 kJh kJs 1111 40000 kJh Loss Q The rate at which this heat pump supplies heat is 192 kW 8 kW 24 COP HP netin W QH That is this heat pump can supply heat at a rate of 192 kJs Taking the house as the system a closed system the energy balance can be written as Win 1 2 out in 1 2 out in 1 2 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T t Q Q T mc T Q Q u m u U Q Q E E E v v 43 42 1 4243 1 Substituting 3 C 2000kg 0718kJkg C 22 1111kJs 192 o o t Solving for t it will take t 3373 s 0937 h for the temperature in the house to rise to 22C PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 650 6126E A refrigerator with a watercooled condenser is considered The cooling load and the COP of a refrigerator are given The power input the exit temperature of water and the maximum possible COP of the refrigerator are to be determined Assumptions The refrigerator operates steadily 25F Water 58F R H Q L Q in W Condenser 24000 Btuh Analysis a The power input is 3974 kW 3600 s h 1 1Btu 1055 kJ 1 77 24000 Btuh COP in QL W b The rate of heat rejected in the condenser is 37560 Btuh h 1 3600 s 1055 kJ 1Btu 3 974 kW 000 Btuh 24 in W Q Q L H The exit temperature of the water is 652F F Btulbm 01 h 1 45 lbms 3600s 1 37560 Btuh F 58 1 2 1 2 p H p H mc Q T T T T mc Q c Taking the temperature of hightemperature medium to be the average temperature of water in the condenser 133 25 65 2 50 58 460 25 COPrev L H L T T T 6127 A Carnot heat engine cycle is executed in a closed system with a fixed mass of R134a The thermal efficiency of the cycle is given The net work output of the engine is to be determined Assumptions All components operate steadily Properties The enthalpy of vaporization of R134a at 50C is hfg 15179 kJkg Table A11 Analysis The enthalpy of vaporization hfg at a given T or P represents the amount of heat transfer as 1 kg of a substance is converted from saturated liquid to saturated vapor at that T or P Therefore the amount of heat transfer to R134a during the heat addition process of the cycle is R134a 1518 kJ 001 kg 15179 kJkg 50 C o fg H mh Q Carnot HE Then the work output of this heat engine becomes 015 1518 kJ th netout 0228 kJ QH W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 651 6128 A heat pump with a specified COP and power consumption is used to heat a house The time it takes for this heat pump to raise the temperature of a cold house to the desired level is to be determined Assumptions 1 Air is an ideal gas with constant specific heats at room temperature 2 The heat loss of the house during the warpup period is negligible 3 The house is wellsealed so that no air leaks in or out Properties The constant volume specific heat of air at room temperature is cv 0718 kJkgC Analysis Since the house is wellsealed constant volume the total amount of heat that needs to be supplied to the house is 16155 kJ 7 C C 22 1500 kg 0718 kJkg house T mc QH v 5 kW House HP The rate at which this heat pump supplies heat is 14 kW 82 5 kW COP HP netin W QH That is this heat pump can supply 14 kJ of heat per second Thus the time required to supply 16155 kJ of heat is 192 min 1154 s 14 kJs kJ 16155 H H Q Q t 6129 A solar pond power plant operates by absorbing heat from the hot region near the bottom and rejecting waste heat to the cold region near the top The maximum thermal efficiency that the power plant can have is to be determined Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency which is determined from 35C 80C HE W 0127 or 127 353 K 308 K 1 1 thC thmax H L T T η η In reality the temperature of the working fluid must be above 35C in the condenser and below 80C in the boiler to allow for any effective heat transfer Therefore the maximum efficiency of the actual heat engine will be lower than the value calculated above PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 652 6130 A Carnot heat engine cycle is executed in a closed system with a fixed mass of steam The net work output of the cycle and the ratio of sink and source temperatures are given The low temperature in the cycle is to be determined Assumptions The engine is said to operate on the Carnot cycle which is totally reversible Analysis The thermal efficiency of the cycle is Carnot HE 0025 kg H2O Also 120 kJ 05 kJ 60 50 2 1 1 1 th th th η η η W Q Q W T T H H H L and 2400 kJkg 0025 kg kJ 60 60 kJ 60 120 L T fg L L H L h m Q q W Q Q since the enthalpy of vaporization hfg at a given T or P represents the amount of heat transfer as 1 kg of a substance is converted from saturated liquid to saturated vapor at that T or P Therefore TL is the temperature that corresponds to the hfg value of 2400 kJkg and is determined from the steam tables Table A4 to be TL 425C PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 653 6131 Problem 6130 is reconsidered The effect of the net work output on the required temperature of the steam during the heat rejection process as the work output varies from 40 kJ to 60 kJ is to be investigated Analysis The problem is solved using EES and the results are tabulated and plotted below Given m0025 kg RatioT05 RatioTTLTH Wnetout60 kJ Properties Fluid steamiapws hfenthalpyFluid TTL x0 hgenthalpyFluid TTL x1 hfghghf Analysis etath1RatioT etathWnetoutQH QLQHWnetout QLmhfg Wout kJ TLC C 40 425 45 475 50 525 55 575 60 2708 2529 2328 2099 184 1544 1208 8317 425 40 44 48 52 56 60 0 50 100 150 200 250 300 Wnetout kJ TL C PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 654 6132 A Carnot refrigeration cycle is executed in a closed system with a fixed mass of R134a The net work input and the ratio of maximumtominimum temperatures are given The minimum pressure in the cycle is to be determined Assumptions The refrigerator is said to operate on the reversed Carnot cycle which is totally reversible Analysis The coefficient of performance of the cycle is T TH 12TL TL 2 1 4 TH 3 5 1 21 1 1 1 COPR L H T T Also 110 kJ 5 22 kJ COP COP in R in R W Q W Q L L and TH fg H H L H h m Q q W Q Q 1375kJkg 096kg kJ 132 132kJ 22 110 v since the enthalpy of vaporization hfg at a given T or P represents the amount of heat transfer per unit mass as a substance is converted from saturated liquid to saturated vapor at that T or P Therefore TH is the temperature that corresponds to the hfg value of 1375 kJkg and is determined from the R134a tables to be Then C 65 2786 K 12 334 3 K 21 334 3 K C 3 61 H L H T T T Therefore 355 kPa sat56C min P P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 655 6133 Problem 6132 is reconsidered The effect of the net work input on the minimum pressure as the work input varies from 10 kJ to 30 kJ is to be investigated The minimum pressure in the refrigeration cycle is to be plotted as a function of net work input Analysis The problem is solved using EES and the results are tabulated and plotted below Analysis The coefficient of performance of the cycle is given by mR134a 096 kg THtoTLRatio 12 TH 12TL Win 22 kJ Depending on the value of Win adjust the guess value of TH COPR 1 THtoTLRatio 1 QL WinCOPR First law applied to the refrigeration cycle yields QL Win QH Steadyflow analysis of the condenser yields mR134ah3 mR134ah4 QH QH mR134ah3h4 and hfg h3 h4 also THT3T4 QHmR134ahfg hfgenthalpyR134aTTHx1 enthalpyR134aTTHx0 THTHtoTLRatioTL The minimum pressure is the saturation pressure corresponding to TL Pmin pressureR134aTTLx0convertkPaMPa TLC TL 273 Win kJ Pmin MPa TH K TL K TLC C 10 15 20 25 30 08673 06837 045 02251 006978 3688 3589 3427 3193 2871 3073 299 2856 2661 2392 3432 2605 1261 6907 3378 10 14 18 22 26 30 0 01 02 03 04 05 06 07 08 09 Win kJ Pmin MPa 10 14 18 22 26 30 40 30 20 10 0 10 20 30 40 Win kJ TLC C PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 656 6134 Two Carnot heat engines operate in series between specified temperature limits If the thermal efficiencies of both engines are the same the temperature of the intermediate medium between the two engines is to be determined Assumptions The engines are said to operate on the Carnot cycle which is totally reversible T TH HE 1 HE 2 TL Analysis The thermal efficiency of the two Carnot heat engines can be expressed as T T T T L H 1 and 1 thII thI η η Equating 1 1 T T T T H L Solving for T 735 K 1800 K300 K THTL T 6135E The thermal efficiency of a completely reversible heat engine as a function of the source temperature is to be calculated and plotted Assumptions The heat engine operates steadily Analysis With the specified sink temperature the thermal efficiency of this completely reversible heat engine is H H L T T T 500 R 1 1 threv η Using EES we tabulate and plot the variation of thermal efficiency with the source temperature TH R ηthrev 500 0 650 02308 800 0375 950 04737 1100 05455 1250 06 1400 06429 1550 06774 1700 07059 1850 07297 2000 075 500 800 1100 1400 1700 2000 0 01 02 03 04 05 06 07 08 TH R ηthrev PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 657 6136 A Carnot heat engine drives a Carnot refrigerator that removes heat from a cold medium at a specified rate The rate of heat supply to the heat engine and the total rate of heat rejection to the environment are to be determined Analysis a The coefficient of performance of the Carnot refrigerator is 6 143 1 300 K 258 K 1 1 1 COP RC L H T T 300 K 900 K HE QL HE QH HE QH R 250 kJmin R 15C Then power input to the refrigerator becomes 407 kJmin 6143 250 kJmin COP RC netin QL W which is equal to the power output of the heat engine netout W The thermal efficiency of the Carnot heat engine is determined from 0 6667 900 K 300 K 1 1 thC H L T T η Then the rate of heat input to this heat engine is determined from the definition of thermal efficiency to be 611 kJmin 06667 kJmin 407 HE th netout HE η W QH b The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine and the heat discarded by the refrigerator LHE Q H R Q 290 7 kJmin 40 7 250 204 kJmin 40 7 1 61 netin R R netout HE HE W Q Q W Q Q L H H L and 311 kJmin 290 7 204 R HE Ambient H L Q Q Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 658 6137 Problem 6136 is reconsidered The effects of the heat engine source temperature the environment temperature and the cooled space temperature on the required heat supply to the heat engine and the total rate of heat rejection to the environment as the source temperature varies from 500 K to 1000 K the environment temperature varies from 275 K to 325 K and the cooled space temperature varies from 20C to 0C are to be investigated The required heat supply is to be plotted against the source temperature for the cooled space temperature of 15C and environment temperatures of 275 300 and 325 K Analysis The problem is solved using EES and the results are tabulated and plotted below QdotLR 250 kJmin Tsurr 300 K TH 900 K TLC 15 C TL TLC 273 Coefficient of performance of the Carnot refrigerator THR Tsurr COPR 1THRTL1 Power input to the refrigerator WdotinR QdotLRCOPR Power output from heat engine must be WdotoutHE WdotinR The efficiency of the heat engine is TLHE Tsurr etaHE 1 TLHETH The rate of heat input to the heat engine is QdotHHE WdotoutHEetaHE First law applied to the heat engine and refrigerator QdotLHE QdotHHE WdotoutHE QdotHR QdotLR WdotinR TH K QHHE kJmin Qsurr kJmin 500 600 700 800 900 1000 3661 3041 2713 251 2372 2272 2866 2804 2771 2751 2737 2727 500 600 700 800 900 1000 0 20 40 60 80 100 120 140 160 180 200 TH K QHHE kJmin Tsurr 325 K Tsurr 300 K Tsurr 275 K TLC C QHHE kJmin Qsurr kJmin 20 18 16 14 12 10 8 6 4 2 0 313 2824 2521 2224 1931 1643 1358 1079 803 5314 2637 2813 2782 2752 2722 2693 2664 2636 2608 258 2553 2526 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 659 500 600 700 800 900 1000 260 280 300 320 340 360 380 400 420 440 TH K Qsurr kJmin Tsurr 325 K Tsurr 300 K Tsurr 275 K 20 16 12 8 4 0 0 5 10 15 20 25 30 35 TLC C QHHE kJmin 20 16 12 8 4 0 250 255 260 265 270 275 280 285 TLC C Qsurr kJmin PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 660 6138 Half of the work output of a Carnot heat engine is used to drive a Carnot heat pump that is heating a house The minimum rate of heat supply to the heat engine is to be determined Assumptions Steady operating conditions exist Analysis The coefficient of performance of the Carnot heat pump is 20C 800C HE HP House 22C 2C 62000 kJh 1475 273 K 273 K 22 2 1 1 1 1 COP HPC TL TH Then power input to the heat pump which is supplying heat to the house at the same rate as the rate of heat loss becomes 4203 kJh 1475 62000 kJh COP HPC netin QH W which is half the power produced by the heat engine Thus the power output of the heat engine is 8406 kJh 24203 kJh 2 netin netout W W To minimize the rate of heat supply we must use a Carnot heat engine whose thermal efficiency is determined from 0 727 1073 K 293 K 1 1 thC H L T T η Then the rate of heat supply to this heat engine is determined from the definition of thermal efficiency to be 11560 kJh 0727 kJh 8406 HE th netout HE η W QH 6139E An extraordinary claim made for the performance of a refrigerator is to be evaluated Assumptions Steady operating conditions exist 35F 75F R H Q QL COP135 Analysis The performance of this refrigerator can be evaluated by comparing it with a reversible refrigerator operating between the same temperature limits 12 4 1 460 460 35 75 1 1 1 COP COP Rrev Rmax L H T T Discussion This is the highest COP a refrigerator can have when absorbing heat from a cool medium at 35F and rejecting it to a warmer medium at 75F Since the COP claimed by the inventor is above this maximum value the claim is false PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 661 6140 A Carnot heat pump cycle is executed in a steadyflow system with R134a flowing at a specified rate The net power input and the ratio of the maximumtominimum temperatures are given The ratio of the maximum to minimum pressures is to be determined Analysis The coefficient of performance of the cycle is TH 12TL TH TL QH T 06 21 1 1 1 1 1 COPHP TL TH and TH fg H H H h m Q q W Q in HP 13636 kJkg 022 kgs kJs 300 30 0 kJs 60 5 kW COP since the enthalpy of vaporization hfg at a given T or P represents the amount of heat transfer per unit mass as a substance is converted from saturated liquid to saturated vapor at that T or P Therefore TH is the temperature that corresponds to the hfg value of 13636 kJkg and is determined from the R134a tables to be v and kPa 1763 3351 K C 0 62 sat620 C max P P TH Also kPa 542 C 18 3 2914 K 12 3351 K 25 1 sat183 C min P P T T H L Then the ratio of the maximum to minimum pressures in the cycle is 325 542 kPa kPa 1763 min max P P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 662 6141 Switching to energy efficient lighting reduces the electricity consumed for lighting as well as the cooling load in summer but increases the heating load in winter It is to be determined if switching to efficient lighting will increase or decrease the total energy cost of a building Assumptions The light escaping through the windows is negligible so that the entire lighting energy becomes part of the internal heat generation Analysis a Efficient lighting reduces the amount of electrical energy used for lighting yeararound as well as the amount of heat generation in the house since light is eventually converted to heat As a result the electrical energy needed to air condition the house is also reduced Therefore in summer the total cost of energy use of the household definitely decreases b In winter the heating system must make up for the reduction in the heat generation due to reduced energy used for lighting The total cost of energy used in this case will still decrease if the cost of unit heat energy supplied by the heating system is less than the cost of unit energy provided by lighting The cost of 1 kWh heat supplied from lighting is 008 since all the energy consumed by lamps is eventually converted to thermal energy Noting that 1 therm 105500 kJ 293 kWh and the furnace is 80 efficient the cost of 1 kWh heat supplied by the heater is 0 060 per kWh heat 293 kWh 1 therm 1kWh080140therm Price Amount of useful energy Cost of 1 kWh heat supplied by furnace furnace η which is less than 008 Thus we conclude that switching to energy efficient lighting will reduce the total energy cost of this building both in summer and in winter Discussion To determine the amount of cost savings due to switching to energy efficient lighting consider 10 h of operation of lighting in summer and in winter for 1 kW rated power for lighting Current lighting Lighting cost Energy usedUnit cost 1 kW10 h008kWh 080 Increase in air conditioning cost Heat from lightingCOPunit cost 10 kWh35008kWh 023 Decrease in the heating cost Heat from lightingEffunit cost1008 kWh140293kWh 060 Total cost in summer 080023 103 Total cost in winter 080060 020 Energy efficient lighting Lighting cost Energy usedUnit cost 025 kW10 h008kWh 020 Increase in air conditioning cost Heat from lightingCOPunit cost 25 kWh35008kWh 006 Decrease in the heating cost Heat from lightingEffunit cost2508 kWh140293kWh 015 Total cost in summer 020006 026 Total cost in winter 020015 005 Note that during a day with 10 h of operation the total energy cost decreases from 103 to 026 in summer and from 020 to 005 in winter when efficient lighting is used PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 663 6142 A heat pump is used to heat a house The maximum money saved by using the lake water instead of outside air as the heat source is to be determined Assumptions 1 Steady operating conditions exist 2 The kinetic and potential energy changes are zero Analysis When outside air is used as the heat source the cost of energy is calculated considering a reversible heat pump as follows 1192 273 273 25 0 1 1 1 1 COPmax TL TH 3 262 kW 1192 140000 3600 kW COPmax inmin QH W 2773 3 262 kW100 h0085kWh Cost air Repeating calculations for lake water 1987 273 273 25 10 1 1 1 1 COPmax TL TH 1 957 kW 1987 140000 3600 kW COPmax inmin QH W 1663 1 957 kW100 h0085kWh Cost lake Then the money saved becomes 1110 1663 2773 Cost Cost Money Saved lake air PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 664 6143 The cargo space of a refrigerated truck is to be cooled from 25C to an average temperature of 5C The time it will take for an 8kW refrigeration system to precool the truck is to be determined Assumptions 1 The ambient conditions remain constant during precooling 2 The doors of the truck are tightly closed so that the infiltration heat gain is negligible 3 The air inside is sufficiently dry so that the latent heat load on the refrigeration system is negligible 4 Air is an ideal gas with constant specific heats Properties The density of air is taken 12 kgm3 and its specific heat at the average temperature of 15C is cp 10 kJkgC Table A2 Analysis The mass of air in the truck is Truck T1 25C T2 5C 116 kg 35 m 23 m 12 kgm 12 m 3 truck air air ρ V m The amount of heat removed as the air is cooled from 25 to 5ºC kJ 2320 5 C 116 kg10 kJkg C25 air coolingair T mc Q p Q Noting that UA is given to be 80 WºC and the average air temperature in the truck during precooling is 2552 15ºC the average rate of heat gain by transmission is determined to be Q UA T transmissionave 80 Wº C25 C 800 W 080kJ s 15º Therefore the time required to cool the truck from 25 to 5ºC is determined to be 54min 322s kJs 80 8 kJ 2320 transmission refrig air cooling transmission coolingair refrig Q Q Q t t Q Q t Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 665 6144 A refrigeration system is to cool bread loaves at a rate of 1200 per hour by refrigerated air at 30C The rate of heat removal from the breads the required volume flow rate of air and the size of the compressor of the refrigeration system are to be determined Assumptions 1 Steady operating conditions exist 2 The thermal properties of the bread loaves are constant 3 The cooling section is wellinsulated so that heat gain through its walls is negligible Properties The average specific and latent heats of bread are given to be 293 kJkgC and 1093 kJkg respectively The gas constant of air is 0287 kPam3kgK Table A1 and the specific heat of air at the average temperature of 30 222 26C 250 K is cp 10 kJkgC Table A2 Air 30C Bread Analysis a Noting that the breads are cooled at a rate of 500 loaves per hour breads can be considered to flow steadily through the cooling section at a mass flow rate of 420 kgh 01167 kgs 1200 breadsh0350 kgbread bread m Then the rate of heat removal from the breads as they are cooled from 30C to 10ºC and frozen becomes 49224 kJh 10 C 420 kgh293 kJkg C30 bread bread T mc Q p 45906 kJh 420 kgh 1093kJkg bread latent freezing mh Q and 95130 kJh 45906 49224 freezing bread total Q Q Q b All the heat released by the breads is absorbed by the refrigerated air and the temperature rise of air is not to exceed 8C The minimum mass flow and volume flow rates of air are determined to be 11891 kgh 10 kJkg C8 C 95130 kJh air air air T c Q m p 3 3 1 453 kgm 0 287 kPam kgK30 273 K 101 3 kPa RT P ρ 8185 m h 3 3 air air air kgm 1453 11891 kgh ρ m V c For a COP of 12 the size of the compressor of the refrigeration system must be 2202 kW 79275 kJh 12 95130 kJh COP refrig refrig Q W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 666 6145 The drinking water needs of a production facility with 20 employees is to be met by a bobbler type water fountain The size of compressor of the refrigeration system of this water cooler is to be determined Assumptions 1 Steady operating conditions exist 2 Water is an incompressible substance with constant properties at room temperature 3 The cold water requirement is 04 Lh per person Properties The density and specific heat of water at room temperature are ρ 10 kgL and c 418 kJkgCC Table A 3 Analysis The refrigeration load in this case consists of the heat gain of the reservoir and the cooling of the incoming water The water fountain must be able to provide water at a rate of 1 kgL04 Lh person20 persons 80 kgh water water V ρ m To cool this water from 22C to 8C heat must removed from the water at a rate of Refrig Water in 22C Water out 8C kJh 130 W since1 W 36 kJh 468 kgh418 kJkg C22 8 C 08 out in p cooling T mc T Q Then total refrigeration load becomes 175 W 45 130 transfer cooling refrig total Q Q Q Noting that the coefficient of performance of the refrigeration system is 29 the required power input is 603 W 29 175 W COP refrig refrig Q W Therefore the power rating of the compressor of this refrigeration system must be at least 603 W to meet the cold water requirements of this office PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 667 6146E A washing machine uses 33year worth of hot water heated by a gas water heater The amount of hot water an average family uses per week is to be determined Assumptions 1 The electricity consumed by the motor of the washer is negligible 2 Water is an incompressible substance with constant properties at room temperature Properties The density and specific heat of water at room temperature are ρ 621 lbmft3 and c 100 BtulbmF Table A3E Analysis The amount of electricity used to heat the water and the net amount transferred to water are 30420 Btuweek 52 weeks 1year 1 therm 1582 thermsyear 1582 thermsyear 100000 Btu 2727 thermsyear EfficiencyTotal energy used 058 energy transfer to water Total 2727 thermsyear 121therm 33year Unit cost of energy Total cost of energy energy used gas Total in E Then the mass and the volume of hot water used per week become 434 6 lbmweek 10 Btulbm F13060 F 30420 Btuweek in out in in out in T c T E m T mc T E and 524 galweek 3 3 3 water ft 1 ft week 74804 gal 07 1 lbmft 62 4346 lbmweek ρ m V Therefore an average family uses about 52 gallons of hot water per week for washing clothes PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 668 6147 A typical heat pump powered water heater costs about 800 more to install than a typical electric water heater The number of years it will take for the heat pump water heater to pay for its cost differential from the energy it saves is to be determined PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Assumptions 1 The price of electricity remains constant 2 Water is an incompressible substance with constant properties at room temperature 3 Time value of money interest inflation is not considered Analysis The amount of electricity used to heat the water and the net amount transferred to water are 2969 kWhyear 3125 kWhyear EfficiencyTotal energy used 095 energy transfer to water Total 3125 kWhyear 0080kWh 250year Unit cost of energy Total cost of energy energy used electrical Total in E Water Heater Cold water Hot water The amount of electricity consumed by the heat pump and its cost are 7197year cost of heat pump Energy usageUnit cost of energy 8996 kWhyear008kWh Energy 899 6 kWhyear 33 2969 kWhyear COP Energy usage of heat pump Energy transfer to water HP Then the money saved per year by the heat pump and the simple payback period become 1780year 449 years 800 Money saved payback period Additional installation cost Simple 7197 1780 250 saved Energy cost of electric heater Energy cost of heat pump Money Discussion The economics of heat pump water heater will be even better if the air in the house is used as the heat source for the heat pump in summer and thus also serving as an airconditioner preparation If you are a student using this Manual you are using it without permission 669 6148 Problem 6147 is reconsidered The effect of the heat pump COP on the yearly operation costs and the number of years required to break even are to be considered Analysis The problem is solved using EES and the results are tabulated and plotted below Energy supplied by the water heater to the water per year is EElecHeater Cost per year to operate electric water heater for one year is CostElectHeater 250 year Energy supplied to the water by electric heater is 90 of energy purchased eta095 EElectHeater etaCostElectHeater UnitCost kWhyear UnitCost008 kWh For the same amont of heated water and assuming that all the heat energy leaving the heat pump goes into the water then Energy supplied by heat pump heater Energy supplied by electric heater EHeatPump EElectHeater kWhyear Electrical Work enegy supplied to heat pump Heat added to waterCOP COP33 WHeatPump EHeatPumpCOP kWhyear Cost per year to operate the heat pump is CostHeatPumpWHeatPumpUnitCost Let NBrkEven be the number of years to break even At the break even point the total cost difference between the two water heaters is zero Years to break even neglecting the cost to borrow the extra 800 to install heat pump CostDifftotal 0 CostDifftotalAddCostNBrkEvenCostHeatPumpCostElectHeater AddCost800 COP BBrkEven years CostHeatPump year CostElektHeater year 2 23 26 29 32 35 38 41 44 47 5 6095 5452 5042 4759 4551 4392 4267 4165 4081 4011 3951 1188 1033 9135 819 7422 6786 625 5793 5398 5053 475 250 250 250 250 250 250 250 250 250 250 250 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 670 2 25 3 35 4 45 5 35 4 45 5 55 6 65 COP NBrkEven years 2 25 3 35 4 45 5 40 80 120 160 200 240 COP Cost year Electric Heat pump PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 671 6149 A home owner is to choose between a highefficiency natural gas furnace and a groundsource heat pump The system with the lower energy cost is to be determined Assumptions The two heater are comparable in all aspects other than the cost of energy Analysis The unit cost of each kJ of useful energy supplied to the house by each system is Natural gas furnace kJ 10 13 8 105500 kJ 1 therm 097 142therm Unit cost of useful energy 6 Heat Pump System kJ 10 37 3600 kJ 1kWh 35 0092kWh Unit cost of useful energy 6 The energy cost of groundsource heat pump system will be lower 6150 The ventilating fans of a house discharge a houseful of warmed air in one hour ACH 1 For an average outdoor temperature of 5C during the heating season the cost of energy vented out by the fans in 1 h is to be determined Assumptions 1 Steady operating conditions exist 2 The house is maintained at 22C and 92 kPa at all times 3 The infiltrating air is heated to 22C before it is vented out 4 Air is an ideal gas with constant specific heats at room temperature 5 The volume occupied by the people furniture etc is negligible Properties The gas constant of air is R 0287 kPam3kgK Table A1 The specific heat of air at room temperature is cp 10 kJkgC Table A2a Analysis The density of air at the indoor conditions of 92 kPa and 22C is 22C 5C 92 kPa Bathroom fan 3 3 1 087 kgm 0 287 kPam kgK22 273 K 92 kPa o o o RT P ρ Noting that the interior volume of the house is 200 28 560 m3 the mass flow rate of air vented out becomes 0169 kgs 608 7 kgh 1087 kgm 560 m h 3 3 air air V ρ m Noting that the indoor air vented out at 22C is replaced by infiltrating outdoor air at 5C this corresponds to energy loss at a rate of 2874 kW 2 874 kJs 5 C 0 169 kgs10 kJkg C22 outdoors indoors air lossfan T T c m Q p Then the amount and cost of the heat vented out per hour becomes 0123 293 kWh 1 therm 2 994 kWh 1 20therm Fuel energy lossUnit cost of energy loss Money 2994 kWh 2 874 kW1 h096 Fuel energy loss furnace lossfan t η Q Discussion Note that the energy and money loss associated with ventilating fans can be very significant Therefore ventilating fans should be used sparingly PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 672 6151 The ventilating fans of a house discharge a houseful of airconditioned air in one hour ACH 1 For an average outdoor temperature of 28C during the cooling season the cost of energy vented out by the fans in 1 h is to be determined Assumptions 1 Steady operating conditions exist 2 The house is maintained at 22C and 92 kPa at all times 3 The infiltrating air is cooled to 22C before it is vented out 4 Air is an ideal gas with constant specific heats at room temperature 5 The volume occupied by the people furniture etc is negligible 6 Latent heat load is negligible Properties The gas constant of air is R 0287 kPam3kgK Table A1 The specific heat of air at room temperature is cp 10 kJkgC Table A2a Analysis The density of air at the indoor conditions of 92 kPa and 22C is 22C 28C 92 kPa Bathroom fan 3 3 1 087 kgm 0 287 kPam kgK22 273 K 92 kPa o o o RT P ρ Noting that the interior volume of the house is 200 28 560 m3 the mass flow rate of air vented out becomes 0169 kgs 608 7 kgh 1087 kgm 560 m h 3 3 air air V ρ m Noting that the indoor air vented out at 22C is replaced by infiltrating outdoor air at 28C this corresponds to energy loss at a rate of 1 014 kJs 1014 kW 22 C 0 169 kgs10 kJkg C28 indoors outdoors air lossfan T T c m Q p Then the amount and cost of the electric energy vented out per hour becomes 0044 0 441 kWh 0 10 kWh Fuel energy lossUnit cost of energy loss Money 0441 kWh 1 014 kW1 h23 Electric energy loss lossfan t COP Q Discussion Note that the energy and money loss associated with ventilating fans can be very significant Therefore ventilating fans should be used sparingly PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 673 6152 A geothermal heat pump with R134a as the working fluid is considered The evaporator inlet and exit states are specified The mass flow rate of the refrigerant the heating load the COP and the minimum power input to the compressor are to be determined Assumptions 1 The heat pump operates steadily 2 The kinetic and potential energy changes are zero 3 Steam properties are used for geothermal water QH 40C Win Condenser Evaporator Compressor Expansion valve 12C x015 QL Sat vap Geo water 60C Properties The properties of R134a and water are Steam and R134a tables 25727 kJkg 1 3 kPa 443 3 kPa 443 55 kJkg 96 15 0 C 12 2 2 1 2 1 1 1 1 h x P P P h x T 16753 kJkg 0 C 40 25118 kJkg 0 C 60 2 2 2 1 1 1 w w w w w w h x T h x T Analysis a The rate of heat transferred from the water is the energy change of the water from inlet to exit 5 437 kW 16753 kJkg 0 065 kgs25118 2 1 w w w L h h m Q The energy increase of the refrigerant is equal to the energy decrease of the water in the evaporator That is 00338 kgs 9655 kJkg 25727 5 437 kW 1 2 1 2 h h Q m h h m Q L R R L b The heating load is 704 kW 61 5 437 Win Q Q L H c The COP of the heat pump is determined from its definition 440 kW 61 7 04 kW COP in W QH d The COP of a reversible heat pump operating between the same temperature limits is 9 51 273 273 60 25 1 1 1 1 COPmax TL TH Then the minimum power input to the compressor for the same refrigeration load would be 0740 kW 9 51 7 04 kW COPmax inmin QH W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 674 6153 A heat pump is used as the heat source for a water heater The rate of heat supplied to the water and the minimum power supplied to the heat pump are to be determined Assumptions 1 Steady operating conditions exist 2 The kinetic and potential energy changes are zero Properties The specific heat and specific volume of water at room temperature are cp 418 kJkgK and v0001 m3kg Table A3 Analysis a An energy balance on the water heater gives the rate of heat supplied to the water 5573 kW 4 18 kJkg C50 10 C m kg 0001 0 02 60 m s 3 3 1 2 1 2 T T c T T mc Q p p H v V b The COP of a reversible heat pump operating between the specified temperature limits is 10 1 273 273 30 1 0 1 1 1 COPmax TL TH Then the minimum power input would be 552 kW 10 1 5573 kW COPmax inmin QH W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 675 6154 A heat pump receiving heat from a lake is used to heat a house The minimum power supplied to the heat pump and the mass flow rate of lake water are to be determined Assumptions 1 Steady operating conditions exist 2 The kinetic and potential energy changes are zero Properties The specific heat of water at room temperature is cp 418 kJkgK Table A3 Analysis a The COP of a reversible heat pump operating between the specified temperature limits is 1429 273 273 27 1 6 1 1 1 COPmax TL TH Then the minimum power input would be 1244 kW 1429 64000 3600 kW COPmax inmin QH W b The rate of heat absorbed from the lake is 1653 kW 1 244 1778 inmin W Q Q H L An energy balance on the heat exchanger gives the mass flow rate of lake water 0791 kgs 4 18 kJkg C 5 C 1653 kJs water T c Q m p L PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 676 6155 It is to be proven that a refrigerators COP cannot exceed that of a completely reversible refrigerator that shares the same thermalenergy reservoirs Assumptions The refrigerator operates steadily Analysis We begin by assuming that the COP of the general refrigerator B is greater than that of the completely reversible refrigerator A COPB COPA When this is the case a rearrangement of the coefficient of performance expression yields A A L B L B W Q Q W COP COP B WB QH B QL TL TH QL WA A QH A That is the magnitude of the work required to drive refrigerator B is less than that needed to drive completely reversible refrigerator A Applying the first law to both refrigerators yields H A H B Q Q since the work supplied to refrigerator B is less than that supplied to refrigerator A and both have the same cooling effect QL Since A is a completely reversible refrigerator we can reverse it without changing the magnitude of the heat and work transfers This is illustrated in the figure below The heat QL which is rejected by the reversed refrigerator A can now be routed directly to refrigerator B The net effect when this is done is that no heat is exchanged with the TL reservoir The magnitude of the heat supplied to the reversed refrigerator A QHA has been shown to be larger than that rejected by refrigerator B There is then a net heat transfer from the TH reservoir to the combined device in the dashed lines of the figure whose magnitude is given by QHA QHB Similarly there is a net work production by the combined device whose magnitude is given by WA WB B QL TL TH A WB QL QHA QHB WA WB The combined cyclic device then exchanges heat with a reservoir at a single temperature and produces work which is clearly a violation of the KelvinPlanck statement of the second law Our assumption the COPB COPA must then be wrong PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 677 6156 It is to be proven that the COP of all completely reversible refrigerators must be the same when the reservoir temperatures are the same Assumptions The refrigerators operate steadily Analysis We begin by assuming that COPA COPB When this is the case a rearrangement of the coefficient of performance expression yields B B L A L A W Q Q W COP COP B WB QH B QL TL TH QL WA A QH A That is the magnitude of the work required to drive refrigerator A is greater than that needed to drive refrigerator B Applying the first law to both refrigerators yields H B H A Q Q since the work supplied to refrigerator A is greater than that supplied to refrigerator B and both have the same cooling effect QL Since A is a completely reversible refrigerator we can reverse it without changing the magnitude of the heat and work transfers This is illustrated in the figure below The heat QL which is rejected by the reversed refrigerator A can now be routed directly to refrigerator B The net effect when this is done is that no heat is exchanged with the TL reservoir The magnitude of the heat supplied to the reversed refrigerator A QHA has been shown to be larger than that rejected by refrigerator B There is then a net heat transfer from the TH reservoir to the combined device in the dashed lines of the figure whose magnitude is given by QHA QHB Similarly there is a net work production by the combined device whose magnitude is given by WA WB B QL TL TH A WB QL QHA QHB WA WB The combined cyclic device then exchanges heat with a reservoir at a single temperature and produces work which is clearly a violation of the KelvinPlanck statement of the second law Our assumption the COPA COPB must then be wrong If we interchange A and B in the previous argument we would conclude that the COPB cannot be less than COPA The only alternative left is that COPA COPB PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 678 6157 An expression for the COP of a completely reversible heat pump in terms of the thermalenergy reservoir temperatures TL and TH is to be derived Assumptions The heat pump operates steadily Analysis Application of the first law to the completely reversible heat pump yields L H Q Q W netin HP TH TL QL QH Wnetin This result may be used to reduce the coefficient of performance H L L H H H Q Q Q Q Q W Q 1 1 COP netin HPrev Since this heat pump is completely reversible the thermodynamic definition of temperature tells us that H L H L T T Q Q When this is substituted into the COP expression the result is L H H H L T T T T T 1 1 COP HPrev 6158 A Carnot heat engine is operating between specified temperature limits The source temperature that will double the efficiency is to be determined Analysis Denoting the new source temperature by TH the thermal efficiency of the Carnot heat engine for both cases can be expressed as thC thC thC 2 1 and 1 η η η H L H L T T T T TL TH HE ηth 2ηth HE TH Substituting H L H L T T T T 2 1 1 Solving for TH T T T T T H H L H L 2 which is the desired relation PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 679 6159 A Carnot cycle is analyzed for the case of temperature differences in the boiler and condenser The ratio of overall temperatures for which the power output will be maximum and an expression for the maximum net power output are to be determined Analysis It is given that PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course H H H H T T hA Q Therefore or 1 1 1 1 1 1 1 th r x T T T T T hA W T T T hA T T T T hA T T Q W H H H L H H H H H H H L H H H H L H η where we defined r and x as r TL TH and x 1 TH TH For a reversible cycle we also have H L H L H L H H H H L L L H H H L H L H T T T T T hA T T T hA T T hA T T hA r Q Q T T 1 1 TL TH HE W TH TL but x r T T T T T T H H H L H L 1 Substituting into above relation yields H L L H T T x r hA x hA r 1 1 Solving for x 2 1 L H H L hA r hA T T r x Substitute 2 into 1 3 1 1 L H H L H H hA r hA T T r r T hA W Taking the partial derivative r W holding everything else constant and setting it equal to zero gives 4 2 1 H L H L T T T T r which is the desired relation The maximum net power output in this case is determined by substituting 4 into 3 It simplifies to 2 max 2 1 1 1 H L L H H H T T hA hA T hA W preparation If you are a student using this Manual you are using it without permission 680 Fundamentals of Engineering FE Exam Problems 6160 The label on a washing machine indicates that the washer will use 85 worth of hot water if the water is heated by a 90 efficiency electric heater at an electricity rate of 009kWh If the water is heated from 18C to 45C the amount of hot water an average family uses per year in metric tons is a 116 tons b 158 tons c 271 tons d 301 tons e 335 tons Answer b 271 tons Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Eff090 C418 kJkgC T118 C T245 C Cost85 Price009 kWh EinCostPrice3600 kJ EinmCT2T1Eff kJ Some Wrong Solutions with Common Mistakes EinW1mCT2T1Eff Multiplying by Eff instead of dividing EinW2mCT2T1 Ignoring efficiency EinW3mT2T1Eff Not using specific heat EinW4mCT2T1Eff Adding temperatures PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 681 6161 A 24m high 200m2 house is maintained at 22C by an airconditioning system whose COP is 32 It is estimated that the kitchen bath and other ventilating fans of the house discharge a houseful of conditioned air once every hour If the average outdoor temperature is 32C the density of air is 120 kgm3 and the unit cost of electricity is 010kWh the amount of money vented out by the fans in 10 hours is a 050 b 160 c 500 d 1100 e 1600 Answer a 050 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values COP32 T122 C T232 C Price010 kWh Cp1005 kJkgC rho120 kgm3 V24200 m3 mrhoV mtotalm10 EinmtotalCpT2T1COP kJ CostEin3600Price Some Wrong Solutions with Common Mistakes W1CostPrice3600mtotalCpT2T1COP Multiplying by Eff instead of dividing W2CostPrice3600mtotalCpT2T1 Ignoring efficiency W3CostPrice3600mCpT2T1COP Using m instead of mtotal W4CostPrice3600mtotalCpT2T1COP Adding temperatures 6162 The drinking water needs of an office are met by cooling tab water in a refrigerated water fountain from 23C to 6C at an average rate of 10 kgh If the COP of this refrigerator is 31 the required power input to this refrigerator is a 197 W b 612 W c 64 W d 109 W e 403 W Answer c 64 W Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values COP31 Cp418 kJkgC T123 C T26 C mdot103600 kgs QLmdotCpT1T2 kW WinQL1000COP W Some Wrong Solutions with Common Mistakes W1WinmdotCpT1T2 1000COP Multiplying by COP instead of dividing W2WinmdotCpT1T2 1000 Not using COP W3WinmdotT1T2 1000COP Not using specific heat W4WinmdotCpT1T2 1000COP Adding temperatures PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 682 6163 A heat pump is absorbing heat from the cold outdoors at 5C and supplying heat to a house at 25C at a rate of 18000 kJh If the power consumed by the heat pump is 19 kW the coefficient of performance of the heat pump is a 13 b 26 c 30 d 38 e 139 Answer b 26 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TL5 C TH25 C QH180003600 kJs Win19 kW COPQHWin Some Wrong Solutions with Common Mistakes W1COPWinQH Doing it backwards W2COPTHTHTL Using temperatures in C W3COPTH273THTL Using temperatures in K W4COPTL273THTL Finding COP of refrigerator using temperatures in K 6164 A heat engine cycle is executed with steam in the saturation dome The pressure of steam is 1 MPa during heat addition and 04 MPa during heat rejection The highest possible efficiency of this heat engine is a 80 b 156 c 202 d 798 e 100 Answer a 80 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values PH1000 kPa PL400 kPa THTEMPERATURESteamIAPWSx0PPH TLTEMPERATURESteamIAPWSx0PPL EtaCarnot1TL273TH273 Some Wrong Solutions with Common Mistakes W1EtaCarnot1PLPH Using pressures W2EtaCarnot1TLTH Using temperatures in C W3EtaCarnotTLTH Using temperatures ratio PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 683 6165 A heat engine receives heat from a source at 1000C and rejects the waste heat to a sink at 50C If heat is supplied to this engine at a rate of 100 kJs the maximum power this heat engine can produce is a 254 kW b 554 kW c 746 kW d 950 kW e 1000 kW Answer c 746 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TH1000 C TL50 C Qin100 kW Eta1TL273TH273 WoutEtaQin Some Wrong Solutions with Common Mistakes W1Wout1TLTHQin Using temperatures in C W2WoutQin Setting work equal to heat input W3WoutQinEta Dividing by efficiency instead of multiplying W4WoutTL273TH273Qin Using temperature ratio 6166 A heat pump cycle is executed with R134a under the saturation dome between the pressure limits of 14 MPa and 016 MPa The maximum coefficient of performance of this heat pump is a 11 b 38 c 48 d 53 e 29 Answer c 48 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values PH1400 kPa PL160 kPa THTEMPERATURER134ax0PPH C TLTEMPERATURER134ax0PPL C COPHPTH273THTL Some Wrong Solutions with Common Mistakes W1COPPHPHPL Using pressures W2COPTHTHTL Using temperatures in C W3COPTLTHTL Refrigeration COP using temperatures in C W4COPTL273THTL Refrigeration COP using temperatures in K PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 684 6167 A refrigeration cycle is executed with R134a under the saturation dome between the pressure limits of 16 MPa and 02 MPa If the power consumption of the refrigerator is 3 kW the maximum rate of heat removal from the cooled space of this refrigerator is a 045 kJs b 078 kJs c 30 kJs d 116 kJs e 146 kJs Answer d 116 kJs Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values PH1600 kPa PL200 kPa Win3 kW THTEMPERATURER134ax0PPH C TLTEMPERATURER134ax0PPL C COPTL273THTL QLWinCOP kW Some Wrong Solutions with Common Mistakes W1QLWinTLTHTL Using temperatures in C W2QLWin Setting heat removal equal to power input W3QLWinCOP Dividing by COP instead of multiplying W4QLWinTH273THTL Using COP definition for Heat pump 6168 A heat pump with a COP of 32 is used to heat a perfectly sealed house no air leaks The entire mass within the house air furniture etc is equivalent to 1200 kg of air When running the heat pump consumes electric power at a rate of 5 kW The temperature of the house was 7C when the heat pump was turned on If heat transfer through the envelope of the house walls roof etc is negligible the length of time the heat pump must run to raise the temperature of the entire contents of the house to 22C is a 135 min b 431 min c 138 min d 188 min e 808 min Answer a 135 min Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values COP32 Cv0718 kJkgC m1200 kg T17 C T222 C QHmCvT2T1 Win5 kW WintimeQHCOP60 Some Wrong Solutions with Common Mistakes WinW1time60mCvT2T1 COP Multiplying by COP instead of dividing WinW2time60mCvT2T1 Ignoring COP WinW3timemCvT2T1 COP Finding time in seconds instead of minutes WinW4time60mCpT2T1 COP Using Cp instead of Cv Cp1005 kJkgK PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 685 6169 A heat engine cycle is executed with steam in the saturation dome between the pressure limits of 7 MPa and 2 MPa If heat is supplied to the heat engine at a rate of 150 kJs the maximum power output of this heat engine is a 81 kW b 197 kW c 386 kW d 107 kW e 130 kW Answer b 197 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values PH7000 kPa PL2000 kPa Qin150 kW THTEMPERATURESteamIAPWSx0PPH C TLTEMPERATURESteamIAPWSx0PPL C Eta1TL273TH273 WoutEtaQin Some Wrong Solutions with Common Mistakes W1Wout1TLTHQin Using temperatures in C W2Wout1PLPHQin Using pressures W3WoutQinEta Dividing by efficiency instead of multiplying W4WoutTL273TH273Qin Using temperature ratio 6170 An airconditioning system operating on the reversed Carnot cycle is required to remove heat from the house at a rate of 32 kJs to maintain its temperature constant at 20C If the temperature of the outdoors is 35C the power required to operate this airconditioning system is a 058 kW b 320 kW c 156 kW d 226 kW e 164 kW Answer e 164 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TL20 C TH35 C QL32 kJs COPTL273THTL COPQLWin Some Wrong Solutions with Common Mistakes QLW1WinTLTHTL Using temperatures in C QLW2Win Setting work equal to heat input QLW3WinCOP Dividing by COP instead of multiplying QLW4WinTH273THTL Using COP of HP PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 686 6171 A refrigerator is removing heat from a cold medium at 3C at a rate of 7200 kJh and rejecting the waste heat to a medium at 30C If the coefficient of performance of the refrigerator is 2 the power consumed by the refrigerator is a 01 kW b 05 kW c 10 kW d 20 kW e 50 kW Answer c 10 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TL3 C TH30 C QL72003600 kJs COP2 QLWinCOP Some Wrong Solutions with Common Mistakes QLW1WinTL273THTL Using Carnot COP QLW2Win Setting work equal to heat input QLW3WinCOP Dividing by COP instead of multiplying QLW4WinTLTHTL Using Carnot COP using C 6172 Two Carnot heat engines are operating in series such that the heat sink of the first engine serves as the heat source of the second one If the source temperature of the first engine is 1300 K and the sink temperature of the second engine is 300 K and the thermal efficiencies of both engines are the same the temperature of the intermediate reservoir is a 625 K b 800 K c 860 K d 453 K e 758 K Answer a 625 K Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TH1300 K TL300 K Setting thermal efficiencies equal to each other 1TmidTH1TLTmid Some Wrong Solutions with Common Mistakes W1TmidTLTH2 Using average temperature PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 687 6173 Consider a Carnot refrigerator and a Carnot heat pump operating between the same two thermal energy reservoirs If the COP of the refrigerator is 34 the COP of the heat pump is a 17 b 24 c 34 d 44 e 50 Answer d 44 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values COPR34 COPHPCOPR1 Some Wrong Solutions with Common Mistakes W1COPCOPR1 Subtracting 1 instead of adding 1 W2COPCOPR Setting COPs equal to each other 6174 A typical new household refrigerator consumes about 680 kWh of electricity per year and has a coefficient of performance of 14 The amount of heat removed by this refrigerator from the refrigerated space per year is a 952 MJyr b 1749 MJyr c 2448 MJyr d 3427 MJyr e 4048 MJyr Answer d 3427 MJyr Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Win68036 MJ COPR14 QLWinCOPR MJ Some Wrong Solutions with Common Mistakes W1QLWinCOPR36 Not using the conversion factor W2QLWin Ignoring COP W3QLWinCOPR Dividing by COP instead of multiplying PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 71 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 7 ENTROPY PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 72 Entropy and the Increase of Entropy Principle 71C No A system may produce more or less work than it receives during a cycle A steam power plant for example produces more work than it receives during a cycle the difference being the net work output 72C The entropy change will be the same for both cases since entropy is a property and it has a fixed value at a fixed state 73C No In general that integral will have a different value for different processes However it will have the same value for all reversible processes 74C That integral should be performed along a reversible path to determine the entropy change 75C No An isothermal process can be irreversible Example A system that involves paddlewheel work while losing an equivalent amount of heat 76C The value of this integral is always larger for reversible processes 77C No Because the entropy of the surrounding air increases even more during that process making the total entropy change positive 78C It is possible to create entropy but it is not possible to destroy it 79C If the system undergoes a reversible process the entropy of the system cannot change without a heat transfer Otherwise the entropy must increase since there are no offsetting entropy changes associated with reservoirs exchanging heat with the system 710C The claim that work will not change the entropy of a fluid passing through an adiabatic steadyflow system with a single inlet and outlet is true only if the process is also reversible Since no real process is reversible there will be an entropy increase in the fluid during the adiabatic process in devices such as pumps compressors and turbines 711C Sometimes PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 73 712C Never 713C Always 714C Increase 715C Increases 716C Decreases 717C Sometimes 718C Greater than 719C Yes This will happen when the system is losing heat and the decrease in entropy as a result of this heat loss is equal to the increase in entropy as a result of irreversibilities 720C They are heat transfer irreversibilities and entropy transport with mass PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 74 721E The source and sink temperatures and the entropy change of the sink for a completely reversible heat engine are given The entropy decrease of the source and the amount of heat transfer from the source are to be determined Assumptions The heat engine operates steadily PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course HE QH Wnet TH TL QL Analysis According to the increase in entropy principle the entropy change of the source must be equal and opposite to that of the sink Hence S 10 BtuR L H S Applying the definition of the entropy to the source gives 15000 Btu 1500 R 10 BtuR H H H S T Q which is the heat transfer with respect to the source not the device 722 The source and sink temperatures and the entropy change of the sink for a completely reversible heat engine are given The amount of heat transfer from the source are to be determined Assumptions The heat engine operates steadily HE QH QL Wnet TH TL Analysis According to the increase in entropy principle the entropy change of the source must be equal and opposite to that of the sink Hence S 20 kJK L H S Applying the definition of the entropy to the source gives 20000 kJ 1000 K 20 kJK H H H S T Q which is the heat transfer with respect to the source not the device 723E The operating conditions of a heat engine are given The entropy change of all components and the work input are to be calculated and it is to be determined if this heat engine is reversible Assumptions The heat engine operates steadily 600 R 1500 R HE QH QL Wne Analysis The entropy change of all the components is S device total L H where the last term is zero each time the engine completes a cycle S S S Applying the definition of the entropy to the two reservoirs reduces this to 333 BtuR 600 R 100000 Btu 1500 R 200000 Btu total L L H H T Q T Q S Since the entropy of everything involved with this engine has increased the engine is not reversible but possible Applying the first law to this engine 100000 Btu 100000 Btu 200000 Btu net L H Q Q W preparation If you are a student using this Manual you are using it without permission 76 726 It is assumed that heat is transferred from a cold reservoir to the hot reservoir contrary to the Clausius statement of the second law It is to be proven that this violates the increase in entropy principle Assumptions The reservoirs operate steadily Analysis According to the definition of the entropy the entropy change of the hightemperature reservoir shown below is 0 08333 kJK 1200 K 100 kJ H H T Q S TL TH Q 100 kJ and the entropy change of the lowtemperature reservoir is 1667 kJK 0 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 600 K L L T 100 kJ Q he total entropy change of everything involved with this system is en S T th 00833 kJK 0 1667 0 08333 total L H S S S which violates the increase in entropy principle since the total entropy change is negative e of two reservoirs is to pump satisfies the increase in entropy principle efficient of erformance expression first Law and thermodynamic temperature scale gives 727 A reversible heat pump with specified reservoir temperatures is considered The entropy chang be calculated and it is to be determined if this heat Assumptions The heat pump operates steadily Analysis Since the heat pump is completely reversible the combination of the co 7C net W L Q 24C HP 300 kW p 1747 1 1 COP HPrev 280 K 297 K 1 1 L TH T he power required to drive this heat pump according to the coefficient of ance is then T perform 1717 kW 300 kW QH W 1747 COP HPrev in 1717 300 kW netin W Q Q H L The rate at which the entropy of the high temperature reservoir changes according to the definition of the entropy is net According to the first law the rate at which heat is removed from the lowtemperature energy reservoir is kW 8 kW 282 101 kWK K 297 300 kW Q H H H T S and that of the lowtemperature reservoir is kWK 101 280 K L L T S 1717 kW QL The net rate of entropy change of everything in this system is as it must be since the heat pump is completely reversible 0kWK 1 01 1 01 total L H S S S preparation If you are a student using this Manual you are using it without permission 77 728E Heat is transferred isothermally from the working fluid of a Carnot engine to a heat sink The entropy change of the working fluid is given The amount of heat transfer the entropy change of the sink and the total entropy change during the process are to be determined Analysis a This is a reversible isothermal process and the entropy change during such a process is given by 95F Carnot heat engine SINK 95F Heat S Q T is equal to the heat ansferred to the sink the heat transfer become out fluid Noting that heat transferred from the working fluid tr 388 07 BtuR 555 R fluid fluid fluid 3885 Btu 5 Btu S T Q Q The entropy change of the sink is determined from b PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 07 BtuR 3885 Btu sinkin Q S 555 R sink sink T c Thus the total entropy change of the process is 0 total gen S S 70 70 sink fluid S S his is expected since all processes of the Carnot cycle are reversible processes and no entropy is generated during a versible process rant othermal internally reversible process Analysis Noting that both the refrigerant and the cooled space undergo reversible isothermal processes the entropy change T re 729 R134a enters an evaporator as a saturated liquidvapor at a specified pressure Heat is transferred to the refrige from the cooled space and the liquid is vaporized The entropy change of the refrigerant the entropy change of the cooled space and the total entropy change for this process are to be determined Assumptions 1 Both the refrigerant and the cooled space involve no internal irreversibilities such as friction 2 Any temperature change occurs within the wall of the tube and thus both the refrigerant and the cooled space remain isothermal during this process Thus it is an is for them can be determined from S Q T pr sure of the refrigerant is maintained constant Therefore the temperature of the refrigerant also remains Table A12 Then a The es constant at the saturation value 2574 K 156 C sat160 kPa T T R134a 160 kPa 5C 180 kJ 0699 kJK 2574 K refrigerant T 180 kJ refrigerantin refrigerant Q S Similarly b 0672 kJK 268 K kJ 180 space spaceout space T Q S c The total entropy change of the process is 0027 kJK 0 672 0 699 space refrigerant total gen S S S S preparation If you are a student using this Manual you are using it without permission 78 Entropy Changes of Pure Substances 730C Yes because an internally reversible adiabatic process involves no irreversibilities or heat transfer 731E A pistoncylinder device that is filled with water is heated The total entropy change is to be determined Analysis The initial specific volume is 1 25 ft lbm 2 lbm ft 52 3 3 1 1 m V v PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course which is between vf and vg for 300 psia The initial quality and the entropy are then Table A5E 1 3334 Btulbm R 0 8075 0 92289 Btulbm R 58818 Btulbm R 0 0 8075 0 01890 ft lbm 5435 1 1 fg x v 0 01890 ft lbm 25 1 1 1 3 3 1 fg f f x s s s v v s P P T ence the change in the total entropy is 1 2 s m s S 732 An insulated rigid tank contains a saturated liquidvapor mixture of water at a specified pressure An electric heater ide is turned on and kept on until all the liquid vaporized The entropy change of the water during this process is to be determined Analysis From the steam tables Tables A4 through A6 H2O 300 psia 2 lbm 25 ft3 The final state is superheated vapor and P v 2 1 1 5706 Btulbm R Table A 6E psia 300 500 F 2 1 2 2 H 04744 BtuR 1 3334 Btulbm R 2 lbm 1 5706 ins 67298 kJkg K vapor sat 28810 kJkg K 5 7894 0 25 1 4337 029065 m kg 0001053 11594 025 We H2O 5 kg 150 kPa 0001053 150 kPa 1 1 1 x P fg f v v v 20 2 1 2 1 1 3 1 s x s s s x fg f v v Then the entropy change of the steam becomes 5 192 kJK 28810 kJkg K 5 kg67298 1 2 s m s S preparation If you are a student using this Manual you are using it without permission 79 733 A rigid tank is divided into two equal parts by a partition One part is filled with compressed liquid water while the other side is evacuated The partition is removed and water expands into the entire tank The entropy change of the water during this process is to be determined Analysis The properties of the water are Table A4 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course kJkg K 08313 0001017m kg C 60 00 kPa 4 60 C 1 3 60 C 1 1 T 1 f f s s P v v Noting that v v 0002034 m kg 0 001017 2 2 3 1 2 1 0278 kJkg K 6 6430 0 0002524 0261 1 0 0002524 0 001026 3 993 0 001026 002034 0 002034 m kg 0 kPa 40 2 2 2 2 3 2 2 Vacuum 25 kg compressed liquid 400 kPa 60C v f v fg f fg x s s s x P v v Then the entropy change of the water becomes 0492 kJK 08313 kJkg K 25 kg 10278 1 2 s m s S preparation If you are a student using this Manual you are using it without permission 710 734 Problem 733 is reconsidered The entropy generated is to be evaluated and plotted as a function of surroundings temperature and the values of the surroundings temperatures that are valid for this problem are to be nalysis The problem is solved using EES and the results are tabulated and plotted below a m3kg e 1 kJkgK kJkgK call this is a constant volume system ue for the surroundings temperature occurs when we set Sgen 0Delta surr press the table The results a rob m determined The surrounding temperature is to vary from 0C to 100C A Input Data P1400 kP T160 C m25 kg P240 kPa FluidSteamIAPWS V1mspv1 spv1volumeFluidTT1 PP1 specific volume of steam at state 1 s1entropyFluidTT1PP1 entropy of steam at stat V22V1 Steam expands to fill entire volume at state 2 State 2 is identified by P2 and spv2 spv2V2m specific volume of steam at state 2 m3kg s2entropyFluidPP2vspv2 entropy of steam at state 2 T2temperatureFluidPP2vspv2 DELTASsysms2s1 Total entopy change of steam kJK What does the first law tell us about this problem Ein Eout DELTAEsys Conservation of Energy for the entire closed system neglecting changes in KE and PE for the system DELTAEsysmintenergyFluid PP2 vspv2 intenergyFluidTT1PP1 Ein 0 How do you interpert the energy leaving the system Eout Re Qout Eout What is the maximum value of the Surroundings temperature The maximum possible val SsyssumDeltaS QnetsurrQout Sgen 0 Sgen DELTASsysQnetsurrTsurr273 Establish a parametric table for the variables Sgen Qnetsurr Tsurr and DELTASsys In the Parametric Table window select Tsurr and insert a range of values Then place and about the Sgen 0 line F3 to solve p re shown in Plot Window 1 What values of Tsurr are valid for this le Tsurr C Sgen kJK 0 10 20 30 40 50 60 70 80 90 100 004235 01222 01005 008033 006145 004378 00272 001162 0003049 001689 002997 0 20 40 60 80 100 016 012 008 004 0 004 008 Tsurr C Sgen kJK PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 711 735E A cylinder is initially filled with R134a at a specified state The refrigerant is cooled and condensed at constant pressure The entropy change of refrigerant during this process is to be determined R134a 120 psia 100F Q Analysis From the refrigerant tables Tables A11E through A13E 006039 Btulbm R psia 120 0 F 5 022361 Btulbm R F 100 psia 120 90 F 2 2 2 1 1 1 o fs s P T s T P Then the entropy change of the refrigerant becomes 03264 BtuR 022361 Btulbm R 2 lbm 006039 1 2 s m s S 736 An insulated cylinder is initially filled with saturated liquid water at a specified pressure The water is heated electrically at constant pressure The entropy change of the water during this process is to be determined Assumptions 1 The kinetic and potential energy changes are negligible 2 The cylinder is wellinsulated and thus heat transfer is negligible 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or expansion process is quasiequilibrium PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis From the steam tables Tables A4 through A6 kJkg K 14337 kJkg 46713 m kg 0001053 kPa 150 150 kPa 1 150 kPa 1 3 150 kPa 1 1 f f f s s h h liquid sat P v v Also 475 kg 0001053 m kg 3 1 m v 0005 m 3 V We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as U W W E E E 43 42 1 4243 1 U b H during a constant pressure quasiequilibrium process Solving for h2 H O 2 150 kPa Sat liquid 2200 kJ bout in e potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 1 2 ein h m h W since W 93033 kJkg 475 kg 2200 kJ 46713 ein 1 2 m W h h Thus 2 6384 kJkg K 5 7894 0 2081 4337 1 0 2081 2226 0 46713 93 33 150 kPa 2 2 2 f h h h x P 0 93033 kJkg 2 2 fg f fg x s s s h Then the entropy change of the water becomes 2 572 kJK 14337 kJkg K 475 kg 26384 1 2 s m s S preparation If you are a student using this Manual you are using it without permission 712 737 Entropy change of water as it is cooled at constant pressure is to be determined using Gibbs equation and to be compared to steam tables Analysis The Gibbs equation is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course dP dh Tds v As water is converted from a saturated liquid to a saturated vapor both the pressure and temperature remain constant Then the Gibbs equation reduces to T ds dh When the result is integrated between the saturated liquid and saturated vapor states the result is P s 1 2 sf sg 5322 kJkg K 13352 273K Tsat300 kPa T T 5 kJkg 2163 h 300 kPa h h h h s s fg f g f g f g here enthalpy and temperature data are obtained from Table A5 The entropy change from the steam tables is K he result is practically the same ressor during which the entropy remains constant The final temperature and nalysis The initial state is saturated vapor a d the properties are Table A11E F 0 1 hg h The final state is superheated vapor and the properties are Table A13E 2 1 2 h s s he change in the enthalpy across the compressor is then W 300 kPa 5320 kJkg s fg Table A 5 T 738E R134a is compressed in a comp enthalpy change are to be determined A n 08 Btulbm 103 T s 2 1 0 22539 Btulbm R F 0 1 sg s 11223 Btulbm 22539 Btulbm R 0 60 psia 2 2 T P F 593 T 915 Btulbm 11223 10308 1 2 h h h preparation If you are a student using this Manual you are using it without permission 713 739 Water vapor is expanded in a turbine during which the entropy remains constant The enthalpy difference is to be determined Analysis The initial state is superheated vapor and thus PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 6 MPa 1 h P The entropy e final state is a mixture since the entropy is between sf and sg for 100 kPa The properties at this state are Table A5 6 5432 kJkg K Table A 6 400 C 1 1 s T is constant during the process Th 3178 3 kJkg 1 2370 9 kJkg 0 86532257 5 41751 2 2 fg f x h h h 0 8653 6 0562 kJkg K 1 3028 kJkg K 6 5432 2 2 fg f s s s x he chang in the enthalpy across the turbine is then s during which the entropy is kept constant The final temperature and internal energy are nalysis The initial entropy is T The entrop is constant during the process The final state is a mixture since the entropy is between f and sg for 100 kPa The properties at this state are Table A12 T s 2 1 T e 8074 kJkg 3178 3 2370 9 1 2 h h h 740 R134a undergoes a proces to be determined A T s 2 1 0 9341 kJkg K Table A 13 600 kPa 1 1 1 s P C 25 y s kJkg 2112 C 2637 0 979919798 21 17 0 9799 0 87995 kJkg K 0 07188 kJkg K 0 9341 2 2 2 2 sat 100 kPa 2 fg f fg f x u u u s s s x T T preparation If you are a student using this Manual you are using it without permission 714 741 Refrigerant134a is is expanded in a turbine during which the entropy remains constant The inlet and outlet velocities are to be determined Analysis The initial state is superheated vapor and thus PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 100 kPa 3 2 v P he inlet and outlet veloicites are T s 2 1 Table A 13 0110 kJkg K 1 0 02997 m kg C 60 kPa 800 1 3 1 1 1 s T P v The entropy is constant during the process The properties at the exit state are 2 0 2098 m kg Table A 13 1 0110 kJkg K 1 2 s s T 0030 ms 2 3 1 1 1 m 50 kgs002997 m kg 50 A m V v 0105 ms 2 3 2 2 2 m 01 kgs02098 m kg 50 A m V v 742 An insulated cylinder is initially filled with superheated steam at a specified state The steam is compressed in a reversible manner until the pressure drops to a specified value The work input during this process is to be determined Assumptions 1 The kinetic and potential energy changes are negligible 2 The cylinder is wellinsulated and thus heat transfer is negligible 3 The thermal energy stored in the cylinder itself is negligible 4 The process is stated to be reversible Analysis This is a reversible adiabatic ie isentropic process and thus s2 s1 From the steam tables Tables A4 through 1 1 3 1 1 1 u s s P s u T P v A6 29216 kJkg MPa 12 73132 kJkg K 26510 kJkg m kg 071643 0 C 20 kPa 300 2 1 2 2 Also 002792 kg m kg 071643 m 002 3 3 1 v V m We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this adiabatic closed system can be expressed as by heat work and mass Net energy transfer u m u U W 43 42 1 4243 Substituting the work input during this adiabatic process is determined to be 1 1 2 in b potential etc energies in internal kinetic Change system out in E E E 755 kJ kJkg 2651 0 002792 kg 29216 1 2 bin u m u W H2O 300 kPa 200C preparation If you are a student using this Manual you are using it without permission 715 743 Problem 742 is reconsidered The work done on the steam is to be determined and plotted as a function of final pressure as the pressure varies from 300 kPa to 12 MPa Analysis The problem is solved using EES and the results are tabulated and plotted below Knowns P1 300 kPa T1 200 C Vsys 002 m3 P2 1200 kPa Analysis FluidSteamIAPWS Treat the pistoncylinder as a closed system with no heat transfer in neglect changes in KE and PE of the Steam The process is reversible and adiabatic thus isentropic The isentropic work is determined from Ein Eout DELTAEsys Eout 0 kJ Ein Workin DELTAEsys msysu2 u1 u1 INTENERGYFluidPP1TT1 v1 volumeFluidPP1TT1 s1 entropyFluidPP1TT1 Vsys msysv1 The process is reversible and adiabatic or isentropic s2 s1 u2 INTENERGYFluidPP2ss2 T2isen temperatureFluidPP2ss2 P2 kPa Workin kJ 300 400 500 600 700 800 900 1000 0 1366 2494 3462 4314 5078 5773 6411 300 400 500 600 700 800 900 1000 0 1 2 3 4 5 6 7 P2 kPa Workin kJ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 716 744 A cylinder is initially filled with saturated water vapor at a specified temperature Heat is transferred to the steam and it expands in a reversible and isothermal manner until the pressure drops to a specified value The heat transfer and the work output for this process are to be determined Assumptions 1 The kinetic and potential energy changes are negligible 2 The cylinder is wellinsulated and thus heat transfer is negligible 3 The thermal energy stored in the cylinder itself is negligible 4 The process is stated to be reversible and isothermal PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 25942 kJkg C 200 2 2 200 C 1 u P u u T g ess can be determined from H2O 200C sat vapor T const Analysis From the steam tables Tables A4 through A6 64302 kJkg K 200 C 1 s s sat vapor g 1 68177 kJkg K 2 1 2 s T T The heat transfer for this reversible isothermal proc 26311 kJkg kPa 800 Q 2199 kJ 64302kJkg K 473 K12 kg68177 Q 1 2 s Tm s TS This is a closed system since no mass enters or leaves The energy r this losed system can be expressed as Substituting the work done during this process is determined to be We take the contents of the cylinder as the system balance fo c 1 2 in out b 1 2 bout in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u m u Q W u m u U W Q E E E 43 42 1 4243 1 1756 kJ 25942 kJkg 12 kg26311 9 kJ 219 Wbout preparation If you are a student using this Manual you are using it without permission 717 745 Problem 744 is reconsidered The heat transferred to the steam and the work done are to be determined and alue of 800 kPa nalysis The problem is solved using EES and the results are tabulated and plotted below kPa a closed system neglect changes in KE and PE of the Steam The process is Esys 10 TT1x10 T2 in T1273msyss2s1 plotted as a function of final pressure as the pressure varies from the initial value to the final v A Knowns T1 200 C x1 10 msys 12 kg P2 800 Analysis FluidSteamIAPWS Treat the pistoncylinder as reversible and isothermal T2 T1 Ein Eout DE 800 1000 1200 1400 1600 0 40 80 120 160 200 Workout KJ LTA Ein Qin Eout Workout DELTAEsys msysu2 u1 P1 pressureFluidTT1x10 u1 INTENERGYFluidTT1x v1 volumeFluidTT1x10 s1 entropyFluid Vsys msysv1 The process is reversible and isothermal Then P2 and T2 specify state 2 u2 INTENERGYFluidPP2T s2 entropyFluidPP2TT2 P2 kPa Q 800 1000 1200 1400 1600 0 40 80 120 160 200 Qin kJ P2 kPa Qin kJ Workout kJ 800 900 1000 1100 1200 1300 1400 1500 1553 04645 03319 2199 1837 1506 120 9 123 6408 382 1332 1757 1447 117 9184 6885 4765 2798 9605 P2 kPa PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 718 746 R134a undergoes an isothermal process in a closed system The work and heat transfer are to be determined Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or expansion process is quasiequilibrium Analysis The energy balance for this system can be expressed as R134a 240 kPa T1 T2 20C T 1 2 1 2 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u m u U Q W E E E 43 42 1 4243 1 s The initial state properties are PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 240 kPa 1 u P For this isothermal process the final state properties are Table A11 C 20 2 2 2 1 2 fg fg f s x x u u u T T he heat t ined from 1 0134 kJkg K Table A 13 20 C 1 1 s T 24674 kJkg 1 0 42497 kJkg K 0 20 0 62172 30063 0 11129 kJkg 0 2016216 86 78 0 20 2 2 fs s x T ransfer is determ 1724 kJkg 1 0134 kJkg K 293 K 0 42497 1 2 0 in s s T q T he negative sign shows that th e heat is actually transferred from the system That is ined from the energy balance to be qout 1724 kJkg The work required is determ 3695 kJkg 24674 kJkg 11129 172 4 kJkg 1 2 out in u u q w 747 The heat transfer during the process shown in the figure is Assumptions The process is reversible Analysis No heat is transferred durin to be de g the process 23 since the area under process line is zero Then the heat transfer is equal to the area under the process line 12 termined T C s kJkgK 200 2 1 03 1 0 600 3 kJkg 471 03kJkg K 10 2 273K 200 273K 600 2 Area 1 2 2 1 2 1 12 s s T T Tds q preparation If you are a student using this Manual you are using it without permission 719 748E The heat transfer during the process shown in the figure is to be determined Assumptions The process is reversible Analysis Heat transfer is equal to the sum of the areas under the process 12 and 23 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 30B 460R 20 360 515 Btulbm R tulbm 10Btulbm R 460R 30 360 460R 55 2 2 3 2 1 2 2 1 3 2 2 1 12 s s T s s T T Tds Tds q 749 The heat transfer during the process shown in the figure is to be determined Assumptions The process is reversible Analysis Heat transfer is equal to the sum of the areas under the process 12 and 23 T F 1 3 2 360 2 55 1 2 3 s BtulbmR 140 kJkg 30 kJkg K 273K 10 600 10kJkg K 03 2 273K 600 273K 200 2 2 3 2 1 2 2 1 3 2 2 1 12 s s T s s T T Tds Tds q T C s kJkgK 200 1 2 03 1 0 600 3 preparation If you are a student using this Manual you are using it without permission 722 752 Water is compressed in a closed system during which the entropy remains constant The final temperature and the work required are to be determined Analysis The initial state is superheated vapor and thus T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4000 kPa 2 u P To determine the work done we take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as ubstitutin 2 1 7 5344 kJkg K Table A 6 2509 4 kJkg 100 C kPa 70 1 1 1 1 s u T P The entropy is constant during the process The properties at the exit state are Table A 6 7 5344 kJkg K 2 1 2 664C T s s 3396 5 kJkg 2 s KE PE 0 1 since 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in Q u u u w E E E 43 42 1 4243 1 S g 8871 kJkg 2509 4 kJkg 33965 1 2 in u u w 753 Refrigerant134a is expanded in a closed system d work production are to be determined uring which the entropy remains constant The heat transfer and the nalysis The initial state is superheated vapor and thus 600 kPa u P The entrop is constant during the process The properties at the exit state are 140 2 1 2 2 u s s P the work done we take the contents of the cylinder as the system This is a closed system since no mass balance for this stationary closed system can be expressed as Substituting A T s 2 1 0 9499 kJkg K Table A 13 30 C 1 1 1 1 s T 22 kJkg 249 y 22075 kJkg Table A 13 0 9499 kJkg K kPa Since the process is isentropic and thus the heat transfer is zero Q 0 kJ To determine enters or leaves The energy KE PE 0 since 1 2 ou potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in Q u m u U W E E E t 43 42 1 4243 1 142 kJ 22075kJkg 05 kg24922 2 1 ou u m u W t preparation If you are a student using this Manual you are using it without permission 724 755 Water vapor is expanded adiabatically in a pistoncylinder device The entropy change is to be determined and it is to be discussed if this process is realistic Analysis a The properties at the initial state are 6 7593 kJkg K Table A 5 2566 8 kJkg 1 kPa 600 1 1 1 1 s u x P We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course E KE since 1 2 ou es netic system out in Q u m u U W E E E t 43 42 1 4243 1 Solving for the final state internal energy potential etc energi Change in internal ki by heat work and mass Net energy transfer 100 kPa 600 kPa 1 2s 2 s T P 0 2216 8 kJkg kg 2 700 kJ Wou t 8 kJkg 2566 1 m u u state is from Table A5 2 The entropy at the final 6 5215 kJkg K 6 0562 0 861 3028 1 0 8617 2088 2 8 kJkg 2216 2 2 2 2 fg f fg xs s s u u 7 417 2216 8 100 kPa 2 u f u x P he entropy change is 40 T 0238 kJkg K 6 7593 6 5215 1 2 s s s b The process is not realistic since entropy cannot decrease during an adiabatic process In the limiting case of a reversible and adiabatic process the entropy would remain constant preparation If you are a student using this Manual you are using it without permission 727 758 Heat is added to a pressure cooker that is maintained at a specified pressure The minimum entropy change of the thermalenergy reservoir supplying this heat is to be determined Assumptions 1 Only water vapor escapes through the pressure relief valve Analysis According to the conservation of mass principle out out in CV m dt dm m m dt dm An entropy balance adapted to this system becomes 0 out dt dt surr s m d ms d hen this is combined with the mass balance it becomes S W 0 surr dt dt dt s dm d ms dS ultiplyin y dt and integrating the result yields The properties at the initial and final states are from Table A5 at P1 175 kPa and P2 150 kPa 0 4644 m kg 0 001053 0 40 1 1594 001053 0 2 0537 kJkg K 0 10 5 6865 4850 1 0 1013 m kg 0 001057 0 10 1 0037 001057 0 3 2 1 3 1 fg f fg f fg f x xs s s x v v v v v v he initial and final masses are M g b 0 1 2 out 1 1 2 2 surr m m s m s m s S 1 4337 2 fg f xs s s 3 7494 kJkg K 0 40 5 7894 T 0 1974 kg 01013 m kg 0 020 m 0 3 3 1 1 v V m PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 04307 kg 4644 m kg 0 0 020 m 3 V 3 2 m v f escaping water vapor is Substituting 2 The entropy o kJkg K 7 2231 150 kPa out sg s 0 8708 0 0 0 1974 7 2231 0 04307 0 1974 2 0537 0 04307 3 7494 0 surr surr 1 2 out 1 1 2 2 surr S m m s The entropy change of the thermal energy reservoir must then satisfy m s m s S S surr 08708 kJK S preparation If you are a student using this Manual you are using it without permission 728 759 Heat is added to a pressure cooker that is maintained at a specified pressure Work is also done on water The minimum entropy change of the thermalenergy reservoir supplying this heat is to be determined Assumptions 1 Only water vapor escapes through the pressure relief valve Analysis According to the conservation of mass principle out out in CV m dt dm m m dt dm An entropy balance adapted to this system becomes 0 out dt dt surr s m d ms d hen this is combined with the mass balance it becomes S W 0 surr dt dt dt s dm d ms dS ultiplyin y dt and integrating the result yields entropy transfer The properties at the initial and final states are from Table A5 at P1 175 kPa and P2 150 kPa 0 4644 m kg 0 001053 0 40 1 1594 001053 0 2 0537 kJkg K 0 10 5 6865 4850 1 0 1013 m kg 0 001057 0 10 1 0037 001057 0 3 2 1 3 1 fg f fg f fg f x xs s s x v v v v v v he initial and final masses are M g b 0 1 2 out 1 1 2 2 surr m m s m s m s S Note that work done on the water has no effect on this entropy balance since work transfer does not involve any 1 4337 2 fg f xs s s 3 7494 kJkg K 0 40 5 7894 T 0 1974 kg 01013 m kg 0 020 m 0 3 3 1 1 v V m PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 04307 kg 4644 m kg 0 0 020 m 3 V 3 2 m v f escaping water vapor is Substituting 2 The entropy o kJkg K 7 2231 150 kPa out sg s 0 8708 0 0 0 1974 7 2231 0 04307 0 1974 2 0537 0 04307 3 7494 0 surr surr 1 2 out 1 1 2 2 surr S m m s The entropy change of the thermal energy reservoir must then satisfy m s m s S S surr 08708 kJK S preparation If you are a student using this Manual you are using it without permission 729 760 A cylinder is initially filled with saturated water vapor mixture at a specified temperature Steam undergoes a reversible heat addition and an isentropic process The processes are to be sketched and heat transfer for the first process and work done during the second process are to be determined Assumptions 1 The kinetic and potential energy changes are negligible 2 The thermal energy stored in the cylinder itself is negligible 3 Both processes are reversible Analysis b From the steam tables Tables A4 through A6 H2O 100C x 05 22479 kJkg kPa 15 7 3542 kJkg K kJkg 25060 kJkg 26756 1 C 100 1547 4 kJkg 50 2256 4 41917 50 C 100 3 2 3 3 2 2 2 2 2 1 1 u s s P s s u u h h x T xh h h x T g g g fg f Q We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this closed system can be expressed as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course ential in internal kinetic ge system out in u m u U W Q E E E 43 42 1 4243 1 pot Chan by heat work and mass Net energy transfer 1 2 bout in etc energies For process 12 it reduces to 5641 kJ 5 kg26756 15474kJkg 1 2 12in h m h Q For process 23 it reduces to c 1291 kJ 5 kg25060 22479kJkg 3 2 23bout u m u W 00 11 22 33 44 55 66 77 88 99 110 0 100 200 300 400 500 600 700 s kJkgK T C 10142 kPa 15 kPa SteamIAPWS 1 2 3 preparation If you are a student using this Manual you are using it without permission 730 761E An insulated rigid can initially contains R134a at a specified state A crack develops and refrigerant escapes slowly The final mass in the can is to be determined when the pressure inside drops to a specified value Assumptions 1 The can is wellinsulated and thus heat transfer is negligible 2 The refrigerant that remains in the can underwent a reversible adiabatic process Analysis Noting that for a reversible adiabatic ie isentropic process s1 s2 the properties of the refrigerant in the can are Tables A11E through A13E PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 01978 ft lbm 001209 01208 15492 001209 0 1208 0 1859 0 03793 0 06039 0 psia 3 006039 Btulbm R 50 F 50 F 1 1 fs s T psia 140 3 2 2 2 2 1 2 2 fg f fg f x s s x s s P P v v v Thus the final mass of the refrigerant in the can is 1 s 404 lbm ft lbm 01978 ft 08 3 3 v 2 V m Leak R134 140 psia 50F preparation If you are a student using this Manual you are using it without permission 731 762E An electric windshield defroster used to remove ice is considered The electrical energy required and the minimum temperature of the defroster are to be determined Assumptions No no heat is transferred from the defroster or ice to the surroundings Analysis The conservation of mass principle is out in cv m m dt dm hich reduces to w PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course mout dt cv dm while the the first law reduces to out out out dmu cv h m dt W Combining these two expressions yield dt d mu dm cv cv dt h W out grated from time when the ice layer is present until it is removed m 0 gives h W out out The original mass of the ice layer is out When this is multiplied by dt and inte m i i mu v v V tA m i required per unit of windshield area is then The work 2 3 out 001602 ft lbm if f i i A v v v That is out 1873 Btuft 144 Btulbm 0 25 12 ft t u u t u h t u W Clasius tells us that the temperature of the defroster cannot be lower than the temperature of the e being melted Then Win 1873 Btuft2 The second law as stated by ic Tmin 32F preparation If you are a student using this Manual you are using it without permission 732 Entropy Change of Incompressible Substances 763C No because entropy is not a conserved property 764 A hot copper block is dropped into water in an insulated tank The final equilibrium temperature of the tank and the total entropy change are to be determined Assumptions 1 Both the water and the copper block are incompressible substances with constant specific heats at room temperature 2 The system is stationary and thus the kinetic and potential energies are negligible 3 The tank is well insulated and thus there is no heat transfer Properties The density and specific heat of water at 25C are ρ 997 kgm3 and cp 418 kJkgC The specific heat of copper at 27C is cp 0386 kJkgC Table A3 Analysis We take the entire contents of the tank water copper block as the system This is a closed system since no mass crosses the system boundary during the process The energy balance for this system can be expressed as U E E E 0 potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 43 42 1 4243 1 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 160 L Copper 75 kg WATER or U 0 water Cu U 0 water 1 2 Cu 1 2 T mc T T mc T where 1595 kg 997 kgm 0160 m 3 3 water V ρ m Using specific heat values for copper and liquid water at room temperature and substituting 0 15 C C 1595 kg418 kJkg 110 C C 75 kg0386 kJkg 2 2 T T T 190 C 292 K 2 The entropy generated during this process is determined from 9 20 kJK 288 K 1595 kg 418 kJkg K ln 2920 K ln 785 kJK 383 K 75 kg 0386 kJkg K ln 2920 K ln 1 2 avg water 1 2 avg copper T T mc S T T mc S Thus 135 kJK 9 20 7 85 water copper total S S S preparation If you are a student using this Manual you are using it without permission 733 765 Computer chips are cooled by placing them in saturated liquid R134a The entropy changes of the chips R134a and the entire system are to be determined Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved 3 There is no heat transfer between the system and the surroundings Analysis a The energy balance for this system can be expressed as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course R134a 1 2 chips 1 2 tial etc energies in internal kinetic system out in 0 u m u u m u U E E E 43 42 1 4243 1 he heat r ased by the chips is poten Change by heat work and mass Net energy transfer R134a 1 2 chips 2 1 u m u u m u T ele 018 kJ 40 K 0 010 kg03 kJkg K 20 2 1 chips T mc T Q The mass of the refrigerant vaporized during this heat exchange process is 0 0008679 kg 20740 kJkg 18 kJ 0 40 C R 134a R 134a g2 fg f g u Q u u Q m Only a small fraction of R134a is vaporized during the process Therefore the temperature of R134a remains constant uring the process The change in the entropy of the R134a is at 40F from Table A11 0 0050 0 00086790 0 005 0 0008679 0 96866 1 1 2 2 f f f f s m s m d 0000841 kJK 2 2 R 134a mg sg S b The entropy change of the chips is 0000687 kJK 273K 20 273K 0 010 kg03 kJkg Kln 40 ln 1 2 chips T T mc S c The total entropy change is 0000154 kJK 0 000687 0 000841 chips R134a gen total S S S S The positive result for the total entropy change ie entropy generation indicates that this process is possible preparation If you are a student using this Manual you are using it without permission 734 766 A hot iron block is dropped into water in an insulated tank The total entropy change during this process is to be determined Assumptions 1 Both the water and the iron block are incompressible substances with constant specific heats at room temperature 2 The system is stationary and thus the kinetic and potential energies are negligible 3 The tank is well insulated and thus there is no heat transfer 4 The water that evaporates condenses back Properties The specific heat of water at 25C is cp 418 kJkgC The specific heat of iron at room temperature is cp 045 kJkgC Table A3 Analysis We take the entire contents of the tank water iron block as the system This is a closed system since no mass crosses the system boundary during the process The energy balance for this system can be expressed as U E E E 0 potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 43 42 1 4243 1 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course r o 0 water iron U U 0 water 1 2 iron 1 2 T mc T T mc T Substituting 2 2 0 18 C K Jkg T T o tropy generated during this process is determined from 267C 2 100 kg418 k 350 C 25 kg045 kJkg K T o The en 12314 kJK 291 K 100 kg 418 kJkg K ln 2997 K ln 8 232 kJK 623 K 1 avg iron T S 25 kg 045 kJkg K ln 2997 K ln 1 2 avg water 2 T T mc S T mc Thus 408 kJK 12314 8 232 water iron total gen S S S S Discussion The results can be improved somewhat by using specific heats at average temperature Iron 350C WATER 18C preparation If you are a student using this Manual you are using it without permission 735 767 An aluminum block is brought into contact with an iron block in an insulated enclosure The final equilibrium temperature and the total entropy change for this process are to be determined Assumptions 1 Both the aluminum and the iron block are incompressible substances with constant specific heats 2 The system is stationary and thus the kinetic and potential energies are negligible 3 The system is wellinsulated and thus there is no heat transfer Properties The specific heat of aluminum at the anticipated average temperature of 400 K is cp 0949 kJkgC The specific heat of iron at room temperature the only value available in the tables is cp 045 kJkgC Table A3 Analysis We take the ironaluminum blocks as the system which is a closed system The energy balance for this system can be expressed as U E E E 0 potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 43 42 1 4243 1 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course or U U Aluminum 30 kg 140C Iron 40 kg 60C 0 0 iron 1 2 alum 1 2 iron alum T mc T T T mc Substituting K 382 0 60 C 40 kg 0 45 kJkg K 140 C 3 0 kg0949 kJkg K 2 2 2 109 C T T T The total entropy change for this process is determined from 2221 kJK 413 K 30 kg 0949 kJkg K ln 382 K ln 2472 kJK 333 K 40 kg 045 kJkg K ln 382 K ln 1 2 avg alum 1 2 avg iron T T mc S T T mc S Thus 0251 kJK 2 221 2 472 alum iron total S S S preparation If you are a student using this Manual you are using it without permission 736 768 Problem 767 is reconsidered The effect of the mass of the iron block on the final equilibrium temperature and the total entropy change for the process is to be studied The mass of the iron is to vary from 10 to 100 kg The equilibrium temperature and the total entropy change are to be plotted as a function of iron mass Analysis The problem is solved using EES and the results are tabulated and plotted below Knowns T1iron 60 C miron 40 kg T1al 140 C mal 30 kg Cal 0949 kJkgK FromTable A3 at the anticipated average temperature of 450 K Ciron 045 kJkgK FromTable A3 at room temperature the only value available Analysis Treat the iron plus aluminum as a closed system with no heat transfer in no work out neglect changes in KE and PE of the system The final temperature is found from the energy balance Ein Eout DELTAEsys Eout 0 Ein 0 DELTAEsys mironDELTAuiron malDELTAual DELTAuiron CironT2iron T1iron DELTAual CalT2al T1al the iron and aluminum reach thermal equilibrium T2iron T2 T2al T2 DELTASiron mironCironlnT2iron273 T1iron273 DELTASal malCallnT2al273 T1al273 DELTAStotal DELTASiron DELTASal 10 20 30 40 50 60 70 80 90 100 005 01 015 02 025 03 035 04 045 miron kg Stotal kJK Stotal kJkg miron kg T2 C 008547 01525 02066 02511 02883 032 03472 03709 03916 041 10 20 30 40 50 60 70 80 90 100 1291 1208 1143 109 1047 1011 9798 9533 9302 91 10 20 30 40 50 60 70 80 90 100 90 95 100 105 110 115 120 125 130 miron kg T2 C PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 737 769 An iron block and a copper block are dropped into a large lake The total amount of entropy change when both blocks cool to the lake temperature is to be determined Assumptions 1 The water the iron block and the copper block are incompressible substances with constant specific heats at room temperature 2 Kinetic and potential energies are negligible Properties The specific heats of iron and copper at room temperature are ciron 045 kJkgC and ccopper 0386 kJkgC Table A3 Analysis The thermalenergy capacity of the lake is very large and thus the temperatures of both the iron and the copper blocks will drop to the lake temperature 15C when the thermal equilibrium is established Then the entropy changes of the blocks become 1571 kJK 353 K 20 kg 0386 kJkg K ln 288 K ln 4579 kJK 353 K 50 kg 045 kJkg K ln 288 K ln 1 2 avg copper 1 2 avg iron T T mc S T T mc S We take both the iron and the copper blocks as the system This is a closed system since no mass crosses the system boundary during the process The energy balance for this system can be expressed as Lake 15C Copper 20 kg 80C Iron 50 kg 80C by heat work Net en copper iron out potential etc energies in internal kinetic Change system and mass nsfer ergy tra out in U U U Q E E E 43 42 1 4243 1 uting or copper 2 1 iron 2 1 out T mc T T mc T Q Substit kJ 1964 288 K 20 kg 0386 kJkg K 353 288 K 50 kg 045 kJkg K 353 out Q Thus 6820 kJK 288 K kJ 1964 lake lakein lake T Q S Then the total entropy change for this process is 0670 kJK 6 820 1 571 4 579 lake copper iron total S S S S PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 738 770 An adiabatic pump is used to compress saturated liquid water in a reversible manner The work input is to be determined by different approaches Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 Heat transfer to or from the fluid is negligible Analysis The properties of water at the inlet and exit of the pump are Tables A4 through A6 001004 m kg 0 20690 kJkg MPa 15 001010 m kg 0 6492 kJkg 0 81 kJkg 191 0 kPa 10 3 2 2 1 2 2 3 1 1 1 1 1 v v h s s P s h x P PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course h h w 10 kPa 15 MPa pump a Using the entropy data from the compressed liquid water table 1510 kJkg 19181 20690 1 2 P b Using inlet specific volume and pressure values 1514 kJkg 0 001010 m kg15000 10kPa 3 1 2 1 P P P w v Error 03 c Using average specific volume and pressure values 1510 kJkg 10kPa 0 001004 m kg 15000 1 2 0 001010 3 1 2 avg P P P w v The results show that any of the method may be used to calculate reversible pump work Error 0 Discussion preparation If you are a student using this Manual you are using it without permission 739 Entropy Changes of Ideal Gases 771C No The entropy of an ideal gas depends on the pressure as well as the temperature 772C The entropy of a gas can change during an isothermal process since entropy of an ideal gas depends on the pressure as well as the temperature 773C The entropy change relations of an ideal gas simplify to s cp lnT2T1 for a constant pressure process and s cv lnT2T1 for a constant volume process Noting that cp cv the entropy change will be larger for a constant pressure process 774 For ideal gases cp cv R and 2 1 1 2 1 2 1 1 1 2 2 2 T P T P T P T P V V V V Thus 2 2 2 ln ln ln P P R T T R T T cv 1 2 1 2 1 1 1 2 1 1 2 1 2 1 2 1 2 1 2 ln ln ln ln ln ln P P R T T c T P T P R T T c R T T c s s p v v V V 775 For an ideal gas dh cp dT and v RTP From the second Tds relation P R dP T dT c T dP P RT T dP c T vdP T dh ds p p Integrating 1 2 1 2 1 2 ln ln P P R T T c s s p Since cp is assumed to be constant PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 740 776 Setting s 0 gives Cp R p p P P T T P P c R T T P P R T T c 1 2 1 2 1 2 1 2 1 2 1 2 ln ln 0 ln ln but k k p p p p P P T T c c k k k k c c c c R 1 1 2 1 2 Thus 1 since 1 1 v v 777 The entropy changes of helium and nitrogen is to be compared for the same initial and final states 1926 kJkgK R 20769 kJkgK Table A2a The specific heat of cp 1056 kJkgK Table A2b The gas constant of Analysis From the entropy change relation of an ideal gas Assumptions Helium and nitrogen are ideal gases with constant specific heats Properties The properties of helium are cp 5 nitrogen at the average temperature of 427272227C500 K is nitrogen is R 02968 kJkgK Table A2a 03826 kJkg K 2000 kPa 20769 kJkg Kln 200 kPa 273K 427 273K kJkg Kln 27 51926 ln ln 1 2 1 2 He P P R T T c s p PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 02113 kJkg K 2000 kPa 02968 kJkg Kln 200 kPa 273K 427 273K kJkg Kln 27 1056 ln ln 1 2 1 2 N2 P R T c s p 778 The entropy difference between the two states of air is to be determined Assumptions Air is an ideal gas with constant specific heats Properties The specific heat of air at the average temperature of 500502275 C 548 K 550 K is cp 1040 kJkgK Table A2b The gas constant of air is R 0287 kJkgK Table A2a Analysis From the entropy change relation of an ideal gas P T Hence helium undergoes the largest change in entropy 00478 kJkg K 2000 kPa 0287 kJkg Kln 100 kPa 273K 500 273K kJkg Kln 50 1040 ln ln 1 2 1 2 air P P R T T c s p preparation If you are a student using this Manual you are using it without permission 741 779E The entropy difference between the two states of air is to be determined Assumptions Air is an ideal gas with constant specific heats Properties The specific heat of air at the average temperature of 702502160F is cp 0241 BtulbmR Table A2Eb The gas constant of air is R 006855 BtulbmR Table A2Ea Analysis From the entropy change relation of an ideal gas 000323 Btulbm R 15 psia 006855 Btulbm Rln 40 psia 460R 70 460R Btulbm Rln 250 0241 ln ln 1 2 1 2 air P P R T T c s p Assumptions Nitrogen is an ideal gas with constant specific heats Properties The specific heat ratio of nitrogen at an anticipated average temperature of 450 K is k 1391 Table A2b 780 The final temperature of nitrogen when it is expanded isentropically is to be determined Analysis From the isentropic relation of an ideal gas under constant specific heat assumption 309 K 0 391 1 391 1 k 1 2 1 2 900 kPa 273 K 100 kPa 300 k P P T T iscussion The average air temperature is 5733092441 K which is sufficiently close to the assumed average mperature of 450 K Assumptions Air is an ideal gas with constant specific heats nalysis From the isentropic relation of an ideal gas under constant specific heat assumption D te 781E The final temperature of air when it is compressed isentropically is to be determined Properties The specific heat ratio of air at an anticipated average temperature of 400F is k 1389 Table A2Eb A 1095 R 0 389 1 389 1 1 2 1 2 15 psia 460 R 200 psia 70 k k P P T T Discussion The average air temperature is 53010952813 K 353F which is sufficiently close to the assumed average temperature of 400F PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 743 783 A cylinder contains N2 gas at a specified pressure and temperature The gas is compressed polytropically until the volume is reduced by half The entropy change of nitrogen during this process is to be determined Assumptions 1 At specified conditions N2 can be treated as an ideal gas 2 Nitrogen has constant specific heats at room temperature Properties The gas constant of nitrogen is R 02968 kJkgK Table A1 The constant volume specific heat of nitrogen at room temperature is cv 0743 kJkgK Table A2 Analysis From the polytropic relation PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 381 7 K 310 K 2 1 31 1 2 1 1 2 2 1 1 2 T T T v v 1 n n T v v Then the entropy change of nitrogen becomes 00384 kJK 02968 kJkg K ln 05 310 K 0743 kJkg K ln 3817 K kg 075 ln ln 1 2 1 2 avg 2 V V v R T T m c S N N2 PV 13 C preparation If you are a student using this Manual you are using it without permission 744 784 Problem 783 is reconsidered The effect of varying the polytropic exponent from 1 to 14 on the entropy v diagram nalysis The problem is solved using EES and the results are tabulated and plotted below atioVV2V1 kgK ation lnT2T1RlnRatioV 1V1mRT1 change of the nitrogen is to be investigated and the processes are to be shown on a common P A Given m075 kg P1140 kPa T137273 K n13 RatioV05 R Properties cv0743 kJkgK R0297 kJ Analysis T2T11RatioVn1 from polytropic rel DELTASmcv P 1 105 11 115 12 125 13 135 14 016 014 012 01 008 006 004 002 0 002 n S kJkg 1 105 11 115 12 125 13 135 14 0000104 01544 01351 01158 009646 007715 005783 003852 001921 S kJK n 0 01 02 03 04 05 06 07 100 200 300 400 500 600 700 800 900 1000 V m3kg P kPa PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 745 785E A fixed mass of helium undergoes a process from one specified state to another specified state The entropy change of helium is to be determined for the cases of reversible and irreversible processes Assumptions 1 At specified conditions helium can be treated as an ideal gas 2 Helium has constant specific heats at room temperature PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The gas constant of helium is R 04961 BtulbmR Table A1E The constant volume specific heat of helium is cv 0753 BtulbmR Table A2E Analysis From the idealgas entropy change relation 971 BtuR ln ln 1 1 ave He v v R T m c S ft lbm 50 04961 Btulbm R ln 10 ft lbm 540 R 15 lbm 0753 Btulbm R ln 660 R 3 3 2 2 v T isothermally until a final pressure The amount of heat transfer is to be Properties The specific heat of air at the given temperature of 127C 400 K is cp 1013 kJkgK Table A2b The gas constant of air is R 0287 kJkgK Table A2a Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as in 1 1 1 2 out in 1 2 out in system out in 0 since since KE PE 0 W Q T T u m u U W Q u m u U W Q E E E 43 42 1 4243 1 work output during this isothermal process is He T1 540 R T2 660 R The entropy change will be the same for both cases 786 Air is expanded in a pistoncylinder device determined Assumptions Air is an ideal gas with constant specific heats potential etc energies Change in internal kinetic by heat work and mass Net energy transfer out The boundary 796 kJ 100 kPa 273 Kln 200 kPa 1 kg0287 kJkg K127 ln 2 1 out P P mRT W Thus 796 kJ out in W Q preparation If you are a student using this Manual you are using it without permission 749 790 Air is compressed in a pistoncylinder device in a reversible and adiabatic manner The final temperature and the work are to be determined for the cases of constant and variable specific heats Assumptions 1 At specified conditions air can be treated as an ideal gas 2 The process is given to be reversible and adiabatic and thus isentropic Therefore isentropic relations of ideal gases apply Properties The gas constant of air is R 0287 kJkgK Table A1 The specific heat ratio of air at the anticipated average temperature of 425 K is k 1393 and R 0730 kJkgK Table A2a Analysis a Assuming constant specific heats the ideal gas isentropic relations give 5649 K 0 393 1 393 1 1 2 1 2 90 kPa 900 kPa K 295 k k P P T T Since av 430 K 565 2 295 g T the assumed average temperature 425 K is close enough to his value We take the air in the cylinder as the system The energy balance for this stationary closed system can be expressed as 1 2 1 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T u m u U W E E E v 43 42 1 4243 1 Thus 1970 kJkg 295 K 0730 kJkg K 5649 1 2 avg in v T T c w b Assuming variable specific heats the final temperature can be determined using the relative pressure data Table A17 and 40809 kJkg 64 9 13068 90 kPa 13068 kPa 900 21049 kJkg 1 3068 K 295 2 2 1 2 1 1 1 2 1 u T P P P P u P T r r r K 5 hen the work input becomes T 1976 kJkg 21049 kJkg 40809 1 2 in u u w AIR Reversible PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 750 791 Problem 790 is reconsidered The work done and final temperature during the compression process are to be calculated and plotted as functions of the final pressure for the two cases as the final pressure varies from 100 kPa to 1200 kPa Analysis The problem is solved using EES and the results are tabulated and plotted below Procedure ConstPropSolP1T1P2GasWorkinConstPropT2ConstProp CPSPECHEATGasT27 MMMOLARMASSGas Ru8314 kJkmolK RRuMM CV CP R k CPCV T2 T1273P2P1k1k T2ConstPropT2273 C DELTAu CvT2T1273 WorkinConstProp DELTAu End Knowns P1 90 kPa T1 22 C P2 900 kPa Analysis Treat the pistoncylinder as a closed system with no heat transfer in neglect changes in KE and PE of the air The process is reversible and adiabatic thus isentropic The isentropic work is determined from ein eout DELTAesys eout 0 kJkg ein Workin DELTAEsys u2 u1 u1 INTENERGYairTT1 v1 volumeairPP1TT1 s1 entropyairPP1TT1 The process is reversible and adiabatic or isentropic s2 s1 u2 INTENERGYairPP2ss2 T2isen temperatureairPP2ss2 Gas air Call ConstPropSolP1T1P2GasWorkinConstPropT2ConstProp PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 751 P2 kPa Workin kJkg WorkinConstProp kJkg T2ConstProp C T2isen C 100 200 300 400 500 600 700 800 900 1000 1100 1200 6467 5429 8709 1128 1343 1529 1694 1843 198 2105 2223 2332 6469 5425 8691 1125 1338 1523 1687 1835 197 2095 2211 232 3101 9759 1431 1787 2085 2342 257 2777 2965 3139 3301 3453 3101 9742 1426 1778 207 2321 2543 2742 2922 3089 3242 3386 0 200 400 600 800 1000 1200 0 50 100 150 200 250 P2 kPa Workin kJkg 0 200 400 600 800 1000 1200 0 50 100 150 200 250 300 350 P2 kPa T2 C PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 752 792 An insulated rigid tank contains argon gas at a specified pressure and temperature A valve is opened and argon escapes until the pressure drops to a specified value The final mass in the tank is to be determined Assumptions 1 At specified conditions argon can be treated as an ideal gas 2 The process is given to be reversible and adiabatic and thus isentropic Therefore isentropic relations of ideal gases apply Properties The specific heat ratio of argon is k 1667 Table A2 Analysis From the ideal gas isentropic relations 219 0 K 450 kPa 303 K 1 1 2 P T T 200 kPa 0667 1667 1 2 k k P The final mass in the tank is determined from the ideal gas relation 246 kg 4 kg 450 kPa 219 K 200 kPa 303 K 1 2 1 2 1 2 2 2 1 1 2 1 P T m P T m m RT m RT P P V V ARGON 4 kg 450 kPa 30C PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 753 793 Problem 792 is reconsidered The effect of the final pressure on the final mass in the tank is to be investigated nalysis The problem is solved using EES and the results are tabulated and plotted below 3kgK a kPa t the final pressure in the tank by using the isentropic relation k1k273 2V2m2RT2273 as the pressure varies from 450 kPa to 150 kPa and the results are to be plotted A Knowns cp 05203 kJkgK cv 03122 kJkgK R02081 kPam P1 450 kP T1 30 C m1 4 kg P2 150 Analysis We assume the mass that stays in the tank undergoes an isentropic expansion process This allows us to determine the final temperature of that gas a k cpcv T2 T1273P2P1 V2 V1 P1V1m1RT1273 P 150 200 250 300 350 400 450 2 24 28 32 36 4 P2 kPa m2 kg 150 200 250 300 350 400 450 3727 4 2069 2459 2811 3136 344 m2 kg P2 kPa PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 757 797 Air is expanded in a pistoncylinder device until a final pressure The maximum work input is given The mass of air in the device is to be determined Assumptions Air is an ideal gas with constant specific heats Properties The properties of air at 300 K is cv 0718 kJkgK and k 14 Table A2a Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course system out in E E E 43 42 1 4243 1 For the minimum work input to the compressor the process must be reversib well as adiabatic ie isentropic This being the case the exit temperature will be KE PE 0 since out potential etc energies Change in internal kinetic by heat work and mass Net energy transfer Q u m u U W T s 2 1 600 kPa 100 kPa 1 2 le as 419 5 K 600 kPa 10 kPa 1 2 k k P 0 273 K 427 41 40 1 1 2 T P T ubstituting into the energy balance equation gives S 497 kg 419 5 K 0 718 kJkg K700 1000 kJ 2 1 out 2 1 2 1 out T T c W m T T mc u m u W v v preparation If you are a student using this Manual you are using it without permission 759 799 Air is expanded adiabatically in a pistoncylinder device The entropy change is to be determined and it is to be discussed if this process is realistic Assumptions 1 Air is an ideal gas with constant specific heats Properties The properties of air at 300 K are cp 1005 kJkgK cv 0718 kJkgK and k 14 Also R 0287 kJkgK Table A2a Analysis a We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as KE PE 0 since 1 2 out 1 2 out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc W Q u m u U W E E E v 43 42 1 4243 1 Solving for the final temperature 532 9 K 5 kg 0 718 kJkg K 600 kJ 273 K 427 out 1 2 2 1 out v v mc W T T T T mc W From the entropy change relation of an ideal gas 0240 kJkg K 600 kPa 0287 kJkg Kln 100 kPa 700 K kJkg Kln 532 9 K 1005 ln ln 1 2 1 2 air P P R T T c s p b Since the entropy change is positive for this adiabatic process the process is irreversible and realistic PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 760 7100 Air contained in a constantvolume tank s cooled to ambient temperature The entropy changes of the air and the universe due to this process are to be determined and the process is to be sketched on a Ts diagram Assumptions 1 Air is an ideal gas with constant specific heats Properties The specific heat of air at room temperature is cv 0718 kJkgK Table A2a Air 5 kg 327C 100 kPa Analysis a The entropy change of air is determined from 2488 kJK K PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 273 327 273 K kg0718 kJkgKln 27 5 ln T2 mc ut T T mc Q v The entropy change of the surroundings is S 1 air T v 1 T s 2 1 2 air surr 327ºC 27ºC b An energy balance on the system gives 27 5 kg0718 kJkgK327 2 1 o 1077 kJ 359 kJK 300 K kJ 1077 surr out surr T Q s The entropy change of universe due to this process is 110 kJK 3 59 2 488 surr air total gen S S S S preparation If you are a student using this Manual you are using it without permission 761 7101 A container filled with liquid water is placed in a room and heat transfer takes place between the container and the air in the room until the thermal equilibrium is established The final temperature the amount of heat transfer between the water and the air and the entropy generation are to be determined Assumptions 1 Kinetic and potential energy changes are negligible 2 Air is an ideal gas with constant specific heats 3 The room is wellsealed and there is no heat transfer from the room to the surroundings 4 Sea level atmospheric pressure is assumed P 1013 kPa Properties The properties of air at room temperature are R 0287 kPam3kgK cp 1005 kJkgK cv 0718 kJkgK The specific heat of water at room temperature is cw 418 kJkgK Tables A2 A3 Analysis a The mass of the air in the room is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 111 5 kg 273 K 0287 kPa m kg K12 3 1 a a RT m 1013 kPa90 m 3 PV An energy balance on the system that consists of the water in the container and e air in the room gives the final equilibrium temperature 2 2 2 1 2 1 2 12 111 5 kg0718 kJkgK 95 0 T T T T T m c T T c m a a w w w v The heat transfer to the air is Water 45 kg 95C Room 90 m3 12C th 45 kg418 kJkgK 0 702C b 4660 kJ 111 5 kg0718 kJkgK702 12 1 2 a a T T m c Q v c The entropy generation associated with this heat transfer process may be obtained by calculating total entropy change hich is the sum of the entropy changes of water and the air w 1311 kJK 273 K 95 273 K 45 kg418 kJkgKln 702 ln 1 2 w w w w T T m c S 122 kPa 90 m 273 K kg0287 kPa m kg K702 1115 3 3 2 2 V m RT P a 1488 kJK 1013 kPa 0287 kJkgKln 122 kPa 273 K 12 273 K kg 1005 kJkgKln 702 1115 ln ln 1 2 1 2 P P R T T c m S a p a a 177 kJK 1311 1488 total gen a w S S S S preparation If you are a student using this Manual you are using it without permission 763 7103E Air is charged to an initially evacuated container from a supply line The minimum temperature of the air in the container after it is filled is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Air is an ideal gas with constant specific heats 3 Kinetic and potential energies are negligible 4 There are no work interactions involved 5 The tank is wellinsulated and thus there is no heat transfer Properties The specific heat of air at room temperature is cp 0240 BtulbmR Table A2Ea Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and entropy balances for this uniformflow system can be expressed as Mass balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m m m m m m m i i Entropy balance i i i i m s s m m s nimum temperature will result when the equal sign applies Noting that P2 Pi we have 2 1 2 system out in Air 200 psia 100F Evacuated m 0 0 2 2 1 1 2 2 mese m s s m Combining the two balances 0 0 2 2 2 2 i i s s m s m s The mi 0 ln 0 ln ln 2 2 2 2 i p i i p i T T c P P R T T c s s Then 100F iT T2 preparation If you are a student using this Manual you are using it without permission 764 Reversible SteadyFlow Work 7104C The work associated with steadyflow devices is proportional to the specific volume of the gas Cooling a gas during compression will reduce its specific volume and thus the power consumed by the compressor 7105C Cooling the steam as it expands in a turbine will reduce its specific volume and thus the work output of the turbine Therefore this is not a good proposal 7106C We would not support this proposal since the steadyflow work input to the pump is proportional to the specific volume of the liquid and cooling will not affect the specific volume of a liquid significantly 7107E Air is compressed isothermally in a reversible steadyflow device The work required is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 There is no heat transfer associated with the process 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with constant specific heats Properties The gas constant of air is R 006855 BtulbmR Table A1E Analysis Substituting the ideal gas equation of state into the reversible steadyflow work expression gives Compressor 80 psia 90F Air 13 psia 90F 685 Btulbm 13 psia 460 Kln 80 psia Btulbm R90 006855 ln 1 2 2 1 2 1 in P P RT P dP RT dP w v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 765 7108 Saturated water vapor is compressed in a reversible steadyflow device The work required is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 There is no heat transfer associated with the process 3 Kinetic and potential energy changes are negligible Analysis The properties of water at the inlet state are PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 150 C 1 P T Noting that the specific volume remains constant the reversible steadyflow work expression gives Compressor 1 MPa Water 150C sat vap Table A 4 0 39248 m kg 1 3 1 1 v x 47616 kPa 1 2056 kJkg 3 3 1 2 1 2 1 in kPa m 1 1kJ m kg1000 47616kPa 039248 P P dP w v v 7109E The reversible work produced during the process shown in the figure is to be determined o the areas to the left of the reversible process line on the Pv diagram The work done during the process 23 is zero Then Assumptions The process is reversible Analysis The work produced is equal t P psia 15 1 1 33 300 2 3 2 v ft3lbm 1134 Btulbm 3 3 1 2 2 1 2 1 12 13 404 psia ft 5 1Btu 15psia 300 2 33 ft lbm 1 2 0 P P dP w w v v v preparation If you are a student using this Manual you are using it without permission 768 7113 Problem 7112 is reconsidered The effect of the quality of the steam at the turbine exit on the net work output is to be investigated as the quality is varied from 05 to 10 and the net work output us to be plotted as a function of this quality Analysis The problem is solved using EES and the results are tabulated and plotted below Knowns WorkFluid SteamIAPWS P1 10 kPa x1 0 P2 5000 kPa x4 10 Pump Analysis T1temperatureWorkFluidPP1x0 v1volumeworkFluidPP1x0 h1enthalpyWorkFluidPP1x0 s1entropyWorkFluidPP1x0 s2 s1 h2enthalpyWorkFluidPP2ss2 T2temperatureWorkFluidPP2ss2 The Volume function has the same form for an ideal gas as for a real fluid v2volumeWorkFluidTT2pP2 Conservation of Energy SSSF energy balance for pump neglect the change in potential energy no heat transfer h1Wpump h2 Also the work of pump can be obtained from the incompressible fluid steadyflow result Wpumpincomp v1P2 P1 Conservation of Energy SSSF energy balance for turbine neglecting the change in potential energy no heat transfer P4 P1 P3 P2 h4enthalpyWorkFluidPP4xx4 s4entropyWorkFluidPP4xx4 T4temperatureWorkFluidPP4xx4 s3 s4 h3enthalpyWorkFluidPP3ss3 T3temperatureWorkFluidPP3ss3 h3 h4 Wturb Wnetout Wturb Wpump PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 769 x4 Wnetout kJkg 05 055 06 065 07 075 08 085 09 095 1 5556 6374 7192 801 8828 9712 1087 1240 1442 1699 2019 05 06 07 08 09 1 700 1050 1400 1750 2100 x4 Wnetout kJkg 00 11 22 33 44 55 66 77 88 99 110 0 100 200 300 400 500 600 700 800 900 1000 1100 s kJkgK T C 5000 kPa 10 kPa 02 04 06 08 Steam IAPWS 3 4 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 771 7115E Helium gas is compressed from a specified state to a specified pressure at a specified rate The power input to the compressor is to be determined for the cases of isentropic polytropic isothermal and twostage compression Assumptions 1 Helium is an ideal gas with constant specific heats 2 The process is reversible 3 Kinetic and potential energy changes are negligible Properties The gas constant of helium is R 26805 psiaft3lbmR 04961 BtulbmR The specific heat ratio of helium is k 1667 Table A2E PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis The mass flow rate of helium is 01095 lbms psia ft lbm R 545 R 26805 psia 10 ft s 16 3 3 1 1 1 RT P m V a Isentropic compression with k 1667 since 1 hp 07068 Btus 74 Btus 91 1 16 psia 120 psia 1 1667 lbms 1667 04961 Btulbm R 545 R 01095 1 1 1 2 1 compin P m k W 7 0667166 1 1298 hp k k P kRT b Polytropic compression with n 12 He 2 1 W 10 ft3s 1 hp 07068 Btu since 120 psia 04961 Btulbm R 545 R 12 1 0212 1 1 2 1003 hp n n P P nRT 1 1 compin m n W 89 Btus 70 1 16 psia 1 12 lbms 01095 s c Isothermal compression 8442 hp W 5967 Btus 16 psia 01095 lbms 04961 Btulbm R 545 R ln 120 psia ln 1 2 compin P P mRT d Ideal twostage compression with intercooling n 12 In this case the pressure ratio across each stage is the same and its value is determined from 4382 psia 16 psia 120 psia 1 2 P P Px The compressor work across each stage is also the same thus total compressor work is twice the compression work for a single stage 1 hp 07068 Btus since 97 Btus 64 1 14 psia 43 82 psia 1 12 01095 lbms 12 04961 Btulbm R 545 R 2 1 1 2 2 0212 1 1 1 compI in comp 9192 hp n n x P P n mnRT mw W preparation If you are a student using this Manual you are using it without permission 772 7116E Problem 7115E is reconsidered The work of compression and entropy change of the helium is to be evaluated and plotted as functions of the polytropic exponent as it varies from 1 to 1667 Analysis The problem is solved using EES and the results are tabulated and plotted below Given P116 psia T185460 V1dot10 ft3s P2120 psia n12 Properties R04961 BtulbmR R126805 psiaft3lbmR k1667 cp125 BtulbmR Analysis mdotP1V1dotR1T1 WdotcompinamdotkRT1k1P2P1k1k1ConvertBtus hp WdotcompinbmdotnRT1n1P2P1n1n1ConvertBtus hp WdotcompincmdotRT1lnP2P1ConvertBtus hp PxsqrtP1P2 Wdotcompind2mdotnRT1n1PxP1n1n1ConvertBtus hp Entropy change T2T1P2P1n1n DELTASHemdotcplnT2T1RlnP2P1 n Wcompina hp Wcompinb hp Wcompinc hp Wcompind hp SHe BtusR 1 11 12 13 14 15 16 1667 1298 1298 1298 1298 1298 1298 1298 1298 8442 9264 1003 1075 1141 1203 1261 1298 8442 8442 8442 8442 8442 8442 8442 8442 8442 8841 9192 9504 9782 1003 1026 104 01095 00844 00635 004582 003066 001753 0006036 00008937 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 773 09 1 11 12 13 14 15 16 17 80 85 90 95 100 105 110 115 120 125 130 135 n Wcompin hp Twostage polytropic isothermal polytropic isentropic 1 11 12 13 14 15 16 17 012 01 008 006 004 002 0 002 n SHe BtusR PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 774 7117 Water mist is to be sprayed into the air stream in the compressor to cool the air as the water evaporates and to reduce the compression power The reduction in the exit temperature of the compressed air and the compressor power saved are to be determined Assumptions 1 Air is an ideal gas with variable specific heats 2 The process is reversible 3 Kinetic and potential energy changes are negligible 3 Air is compressed isentropically 4 Water vaporizes completely before leaving the compressor 4 Air properties can be used for the airvapor mixture Properties The gas constant of air is R 0287 kJkgK Table A1 The specific heat ratio of air is k 14 The inlet enthalpies of water and air are Tables A4 and A17 hw1 hf20C 8329 kJkg hfg20C 24539 kJkg and ha1 h300 K 30019 kJkg Analysis In the case of isentropic operation thus no cooling or water spray the exit temperature and the power input to the compressor are 6102 K 100 kPa 300 K 1200 kPa 41 1 41 2 1 1 2 1 2 T P P T T k k 1200 kPa100 kPa 1 14 kgs 12 When water is sprayed we first need to check the accuracy of the assumption that the water vaporizes completely in the compressor In the limiting case the compression will be isothermal at the compressor inlet temperature and the water will be a saturated vapor To avoid the complexity of dealing with two fluid streams and a gas mix 654 3 kW 1 0287 kJkg K 300 K 14 1 1 0414 1 1 2 1 in comp k P k P k mkRT W ture we disregard water in the air stream other than the he The rate of heat absorption of water as it evaporates at the inlet mperatur completely is he minimum power input to the compressor is He 2 1 W 100 kPa 300 K 1200 kPa Water 20C mass flow rate and assume air is cooled by an amount equal to t enthalpy change of water te e kgs24539 kJkg 4908 kW 20 20 C coolingmax mwhfg Q T 449 3 kW 100 kPa kgs028 7 kJkg K300 K ln 1200 kPa 12 ln 1 2 compinmin P P mRT W This corresponds to maximum cooling from the air since at constant temperature h 0 and thus hich is close to 4908 kW Therefore the assumption that all the water vapori s approximately valid Then the be polytropic and the water to be a saturated vapor at t a unique solution and we ytropic exponent to obtain a solution Of ection Analysis We take the compressor exit temperature to be T2 200C 473 K Then hw2 hg200C 27920 kJkg and ha2 h473 K 4753 kJkg Then 3 kW 449 in out W Q w zes i reduction in required power input due to water spray becomes 205 kW 449 3 654 3 comp isothermal comp isentropic compin W W W Discussion can be ignored At constant temperature h 0 and thus 449 3 kW in out W Q corresponds to maximum cooling from the air which is less than 4908 kW Therefore the assumption that all the water vaporizes is only roughly valid As an alternative we can assume the compression process to the compressor exit temperature and disregard the remaining liquid But in this case there is no will have to select either the amount of water or the exit temperature or the pol course we can also tabulate the results for different cases and then make a sel Sample PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 775 1 224 100 kPa 1200 kPa 300 K 473 K 1 1 1 2 1 2 n P P T T n n n n PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 570 kW 300K kgs 1224 0287 kJkg K 473 12 1 1224 1 1 1 1 2 1 2 T T mn P P m n W comp in Energy balance h m h W Q h m h Q W ting that this heat is absorbed by water the rate at which water evaporates in the compressor becomes 1 1 nR nRT n n 202 30019 kgs4753 12 569 7 kW 1 2 compin out 1 2 out compin kW 0 No 0 0746 kgs 8329 kJkg 2792 0 202 0 kJs 1 2 w w Then the reductions in the exit temperature and compressor power input become T T T W W W comp in comp com 2 2 2 610 2 473 inwater 1 2 inwater outair w w w w h h Q m h h m Q Q 6543 570 p isentropic water cooled isentropic 1372 C water cooled 843 kW ote that selecting a different compressor exit temperature T2 will result in different values 7118 A waterinjected compressor is used in a gas turbine power plant It is claimed that the power output of a gas turbine will increase when water is injected into the compressor because of the increase in the mass flow rate of the gas air water vapor through the turbine This however is not necessarily right since the compressed air in this case enters the combustor at a low temperature and thus it absorbs much more heat In fact the cooling effect will most likely dominate and cause the cyclic efficiency to drop N preparation If you are a student using this Manual you are using it without permission 776 Isentropic Efficiencies of SteadyFlow Devices 7119C The ideal process for all three devices is the reversible adiabatic ie isentropic process The adiabatic efficiencies of these devices are defined as insentropi c exit kinetic energy actual exit kineticenergy and actual work input insentropi c work input insentropi c work output actual work output N C T η η η 7120C No because the isentropic process is not the model or ideal process for compressors that are cooled intentionally f reversibilities Therefore the actual exit state has to be on the righthand side of the isentropic exit state 7121C Yes Because the entropy of the fluid must increase during an actual adiabatic process as a result o ir PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 777 7122E Steam is compressed in an adiabatic closed system with an isentropic efficiency of 80 The work produced and the final temperature are to be determined Assumptions 1 Kinetic and potential energy changes are negligible 2 The device is adiabatic and thus heat transfer is negligible Analysis We take the steam as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as 2 1 out 1 2 out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u u w u m u U W E E E 43 42 1 4243 1 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course From the steam tables Tables A5 and A6 1068 4 Btulbm 0 9961 80729 19 298 0 9961 1 12888 0 47427 1 7816 MPa 1 1 7816 Btulbm R 650 F 1 1 s T 1233 7 Btulbm psia 100 2 2 2 1 fg s f s fg f s u x u u s s s x P u P T s 1 2s 100 psia 10 psia 2 1 2 1 2 2 s s s s The work input during the isentropic process is 165 3 Btulbm 1068 4 Btulbm 1233 7 2 1 out s s u u w The actual work input is then 1322 Btulbm 0 80165 3 Btulbm out isen aout ws w η he internal energy at the final state is determined from Using this internal energy and the pressure at the final state the temperature is determined from Table A6 to be T 1101 4 Btulbm 132 2 Btulbm 1233 7 out 1 2 2 1 out w u u u u w 2746F 2 2 2 4 Btulbm 1101 10 psia T u P preparation If you are a student using this Manual you are using it without permission 781 7126E Combustion gases enter an adiabatic gas turbine with an isentropic efficiency of 82 at a specified state and leave at a specified pressure The work output of the turbine is to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 The device is adiabatic and thus heat transfer is negligible 4 Combustion gases can be treated as air that is an ideal gas with variable specific heats Analysis From the air table and isentropic relations T h Pr 1 1 1 174 0 2000 R 50471 Btu lbm 4173 Btulbm 870 1740 120 psia 60 psia 2s 1 2 1 2 h P P P P r r There is only one inlet and one exit and thus m m m 1 2 We take the actual turbine as the sy th f stem which is a control volume since mass crosses the boundary The energy balance for is steady low system can be expressed as h m h W Q mh W h m oting that wa ηTws the work output of the turbine per unit mass is determined from out in E E 0 pe ke since 2 1 out a 2 aout 1 N 717 Btulbm 4173 Btulbm 50471 082 wa AIR ηT 82 1 2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 786 7132 Problem 7131 is reconsidered The problem is to be solved by considering the kinetic energy and by ssor exit pipe inside diameter is 2 cm nalysis The problem is solved using EES and the solution is given below agram window mpressor adiabatic efficiency e diagram window a dyflow adiabatic process x1Real fluid equ at the sat vapor state 2 perature of ideal state needed only for plot on of mass Vel2v2 Mass flow rate AVelv Aratio A1A2 window 21000 ntropic efficiency EtacWisenWact nd P d equ at the known outlet h and P mdot1h1 Wdotcnoke mdot2h2 assuming an inlettoexit area ratio of 15 for the compressor when the compre A Input Data from di P1 100 kPa P2 1000 kPa Voldot1 07 m3min Etac 087 Co Aratio 15 d2 002 m System Control volume containing the compressor see th Property Relation Use the real fluid properties for R134 Process Steadystate stea FluidR134a Property Data for state 1 T1temperatureFluidPP1x1Real fluid equ at the sat vapor state h1enthalpyFluid PP1 x1Real fluid equ at the sat vapor state s1entropyFluid PP1 x1Real fluid equ at the sat vapor state v1volumeFluid PP1 Property Data for state 2 ss1s1 Ts1T1 needed for plot ss2s1 for the ideal isentropic process across the compressor hs2ENTHALPYFluid PP2 sss2Enth Fluid PP2 sss2Tem alpy 2 at the isentropic state 2s and pressure P Ts2Temperature Steadystate steadyflow conservati mdot1 mdot2 mdot1 Voldot1v160 Voldot1v1Voldot2v2 Vel2Voldot2A260 A2 pid224 AratioVel1v1 AratioA1A2 Steadystate steadyflow conservation of energy adiabatic compressor see diagram mdot1h1Vel1221000 Wdotc mdot2h2Vel22 Definition of the compressor ise Etac hs2h1h2h1 Knowing h2 the other properties at state 2 can be found v2volumeFluid PP2 hh2v2 is found at the actual state 2 knowing P an atureFluid PP2hh2Real fluid equ for T at the known outlet h a d h T2temper s2entropyFluid PP2 hh2 Real flui TexitT2 Neglecting the kinetic energies the work is PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 787 SOLUTION Aratio15 d2002 m Etac087 FluidR134a mdot1006059 kgs mdot2006059 kgs Texit5651 C Voldot107 m3 min Voldot2008229 m3 min Wdotc333 kW Wdotcnoke3348 kW 025 000 025 050 075 100 125 150 175 90 60 30 0 30 60 90 120 150 s kJkgK T C 1000 kPa 100 kPa R134a 1 2 2s PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 789 7134 Air is compressed by an adiabatic compressor from a specified state to another specified state The isentropic efficiency of the compressor and the exit temperature of air for the isentropic case are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 The device is adiabatic and thus heat transfer is negligible 4 Air is an ideal gas with variable specific heats Analysis a From the air table Table A17 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course P AIR 2 1 T h T h r a 1 1 2 2 1 1 386 300 K 30019 kJ kg 550 K 55474 kJ kg From the isentropic relation 50872 kJkg 8754 1386 95 kPa 1 1 2 r r P P P 600 kPa 2 2 h s P hen the isentropic efficiency becomes T η C h h a 2 1 554 74 30019 0819 819 h s h 2 1 508 72 30019 If the ocess were isentropic the exit temperature would be b pr 5055 K s s T h 2 2 50872 kJkg preparation If you are a student using this Manual you are using it without permission 792 7137E Problem 7136E is reconsidered The effect of varying the nozzle isentropic efficiency from 08 to 10 on the nalysis The problem is solved using EES and the results are tabulated and plotted below ergy balance for turbine neglecting the change in potential energy no heat orkFluidPP1TT1 uidPP2sss2 kes2 hs2 kes2convertft2s2Btulbm orkFluidhh2 2answer T2 η exit temperature and pressure of the air is to be investigated and the results are to be plotted A Knowns WorkFluid Air P1 45 psia T1 940 F Vel2 650 fts Vel1 0 fts etanozzle 085 Conservation of Energy SSSF en transfer h1enthalpy pyW WorkFluidTT1 s1entro Ts1 T1 s2 s1 ss2 s1 hs2enthalpyWorkFluidTTs2 Ts2temperatur eWorkFl etanozzle ke2 ke1 Vel122 ke2Vel222 h1ke1convertft2s2Btulbm h2 ke2convertft2s2Btulbm h1 ke1convertft2s2Btulbm T2temperatureW P2answer P2 T nozzle P2 psia T2 F Ts2 F 08 082 084 0 86 088 09 092 094 096 098 1 4117 9076 9076 4025 4036 4047 4057 4067 4076 4085 4093 4101 4109 9076 9076 9076 9076 9076 9076 9076 9076 9076 9076 8995 9005 9014 9023 9032 904 9048 9056 9063 907 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 793 08 084 088 092 096 1 402 404 406 408 41 412 ηnozzle P2 psia 08 084 088 092 096 1 899 900 901 902 903 904 905 906 907 908 909 910 ηnozzle T2 F Ts2 T2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 7103 7147 Heat is lost from Refrigerant134a as it is throttled The exit temperature of the refrigerant and the entropy generation are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis The properties of the refrigerant at the inlet of the device are Table A13 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 120 0 kPa 1 h P he enthalpy of the refrigerant at the exit of the device is ow the properties at the exit state may be obtained from the R134a tables 2 2 s h ropy generation associated with this process may be obtained by adding the entropy change of R134a as it flows in e device and the entropy change of the surroundings 0 39424 kJkg 40 C 1 1 s T 10823 kJkg 1 R134a 1200 kPa 40C 200 kPa q T 10773 kJkg 50 10823 out 1 2 q h h N 20 0 kPa 2 2 T P 1009 C 0 41800 kJkgK 73 kJkg 107 The ent th 0 02375 kJkgK 0 39424 0 41800 1 2 R134a s s s 0 001678 kJkgK 273 K 25 kJkg 50 surr out surr T q s 002543 kJkgK 0 001678 0 02375 surr R 134a total gen s s s s preparation If you are a student using this Manual you are using it without permission 7108 7152 Stainless steel ball bearings leaving the oven at a uniform temperature of 900C at a rate of 1100 min are exposed to air and are cooled to 850C before they are dropped into the water for quenching The rate of heat transfer from the ball to the air and the rate of entropy generation due to this heat transfer are to be determined Assumptions 1 The thermal properties of the bearing balls are constant 2 The kinetic and potential energy changes of the balls are negligible 3 The balls are at a uniform temperature at the end of the process Properties The density and specific heat of the ball bearings are given to be ρ 8085 kgm3 and cp 0480 kJkgC Analysis a We take a single bearing ball as the system The energy balance for this closed system can be expressed as 2 1 out 1 2 ball out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T Q u m u U Q E E E 43 42 1 4243 1 Furnace Steel balls 900C The total amount of heat transfer from a ball is 0 5925 kJball 850 C 0 02469 kg 0 480 kJkg C900 0 02469 kg 6 8085 kgm 6 m ρ ρV 0 018 m 2 1 out 3 3 3 T mc T Q D π π ecomes b We again take a single bearing ball as the system The entropy generated during this process can be determined by pplying an entropy balance on an extended system that includes the ball and its immediate surroundings so that the ry temperature of the extended system is at 20C at all times Then the rate of heat transfer from the balls to the air b 1086 kW 651 8 kJmin 0 5925 kJball 1100 ballsmin out per ball ball total Q n Q Therefore heat is lost to the air at a rate of 1086 kW a bounda system out gen system gen out in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in S T Q S S S T Q S S S S b b 43 42 1 4243 1 where 0 000516 2 kJK 900 273 0 02469 kg 0 480 kJkgKln 850 273 ln 1 2 avg 1 2 system T T mc s m s S Substituting 0 001506 kJK per ball 0 0005162 kJK 293 K 05925 kJ system out gen S T Q S b Then the rate of entropy generation becomes 0 001506 kJK ball1100 ballsmin 1657 kJminK 002761 kWK ball gen gen n S S PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 7109 7153 An egg is dropped into boiling water The amount of heat transfer to the egg by the time it is cooked and the amount of entropy generation associated with this heat transfer process are to be determined Assumptions 1 The egg is spherical in shape with a radius of r0 275 cm 2 The thermal properties of the egg are constant 3 Energy absorption or release associated with any chemical andor phase changes within the egg is negligible 4 There are no changes in kinetic and potential energies Properties The density and specific heat of the egg are given to be ρ 1020 kgm3 and cp 332 kJkgC Analysis We take the egg as the system This is a closes system since no mass enters or leaves the egg The energy balance for this closed system can be expressed as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course system out in T mc T u m u U Q E E E 43 42 1 4243 1 he mass of the egg and the amount of heat transfer become Boiling Water potential etc energies Change in internal kinetic by heat work and mass Net energy transfer 1 2 1 2 egg in Then t 183 kJ 8 C 0 0889 kg 3 32 kJkg C70 0 0889 kg 6 0 055 m 1020 kgm 6 1 2 in 3 T T mc Q D m p π ρ π ρV 3 3 e again take a single egg as the system The entropy generated during this process can be determined by applying an balance on an extended system that includes the egg and its immediate surroundings so that the boundary ture of the extended system is at 97C at all times Egg 8C W entropy tempera system in gen system gen in in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in S T Q S S S T Q S S S S b b 43 42 1 4243 1 where 0 0588 kJK 8 273 0 0889 kg 3 32 kJkgK ln 70 273 ln 1 2 avg 1 2 system T T mc s m s S Substituting per egg 0 0588 kJK 370 K 183 kJ system in gen 000934 kJK S T Q S b preparation If you are a student using this Manual you are using it without permission 7110 7154 Long cylindrical steel rods are heattreated in an oven The rate of heat transfer to the rods in the oven and the rate of entropy generation associated with this heat transfer process are to be determined Assumptions 1 The thermal properties of the rods are constant 2 The changes in kinetic and potential energies are negligible Properties The density and specific heat of the steel rods are given to be ρ 7833 kgm3 and cp 0465 kJkgC Analysis a Noting that the rods enter the oven at a velocity of 3 mmin and exit at the same velocity we can say that a 3 m long section of the rod is heated in the oven in 1 min Then the mass of the rod heated in 1 minute is 184 6 kg m 4 10 7833 kgm 3m 4 2 3 2 π π ρ ρ ρ D L LA m V PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course nsfer energy tra out in T mc T u m u U Q E E E 43 42 1 4243 1 Steel rod 30C Oven 900C We take the 3m section of the rod in the oven as the system The energy balance for this closed system can be expressed as 1 2 1 2 rod in potential etc energies in internal kinetic Change system by heat work and mass Net Substituting 57512 kJ 30 C 184 6 kg 0 465 kJkg C700 1 2 in T mc T Q b We again take the 3m long section of the rod as the system The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the rod and its immediate surroundings so at the boundary temperature of the extended system is at 900C at all times Noting that this much heat is transferred in 1 min the rate of heat transfer to the rod becomes min 57512 kJmin 9585 kW 57512 kJ1 in in t Q Q th system in gen system gen in in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in S T Q S S S T Q S S S S b b 43 42 1 4243 1 where 100 1 kJK 30 273 184 6 kg 0 465 kJkgKln 700 273 ln 2 avg 1 2 system T T mc s m s S 1 Substitutin g 51 1 kJK 100 1 kJK 900 273 K 57512 kJ system in gen S T Q S b Noting that this much entropy is generated in 1 min the rate of entropy generation becomes 51 1 kJminK 085 kWK 1min 1 kJK 51 gen gen t S S preparation If you are a student using this Manual you are using it without permission 7113 7158E A cylinder contains saturated liquid water at a specified pressure Heat is transferred to liquid from a source and some liquid evaporates The total entropy generation during this process is to be determined Assumptions 1 No heat loss occurs from the water to the surroundings during the process 2 The pressure inside the cylinder and thus the water temperature remains constant during the process 3 No irreversibilities occur within the cylinder during the process Analysis The pressure of the steam is maintained constant Therefore the temperature of the steam remains constant also at PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 72 T Table A5E 26 sat40 psia T 727 2 R F Taking the contents of the cylinder as the system and noting that the temperature of water remains constant the entropy change of the system during this isothermal internally reversible process becomes 0 8251 BtuR 7272 R Btu 600 sys sysin system T Q S Similarly the entropy change of the heat source is determined from 04110 BtuR 1000 460 R source sourceout source T S 600 Btu Q Now consider a combined system that includes the cylinder and the source Noting that no heat or mass crosses the boundaries of this combined system the entropy balance for it can be expressed as in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in S S S S 43 42 1 4243 1 e total entropy generated during this process is source water gentotal 0 S S S Therefore th 0414 BtuR 0 4110 0 8251 source water gentotal S S S Discussion The entropy generation in this case is entirely due to the irreversible heat transfer through a finite temperature difference We could also determine the total entropy generation by writing an energy balance on an extended system that cludes the system and its immediate surroundings so that part of the boundary of the extended system where heat transfer occurs is at the source temperature in H2O 40 psia Source 1000F 600 Btu preparation If you are a student using this Manual you are using it without permission 7120 7165 A rigid tank initially contains saturated liquid water A valve at the bottom of the tank is opened and half of mass in liquid form is withdrawn from the tank The temperature in the tank is maintained constant The amount of heat transfer and the entropy generation during this process are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid leaving the device remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The direction of heat transfer is to the tank will be verified PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m kg 0001060 120 C 120 C 3 120 C 1 o o o f e f e f s s h h v v ake the tank as the system which is a control volume since mass crosses the boundary Noting that the ies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively e mass and energy balances for this uniformflow system can be expressed as H2O 018 m3 120C T const me Properties The properties of water are Tables A4 through A6 Q 5279 kJkg K 1 50360 kJkg liquid sat 0 C 12 120 C 1 120 C 1 1 o o f f s s u u T 5279 kJkg K 1 81 kJkg 503 liquid sat 0 C e 12 T Analysis a We t microscopic energ th Mass balance 2 1 system out in m m m m m m e Energy balance system out in W m u m u m h Q E E E e e 43 42 1 4243 1 potential etc energies Change in internal kinetic by heat work and mass Net energy transfer 0 pe ke 1 1 since 2 2 in The initial and the final masses in the tank are me m m 8488 kg 2 16976 kg 2 1 2 Now we determine the final internal energy and entropy m 1 1 16976 kg m kg 0001060 m 018 3 3 1 1 v V 15346 kJkg K 0001191 56013 15279 50601 kJkg 0001191 20253 60 503 001191 0 0 C 12 0001191 0001060 08913 0001060 0002121 0002121 m kg 8488 kg 2 m v v m 018 2 2 2 2 2 2 2 2 3 3 fg f fg f fg f x s s s x u u u x T x v V v The heat transfer during this process is determined by substituting these values into the energy balance equation b The total entropy generation is determined by considering a combined system that includes the tank and the heat source Noting that no heat crosses the boundaries of this combined system and no mass enters the entropy balance for it can be expressed as 2 2226 kJ 16976 kg 50360 kJkg 8488 kg 50601 kJkg 8488 kg 50381 kJkg 1 1 2 2 in m u m u m h Q e e preparation If you are a student using this Manual you are using it without permission 7121 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course generation source tank gen Change system Entropy gen by heat and mass entropy transfer Net out in S S S m s S S S S e e 43 42 1 4243 1 entropy in Therefore the total entropy generated during this process is 01237 kJK 273 K 230 kg 15279 kJkg K 16976 2226 kJ 8488 kg 15346 kJkg K kg 15279 kJkg K 8488 source T sourceout 1 1 2 2 m s m s e se source tank gen Q m S S m s S e e preparation If you are a student using this Manual you are using it without permission 7122 7166E An unknown mass of iron is dropped into water in an insulated tank while being stirred by a 200W paddle wheel Thermal equilibrium is established after 10 min The mass of the iron block and the entropy generated during this process are to be determined PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course cific heats at room roperties he specific heats of water and the iron block at room temperature are cp water 100 BtulbmF and cp iron tulbmF Table A3E The density of water at room temperature is 621 lbmft³ the system This is a closed system since no mass crosses the system boundary during the process The energy balance on the system can be expressed as potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 43 42 1 4243 1 or Assumptions 1 Both the water and the iron block are incompressible substances with constant spe temperature 2 The system is stationary and thus the kinetic and potential energy changes are zero 3 The system is well insulated and thus there is no heat transfer P T 0107 B Analysis a We take the entire contents of the tank water iron block as E E E WATER 70F Iron 185F 200 W U W in pw W U U pwin iron water 1 water 2 iron 1 2 pwin T mc T T mc T W where 1137 Btu 1055 kJ Btu 1 621 water m ρV Using specific heat values for iron and liquid water and substituting 497 lbm 08 ft lbmft 3 3 60 s 02 kJs10 pw pw t W W iron 70 F 497 lbm 100 Btulbm F75 185 F 0107 Btulbm F75 1137 Btu m o o o o iron 114 lbm m e tank to be the system Noting that no heat or mass crosses the boundaries of this for it can be expressed as by heat and mass 0 S S S S S S 43 42 1 where b Again we take the iron water in th combined system the entropy balance Entropy gen entropy transfer Net out in 4243 1 S water iron total gen in entropy Change system generation 0466 BtuR 530 R 496 lbm 10 Btulbm R ln 535 R ln 0228 BtuR 645 R 114 lbm 0107 Btulbm R ln 535 R ln 1 2 avg water 1 2 avg iron T T mc S T T mc S Therefore the entropy generated during this process is 0238 BtuR 0 466 0 228 water iron gen total S S S S preparation If you are a student using this Manual you are using it without permission 7123 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course contains saturated water mixture until no liquid is left in the nk The quality of steam in the tank at the initial state the amount of mass that has escaped and the entropy generation uring this process are to be determined operties of the steam in the tank at the final state and the properties of exiting steam are Tables A4 e are 7167 Liquid water is withdrawn from a rigid tank that initially ta d Assumptions 1 Kinetic and potential energy changes are zero 2 There are no work interactions Analysis a The pr through A6 2553 1 kJkg 400 kPa 2 u P s2 6 8955 kJkgK 0 46242 m kg sat vap 1 3 2 2 2 x v Water mixture 75 kg 400 kPa Q 1 7765 kJkgK 66 kJkg 604 sat liq 0 400 kPa 2 e e s h x P e The relations for the volume of the tank and the final mass in the tank 1 1 1 kg 57 v v V m 1 1 2 16219 kg 57 v v V m 3 2 046242 m kg v The mass energy and entropy balances may be written as 2 1 m m me 1 1 2 2 in m u m u m h Q e e 1 1 2 2 source m s m s S m s T gen e e in Q Substituting 1 16219 1 57 v e m 1 1 1 57 2553 1 16219 60466 16219 57 5 u v v 2 1 1 gen 1 57 6 8955 16219 1 7765 16219 57 273 500 5 s S v v 3 Eq 2 may be solved by a trialerror approach by trying different qualities at the inlet state Or we can use EES to solve the equations to find x1 08666 Other properties at the initial state are Substituting into Eqs 1 and 3 b c 2129 kJkgK 6 0 40089 m kg 2 kJkg 2293 8666 0 kPa 400 1 3 1 1 1 1 s u x P v 0998 kg 16219 0 40089 57 e m 000553 kJK gen gen 57 6 2129 16219 0 40089 6 8955 1 7765 16219 0 40089 57 273 500 5 S S preparation If you are a student using this Manual you are using it without permission 7124 Special Topic Reducing the Cost of Compressed Air 7168 The total installed power of compressed air systems in the US is estimated to be about 20 million horsepower The amount of energy and money that will be saved per year if the energy consumed by compressors is reduced by 5 percent is to be determined PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course third of the time on average and are shut down the rest of nalysis En gy consumed Motor efficiency hp1336524 hoursyear085 corresponding to a 5 reduction in energy use for compressed air become Energy Savings Energy consumedFraction saved 51251010 kWh005 2563109 kWhyear Cost Savings Energy savingsUnit cost of energy will save 179 million a year Assumptions 1 The compressors operate at full load during one the time 2 The average motor efficiency is 85 percent A The electrical energy consumed by compressors per year is er Power ratingLoad factorAnnual Operating Hours 20106 hp0746 kW 51251010 kWhyear Then the energy and cost savings Air Compressor 1 2 W20106 hp 2563109 kWhyear007kWh 0179109 year Therefore reducing the energy usage of compressors by 5 preparation If you are a student using this Manual you are using it without permission 7125 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course a be The re absolute pressure rather than gage pressure ropertie alysis ergy consumed Power ratingLoad factorAnnual Operating HoursMotor efficiency 90 hp0746 kWhp0753500 hoursyear094 he fract n of en rgy sav e tting o he com 7169 The compressed air requirements of a plant is being met by a 90 hp compressor that compresses air from 1013 kP to 1100 kPa The amount of energy and money saved by reducing the pressure setting of compressed air to 750 kPa is to determined Assumptions 1 Air is an ideal gas with constant specific heats 2 Kinetic and potential energy changes are negligible 3 load factor of the compressor is given to be 075 4 The pressures given a P s The specific heat ratio of air is k 14 Table A2 An The electrical energy consumed by this compressor per year is En 187420 kWhyear T io e ed as a result of reducing the pressur se f t pressor is 0 2098 1 1100101 3 1 41 1 41 1 750 101 3 1 41 1 41 1 2 1 k k P the energy consumed by the compressor ings in this case become Power reduction factor t cost of energy h that is worth 3342 in this ase Air Compressor 2 90 hp W 1100 kPa 1 1 Power Reduction Factor 1 1 2 reduced k k P P P 101 kPa 15C 1 That is reducing the pressure setting will result in about 11 percent savings from and the associated cost Therefore the energy and cost sav Energy Savings Energy consumed 187420 kWhyear02098 39320 kWhyear Cost Savings Energy savingsUni 39320 kWhyear0085kWh 3342year Therefore reducing the pressure setting by 250 kPa will result in annual savings of 39320 kW c Discussion Some applications require very low pressure compressed air In such cases the need can be met by a blower instead of a compressor Considerable energy can be saved in this manner since a blower requires a small fraction of the power needed by a compressor for a specified mass flow rate preparation If you are a student using this Manual you are using it without permission 7126 7170 A 150 hp compressor in an industrial facility is housed inside the production area where the average temperature during operating hours is 25C The amounts of energy and money saved as a result of drawing cooler outside air to the compressor instead of using the inside air are to be determined Assumptions 1 Air is an ideal gas with constant specific heats 2 Kinetic and potential energy changes are negligible Analysis The electrical energy consumed by this compressor per year is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course nergy consume 854500 hoursyear lso ost of Energy t cost of energy he fract sav ler outside air is E d Power ratingLoad factorAnnual Operating HoursMotor efficiency 150 hp0746 kWhp0 09 475384 kWhyear A C Energy consumedUni 475384 kWhyear007kWh 33277year T ion of energy ed as a result of drawing in coo Power Reduction Factor outside 1 T inside 25 273 T 1 10 273 0 0503 hat is d g ercent savings from the energy nsumed by the savings in this case become 23929 kWhyear Cost Savings Energy savingsUnit cost of energy 23929 kWhyear007kWh ation of this measure requires the installation of an ordinary sheet metal or PVC duct from the mpresso re is relatively low an he pressure drop have visited especially the newer ones the co dy taking advantage of the savings easure Air Compressor 101 kPa 25C 1 2 W150 hp T0 10C T rawin in air which is 15C cooler will result in 503 p c o compressor and the associated cost Therefore the energy and cost Energy Savings Energy consumedPower reduction factor 475384 kWhyear00503 1675year Therefore drawing air in from the outside will result in annual savings of 23929 kWh which is worth 1675 in this case Discussion The price of a typical 150 hp compressor is much lower than 50000 Therefore it is interesting to note that the cost of energy a compressor uses a year may be more than the cost of the compressor itself The implement co r intake to the outside The installation cost associated with this measu in the duct in most cases is negligible About half of the manufacturing facilities we d t have the duct from mpressor intake to the outside in place and they are alrea associated with this m preparation If you are a student using this Manual you are using it without permission 7127 7171 The compressed air requirements of the facility during 60 percent of the time can be met by a 25 hp reciprocating compressor instead of the existing 100 hp compressor The amounts of energy and money saved as a result of switching to the 25 hp compressor during 60 percent of the time are to be determined PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course ed by each compressor per year is determined from 060820900409 185990 kWhyear Energy consumedSmall PowerHoursLFxTFηmotorUnloaded LFxTFηmotorLoaded ear0001509 088 year Therefore the energy and cost Small hyear ear0075kWh 7172 A facility stops production for one hour every day including weekends for lunch break but the 125 hp compressor is kept operating If the compressor consumes 35 percent of the rated power when idling the amounts of energy and money saved per year as a result of turning the compressor off during lunch break are to be determined Analysis It seems like the compressor in this facility is kept on unnecessarily for one hour a day and thus 365 hours a year and the idle factor is 035 Then the energy and cost savings associated with turning the compressor off during lunch break are determined to be Energy Savings Power RatingTurned Off HoursIdle Factorηmotor 125 hp0746 kWhp365 hoursyear035084 14182 kWhyear Cost Savings Energy savingsUnit cost of energy 14182 kWhyear009kWh 1276year Discussion Note that the simple practice of turning the compressor off during lunch break will save this facility 1276 a year in energy costs There are also side benefits such as extending the life of the motor and the compressor and reducing the maintenance costs Analysis Noting that 1 hp 0746 kW the electrical energy consum Energy consumedLarge PowerHoursLFxTFηmotorUnloaded LFxTFηmotorLoaded 100 hp0746 kWhp3800 hoursyear035 25 hp0746 kWhp3800 hoursy 65031 kWh 5085 savings in this case become Energy Savings Energy consumedLarge Energy consumed 185990 65031 kW 120959 kWhyear Cost Savings Energy savingsUnit cost of energy 120959 kWhy 9072year Discussion Note that utilizing a small compressor during the times of reduced compressed air requirements and shutting down the large compressor will result in annual savings of 120959 kWh which is worth 9072 in this case Air Compressor 1 2 W100 hp Air Compressor 1 2 W125 hp preparation If you are a student using this Manual you are using it without permission 7128 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course sult oad when operating load conditions is sor i d to h gerat gs COP hoursyear25 C saved l energy worth 2232 per year The ower consumed by the fans andor r However if the heat removed by the aftercooler is utilized for some useful purpose such as space heating or process heating then the actual savings will be much more 7173 It is determined that 25 percent of the energy input to the compressor is removed from the compressed air as heat in the aftercooler with a refrigeration unit whose COP is 25 The amounts of the energy and money saved per year as a re of cooling the compressed air before it enters the refrigerated dryer are to be determined Assumptions The compressor operates at full l Analysis Noting that 25 percent of the energy input to the compressor is removed by the aftercooler the rate of heat removal from the compressed air in the aftercooler under full 150 hp0746 kWhp10025 2796 kW Rated Power of CompressorLoad FactorAftercooling Fraction aftercooli Q ng The compres s sai operate at full load for 2100 hours a year and the COP of t e refri ion unit is 25 Then the energy and cost savings associated with this measure become Energy Savin Q Annual Operating Hours aftercooling 2796 kW2100 23490 kWhyear ost Savings Energy savingsUnit cost of energy 23490 kWhyear0095kWh 2232year Discussion Note that the aftercooler will save this facility 23490 kWh of electrica actual savings will be less than indicated above since we have not considered the p pumps of the aftercoole Air Compressor W After cooler Qaftercooling 150 hp preparation If you are a student using this Manual you are using it without permission 7129 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course tandard motor are to be determined It is also to be determined if the savings from the high nalysis ciency motor in this case are etermin ad Factor1ηstandard 1ηefficient rsyear1010930 10962 tor will pay for this cost ue saving the facility money e the use of the high efficiency motor is local uti 7174 The motor of a 150 hp compressor is burned out and is to be replaced by either a 93 efficient standard motor or a 962 efficient high efficiency motor The amount of energy and money the facility will save by purchasing the high efficiency motor instead of s efficiency motor justify the price differential Assumptions 1 The compressor operates at full load when operating 2 The life of the motors is 10 years 3 There are no rebates involved 4 The price of electricity remains constant A The energy and cost savings associated with the installation of the high effi d ed to be Energy Savings Power RatingOperating HoursLo 150 hp0746 kWhp4368 hou 17483 kWhyear Air Compressor 150 hp Cost Savings Energy savingsUnit cost of energy 17483 kWhyear0075kWh 1311year The additional cost of the energy efficient motor is Cost Differential 10942 9031 1911 Discussion The money saved by the high efficiency mo difference in 19111311 15 years and will contin for the rest of the 10 years of its lifetime Therefor recommended in this case even in the absence of any incentives from the lity company preparation If you are a student using this Manual you are using it without permission 7130 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course ility in the mass flow gm33 ms10 m2 36 kgs 12960 kgh Noting that the temperature rise of air is 32C the rate at which heat can be recovered or the rate at which heat is transferred to air is Rate of Heat Recovery Mass flow rate of airSpecific heat o perature rise 12960 kgh10 kJkgC32C ng the heating season is year e Energy Savings Rate of Heat RecoveryAnnual Operating HoursE 414720 kJh2600 hoursyear08 1347840000 kJyear 12776 thermsyear Cost Savings Energy savingsUnit cost of energy saved 12776 thermsyear10therm 12776year 12776 per year from the heating costs e implementation of this measure requires the installation of an ordinary sheet metal duct from the outlet of r rcooled greater than 50 hp nd wate cooled greater 7175 The compressor of a facility is being cooled by air in a heatexchanger This air is to be used to heat the fac winter The amount of money that will be saved by diverting the compressor waste heat into the facility during the heating season is to be determined Assumptions The compressor operates at full load when operating Analysis Assuming cp 10 kJkgC and operation at sea level and taking the density of air to be 12 kgm3 rate of air through the liquidtoair heat exchanger is determined to be Mass flow rate of air Density of airAverage velocityFlow area 12 k f airTem 414720 kJh ber of operating hours of this compressor duri The num Operating hours 20 hoursday5 daysweek26 weeks 2600 hoursyear Then the annual energy and cost savings becom fficiency Air 20C 3 ms Hot Compressed air 52C Therefore utilizing the waste heat from the compressor will save Discussion Th the heat exchanger into the building The installation cost associated with this measure is relatively low A few of the manufacturing facilities we have visited already have this conservation system in place A damper is used to direct the air into the building in winter and to the ambient in summe Combined compressorheatrecovery systems are available in the market for both ai a r than 125 hp systems preparation If you are a student using this Manual you are using it without permission 7131 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course ing the hole on the compressed air line ic and potential energy changes are negligible c heat ratio of air is k 14 Table A2 is lute pressure is the sum of the gage pressure and the tmosphe ic pres the air at 15C from the spheric pressure of 856 kPa to 00856 7856 is d 7176 The compressed air lines in a facility are maintained at a gage pressure of 700 kPa at a location where the atmospheric pressure is 856 kPa There is a 3mm diameter hole on the compressed air line The energy and money saved per year by seal Assumptions 1 Air is an ideal gas with constant specific heats 2 Kinet Properties The gas constant of air is R 0287 kJkgK The specifi Analys Disregarding any pressure losses and noting that the abso a r sure work needed to compress a unit mass of 7 kPa etermined to be atmo 1 1 2 1 comp in P P k kRT w η 1 k k 6 kJkg 319 856 kPa 1 80 41 1 785 6 kPa 41 1 4 he crosssectional area of the 5mm diameter hole is oting that the line conditions are T0 298 K and P0 7856 kPa the mass flow rate of the air leaking through the hole is determined to be 0 287 kJkgK288 K 41 1 1 comp T 2 6 2 3 2 m 7 069 10 m 4 3 10 4 π πD A N 008451 kgs 0 1 298 K 41 2 1kJkg 0 287 kJkgK 1000 m s 41 m 7 069 10 kPam kgK298 K 0287 785 6 kPa 1 41 2 0 65 1 2 1 2 2 2 2 6 3 1 41 1 0 0 0 1 1 loss air T A kR k RT P k C m k Then the power wasted by the leaking compressed air becomes 2 701 kW kgs319 6 kJkg 0 008451 Power wasted air compin w m Noting that the compressor operates 4200 hours a year and the motor efficiency is 093 the annual energy and cost savings resulting from repairing this leak are determined to be Energy Savings Power wastedAnnual operating hoursMotor efficiency 2701 kW4200 hoursyear093 12200 kWhyear Cost Savings Energy savingsUnit cost of energy 12200 kWhyear007kWh 854year Therefore the facility will save 12200 kWh of electricity that is worth 854 a year when this air leak is sealed Compressed air line 700 kPa 25C Air leak Patm 856 kPa 15C preparation If you are a student using this Manual you are using it without permission 7132 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course used to compress air in the US is estimated to be 051015 kJ per year About 20 of the ompressed air is estimated to be lost by air leaks The amount and cost of electricity wasted per year due to air leaks is to e determined y air leaks are 051015 kJ1 kWh3600 kJ020 9 ergy wastedUnit cost of energy 2778109 kWhyear007kWh herefore air leaks are costing almost 2 billion a year in electricity costs The e of the pollution associated with the generation of this much electricity 7177 The total energy c b Assumptions About 20 of the compressed air is lost by air leaks Air Compressor 2 W051015 kJ Analysis The electrical energy and money wasted b Energy wasted Energy consumedFraction wasted 277810 kWhyear Money wasted En 1945109 year 1 T nvironment also suffers from this because preparation If you are a student using this Manual you are using it without permission 7133 Review Problems PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course al efficiency of a heat engine are given The entropy change of the atisfies the increase of entropy principle ne operates steadily nalysis According to the first law and the definition of he thermal efficiency 7178E The source and sink temperatures and the therm two reservoirs is to be calculated and it is to be determined if this engine s Assumptions The heat engi A t 06 Btu 40 1Btu 1 1 H L Q Q η when the thermal efficiency is 40 The entropy change of everything ved i his process is then invol n t 0000431 BtuR total L H S S S 500 R Btu 60 1300 R Btu 1 L L H H T Q T Q g has increased this engine is possible When the thermal efficiency of the engine is 70 Q Q HE QH TH Wnet QL TL Since the entropy of everythin L 03 Btu 70 1Btu 1 1 H η The total entropy change is then 0000169 BtuR 500 R Btu 30 1300 R Btu 1 total L L H H L H T Q T Q S S S which is a decrease in the entropy of everything involved with this engine Therefore this engine is now impossible preparation If you are a student using this Manual you are using it without permission 7134 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course o o be calculated and it is to be determined if this refrigerator satisfies the second law Analysis Combining the first law and the definition of the coefficient of erformance produces 7179 The source and sink temperatures and the COP of a refrigerator are given The total entropy change of the tw reservoirs is t Assumptions The refrigerator operates steadily p 1 25 kJ 4 COPR when C 1 1 kJ 1 1 1 L H Q Q OP 4 The entropy change of everything is then 0000173 kJK 253 K 1kJ 303 K 25 kJ 1 total L L H H L H T Q T Q S S S Since the entropy increases a refrigerator with COP 4 is possible When the coefficient of performance is increased to 6 20C 30C R QH 1 kJ Wnet 1 167 kJ 6 1 1 kJ 1 COP 1 1 R L H Q Q and the net entropy change is 0000101 kJK 253 K 1kJ 303 K 167 kJ 1 total L L H H L H T Q T Q S S S and the refrigerator can no longer be possible 7180 The operating conditions of a refrigerator are given The rate of entropy changes of all components and the rate of cooling are to be calculated and it is to be determined if this refrigerator is reversible Assumptions The refrigerator operates steadily Analysis Applying the first law to the refrigerator below the rate of cooling is The rate of entropy change for the lowtemperature reservoir according to the definition of the entropy is 4 kW 10 kW 14 kW Wnetin Q Q H L 002 kWK 200 K 4 kW L L L T Q S The rate at which the entropy of the hightemperature energy reservoir is changing is 200 K 400 K R QH L Q netin W 0035 kWK 400 K 14 kW H H H T Q S Since the working fluid inside the refrigerator is constantly returning to its original state the entropy of the device does not change Summing the rates at which the entropy of everything involved with this device changes produces Hence the increase in entropy principle is satisfied and this refrigerator is possible but not completely reversible 0015 kWK 0 0 020 0 035 device total S S S S L H preparation If you are a student using this Manual you are using it without permission 7135 7181 R134a is compressed in a compressor adiabatically The minimum internal energy at the final state is to be determined PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course fg f For minimum internal energy at the final state the process should be isentropic Then Analysis The initial state is saturated mixture and the properties are Table A12 xs s s T 8203 kJkg K 0 0 85 0 78316 15457 0 200 kPa 200 kPa 1 s 2 1 2204 kJkg 00 2 0 8262 0 56431 0 35404 08203 8203 kJkg K 0 800 kPa 2 2 2 fg f s s s x s s P 0 826215 9479 2 2 1 2 fg f x u u u preparation If you are a student using this Manual you are using it without permission 7136 7182 R134a is condensed in a pistoncylinder device in an isobaric and reversible process It is to be determined if the process described is possible PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as 2 1 out 1 2 bout out potential etc energies Change in internal kinetic by heat work and mass Net energy transfer h m h Q u m u U W Q uasiequilibrium process The initial and final state properties are Tables A12 and A13 100 C 1 1 s T Analysis We take the R134a as the system This is a closed system out in E E E 43 42 1 4243 1 R134a 1 MPa 1 2 out 1 2 bout out h m h Q u m u W Q 180C Q since U Wb H during a constant pressure q T v 3124 kJkg K 1 42136 kJkg 1000 kPa 1 1 h P 2 1 3919 kJkg K 0 10732 kJkg 0 kPa 1000 2 2 2 2 s h x P Substituting 314 0 kJkg 10732 42136 2 1 out h h q The entropy change of the energy reservoir as it undergoes a reversible isothermal process is 0 8419 kJkg K 273 K 100 res surr T 0 kJkg 314 out q s where the gn of heat transfer is taken positive as the reservoir receives heat The entropy change of R134a during the si process is 0 1 3124 0 3919 1 2 R134a s s s 9205 kJkg K e total entropy change is then Th 00786 kJkg K 0 8419 0 9205 surr R134a total s s s Since the total entropy change ie entropy generation is negative this process is impossible preparation If you are a student using this Manual you are using it without permission 7137 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 7183 Air is first compressed adiabatically and then expanded adiabatically to the initial pressure It is to be determined if the air can be cooled by this process Analysis From the entropy change relation of an ideal gas 1 2 1 2 air ln ln P R T c s p Since the initial and final pressures are the same the equation reduces to P T 1 T 2 air c ln T s p As there are no heat transfer the total entropy change ie entropy generation for this process is equal to the entropy change of air Therefore we must have 0 ln 2 air T c s p 1 T e only way this result can be satisfied is if ssible to create a cooling effect T2 T1 in the manner proposed in a closed system It is to be determined if this process is possible ssumptio s 1 Changes in the kinetic and potential energies are negligible 4 Air is an ideal gas with constant specific roperties The properties of air at room temperature are R 03704 psiaft mR cp 0240 BtulbmR k 14 Table A 2Ea Th 1 2 T T It is therefore impo 7184E Air is compressed adiabatically A n heats P 3lb Analysis The specific volume of air at the initial state is 1037 ft lbm 20 psia 0 3704 psia ft lbm R560 R 3 3 1 1 1 P RT v The volume at the final state will be minimum if the process is isentropic The specific volume for this case is determined from the isentropic relation of an ideal gas to be 2 884 ft lbm 120 psia 1037 ft lbm 20 psia 3 41 1 3 1 2 1 1 2 min k P P v v and the minimum volume is which is smaller than the proposed volume 3 ft3lbm Hence it is possible to compress this air into 3 ft3lbm 288 ft3 ft lbm 1 lbm2884 3 2 2 v V m preparation If you are a student using this Manual you are using it without permission 7138 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course er and the entropy s the kinetic and potential energy changes are zero 2 There are no work hat the ia The heat picked up by the block is 7185E A solid block is heated with saturated water vapor The final temperature of the block and wat changes of the block water and the entire system are to be determined Assumptions 1 The system is stationary and thu interactions involved 3 There is no heat transfer between the system and the surroundings Analysis a As the block is heated some of the water vapor will be condensed We will assume will be checked later t the water is a mixture of liquid and vapor at the end of the process Based upon this assumption the final temperature of water and solid block is 212F The saturation temperature at 147 ps 7100 Btu 70R 100 lbm05 Btulbm R212 1 2 block T mc T Q The water properties at the initial state are Table A5E he heat r ased by the water is equal to the heat picked up by the block Also noting that the pressure of water remains e process is determined from 1 7566 Btulbm R 3 Btulbm 1150 212 F 1 7 psia 14 1 1 1 1 1 s h T x P T ele constant the enthalpy of water at the end of the heat exchang 440 3 Btulbm 10 lbm 7100 Btu 3 Btulbm 1150 water 1 2 mw Q h h The state of water at the final state is saturated mixture Thus our initial assumption was correct The properties of water at the final state are 0 69947 Btulbm R 0 2682 1 44441 0 31215 02682 97012 4403 18016 3 Btulbm 440 7 psia 14 2 2 2 2 fg f h h h x h P 2 2 fg f x s s s The entropy change of the water is then 1057 BtuR 1 7566Btulbm 10 lbm069947 1 2 water s s m S w b The entropy change of the block is 1187 BtuR 460R 70 460R 100 lbm05 Btulbm Rln 212 ln 1 2 block T T mc S c The total entropy change is 130 BtuR 1057 1187 block water gen total S S S S The positive result for the total entropy change ie entropy generation indicates that this process is possible preparation If you are a student using this Manual you are using it without permission 7139 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course is reversible K Table A2a s or leaves The nergy bal ce for this stationary closed system can be expressed as 1 2 out in potential etc energies Change in internal kinetic by heat work and mass Net energy transfer since 0 b b T T Q W u m u U Q W 43 42 1 thermal reversible process is 7186 Air is compressed in a pistoncylinder device It is to be determined if this process is possible Assumptions 1 Changes in the kinetic and potential energies are negligible 4 Air is an ideal gas with constant specific heats 3 The compression process Properties The properties of air at room temperature are R 0287 kPam3kgK c 1005 kJkg p Analysis We take the contents of the cylinder as the system This is a closed system since no mass enter e an system out in E E E 43 42 1 Air 100 kPa 27C Heat 1 2 out in p b T T mc Q W 1 2 out in in out Wb Q The work input for this iso 7889 kJkg 100 kPa 0 287 kJkg K300 Kln 250 kPa ln 1 2 in P P RT w That is The entropy change of air during this isothermal process is 7889 kJkg in out w q 02630 kJkg K 100 kPa 0 287 kJkg Kln 250 kPa ln ln ln 1 2 1 2 1 2 air P P R P P R T T c s p The entropy change of the reservoir is 02630 kJkg K 300 K 7889 kJkg R R TR q s Note that the sign of heat transfer is taken with respect to the reservoir The total entropy change ie entropy generation is the sum of the entropy changes of air and the reservoir 0 kJkg K 0 2630 0 2630 R air total s s s Not only this process is possible but also completely reversible preparation If you are a student using this Manual you are using it without permission 7140 7187 A paddle wheel does work on the water contained in a rigid tank For a zero entropy change of water the final pressure in the tank the amount of heat transfer between the tank and the surroundings and the entropy generation during the process are to be determined PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course initial state Table A4 liq 1 s ve 2 2 P 474 kPa b The heat transfer can be determined from an energy balance on the tank Assumptions The tank is stationary and the kinetic and potential energy changes are negligible Analysis a Using saturated liquid properties for the compressed liquid at the 76 kJkg 588 sat 0 140 C 1 1 1 u x T 7392 kJkgK 1 Wpw Water 140C 400 kPa The entropy change of water is zero and thus at the final state we ha T C 80 55293 kJkg 1 7392 kJkgK 2 1 2 u s s 163 kJ 58876kJkg kg55293 23 48 kJ 1 2 Pwin out u m u W Q c Since t ropy change of water is zero the entropy generation is only due to the entropy increase of the surroundings he ent which is determined from 0565 kJK 273 K 15 kJ 163 surr out surr gen T Q S S preparation If you are a student using this Manual you are using it without permission 7141 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course elium the final volume of the nitrogen potential energy changes are negligible 2 Nitrogen and helium are ideal gases with constant m temperature are R 02968 kPam kgK cp 1039 kJkgK cv 0743 3kgK cp 51926 kJkgK cv 31156 kJkgK k nalysis rgoes an isentropic compression process nd thus th perature is determined from 7188 A horizontal cylinder is separated into two compartments by a piston one side containing nitrogen and the other side containing helium Heat is added to the nitrogen side The final temperature of the h the heat transferred to the nitrogen and the entropy generation during this process are to be determined Assumptions 1 Kinetic and specific heats at room temperature 3 The piston is adiabatic and frictionless Properties The properties of nitrogen at roo 3 kJkgK k 14 The properties for helium are R 20769 kPam 1667 Table A2 Q He 01 kg N2 02 m3 A a Helium unde a e final helium tem 3217 K 61 1 1 667 1 1 2 1 He2 95 kPa 273K 120 kPa 20 k k P P T T The in d final volumes of the helium are 67 b itial an 3 3 1 He1 273 K 01 kg20769 kPa m kg K20 mRT V 1 0 6406 m 95 kPa P 3 3 2 2 He2 0 5568 m 120 kPa 01 kg20769 kPa m kg K3217 K P mRT V Then the final volume of nitrogen becomes The mass and final temperature of nitrogen are 02838 m3 0 5568 0 6406 20 He2 He1 N21 N22 V V V V c 0 2185 kg 273 K kPa m kg K20 02968 kPa02 m 95 3 3 1 1 1 N2 RT P m V 525 1 K kg02968 kPa m kg K 02185 kPa0283 8 m 120 3 3 2 2 N22 mR P T V The heat transferred to the nitrogen is determined from an energy balance d Noting that helium undergoes an isentropic process the entropy generation is determined to be 466 kJ 293 kg31156 kJkgK3217 10 0 2185 kg0743 kJkgK5251 293 He 1 2 N2 1 2 He N2 in T T mc T T mc U U Q v v 0057 kJK 273 K 500 46 6 kJ 95 kPa 02968 kJkgKln 120 kPa 293 K kg 1039 kJkgKln 5251 K 02185 ln ln R in 1 2 1 2 N2 surr N2 gen T Q P P R T T c m S S S p preparation If you are a student using this Manual you are using it without permission 7142 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course entropy generation are to be determined of 350 K are R 01889 kPam3kgK cp 0895 able A2b 7189 An electric resistance heater is doing work on carbon dioxide contained an a rigid tank The final temperature in the tank the amount of heat transfer and the Assumptions 1 Kinetic and potential energy changes are negligible 2 Carbon dioxide is ideal gas with constant specific heats at room temperature Properties The properties of CO2 at an anticipated average temperature kJkgK cv 0706 kJkgK T Analysis a The mass and the final temperature of CO2 may be determined from ideal gas equation We CO2 250 K 100 kPa 694 kg 1 01889 kPa m kg K250 K 3 1 m RT 100 kPa08 m 3 1 PV 4375 K kg01889 kPa m kg K 1694 kPa08 m 175 3 3 2 2 mR P T V b The amount of heat transfer may be determined from an energy balance on the system 1 2 ein out T mc T t E Q v c The ent opy generation associated with this process may be obtained by calculating total entropy change which is the sum of the entropy changes of CO2 and the surroundings 9758 kJ 60 s 1694 kg0706 kJkgK4375 250K kW40 50 r 392 kJK 300 K 975 8 kJ 100 kPa 01889 kJkgKln 175 kPa 250 K kg 0895 kJkgKln 4375 K 1694 ln ln surr out 1 2 1 2 surr CO2 gen T Q P P R T T m c S S S p preparation If you are a student using this Manual you are using it without permission 7143 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course e Helium is an Properties The properties of helium are R 20769 kPam kgK cp 51926 kJkgK Table A2a lysis a The final temperature of helium may be determined from an energy balance on the control volume 7190 Heat is lost from the helium as it is throttled in a throttling valve The exit pressure and temperature of helium and th entropy generation are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 ideal gas with constant specific heats 3 Helium 300 kPa 50C q Ana T T T T c q 497C 322 7 K 51926 kJkg C 175 kJkg 50 C out 1 2 2 1 t p p c q change of helium ou The final pressure may be determined from the relation for the entropy 265 kPa 2 2 1 2 1 2 He 20769 kJkgKln 300 kPa K 323 51926 kJkgKln 3227 K 25 kJkgK 0 ln ln P P P P R T T c s p ted with this process may be obtained by adding the entropy change of helium as it flows he surroundings b The entropy generation associa in the valve and the entropy change of t 0256 kJkgK 273 K 25 1 75 kJkg 0 25 k JkgK surr out He surr He gen T q s s s s preparation If you are a student using this Manual you are using it without permission 7144 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course f the rate of entropy generation are to be determined tial energy changes are negligible 1 1 1 1 1 s h x The mass flow rate of the refrigerant is 7191 Refrigerant134a is compressed in a compressor The rate of heat loss from the compressor the exit temperature o R134a and Assumptions 1 Steady operating conditions exist 2 Kinetic and poten Analysis a The properties of R134a at the inlet of the compressor are Table A12 0 93773 kJkgK 46 kJkg 244 0 09987 m kg 200 kPa 3 1 P v Compressor R134a 200 kPa sat vap 700 kPa Q 0 3004 kgs m kg 009987 03 m s 0 3 3 1 1 v V m Given the entropy increase of the surroundings the heat lost from the compressor is 2344 kW 273 K0008 kWK 20 surr surr out surr out surr S T Q T Q S b An energy balance on the compressor gives The exit state is now fixed Then 2 2 s h in the compressor and the entropy change of the surroundings 26994 kJkg 24446 kJkg 03004 kgs kW 2344 kW 10 2 2 1 2 out in h h h m h Q W 2 2 700 kPa T P C 315 0 93620 kJkgK 94 kJkg 269 c The entropy generation associated with this process may be obtained by adding the entropy change of R134a as it flows 000754 kJK 0 008 kWK kgs093620 093773 kJkgK 03004 surr 1 2 surr R gen S s m s S S S preparation If you are a student using this Manual you are using it without permission 7145 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 7 kJkgK Table A1 lows 7192 Air flows in an adiabatic nozzle The isentropic efficiency the exit velocity and the entropy generation are to be determined Properties The gas constant of air is R 028 Assumptions 1 Steady operating conditions exist 2 Potential energy changes are negligible Analysis a b Using variable specific heats the properties can be determined from air table as fol 34631 kJkg 2 2836 500 kPa 3806 kPa 300 85708 kJkgK 1 35 49 kJkg 350 K 2 2 h T 0 806 3 1 99194 kJkgK 2 1 1 2 2 0 2 1 0 1 s r r r h P P P P s P s balances on the control volume for the actual and isentropic processes give 40098 kJkg K 400 1 1 h T Energy 3191 m V2 s 2 2 2 2 2 2 2 2 2 2 1 1 m s 1000 1kJkg 2 49 kJkg 1000 m s 2 2 2 V V h V h 2 350 1kJkg 30 ms 98 kJkg 400 8 ms 331 m s 1000 1kJkg 2 31 kJkg 346 m s 1000 1kJkg 2 30 ms 98 kJkg 400 2 2 2s 2 2 2 2s 2 2 2 2s 2 1 1 V V h h s isentr ic efficiency is determined from its definition 2 2 V V The op 0925 2 2 2 2s 2 2 N 331 8 ms 319 1 ms V V η c Since the nozzle is adiabatic the entropy generation is equal to the entropy increase of the air as it flows in the nozzle 00118 kJkgK ln 2 0 1 0 2 air gen P P R s s s s 500 kPa 0287 kJkgKln 300 kPa 1 99194kJkgK 85708 1 1 Air 500 kPa 400 K 30 ms 300 kPa 350 K preparation If you are a student using this Manual you are using it without permission 7146 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course team in the tank is allowed to flow into the cylinder The final temperatures in the linder are wellinsulated and thus heat transfer is negligible 2 The water that remains k and cylinder themselves is ligible s 7193 An insulated rigid tank is connected to a pistoncylinder device with zero clearance that is maintained at constant pressure A valve is opened and some s tank and the cylinder are to be determined Assumptions 1 Both the tank and cy in the tank underwent a reversible adiabatic process 3 The thermal energy stored in the tan negligible 4 The system is stationary and thus kinetic and potential energy changes are neg Analysis a The steam in tank A undergoes a reversible adiabatic process and thus s2 s1 From the steam tables Table A4 through A6 24615 kJkg 0966620246 kJkg 50450 0 85626 m kg 0001061 09666 0 88578 0001061 0 9666 5 5968 1 5302 9402 6 0 kPa 20 69402 kJkg K 25485 kJkg m kg 052422 kPa 350 2 2 3 2 2 2 2 200 kPa 2 2 1 2 x mixture sat s s v v v 35 0 kPa 1 35 0 kPa 1 3 350 kPa 1 1 fg A f A fg A f A fg f A A sat A g g g u x u u s s s x T T P s s u u v vapor sat P v 1202 C The initial and the final masses are 01479 kg 02336 3815 0 02336 kg m kg 085626 m 02 0 3815 kg m kg 052422 m 02 2 1 2 3 3 2 2 3 3 1 1 A A B A A A A A A m m m m m v V v V Sat vapor 350 kPa 02 m3 200 kPa b The boundary work done during this process is Taking the contents of both the tank and the cylinder to be the system the energy balance for this closed system can be expressed as or B B B B B b out P m P Pd W 2 2 2 2 1 0 v V V B A U U U W E E E out b potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 43 42 1 4243 1 0 0 0 1 1 2 2 2 2 2 2 1 1 2 2 2 2 out b A B B B A B B B B A m u m u h m m u m u m u m P U U W v Thus 2685 8 kJkg 0 1479 0 2336 2461 5 3815 2548 5 0 2 2 2 1 1 2 B A B m m u m u h At 200 kPa hf 50471 and hg 27063 kJkg Thus at the final state the cylinder will contain a saturated liquidvapor mixture since hf h2 hg Therefore 12025C sat200 kPa 2 T T B preparation If you are a student using this Manual you are using it without permission 7147 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course e ed e properties of helium are cv 31156 kJkgK and k 1667 Table A2b Analysis Analysis We take the helium as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as 7194 Helium gas is compressed in an adiabatic closed system with an isentropic efficiency of 80 The work input and th final temperature are to be determin Assumptions 1 Kinetic and potential energy changes are negligible 3 The device is adiabatic and thus heat transfer is negligible 4 Helium is an ideal gas Properties Th 1 2 1 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc u m u U W E E E v 43 42 1 4243 1 Helium 100 kPa 27C The isentropic exit temperature is 722 7 K 100 kPa 300 K 900 kPa 06671667 1 1 2 1 2 k k s P P T T The work input during isentropic process would be 3950 kJ 300K 3 kg 3 1156 kJkg K7227 1 2 sin T T mc W s v The work put during the actual process is in 4938 kJ 080 3950 kJ sin W in η W preparation If you are a student using this Manual you are using it without permission 7150 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course be 718 kJkgK and k 14 The properties of neon at room temperature are cv 06179 kJkgK and k 1667 Table A2a he The isentropic exit temperature is 7197 A gas is adiabatically expanded in a pistoncylinder device with a specified isentropic efficiency It is to determined if air or neon will produce more work Assumptions 1 Kinetic and potential energy changes are negligible 3 The device is adiabatic and thus heat transfer is negligible 4 Air and helium are ideal gases Properties The properties of air at room temperature are cv 0 Analysis We take the gas as the system This is a closed system since no mass crosses the boundaries of the system T energy balance for this system can be expressed as 2 1 out 1 2 1 2 out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc W T T mc u m u U W E E E v v 43 42 1 4243 1 Air 3 MPa 300C 203 4 K 3000 kPa 80 kPa 273 K 300 0414 1 1 2 1 2 k k s P P T T The work output during the actual process is 239 kJkg 2034K 0 90 0 718 kJkg K573 2 1 in s v T T c w η Repeating the same calculations for neon 134 4 K 3000 kPa 80 kPa 273 K 300 06671667 1 1 2 1 2 k k s s P P T T 217 kJkg 1344K 0 80 0 6179 kJkg K573 2 1 in s v T T c w η Air will produce more work preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7153 1 The compressor operates steadily 2 Kinetic and potential energies are negligible 3 The compression deal gas with constant specific heats at room temperature able A1 The specific heat ratio of air is k 14 Table A s is 7200 Air is compressed in a twostage ideal compressor with intercooling For a specified mass flow rate of air the power input to the compressor is to be determined and it is to be compared to the power input to a singlestage compressor Assumptions process is reversible adiabatic and thus isentropic 4 Air is an i Properties The gas constant of air is R 0287 kPam3kgK T 2 Analysis The intermediate pressure between the two stage 25 0 kPa 100 kPa 625 kPa 1 2 P P Px compressor work is twice the compression work for a single stage The compressor work across each stage is the same thus total 4 kJkg 180 1 100 kPa 250 kPa 14 1 14 0287 kJkg K 300 K 2 he work put to a singlestage compressor operating between the same pressure limits would be 1 1 2 2 0414 1 1 1 compinI in comp k k Px P k kRT w w and 015 kgs 1804 kJkg compin in mw W 271 kW T in 207 4 kJkg 1 100 kPa 14 1 1 4 1 2 compin k nd Discussion Note that the power consumption of the compressor decreases significantly by using 2stage compression with intercooling 625 kPa 14 0287 kJkg K 300 K 1 041 1 1 k k P P kRT w a 311 kW 015 kgs 2074 kJkg compin in mw W Stage II 625 kPa Stage I 27C 100 kPa 27C Heat W 7158 7205 Problem 7204 is reconsidered The isentropic efficiencies for the compressor and turbine are to be turbine efficiency to be investigated The and 09 e problem is solved using EES and the results are tabulated and plotted below In ata mdotair 10 kgs air compressor air data T 29 C e tem eratur P 98 T 70 C P 10 mdotst25 kgs s u data st1500 C st210 kPa xst2092 quality Compressor Analysis Conservation of mass for the compressor mdotairin mdotairout mdotair Conservation of energy for the compressor is Edotcompin Edotcompout DELTAEdotcomp DELTAEdotcomp 0 Steady flow requirement EdotcompinmdotairenthalpyairTTair1 Wdotcompin EdotcompoutmdotairenthalpyairTTair2 Compressor adiabatic efficiency EtacompWdotcompinisenWdotcompin WdotcompinisenmdotairenthalpyairTTairisen2enthalpyairTTair1 sair1entropyairTTair1PPair1 sair2entropyairTTair2PPair2 sairisen2entropyair TTairisen2PPair2 sairisen2sair1 rbine Analysis onservation of mass for the turbine mdotstin mdotstout mdotst Conservation of energy for the turbine is turbin Edotturbout DELTAEdotturb DELTAEdotturb 0 Steady flow requirement Edotturbinmdotsthst1 hst1enthalpysteamTTst1 PPst1 Edotturboutmdotsthst2Wdotturbout hst2enthalpysteamPPst2 xxst2 Turbine adiabatic efficiency EtaturbWdotturboutWdotturboutisen Wdotturboutisenmdotsthst1hstisen2 sst1entropysteamTTst1PPst1 hstisen2enthalpysteam PPst2ssst1 Note When Etaturb is specified as an independent variable in the Parametric Table the iteration process may put the steam state 2 in the superheat region where the quality is undefined Thus sst2 Tst2 are calculated at Pst2 hst2 and not Pst2 and xst2 sst2entropysteamPPst2hhst2 Tst2temperaturesteamPPst2 hhst2 sstisen2sst1 Net work done by the process WdotnetWdotturboutWdotcompin Entropy generation Since both the compressor and turbine are adiabatic and thus there is no heat transfer to the surroundings the entropy generation for the two steady flow devices becomes Sdotgencompmdotair sair2sair1 determined and then the effect of varying the compressor efficiency over the range 06 to 08 and the over the range 07 to 095 on the net work for the cycle and the entropy generated for the process is net work is to be plotted as a function of the compressor efficiency for turbine efficiencies of 07 08 Analysis Th put D air1 5273 W will input p e in C air1 kPa air2 0273 air2 00 kPa team t rbine st T Pst112500 kPa P Tu C Edot PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 7159 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course ine Tair2700273 C comp Parmetric Table then press F3 to solve the table EES then solves for the hst2 for each Etaturb Wnet kW Sgentotal kWK ηturb ηcomp Wnet kW Sgentotal kWK ηturb ηcomp Sdotgenturbmdotstsst2sst1 SdotgentotalSdotgencompSdotgenturb To generate the data for Plot Window 1 Comment out the l and select values for Etacomp in the Parmetric Table then press F3 to solve the table EES then solves for the unknown value of Tair2 for each Eta To generate the data for Plot Window 2 Comment out the two lines xst2092 quality and hst2enthalpysteamPPst2 xxst2 and select values for Etaturb in the 20124 2759 075 06665 19105 30 07327 06 21745 2251 08 06665 19462 2951 07327 065 23365 1744 085 06665 19768 2907 07327 07 24985 1236 09 06665 20033 2867 07327 075 26606 7281 095 06665 20265 2832 07327 08 060 065 070 075 080 19200 19600 20400 20000 284 288 292 296 300 ηcompb Wnet k Sgentotal kWK ηturb 07333 Effect of Compressor Efficiency on Net Work and Entropy Generated W 075 078 081 084 087 090 093 20000 22000 24000 26000 5 10 15 20 25 30 ηturbb Wnet kW Sgentotal kWK ηcomp 06665 Effect of Turbine Efficiency on Net Work and Entropy Generated preparation If you are a student using this Manual you are using it without permission 7160 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course T equivalence 7206 The pressure in a hot water tank rises to 2 MPa and the tank explodes The explosion energy of the water is to be determined and expressed in terms of its TN Assumptions 1 The expansion process during explosion is isentropic 2 Kinetic and potential energy changes are negligible 3 Heat transfer with the surroundings during explosion is negligible Properties The explosion energy of TNT is 3250 kJkg From the steam tables Tables A4 through 6 kJkg K 24467 90612 kJkg liquid sat MPa 2 2 MPa 1 2 MPa 1 1 f f s s u u P 81183 kJkg 0 1889 2088 2 40 417 0 1889 6 0562 1 3028 4467 2 kJkg K 60562 kJkg 20882 3028 1 41740 kPa 100 m kg 0001177 2 2 2 2 1 2 2 3 2 MPa 1 fg f fg f fg fg f f f x u u u s s s x s u s u s s P v v We idealize the water tank as a closed system that undergoes a reversible adiabatic process with negligible changes in kinetic and potential energies The work done during this idealized process represents the explosive energy of s determined from the closed system energy balance to be Net energy transfer out in u m u W E u m u U W E E E 43 42 1 4243 1 where Water Tank 2 MPa Analysis the tank and i 2 1 bout exp 1 2 out b potential etc energies in internal kinetic Change system heat work and mass by 6799 kg 0001177 m kg m 0080 3 3 1 v V m Substituting 6410 kJ 81183 kJkg 6799 kg 90612 exp E which is equivalent to 1972 kg TNT 3250 kJkg 6410 kJ mTNT preparation If you are a student using this Manual you are using it without permission 7161 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course d uivalence Tables A4 through 6 7207 A 035L canned drink explodes at a pressure of 12 MPa The explosive energy of the drink is to be determined an expressed in terms of its TNT eq Assumptions 1 The expansion process during explosion is isentropic 2 Kinetic and potential energy changes are negligible 3 Heat transfer with the surroundings during explosion is negligible 4 The drink can be treated as pure water Properties The explosion energy of TNT is 3250 kJkg From the steam tables kJkg K 22159 79696 kJkg liquid Comp MPa 12 12 MPa 1 12 MPa 1 12 MPa 1 1 f f f s s u u P 73226 kJkg 0 1508 2088 2 40 417 0 1508 6 0562 1 3028 2159 2 kJkg K 60562 kJkg 20882 3028 1 41740 kPa 100 kg 2 2 2 2 1 2 2 fg f fg f fg fg f f x u u u s s s x s u s u s s P that undergoes a reversible adiabatic process with negligible rgies The work done during this idealized process represents the explosive energy of the can and is determined from the closed system energy balance to be potential etc energies Change in internal kinetic by heat work and mass Net energy transfer u m u W E u u where 0001138 m 3 v v COLA 12 MPa Analysis We idealize the canned drink as a closed system changes in kinetic and potential ene system out in E E E 43 42 1 4243 1 bout m U W 2 1 bout exp 1 2 03074 kg m kg 0001138 m 000035 3 3 1 v V m Substituting 199 kJ 73226 kJkg 03074 kg 79696 Eexp which is equivalent to 000612 kg TNT 3250 kJkg 199 kJ mTNT preparation If you are a student using this Manual you are using it without permission 7165 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course ksa re to be determined ssumptio s 1 Changes in the kinetic and potential energies are negligible 4 Oxygen is an ideal gas with constant specific roperties The properties of oxygen at room temperature are R 02598 kPam3kgK cp 0918 kJkgK cv 0658 JkgK k 1395 Table A2a alysis As the tank is being filled the pressure in the tank increases as time passes but the temperature does not In the ne betwe the compressor and tank the pressure always matches that in the compressor and as a result the temperature time Applying the isentropic process relations to the compressor yields the temperature in this line 7211 Oxygen tanks are filled by an isentropic compressor The work required by the compressor and the heat transfer from the tan A n heats P k An li en changes in this line with as k k P P T T 1 1 2 1 2 Reducing the mass balance to the conditions of the tank gives min dt dmcv but RT P mcv V Combining these two results produces dt dP RT RT d P m V dt V in where the last step incorporates the fact that the tank volume and temperature do not change as time passes The mass in the tank at the end of the compression is 170 8 kg 0 2598 kPa m kg K293 K 13000 kPa1 m 3 3 final final RT P m V Adapting the first law to the tank produces dt dP RT P P c T dt T dm c c T m dt c T dm m h dt d mu Q k k p v p v V 1 1 1 in 2 in in Integrating this result from the beginning of the compression to the end of compression yields 93720 kJ 2 1 395 1 170 8 1 395 150 0 918293 13000 0 658293170 8 1 2 1 2 1 1 395 1 395 0 1 1 1 1 1 1 1 1 1 1 1 in f k k f p f v f k k f k k p f v f i k k k k p f i v m k k P P c T Tm c P P k k RT P c T Tm c dP P RT P c T dm c T Q V V preparation If you are a student using this Manual you are using it without permission 7166 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course The negative sign indicates that heat is lost from the tank Adopting the first law to the compressor and tank as the system gives in in in in m h d mu d t W Q Recognizing that the enthalpy of the oxygen entering the compressor remains constant this results integrates to f f f h m m u W Q 1 in in or 80710 kJ 293 0 658 170 8 0 918 720 93 in 1 in 1 in in T c c m Q c T c T m Q m u m h Q W v p f f v p f f f f preparation If you are a student using this Manual you are using it without permission 7167 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 7212 Two rigid tanks that contain water at different states are connected by a valve The valve is opened and steam flows from tank A to tank B until the pressure in tank A drops to a specified value Tank B loses heat to the surroundings The nal temperature in each tank and the entropy generated during this process are to be determined Assumptions 1 Tank A is insulated and thus heat transfer is negligible 2 The water that remains in tank A undergoes a reversible adiabatic process 3 The thermal energy stored in the tanks themselves is negligible 4 The system is stationary Analysis in tank A undergoes a reversible adiabatic process and thus s2 s1 From the steam tables Tables A4 through A6 ank A fi and thus kinetic and potential energy changes are negligible 5 There are no work interactions a The steam T 1704 7 kJkg 0 5928 20246 kJkg 50 504 052552 m kg 0 001061 0 8858 0 5928 001061 0 0 5928 5 59680 1 5302 8479 4 mixture sat 00 kPa 2 48479 kJkg K 5 1191 60 7765 1 1773 6 kJkg 1948 9 60 60422 027788 m kg 0 001084 0 46242 60 001084 0 60 kPa 400 2 2 3 2 2 2 2 sat200 kPa 2 1 2 1 1 1 1 1 3 1 1 1 1 fg A f A fg A f A fg f A A A fg f A fg f A fg f A u x u u x s s s x T T s s P x s s s x u u u x x P v v v v v v 1202 C Tank B The initial and the final masses in tank A are and A steam V 03 m3 P 400 kPa x 06 B steam m 2 kg T 250C P 200 kPa 300 kJ kJkg K 77100 27314 kJkg m kg 11989 C 250 kPa 200 1 1 3 1 1 1 B B B s u T P v 05709 kg m kg 052552 m 03 1080 kg m kg 027788 m 03 3 3 2 2 3 3 1 1 A A A A A A m m v V v V Thus 1080 05709 05091 kg of mass flows into tank B Then 2509 kg 0 5091 2 0 5091 1 2 B B m m The final specific volume of steam in tank B is determined from 0 9558 m kg 2509 kg 2 kg 11989 m kg 3 3 2 1 1 2 2 B B B B B m m m v V v We take the entire contents of both tanks as the system which is a closed system The energy balance for this stationary closed system can be expressed as Substituting B A B A m u m u m u m u Q W U U U Q E E E 1 1 2 2 1 1 2 2 out out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in KE PE 0 since 43 42 1 4243 1 24333 kJkg 2 2731 4 2 509 1 080 1773 6 0 5709 1704 7 300 2 2 B B u u preparation If you are a student using this Manual you are using it without permission 7168 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course T C 1161 mined by applying the entropy balance on an extended system temperature of the extended system is the Thus 09558 m kg 3 v 69156 kJkg K 24333 kJkg 2 2 2 2 B B B B s u b The total entropy generation during this process is deter that includes both tanks and their immediate surroundings so that the boundary temperature of the surroundings at all times It gives B A gen surr b out in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in S S S T Q S S S S 43 42 1 4243 1 Rearranging and substituting the total entropy generated during this process is determined to be 0498 kJK K 290 kJ 300 bsurr surr b 7 7100 2 6 9156 2 509 4 8479 1 080 4 8479 5709 0 out 1 1 2 2 1 1 2 2 out gen T Q m s m s m s m s T Q S S S B A B A preparation If you are a student using this Manual you are using it without permission 7171 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 7216 Problem 215 is rec he effe e state of the steam at the inlet to the feedwater heater is to be investigated The f the ex eam is ass to be constant at the value for 1 MPa 200C and the extraction steam pressure is to be varied from 1 MPa to 100 kPa Both the ratio of the mass flow rates of the extracted steam and the feed eater and change for this process per unit mass of the feedwater are to be plotted as functions of the on press Analysis The prob olved usi nd the results are tabulated and plotted below Knowns WorkFluid S pws P3 1000 kP e aro and T3 ve the table 3 200 C P3 40 emperatureWorkFluidPP4xx4 1 50 C P2 2500 kPa T2 T4 10C Since we dont know the mass flow rates and we want to determine the ratio of mass flow rate of the extracted steam and the feedwater we can assume the mass flow rate of the feedwater is 1 kgs without loss of generality We write the conservation of energy Conservation of mass for the steam extracted from the turbine mdotsteam3 mdotsteam4 Conservation of mass for the condensate flowing through the feedwater heater mdotfw1 1 mdotfw2 mdotfw1 Conservation of Energy SSSF energy balance for the feedwater heater neglecting the change in potential energy no heat transfer no work h3enthalpyWorkFluidPP3TT3 To solve the table place around s3 and remove them from the 2nd and 3rd equations s3entropyWorkFluidPP3TT3 s3 6693 kJkgK This s3 is for the initial T3 P3 T3temperatureWorkFluidPP3ss3 Use this equation for T3 only when s3 is given h4enthalpyWorkFluidPP4xx4 s4entropyWorkFluidPP4xx4 h1enthalpyWorkFluidPP1TT1 s1entropyWorkFluidPP1TT1 h2enthalpyWorkFluidPP2TT2 s2entropyWorkFluidPP2TT2 For the feedwater heater Edotin Edotout Edotin mdotsteam3h3 mdotfw1h1 Edotout mdotsteam4h4 mdotfw2h2 mratio mdotsteam3 mdotfw1 Second Law analysis Sdotin Sdotout Sdotgen DELTASdotsys DELTASdotsys 0 KWK steadyflow result Sdotin mdotsteam3s3 mdotfw1s1 Sdotout mdotsteam4s4 mdotfw2s2 SgenPerUnitMassFWH Sdotgenmdotfw1kJkgfwK 7 onsidered T ct of th entropy o traction st umed water h the total entropy extracti ure lem is s ng EES a teamia a plac und P3 eqations to sol T P4 x T4 t P1 2500 kPa T preparation If you are a student using this Manual you are using it without permission 7172 mratio SgenPerUnitMass kJkgK P3 kPa 02109 02148 0219 0223 01811 0185 0189 01929 732 760 790 820 0227 02309 01968 02005 850 880 02347 02385 02422 02459 02042 02078 02114 02149 910 940 970 1000 021 022 023 024 025 018 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 019 02 021 022 mratio enPerUni J Sg tMassFWH k kgfwK P3 732 kPa P3 1000 kPa For P 732 k 3 Pa Sgen 0 preparation If you are a student using this Manual you are using it without permission 7173 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course ntil r with ied are Tables A11 through A13 10335 Btulbm bm 059750 ft 0 psia 8 80 F 120 psia 2 120 psia 2 3 120 psia 2 2 3 80 psia 1 1 f i f f f g s s h h s s u u P u u P v v v v Analysis Noting that the of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and nergy b flow system can be expressed as m m m m 7217E A rigid tank initially contains saturated R134a vapor The tank is connected to a supply line and is charged u the tank contains saturated liquid at a specified pressure The mass of R134a that entered the tank the heat transfe the surroundings at 100F and the entropy generated during this process are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The direction of heat transfer is to the tank will be verif Properties The properties of R134a l 160 psia 80F R134a 5 ft3 R134a Q 100F 022040 Btulbm R vapor sat 80 psia 1 80 psia 1 g g s s 007934 Btulbm R Btulbm 3817 F 80 psia 160 008589 Btulbm R 4149 Btulbm ft lbm 001360 liquid sat psia 120 f 80 F i i a We take the tank as the system which is a control volume since mass crosses the boundary i T P energies e alances for this uniform Mass balance m m 1 2 system out in i netic by heat work and mass W m u m u m h Q Energy balance in internal ki Change system Net energy transfer out in E E E 43 42 1 4243 1 0 pe ke 1 1 since 2 2 in potential etc energies i i The initial and the final masses in the tank are 36758 lbm ft lbm 001360 ft 5 8 368 lbm ft lbm 059750 ft 5 3 3 2 2 3 3 1 1 v V v V m m Then from the mass balance b The heat transfer during this process is determined from the energy balance to be c The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the tank and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times The entropy balance for it can be expressed as 3592 lbm 8 368 36758 1 2 m m mi 673 Btu 8368 lbm 10335 Btulbm 36758 lbm 4149 Btulbm lbm 3817 Btulbm 3592 1 1 2 2 in m u m u m h Q i i 1 1 2 2 tank gen in b in in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in m s m s S S m s T Q S S S S i i 43 42 1 4243 1 Therefore the total entropy generated during this process is 00264 BtuR 560 R 673 Btu 0 22040 8 368 0 08589 36758 0 07934 2 359 in b in 1 1 2 2 gen T Q m s m s m s S i i preparation If you are a student using this Manual you are using it without permission 7176 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7176 7220 An insulated cylinder is divided into two parts One side of the cylinder contains N2 gas and the other side contains The energy stored in the container itself is tant volume specific heats are R 02968 kPam3kgK cv 0743 kJkgC and and R 20769 kPam3kgK cv 31156 kJkgC and cp 51926 kJkgC for He Tables He gas at different states The final equilibrium temperature in the cylinder and the entropy generated are to be determined for the cases of the piston being fixed and moving freely Assumptions 1 Both N2 and He are ideal gases with constant specific heats 2 negligible 3 The cylinder is wellinsulated and thus heat transfer is negligible Properties The gas constants and the cons c 1039 kJ p kgC for N2 A1 and A2 Analysis The mass of each gas in the cylinder is 04039 kg 1 m kPa 250 4516 kg kPa m kg K 373 K 02968 2 m kPa 250 3 1 3 3 1 V V ents of the cylinder as our system the 1st law relation can be written as at Net energy transfer 0 0 2 2 T mc T T mc T U U U E v v 43 42 1 N2 2 m3 250 kPa 100C He 1 m3 250 kPa 25C 1 N2 RT 1 P m 2 N 20769 kPa m kg K 298 K 3 1 He He RT m 1 P Taking the entire cont He 1 2 N 1 2 He N potential etc energies in internal kinetic Change system and mass work out in E E 43 42 1 he by 0 25 C C 04039 kg 31156 kJkg 100 C Substituting C 4516 kg 0743 kJkg f f T T gives nswer would be the same if the piston were not free to move since it would effect only pressure and not the the cylinder is wellinsulated and thus as ation opy gen 2 0 S S S S S S S 43 42 1 It Tf 795C where T is the final equilibrium temperature in the cylinder f The a specific heats b We take the entire cylinder as our system which is a closed system Noting that essed there is no heat transfer the entropy balance for this closed system can be expr gener Entr by heat and mass Net entropy transfer out in 4243 1 He N gen in entropy Change system But first we determine the final pressure in the cylinder 3 kPa 256 3 m 3 total 2 V P u kmol 8314 kPa m kmol K 3525 K 02623 kmol 04039 kg kg 4516 3 R T N m 02623 4 kgkmol 28 kgkmol He He N total 2 M M N N N m N2 total Then kJK 02978 250 kPa 2563 kPa ln 2 N ln ln 2 1 2 1 2 N P P R T T m c S p 02968 kJkg K 1039 kJkg K ln 373 K 4516 kg 3525 K preparation If you are a student using this Manual you are using it without permission 7178 7221 Problem 7220 is reconsidered The results for constant specific heats to those obtained using variable specific heats are to be compared using builtin EES or other functi ons TN21100 C PN21250 kPa From Table A2a at 27C THe125 C PHe1250 kPa RHe20769 kJkgK From Table A2a CpHeRHeCvHe Solution calculations N21 1273 The entir cylinder is considered to be a closed system allowing the piston to move Conservation of Energy for the closed system out DELTAE we neglect DELTA KE and DELTA PE for the cylinder AE in 0 N2CvN2T2TN21mHeCvHeT2THe1 MMHe 4 kgkmol MN2 28 kgkmol Ntotal mHeMMHemN2MMN2 Final pressure at equilibrium ides is the same P2 is NtotalRuT2273 SgenPistonMoving DELTASHePMDELTASN2PM DELTASHePMmHeCpHelnT2273THe1273RHelnP2PHe1 DELTASN2PMmN2CpN2lnT2273TN21273RN2lnP2PN21 The final temperature of the system when the piston does not move will be the same as when it does move The volume of the gases remain constant and the entropy changes are given by SgenPistNotMoving DELTASHePNMDELTASN2PNM DELTASHePNMmHeCvHelnT2273THe1273 DELTASN2PNMmN2CvN2lnT2273TN21273 The following uses the EES functions for the nitrogen Since helium is monatomic we use the constant specific heat approach to find its property changes Analysis The problem is solved using EES and the results are given below Knowns Ru8314 kJkmolK VN212 m3 CvN20743 kJkgK From Table A2a at 27C RN202968 kJkgK From Table A2a CpN2RN2CvN2 VHe11 m3 CvHe31156 kJkgK mass P VN21mN2RN2TN2 PHe1VHe1mHeRHeTHe1273 e Ein E Ein Eout DELT E kJ Eout 0 kJ At the final equilibrium state N2 and He will have a common temperature DELTAE m Total volume of gases VtotalVN21VHe1 M Allowing the piston to move the pressure on both s P2Vtotal PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 7179 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Eout DELTAEVP ENERGYN2TTN21mHeCvHeT2VP THe1 Final Pressure for moving piston P2VPVtotalNtotalRuT2VP273 SgenPistMovingVP DELTASHePMVPDELTASN2PMVP DELTASN2PMVPmN2ENTROPYN2TT2VPPP2VPENTROPYN2TTN21PPN21 DELTASHePMVPmHeCpHelnT2273THe1273RHelnP2PHe1 Fianl N2 Pressure for piston not moving P2N2VPVN21mN2RN2T2VP273 SgenPistNotMovingVP DELTASHePNMVPDELTASN2PNMVP DELTASN2PNMVP mN2ENTROPYN2TT2VPPP2N2VP ENTROPYN2TTN21PPN21 DELTASHePNMVPmHeCvHelnT2VP273THe1273 193 kJkgK pN2104 kJkgK vHe3116 kJkgK vN20743 kJkgK DELTAE0 kJ DELTAEVP0 kJ DELTASHePM03318 kJK DELTASHePMVP03318 kJK DELTASHePNM02115 kJK DELTASHePNMVP02116 kJK DELTASN2PM0298 kJK DELTASN2PMVP0298 kJK DELTASN2PNM01893 kJK DELTASN2PNMVP01893 kJK Ein0 kJ Eout0 kJ MMHe4 kgkmol MMN228 kgkmol mHe04039 kg mN24516 kg Ntotal02623 kmol P22563 kPa P2N2VP2363 kPa P2VP2563 kPa RHe2077 kJkgK RN202968 kJkgK Ru8314 kJkmolK SgenPistMovingVP00338 kJK SgenPistNotMoving002226 kJK SgenPistNotMovingVP002238 kJK SgenPistonMoving003379 kJK T27954 C T2VP7958 C Vtotal3 m3 Ein DELTAEVP mN2INTENERGYN2TT2VPINT SOLUTION CpHe5 C C C preparation If you are a student using this Manual you are using it without permission 7180 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course tants and the constant volume specific heats are R 02968 kPam3kgK c 0743 kJkgC and or He Tables A1 and A2 The specific heat of the copper at room temperature is c 0386 kJkgC Table A3 lysis The mass of each gas in the cylinder is 7222 An insulated cylinder is divided into two parts One side of the cylinder contains N2 gas and the other side contains He gas at different states The final equilibrium temperature in the cylinder and the entropy generated are to be determined for the cases of the piston being fixed and moving freely Assumptions 1 Both N2 and He are ideal gases with constant specific heats 2 The energy stored in the container itself except the piston is negligible 3 The cylinder is wellinsulated and thus heat transfer is negligible 4 Initially the piston is at the average temperature of the two gases Properties The gas cons v cp 1039 kJkgC for N2 and R 20769 kPam3kgK cv 31156 kJkgC and cp 51926 kJkgC f Ana 0808 kg 20769 kPa m kg K 298 K 500 kPa 1m 3 3 He 1 1 1 He 2 RT P m V 477 kg kPa m kg K 353 K 02968 1 m kPa 500 3 3 N 1 1 1 N2 RT P m V tents of the cylinder as our system the 1st law relation can be written as 2 2 T mc T T mc T T T mc U U U U E E v v 43 42 1 43 T u 80 25 2 525C N2 1 m3 500 kPa 80C 500 kPa 25C He 1 m3 Copper Taking the entire con 0 0 Cu 1 2 He 1 2 N 1 2 Cu He N potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out Ein 42 1 where 1 C Substituting 0 525 C C 50 kg 0386 kJkg 25 C C 0808 kg 31156 kJkg 80 C C kg 0743 kJkg 477 f f f T T T where Tf is the final equilibrium temperature in the cylinder The answer would be the same if the piston were not free to move since it would effect only pressure and not the specific heats b We take the entire cylinder as our system which is a closed system Noting that the cylinder is wellinsulated and thus is n heat transfer the entropy balance for this closed system can be expressed as system gen out in 2 0 S S S S S S S But first we determine the final pressure in the cylinder It gives Tf 560C there o Change generation Entropy by heat and mass Net entropy transfer 43 42 1 4243 1 S piston He N gen entropy in 5094 kPa m 2 03724 kmol 4 kgkmol 0808 kg 28 kgkmol kg 477 3 total He N He N total 2 2 M m M m N N N kmol 8314 kPa m kmol K 329 K 03724 3 total 2 V R T N P u preparation If you are a student using this Manual you are using it without permission 7182 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course ulated rigid tank equipped with an electric heater initially contains pressurized air A valve is opened and air is allowed to escape at constant temperature until the pressure inside drops to a specified value The amount of electrical work done during this process and the total entropy change are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the exit temperature and enthalpy of air remains constant 2 Kinetic and potential energies are negligible 3 The tank is insulated and thus heat transfer is negligible 4 Air is an ideal gas with variable specific heats Properties The gas constant is R 0287 kPam3kgK Table A1 The properties of air are Table A17 Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance 7223 An ins 23561 kJkg K 330 23561 kJkg K 330 33034 kJkg K 330 2 2 1 1 u T u T h T e e AIR 5 m3 500 kPa 57C We 2 1 system out in m m m m m m e Energy balance The initial and the final masses of air in the tank are 0 pe ke 1 1 since 2 2 in e potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in Q m u m u m h W E E E e e 43 42 1 4243 1 1056 kg 0287 kPa m kg K 330 K 5 m kPa 200 2640 kg kPa m kg K 330 K 0287 5 m kPa 500 3 3 2 2 2 3 3 1 1 1 RT P m RT P m V V Then from the mass and energy balances b The total entropy change or the total entropy generation within the tank boundaries is determined from an entropy balance on the tank expressed as or 1584 kg 1056 2640 2 1 m m me 1501 kJ 2640 kg 23561 kJkg 1056 kg 23561 kJkg kg 33034 kJkg 1584 1 1 2 2 ein m u m u m h W e e tank gen in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in S S s m S S S S e e 43 42 1 4243 1 e e e e e e e s m s s s m m s m s s m m m s m s m s S m s S 1 1 2 2 1 1 2 2 2 1 1 1 2 2 tank gen Assuming a constant average pressure of 500 2002 350 kPa for the exit stream the entropy changes are determined to be preparation If you are a student using this Manual you are using it without permission 7184 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course l properties of the ice are constant 2 The cylinder is wellinsulated and thus heat transfer is ting thus the pressure remains constant during this phase change process and thus Wb U H the work and mass b H H H U W b in 7224 An insulated cylinder initially contains a saturated liquidvapor mixture of water at a specified temperature The entire vapor in the cylinder is to be condensed isothermally by adding ice inside the cylinder The amount of ice added and the entropy generation are to be determined Assumptions 1 Therma negligible 3 There is no stirring by hand or a mechanical device it will add energy Properties The specific heat of ice at about 0C is c 211 kJkgC Table A3 The melting temperature and the heat of fusion of ice at 1 atm are 0C and 3337 kJkg Analysis a We take the contents of the cylinder ice and saturated water as our system which is a closed system No that the temperature and energy balance for this system can be written as in internal kinetic Change system Net energy transfer E E E out in 43 42 1 4243 1 0 0 water ice potential etc energies heat y or 0 C 0 C water 1 2 liquid ice 2 1 solid h m h mc T mh T 0 mc if ice 18C WATER 002 m3 100C The properties of water at 100C are Table A4 22564 kJkg 17 419 16720 m kg 0 001043 3 fg f g f h h v v 6 0490 kJkgK 1 3072 fg f s s 0119 kg m kg 016814 m 002 kJkg K 13072 kJkg 41917 kJkg K 9119 64481 kJkg 01 22564 41917 m kg 6814 3 1 eam 100 C 2 1 3 v V m s s h h x h h h f fg f o JkgK018 3337 kJkg 418 kJkgC1000C 0119 kg41917 64481 kJkg 0 m 340 g ice as our system which is a closed system Considering that the tank is wellinsulated and is no h t transfer the entropy balance for this closed system can be expressed as system gen out in 0 S S S S S S S 43 42 1 4243 1 01 0001043 01 16720 0001043 1 1 v v v x fg f 1 1 01 60470 13072 1 1 x s s s fg f 3 100 C 2 f o 1 Noting that T st 1 ice 18C and T2 100C and substituting gives m211 k 0034 kg b We take the ice and the steam thus there ea in entropy Change generation Entropy by heat and mass Net entropy transfer steam ice gen preparation If you are a student using this Manual you are using it without permission 7185 00907 kJK 27315 K 418 kJkg K ln 37315 K 27315 K 3337 kJkg 25515 K 211 kJkg K ln 27315 K kg 0034 ln ln 00719 kJK 19119 kJkg K kg 13072 0119 2 melting liquid ice melting solid ice 1 2 steam T mc mh T mc S S S S s m s S if ice liquid T T T 1 melting solid 1 Then 00188 kJK ice steam gen S S S 0 0907 0 0719 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 7188 3005K 275 C cout o T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course tion in this case is determined similarly to be The rate of entropy generation in the heating sec 00350 kJK heating section is S duction 3005 K 1060 kgs 418 kJkg K ln 316 K ln 2 gen T T mc S 1 Thus the reduction in the rate of entropy generation within the 00272 kWK 0 0350 0 0622 re preparation If you are a student using this Manual you are using it without permission 7189 7227 Using EES or other software the work input to a multistage compressor is to be determined for a given set ed to be identical and the be tabulated and plotted against the number of stages for 35 for air Analysis The problem is solved using EES and the results are tabulated and plotted below AS Air stage 2 number of stages of compression with intercooling each having same pressure ratio n135 K RuM 21000 kPa Rp P2P11Nstage Wdotcomp NstagenRT1273n1Rpn1n 1 Nstage w of inlet and exit pressures for any number of stages The pressure ratio across each stage is assum compression process to be polytropic The compressor work is to P1 100 kPa T1 25C P2 1000 kPa and n 1 G N MMMOLARMASSGAS Ru 8314 kJkmol R M k14 P1100 kPa T125 C P 1 2 3 4 5 6 7 8 9 10 200 210 220 230 240 250 260 270 Nstage wcomp kJkg comp kJkg 1 2 3 4 5 6 7 8 9 10 2694 2295 2179 2124 2092 2071 2056 2045 2036 2029 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 7190 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course of ir properties and air Properties he specific heat of exhaust gases at the average temperature at an anticipated average temperature of 100ºC are cp 1011 kJkgK and k 1397 Table 2 Analysis a The turbine power output is determined from p Then the air temperature at the compressor exit becomes 2 1 2 air C 70 C 0 018 kgs101 1kJkg C 1kW 1 T T T c m p The air temperature at the comp for the case of isentropic process is 7228 The turbocharger of an internal combustion engine consisting of a turbine driven by hot exhaust gases and a compressor driven by the turbine is considered The air temperature at the compressor exit and the isentropic efficiency the compressor are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential Compressor Turbine Exh gas 450C 002 kgs Air 70C 95 kPa 0018 kgs 400C 135 kPa energy changes are negligible 3 Exhaust gases have a is an ideal gas with constant specific heats T of 425ºC is cp 1075 kJkgK and properties of air A 1 075 kW 0 02 kgs107 5 kJkg C450 400 C 2 1 exh T T T c m W For a mechanical efficiency of 95 between the turbine and the compressor 1 021 kW 0 95 1 075 kW T C W W m η W 02 2 1261C T b ressor exit 106 C 379 K 95 kPa 273 K 135 kPa 70 1 2 1 2 s P T 139711397 1 k k P T ined to be The isentropic efficiency of the compressor is determ 0642 70 126 1 1 2 C T T η 106 1 2 T T s 70 preparation If you are a student using this Manual you are using it without permission 7191 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 229 Air is allowed to enter an insulated pistoncylinder device until the volume of the air increases by 50 The final mperature in the cylinder the amount of mass that has entered the work done and the entropy generation are to be determined Assumptions 1 Kinetic and potential energy changes are negligible 2 Air is an ideal gas with constant specific heats Properties The gas constant of air is R 0287 kJkgK and the specific heats of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK Table A2 Analysis The initial pressure in the cylinder is 7 te 235 5 kPa m 025 273 K kg0287 kPa m kg K20 07 3 3 1 1 1 1 V m RT P Air 025 m3 07 kg 20C Air 500 kPa 70C 2 2 3 3 2 2 2 2 71 307 kPa m kg K 0287 2355 kPa15 025 m T T RT P m V A mass balance on the system gives the expression for the mass entering the cylinder 70 71 307 2 1 2 T m m mi c Noting that the pressure remains constant the boundary work is determined to be a An energy balance on the system may be used to determine the final temperature 2943 kJ 3 1 2 1 bout 235 5 kPa15 02505m V P V W 273 70 0 71820 30771 0 718 2943 273 100570 70 30771 2 2 2 1 1 2 2 out b 1 1 2 2 out b T T T m c T m c T W c T m m u m u W h m p i i i i v v There is only one unknown which is the final temperature By a trialerror approach or using EES we find T2 3080 K b The final mass and the amount of mass that has entered are 0 999 kg 308 0 30771 2 m d The rate of entropy generation is determined from 0299 kg 70 0 999 1 2 m m mi 00673 kJK 500 kPa 0287 kJkgKln 235 5 kPa 343 K kg 1005 kJkgKln 293 K 70 500 kPa 0287 kJkgKln 235 5 kPa 343 K 0 999 kg 1005 kJkgKln 308 K ln ln ln ln 1 1 1 2 2 2 1 1 2 2 1 2 1 1 2 2 1 1 2 2 gen i i p i i p i i i i i P P R T T c m P P R T T c m s s m s s m s m m m s m s m s m s m s S preparation If you are a student using this Manual you are using it without permission 7192 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course be determined Analysis The maximum possible power that can be obtained from this turbine for the given inlet and exit pressures can be deter 7230 A cryogenic turbine in a natural gas liquefaction plant produces 115 kW of power The efficiency of the turbine is to Assumptions 1 The turbine operates steadily Properties The density of natural gas is given to be 4238 kgm3 Cryogenic turbine LNG 30 bar 160C 20 kgs 3 bar mined from 127 4 kW 300kPa 3000 423 8 kgm 3 out in max P P W ρ 20 kgs m iven the tual power the efficiency of this cryogenic turbine becomes G ac 903 0903 127 4 kW kW 115 Wmax W η T n his efficiency is also known as hydraulic efficiency since the cryogenic turbine handles atural gas in liquid state as the hydraulic turbine handles liquid water preparation If you are a student using this Manual you are using it without permission 7193 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course ure of e to be obtained and plotted against initial temperature in the tank Properties The constantvolume specific heat of air at 300 K is cv 0718 kJkgK Table A2a Analysis The entropy change of air in the tank is 7231 Heat is transferred from a tank to a heat reservoir until the temperature of the tank is reduced to the temperat reservoir The expressions for entropy changes ar 1 2 1 2 1 2 air ln ln ln T T mc R T T m c S v v V V The entropy change of heat reservoir is L v L T T mc T T Q S 2 1 HR The total entropy change ie entropy generation is L v v T T T mc T T mc S S S S ln 2 1 1 2 HR air total gen The heat transfer will continue until T2 TL Now using m 2 kg cv 0718 kJkgK and TL 300 K we plot entropy hange terms against initial temperature as shown in the figure c 100 150 200 250 300 350 400 450 500 1 08 06 04 02 0 16 14 02 04 06 08 1 12 T1 K a opy ch nge kJK Entr Stank SHR Sgen preparation If you are a student using this Manual you are using it without permission 7194 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course r The tropy change work produced and Properties The constantvolume specific heat of air at 300 K is cv 0718 kJkgK Table A2a Analysis a The entropy change of air in the tank is 7232 Heat is transferred from a tank to a heat engine which produces work This work is stored in a work reservoi initial temperature of the air for maximum work and thermal efficiency the total en thermal efficiency are to be determined for three initial temperatures 1 2 1 2 1 2 tank ln ln ln T T mc R T T m c S v v V V he heat transfer will continue until T2 TL Thus T 1 tank mc ln T S L v T tropy change of heat reservoir is The en L L T Q S HR The entropy change of heat engine is zero since the engine is reversible and produces maximum work The work reservoir involves no entropy change Then the total entropy change ie entropy generation is en 0 0 HR tank WR HE HR tank total S S S S S S es g S S which becom L L L v T Q T T mc S 1 gen ln 1 The expression for the thermal efficiency is QH ηth 2 W r H L Q ηth 1 Q o 3 T T mc Q 4 eration will be zero Then using m 2 kg cv 0718 kJkgK TL 300 K QL 400 kJ and solving equations 1 2 3 and 4 simultaneously using an equation solver such as EES we obtain 1 H Q T 03934 K 7592 η b At the initial air temperature of 7592 K the entropy generation is zero and At the initial air temperature of 7592100 6592 K we obtain Heat transfer from the tank is expressed as 1 L v H In ideal operations the entropy gen 4 kJ 259 th W 4 kJ 659 kJ 2594 03934 0 W S th gen η kJ 1158 02245 kJK 02029 W S th gen η preparation If you are a student using this Manual you are using it without permission 7195 At the initial air temperature of 7592 100 8592 K we obtain 01777 kJK Sgen ηth 05019 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course W A negative value for entropy generation indicates that this process is not possible with the given values c The thermal efficiency and the entropy generation as functions of the initial temperature of the air are plotted below kJ 4030 500 600 700 800 900 1000 04 04 0 02 02 06 08 04 02 0 02 04 06 T1 K ηth Sgen kJK preparation If you are a student using this Manual you are using it without permission 7196 7233 It is to be shown that for an ideal gas with constant specific heats the compressor and turbine isentropic efficiencies PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course may be written as 1 1 2 T T ηC 1 1 2 P P k k and 1 1 3 4 T T 1 3 4 k k T P P η Analysis The isentropic efficiency of a compressor for an ideal gas with constant specific heats is given by 1 2 1 2 1 2 1 2 1 2 1 2 T T T T T T c T T c h h h h s p s p s C η The temperature at the compressor exit fort he isentropic case is expressed a s k k s P P T T 1 1 2 1 2 Substituting 1 1 1 1 1 1 1 T T T 2 1 2 1 1 1 2 1 1 1 1 2 1 2 T P T P T T T P T T T T T s ηC The isentropic efficiency of a turbine for an ideal gas with constant specific heats is given by 1 2 1 2 1 2 P P T P k k k k k k 3 4 3 4 3 4 3 4 3 4 3 4 T T T T T T c T T c h h h h s s p p s T η The temperature at the turbine exit fort he isentropic case is expressed as k k P T T 1 4 s P3 3 4 Substituting 1 1 1 1 1 3 4 3 4 1 3 4 3 3 4 3 3 1 3 4 3 3 4 3 4 3 4 k k k k k k s C P P T T P P T T T T T P P T T T T T T T η preparation If you are a student using this Manual you are using it without permission 7197 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course l gas during the isentropic process is to be obtained nalysis The expression is obtained as follows 7234 An expression for the internal energy change of an idea A 2 2 2 1 Pv RT R u T T 2 2 1 1 1 1 1 1 1 2 1 1 1 1 1 1 k k k k Pd on v Pv v u const v dv const k k k k k C T T s of entropy and specific volume The Tv relation for this ideal 0 k du Tds Pdv Pdv u v where P c st v 2 u V 7235 The temperature of an ideal gas is given as function gas undergoing an isentropic process is to be obtained Analysis The expressions for temperatures T1 and T2 are exp exp and 2 1 2 2 1 1 1 1 2 2 2 1 1 1 v k v k c s A T c s A T T s T T s T v v v v Now divide T by T 2 1 exp exp 1 1 2 1 2 k A A T T v v 1 2 v v c s c s 1k Since A and cv are constants and the process is isentropic s2 s1 the temperature ratio reduces to 1 1 2 k T v 2 1 T v preparation If you are a student using this Manual you are using it without permission 7198 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course ss and energy for the steadyflow process are m 7236 An ideal gas undergoes a reversible steadyflow process An expression for heat transfer is to be obtained Analysis a The conservation of ma exits inlets e mi exits net inlets net 2 2 e e i i gz h m W gz h m Q 2 2 V V here the sign of heat transfer is to the system and work is from the system For one entrance one exit neglecting kinetic s and unit mass flow the conservation of mass and energy reduces to i e h h w the isothermal process is found from w and potential energie qnet net The steadyflow reversible work for P RT P dP dP w ln netsfrev v e e Pe RT i i i Then ln net net i e p i e i e T T c P P RT RT h h w q However i e T T Thus RT q ln net i e P P i s T s ds T Tds q change of an ideal gas is found from b For the isothermal process the heat transfer may be determined from e e net i e i The entropy i e T p T dT c s s e T i e P P R T i ln For the isothermal process the entropy change is i e i e P P R s s ln The expression for heat transfer is i e i e P P RT s T s q ln net which is the same as part a preparation If you are a student using this Manual you are using it without permission 7199 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course sfer ared to the result by a first law analysis 1 p k k s c P 7237 The temperature of an ideal gas is given as functions of entropy and specific volume The expression for heat tran is to be determined and comp Analysis The heat transfer may be expressed as A T s P exp 2 2 exp 1 s k k s p ds s c AP Tds Q q 1 1 exp 1 s p s k k m s c AP T s v integral becomes For P constant the exp 2 1 p p k k c s c AP 1 1s exp exp 1 1 2 p s s p p k k p c s s c c AP ds s c oting that his result is the same as that given by applying the first law to a closed system undergoing a constant pressure process exp 1 2s k q AP k N exp 1 k k c s AP T exp 1 1 1 p k k c s AP T 2 2 p We obtain 1 2 T T c q p T preparation If you are a student using this Manual you are using it without permission 7200 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course sion um intercooling pressure is to be obtained s 7238 A twostage compression with intercooling process is considered The isentropic efficiencies for the two compres processes are given The optim Analysis The work of compression i c 1 1 1 1 1 1 1 1 1 3 4 3 1 1 2 1 3 4 3 1 2 1 3 4 1 k k H C k k C L p s C H s C L p p P P T P P T T T T T T T c T T c T T c T T c T η η η η e that intercooling takes place at constant pressure and the gases can be cooled to the inlet temperature for uch that P3 P2 and T3 T1 Then the total work supplied to the compressor becomes 2 comp cp T w 1 1 3 4 1 2 s C H p s C L p η η We are to assum ressor s the comp C H C L P P 2 1 η η cpT 2 1 C H L C k k k k H C k k L C P P P P P P 1 4 1 2 4 1 1 2 1 1 1 1 1 1 1 η η η η ut for fixed compressor inlet conditions T1 P1 exit iencies we set cpT w 1 comp 1 k k 1 To find the unknown pressure P2 that gives the minimum work inp pressure P4 and isentropic effic 0 2 2 comp dP P dw This yields 1 2 4 1 2 k C L P P η 1 k k C H P P η P2 P3 the pressure ratios across the two compressors are related by 1 2 4 1 2 k k C L P P P η η k H C Since 1 3 1 P P 4 1 2 k k C H k k C L P P η η preparation If you are a student using this Manual you are using it without permission 7201 Fundamentals of Engineering FE Exam Problems PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course S erature of 30 rejecting W c as h the cond b 018 MWK c 0 MWK d 056 MWK e 122 MWK nswer b 018 MWK olution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES T1273 MWK 3SchangeQoutT1273 Wrong sign angesfg Taking entropy of vaporization SteamIAPWSTT1x1ENTROPYSteamIAPWSTT1x0 MPa and 300C to 10 MPa isentropically The final temperature of the steam is 290C b 300C c 311C d 371C e 422C y modifying numerical values 16000 kPa T1300 C P210000 kPa s2s1 s1ENTROPYSteamIAPWSTT1PP1 T2TEMPERATURESteamIAPWSss2PP2 Some Wrong Solutions with Common Mistakes W1T2T1 Assuming temperature remains constant W2T2TEMPERATURESteamIAPWSx0PP2 Saturation temperature at P2 W3T2TEMPERATURESteamIAPWSx0PP2 Saturation temperature at P1 7239 team is condensed at a constant temp C as it flows through the condenser of a power plant by heat at a rate of 55 M The rate of entropy hange of steam it flows throug enser is a 183 MWK A S screen Similar problems and their solutions can be obtained easily by modifying numerical values T130 C Qout55 MW SchangeQout Some Wrong Solutions with Common Mistakes W1Schange0 Assuming no change W2SchangeQoutT1 Using temperature in C W W4Sch sfgENTROPY 7240 Steam is compressed from 6 a Answer d 371C Solution Solved by EES Software Solutions can be verified by copy screen Similar problems and their solutions can be obtained easily b ingandpasting the following lines on a blank EES P preparation If you are a student using this Manual you are using it without permission 7202 7241 An apple with an average mass of 012 kg and average specific heat of 365 kJkgC is cooled from 25C to 5C The entropy change of the apple is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course K K 1273 242 A pistoncylinder device contains 5 kg of saturated water vapor at 3 MPa Now heat is rejected from the cylinder at the JK kJK kJK kJK kJK g PYSteamIAPWSPP1x1ENTROPYSteamIAPWSPP1x0 msfg kJK a 0705 kJK b 0254 kJK c 00304 kJ d 0 kJK e 0348 kJK Answer c 00304 kJK Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values C365 kJkg m012 kg T125 C T25 C SchangemClnT2273T Some Wrong Solutions with Common Mistakes W1SchangeClnT2273T1273 Not using mass W2SchangemClnT2T1 Using C W3SchangemCT2T1 Using Wrong relation 7 constant pressure until the water vapor completely condenses so that the cylinder contains saturated liquid at 3 MPa at end of the process The entropy change of the system during this process is a 0 kJK b 35 k c 125 d 177 e 195 Answer d 177 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P13000 kPa m5 k sfgENTRO Schange preparation If you are a student using this Manual you are using it without permission 7203 7243 Helium gas is compressed from 1 atm and 25C to a pressure of 10 atm adiabatically The lowest temperature of helium after compression is a 25C b 63C c 250C d 384C e 476C Answer e 476C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 25 kPa 25 kPa erature will be lowest for isentropic compression C with Common Mistakes 1T2T1 Assuming temperature remains constant al to P using C and 500C to 01 MPa at a rate of 2 kgs If steam leaves the rbine as saturated vapor the power output of the turbine is a 2058 kW b 1910 kW c 1780 kW d 1674 kW e 1542 kW Answer e 1542 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P14000 kPa T1500 C P2100 kPa x21 m2 kgs h1ENTHALPYSteamIAPWSTT1PP1 h2ENTHALPYSteamIAPWSxx2PP2 Woutmh1h2 Some Wrong Solutions with Common Mistakes s1ENTROPYSteamIAPWSTT1PP1 h2sENTHALPYSteamIAPWS ss1PP2 W1Woutmh1h2s Assuming isentropic expansion k1667 P11013 T125 C P2101013 s2s1 The exit temp T2T1273P2P1k1k K T2C T2273 Some Wrong Solutions W W2T2T1P2P1k1k Using C instead of K W3T2T1273P2P1273 Assuming T is proportional to P W4T2T1P2P1 Assuming T is proportion 7244 Steam expands in an adiabatic turbine from 4 MPa tu preparation If you are a student using this Manual you are using it without permission 7204 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course c 143 MW d 176 MW e 208 MW 176 MW olution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES p05203 3 K a 1WmaxmCvT1T2 Using Cv e amount of q If the entropy of the substance is s1 at state 1 and s2 at state 2 the entropy s s2 s1 c s s2 s1 d s s2 s1 qT e s s2 s1 qT nswer c s s2 s1 an ideal gas at temperature T undergoes a reversible isothermal process from pressure P1 to pressure o the surroundings at temperature T in the amount of q If the gas constant of the gas is R the entropy of the gas s during this process is 1 b s R lnP2P1 qT c s R lnP1P2 d s R lnP1P2qT e s 0 R lnP1P2 7245 Argon gas expands in an adiabatic turbine from 3 MPa and 750C to 02 MPa at a rate of 5 kgs The maximum power output of the turbine is a 106 MW b 129 MW Answer d S screen Similar problems and their solutions can be obtained easily by modifying numerical values C k1667 P13000 kPa T175027 m5 kgs P2200 kP s2s1 T2T1P2P1k1k WmaxmCpT1T2 Some Wrong Solutions with Common Mistakes Cv02081kJkgK W T22T1P2P1k1k Using C instead of K W2WmaxmCpT1T22 W3WmaxCpT1T2 Not using mass flow rate T24T1P2P1 Assuming T is proportional to P using C W4WmaxmCpT1T24 7246 A unit mass of a substance undergoes an irreversible process from state 1 to state 2 while gaining heat from the surroundings at temperature T in th change of the substance s during this process is a s s2 s1 b A 7247 A unit mass of P2 while loosing heat t change a s R lnP2P Answer c s preparation If you are a student using this Manual you are using it without permission 7205 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course versible isothermal compression and g the rever atic compre rding entrop air per unit mass is b sisot sadia0 c sadia 0 d sisot 0 e sisot 0 nswer d sisot 0 as is compressed from 27C and 35 m3kg to 0775 m3kg in a reversible adiabatic manner The temperature on is a 74C b 122C c 547C d 709C e 1082C olution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES reen Similar problems and their solutions can be obtained easily by modifying numerical values 3kg om isentropic compression relation 2T1273v1v2k1 K 1T2T1 Assuming temperature remains constant 1v1v2k1 Using C instead of K v2273 Assuming T is proportional to v ssuming T is proportional to v using C 7248 Air is compressed from room conditions to a specified pressure in a reversible manner by two compressors one isothermal and the other adiabatic If the entropy change of air is sisot during the re sadia durin sible adiab ssion the correct statement rega y change of a sisot sadia0 A 7249 Helium g of helium after compressi Answer c 547C S sc k1667 v135 m3kg T127 C v20775 m s2s1 The exit temperature is determined fr T T2C T2273 C Some Wrong Solutions with Common Mistakes W W2T2T W3T2T1273v1 W4T2T1v1v2 A preparation If you are a student using this Manual you are using it without permission 7206 7250 Heat is lost through a plane wall steadily at a rate of 600 W If the inner and outer surface temperatures of the wall are 20C and 5C respectively the rate of entropy generation within the wall is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course K K W ce Sin Sout Sgen DSsystem for the wall for steady operation gives ome Wrong Solutions with Common Mistakes stead of K in K difference in K K C P2P1k1k273 tacompCpT2sT1CpT2T1 Some Wrong Solutions with Common Mistakes T2sW1T1P2P1k1k Using C instead of K in finding T2s W1EtacompCpT2sW1T1CpT2T1 W2EtacompT2sT2 Using wrong definition for isentropic efficiency and using C W3EtacompT2s273T2273 Using wrong definition for isentropic efficiency with K a 011 WK b 421 WK c 210 WK d 421 WK e 900 W Answer a 011 W Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Q600 T120273 K T25273 K Entropy balan QT1QT2Sgen0 WK S QT1273QT2273W1Sgen0 Using C in W2SgenQT1T22 Using avegage temperature W3SgenQT1T22273 Using avegage temperature in C W4SgenQT1T2273 Using temperature 7251 Air is compressed steadily and adiabatically from 17C and 90 kPa to 200C and 400 kPa Assuming constant specific heats for air at room temperature the isentropic efficiency of the compressor is a 076 b 094 c 086 d 084 e 100 Answer d 084 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cp1005 kJkg k14 P190 kPa T117 P2400 kPa T2200 C T2sT1273 E preparation If you are a student using this Manual you are using it without permission 7207 7252 Argon gas expands in an adiabatic turbine steadily from 600C and 800 kPa to 80 kPa at a rate of 25 kgs For an isentropic efficiency of 88 the power produced by the turbine is a 240 kW b 361 kW c 414 kW d 602 kW e 777 kW Answer d 602 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 280 kPa 2sW1T1P2P1k1k Using C instead of K to find T2s taturbCpT2W1T1CpT2sW1T1 kJkgK T2 Using Cv instead of Cp Cp05203 kJkgK k1667 m25 kgs T1600 C P1800 kPa P T2sT1273P2P1k1k273 Etaturb088 EtaturbCpT2T1CpT2sT1 WoutmCpT1T2 Some Wrong Solutions with Common Mistakes T E W1WoutmCpT1T2W1 EtaturbCpT2sT1CpT2W2T1 Using wrong definition for isentropic efficiency and using C W2WoutmCpT1T2W2 W3WoutCpT1T2 Not using mass flow rate Cv03122 W4WoutmCvT1 preparation If you are a student using this Manual you are using it without permission 7208 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course where the discharge pressure is measured is 61 m above the inlet section The c 27 kW d 52 kW e 44 kW nswer c 27 kW es on a blank EES reen Similar problems and their solutions can be obtained easily by modifying numerical values 3s 2 um when compression is reversible and thus wvP2P1Dpe 1VOLUMESteamIAPWSTT1PP1 kW The effect of 61 m elevation difference turns out to be small ome Wrong Solutions with Common Mistakes 1Winmv1P2P1 Disregarding potential energy 2Winmv1P2P1mgh1000 Subtracting potential energy instead of adding 10 kW b 112 kW c 258 kW d 193 kW e 161 kW by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES s and their solutions can be obtained easily by modifying numerical values ome Wrong Solutions with Common Mistakes W1WinmRTlnP2P1 Using C instead of K W2WinmTP2P1 Using wrong relation W3WinRT273lnP2P1 Not using mass flow rate 7253 Water enters a pump steadily at 100 kPa at a rate of 35 Ls and leaves at 800 kPa The flow velocities at the inlet and the exit are the same but the pump exit minimum power input to the pump is a 34 kW b 22 kW A Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lin sc V0035 m g981 ms h61 m P1100 kPa T120 C P2800 kPa Pump power input is minim v mVv1 Wminmv1P2P1mgh1000 kPam3s S W W W3Winmv1P2P1mgh Not using the conversion factor 1000 in PE term W4Winmv1P2P1mgh1000 Adding pressures instead of subtracting 7254 Air at 15C is compressed steadily and isothermally from 100 kPa to 700 kPa at a rate of 012 kgs The minimum power input to the compressor is a Answer d 193 kW Solution Solved screen Similar problem Cp1005 kJkgK R0287 kJkgK Cv0718 kJkgK k14 P1100 kPa T15 C m012 kgs P2700 kPa WinmRT273lnP2P1 S preparation If you are a student using this Manual you are using it without permission 7209 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course ally from 1 atm to 16 atm by a twostage compressor To minimize the total compression work the intermediate pressure between the two stages must be a 3 atm b 4 atm c 85 atm d 9 atm e 12 atm Answer b 4 atm Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P11 atm P216 atm PmidSQRTP1P2 Some Wrong Solutions with Common Mistakes W1PP1P22 Using average pressure W2PP1P22 Half of product 7256 Helium gas enters an adiabatic nozzle steadily at 500C and 600 kPa with a low velocity and exits at a pressure of 90 kPa The highest possible velocity of helium gas at the nozzle exit is a 1475 ms b 1662 ms c 1839 ms d 2066 ms e 3040 ms Answer d 2066 ms Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values k1667 Cp51926 kJkgK Cv31156 kJkgK T1500 C P1600 kPa Vel10 P290 kPa s2s1 for maximum exit velocity The exit velocity will be highest for isentropic expansion T2T1273P2P1k1k273 C Energy balance for this case is hkeconstant for the fluid stream QWpe0 05Vel121000CpT105Vel221000CpT2 Some Wrong Solutions with Common Mistakes T2aT1P2P1k1k Using C for temperature 05Vel121000CpT105W1Vel221000CpT2a T2bT1P2P1k1k Using Cv 05Vel121000CvT105W2Vel221000CvT2b T2cT1P2P1k Using wrong relation 05Vel121000CpT105W3Vel221000CpT2c 7255 Air is to be compressed steadily and isentropic preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7210 7257 Combustion gases with a specific heat ratio of 13 enter an adiabatic nozzle steadily at 800C and 800 kPa with a low velocity and exit at a pressure of 85 kPa The lowest possible temperature of combustion gases at the nozzle exit is a 43C b 237C c 367C d 477C e 640C Answer c 367C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values k13 T1800 C P1800 kPa P285 kPa Nozzle exit temperature will be lowest for isentropic operation T2T1273P2P1k1k273 Some Wrong Solutions with Common Mistakes W1T2T1P2P1k1k Using C for temperature W2T2T1273P2P1k1k Not converting the answer to C W3T2T1P2P1k Using wrong relation 7258 Steam enters an adiabatic turbine steadily at 400C and 5 MPa and leaves at 20 kPa The highest possible percentage of mass of steam that condenses at the turbine exit and leaves the turbine as a liquid is a 4 b 8 c 12 d 18 e 0 Answer d 18 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P15000 kPa T1400 C P220 kPa s2s1 s1ENTROPYSteamIAPWSTT1PP1 x2QUALITYSteamIAPWSss2PP2 moisture1x2 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7211 7259 Liquid water enters an adiabatic piping system at 15C at a rate of 8 kgs If the water temperature rises by 02C during flow due to friction the rate of entropy generation in the pipe is a 23 WK b 55 WK c 68 WK d 220 WK e 443 WK Answer a 23 WK Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cp4180 JkgK m8 kgs T115 C T2152 C SgenmCplnT2273T1273 WK Some Wrong Solutions with Common Mistakes W1SgenmCplnT2T1 Using deg C W2SgenCplnT2T1 Not using mass flow rate with deg C W3SgenCplnT2273T1273 Not using mass flow rate with deg C 7260 Liquid water is to be compressed by a pump whose isentropic efficiency is 75 percent from 02 MPa to 5 MPa at a rate of 015 mP3 Pmin The required power input to this pump is a 48 kW b 64 kW c 90 kW d 160 kW e 120 kW Answer d 160 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values V01560 m3s rho1000 kgm3 v11rho mrhoV kgs P1200 kPa Etapump075 P25000 kPa Reversible pump power input is w mvP2P1 VP2P1 Wrevmv1P2P1 kPam3skW WpumpWrevEtapump Some Wrong Solutions with Common Mistakes W1WpumpWrevEtapump Multiplying by efficiency W2WpumpWrev Disregarding efficiency W3Wpumpmv1P2P1Etapump Adding pressures instead of subtracting PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7212 7261 Steam enters an adiabatic turbine at 8 MPa and 500C at a rate of 18 kgs and exits at 02 MPa and 300C The rate of entropy generation in the turbine is a 0 kWK b 72 kWK c 21 kWK d 15 kWK e 17 kWK Answer c 21 kWK Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P18000 kPa T1500 C m18 kgs P2200 kPa T2300 C s1ENTROPYSteamIAPWSTT1PP1 s2ENTROPYSteamIAPWSTT2PP2 Sgenms2s1 kWK Some Wrong Solutions with Common Mistakes W1Sgen0 Assuming isentropic expansion 7262 Helium gas is compressed steadily from 90 kPa and 25C to 800 kPa at a rate of 2 kgmin by an adiabatic compressor If the compressor consumes 80 kW of power while operating the isentropic efficiency of this compressor is a 540 b 805 c 758 d 901 e 100 Answer d 901 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cp51926 kJkgK Cv31156 kJkgK k1667 m260 kgs T125 C P190 kPa P2800 kPa Wcomp80 kW T2sT1273P2P1k1k273 WsmCpT2sT1 EtacompWsWcomp Some Wrong Solutions with Common Mistakes T2sAT1P2P1k1k Using C instead of K W1EtacompmCpT2sAT1Wcomp W2EtacompmCvT2sT1Wcomp Using Cv instead of Cp 81 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 8 EXERGY A MEASURE OF WORK POTENTIAL PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 82 Exergy Irreversibility Reversible Work and SecondLaw Efficiency 81C Reversible work and irreversibility are identical for processes that involve no actual useful work 82C The dead state 83C Yes exergy is a function of the state of the surroundings as well as the state of the system 84C Useful work differs from the actual work by the surroundings work They are identical for systems that involve no surroundings work such as steadyflow systems 85C Yes 86C No not necessarily The well with the higher temperature will have a higher exergy 87C The system that is at the temperature of the surroundings has zero exergy But the system that is at a lower temperature than the surroundings has some exergy since we can run a heat engine between these two temperature levels 88C They would be identical 89C The secondlaw efficiency is a measure of the performance of a device relative to its performance under reversible conditions It differs from the first law efficiency in that it is not a conversion efficiency 810C No The power plant that has a lower thermal efficiency may have a higher secondlaw efficiency 811C No The refrigerator that has a lower COP may have a higher secondlaw efficiency PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 83 812C A processes with Wrev 0 is reversible if it involves no actual useful work Otherwise it is irreversible 813C Yes 814 Windmills are to be installed at a location with steady winds to generate power The minimum number of windmills that need to be installed is to be determined Assumptions Air is at standard conditions of 1 atm and 25C Properties The gas constant of air is 0287 kPam3kgK Table A1 Analysis The exergy or work potential of the blowing air is the kinetic energy it possesses 0 0180 kJkg m s 1000 1kJkg 2 6 ms 2 ke Exergy 2 2 2 2 V At standard atmospheric conditions 25C 101 kPa the density and the mass flow rate of air are 1 18 m kg kPa m kg K298 K 0287 kPa 101 3 3 RT P ρ and Thus 2225 kgs00180 kJkg 4005 kW ke Power Available 420 m 6 ms 2225 kgs 1 18 kgm 4 2 3 1 2 1 m D V AV m π ρ π ρ The minimum number of windmills that needs to be installed is 23 windmills 22 5 4005 kW 900 kW total W W N PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 84 815E Saturated steam is generated in a boiler by transferring heat from the combustion gases The wasted work potential associated with this heat transfer process is to be determined Also the effect of increasing the temperature of combustion gases on the irreversibility is to be discussed Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis The properties of water at the inlet and outlet of the boiler and at the dead state are Tables A4E through A6E 09328 Btulbm R 0 07 Btulbm 48 7 psia 14 F 80 1 5460 Btulbm R 8 Btulbm 1198 sat vap 1 psia 200 0 54379 Btulbm R 46 Btulbm 355 sat liq 0 psia 200 80 F 0 80 F 0 0 0 2 2 2 2 1 1 1 1 f f g g f f s s h h P T s s h h x P s s h h x P Water 200 psia sat liq q 200 psia sat vap The heat transfer during the process is 843 3 Btulbm 35546 1198 8 1 2 in h h q The entropy generation associated with this process is 0 12377 Btulbm R 460R 500 843 3 Btulbm 0 54379Btulbm R 5460 1 in 1 2 gen R R w T q s s s s s The wasted work potential exergy destruction is 668 Btulbm 460 R012377 Btulbm R 80 0 gen dest T s x The work potential exergy of the steam stream is 1 Btulbm 302 0 54379Btulbm R 540 R 1 5460 35546Btulbm 1198 8 1 2 0 1 2 s T s h h w ψ Increasing the temperature of combustion gases does not effect the work potential of steam stream since it is determined by the states at which water enters and leaves the boiler Discussion This problem may also be solved as follows Exergy transfer by heat transfer 368 9 Btulbm 960 540 1 843 3 1 0 heat TR T q x Exergy increase of steam 302 1 Btulbm ψw The net exergy destruction 668 Btulbm 302 1 368 9 heat dest w x x ψ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 85 816 Water is to be pumped to a high elevation lake at times of low electric demand for use in a hydroelectric turbine at times of high demand For a specified energy storage capacity the minimum amount of water that needs to be stored in the lake is to be determined Assumptions The evaporation of water from the lake is negligible 75 m Analysis The exergy or work potential of the water is the potential energy it possesses Exergy PE mgh Thus kg 10 245 10 1kW skg 1000 m s h 1 s 3600 ms 75 m 89 10 kWh 5 2 2 2 6 gh PE m 817 A body contains a specified amount of thermal energy at a specified temperature The amount that can be converted to work is to be determined Analysis The amount of heat that can be converted to work is simply the amount that a reversible heat engine can convert to work 298 K 800 K HE 100 kJ 6275 kJ 0 6275100 kJ 0 6275 800 K 298 K 1 1 in threv revout out max 0 rev th Q W W T T H η η 818 The thermal efficiency of a heat engine operating between specified temperature limits is given The secondlaw efficiency of a engine is to be determined Analysis The thermal efficiency of a reversible heat engine operating between the same temperature reservoirs is 20C 1200C HE η th 040 Thus 499 0 801 40 0 0 801 273 K 1200 293 K 1 1 rev th th II 0 rev th η η η η TH T PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 86 819 A heat reservoir at a specified temperature can supply heat at a specified rate The exergy of this heat supplied is to be determined Analysis The exergy of the supplied heat in the rate form is the amount of power that would be produced by a reversible heat engine 298 K 1500 K HE 334 kW 0 8013150000 3600 kJs Exergy 0 8013 1500 K 298 K 1 1 in threv revout out max 0 threv max th Q W W T T H η η η Wrev 820 A heat engine receives heat from a source at a specified temperature at a specified rate and rejects the waste heat to a sink For a given power output the reversible power the rate of irreversibility and the 2nd law efficiency are to be determined Analysis a The reversible power is the power produced by a reversible heat engine operating between the specified temperature limits 0 7091400 kJs 2836 kW 0 7091 1100 K 320 K 1 1 in threv out rev threv max th Q W T T H L η η η 320 K 1100 K HE 400 kJs 120 kW b The irreversibility rate is the difference between the reversible power and the actual power output 1636 kW 120 283 6 uout revout W W I c The second law efficiency is determined from its definition 423 0 423 283 6 kW kW 120 out rev uout II W W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 87 821 Problem 820 is reconsidered The effect of reducing the temperature at which the waste heat is rejected on the reversible power the rate of irreversibility and the second law efficiency is to be studied and the results are to be plotted Analysis The problem is solved using EES and the solution is given below Input Data TH 1100 K QdotH 400 kJs TL320 K Wdotout 120 kW TLsurr 25 C The reversible work is the maximum work done by the Carnot Engine between TH and TL EtaCarnot1 TLTH WdotrevQdotHEtaCarnot The irreversibility is given as Idot WdotrevWdotout The thermal efficiency is in percent Etath EtaCarnotConvert The second law efficiency is in percent EtaII WdotoutWdotrevConvert TL K Wrev kJs I kJs ηII 500 4776 4551 4327 4102 3878 3653 3429 3204 298 2182 2263 2345 2427 2508 259 2672 2753 2835 2916 9818 1063 1145 1227 1308 139 1472 1553 1635 1716 55 5302 5117 4945 4784 4633 4492 4359 4233 4115 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 88 275 320 365 410 455 500 210 220 230 240 250 260 270 280 290 300 TL K Wrev kJs 275 320 365 410 455 500 100 120 140 160 180 TL K I kJs 275 320 365 410 455 500 40 42 44 46 48 50 52 54 56 TL K ηII PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 89 822E The thermal efficiency and the secondlaw efficiency of a heat engine are given The source temperature is to be determined 530 R TH HE η th 36 η II 60 Analysis From the definition of the second law efficiency Thus 530 R040 1325 R 1 1 0 60 0 60 36 0 threv rev th II th threv rev th th II η η η η η η η η L H H L T T T T 823 A house is maintained at a specified temperature by electric resistance heaters The reversible work for this heating process and irreversibility are to be determined Analysis The reversible work is the minimum work required to accomplish this process and the irreversibility is the difference between the reversible work and the actual electrical work consumed The actual power input is 50000 kJh 1389 kW out in QH Q W W 50000 kJh 4 C House 25 C The COP of a reversible heat pump operating between the specified temperature limits is 1420 27715 29815 1 1 1 1 COP HPrev TL TH Thus and kW 1291 kW 0978 0 978 89 13 1420 1389 kW COP revin in u HPrev in rev W W I Q W H PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 810 824E A freezer is maintained at a specified temperature by removing heat from it at a specified rate The power consumption of the freezer is given The reversible power irreversibility and the secondlaw efficiency are to be determined Analysis a The reversible work is the minimum work required to accomplish this task which is the work that a reversible refrigerator operating between the specified temperature limits would consume 8 73 535 480 1 1 1 1 COP Rrev L H T T Freezer 20F 75F R 75 Btumin 070 hp 020 hp 4241 Btumin 1hp 8 73 Btumin 75 Rrev revin COP Q W L b The irreversibility is the difference between the reversible work and the actual electrical work consumed 050 hp 0 20 0 70 revin uin W W I c The second law efficiency is determined from its definition 289 hp 70 0 20 hp rev II u W W η 825 A geothermal power produces 51 MW power while the exergy destruction in the plant is 75 MW The exergy of the geothermal water entering to the plant the secondlaw efficiency of the plant and the exergy of the heat rejected from the plant are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 Water properties are used for geothermal water Analysis a The properties of geothermal water at the inlet of the plant and at the dead state are Tables A4 through A6 0 36723 kJkgK 83 kJkg 104 0 C 25 8418 kJkgK 1 18 kJkg 632 0 C 150 0 0 0 0 1 1 1 1 s h x T s h x T The exergy of geothermal water entering the plant is 1846 MW 460 kW 18 0 36723kJkgK 273 K 1 8418 25 10483 kJkg 210 kgs 63218 0 1 0 0 1 in s s T h m h X b The secondlaw efficiency of the plant is the ratio of power produced to the exergy input to the plant 276 0 276 18460 kW kW 5100 in out X W II η c The exergy of the heat rejected from the plant may be determined from an exergy balance on the plant 586 MW 5864 kW 7500 5100 18460 dest out in heatout X W X X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 811 826 It is to be shown that the power produced by a wind turbine is proportional to the cube of the wind velocity and the square of the blade span diameter Analysis The power produced by a wind turbine is proportional to the kinetic energy of the wind which is equal to the product of the kinetic energy of air per unit mass and the mass flow rate of air through the blade span area Therefore 2 3 2 3 wind 2 2 wind 2 wind Constant 8 4 2 2 power EfficiencyKinetic energyMass flow rate of air Wind V D D V V D V AV V ρ π η ρ π η ρ η which completes the proof that wind power is proportional to the cube of the wind velocity and to the square of the blade span diameter PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 812 Exergy Analysis of Closed Systems 827C Yes it can For example the 1st law efficiency of a reversible heat engine operating between the temperature limits of 300 K and 1000 K is 70 However the second law efficiency of this engine like all reversible devices is 100 828 A fixed mass of helium undergoes a process from a specified state to another specified state The increase in the useful energy potential of helium is to be determined Assumptions 1 At specified conditions helium can be treated as an ideal gas 2 Helium has constant specific heats at room temperature Properties The gas constant of helium is R 20769 kJkgK Table A1 The constant volume specific heat of helium is cv 31156 kJkgK Table A2 He 8 kg 288 K Analysis From the idealgas entropy change relation 3087 kJkg K m kg 3 20769 kJkg K ln 05 m kg 288 K kJkg K ln 353 K 31156 ln ln 3 3 1 2 1 2 avg 1 2 v v v R T T c s s The increase in the useful potential of helium during this process is simply the increase in exergy 6980 kJ Φ Φ 50 m kgkJkPa m 100 kPa3 298 K3087 kJkg K 353 K 8 kg3115 6 kJkg K288 3 3 2 1 0 2 1 0 2 1 1 2 v P v s s T u u m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 813 829E Air is expanded in an adiabatic closed system with an isentropic efficiency of 95 The second law efficiency of the process is to be determined Assumptions 1 Kinetic and potential energy changes are negligible 2 The process is adiabatic and thus there is no heat transfer 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 0240 BtulbmR cv 0171 BtulbmR k 14 and R 006855 BtulbmR Table A2Ea Analysis We take the air as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as Air 150 psia 100F 1 2 out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc U W E E E b v 43 42 1 4243 1 The final temperature for the isentropic case is 290 1 R 150 psia R 15 psia 560 41 40 1 1 2 1 2 k k s P P T T 1 2 2s 15 psia 150 psia s T The actual exit temperature from the isentropic relation is 3036 R 290 1 0 95560 560 2 1 1 2 2s 1 2 1 T s T T T T T T T η η The boundary work output is 4384 Btulbm 3036R 0 171 Btulbm R560 2 1 out T T c wb v The entropy change of air is 001091 Btulbm R 150 psia 006855 Btulbm Rln 15 psia 560 R Btulbm Rln 303 6 R 0240 ln ln 1 2 1 2 air P P R T T c s p The exergy difference between states 1 and 2 is 07 Btulbm 33 Btulbm R 537 R 001091 15 psia 3036 R 150 psia 560 R 14 7 psia006855 Btulbm R 84 Btulbm 43 2 1 0 2 2 1 1 0 2 1 2 1 0 2 1 0 2 1 2 1 s T s P T P P R T T T c s T s P u u v v v φ φ The useful work is determined from 21 Btulbm 27 150 psia 560 R 15 psia 14 7 psia006855 Btulbm R 3036 R 84 Btulbm 43 1 1 2 2 0 2 1 1 2 0 2 1 surr out P T P P R T T c T P T c T w w w b u v v v v The second law efficiency is then 0823 3307 Btulbm 2721 Btulbm II φ η wu PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 814 830E Air and helium at specified states are considered The gas with the higher exergy content is to be identified Assumptions 1 Kinetic and potential energy changes are negligible 2 Air and helium are ideal gases with constant specific heats Properties The properties of air at room temperature are cp 0240 BtulbmR cv 0171 BtulbmR k 14 and R 006855 BtulbmR 03704 psiaft3lbmR For helium cp 125 BtulbmR cv 0753 BtulbmR k 1667 and R 04961 BtulbmR 26809 psiaft3lbmR Table A2E Analysis The mass of air in the system is Air 15 ft3 100 psia 250F 5 704 lbm psia ft lbm R710 R 03704 psia15 ft 100 3 3 RT P m V The entropy change of air between the given state and the dead state is 006441 Btulbm R 147 psia 006855 Btulbm Rln 100 psia 537 R Btulbm Rln 710 R 0240 ln ln 0 0 0 P P R T T c s s p The airs specific volumes at the given state and dead state are 2 630 ft lbm 100 psia 03704 psia ft lbm R710 R 3 3 P RT v 1353 ft lbm 147 psia 03704 psia ft lbm R537 R 3 3 0 0 0 P RT v The specific closed system exergy of the air is then 52 Btulbm 34 537 R 006441 Btulbm R psia ft 5404 1Btu 1353ft lbm 14 7 psia2630 77R 0 171 Btulbm R300 3 3 0 0 0 0 0 0 0 0 0 0 s s T P T T c s s T P u u v v v v v φ The total exergy available in the air for the production of work is then 197 Btu Φ 5 704 lbm3452 Btulbm mφ We now repeat the calculations for helium 0 6782 lbm psia ft lbm R660 R 26809 psia20 ft 60 3 3 RT P m V Helium 20 ft3 60 psia 200F 04400 Btulbm R 147 psia 04961 Btulbm Rln 60 psia 537 R Btulbm Rln 660 R 125 ln ln 0 0 0 P P R T T c s s p 2949 ft lbm 60 psia 26809 psia ft lbm R660 R 3 3 P RT v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 815 9793 ft lbm 147 psia 26809 psia ft lbm R537 R 3 3 0 0 0 P RT v 7 Btulbm 142 537 R 04400 Btulbm R psia ft 5404 1Btu 9793ft lbm 14 7 psia2949 77R 0 753 Btulbm R200 3 3 0 0 0 0 0 0 0 0 0 0 s s T P T T c s s T P u u v v v v v φ 968 Btu Φ 0 6782 lbm1427 Btulbm mφ Comparison of two results shows that the air system has a greater potential for the production of work PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 816 831 Steam and R134a at the same states are considered The fluid with the higher exergy content is to be identified Assumptions Kinetic and potential energy changes are negligible Analysis The properties of water at the given state and at the dead state are Steam 1 kg 800 kPa 180C Table A 4 3672 kJkg K 0 001003 m kg 0 83 kJkg 104 kPa 100 C 25 Table A 6 7155 kJkg K 6 24720 m kg 0 7 kJkg 2594 C 180 kPa 800 25 C 0 3 25 C 0 25 C 0 0 0 3 f f f s s u u P T s u T P v v v The exergy of steam is 6227 kJ Φ 0 3672kJkg K 298 K67155 kPa m 1 1kJ 0 001003m kg 100 kPa024720 10483kJkg kg 2594 7 1 3 3 0 0 0 0 0 s s T P u m u v v For R134a Table A 11 32432 kJkg K 0 0008286 m kg 0 85 kJkg 85 kPa 100 C 25 Table A 13 3327 kJkg K 1 044554 m kg 0 99 kJkg 386 C 180 kPa 800 25 C 0 3 25 C 0 25 C 0 0 0 3 f f f s s u u P T s u T P v v v R134a 1 kg 800 kPa 180C 502 kJ Φ 0 32432kJkg K 298 K13327 kPa m 1 1kJ 0 0008286m kg 100 kPa0044554 8585kJkg kg 38699 1 3 3 0 0 0 0 0 s s T P u m u v v The steam can therefore has more work potential than the R134a PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 818 833E An insulated rigid tank contains saturated liquidvapor mixture of water at a specified pressure An electric heater inside is turned on and kept on until all the liquid is vaporized The exergy destruction and the secondlaw efficiency are to be determined Assumptions Kinetic and potential energies are negligible Properties From the steam tables Tables A4 through A6 0 70751 Btu lbm R 1 30632 0 25 38093 0 44347 Btu lbm 0 25 86219 22792 2 9880 ft lbm 0 01708 0 25 11901 01708 0 025 psia 35 1 1 1 1 3 1 1 1 1 fg f fg f fg f x s s s x u u u x x P v v v Btulbm R 15692 11109 Btulbm vapor sat 29880 ft lbm 2 29880 ft lbm 2 1 2 3 3 g g g g s s u u v v v v H2O 35 psia We Analysis a The irreversibility can be determined from its definition Xdestroyed T0Sgen where the entropy generation is determined from an entropy balance on the tank which is an insulated closed system 1 2 system gen in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in s m s S S S S S S 43 42 1 4243 1 Substituting 6 lbm535 R15692 070751Btulbm R 2766 Btu 1 2 0 gen 0 destroyed s s mT T S X b Noting that V constant during this process the W and Wu are identical and are determined from the energy balance on the closed system energy equation 1 2 in e potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u m u U W E E E 43 42 1 4243 1 or 6 lbm11109 44347Btulbm 4005 Btu ein W Then the reversible work during this process and the secondlaw efficiency become 1239 Btu 2766 4005 destroyed uin revin X W W Thus 309 4005 Btu 1239 Btu rev II Wu W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 819 834 A rigid tank is divided into two equal parts by a partition One part is filled with compressed liquid while the other side is evacuated The partition is removed and water expands into the entire tank The exergy destroyed during this process is to be determined Assumptions Kinetic and potential energies are negligible Analysis The properties of the water are Tables A4 through A6 Vacuum 4 kg 200 kPa 80C WATER 1 0756 kJkg K 33497 kJkg 0 001029 m kg C 80 kPa 200 80 C 1 80 C 1 3 80 C 1 1 1 f f f s s u u T P v v Noting that 0 002058 m kg 0 001029 2 2 3 1 2 v v 1 0278 kJkg K 6 6430 0 0002584 0261 1 31814 kJkg 2158 8 0 0002584 58 317 0 0002584 0 001026 3 9933 0 001026 002058 0 002058 m kg 0 kPa 40 2 2 2 2 2 2 3 2 2 fg f fg f fg f x s s s x u u u x P v v v v Taking the direction of heat transfer to be to the tank the energy balance on this closed system becomes 1 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u m u U Q E E E 43 42 1 4243 1 or 6730 kJ 33497kJkg 6730 kJ 4 kg31814 out in Q Q The irreversibility can be determined from its definition Xdestroyed T0Sgen where the entropy generation is determined from an entropy balance on an extended system that includes the tank and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times surr out 1 2 gen 1 2 system gen out b out in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in T Q s m s S s m s S S T Q S S S S 43 42 1 4243 1 Substituting 103 kJ 298 K 1 0756kJkg K 6730 kJ 298 K 4 kg10278 surr out 1 2 0 gen 0 destroyed T Q s m s T T S X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 820 835 Problem 834 is reconsidered The effect of final pressure in the tank on the exergy destroyed during the process is to be investigated Analysis The problem is solved using EES and the solution is given below T180 C P1200 kPa m4 kg P240 kPa To25 C Po100 kPa Tsurr To Conservation of energy for closed system is Ein Eout DELTAE 5 10 15 20 25 30 35 40 45 0 100 200 300 400 500 600 700 800 P2 kPa Qout kJ DELTAE mu2 u1 Ein0 Eout Qout u1 intenergysteamiapwsPP1TT1 v1 volumesteamiapwsPP1TT1 s1 entropysteamiapwsPP1TT1 v2 2v1 u2 intenergysteamiapws vv2PP2 s2 entropysteamiapws vv2PP2 Sin SoutSgenDELTASsys Sin0 kJK SoutQoutTsurr273 DELTASsysms2 s1 Xdestroyed To273Sgen P2 kPa Xdestroyed kJ Qout kJ 5 10 15 20 25 30 35 40 45 7441 644 538 4385 3461 2599 1791 103 3091 7884 5719 4351 3329 2505 181 1207 6715 1895 5 10 15 20 25 30 35 40 45 0 10 20 30 40 50 60 70 80 P2 kPa Xdestroyed kJ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 822 837 Problem 836 is reconsidered The effect of the amount of electrical work on the minimum work and the exergy destroyed is to be investigated Analysis The problem is solved using EES and the solution is given below x10 P1120 kPa V8 L P2P1 WEle 1400 kJ To25 C Po100 kPa Conservation of energy for closed system is Ein Eout DELTAE DELTAE mu2 u1 EinWEle Eout Wb Wb mP1v2v1 u1 intenergysteamiapwsPP1xx1 0 400 800 1200 1600 2000 250 150 550 950 1350 1750 WEle kJ Xdestroyed kJ v1 volumesteamiapwsPP1xx1 s1 entropysteamiapwsPP1xx1 u2 intenergysteamiapws vv2PP2 s2 entropysteamiapws vv2PP2 mVconvertLm3v1 Wrevinmu2 u1 To27315 s2s1Pov2v1 Entropy Balance Sin SoutSgen DELTASsys DELTASsys ms2 s1 Sin0 kJK Sout 0 kJK The exergy destruction or irreversibility is Xdestroyed To27315Sgen 0 400 800 1200 1600 2000 0 50 100 150 200 250 300 350 400 WEle kJ Wrevin kJ WEle kJ Wrevin kJ Xdestroyed kJ 0 200 400 600 800 1000 1200 1400 1600 1800 2000 0 3968 7935 119 1587 1984 2381 2777 3174 3571 3968 0 1578 3156 4733 6311 7889 9467 1104 1262 1420 1578 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 829 844E A hot copper block is dropped into water in an insulated tank The final equilibrium temperature of the tank and the work potential wasted during this process are to be determined Assumptions 1 Both the water and the copper block are incompressible substances with constant specific heats at room temperature 2 The system is stationary and thus the kinetic and potential energies are negligible 3 The tank is well insulated and thus there is no heat transfer Properties The density and specific heat of water at the anticipated average temperature of 90F are ρ 621 lbmft3 and cp 100 BtulbmF The specific heat of copper at the anticipated average temperature of 100F is cp 00925 BtulbmF Table A3E Analysis a We take the entire contents of the tank water copper block as the system which is a closed system The energy balance for this system can be expressed as U E E E 0 potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 43 42 1 4243 1 Copper 180F Water 75F or 0 water Cu U U 0 water 1 2 Cu 1 2 T mc T T mc T where 7452 lbm ft 21 62 1 lbmft 3 3 V ρ w m Substituting 7 R 541 75 F F 7452 lbm10 Btulbm 180 F F 55 lbm0092 5 Btulbm 0 2 2 2 817 F T T T b The wasted work potential is equivalent to the exergy destruction or irreversibility and it can be determined from its definition Xdestroyed T0Sgen where the entropy generation is determined from an entropy balance on the system which is an insulated closed system copper water system gen in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in S S S S S S S S 43 42 1 4243 1 where 0 9250 BtuR 535 R 7452 lbm10 Btulbm Rln 5417 R ln 0 8483 BtuR 640 R 55 lbm0092 Btulbm Rln 5417 R ln 1 2 avg water 1 2 avg copper T T mc S T T mc S Substituting 0 9250BtuR 431 Btu 0 8483 535 R destroyed X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 830 845 A hot iron block is dropped into water in an insulated tank that is stirred by a paddlewheel The mass of the iron block and the exergy destroyed during this process are to be determined Assumptions 1 Both the water and the iron block are incompressible substances with constant specific heats at room temperature 2 The system is stationary and thus the kinetic and potential energies are negligible 3 The tank is well insulated and thus there is no heat transfer Properties The density and specific heat of water at 25C are ρ 997 kgm3 and cp 418 kJkgF The specific heat of iron at room temperature the only value available in the tables is cp 045 kJkgC Table A3 Analysis We take the entire contents of the tank water iron block as the system which is a closed system The energy balance for this system can be expressed as water iron in pw potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in U U U W E E E 43 42 1 4243 1 Wpw Water Iron 85C 100 L 20C water 1 2 iron 1 2 pwin T mc T T mc T W where 240 kJ kJs20 60 s 20 99 7 kg m 10 997 kgm pwin pw 3 3 water t W W m V ρ Substituting 520 kg iron iron 20 C C24 99 7 kg418 kJkg 85 C C24 0 45 kJkg kJ 240 m m b The exergy destruction or irreversibility can be determined from its definition Xdestroyed T0Sgen where the entropy generation is determined from an entropy balance on the system which is an insulated closed system water iron system gen in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in S S S S S S S S 43 42 1 4243 1 where 5 651 kJK 293 K 99 7 kg418 kJkg Kln 297 K ln 4 371 kJK 358 K 52 0 kg045 kJkg K ln 297 K ln 1 2 avg water 1 2 avg iron T T mc S T T mc S Substituting 5 651 kJK 3750 kJ 4 371 293 K gen 0 destroyed T S X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 831 846 An iron block and a copper block are dropped into a large lake where they cool to lake temperature The amount of work that could have been produced is to be determined Assumptions 1 The iron and copper blocks and water are incompressible substances with constant specific heats at room temperature 2 Kinetic and potential energies are negligible Properties The specific heats of iron and copper at room temperature are cp iron 045 kJkgC and cpcopper 0386 kJkgC Table A3 Analysis The thermalenergy capacity of the lake is very large and thus the temperatures of both the iron and the copper blocks will drop to the lake temperature 15C when the thermal equilibrium is established We take both the iron and the copper blocks as the system which is a closed system The energy balance for this system can be expressed as copper iron out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in U U U Q E E E 43 42 1 4243 1 Copper Iron Lake 15C Iron 85C or copper out T mc T T mc T Q 2 1 iron 2 1 Substituting kJ 1964 288 K 20 kg 0386 kJkg K 353 288 K 50 kg 045 kJkg K 353 out Q The work that could have been produced is equal to the wasted work potential It is equivalent to the exergy destruction or irreversibility and it can be determined from its definition Xdestroyed T0Sgen The entropy generation is determined from an entropy balance on an extended system that includes the blocks and the water in their immediate surroundings so that the boundary temperature of the extended system is the temperature of the lake water at all times lake out copper iron gen copper iron system gen out b out in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in T Q S S S S S S S T Q S S S S 43 42 1 4243 1 where 1571 kJK 353 K 20 kg 0386 kJkg K ln 288 K ln 4579 kJK 353 K 50 kg 045 kJkg K ln 288 K ln 1 2 avg copper 1 2 avg iron T T mc S T T mc S Substituting kJK 196 kJ 288 K 1964 kJ 1 571 4 579 293 K gen 0 destroyed T S X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 832 847E A rigid tank is initially filled with saturated mixture of R134a Heat is transferred to the tank from a source until the pressure inside rises to a specified value The amount of heat transfer to the tank from the source and the exergy destroyed are to be determined Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero 2 There is no heat transfer with the environment Properties From the refrigerant tables Tables A11E through A13E 0 65234 ft lbm 1 16368 0 55 0 01232 v 0 1436 Btu lbm R 0 17580 0 55 0 04688 6376 Btu lbm 0 55 77307 21246 055 psia 40 3 1 1 1 1 1 1 1 1 fg f fg f fg f x x s s s x u u u x P v v 8803 Btulbm 0 8191 73360 939 27 0 1922 Btulbm R 0 16098 0 8191 06029 0 0 8191 0 01270 0 79361 0 01270 65234 0 psia 60 2 2 2 2 2 2 1 2 2 fg f fg f fg f x u u u x s s s x P v v v v v Analysis a The mass of the refrigerant is Q Source 120C R134a 40 psia 1840 lbm 65234 ft lbm 0 ft 12 3 3 1 v V m We take the tank as the system which is a closed system The energy balance for this stationary closed system can be expressed as 1 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u m u U Q E E E 43 42 1 4243 1 Substituting 1840 lbm88036376 Btulbm 4463 Btu 1 2 in u m u Q b The exergy destruction or irreversibility can be determined from its definition Xdestroyed T0Sgen The entropy generation is determined from an entropy balance on an extended system that includes the tank and the region in its immediate surroundings so that the boundary temperature of the extended system where heat transfer occurs is the source temperature source in 1 2 gen 1 2 system gen in b in in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in T Q s m s S s m s S S T Q S S S S 43 42 1 4243 1 Substituting 665 Btu 580 R 446 3 Btu 0 1436Btulbm R 535 R 1840 lbm 0 1922 gen 0 destroyed T S X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 834 849 Carbon steel balls are to be annealed at a rate of 2500h by heating them first and then allowing them to cool slowly in ambient air at a specified rate The total rate of heat transfer from the balls to the ambient air and the rate of exergy destruction due to this heat transfer are to be determined Assumptions 1 The thermal properties of the balls are constant 2 There are no changes in kinetic and potential energies 3 The balls are at a uniform temperature at the end of the process Properties The density and specific heat of the balls are given to be ρ 7833 kgm3 and cp 0465 kJkgC Analysis a We take a single ball as the system The energy balance for this closed system can be expressed as 2 1 out 1 2 ball out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc Q u m u U Q E E E p 43 42 1 4243 1 The amount of heat transfer from a single ball is 0781 kJ per ball J 781 0 0021 kg 0 465 kJkg C900 100 C 0 00210 kg 6 0 008 m 7833 kgm 6 2 1 out 3 3 3 T T mc Q D m p π ρ π ρV Then the total rate of heat transfer from the balls to the ambient air becomes 260 W 936 kJh 1200 ballsh 0 781 kJball out ball out Q n Q b The exergy destruction or irreversibility can be determined from its definition Xdestroyed T0Sgen The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the ball and its immediate surroundings so that the boundary temperature of the extended system is at 35C at all times system out gen system gen out in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in S T Q S S S T Q S S S S b b 43 42 1 4243 1 where 0 00112 kJK 900 273 0 00210 kg 0 465 kJkgKln 100 273 ln 1 2 1 2 system T T mc s m s S p Substituting 0 00142 kJK per ball 0 00112 kJK 308 K kJ 0781 system out gen S T Q S b Then the rate of entropy generation becomes 0 00142 kJK ball1200 ballsh 1704 kJhK 0000473 kWK ball gen gen n S S Finally 146 W 0 146 kW 308 K0000473 kW K gen 0 destroyed T S X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 835 850 Heat is transferred to a pistoncylinder device with a set of stops The work done the heat transfer the exergy destroyed and the secondlaw efficiency are to be determined Assumptions 1 The device is stationary and kinetic and potential energy changes are zero 2 There is no friction between the piston and the cylinder 3 Heat is transferred to the refrigerant from a source at 150C Analysis a The properties of the refrigerant at the initial and final states are Tables A11 through A13 1 2553 kJkgK 38 kJkg 305 20847 m kg 0 C 90 kPa 140 1 0760 kJkgK 51 kJkg 248 19390 m kg 0 C 20 kPa 120 2 2 3 2 2 2 1 1 3 1 1 1 s u T P s u T P v v R134a 075 kg 120 kPa 20C Q Noting that pressure remains constant at 140 kPa as the piston moves the boundary work is determined to be 153 kJ 019390m kg 0 75 kg140 kPa020847 3 1 2 2 bout v mP v W b The heat transfer can be determined from an energy balance on the system 442 kJ 153 kJ 24851kJkg 0 75 kg30538 bout 1 2 in W u m u Q c The exergy destruction associated with this process can be determined from its definition Xdestroyed T0Sgen The entropy generation is determined from an entropy balance on an extended system that includes the pistoncylinder device and the region in its immediate surroundings so that the boundary temperature of the extended system where heat transfer occurs is the source temperature source in 1 2 gen 1 2 system gen in b in in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in T Q s m s S s m s S S T Q S S S S 43 42 1 4243 1 Substituting 8935 kJ 273 K 150 44 2 kJ 1 0760kJkg K 298 K 075 kg12553 gen 0 destroyed T S X d Exergy expended is the work potential of the heat extracted from the source at 150C 1306 kJ 273 K 44 2 kJ 150 273 K 25 1 1 threv expended Q T T Q X X H L Q η Then the 2nd law efficiency becomes 0 316 or 316 1306 kJ 8 935 kJ 1 1 expended destroyed expended recovered II X X X X η Discussion The secondlaw efficiency can also be determine as follows The exergy increase of the refrigerant is the exergy difference between the initial and final states 666 kJ 3 019390m kg 100 kPa020847 298 K12553 10760kgK 24851kJkg kg 30538 075 3 1 2 0 1 2 0 1 2 v P v s s T u m u X The useful work output for the process is 0 437 kJ 0 19390m kg 0 75 kg100 kPa020847 1 53 kJ 3 1 2 0 bout uout v mP v W W The exergy recovered is the sum of the exergy increase of the refrigerant and the useful work output 4 103 kJ 0 437 3 666 uout recovered W X X Then the secondlaw efficiency becomes 0 314 or 314 1306 kJ 103 kJ 4 expended recovered II X X η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 836 851 A tank containing hot water is placed in a larger tank The amount of heat lost to the surroundings and the exergy destruction during the process are to be determined Assumptions 1 Kinetic and potential energy changes are negligible 2 Air is an ideal gas with constant specific heats 3 The larger tank is wellsealed Properties The properties of air at room temperature are R 0287 kPam3kgK cp 1005 kJkgK cv 0718 kJkgK Table A2 The properties of water at room temperature are ρ 997 kgm3 cw 418 kJkgK Table A3 Analysis a The final volume of the air in the tank is Q Water 85C 15 L Air 22C 3 1 2 0 025 m 0 015 0 04 w a a V V V The mass of the air in the large tank is 0 04724 kg 273 K kPa m kg K22 0287 kPa004 m 100 3 3 1 1 1 a a a RT P m V The pressure of air at the final state is 171 9 kPa m 0025 273 K kg0287 kPa m kg K44 004724 3 3 2 2 2 a a a a m RT P V The mass of water is 1496 kg 997 kgm 0015 m 3 3 w w mw V ρ An energy balance on the system consisting of water and air is used to determine heat lost to the surroundings 2563 kJ 22 0 04724 kg0718 kJkgK44 85 1496 kg418 kJkgK44 1 2 1 2 out a a w w w T T m c T T m c Q v b An exergy balance written on the system immediate surroundings can be used to determine exergy destruction But we first determine entropy and internal energy changes 7 6059 kJK 273 K 44 273 K 1496 kg418 kJkgKln 85 ln 2 1 T T m c S w w w w 003931 kJK 0 1719 kPa 0287 kJkgKln 100 kPa 273 K 44 273 K kg 1005 kJkgKln 22 004724 ln ln 2 1 2 1 P P R T T c m S a a p a a 0 7462 kJ 004724 kg0718 kJkgK22 44K 2564 kJ 1496 kg418 kJkgK85 44K 2 1 2 1 T T m c U T T m c U a a a w w w w v 3184 kJ 295 K0003931 kJK 0 7462 kJ 295 K76059 kJK kJ 2564 0 0 dest a a w w a w S T U S T U X X X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 837 Exergy Analysis of Control Volumes 852 R134a is is throttled from a specified state to a specified pressure The temperature of R134a at the outlet of the expansion valve the entropy generation and the exergy destruction are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Heat transfer is negligible a The properties of refrigerant at the inlet and exit states of the throttling valve are from R134a tables 4244 kJkg K 0 77 kJkg 117 0 kPa 1200 1 1 1 1 s h x P 0 4562 kJkg K 77 kJkg 117 kPa 200 2 2 1 2 2 s T h h P 101 C b Noting that the throttling valve is adiabatic the entropy generation is determined from 003176 kJkg K 0 4244kJkg K 04562 1 2 gen s s s Then the irreversibility ie exergy destruction of the process becomes 9464 kJkg 298 K003176 kJkg K 0 gen dest T s ex PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 840 855 Problem 854 is reconsidered The problem is to be solved and the actual heat transfer its direction the minimum power input and the compressor secondlaw efficiency are to be determined Analysis The problem is solved using EES and the solution is given below Function DirectionQ If Q0 then Directionout else Directionin end Function Violationeta If eta1 then ViolationYou have violated the 2nd Law else Violation end Input Data from the Diagram Window T117 C P1100 kPa Wdotc 8 kW P2600 kPa Sdotgen0 Qdotnet0 Special cases T2167 C mdot21 kgmin ToT1 PoP1 mdotinmdotConvertkgmin kgs Steadyflow conservation of mass mdotin mdotout Conservation of energy for steadyflow is Edotin Edotout DELTAEdot DELTAEdot 0 EdotinQdotnet mdotinh1 Wdotc If Qdotnet 0 heat is transferred from the compressor Edotout mdotouth2 h1 enthalpyairTT1 h2 enthalpyair TT2 WdotnetWdotc Wdotrevmdotinh2 h1 T127315s2s1 Irreversibility entropy generated second law efficiency and exergy destroyed s1entropyair TT1PP1 s2entropyairTT2PP2 s2sentropyairTT2sPP2 s2ss1This yields the isentropic T2s for an isentropic process bewteen T1 P1 and P2IdotTo27315SdotgenIrreversiblility for the Process KW SdotgenQdotnetTo27315 mdotins2s1 Entropy generated kW EtaIIWdotrevWdotnetDefinition of compressor second law efficiency Eq 76 hoenthalpyairTTo soentropyairTToPPo Psiinh1hoTo27315s1so availability function at state 1 Psiouth2hoTo27315s2so availability function at state 2 XdotinPsiinmdotin XdotoutPsioutmdotout DELTAXdotXdotinXdotout General Exergy balance for a steadyflow system Eq 747 1To27315To27315QdotnetWdotnetmdotinPsiin mdotoutPsiout Xdotdest For the Diagram Window TextDirectionQdotnet Text2ViolationEtaII PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 841 ηII I kW Xdest kW T2s C T2 C Qnet kW 07815 1748 1748 209308 167 27 08361 1311 1311 209308 2006 1501 08908 0874 0874 209308 2305 04252 09454 0437 0437 209308 2581 05698 1 1425E13 5407E15 209308 2839 1506 50 55 60 65 0 50 100 150 200 250 s kJkgK T C 100 kPa 600 kPa 1 2 2s ideal actual How can entropy decrease PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 075 080 085 090 095 100 160 180 200 220 240 260 280 300 00 05 10 15 20 ηII Xdest T2 075 080 085 090 095 100 3 2 1 0 1 2 00 05 10 15 20 ηII Qnet Xdest preparation If you are a student using this Manual you are using it without permission 845 858 Problem 857 is reconsidered The effect of varying the nozzle exit velocity on the exit temperature and exergy destroyed is to be investigated Analysis The problem is solved using EES and the solution is given below Knowns WorkFluid Air P1 200 kPa T1 65 C P2 95 kPa Vel1 35 ms Vel2 240 ms To 17 C Tsurr To qloss 3 kJkg Conservation of Energy SSSF energy balance for nozzle neglecting the change in potential energy h1enthalpyWorkFluidTT1 s1entropyWorkFluidPP1TT1 ke1 Vel122 ke2Vel222 h1ke1convertm2s2kJkg h2 ke2convertm2s2kJkgqloss T2temperatureWorkFluidhh2 s2entropyWorkFluidPP2hh2 The entropy generated is detemined from the entropy balance s1 s2 qlossTsurr273 sgen 0 xdestroyed To273sgen Vel2 ms T2 C xdestroyed kJkg 100 140 180 220 260 300 5766 5289 4653 3858 2902 1787 5856 5432 4856 412 3212 2116 100 140 180 220 260 300 15 20 25 30 35 40 45 50 55 60 Vel2 ms T2 C PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 846 100 140 180 220 260 300 20 25 30 35 40 45 50 55 60 Vel2 ms xdestroyed kJkg PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 854 866 Refrigerant134a enters an adiabatic compressor at a specified state with a specified volume flow rate and leaves at a specified state The power input the isentropic efficiency the rate of exergy destruction and the secondlaw efficiency are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 The device is adiabatic and thus heat transfer is negligible Analysis a The properties of refrigerant at the inlet and exit states of the compressor are obtained from R134a tables 18946 m kg 0 0 9514 kJkg K 68 kJkg 234 1 C 26 3 1 1 1 1 1 v s h x T 9802 kJkg K 0 69 kJkg 286 C 50 kPa 800 2 2 2 2 s h T P 27753 kJkg 9514 kJkg K 0 800 kPa 2 1 2 2 h s s s P The mass flow rate of the refrigerant and the actual power input are 0 03959 kgs m kg 018946 0 45 60 m s 3 3 1 1 v V m 2059 kW 23468kJkg 0 03959 kgs28669 1 2 act h m h W b The power input for the isentropic case and the isentropic efficiency are 1 696 kW 23468kJkg 0 03959 kgs27753 1 2 isen h m h W s 824 08238 2 059 kW 696 kW 1 act isen Compisen W W η c The exergy destruction is 03417 kW 0 9514kJkg K 0 03959 kgs300 K09802 1 2 0 dest s s mT X The reversible power and the secondlaw efficiency are 1 717 kW 0 3417 2 059 dest act rev X W W 834 08341 2 059 kW 717 kW 1 act rev CompII W W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 855 867 Refrigerant134a is condensed in a refrigeration system by rejecting heat to ambient air The rate of heat rejected the COP of the refrigeration cycle and the rate of exergy destruction are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible Analysis a The properties of refrigerant at the inlet and exit states of the condenser are from R134a tables 9954 kJkg K 0 28853 kJkg C 50 kPa 700 1 1 1 1 s h T P 3323 kJkg K 0 8882 kJkg 0 kPa 700 2 2 2 2 s h x P The rate of heat rejected in the condenser is 9985 kW 8882kJkg 0 05 kgs28853 2 1 h h m Q R H b From the definition of COP for a refrigerator 1506 6 kW 9 985 6 kW COP in L H L L Q Q Q W Q c The entropy generation and the exergy destruction in the condenser are 0 0003516 kWK 298 K 9 985 kW 0 9954 kJkg K 0 05 kgs 0 3323 1 2 gen H H R T Q s s m S 01048 kW 298 K00003516 kJkg K gen 0 dest T S X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 856 868E Refrigerant134a is evaporated in the evaporator of a refrigeration system the rate of cooling provided the rate of exergy destruction and the secondlaw efficiency of the evaporator are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 Kinetic and potential energy changes are negligible Analysis a The rate of cooling provided is 18580 Btuh 5162 Btus 0 08 lbms1721 107 5 Btulbm 1 2 h m h QL b The entropy generation and the exergy destruction are 0 0008691 Btus R 460 R 50 5 162 Btus 0 2851 Btulbm R 0 08 lbms 0 4225 1 2 gen L L T Q s m s S 04667 Btus 537 R00008691 Btus R gen 0 dest T S X c The exergy supplied or expended during this cooling process is the exergy decrease of the refrigerant as it evaporates in the evaporator 7400 Btus 0 0 4225 Btulbm R 0 08 lbms537 R 0 2851 162 5 2 1 0 2 1 2 1 s s mT h m h X X The exergy efficiency is then 369 0 3693 0 7400 0 4667 1 1 2 1 dest IIEvap X X X η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 858 870 Problem 869 is reconsidered The effect of compressor exit pressure on reversible power is to be investigated Analysis The problem is solved using EES and the solution is given below T127 C P1101 kPa mdot 015 kgs P2400 kPa T2220 C To25 C Po100 kPa mdotinmdot Steadyflow conservation of mass mdotin mdotout h1 enthalpyairTT1 h2 enthalpyair TT2 Wdotrevmdotinh2 h1 T127315s2s1 s1entropyair TT1PP1 s2entropyairTT2PP2 P2 kPa Wrev kW 200 250 300 350 400 450 500 550 600 1555 1844 2079 2279 2451 2603 274 2863 2975 200 250 300 350 400 450 500 550 600 16 18 20 22 24 26 28 30 P2 kPa Wrev kW PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 859 871 A rigid tank initially contains saturated liquid of refrigerant134a R134a is released from the vessel until no liquid is left in the vessel The exergy destruction associated with this process is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process It can be analyzed as a uniformflow process since the state of fluid leaving the device remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved Properties The properties of R134a are Tables A11 through A13 26159 kJkg 092234 kJkg K 24102 kJkg 0035969 m kg vapor sat C 20 030063 kJkg K 7886 kJkg 00008161 m kg liquid sat C 20 20 C 20 C 2 20 C 2 3 20 C 2 2 20 C 1 20 C 1 3 20 C 1 1 g e g e g g f f f h h s s s u u T s s u u T v v v v R134a 1 kg 20C sat liq me Analysis The volume of the container is 3 3 1 1 m 0 0008161 m kg 1 kg 0 0008161 v V m The mass in the container at the final state is 0 02269 kg 035969 m kg 0 0008161 m 0 3 3 2 2 v V m The amount of mass leaving the container is 0 9773 kg 0 02269 1 2 1 m m me The exergy destroyed during a process can be determined from an exergy balance or directly from its definition gen The entropy generation S 0 destroyed T S X gen in this case is determined from an entropy balance on the system e e e e m s m s m s S m s m s S S s m S S S S 1 1 2 2 gen tank 1 1 2 2 tank gen in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in 43 42 1 4243 1 Substituting 1822 kJ 0 92234 0 9773 0 30063 0 92234 1 293 K002269 1 1 2 2 0 gen 0 destroyed mese m s m s T T S X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 860 872E An adiabatic rigid tank that is initially evacuated is filled by air from a supply line The work potential associated with this process is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process It can be analyzed as a uniformflow process since the state of fluid entering the device remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 0240 BtulbmR k 14 and R 006855 BtulbmR 03704 kPam3lbmR Table A2Ea Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance 40 ft3 150 psia 90F Air 2 system out in m m m m m i Energy balance 2 2 potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in m u h m E E E i i 43 42 1 4243 1 Combining the two balances i i p i p i kT T c c T c T c T u h v v 2 2 2 Substituting 770 R 550 R 41 2 kTi T The final mass in the tank is 2104 lbm 03704 psia ft lbm R770 R 15 0 psia40 ft 3 3 2 2 RT P m m i V The work potential associated with this process is equal to the exergy destroyed during the process The exergy destruction during a process can be determined from an exergy balance or directly from its definition The entropy generation S gen 0 destroyed T S X gen in this case is determined from an entropy balance on the system 2 2 gen 2 2 gen 2 2 tank gen in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in i i i i i s s m S m s m s S m s S S s m S S S S 43 42 1 4243 1 Substituting 917 Btu 550 R 2104 lbm540 R 0240 Btulbm Rln 770 R ln 2 0 2 2 0 2 destyroyed rev i p i T T c m T s s m T X W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 861 873E An rigid tank that is initially evacuated is filled by air from a supply line The work potential associated with this process is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process It can be analyzed as a uniformflow process since the state of fluid entering the device remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 0240 BtulbmR and R 006855 BtulbmR 03704 kPam3lbmR Table A2Ea Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance 2 system out in m m m m m i Energy balance 10 ft3 200 psia 100F Air 2 2 out 2 2 out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in m u m h Q m u Q h m E E E i i i i 43 42 1 4243 1 Combining the two balances 2 2 out u h m Q i The final mass in the tank is 2945 lbm 03704 psia ft lbm R550 R 15 0 psia40 ft 3 3 2 2 RT P m m i V Substituting Btu 1110 2945 lbm550 R006855 Btulbm R 2 2 2 2 2 out m T R c m T c c T c T m u h m Q i v p i i v i p i The work potential associated with this process is equal to the exergy destroyed during the process The exergy destruction during a process can be determined from an exergy balance or directly from its definition The entropy generation S gen 0 destroyed T S X gen in this case is determined from an entropy balance on the system 0 out 2 2 gen 0 out 2 2 gen 2 2 tank gen 0 out in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in T Q s s m S T Q m s m s S m s S S T Q s m S S S S i i i i i 43 42 1 4243 1 Noting that both the temperature and pressure in the tank is same as those in the supply line at the final state substituting gives 1110 Btu out 0 out 0 0 out 0 0 out 2 2 0 destroyed rev 0 Q T Q T T Q T T Q s s m T X W i PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 867 879 A rigid tank initially contains saturated R134a vapor The tank is connected to a supply line and R134a is allowed to enter the tank The mass of the R134a that entered the tank and the exergy destroyed during this process are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The direction of heat transfer is to the tank will be verified Properties The properties of refrigerant are Tables A11 through A13 091303 kJkg K 25381 kJkg 001672 m kg vapor sat MPa 21 MPa 21 1 MPa 21 1 3 MPa 21 1 1 g g g s s u u P v v Q R134a 01 m3 12 MPa Sat vapor 16 MPa 30C R134a 045315 kJkg K 12594 kJkg 00009166 m kg liquid sat MPa 41 MPa 41 2 MPa 41 2 3 MPa 41 2 2 f f f s s u u T v v 34554 kJkg K 0 56 kJkg 93 C 30 MPa 61 i i i i s h T P Analysis We take the tank as the system which is a control volume Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance 1 2 system out in m m m m m m i Energy balance 0 pe ke 1 1 since 2 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in W m u m u m h Q E E E i i 43 42 1 4243 1 a The initial and the final masses in the tank are 10910 kg m kg 00009166 m 01 5983 kg m kg 001672 m 01 3 3 2 2 2 3 3 1 1 1 v V v V m m Then from the mass balance 10311 kg 5 983 10910 1 2 m m mi The heat transfer during this process is determined from the energy balance to be kJ 2573 5983 kg 25381 kJkg 10910 12594 kJkg kg 9356 kJkg 10311 1 1 2 2 in m u m u m h Q i i PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 868 b The exergy destroyed during a process can be determined from an exergy balance or directly from its definition gen The entropy generation S 0 destroyed T S X gen in this case is determined from an entropy balance on an extended system that includes the tank and its immediate surroundings so that the boundary temperature of the extended system is the surroundings temperature Tsurr at all times It gives 0 in 1 1 2 2 gen 1 1 tank 2 2 tank gen in b in in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in T Q m s m s m s S m s m s S S m s T Q S S S S i i i i 43 42 1 4243 1 Substituting the exergy destruction is determined to be 803 kJ 2573 kJ318 K 0 91303 10311 034554 5 983 0 45315 318 K 10910 0 in 1 1 2 2 0 gen 0 destroyed T Q m s m s m s T T S X i i PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 869 880 A rigid tank initially contains saturated liquid water A valve at the bottom of the tank is opened and half of mass in liquid form is withdrawn from the tank The temperature in the tank is maintained constant The amount of heat transfer the reversible work and the exergy destruction during this process are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid leaving the device remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The direction of heat transfer is to the tank will be verified H2O 06 m3 170C T const me Properties The properties of water are Tables A4 through A6 Q kJkg K 20417 08 kJkg 719 liquid sat 0 C 17 kJkg K 20417 20 kJkg 718 m kg 0001114 liquid sat 0 C 17 170 C 170 C 170 C 1 170 C 1 3 170 C 1 1 o o o o o f e f e e f f f s s h h T s s u u T v v Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance 2 1 system out in m m m m m m e Energy balance 0 pe ke 1 1 since 2 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in W m u m u m h Q E E E e e 43 42 1 4243 1 The initial and the final masses in the tank are me m m m 26924 kg 2 53847 kg 1 2 1 53847 kg m kg 0001114 m 06 1 2 3 3 1 1 v V Now we determine the final internal energy and entropy 20630 kJkg K 0004614 46233 20417 72677 kJkg 0004614 18575 20 718 004614 0 0 C 17 0004614 0001114 024260 0001114 0002229 0002229 m kg 26924 kg m 06 2 2 2 2 2 2 2 2 3 3 2 2 fg f fg f fg f x s s s x u u u x T x m v v v V v The heat transfer during this process is determined by substituting these values into the energy balance equation 2545 kJ 53847 kg 71820 kJkg 26924 kg 72677 kJkg 26924 kg 71908 kJkg 1 1 2 2 in m u m u m h Q e e PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 870 b The exergy destroyed during a process can be determined from an exergy balance or directly from its definition gen The entropy generation S 0 destroyed T S X gen in this case is determined from an entropy balance on an extended system that includes the tank and the region between the tank and the source so that the boundary temperature of the extended system at the location of heat transfer is the source temperature Tsource at all times It gives source in 1 1 2 2 gen 1 1 tank 2 2 tank gen in b in in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in S S T Q m s m s m s S m s m s S S m s T Q S S e e e e 43 42 1 4243 1 Substituting the exergy destruction is determined to be 1412 kJ 2545 kJ523 K 2 0417 2 0417 26924 53847 2 0630 298 K 26924 source in 1 1 2 2 0 gen 0 destroyed T Q m s m s m s T T S X e e For processes that involve no actual work the reversible work output and exergy destruction are identical Therefore 1412 kJ destroyed revout actout revout destroyed X W W W X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 872 882 A cylinder initially contains helium gas at a specified pressure and temperature A valve is opened and helium is allowed to escape until its volume decreases by half The work potential of the helium at the initial state and the exergy destroyed during the process are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process by using constant average properties for the helium leaving the tank 2 Kinetic and potential energies are negligible 3 There are no work interactions involved other than boundary work 4 The tank is insulated and thus heat transfer is negligible 5 Helium is an ideal gas with constant specific heats Properties The gas constant of helium is R 20769 kPam3kgK 20769 kJkgK The specific heats of helium are cp 51926 kJkgK and cv 31156 kJkgK Table A2 Analysis a From the ideal gas relation the initial and the final masses in the cylinder are determined to be 0 0493 kg 2 0769 kPa m kg K293 K 300 kPa01 m 3 3 1 1 1 RT P m V PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 0247 kg 0 0493 2 1 2 2 m m me The work potential of helium at the initial state is simply the initial exergy of helium and is determined from the closedsystem exergy relation 0 1 0 0 1 0 0 1 1 1 1 v v Φ P s s T u u m m φ HELIUM 300 kPa 01 m3 20C Q where 6 405 m kg 95 kPa 2 0769 kPa m kg K293 K 2 0284 m kg 300 kPa 2 0769 kPa m kg K293 K 3 3 0 0 0 3 3 1 1 1 P RT P RT v v and 2 388 kJkg K 95 kPa 2 0769 kJkg Kln 300 kPa 293 K 5 1926 kJkg K ln 293 K ln ln 0 1 0 1 0 1 P P R T T c s s p Thus 140 kJ Φ 6 405m kgkJkPa m 95 kPa20284 2 388 kJkg K 293 K 20 C 0 0493 kg3115 6 kJkg K20 3 3 1 b We take the cylinder as the system which is a control volume Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance 2 1 system out in m m m m m m e Energy balance 1 1 2 2 bin in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in m u m u W m h Q E E E e e 43 42 1 4243 1 Combining the two relations gives preparation If you are a student using this Manual you are using it without permission 873 0 1 1 2 2 1 1 1 2 2 2 1 bin 1 1 2 2 2 1 in m h m m m m h m h h m m W m u m u h m m Q e e since the boundary work and U combine into H for constant pressure expansion and compression processes The exergy destroyed during a process can be determined from an exergy balance or directly from its definition gen where the entropy generation S 0 destroyed T S X gen can be determined from an entropy balance on the cylinder Noting that the pressure and temperature of helium in the cylinder are maintained constant during this process and heat transfer is zero it gives 0 1 2 1 1 2 2 1 1 1 2 2 1 1 2 2 gen 1 1 cylinder 2 2 cylinder gen in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in s m m m m s m m m s s m m s m s m s S m s m s S S s m S S S S e e e e e 43 42 1 4243 1 since the initial final and the exit states are identical and thus se s2 s1 Therefore this discharge process is reversible and 0 0 gen destroyed T S X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 874 883 A rigid tank initially contains saturated R134a vapor at a specified pressure The tank is connected to a supply line and R134a is allowed to enter the tank The amount of heat transfer with the surroundings and the exergy destruction are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The direction of heat transfer is from the tank will be verified Properties The properties of refrigerant are Tables A11 through A13 0020313 m kg 091558 kJkg K 25068 kJkg satvapor MPa 1 3 1MPa 1 1MPa 1 1MPa 1 1 g g g s s u u P v v PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 93889 kJkg K 47 kJkg 285 C 60 MPa 41 i i i i s h T P Analysis a We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as R134a 02 m3 1 MPa Sat vapor 14 MPa 60C R134a Q Mass balance 1 2 system out in m m m m m m i Energy balance 0 pe ke 1 1 since 2 2 out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in W m u m u Q h m E E E i i 43 42 1 4243 1 The initial and the final masses in the tank are 9 846 kg 020313 m kg 0 m 20 3 3 1 1 v V m 11791 kg 5 983 11193 016715 m kg 0 m 10 0008934 m kg 0 m 10 3 3 3 3 2 g g f f g f m m m v V v V 52967 kJK 0 91303 5 983 0 42441 11193 14581 kJ 25381 5 983 93 11670 111 2 2 2 2 2 2 g g f f g g f f m s m s m s S m u m u m u U Then from the mass and energy balances 10806 kg 9 846 11791 1 2 m m mi The heat transfer during this process is determined from the energy balance to be 18737 kJ 10806 28547 14581 9846 25068 1 1 2 2 out m u m u m h Q i i b The exergy destroyed during a process can be determined from an exergy balance or directly from its definition gen The entropy generation S 0 destroyed T S X gen in this case is determined from an entropy balance on an extended system that includes the cylinder and its immediate surroundings so that the boundary temperature of the extended system is the surroundings temperature Tsurr at all times It gives 0 out 1 1 2 2 gen 1 1 tank 2 2 tank gen out b out in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in T Q m s m s m s S m s m s S S m s T Q S S S S i i i i 43 42 1 4243 1 Substituting the exergy destruction is determined to be 1599 kJ 0 91558 10806 093889 18737 298 9 846 298 K52967 0 out 1 1 2 2 0 gen 0 destroyed T Q m s m s m s T T S X i i preparation If you are a student using this Manual you are using it without permission 875 884 An insulated cylinder initially contains saturated liquidvapor mixture of water The cylinder is connected to a supply line and the steam is allowed to enter the cylinder until all the liquid is vaporized The amount of steam that entered the cylinder and the exergy destroyed are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant 2 The expansion process is quasiequilibrium 3 Kinetic and potential energies are negligible 4 The device is insulated and thus heat transfer is negligible Properties The properties of steam are Tables A4 through A6 6 2824 kJkg K 5 3200 0 8667 1 6716 2436 5 kJkg 2163 5 0 8667 43 561 0 8667 15 13 kPa 300 1 1 1 1 1 1 fg f fg f x s s s x h h h x P 6 9917 kJkg K 9 kJkg 2724 satvapor kPa 300 300 kPa 2 300 kPa 2 2 g g s s h h P H2O 300 kPa P const 2 MPa 400C 7 1292 kJkg K 4 kJkg 3248 C 400 MPa 2 i i i i s h T P Analysis a We take the cylinder as the system which is a control volume Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this unsteadyflow system can be expressed as Mass balance 1 2 system out in m m m m m m i Energy balance 0 pe ke 1 1 since 2 2 out b potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in Q m u m u W h m E E E i i 43 42 1 4243 1 Combining the two relations gives 1 1 2 2 1 2 bout 0 m u m u m h m W i or 1 1 2 2 1 2 0 m h m h m h m i since the boundary work and U combine into H for constant pressure expansion and compression processes Solving for m2 and substituting 2724 9 kJkg 15 kg 2327 kg 3248 4 2436 5 kJkg 3248 4 1 2 1 2 h m h h h m i i Thus 827 kg 15 2327 1 2 m m mi b The exergy destroyed during a process can be determined from an exergy balance or directly from its definition gen where the entropy generation S 0 destroyed T S X gen is determined from an entropy balance on the insulated cylinder i i i i m s m s m s S m s m s S S s m S S S S 1 1 2 2 gen 1 1 2 2 system gen in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in 43 42 1 4243 1 Substituting the exergy destruction is determined to be 2832 kJ 7 1292 8 27 6 2824 15 6 9917 298 K2327 1 1 2 2 0 gen 0 destroyed misi m s T m s T S X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 881 890 Steam expands in a turbine which is not insulated The reversible power the exergy destroyed the secondlaw efficiency and the possible increase in the turbine power if the turbine is well insulated are to be determined Assumptions 1 Steady operating conditions exist 2 Potential energy change is negligible Analysis a The properties of the steam at the inlet and exit of the turbine are Tables A4 through A6 5535 kJkgK 7 1 kJkg 2491 95 0 kPa 20 9605 kJkgK 6 1 kJkg 3634 C 600 MPa 9 2 2 2 2 1 1 1 1 s h x P s h T P 20 kPa 130 ms x 095 Q Steam 9 MPa 600C 60 ms Turbine The enthalpy at the dead state is 10483 kJkg 0 25 C 0 0 h x T The mass flow rate of steam may be determined from an energy balance on the turbine 4 137 kgs 4500 kW kW 220 m s 1000 1kJkg 2 130 ms 2491 1 kJkg m s 1000 1kJkg 2 60 ms 1 kJkg 3634 2 2 2 2 2 2 2 2 out 2 2 2 2 1 1 m m m W Q V m h V m h a The reversible power may be determined from kW 5451 2 2 2 2 2 2 2 1 2 1 0 2 1 rev m s 1000 1kJkg 2 130 ms 60 ms 2986960575535 2491 1 2 693 3634 1 2 V V s s T h m h W b The exergy destroyed in the turbine is kW 951 4500 5451 a rev dest W W X c The secondlaw efficiency is 0826 5451 kW kW 4500 rev a W W II η d The energy of the steam at the turbine inlet in the given dead state is 14602 kW 4 137 kgs3634110483kJkg 0 1 h m h Q The fraction of energy at the turbine inlet that is converted to power is 0 3082 14602 kW 4500 kW a Q W f Assuming that the same fraction of heat loss from the turbine could have been converted to work the possible increase in the power if the turbine is to be wellinsulated becomes 678 kW 0 3082220 kW out increase fQ W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 882 891 Air is compressed in a compressor that is intentionally cooled The actual and reversible power inputs the second law efficiency and the mass flow rate of cooling water are to be determined Assumptions 1 Steady operating conditions exist 2 Potential energy change is negligible 3 Air is an ideal gas with constant specific heats Air 100 kPa 20C Q 900 kPa 60C 80 ms Compressor Properties The gas constant of air is R 0287 kJkgK and the specific heat of air at room is cp 1005 kJkgK the specific heat of water at room temperature is cw 418 kJkgK Tables A2 A3 Analysis a The mass flow rate of air is 5 351 kgs m s 54 273 K 0 287 kJkgK20 100 kPa 3 1 1 1 1 V V RT P m ρ The power input for a reversibleisothermal process is given by 9888 kW 100 kPa 273 Kln 900 kPa 5 351 kgs028 7 kJkgK20 ln 1 2 1 rev P P mRT W Given the isothermal efficiency the actual power may be determined from 1413 kW 070 8 kW 988 rev actual T W W η b The given isothermal efficiency is actually the secondlaw efficiency of the compressor 070 ηT η II c An energy balance on the compressor gives kW 1181 1413 kW m s 1000 1kJkg 2 80 ms 0 60 C 5 351 kgs 1005 kJkg C20 2 2 2 2 in actual 2 2 2 1 2 1 out W V V T T m c Q p The mass flow rate of the cooling water is 2825 kgs 418 kJkg C10 C kW 1181 out T c Q m w w PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 883 892 Water is heated in a chamber by mixing it with saturated steam The temperature of the steam entering the chamber the exergy destruction and the secondlaw efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 Heat loss from the chamber is negligible Analysis a The properties of water are Tables A4 through A6 Mixture 45C Sat vap 023 kgs Mixing chamber Water 15C 46 kgs 63862 kJkgK 0 44 kJkg 188 0 C 45 22447 kJkgK 0 98 kJkg 62 0 C 15 3 3 1 3 0 1 0 1 1 1 s h x T s s h h x T An energy balance on the chamber gives 5 kJkg 2697 0 23 kgs18844 kJkg 64 0 23 kgs kgs6298 kJkg 64 2 2 3 2 1 3 3 2 2 1 1 h h m h m m h m h m h The remaining properties of the saturated steam are 7 1907 kJkgK 1 5 kJkg 2697 2 2 2 2 s T x h C 1143 b The specific exergy of each stream is ψ1 0 62828 kJkg 0 22447kJkgK 273 K71907 15 6298kJkg 2697 5 0 2 0 0 2 2 s T s h h ψ 6 18 kJkg 0 22447kJkgK 273 K063862 15 6298kJkg 18844 0 3 0 0 3 3 s T s h h ψ The exergy destruction is determined from an exergy balance on the chamber to be 1147 kW 0 23 kgs 6 18 kJkg 64 0 23 kgs62828 kJkg 0 3 2 1 2 2 1 1 dest ψ ψ ψ m m m m X c The secondlaw efficiency for this mixing process may be determined from 0207 0 23 kgs62828 kJkg 0 0 23 kgs 6 18 kJkg 64 2 2 1 1 3 2 1 II ψ ψ ψ η m m m m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 884 893 An expression is to be derived for the work potential of the singlephase contents of a rigid adiabatic container when the initially empty container is filled through a single opening from a source of working fluid whose properties remain fixed Analysis The conservation of mass principle for this system reduces to i CV m dt dm where the subscript i stands for the inlet state When the entropy generation is set to zero for calculating work potential and the combined first and second law is reduced to fit this system it becomes T S i mi h dt T S d U W 0 0 rev When these are combined the result is dt dm T S h dt T S d U W i CV 0 0 rev Recognizing that there is no initial mass in the system integration of the above equation produces 2 0 2 2 rev 2 0 2 2 2 0 rev s s T h h m W T s h m m T s h W i i i where the subscript 2 stands for the final state in the container PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 885 Review Problems 894E The 2ndlaw efficiency of a refrigerator and the refrigeration rate are given The power input to the refrigerator is to be determined Analysis From the definition of the second law efficiency the COP of the refrigerator is determined to be 2 089 7 462 0 28 COP COP COP COP 7 462 1 550 485 1 1 1 COP Rrev II R rev R R rev R η η II L H T T 25F 90F R 800 Btumin η II 028 Thus the power input is 903 hp 4241 Btumin 1hp 2 089 80 0 Btumin COPR in QL W 895 Refrigerant134a is expanded adiabatically in an expansion valve The work potential of R134a at the inlet the exergy destruction and the secondlaw efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a The properties of the refrigerant at the inlet and exit of the valve and at dead state are Tables A11 through A 13 1 0918 kJkgK 17 kJkg 272 C 20 kPa 100 0 37614 kJkgK 57 kJkg 93 kPa 120 0 34751 kJkgK 57 kJkg 93 C 30 MPa 90 0 0 0 0 2 1 2 2 1 1 1 1 s h T P s h h P s h T P 120 kPa R134a 09 MPa 30C The specific exergy of the refrigerant at the inlet and exit of the valve are 3959 kJkg 27315 K034751 10918kJkg K 20 27217kJkg 9357 0 1 0 0 1 1 s s T h h ψ kJkg 3120 27315 K037614 10918 kJkgK 20 27217kJkg 9357 0 2 0 0 2 2 s s T h h ψ b The exergy destruction is determined to be 839 kJkg 034751kJkg K 27315 K037614 20 1 2 0 dest s s T x c The secondlaw efficiency for this process may be determined from 788 0 788 3959 kJkg 20 kJkg 31 1 2 II ψ ψ η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 886 896 Steam is accelerated in an adiabatic nozzle The exit velocity the rate of exergy destruction and the secondlaw efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 Potential energy changes are negligible Analysis a The properties of the steam at the inlet and exit of the turbine and at the dead state are Tables A4 through A 6 0 2678 kJkgK 54 kJkg 75 0 C 18 6 6753 kJkgK 9 kJkg 2919 C 250 kPa 61 6 4484 kJkgK 4 kJkg 2978 C 300 MPa 53 0 0 0 2 2 2 2 1 1 1 1 s h x T s h T P s h T P Steam 35 MPa 300C 16 MPa 250C V2 The exit velocity is determined from an energy balance on the nozzle 3420 ms 2 2 2 2 2 2 2 2 2 2 2 2 1 1 m s 1000 1kJkg 2 V 2919 9 kJkg m s 1000 1kJkg 2 0 ms 4 kJkg 2978 2 2 V V h V h b The rate of exergy destruction is the exergy decrease of the steam in the nozzle 2641 kW 6 4484kJkgK 291 K 6 6753 m s 1000 1kJkg 2 0 342 ms 29784kJkg kgs 29199 40 2 2 2 2 1 2 0 2 1 2 2 1 2 dest s s T V V h m h X c The exergy of the refrigerant at the inlet is 44172 kW 0 2678kJkgK 291 K 6 4484 0 7554 kJkg kgs 29784 40 2 0 1 0 2 1 0 1 1 s T s V h m h X The secondlaw efficiency for this device may be defined as the exergy output divided by the exergy input 0940 44172 kW 2641 kW 1 1 1 dest 1 2 II X X X X η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 887 897 R134a is expanded in an adiabatic process with an isentropic efficiency of 085 The second law efficiency is to be determined Assumptions 1 Kinetic and potential energy changes are negligible 2 The device is adiabatic and thus heat transfer is negligible Analysis We take the R134a as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as 1 2 out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u m u U W E E E 43 42 1 4243 1 T 1 2s 2 100 kPa 16 MPa s From the R134a tables Tables A11 through A13 22316 kJkg kPa 100 0 9875 kJkg K 09 kJkg 282 014362 m kg 0 0 C 8 kPa 1600 2 1 2 2 1 1 3 1 1 1 s s u s s P s u T P v The actual work input is 5009 kJkg 22316kJkg 0 8528209 2 1 out out s T s T a u u w w η η The actual internal energy at the end of the expansion process is 23200 kJkg 5009 28209 out 1 2 2 1 out a a w u u u u w Other actual properties at the final state are Table A13 0251 kJkg K 1 0 2139 m lbm 00 kJkg 232 0 kPa 10 2 3 2 2 2 s u P v The useful work is determined from 3014 kJkg kPa m 1 1kJ 0 014362 m kg 100 kPa 0 2139 09 kJkg 50 3 3 1 2 0 out surr out v P v w w w w a a u The exergy change between initial and final states is 34 kJkg 41 1 0251kJkg K 298 K 0 9875 kPa m 1 1kJ 0 2139 m kg 100 kPa 0 014362 23200kJkg 28209 3 3 2 1 0 2 1 0 2 1 2 1 s s T P u u v v φ φ The second law efficiency is then 0729 4134 kJkg 3014 kJkg II φ η wu PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 888 898 Steam is condensed in a closed system at a constant pressure from a saturated vapor to a saturated liquid by rejecting heat to a thermal energy reservoir The second law efficiency is to be determined Assumptions 1 Kinetic and potential energy changes are negligible Analysis We take the steam as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as q Steam 75 kPa Sat vapor 1 2 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u m u U Q W E E E b 43 42 1 4243 1 From the steam tables Table A5 2132 kJkg K 1 38436 kJkg 0 001037 m kg liquid Sat kPa 75 7 4558 kJkg K 2496 1 kJkg 2 2172 m kg vapor Sat kPa 75 2 2 3 2 2 1 1 3 1 1 f f f g g g s s u u P s s u u P v v v v T 75 kPa 2 1 s The boundary work during this process is 166 2 kJkg kPa m 1 1kJ 0 001037 m kg 75 kPa 2 2172 3 3 2 1 in v P v wb The heat transfer is determined from the energy balance 2278 kJkg kJkg 2496 1 38436 166 2 kJkg 1 2 in out u u w q b The exergy change between initial and final states is 384 9 kJkg 310 K 298 K 2278 kJkg 1 1 2132kJkg K 298 K 7 4558 kPa m 1 1kJ 0 001037 m kg 100 kPa 2 2172 38436kJkg 2496 1 1 3 3 0 out 2 1 0 2 1 0 2 1 2 1 TR T q s s T P u u v v φ φ The second law efficiency is then 432 0 432 384 9 kJkg 2 kJkg 166 in II φ η wb PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 889 899 R134a is vaporized in a closed system at a constant pressure from a saturated liquid to a saturated vapor by transferring heat from a reservoir at two pressures The pressure that is more effective from a secondlaw point of view is to be determined Assumptions 1 Kinetic and potential energy changes are negligible Analysis We take the R134a as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as q R134a 100 kPa sat liquid 1 2 in out in 1 2 out in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in h m h H Q U W Q u m u U W Q E E E b b 43 42 1 4243 1 At 100 kPa T From the R134a tables Table A12 100 kPa 1 2 s 0 19181 m kg 0 0007259 19254 0 87995 kJkg K 0 16 kJkg 217 98 kJkg 197 3 100 kPa 100 kPa 100 kPa 100 kPa f g fg fg fg fg s h u v v v The boundary work during this process is 1918 kJkg kPa m 1 1kJ 100 kPa 0 19181 m kg 3 3 1 2 out fg b P P w v v v The useful work is determined from 0 kJkg 1 2 0 1 2 surr out v v v v P P w w w b u since P P0 100 kPa The heat transfer from the energy balance is 21716 kJkg in hfg q The exergy change between initial and final states is 2518 kJkg 273 K 298 K kJkg 1 21716 298 K 0 87995 kJkg K kPa m 1 1kJ 100 kPa 0 19181 m kg 98 kJkg 197 1 1 3 3 0 in 0 0 0 in 2 1 0 2 1 0 2 1 2 1 R fg fg fg R T T q T s P u T T q s s T P u u v v v φ φ The second law efficiency is then 0 2518 kJkg 0 kJkg II φ η wu PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 890 At 200 kPa 0 099114 m kg 0 0007533 099867 0 78316 kJkg K 0 03 kJkg 206 21 kJkg 186 3 200 kPa 200 kPa 200 kPa 200 kPa f g fg fg fg fg s h u v v v 1982 kJkg kPa m 1 1kJ 200 kPa 0 099114 m kg 3 3 1 2 out fg b P P w v v v 9 911 kJkg kPa m 1 1kJ 200 100 kPa 0 099114 m kg 3 3 0 1 2 0 1 2 surr out fg b u P P P P w w w v v v v v 20603 kJkg in hfg q 1839 kJkg 273 K 298 K kJkg 1 20603 298 K 0 78316 kJkg K kPa m 1 1kJ 100 kPa 0 099114 m kg 21 kJkg 186 1 1 3 3 0 in 0 0 0 in 2 1 0 2 1 0 2 1 2 1 R fg fg fg R T T q T s P u T T q s s T P u u v v v φ φ 0539 1839 kJkg 9 911 kJkg II φ η wu The process at 200 kPa is more effective from a work production standpoint PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 891 8100 An electrical radiator is placed in a room and it is turned on for a period of time The time period for which the heater was on the exergy destruction and the secondlaw efficiency are to be determined Assumptions 1 Kinetic and potential energy changes are negligible 2 Air is an ideal gas with constant specific heats 3 The room is wellsealed 4 Standard atmospheric pressure of 1013 kPa is assumed Properties The properties of air at room temperature are R 0287 kPam3kgK cp 1005 kJkgK cv 0718 kJkgK Table A2 The properties of oil are given to be ρ 950 kgm3 coil 22 kJkgK Analysis a The masses of air and oil are 6C Q Radiator Room 9488 kg 273 K kPa m kg K6 0287 kPa75 m 1013 3 3 1 1 RT P ma V 4750 kg 950 kgm 0050 m 3 3 oil oil oil ρ V m An energy balance on the system can be used to determine time period for which the heater was kept on 666 min s 3988 6 C 4750 kg22 kJkg C60 6 C 9488 kg0718 kJkg C20 0 75 kW 42 oil 1 2 1 2 out in t t T mc T T T mc t Q W a v b The pressure of the air at the final state is 106 4 kPa m 75 273 K kg0287 kPa m kg K20 9488 3 3 2 2 V a a a m RT P The amount of heat transfer to the surroundings is 2999 kJ 075 kJs3988 s out out t Q Q The entropy generation is the sum of the entropy changes of air oil and the surroundings 335 kJK 3 1013 kPa 0287 kJkgKln 1064 kPa 273 K 6 273 K kg 1005 kJkgKln 20 9488 ln ln 1 2 1 2 P P R T T m c S p a 1849 kJK 273 K 6 273 K 4750 kg22 kJkgKln 60 ln 1 2 oil T T mc S 1075 kJK 273 K 6 kJ 2999 surr out surr T Q S 3257 kJK 1075 1849 3 335 surr oil a gen S S S S The exergy destruction is determined from 909 MJ 9088 kJ 273 K3257 kJK 6 gen 0 dest T S X c The secondlaw efficiency may be defined in this case as the ratio of the exergy recovered to the exergy input That is 2316 kJ 273 K3335 kJK 6 6 C kg 0718 kJkg C20 9488 0 1 2 2 a a S T T T m c X v 484 5 kJ 273 K1849 kJK 6 6 C kg 22 kJkg C60 4750 0 1 2 oil2 Sa T T m C T X 53 00529 24 kJs3998 s 484 5 kJ 2316 in oil2 2 supplied recovered t W X X X X a II η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 892 8101 Hot exhaust gases leaving an internal combustion engine is to be used to obtain saturated steam in an adiabatic heat exchanger The rate at which the steam is obtained the rate of exergy destruction and the secondlaw efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 Air properties are used for exhaust gases 4 Pressure drops in the heat exchanger are negligible Properties The gas constant of air is R 0287 kJkgK The specific heat of air at the average temperature of exhaust gases 650 K is cp 1063 kJkgK Table A2 Analysis a We denote the inlet and exit states of exhaust gases by 1 and 2 and that of the water by 3 and 4 The properties of water are Table A4 Heat Exchanger Sat vap 200C Exh gas 400C 150 kPa 4302 kJkgK 6 0 kJkg 2792 1 C 200 29649 kJkgK 0 91 kJkg 83 0 C 20 4 4 4 4 3 3 3 3 s h x T s h x T 350C Water 20C An energy balance on the heat exchanger gives 001570 kgs w w w p a w a w a m m h h m T T c m m h m h m h h m 8391kJkg 2792 0 350 C kgs106 3 kJkg C400 80 3 4 2 1 4 2 3 1 b The specific exergy changes of each stream as it flows in the heat exchanger is 0 08206 kJkgK 273 K 400 273 K 08 kgs106 3 kJkgKln 350 ln 1 2 T T c s p a 106 kJkg 29 273 K00820 6 kJkgK 20 063 kJkg C350 400 C 1 0 1 2 a p a s T T T c ψ 910913 kJkg 0 29649kJkgK 273 K64302 20 8391kJkg 2792 0 3 4 0 3 4 s T s h h w ψ The exergy destruction is determined from an exergy balance on the heat exchanger to be or 898 kW dest dest 8 98 kW 0 01570 kgs910913 kJkg kgs29106 kJkg 80 X m m X w w a a ψ ψ c The secondlaw efficiency for a heat exchanger may be defined as the exergy increase of the cold fluid divided by the exergy decrease of the hot fluid That is 0614 kgs29106 kJkg 80 0 01570 kgs910913 kJkg II a a w w m m ψ ψ η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 895 8105 An electric resistance heater is immersed in water The time it will take for the electric heater to raise the water temperature to a specified temperature the minimum work input and the exergy destroyed during this process are to be determined Assumptions 1 Water is an incompressible substance with constant specific heats 2 The energy stored in the container itself and the heater is negligible 3 Heat loss from the container is negligible 4 The environment temperature is given to be T0 20C Properties The specific heat of water at room temperature is c 418 kJkgC Table A3 Analysis Taking the water in the container as the system which is a closed system the energy balance can be expressed as 1 water 2 in e water in e potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T t W U W E E E 43 42 1 4243 1 Heater Water 40 kg Substituting 800 Jst 40 kg4180 JkgC80 20C Solving for t gives t 12540 s 209 min 348 h Again we take the water in the tank to be the system Noting that no heat or mass crosses the boundaries of this system and the energy and entropy contents of the heater are negligible the entropy balance for it can be expressed as water gen in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in 0 S S S S S S 43 42 1 4243 1 Therefore the entropy generated during this process is 3115 kJK 293 K 40 kg 418 kJkg K ln 353 K ln 1 2 water gen T T mc S S The exergy destroyed during a process can be determined from an exergy balance or directly from its definition gen 0 destroyed T S X 9127 kJ 293 K3115 kJK gen 0 destroyed T S X The actual work input for this process is kJs12540 s 10032 kJ 80 actin actin t W W Then the reversible or minimum required work input becomes 906 kJ 9127 10032 destroyed actin revin X W W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8100 8110 A throttle valve is placed in the steam line supplying the turbine inlet in order to control an isentropic steam turbine The secondlaw efficiency of this system when the valve is partially open to when it is fully open is to be compared Assumptions 1 This is a steadyflow process since there is no change with time 2 The turbine is wellinsulated and there is no heat transfer from the turbine Analysis Valve is fully open T s 3p 1 3 MPa 70 kPa 3 2 6 MPa The properties of steam at various states are 3672 kJkg K 0 104 8 kJkg C 25 kPa 100 25 C 0 25 C 0 1 0 f f s s h h T P 4247 kJkg K 7 3894 3 kJkg C 700 MPa 6 2 1 2 1 2 1 2 1 s s h h T T P P 7 kJkg 2639 0 9914 kPa 70 3 3 1 2 3 h x s s P The stream exergy at the turbine inlet is 1686 kJkg 0 3672 298 7 4247 104 8 3894 3 0 1 0 0 1 1 s s T h h ψ The second law efficiency of the entire system is then 10 3 1 3 1 3 1 0 3 1 3 1 rev out II h h h h s s T h h h h w w η since s1 s3 for this system Valve is partly open 7 7405 kJkg K 3 kJkg 3894 3 MPa 2 1 2 2 s h h P from EES 2760 8 kJkg 70 kPa 3 2 3 3 h s s P from EES 1592 kJkg 0 3672 298 7 7405 104 8 3894 3 0 2 0 0 2 2 s s T h h ψ 10 7 7405 298 7 7405 2760 8 3894 3 2760 8 3894 3 3 2 0 3 2 3 2 rev out II s s T h h h h w w η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8101 8111 Two rigid tanks that contain water at different states are connected by a valve The valve is opened and steam flows from tank A to tank B until the pressure in tank A drops to a specified value Tank B loses heat to the surroundings The final temperature in each tank and the work potential wasted during this process are to be determined Assumptions 1 Tank A is insulated and thus heat transfer is negligible 2 The water that remains in tank A undergoes a reversible adiabatic process 3 The thermal energy stored in the tanks themselves is negligible 4 The system is stationary and thus kinetic and potential energy changes are negligible 5 There are no work interactions Analysis a The steam in tank A undergoes a reversible adiabatic process and thus s2 s1 From the steam tables Tables A4 through A6 kJkg K 77100 kJkg 27314 m kg 11989 C 250 kPa 200 21259 kJkg 0 7895 19821 kJkg 11 561 047850 m kg 0 001073 0 60582 0 7895 001073 0 0 7895 5 3200 1 6717 8717 5 mixture sat kPa 300 58717 kJkg K 5 1191 80 7765 1 21633 kJkg 1948 9 80 22 604 037015 m kg 0 001084 0 46242 80 001084 0 80 kPa 400 1 1 3 1 1 1 2 2 3 2 2 2 2 300 2 1 2 2 1 1 1 1 3 1 1 1 1 B B B fg A f A fg A f A fg f A A kPa sat A fg f A fg f A fg f A s u T P u x u u x s s s x T T s s P x s s s x u u u x x P v v v v v v v B Tank A Tank C 13352 900 kJ B m 3 kg steam T 250C P 200 kPa A V 02 m3 steam P 400 kPa x 08 The initial and the final masses in tank A are and 04180 kg m kg 0479 m 02 05403 kg m kg 037015 m 02 3 3 2 2 3 3 1 1 A A A A A A m m v V v V Thus 0540 0418 0122 kg of mass flows into tank B Then m m B B 2 1 0122 3 0122 3122 kg The final specific volume of steam in tank B is determined from 1152 m kg m 3122 3 kg 11989 m kg 3 3 3 2 1 1 2 2 B B B B B m v m m v V We take the entire contents of both tanks as the system which is a closed system The energy balance for this stationary closed system can be expressed as B A B A m u m u m u m u Q W U U U Q E E E 1 1 2 2 1 1 2 2 out out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in KE PE 0 since 43 42 1 4243 1 Substituting 24259 kJkg 3 2731 4 3 122 0 5403 2163 3 0 418 2125 9 900 2 2 B B u u PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8102 Thus 69772 kJkg K kJkg 24259 m kg 1152 2 2 2 3 2 B B B B s T u 1101 C v b The total entropy generation during this process is determined by applying the entropy balance on an extended system that includes both tanks and their immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times It gives B A gen surr b out in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in S S S T Q S S S S 43 42 1 4243 1 Rearranging and substituting the total entropy generated during this process is determined to be 1 234 kJK 273 K 900 kJ 7 7100 3 6 9772 3 122 5 8717 0 5403 5 8717 418 0 surr b out 1 1 2 2 1 1 2 2 surr b out gen T Q m s m s m s m s T Q S S S B A B A The work potential wasted is equivalent to the exergy destroyed during a process which can be determined from an exergy balance or directly from its definition gen 0 destroyed T S X 337 kJ 273 K 1 234 kJK gen 0 destroyed T S X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8103 8112E A cylinder initially filled with helium gas at a specified state is compressed polytropically to a specified temperature and pressure The actual work consumed and the minimum useful work input needed are to be determined Assumptions 1 Helium is an ideal gas with constant specific heats 2 The cylinder is stationary and thus the kinetic and potential energy changes are negligible 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or expansion process is quasiequilibrium 5 The environment temperature is 70F Properties The gas constant of helium is R 26805 psiaft3lbmR 04961 BtulbmR Table A1E The specific heats of helium are cv 0753 and cp 125 BtulbmR Table A2E Analysis a Helium at specified conditions can be treated as an ideal gas The mass of helium is 02252 lbm 26805 psia ft lbm R530 R 40 psia 8 ft 3 3 1 1 1 RT P m V HELIUM 8 ft3 PV n const Q The exponent n and the boundary work for this polytropic process are determined to be 1 446 3 364 8 40 140 3 364 ft 530 R14 0 psia 8 ft 780 R40 psia 2 1 1 2 1 1 2 2 3 3 1 2 1 1 2 2 2 2 2 1 1 1 n P P P P P P T T T P T P n n n n V V V V V V V V Then the boundary work for this polytropic process can be determined from 6262 Btu 1 1446 530 R lbm 04961 Btulbm R 780 02252 1 1 1 2 1 1 2 2 2 1 in b n T mR T n P P Pd W V V V Also Thus 500 Btu 1261 62 62 1261 Btu psia ft 54039 1Btu 8ft 14 7 psia3364 surrin bin in u 3 3 1 2 0 in surr W W W P W V V b We take the helium in the cylinder as the system which is a closed system Taking the direction of heat transfer to be from the cylinder the energy balance for this stationary closed system can be expressed as 1 2 bin out bin 1 2 out 1 2 bin out potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T mc W Q W u m u Q u m u U W Q E E E v 43 42 1 4243 1 Substituting 2069 Btu 530 R 02252 lbm 0753 Btulbm R 780 6262 Btu out Q The total entropy generation during this process is determined by applying the entropy balance on an extended system that includes the cylinder and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times It gives sys gen surr b out in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in S S T Q S S S S 43 42 1 4243 1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8107 8115 An insulated cylinder is divided into two parts One side of the cylinder contains N2 gas and the other side contains He gas at different states The final equilibrium temperature in the cylinder and the wasted work potential are to be determined for the cases of piston being fixed and moving freely Assumptions 1 Both N2 and He are ideal gases with constant specific heats 2 The energy stored in the container itself is negligible 3 The cylinder is wellinsulated and thus heat transfer is negligible Properties The gas constants and the specific heats are R 02968 kPam3kgK cp 1039 kJkgC and cv 0743 kJkgC for N2 and R 20769 kPam3kgK cp 51926 kJkgC and cv 31156 kJkgC for He Tables A1 and A 2 He 1 m3 500 kPa 25C N2 1 m3 500 kPa 80C Analysis The mass of each gas in the cylinder is 08079 kg kPa m kg K 298 K 20769 1 m kPa 500 4772 kg kPa m kg K 353 K 02968 1 m kPa 500 3 3 He 1 1 1 He 3 3 N 1 1 1 N 2 2 RT P m RT P m V V Taking the entire contents of the cylinder as our system the 1st law relation can be written as He 1 2 N 1 2 He N potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 0 0 2 2 T T mc T T mc U U U E E E v v 43 42 1 4243 1 Substituting 0 25 C C 08079 kg 31156 kJkg 80 C C 4772 kg 0743 kJkg o o o o f f T T It gives Tf 572C where Tf is the final equilibrium temperature in the cylinder The answer would be the same if the piston were not free to move since it would effect only pressure and not the specific heats b We take the entire cylinder as our system which is a closed system Noting that the cylinder is wellinsulated and thus there is no heat transfer the entropy balance for this closed system can be expressed as He N gen in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in 2 0 S S S S S S S 43 42 1 4243 1 But first we determine the final pressure in the cylinder 5111 kPa m 2 kmol 8314 kPa m kmol K 3302 K 03724 03724 kmol 4 kgkmol 08079 kg 28 kgkmol kg 4772 3 3 total total 2 He N He N total 2 2 V R T N P M m M m N N N u Then 03628 kJK 500 kPa 02968 kJkg K ln 5111 kPa 353 K 1039 kJkg K ln 3302 K kg 4772 ln ln 2 2 N 1 2 1 2 N P P R T T m c S p PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8109 8116 An insulated cylinder is divided into two parts One side of the cylinder contains N2 gas and the other side contains He gas at different states The final equilibrium temperature in the cylinder and the wasted work potential are to be determined for the cases of piston being fixed and moving freely Assumptions 1 Both N2 and He are ideal gases with constant specific heats 2 The energy stored in the container itself except the piston is negligible 3 The cylinder is wellinsulated and thus heat transfer is negligible 4 Initially the piston is at the average temperature of the two gases Properties The gas constants and the specific heats are R 02968 kPam3kgK cp 1039 kJkgC and cv 0743 kJkgC for N2 and R 20769 kPam3kgK cp 51926 kJkgC and cv 31156 kJkgC for He Tables A1 and A 2 The specific heat of copper piston is c 0386 kJkgC Table A3 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis The mass of each gas in the cylinder is 08079 kg kPa m kg K 353 K 20769 1 m kPa 500 4772 kg kPa m kg K 353 K 02968 1 m kPa 500 3 3 He 1 1 1 He 3 3 N 1 1 1 N 2 2 RT P m RT P m V V Taking the entire contents of the cylinder as our system the 1st law relation can be written as He 1 m3 500 kPa 25C N2 1 m3 500 kPa 80C Copper 0 0 Cu 1 2 He 1 2 N 1 2 Cu He N potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 2 2 T mc T T T mc T T mc U U U U E E E v v 43 42 1 4243 1 where T1 Cu 80 25 2 525C Substituting 0 525 C 50 kg 0386 kJkg C 25 C 08079 kg 31156 kJkg C 80 C kg 0743 kJkg C 4772 o o o o o o f f f T T T It gives Tf 560C where Tf is the final equilibrium temperature in the cylinder The answer would be the same if the piston were not free to move since it would effect only pressure and not the specific heats b We take the entire cylinder as our system which is a closed system Noting that the cylinder is wellinsulated and thus there is no heat transfer the entropy balance for this closed system can be expressed as piston He N gen in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in 2 0 S S S S S S S S 43 42 1 4243 1 But first we determine the final pressure in the cylinder 5094 kPa m 2 kmol 8314 kPa m kmol K 329 K 03724 03724 kmol 4 kgkmol 08079 kg 28 kgkmol kg 4772 3 3 total total 2 He N He N total 2 2 V R T N P M m M m N N N u preparation If you are a student using this Manual you are using it without permission 8112 The exergy destroyed during a process can be determined from an exergy balance or directly from its definition gen 0 destroyed T S X 14 3 hp 4241 Btumin 1hp 40 lbmmin537 R002816 Btulbm R 1 2 0 gen 0 destroyed s mT s T S X Then the reversible power and secondlaw efficiency become 109 3 hp 14 3 95 destroyed out revout X W W a and 869 109 3 hp hp 95 out rev aout II W W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8114 8119 Problem 8118 is reconsidered The effect of the state of the steam at the inlet of the feedwater heater on the ratio of mass flow rates and the reversible power is to be investigated Analysis Using EES the problem is solved as follows Input Data Steam let ststeam data FluidSteamIAPWS Tst1250 C Pst11600 kPa Pst2 Pst1 xst20 saturated liquid quality 0 Tst2temperaturesteam PPst2 xxst2 Feedwater let fwfeedwater data Tfw130 C Pfw14000 kPa Pfw2Pfw1 assume no pressure drop for the feedwater Tfw2Tst210 Surroundings To 25 C Po 100 kPa Assumed value for the surrroundings pressure Conservation of mass There is one entrance one exit for both the steam and feedwater Steam mdotst1 mdotst2 Feedwater mdotfw1 mdotfw2 Let mratio mdotstmdotfw Conservation of Energy We write the conservation of energy for steadyflow control volume having two entrances and two exits with the above assumptions Since neither of the flow rates is know or can be found write the conservation of energy per unit mass of the feedwater Ein Eout DELTAEcv DELTAEcv0 Steadyflow requirement Ein mratiohst1 hfw1 hst1enthalpyFluid TTst1 PPst1 hfw1enthalpyFluidTTfw1 PPfw1 Eout mratiohst2 hfw2 hfw2enthalpyFluid TTfw2 PPfw2 hst2enthalpyFluid xxst2 PPst2 The reversible work is given by Eq 747 where the heat transfer is zero the feedwater heater is adiabatic and the Exergy destroyed is set equal to zero Wrev mratioPsist1Psist2 Psifw1Psifw2 Psist1hst1hsto To 273sst1ssto sst1entropyFluidTTst1 PPst1 hstoenthalpyFluid TTo PPo sstoentropyFluid TTo PPo Psist2hst2hsto To 273sst2ssto sst2entropyFluidxxst2 PPst2 Psifw1hfw1hfwo To 273sfw1sfwo hfwoenthalpyFluid TTo PPo sfw1entropyFluidTTfw1 PPfw1 sfwoentropyFluid TTo PPo Psifw2hfw2hfwo To 273sfw2sfwo sfw2entropyFluidTTfw2 PPfw2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8115 Pst1 kPa mratio kgkg Wrev kJkg 200 400 600 800 1000 1200 1400 1600 1800 2000 01361 01843 02186 02466 0271 0293 03134 03325 03508 03683 4207 598 7221 8206 9035 9758 104 1099 1153 1203 200 400 600 800 1000 1200 1400 1600 1800 2000 01 015 02 025 03 035 04 Pst1 kPa mratio kgkg 200 400 600 800 1000 1200 1400 1600 1800 2000 40 50 60 70 80 90 100 110 120 130 Pst1 kPa Wrev kJkg PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8116 8120 A 1ton 1000 kg of water is to be cooled in a tank by pouring ice into it The final equilibrium temperature in the tank and the exergy destruction are to be determined Assumptions 1 Thermal properties of the ice and water are constant 2 Heat transfer to the water tank is negligible 3 There is no stirring by hand or a mechanical device it will add energy Properties The specific heat of water at room temperature is c 418 kJkgC and the specific heat of ice at about 0C is c 211 kJkgC Table A3 The melting temperature and the heat of fusion of ice at 1 atm are 0C and 3337 kJkg Analysis a We take the ice and the water as the system and disregard any heat transfer between the system and the surroundings Then the energy balance for this process can be written as ice 5C 80 kg WATER 1 ton water ice potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in 0 0 U U U E E E 43 42 1 4243 1 0 0 C 0 C water 1 2 ice liquid 2 1 solid T mc T mc T mh T mc if o o Substituting PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course o 80 2 2 kg211 kJ kg C0 5 C 3337 kJ kg 418 kJ kg C 0 C 1000 kg 418 kJ kg C 20 C 0 o o o o o T T It gives T2 1242C which is the final equilibrium temperature in the tank b We take the ice and the water as our system which is a closed system Considering that the tank is wellinsulated and thus there is no heat transfer the entropy balance for this closed system can be expressed as Swater S S S S S S ice gen in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in 0 43 42 1 4243 1 where kJK 115783 273 K 418 kJkg K ln 28542 K 273 K 3337 kJkg 268 K 211 kJkg K ln 273 K kg 80 ln ln 109590 kJK 293 K 1000 kg 418 kJkg K ln 28542 K ln ice liquid 1 2 melting solid 1 melting liquid ice melting solid ice water 1 2 water T T mc T mh T T mc S S S S T T mc S ig Then 6193 kJK 109590 115783 ice water gen S S S The exergy destroyed during a process can be determined from an exergy balance or directly from its definition gen 0 destroyed T S X 293 K6193 kJK 1815 kJ gen 0 destroyed T S X preparation If you are a student using this Manual you are using it without permission 8119 8123 Argon gas in a pistoncylinder device expands isothermally as a result of heat transfer from a furnace The useful work output the exergy destroyed and the reversible work are to be determined Assumptions 1 Argon at specified conditions can be treated as an ideal gas since it is well above its critical temperature of 151 K 2 The kinetic and potential energies are negligible Analysis We take the argon gas contained within the pistoncylinder device as the system This is a closed system since no mass crosses the system boundary during the process We note that heat is transferred to the system from a source at 1200 K but there is no heat exchange with the environment at 300 K Also the temperature of the system remains constant during the expansion process and its volume doubles that is T2 T1 and V2 2V1 a The only work interaction involved during this isothermal process is the quasiequilibrium boundary work which is determined from 3 kJ 42 3 kPa m 42 m 001 350 kPa001 m ln 002 m ln 3 3 3 3 1 2 1 1 2 1 b V V V V P Pd W W This is the total boundary work done by the argon gas Part of this work is done against the atmospheric pressure P0 to push the air out of the way and it cannot be used for any useful purpose It is determined from 1kJ 1kPa m 0 01m 100 kPa002 3 3 1 2 0 surr V P V W The useful work is the difference between these two 143 kJ 1 2 43 surr u W W W That is 143 kJ of the work done is available for creating a useful effect such as rotating a shaft Also the heat transfer from the furnace to the system is determined from an energy balance on the system to be 43 kJ 2 0 bout in bout in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in Q Q T mc U Q Q E E E v 43 42 1 4243 1 b The exergy destroyed during a process can be determined from an exergy balance or directly from Xdestroyed T0Sgen We will use the second approach since it is usually easier But first we determine the entropy generation by applying an entropy balance on an extended system system immediate surroundings which includes the temperature gradient zone between the cylinder and the furnace so that the temperature at the boundary where heat transfer occurs is TR 1200 K This way the entropy generation associated with the heat transfer is included Also the entropy change of the argon gas can be determined from QTsys since its temperature remains constant sys R T Q S S T Q S S S S sys gen in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in 43 42 1 4243 1 Therefore 0 00405 kJK 1200 K 2 43 kJ 400 K 2 43 kJ gen R sys T Q T Q S and 122 kJ 300 K 0 00405 kJK gen 0 dest T S X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8123 8126 A rigid tank containing nitrogen is considered Heat is now transferred to the nitrogen from a reservoir and nitrogen is allowed to escape until the mass of nitrogen becomes onehalf of its initial mass The change in the nitrogens work potential is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 Nitrogen is an ideal gas with constant specific heats Properties The properties of nitrogen at room temperature are cp 1039 kJkgK cv 0743 kJkgK and R 02968 kJkgK Table A2a Analysis The initial and final masses in the tank are Nitrogen 50 L 1200 kPa 20C me 0 690 kg 02968 kPa m kg K293 K 120 0 kPa 0 050 m 3 3 1 1 RT P m V Q 0 345 kg 2 0 690 kg 2 1 2 m m m e The final temperature in the tank is 586 K 0 345 kg02968 kPa m kg K 120 0 kPa 0 050 m 3 3 2 2 m R P T V We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance 2 system out in m m m m m e Energy balance 1 1 2 2 out 1 1 2 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in m u m u m h Q m u m u m h Q E E E e e e e 43 42 1 4243 1 Using the average of the initial and final temperatures for the exiting nitrogen 439 5 K this energy balance equation becomes 586 50 293 50 2 1 T T Te 157 5 kJ 0 690 0 743293 0 345 0 743586 0 345 1 039439 5 1 1 2 2 1 1 2 2 out m c T m c T c T m m u m u m h Q e p e e e v v The work potential associated with this process is equal to the exergy destroyed during the process The exergy destruction during a process can be determined from an exergy balance or directly from its definition The entropy generation S gen 0 destroyed T S X gen in this case is determined from an entropy balance on the system R e e e e R T Q m s m s m s S m s m s S S m s T Q S S S S in 1 1 2 2 gen 1 1 2 2 tank gen in in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in 43 42 1 4243 1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8124 Noting that pressures are same rearranging and substituting gives 0 190 kJK 773 157 5 0 345 1 039ln439 5 0 690 1 039ln293 0 345 1 039ln586 ln ln ln in 1 1 2 2 in 1 1 2 2 gen R e p e p p R e e T Q T m c T m c T c m T Q m s m s m s S Then 557 kJ 293 K 0 190 kJK gen 0 destroyed rev T S X W Alternative More Accurate Solution This problem may also be solved by considering the variation of gas temperature at the outlet of the tank The mass and energy balances are dt c T dm dt c d mT dt h dm dt d mu Q dt dm m p v e Combining these expressions and replacing T in the last term gives dt dm Rm P c dt d mT c Q p v V Integrating this over the time required to release onehalf the mass produces 1 2 1 1 2 2 ln m m R c P m T m T c Q p v V The reduced combined first and second law becomes dt T s dm h dt T S d U T T Q W R 1 0 0 0 rev when the mass balance is substituted and the entropy generation is set to zero for maximum work production Expanding the system time derivative gives dt dh T m T dt h dm dt d mu T T Q dt T s dm h dt T s dm dt T m ds dt d mu T T Q dt T s dm h dt T ms d mu T T Q W R R R 0 0 0 0 0 0 0 0 0 rev 1 1 1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8125 Substituting from the first law Q dt dT mc dt c T dm dt d mT c T T dt m dh dt h dm dt d mu T T dt dh T m T dt h dm dt d mu T T dt h dm dt d mu W p p v R R R 1 0 0 0 0 rev At any time mR P T V which further reduces this result to dt dP P R dt dT T c T m dt dm mR P c T T W p p R 0 0 rev V When this integrated over the time to complete the process the result is 498 kJ 586 1 293 1 0 2968 1 0391200 0 050 293 2 ln 1 0 2968 1 0391200 0 050 773 293 1 1 ln 2 1 0 1 2 0 rev T T R c P T m m R P c T T W p p R V V PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8126 8127 A rigid tank containing nitrogen is considered Nitrogen is allowed to escape until the mass of nitrogen becomes one half of its initial mass The change in the nitrogens work potential is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 Nitrogen is an ideal gas with constant specific heats Properties The properties of nitrogen at room temperature are cp 1039 kJkgK cv 0743 kJkgK k 14 and R 02968 kJkgK Table A2a Nitrogen 100 L 1000 kPa 20C me Analysis The initial and final masses in the tank are 1 150 kg 02968 kPa m kg K293 K 100 0 kPa 0 100 m 3 3 1 1 RT P m V 0 575 kg 2 1 150 kg 2 1 2 m m m e We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance 2 system out in m m m m m e Energy balance 1 1 2 2 potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in m u m u h m E E E e e 43 42 1 4243 1 Using the average of the initial and final temperatures for the exiting nitrogen this energy balance equation becomes 1 150 0 743293 0 575 0 743 0 575 1 039 50 293 2 2 1 1 2 2 1 1 2 2 T T m c T m c T c T m m u m u h m e p e e e v v Solving for the final temperature we get 3 K 2 224 T The final pressure in the tank is 382 8 kPa 100 m 0 0 575 kg02968 kPa m kg K224 3 K 3 3 2 2 2 V m RT P The average temperature and pressure for the exiting nitrogen is 258 7 K 224 3 50 293 50 2 1 T T Te 691 4 kPa 382 8 50 1000 50 2 1 P P Pe The work potential associated with this process is equal to the exergy destroyed during the process The exergy destruction during a process can be determined from an exergy balance or directly from its definition The entropy generation S gen 0 destroyed T S X gen in this case is determined from an entropy balance on the system e e e e m s m s m s S m s m s S S s m S S S S 1 1 2 2 gen 1 1 2 2 tank gen in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in 43 42 1 4243 1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8127 Rearranging and substituting gives PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 007152 kJK 2 2032 4 4292 2 2188 0 2968 ln691 4 0 575 1 039ln258 7 0 2968 ln1000 1 15 1 039ln293 0 2968 ln382 8 0 575 1 039ln224 3 ln ln ln ln ln ln 1 1 1 2 2 2 1 1 2 2 gen e e p e p p e e P R T c m P R T m c P R T c m m s m s m s S Then 210 kJ 0 007152 kJK 293 K gen 0 destroyed rev T S X W The entropy generation cannot be negative for a thermodynamically possible process This result is probably due to using average temperature and pressure values for the exiting gas and using constant specific heats for nitrogen This sensitivity occurs because the entropy generation is very small in this process Alternative More Accurate Solution This problem may also be solved by considering the variation of gas temperature and pressure at the outlet of the tank The mass balance in this case is dt dm me which when combined with the reduced first law gives dt h dm dt d mu Using the specific heats and the ideal gas equation of state reduces this to dt c T dm dt dP R c p v V which upon rearrangement and an additional use of ideal gas equation of state becomes dt dm m c c dt dP P v p 1 1 When this is integrated the result is 378 9 kPa 2 1 1000 41 1 2 1 2 k m m P P The final temperature is then 222 0 K 0 575 kg02968 kPa m kg K 378 9 kPa 0 100 m 3 3 2 2 2 m R P T V The process is then one of P const mk or const 1 T mk The reduced combined first and second law becomes dt T s dm h dt T S d U W 0 0 rev when the mass balance is substituted and the entropy generation is set to zero for maximum work production Replacing the enthalpy term with the first law result and canceling the common dUdt term reduces this to preparation If you are a student using this Manual you are using it without permission 8128 dt T s dm dt T d ms W 0 0 rev Expanding the first derivative and canceling the common terms further reduces this to dt T m ds W rev 0 Letting and the pressure and temperature of the nitrogen in the system are related to the mass by mk P a 1 1 1 1 1 mk T b and P amk 1 bmk T according to the first law Then and dm akm dP k 1 dm m b k dT k 2 1 The entropy change relation then becomes m Rk dm c k P R dP T dT c ds p p 1 Now multiplying the combined first and second laws by dt and integrating the result gives 00135 kJ 1 15 0 2968 41 0 575 1 1 039 41 293 1 1 1 2 0 2 1 0 2 1 0 rev m Rk m c k T Rk dm c mds k T mds T W p p Once again the entropy generation is negative which cannot be the case for a thermodynamically possible process This is probably due to using constant specific heats for nitrogen This sensitivity occurs because the entropy generation is very small in this process PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8130 8129 A system consisting of a compressor a storage tank and a turbine as shown in the figure is considered The change in the exergy of the air in the tank and the work required to compress the air as the tank was being filled are to be determined Assumptions 1 Changes in the kinetic and potential energies are negligible 4 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are R 0287 kPam3kgK cp 1005 kJkgK cv 0718 kJkgK k 14 Table A2a Analysis The initial mass of air in the tank is 0 5946 10 kg 0 287 kPa m kg K293 K 100 kPa5 10 m 6 3 3 5 initial initial initial RT P m V and the final mass in the tank is 3 568 10 kg 0 287 kPa m kg K293 K 600 kPa5 10 m 6 3 3 5 final final final RT P m V Since the compressor operates as an isentropic device k k P P T T 1 1 2 1 2 The conservation of mass applied to the tank gives min dt dm while the first law gives dt h dm dt d mu Q Employing the ideal gas equation of state and using constant specific heats expands this result to dt dP RT c T dt dP R c Q p V V v 2 Using the temperature relation across the compressor and multiplying by dt puts this result in the form RT dP P P c T R dP c dt Q k k p V V v 1 1 1 When this integrated it yields i and f stand for initial and final states 10 kJ 6017 100 100 600 600 0 287 1 0055 10 1 41 2 41 600 100 0 287 10 0718 5 1 2 8 41 40 5 5 1 i k k i f f p i f P P P P R c k k P P R c Q V V v The negative result show that heat is transferred from the tank Applying the first law to the tank and compressor gives h dm d mu dt W Q 1 out which integrates to 1 out i f i i f f m h m m u m u Q W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8131 Upon rearrangement 10 kJ 3516 8 0 5946 10 0 718293 3 568 1 005 017 10 6 6 8 out i f p m T m c c Q W v The negative sign shows that work is done on the compressor When the combined first and second laws is reduced to fit the compressor and tank system and the mass balance incorporated the result is dt T s dm h dt T S d U T T Q W R 1 0 0 0 rev which when integrated over the process becomes 10 kJ 7875 8 100 293 0 287 ln 600 1 005293 0 718 568 10 3 1 005293 0 718 0 5946 10 293 293 1 017 10 6 ln 1 1 6 6 8 0 0 1 0 1 1 0 1 0 rev i f p v f f p v i i R f f f i i i R P P T R c c T m c m T c T T Q s s T h u m s s T h u m T T Q W This is the exergy change of the air stored in the tank PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8132 8130 The air stored in the tank of the system shown in the figure is released through the isentropic turbine The work produced and the change in the exergy of the air in the tank are to be determined Assumptions 1 Changes in the kinetic and potential energies are negligible 4 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are R 0287 kPam3kgK cp 1005 kJkgK cv 0718 kJkgK k 14 Table A2a Analysis The initial mass of air in the tank is 3 568 10 kg 0 287 kPa m kg K293 K 600 kPa5 10 m 6 3 3 5 initial initial initial RT P m V and the final mass in the tank is 0 5946 10 kg 0 287 kPa m kg K293 K 100 kPa5 10 m 6 3 3 5 final final final RT P m V The conservation of mass is min dt dm while the first law gives dt h dm dt d mu Q Employing the ideal gas equation of state and using constant specific heats expands this result to dt dP dt dP R c c dt dP c T RT dt dP R c Q p p V V V V v v When this is integrated over the process the result is i and f stand for initial and final states 10 kJ 52 600 5 10 100 8 5 i f P P Q V Applying the first law to the tank and compressor gives hdm d mu dt W Q out which integrates to 10 kJ 300 8 0 5946 10 005293 3 568 10 1 3 568 10 1 005293 0 5946 10 0 718293 10 52 6 6 6 6 8 out out out f i p p i v f f i i i f f f i i i f f f i i i f f m c T m m c T m c T Q m h m m u m u Q W m h m m u m u Q W m h m m u m u W Q This is the work output from the turbine When the combined first and second laws is reduced to fit the turbine and tank system and the mass balance incorporated the result is PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8133 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 rev i f v p R v p R R R P P T T dt T dm c c T T Q dt mT ds dt T dm c c T T Q dt T s dm h dt T s m d u dt T s dm u T T Q dt T s dm h dt T S d U T T Q W V where the last step uses entropy change equation When this is integrated over the process it becomes 10 kJ 500 8 8 8 5 6 8 0 0 rev 10 52 2 500 10 0 600 293 100 293 5 10 3 568 10 0 718293 0 5946 1 005 293 293 1 00 10 3 1 i f i f v p R P P T T m T m c c T T Q W V This is the exergy change of the air in the storage tank PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8135 8132E Large brass plates are heated in an oven at a rate of 300min The rate of heat transfer to the plates in the oven and the rate of exergy destruction associated with this heat transfer process are to be determined Assumptions 1 The thermal properties of the plates are constant 2 The changes in kinetic and potential energies are negligible 3 The environment temperature is 75F Properties The density and specific heat of the brass are given to be ρ 5325 lbmft3 and cp 0091 BtulbmF Analysis We take the plate to be the system The energy balance for this closed system can be expressed as 1 2 1 2 plate in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T u m u U Q E E E 43 42 1 4243 1 The mass of each plate and the amount of heat transfer to each plate is 213 lbm 12 ft 2 ft2 ft 532 5 lbmft 21 3 LA m ρ ρV 17930 Btuplate 75 F 213 lbmplate 0 091 Btulbm F1000 1 2 in T mc T Q Then the total rate of heat transfer to the plates becomes 5379000 Btumin 89650 Btus 17930 Btuplate 300 platesmin in per plate plate total Q n Q We again take a single plate as the system The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the plate and its immediate surroundings so that the boundary temperature of the extended system is at 1300F at all times system in gen system gen in in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in S T Q S S S T Q S S S S b b 43 42 1 4243 1 where 1946 BtuR 75 460 R 213 lbm 0 091 BtulbmR ln 1000 460 R ln 1 2 avg 1 2 system T T mc s m s S Substituting 9 272 BtuR per plate 1946 BtuR 1300 460 R 17930 Btu system in gen S T Q S b Then the rate of entropy generation becomes 9 272 BtuR plate300 platesmin 2781 BtuminR 4635 BtusR ball gen gen n S S The exergy destroyed during a process can be determined from an exergy balance or directly from its definition gen 0 destroyed T S X 535 R4635 BtusR 24797 Btus gen 0 destroyed T S X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8136 8133 Long cylindrical steel rods are heattreated in an oven The rate of heat transfer to the rods in the oven and the rate of exergy destruction associated with this heat transfer process are to be determined Assumptions 1 The thermal properties of the rods are constant 2 The changes in kinetic and potential energies are negligible 3 The environment temperature is 30C Properties The density and specific heat of the steel rods are given to be ρ 7833 kgm3 and cp 0465 kJkgC Analysis Noting that the rods enter the oven at a velocity of 3 mmin and exit at the same velocity we can say that a 3m long section of the rod is heated in the oven in 1 min Then the mass of the rod heated in 1 minute is m V LA L D ρ ρ ρ π π 2 2 4 7833 3 01 4 184 6 kg m m m kg 3 We take the 3m section of the rod in the oven as the system The energy balance for this closed system can be expressed as 1 2 1 2 rod in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T mc T u m u U Q E E E 43 42 1 4243 1 Substituting 57512 kJ 30 C 184 6 kg 0 465 kJkg C700 1 2 in T mc T Q Noting that this much heat is transferred in 1 min the rate of heat transfer to the rod becomes 57512 kJ1 min 57512 kJmin 9585 kW in in t Q Q We again take the 3m long section of the rod as the system The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the rod and its immediate surroundings so that the boundary temperature of the extended system is at 900C at all times system in gen system gen in in entropy Change system generation Entropy gen by heat and mass entropy transfer Net out in S T Q S S S T Q S S S S b b 43 42 1 4243 1 where 100 1 kJK 30 273 184 6 kg 0 465 kJkgK ln 700 273 ln 1 2 avg 1 2 system T T mc s m s S Substituting 51 1 kJK 100 1 kJK 900 273 R 57512 kJ system in gen S T Q S b Noting that this much entropy is generated in 1 min the rate of entropy generation becomes 51 1 kJminK 0852 kWK 1min 1 kJK 51 gen gen t S S The exergy destroyed during a process can be determined from an exergy balance or directly from its definition gen 0 destroyed T S X 298 K0852 kWK 254 kW gen 0 destroyed T S X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8139 Assumptions 1 Steady operating conditions exist 2 Potential energy change is negligible 3 Air is an ideal gas with constant specific heats Properties The gas constant of air is R 0287 kJkgK and the specific heat of air at the average temperature of 6275272 577ºC 850 K is cp 111 kJkgºC Table A2 Analysis a The enthalpy and entropy changes of air across the turbine are 527C 500 kPa Q Exh gas 627C 12 MPa Turbine 111 kJkg 627 C kJkg C527 111 1 2 T T c h p 0 1205 kJkgK 1200 kPa 0287 kJkgK ln 500 kPa 273 K 627 273 K 111 kJkgKln 527 ln ln 1 2 1 2 P P R T T c s p The actual and reversible power outputs from the turbine are kW 3673 kW 2575 273 K01205 kJkgK 25 kJkg kgs111 52 20 kW kJkg kgs111 52 0 out rev out out a s T h m W Q m h W or kW 3673 kW 2575 out rev out a W W b The exergy destroyed in the turbine is 1098 kW 257 5 367 3 a rev dest W W X c The secondlaw efficiency is 701 0 701 367 3 kW 5 kW 257 rev a II W W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8140 8137 Refrigerant134a is compressed in an adiabatic compressor whose secondlaw efficiency is given The actual work input the isentropic efficiency and the exergy destruction are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a The properties of the refrigerant at the inlet of the compressor are Tables A11 through A13 95153 kJkgK 0 60 kJkg 243 3 C 1560 kPa 160 60 C 15 1 1 1 1 kPa sat160 s h T P T 1 MPa R134a 160 kPa Compressor The enthalpy at the exit for if the process was isentropic is 28241 kJkg 95153 kJkgK 0 1MPa 2 1 2 2 h s s s P The expressions for actual and reversible works are 24360kJkg 2 1 2 a h h h w 095153kJkgK 273 K 25 24360kJkg 2 2 1 2 0 1 2 rev s h s s T h h w Substituting these into the expression for the secondlaw efficiency 24360 095153 298 24360 0 80 2 2 2 a rev II h s h w w η The exit pressure is given 1 MPa We need one more property to fix the exit state By a trialerror approach or using EES we obtain the exit temperature to be 60ºC The corresponding enthalpy and entropy values satisfying this equation are 98492 kJkgK 0 36 kJkg 293 2 2 s h Then 4976kJkg 24360 29336 1 2 a h h w 3981 kJkg 09515kJkg K 273 K098492 25 24360kJkg 29336 1 2 0 1 2 rev s T s h h w b The isentropic efficiency is determined from its definition 0780 24360kJkg 29336 24360kJkg 28241 1 2 1 2s h h h h ηs b The exergy destroyed in the compressor is 995 kJkg 3981 4976 rev a dest w w x PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8141 8138 The isentropic efficiency of a water pump is specified The actual power output the rate of frictional heating the exergy destruction and the secondlaw efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a Using saturated liquid properties at the given temperature for the inlet state Table A4 4 MPa PUMP Water 100 kPa 30C 135 kgs 001004 m kg 0 0 4367 kJkgK 82 kJkg 125 0 C 30 3 1 1 1 1 1 v s h x T The power input if the process was isentropic is 5 288 kW 100kPa 1 35 kgs0001004 m kg4000 3 1 2 1 s P P m W v Given the isentropic efficiency the actual power may be determined to be 7554 kW 0 70 288 kW 5 s η s a W W b The difference between the actual and isentropic works is the frictional heating in the pump 2266 kW 5 288 7 554 frictional s a W W Q c The enthalpy at the exit of the pump for the actual process can be determined from 13142 kJkg 12582kJkg 1 35 kgs 7 554 kW 2 2 1 2 a h h h m h W The entropy at the exit is 0 4423 kJkgK 42 kJkg 131 4 MPa 2 2 2 s h P The reversible power and the exergy destruction are 5 362 kW 04367kJkgK 273 K04423 20 12582kJkg 35 kgs 13142 1 1 2 0 1 2 rev s T s h m h W 2193 kW 5 362 7 554 rev a dest W W X d The secondlaw efficiency is 0710 7 554 kW 362 kW 5 a rev II W W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8142 8139 Argon gas is expanded adiabatically in an expansion valve The exergy of argon at the inlet the exergy destruction and the secondlaw efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are zero 3 Argon is an ideal gas with constant specific heats Properties The properties of argon gas are R 02081 kJkgK cp 05203 kJkgºC Table A2 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis a The exergy of the argon at the inlet is 2247 kJkg 100 kPa 02081 kJkgKln 3500 kPa 298 K 298 K 05203 kJkgKln 373 K 25 C 05203 kJkgK100 ln ln 0 1 0 1 0 0 1 0 1 0 0 1 1 P P R T T c T T T c s s T h h x p p 500 kPa Argon 35 MPa 100C b Noting that the temperature remains constant in a throttling process of an ideal gas the exergy destruction is determined from 1207 kJkg 3500 kPa 500 kPa 02081 kJkgKln 298 K ln 0 1 0 1 2 0 gen 0 dest P P R T s s T T s x c The secondlaw efficiency is 0463 224 7 kJkg 120 7 kJkg 224 7 1 dest 1 II x x x η preparation If you are a student using this Manual you are using it without permission 8143 8140 Heat is lost from the air flowing in a diffuser The exit temperature the rate of exergy destruction and the second law efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 Potential energy change is negligible 3 Nitrogen is an ideal gas with variable specific heats Properties The gas constant of nitrogen is R 02968 kJkgK Analysis a For this problem we use the properties from EES software Remember that for an ideal gas enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure At the inlet of the diffuser and at the dead state we have 101 kJkg K 7 39 kJkg 88 kPa 100 383 K C 110 1 1 1 1 s h P T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1 93 kJkg 300 K 0 h T An energy balance on the diffuser gives q 110 kPa 45 ms Nitrogen 100 kPa 110C 205 ms 6 846 kJkg 100 kPa 0 1 s P K 1 9 kJ 2 105 h kg kJkg 52 m s 1000 1kJkg 2 ms 45 m s 1000 1kJkg 2 205 ms 39 kJkg 88 2 2 2 2 2 2 2 2 2 out 2 2 2 2 1 1 h q V h V h he corres nding properties at the exit of the diffuser are 110 kPa 2 2 1 2 s P b The mass flow rate of the nitrogen is determined to be T po 117 kJkg K 7 400 K 105 9 kJkg T h C 127 1 669 kgs 45 ms 0 2968 kJkgK400 K 0 04 m 110 kPa P 2 2 2 2 2 2 2 2 A V RT A V m ρ The exergy destruction in the nozzle is the exergy difference between the inlet and exit of the diffuser 124 kW 7 117kJkgK 300 K 7 101 m s 1000 1kJkg 2 45 ms 205 ms 105 9 kJkg 669 kgs 8839 1 2 2 2 2 2 2 1 0 2 2 2 1 2 1 dest s s T V V h m h X c The secondlaw efficiency for this device may be defined as the exergy output divided by the exergy input 96 kW 51 6 846kJkgK 300 K 7 101 m s 1000 1kJkg 2 205 ms 1 93 kJkg 669 kgs 8839 1 2 2 2 2 0 1 0 2 1 0 1 1 s s T V h m h X 761 0 761 5196 kW 12 4 kW 1 1 1 dest 1 2 II X X X X η preparation If you are a student using this Manual you are using it without permission 8144 8141 Using an incompressible substance as an example it is to be demonstrated if closed system and flow exergies can be negative Analysis The availability of a closed system cannot be negative However the flow availability can be negative at low pressures A closed system has zero availability at dead state and positive availability at any other state since we can always produce work when there is a pressure or temperature differential To see that the flow availability can be negative consider an incompressible substance The flow availability can be written as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 ψ ξ h h T s s u u P P T s s P P 0 0 0 0 0 0 0 v v The closed system availability ξ is always positive or zero and the flow availability can be negative when P P0 8142 A relation for the secondlaw efficiency of a heat engine operating between a heat source and a heat sink at specified temperatures is to be obtained HE TL Sink QL Source TH QH W Analysis The secondlaw efficiency is defined as the ratio of the availability recovered to availability supplied during a process The work W produced is the availability recovered The decrease in the availability of the heat supplied QH is the availability supplied or invested Therefore 1 1 0 0 II W Q T T Q T T H L H H Note that W η the first term in the denominator is the availability of heat supplied to the heat ngine whereas the second term is the availability of the heat rejected by the heat ngine The difference between the two is the availability consumed during the rocess e e p preparation If you are a student using this Manual you are using it without permission 8145 8143 Writing energy and entropy balances a relation for the reversible work is to be obtained for a closed system that exchanges heat with surroundings at T0 in the amount of Q0 as well as a heat reservoir at temperature TR in the amount QR Assumptions Kinetic and potential changes are negligible Analysis We take the direction of heat transfers to be to the system heat input and the direction of work transfer to be from the system work output The result obtained is still general since quantities wit opposite directions can be handled the same way by using negative signs The energy and entropy balances for this stationary closed system can be expressed as Energy balance 1 R R Q Q U U W U U W Q Q E E E 0 2 1 1 2 0 system out in Entropy balance 0 0 1 2 gen system gen out in T Q T Q S S S S S S R R S 2 Solving for Q0 from 2 and substituting in 1 yields System Source TR QR gen 0 2 1 0 2 1 TR R 0 1 T S T Q S S T U U W The useful work relation for a closed system is obtained from surr W W Wu 1 1 0 V2 gen 0 0 2 1 0 2 1 V P T S T T Q S S T U U R R Then the reversible work relation is obtained by substituting Sgen 0 T R R T Q P S S T U U W 0 2 1 0 2 1 0 2 1 rev 1 V V A positive result for Wrev indicates work output and a negative result work input Also the QR is a positive quantity for heat transfer to the system and a negative quantity for heat transfer from the system PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8146 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course same way by using negative signs The energy and entropy balances for this stationary closed system can be expressed Energy balance 8144 Writing energy and entropy balances a relation for the reversible work is to be obtained for a steadyflow system that exchanges heat with surroundings at T0 at a rate of 0 as well as a heat reservoir at temperature T Q R in the amount QR Analysis We take the direction of heat transfers to be to the system heat input and the direction of work transfer to be from the system work output The result obtained is still general since quantities wit opposite directions can be handled the as out in system out in E E E E E System 2 2 2 2 0 QR i i i i e e e e gz V m h gz V h m W Q or R e e e e i i i i Q Q gz V h m gz V m h W 0 2 2 2 2 1 Entropy balance 0 S S S system gen in out gen S S S out in S 0 gen T T m s m s S R i i e e Q0 2 QR olving fo r Q0 from 2 and substituting in 1 yields S R R e e e e e i i i i i T Q T S T s gz h m T s gz m h W 0 gen 0 0 0 1 2 2 T V V 2 2 hen the reversible work relation is obtained by substituting Sgen 0 T R R e e e e e i i i i i T T Q T s gz V h m T s gz V m h W 0 0 2 0 2 rev 1 2 2 A positive result for Wrev indicates work output and a negative result work input Also the QR is a positive quantity for heat transfer to the system and a negative quantity for heat transfer from the system preparation If you are a student using this Manual you are using it without permission 8147 8145 Writing energy and entropy balances a relation for the reversible work is to be obtained for a uniformflow system that exchanges heat with surroundings at T0 in the amount of Q0 as well as a heat reservoir at temperature TR in the amount QR Assumptions Kinetic and potential changes are negligible Analysis We take the direction of heat transfers to be to the system heat input and the direction of work transfer to be from the system work output The result obtained is still general since quantities wit opposite directions can be handled the same way by using negative signs The energy and entropy balances for this stationary closed system can be expressed as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course E E E Energy balance in system out cv i i i i e e e e R U U gz V m h gz V h m W Q Q 2 2 1 2 2 2 0 R cv e e e e i i i i Q Q U U gz V h m gz V m h W or 0 1 2 2 2 2 2 1 tr py system gen out in S S S S En o balance System Q me Source TR 0 0 1 2 gen T Q Q T m s m s S S S R R i i e e cv 2 Solving for Q0 from 2 and substituting in 1 yields T 2 2 R R gen cv e e e e e i i i i i T Q T S S S T U U T s gz V h m T s gz V m h W 0 0 2 1 0 2 1 0 2 0 2 1 The useful work relation for a closed system is obtained from 1 2 2 1 2 0 0 gen 0 2 1 0 2 1 0 2 0 2 surr V V P T T Q T S S S T U U T s gz V h m T s gz V m h W W W R R cv e e e e e i i i i i u Then the reversible work relation is obtained by substituting Sgen 0 R R cv e e e e e i i i i i T T Q P S S T U U T s gz V h m T s gz V m h W 0 2 1 0 2 1 0 2 1 0 2 0 2 rev 1 2 2 V V A positive result for Wrev indicates work output and a negative result work input Also the QR is a positive quantity for heat transfer to the system and a negative quantity for heat transfer from the system preparation If you are a student using this Manual you are using it without permission 8148 Fundamentals of Engineering FE Exam Problems 8146 Heat is lost through a plane wall steadily at a rate of 800 W If the inner and outer surface temperatures of the wall are 20C and 5C respectively and the environment temperature is 0C the rate of exergy destruction within the wall is a 40 W b 17500 W c 765 W d 32800 W e 0 W Answer a 40 W Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Q800 W T120 C T25 C To0 C Entropy balance Sin Sout Sgen DSsystem for the wall for steady operation gives QT1273QT2273Sgen0 WK XdestTo273Sgen W Some Wrong Solutions with Common Mistakes QT1QT2Sgen10 W1XdestTo273Sgen1 Using C instead of K in Sgen Sgen2QT1T22 W2XdestTo273Sgen2 Using avegage temperature in C for Sgen Sgen3QT1T22273 W3XdestTo273Sgen3 Using avegage temperature in K W4XdestToSgen Using C for To 8147 Liquid water enters an adiabatic piping system at 15C at a rate of 3 kgs It is observed that the water temperature rises by 03C in the pipe due to friction If the environment temperature is also 15C the rate of exergy destruction in the pipe is a 38 kW b 24 kW c 72 kW d 98 kW e 124 kW Answer a 38 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cp418 kJkgK m3 kgs T115 C T2153 C To15 C SgenmCplnT2273T1273 kWK XdestTo273Sgen kW Some Wrong Solutions with Common Mistakes W1XdestTo273mCplnT2T1 Using deg C in Sgen W2XdestTomCplnT2T1 Using deg C in Sgen and To W3XdestTo273CplnT2T1 Not using mass flow rate with deg C W4XdestTo273CplnT2273T1273 Not using mass flow rate with K PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8149 8148 A heat engine receives heat from a source at 1500 K at a rate of 600 kJs and rejects the waste heat to a sink at 300 K If the power output of the engine is 400 kW the secondlaw efficiency of this heat engine is a 42 b 53 c 83 d 67 e 80 Answer c 83 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Qin600 kJs W400 kW TL300 K TH1500 K Etarev1TLTH EtathWQin EtaIIEtathEtarev Some Wrong Solutions with Common Mistakes W1EtaIIEtath1Etarev Etath11WQin Using wrong relation for thermal efficiency W2EtaIIEtath Taking secondlaw efficiency to be thermal efficiency W3EtaIIEtarev Taking secondlaw efficiency to be reversible efficiency W4EtaIIEtathEtarev Multiplying thermal and reversible efficiencies instead of dividing 8149 A water reservoir contains 100 tons of water at an average elevation of 60 m The maximum amount of electric power that can be generated from this water is a 8 kWh b 16 kWh c 1630 kWh d 16300 kWh e 58800 kWh Answer b 16 kWh Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values m100000 kg h60 m g981 ms2 Maximum power is simply the potential energy change Wmaxmgh1000 kJ WmaxkWhWmax3600 kWh Some Wrong Solutions with Common Mistakes W1Wmax mgh3600 Not using the conversion factor 1000 W2Wmax mgh1000 Obtaining the result in kJ instead of kWh W3Wmax mgh361000 Using worng conversion factor W4Wmax mh3600Not using g and the factor 1000 in calculations PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8150 8150 A house is maintained at 21C in winter by electric resistance heaters If the outdoor temperature is 9C the second law efficiency of the resistance heaters is a 0 b 41 c 57 d 25 e 100 Answer b 41 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TL9273 K TH21273 K ToTL COPrevTHTHTL COP1 EtaIICOPCOPrev Some Wrong Solutions with Common Mistakes W1EtaIICOPCOPrev1 COPrev1TLTHTL Using wrong relation for COPrev W2EtaII1TL273TH273 Taking secondlaw efficiency to be reversible thermal efficiency with C for temp W3EtaIICOPrev Taking secondlaw efficiency to be reversible COP W4EtaIICOPrev2COP COPrev2TL273THTL Using C in COPrev relation instead of K and reversing 8151 A 10kg solid whose specific heat is 28 kJkgC is at a uniform temperature of 10C For an environment temperature of 25C the exergy content of this solid is a Less than zero b 0 kJ c 223 kJ d 625 kJ e 980 kJ Answer d 625 kJ Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values m10 kg Cp28 kJkgK T110273 K To25273 K Exergy content of a fixed mass is x1u1uoTos1soPov1vo exmCpT1ToToCplnT1To Some Wrong Solutions with Common Mistakes W1exmCpToT1 Taking the energy content as the exergy content W2exmCpT1ToToCplnT1To Using for the second term instead of W3exCpT1ToToCplnT1To Using exergy content per unit mass W4ex0 Taking the exergy content to be zero PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8151 8152 Keeping the limitations imposed by the secondlaw of thermodynamics in mind choose the wrong statement below a A heat engine cannot have a thermal efficiency of 100 b For all reversible processes the secondlaw efficiency is 100 c The secondlaw efficiency of a heat engine cannot be greater than its thermal efficiency d The secondlaw efficiency of a process is 100 if no entropy is generated during that process e The coefficient of performance of a refrigerator can be greater than 1 Answer c The secondlaw efficiency of a heat engine cannot be greater than its thermal efficiency 8153 A furnace can supply heat steadily at a 1300 K at a rate of 500 kJs The maximum amount of power that can be produced by using the heat supplied by this furnace in an environment at 300 K is a 115 kW b 192 kW c 385 kW d 500 kW e 650 kW Answer c 385 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Qin500 kJs TL300 K TH1300 K WmaxQin1TLTH kW Some Wrong Solutions with Common Mistakes W1WmaxWmax2 Taking half of Wmax W2WmaxQin1TLTH Dividing by efficiency instead of multiplying by it W3Wmax QinTLTH Using wrong relation W4WmaxQin Assuming entire heat input is converted to work PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8152 8154 Air is throttled from 50C and 800 kPa to a pressure of 200 kPa at a rate of 05 kgs in an environment at 25C The change in kinetic energy is negligible and no heat transfer occurs during the process The power potential wasted during this process is a 0 b 020 kW c 47 kW d 59 kW e 119 kW Answer d 59 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values R0287 kJkgK Cp1005 kJkgK m05 kgs T150273 K P1800 kPa To25 C P2200 kPa Temperature of an ideal gas remains constant during throttling since hconst and hhT T2T1 dsCplnT2T1RlnP2P1 XdestTo273mds kW Some Wrong Solutions with Common Mistakes W1dest0 Assuming no loss W2destTo273ds Not using mass flow rate W3destTomds Using C for To instead of K W4destmP1P2 Using wrong relations PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 91 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 9 GAS POWER CYCLES PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 92 Actual and Ideal Cycles Carnot cycle AirStandard Assumptions Reciprocating Engines 91C It is less than the thermal efficiency of a Carnot cycle 92C It represents the net work on both diagrams 93C The air standard assumptions are 1 the working fluid is air which behaves as an ideal gas 2 all the processes are internally reversible 3 the combustion process is replaced by the heat addition process and 4 the exhaust process is replaced by the heat rejection process which returns the working fluid to its original state 94C The cold air standard assumptions involves the additional assumption that air can be treated as an ideal gas with constant specific heats at room temperature 95C The clearance volume is the minimum volume formed in the cylinder whereas the displacement volume is the volume displaced by the piston as the piston moves between the top dead center and the bottom dead center 96C It is the ratio of the maximum to minimum volumes in the cylinder 97C The MEP is the fictitious pressure which if acted on the piston during the entire power stroke would produce the same amount of net work as that produced during the actual cycle 98C Yes 99C Assuming no accumulation of carbon deposits on the piston face the compression ratio will remain the same otherwise it will increase The mean effective pressure on the other hand will decrease as a car gets older as a result of wear and tear 910C The SI and CI engines differ from each other in the way combustion is initiated by a spark in SI engines and by compressing the air above the selfignition temperature of the fuel in CI engines PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 93 911C Stroke is the distance between the TDC and the BDC bore is the diameter of the cylinder TDC is the position of the piston when it forms the smallest volume in the cylinder and clearance volume is the minimum volume formed in the cylinder 912E The maximum possible thermal efficiency of a gas power cycle with specified reservoirs is to be determined Analysis The maximum efficiency this cycle can have is 0643 1 L T 460 R 940 460 R 40 1 thCarnot TH η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 94 913 An airstandard cycle executed in a pistoncylinder system is composed of three specified processes The cycle is to be sketcehed on the Pv and Ts diagrams and the back work ratio are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air are given as R 0287 kPam3kgK cp 1005 kJkgK cv 0718 kJkgK and k 14 Analysis a The Pv and Ts diagrams of the cycle are shown in the figures b Process 12 Isentropic compression 1 2 1 2 T T mc w in v s T 3 2 1 v P 3 2 1 1 1 2 1 2 T r T T v 1 1 k k v rocess 23 Constant pressure heat addition T mR T P V V v he back w rk ratio is P 2 2 3 Pd w out 2 3 2 3 2 3 T o 2 3 1 2 3 2 1 2 T mR T T T mc w w r out in bw v Noting that 1 and thus and k R c c c k c c R p p v v v From ideal gas relation r T T 2 1 2 3 2 3 v v v v Substituting these into back work relation 0256 1 6 6 1 1 41 1 1 1 1 1 1 1 1 1 1 1 1 1 1 40 1 1 2 3 2 1 2 2 r r k r r k T T T T T T R k R r k k bw PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 95 914 The three processes of an airstandard cycle are described The cycle is to be shown on the Pv and Ts diagrams and the back work ratio and the thermal efficiency are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air are given as R 0287 kJkgK cp 1005 kJkgK cv 0718 kJkgK and k 14 Analysis a The Pv and Ts diagrams of the cycle are shown in the figures b The temperature at state 2 is 2100 K 100 kPa 300 K 700 kPa 1 2 1 2 P T P T 2100 K 2 3 T T During process 13 we have 3 1 3 T R T P Pd w in V V v During process 23 we have 5166 kJkg 2100K 0 287 kJkg K300 3 1 3 1 1 1 s T 3 2 1 v P 3 2 1 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1172 8 kJkg 0 287 kJkg K2100Kln7 2 2 ln 7 ln 7 ln 2 2 2 3 3 3 RT RT RT RT d Pd w out V V V V v V v The back work ratio is then 3 2 0440 1172 8 kJkg 6 kJkg 516 3 2 1 3 out in bw w w r Heat input is determined from an energy balance on the cycle during process 13 3 2 1 3 2 1 3 3 1 3 1 out out in ut w T w u q u The net work output is 3 2 1 3 o in w q 0 718 3 v T c 2465 kJkg 11728 kJkg 300 kJkg K2100 656 2 kJkg 516 6 1172 8 1 3 2 3 in out net w w w c The thermal efficiency is then 266 0 266 2465 kJ kJ 6562 in net th q w η preparation If you are a student using this Manual you are using it without permission 96 915 The three processes of an ideal gas power cycle are described The cycle is to be shown on the Pv and Ts diagrams and the maximum temperature expansion and compression works and thermal efficiency are to be determined Assumptions 1 Kinetic and potential energy changes are negligible 2 The ideal gas has constant specific heats Properties The properties of ideal gas are given as R 03 kJkgK cp 09 kJkgK cv 06 kJkgK and k 15 Analysis a The Pv and Ts diagrams of the cycle are shown in the figures b The maximum temperature is determined from PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 7348 K 1 1 2 1 1 2 max 27 T r T T T V 15 1 1 273 K6 k k V An energy balance during process 23 gives v out in T T T T c u w q 2 3 2 3 2 3 3 2 2 3 0 since Then the work of compression is s T 3 2 1 v P 3 2 1 c ou in w q 3 2 3 2 t 3950 kJkg kJkg K7348 Kln6 30 ln ln 2 2 3 2 2 2 3 2 2 3 r RT RT RT d Pd out in V V v V v d The work during isentropic compression is determined from an energy 12 300 1 2 1 2 2 T T c u w v in Net work output is 3 3 w q balance during process 1 2609 kJkg kJkg K7348 60 e 134 1 kJkg 260 9 395 0 2 1 2 3 in out net w w w The thermal efficiency is then 339 0 339 3950 kJ kJ 1341 in net th q w η preparation If you are a student using this Manual you are using it without permission 97 916 The four processes of an airstandard cycle are described The cycle is to be shown on Pv and Ts diagrams and the net work output and the thermal efficiency are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable specific heats Properties The properties of air are given in Table A17 Analysis b The properties of air at various states are v P 1 2 4 3 qin qout s T 1 2 4 3 qin qout 73971 kJkg 3279 18356 kPa 6019 kPa 100 1835 6 kPa 4903 K 600 kPa K 1500 9 601 120541 kJkg K 1500 3 K 490 35229 kJkg 7 841 100 kPa 13068 kPa 600 3068 1 17 kJkg 295 295 K 1 T 4 3 4 2 2 3 2 3 3 3 2 2 1 2 1 3 4 3 1 2 1 h P P P P T T P u T T u P P P P P h r r r r r r From energy balances c Then the thermal efficiency becomes 3 2 2 3 3 T P P P P v v T 4086 kJkg 444 5 1 853 4445 kJkg 29517 71 739 853 1 kJkg 35229 41 1205 out in out net 1 4 out 2 3 in q q w h h q u u q 479 0 479 8531 kJkg kJkg 4086 in netout th q w η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 98 917 Problem 916 is reconsidered The effect of the maximum temperature of the cycle on the net work output and thermal efficiency is to be investigated Also Ts and Pv diagrams for the cycle are to be plotted Analysis Using EES the problem is solved as follows Input Data T1295 K P1100 kPa P2 600 kPa T31500 K P4 100 kPa Process 12 is isentropic compression s1entropyairTT1PP1 s2s1 T2temperatureair ss2 PP2 P2v2T2P1v1T1 P1v1RT1 R0287 kJkgK Conservation of energy for process 1 to 2 q12 w12 DELTAu12 q12 0isentropic process DELTAu12intenergyairTT2intenergyairTT1 Process 23 is constant volume heat addition s3entropyair TT3 PP3 P3v3T3P2v2T2 P3v3RT3 v3v2 Conservation of energy for process 2 to 3 q23 w23 DELTAu23 w23 0constant volume process DELTAu23intenergyairTT3intenergyairTT2 Process 34 is isentropic expansion s4entropyairTT4PP4 s4s3 P4v4T4P3v3T3 P4v40287T4 Conservation of energy for process 3 to 4 q34 w34 DELTAu34 q34 0isentropic process DELTAu34intenergyairTT4intenergyairTT3 Process 41 is constant pressure heat rejection P4v4T4P1v1T1 Conservation of energy for process 4 to 1 q41 w41 DELTAu41 w41 P1v1v4 constant pressure process DELTAu41intenergyairTT1intenergyairTT4 qintotalq23 wnet w12w23w34w41 Etathwnetqintotal100 Thermal efficiency in percent PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 99 T3 K ηth qintotal kJkg Wnet kJkg 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 4791 4831 4868 4903 4935 4966 4995 5022 5048 5072 5095 8529 9457 1040 1134 1229 1325 1422 1519 1617 1715 1813 4086 4569 5061 556 6067 6581 7105 763 8161 8698 924 1500 1700 1900 2100 2300 2500 475 48 485 49 495 50 505 51 T3 K ηth 1500 1700 1900 2100 2300 2500 800 1020 1240 1460 1680 1900 T3 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course K qintotal kJk g preparation If you are a student using this Manual you are using it without permission 910 1500 1700 1900 2100 2300 2500 400 500 600 700 800 900 1000 T3 K wnet kJkg 50 55 60 65 70 75 80 85 0 500 1000 1500 2000 s kJkgK T K 100 kPa 600 kPa Air 1 2 3 4 102 101 100 101 102 101 102 103 104 v m3kg P kPa 295 K 1500 K Air 1 2 3 4 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 911 918 The three processes of an airstandard cycle are described The cycle is to be shown on Pv and Ts diagrams and the heat rejected and the thermal efficiency are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK and k 14 Table A 2 Analysis b The temperature at state 2 and the heat input are PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5792 K 100 kPa K 1000 kPa 300 0414 1 1 2 1 2 k k P P T T 579 2 0004 kg 1005 kJkg K 276 kJ 3 T P 1266 K 3 2 3 2 3 in T T T mc h m h Q p Process 31 is a straight line on the v diagram thus the w31 is simply the area under the process curve v P 3 2 1 qin qout s T 3 2 1 qout qin 273 7 kJkg 0287 kJkg K 1000 kPa 1266 K 100 kPa 300 K 2 100 kPa 1000 2 2 area 3 3 1 1 1 3 3 1 1 3 31 P RT P RT P P P P w v v Energy balance for process 31 gives 0718 kJkg K 3001266 K kg 2737 0004 3 1 31out 3 1 31out out 31 3 1 31out 31out system out in T T c m w T T mc mw Q u m u W Q v v The thermal efficiency is then 1679 kJ E E E c 392 276 kJ 1679 kJ 1 1 in out th Q Q η preparation If you are a student using this Manual you are using it without permission 912 919E The four processes of an airstandard cycle are described The cycle is to be shown on Pv and Ts diagrams and the total heat input and the thermal efficiency are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable specific heats Properties The properties of air are given in Table A17E v P 3 4 2 1 q12 q23 qout s T 3 4 2 1 qout q23 q12 Analysis b The properties of air at various states are 12906 Btulbm 9204 Btulbm 540 R 1 1 1 h u T 59322 Btulbm 3170 576 psia 1242 psia 147 1242 84948 Btulbm R 3200 576 psia 540 R 147 psia R 2116 537 1 Btulbm R 2116 39204 Btulbm 300 04 92 4 3 4 3 3 1 1 2 2 1 1 1 2 2 2 2 2 in12 1 2 1 2 in12 3 4 3 h P P P P P h T T P T P T P T P h T q u u u u q r r r v v From energy balance 46416 Btulbm 22 12906 593 31 38 300 23in 12in in 2 3 23in q q q 2 31238 Btulbm 537 1 48 849 1 4 out h h q h h q 61238 Btulbm c Then the thermal efficiency becomes 242 61238Btulbm 46416Btulbm 1 1 in out th q q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 913 920E The four processes of an airstandard cycle are described The cycle is to be shown on Pv and Ts diagrams and the total heat input and the thermal efficiency are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 0240 BtulbmR cv 0171 BtulbmR and k 14 Table A2E Analysis b The temperature at state 2 and the heat input are PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2174 Btulbm 2294 R Btulbm R 3200 024 6246 psia 540 R 147 psia R 2294 R 2294 540 R BtulbmR 171 2 3 2 3 in23 1 1 2 2 1 1 1 2 2 2 2 2 1 2 1 2 in12 T T c h h q T P T P T P T P T T T T c u u q P v v v Process 34 is isentropic 0 300 Btulbm 3785 Btulbm 540 BtulbmR 2117 0240 217 4 300 2117 R 6246 psia 147 psia R 3200 1 4 1 4 out in23 in12 in 0414 3 4 3 4 T T c h h q q q q P P T p 5174 Btulbm c The thermal efficiency is then 1 k k T 268 5174 Btulbm 3785 Btulbm 1 1 in out th q q η v P 3 4 2 1 q12 q23 qout s T 3 4 2 1 qout q23 q12 preparation If you are a student using this Manual you are using it without permission 916 923 An ideal gas Carnot cycle with air as the working fluid is considered The maximum temperature of the low temperature energy reservoir the cycles thermal efficiency and the amount of heat that must be supplied per cycle are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK and k 14 Table A 2a Analysis The temperature of the lowtemperature reservoir can be found by applying the isentropic expansion process relation PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course K 4811 1 2 1 273 K 12 1027 T T v 1 41 1 2 1 k v ince the Carnot engine is completely reversible its efficiency is s T 3 2 qin qout 4 1 1300 K S 0630 1 1 thCarnot H L T η 273 K 1027 4811 K T he work tput per cycle is T ou 20 kJcycle 1min 60s 1500 cyclemin 500 kJs net net n W W According to the definition of the cycle efficiency 3175 kJcycle 063 kJcycle 20 Carnot th net in in net thCarnot η η W Q Q W preparation If you are a student using this Manual you are using it without permission 917 924 An airstandard cycle executed in a pistoncylinder system is composed of three specified processes The cycle is to be sketcehed on the Pv and Ts diagrams the heat and work interactions and the thermal efficiency of the cycle are to be determined and an expression for thermal efficiency as functions of compression ratio and specific heat ratio is to be obtained Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air are given as R 03 kJkgK and cv 03 kJkgK Analysis a The Pv and Ts diagrams of the cycle are shown in the figures b Noting that PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1429 01 70 kJkg K 01 30 70 v v c R c c Process 12 Isentropic compression s T 3 2 1 v P 3 2 1 p c k p 584 4 K 293 K5 0 429 1 1 1 2 1 1 2 k k T r T T v v 2 1 2in T c w v 2040 kJkg 293 K kJkg K584 4 70 T1 q 2 0 1 From ideal gas relation 2922 584 4 5 3 2 1 2 3 2 3 T r T T v v v v Process 23 Constant pressure heat addition 584 4 K kJkg K2922 30 2 3 3 2 T T 584 4 K 1 kJkg K2922 2 3 2 3 2 3 3 2 3in T T c h u w q p out Process 31 Constant volume heat rejection 2 3 2 2 3out R P Pd w v v v 7013 kJkg 2338 kJkg 2 1 3 3 1out T c u q v 18403 kJkg kJkg K2922 293 K 70 1 3 T w c Net work is 1 0 3 497 3 kJkg K 204 0 701 3 1 2in 2 3out net w w w The thermal efficiency is then 213 0 213 2338 kJ kJ 4973 in net th q w η preparation If you are a student using this Manual you are using it without permission 918 d The expression for the cycle thermal efficiency is obtained as follows 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 3 1 2 2 3 in 1 2in 3out 2 in net th 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 k k p k p k p k k v p k k p k v p p v r k r k r k r c R r T T k r c R r T r c r T T T r c c R T r rT r c T T r c c R T T c T T c T T R q w w q w η since 1 1 1 k c c c c c c R p v p v p p PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 919 Otto Cycle 925C For actual fourstroke engines the rpm is twice the number of thermodynamic cycles for twostroke engines it is equal to the number of thermodynamic cycles 926C The ideal Otto cycle involves external irreversibilities and thus it has a lower thermal efficiency 927C The four processes that make up the Otto cycle are 1 isentropic compression 2 v constant heat addition 3 isentropic expansion and 4 v constant heat rejection 928C They are analyzed as closed system processes because no mass crosses the system boundaries during any of the processes 929C It increases with both of them 930C Because high compression ratios cause engine knock 931C The thermal efficiency will be the highest for argon because it has the highest specific heat ratio k 1667 932C The fuel is injected into the cylinder in both engines but it is ignited with a spark plug in gasoline engines PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 920 933 An ideal Otto cycle is considered The thermal efficiency and the rate of heat input are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK and k 14 Table A 2a Analysis The definition of cycle thermal efficiency reduces to PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 610 0 6096 105 1 1 1 1 14 1 1 th rk η The rate of heat addition is then 148 kW 06096 kW 90 th net in η W Q 2 Kinetic and potential energy changes are negligible 3 Air is ies The properties of air at room temperature are c 1005 kJkgK cv 0718 kJkgK and k 14 Table A nalysis The definition of cycle thermal efficiency reduces to v P 4 1 3 2 qin qout 934 An ideal Otto cycle is considered The thermal efficiency and the rate of heat input are to be determined Assumptions 1 The airstandard assumptions are applicable an ideal gas with constant specific heats Propert p 2a A v P 4 1 3 2 qin qout 575 85 1 1 1 1 th r k η 0 5752 1 1 4 1 he rate o eat addition is then T f h 157 kW 05752 kW 90 th net in η W Q preparation If you are a student using this Manual you are using it without permission 921 935 The two isentropic processes in an Otto cycle are replaced with polytropic processes The heat added to and rejected from this cycle and the cycles thermal efficiency are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are R 0287 kPam3kgK cp 1005 kJkgK cv 0718 kJkgK and k 14 Table A2a Analysis The temperature at the end of the compression is v P 4 1 3 2 537 4 K 288 K8 1 31 1 1 1 2 1 1 2 n n T r T T v v And the temperature at the end of the expansion is 789 4 K 1473 K 8 3 4 3 4 r T T T v 1 1 1 31 1 1 3 n n v for the polytropic compression gives The integral of the work expression PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 238 6 kJkg 1 8 1 31 1 1 2 1 2 n w v 0 287 kJkg K288 K 1 31 1 1 1 n RT v imilarly the work produced during the expansion is S 654 0 kJkg 1 8 1 1 31 0 287 kJkg K1473 K 1 1 1 31 1 n 4 3 3 3 4 n RT w v v pplication of the first law to each of the four processes gives A 1 5953 kJkg 288K 0 718 kJkg K537 4 238 6 kJkg 1 2 1 2 2 T T c w q v 0 718 kJkg K1473 2 3 2 3 T T c q v 671 8 kJkg 537 4 K 163 2 kJkg 789 4 K 0 718 kJkg K1473 654 0 kJkg 4 3 3 4 3 4 T T c w q v 360 0 kJkg 288K 0 718 kJkg K789 4 1 4 4 1 T T c q v The head added and rejected from the cycle are The thermal efficiency of this cycle is then kJkg 4195 kJkg 8350 360 0 53 59 163 2 8 671 4 1 1 2 out 3 4 2 3 in q q q q q q 0498 835 0 419 5 1 1 in out th q q η preparation If you are a student using this Manual you are using it without permission 922 936 An ideal Otto cycle is considered The heat added to and rejected from this cycle and the cycles thermal efficiency are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are R 0287 kPam3kgK cp 1005 kJkgK cv 0718 kJkgK and k 14 Table A2a Analysis The temperature at the end of the compression is v P 4 1 3 2 661 7 K 288 K8 1 41 1 1 1 2 1 1 2 k k T r T T v v and the temperature at the end of the expansion is 2 K 64 1473 K 8 3 4 3 4 r T T T v 1 1 1 1 41 1 1 3 k k v pplication of the first law to the heat addition process gives A 2 3 in T T c q v PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5825 kJkg 661 7 K 0 718 kJkg K1473 imilarly the heat rejected is S 0 718 kJkg K T T c q 2536 kJkg 288K 641 2 1 4 out v he thermal efficiency of this cycle is then T 0565 582 5 253 6 1 1 in out th q q η preparation If you are a student using this Manual you are using it without permission 923 937E A sixcylinder fourstroke sparkignition engine operating on the ideal Otto cycle is considered The power produced by the engine is to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are R 03704 psiaft3lbmR Table A1E cp 0240 BtulbmR cv 0171 BtulbmR and k 14 Table A2Ea Analysis From the data specified in the problem statement v P 4 1 3 2 143 7 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 014 1 2 v v 1 1 v v r Since the compression and expansion processes are isentropic 1153 R 525 R 7 143 1 41 1 1 1 2 1 1 2 k k T r T T v v 938 2 R 7143 1 1 1 41 1 1 3 k k v 2060 R 3 4 3 4 r T T T v mpression and expansion processes gives Application of the first law to the co 0 171 938 2 R 0 171 Btulbm R2060 1 2 4 3 net T T c T T c w v v 8444 Btulbm 525R Btulbm R1153 When each cylinder is charged with the airfuel mixture 1389 ft lbm 14 psia 0 3704 psia ft lbm R525 R 3 3 1 1 1 P RT v The total air mass taken by all 6 cylinders when they are charged is 0 009380 lbm 89 ft lbm 13 12 ft4 12 ft 93 53 6 4 cyl cyl π v v V B S N N m 3 2 1 2 1 π he net work produced per cycle is T 0 7920 Btucycle 0009380 lbm8444 Btulbm net net mw W he power produced is determined from T 233 hp 0 7068 Btus 1hp 2 revcycle Btucycle250060 revs 07920 rev net net N n W W since there are two revolutions per cycle in a fourstroke engine preparation If you are a student using this Manual you are using it without permission 924 938E An Otto cycle with nonisentropic compression and expansion processes is considered The thermal efficiency the heat addition and the mean effective pressure are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are R 03704 psiaft3lbmR Table A1E cp 0240 BtulbmR cv 0171 BtulbmR and k 14 Table A2Ea Analysis We begin by determining the temperatures of the cycle states using the process equations and component efficiencies The ideal temperature at the end of the compression is then v P 4 1 3 2 qout qin PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1195 R 520 R 8 1 2 1 2 s T r T T v 1 41 1 1 1 k k v With the isentropic compression efficiency the actual temperature at e end of the compression is th 1314 R 085 520 R 1195 520 R 1 2 1 2 1 2 1 2 η T T T s η T T T T T s imilarly for the expansion S 1201 R 8 460 R 1 2300 1 1 41 1 3 1 4 3 3 4 k k s r T T T v v 1279 R 0 952760 1201 R 2760 R 4 3 3 4 4 3 T s T 4 3 T s T T T T T η η The specific heat addition is that of process 23 2473 Btulbm 0 171 Btulbm R2760 1314R 2 3 in T T c q v The net work production is the difference between the work produced by the expansion and that used by the compression The thermal efficiency of this cycle is then 1 2 4 3 net T T c T T c w v v 520R 0 171 Btulbm R1314 1279R 0 171 Btulbm R2760 117 5 Btulbm 0475 2473 Btulbm in q 1175 Btulbm net th w η At the beg ning of compression the maximum specific volume of this cycle is in 1482 ft lbm 13 psia 0 3704 psia ft lbm R520 R 3 1 1 v RT 3 1 P hile the minimum specific volume of the cycle occurs at the end of the compression w 1 852 ft lbm 8 1482 ft lbm 3 3 1 2 r v v The engines mean effective pressure is then 490 psia 1Btu 404 psia ft 5 1 852 ft lbm 1482 117 5 Btulbm MEP 3 3 2 1 net v v w preparation If you are a student using this Manual you are using it without permission 925 939 An ideal Otto cycle with air as the working fluid has a compression ratio of 95 The highest pressure and temperature in the cycle the amount of heat transferred the thermal efficiency and the mean effective pressure are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK R 0287 kJkgK and k 14 Table A2 Analysis a Process 12 isentropic compression v P 4 1 3 2 Qin Qout 2338 kPa 100 kPa K PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 308 7579 K 95 7579 K K 95 1 1 2 2 1 2 1 1 1 2 2 2 04 1 T P T P T P T P k v v v v v 308 2 1 1 2 T T v Process 34 isentropic expansion 1969 K 04 1 3 4 4 3 K 95 800 k T T v v Process 23 v constant heat addition 6072 kPa 2338 kPa 7579 K K 1969 2 2 3 2 3 T T T 3 2 2 3 P T P P P v v b 3 kg 6 788 10 kPa m kg K 308 K 0287 100 kPa 00006 m 4 3 3 1 1 1 RT P m V 0590 kJ 7579 K kg 0718 kJkg K 1969 10 6788 4 2 3 2 3 in T T mc u m u Q v c Process 41 v constant heat rejection 0240 kJ 308 K kg 0718 kJkg K 800 6788 10 4 1 4 1 4 out T T mc u m u Q v 0350 kJ 0 240 0 590 out in net Q Q W 594 0590 kJ kJ 0350 in netout th Q W η 652 kPa kJ m kPa 1 195 m 00006 0350 kJ 1 1 MEP 3 3 1 out net 2 1 out net max 2 min r W W r V V V V V V d preparation If you are a student using this Manual you are using it without permission 926 940 An Otto cycle with air as the working fluid has a compression ratio of 95 The highest pressure and temperature in the cycle the amount of heat transferred the thermal efficiency and the mean effective pressure are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK R 0287 kJkgK and k 14 Table A2 Analysis a Process 12 isentropic compression 2338 kPa 100 kPa 308 K 7579 K 95 7579 K K 95 308 1 1 2 2 1 2 1 1 1 2 2 2 04 1 2 1 1 2 T P T P T P T P T T k v v v v v v v P 4 1 3 2 Qin Qout Polytropic 800 K 308 K Process 34 polytropic expansion kg 6 788 10 kPa m kg K 308 K 0287 100 kPa 00006 m 4 3 3 1 1 1 RT P m V 05338 kJ 1 135 kJkg K 800 1759 K 0287 6788 10 1 3 4 34 n T mR T W K 95 800 4 0 35 1 3 4 4 3 T T n 1759 K v v Q W T T mc W u m u Q v That is 0066 kJ of heat is added to the air during the expansion process This is not realistic and probably is due to nstant specific heats at room temperature v constant heat addition T hen energy balance for process 34 gives out in E E E system 3 4 34out 34in u m u W Q 00664 kJ 05338 kJ kg 0718 kJkg K 800 1759 K 10 6788 4 in 34 34out 3 4 34out 3 4 in 34 assuming co b Process 23 5426 kPa T P v v 2338 kPa 7579 K 1759 K 2 2 3 3 2 2 2 3 3 P T P T P Q T T mc u m u Q v efore 3 T 04879 kJ 7579 K kg 0718 kJkg K 1759 10 6788 4 in 23 2 3 2 3 in 23 05543 kJ 0 0664 0 4879 34in 23in in Q Q Q Ther c Process 41 v constant heat rejection 02398 kJ 308 K kg 0718 kJkg K 800 10 6788 4 1 4 1 4 out T T mc u m u Q v 03145 kJ 0 2398 0 5543 out in netout Q Q W 567 05543 kJ kJ 03145 in netout th Q W η 586 kPa kJ m kPa 1 195 m 00006 03145 kJ 1 1 MEP 3 3 1 out net 2 1 out net max 2 min r W W r V V V V V V d preparation If you are a student using this Manual you are using it without permission 927 941E An ideal Otto cycle with air as the working fluid has a compression ratio of 8 The amount of heat transferred to the air during the heat addition process the thermal efficiency and the thermal efficiency of a Carnot cycle operating between the same temperature limits are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable specific heats PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course v P 4 1 3 2 qin qout 2400 R 540 R Properties The properties of air are given in Table A17E Analysis a Process 12 isentropic compression 32 144 9204Btulbm R 540 1 1 1 r u T v 21128 Btulbm 1804 14432 8 1 2 2 2 r r r r v v v v 1 1 2 2 u v rocess 2 v constant heat addition 21128 70 452 419 2 0 Btulbm R 2400 2 3 3 3 u u q T in v r P 3 3 4527 u 24142 Btulbm b Process 34 isentropic expansion 20554 Btulbm 1935 2 419 8 4 3 4 3 3 4 u r r r r v v v v v Process 41 v constant heat rejection 11350 Btulbm 9204 20554 1 4 out u u q 530 24142 Btulbm 11350 Btulbm 1 1 in out th q q η c The thermal efficiency of a Carnot cycle operating between the same temperature limits is 775 2400 R 540 R 1 1 thC H L T T η preparation If you are a student using this Manual you are using it without permission 928 942E An ideal Otto cycle with argon as the working fluid has a compression ratio of 8 The amount of heat transferred to the argon during the heat addition process the thermal efficiency and the thermal efficiency of a Carnot cycle operating between the same temperature limits are to be determined Assumptions 1 The airstandard assumptions are applicable with argon as the working fluid 2 Kinetic and potential energy changes are negligible 3 Argon is an ideal gas with constant specific heats Properties The properties of argon are cp 01253 BtulbmR cv 00756 BtulbmR and k 1667 Table A2E Analysis a Process 12 isentropic compression 2161 R 540 R 8 0667 1 2 1 1 2 k T T v v v P 4 1 3 2 qin qout Process 23 v constant heat addition PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 00756 BtulbmR 2400 1807 Btulbm R 2161 2 3 2 3 T T c u u q v Process 34 isentropic expansion in b 600 R 8 0667 4 v Process 41 v constant heat rejection 1 R 2400 1 3 3 4 k T T v 4536 Btulbm 540 R 00756 BtulbmR 600 1 4 1 4 out T T c u u q v 749 4536 Btulbm 1 1 qout η 1807 Btulbm in th q c The thermal efficiency of a Carnot cycle operating between the same temperature limits is 775 2400 R 540 R 1 1 thC H L T T η preparation If you are a student using this Manual you are using it without permission 929 943 A gasoline engine operates on an Otto cycle The compression and expansion processes are modeled as polytropic The temperature at the end of expansion process the net work output the thermal efficiency the mean effective pressure the engine speed for a given net power and the specific fuel consumption are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at 850 K are cp 1110 kJkgK cv 0823 kJkgK R 0287 kJkgK and k 1349 Table A2b Analysis a Process 12 polytropic compression 1 Qin 2 3 4 P V Qout 2258 kPa kPa 11 100 636 5 K K 11 310 13 2 1 1 2 13 1 1 2 1 1 2 n n P P T T v v v v 312 3 kJkg 13 1 1 12 n 310K 0287 kJkg K6365 1 2 T R T w Process 23 constant volume heat addition 2255 K 2258 kPa 8000 kPa 6365 K 2 3 2 3 P P T T 2 3 2 3 in T T c u u q v 1332 kJkg 6365 K 0823 kJkg K 2255 rocess 34 polytropic expansion P 1098 K 13 1 1 4 3 3 4 11 1 K 2255 n T T v v 354 2 kPa 1 13 n v 11 8000 kPa 1 2 3 4 P P v 1106 kJkg 13 1 2255K 0287 kJkg K1098 T T R 1 3 4 n w Process 41 constant volume heat rejection b The net work output and the thermal efficiency are 34 794 kJkg 312 3 1106 12 34 netout w w w 596 0 596 1332 kJkg kJkg 794 in netout th q w η c The mean effective pressure is determined as follows 982 kPa kJ m kPa m kg 1 111 08897 794 kJkg 1 1 MEP 08897 m kg 100 kPa kPa m kg K 310 K 0287 3 3 1 out net 2 1 out net max 2 min max 3 3 1 1 1 r w w r P RT v v v v v v v v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 930 d The clearance volume and the total volume of the engine at the beginning of compression process state 1 are 3 3 0 00016 m 0 0016 m 11 c c c c d c r V V V V V V 3 1 0 00176 m 0 0016 0 00016 d c V V V The total mass contained in the cylinder is 0001978 kg kPa m kg K 310 K 0287 100 kPa000176 m 3 3 1 1 1 RT P mt V The engine speed for a net power output of 50 kW is 3820 revmin 1min 2 revcycle 0 001978 kg794 kJkg cycle 2 net net m w n t 60 s 50 kJs W n fourstroke engines The mass of fuel burned during one cycle is Note that there are two revolutions in one cycle i e 0 0001164 kg 0 001978 kg 16 AF f f a m m f f f f t m m m m m m inally the specific fuel consumption is F 267 gkWh 1kWh 3600 kJ 1kg 1000 g 0 001978 kg794 kJkg 0 0001164 kg sfc wnet m m t f PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 931 944 The expressions for the maximum gas temperature and pressure of an ideal Otto cycle are to be determined when the compression ratio is doubled Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Analysis The temperature at the end of the compression varies with the compression ratio as 1 1 1 2 1 1 2 k k T r T T v v v P 4 1 3 2 qout qin since T1 is fixed The temperature rise during the combustion remains constant since the amount of heat addition is fixed Then the maximum cycle temperature is given by 1 1 in 2 in 3 T kr c q T c q T v v The smallest gas specific volume during the cycle is r 1 3 v v When this is combined with the maximum temperature the maximum pressure is given by 1 1 in 1 3 3 3 T kr c Rr q RT P v v v 945 It is to be determined if the polytropic exponent to be used in an Otto cycle model will be greater than or less than the isentropic exponent Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats nalysis During a polytropic process T Pv is lost during the expansion of the gas here T4s is the temperature that would occur if the expansion were reversible and adiabatic nk This can only occur when A constant 1 n n Pv n P 4 1 3 2 qout qin P constant and for an isentropic process k constant constant 1 k k TP If heat v 4 T s T 4 w n k PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 932 Diesel Cycle 946C A diesel engine differs from the gasoline engine in the way combustion is initiated In diesel engines combustion is initiated by compressing the air above the selfignition temperature of the fuel whereas it is initiated by a spark plug in a gasoline engine 947C The Diesel cycle differs from the Otto cycle in the heat addition process only it takes place at constant volume in the Otto cycle but at constant pressure in the Diesel cycle 948C The gasoline engine 949C Diesel engines operate at high compression ratios because the diesel engines do not have the engine knock problem 950C Cutoff ratio is the ratio of the cylinder volumes after and before the combustion process As the cutoff ratio decreases the efficiency of the diesel cycle increases 951 An ideal diesel cycle has a compression ratio of 20 and a cutoff ratio of 13 The maximum temperature of the air and the rate of heat addition are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK R 0287 kJkgK and k 14 Table A2a Analysis We begin by using the process types to fix the temperatures of the states PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 954 6 K 288 K 20 1 41 1 1 1 2 1 1 2 k k T r T T v v 1241 K 954 6 K13 2 2 2 3 cr T T T v 3 v ombining the first law as applied to the various processes with the process equations gives v P 4 1 2 3 qin qout C 0 6812 1 31 41 20 1 1 1 14 1 1 th c k k r r η 1 31 1 1 1 41 cr k ccording the definition of the thermal efficiency A to 367 kW 06812 kW 250 th net in η W Q preparation If you are a student using this Manual you are using it without permission 933 952E An ideal diesel cycle has a a cutoff ratio of 14 The power produced is to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are R 03704 psiaft3lbmR Table A1E cp 0240 BtulbmR cv 0171 BtulbmR and k 14 Table A2Ea Analysis The specific volume of the air at the start of the compression is v P 4 1 2 3 qin qout 1312 ft lbm 14 4 psia 0 3704 psia ft lbm R510 R 3 3 1 1 1 P RT v The total air mass taken by all 8 cylinders when they are charged is 01774 lbm 0 1312 ft lbm 8 3 1 cyl 1 cyl v v N N m 4 12 ft 4 12 ft4 4 2 2 π π V B S he rate at which air is processed by the engine is determined from T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 958 0 lbmh 0 2661 lbms 2 revcycle rev m N 001774 lbmcycle180060 revs n m nce there re two revolutions per cycle in a fourstroke engine The compression ratio is si a 2222 0 045 1 r T r k T work integral to the constant pressure heat addition gives At the end of the compression the air temperature is 1763 R 510 R 2222 1 41 1 1 2 Application of the first law and 2393 Btulbm 0 240 Btulbm R2760 1763R 2 3 in T T c q p while the thermal efficiency is 0 6892 1 41 41 1 41 2222 1 1 1 r k 1 1 1 41 14 1 1 c c k k r r η he power produced by this engine is then th T 621 hp 2544 5 Btuh 1hp lbmh068922393 Btulbm 9580 in th net net q m mw W η preparation If you are a student using this Manual you are using it without permission 934 953 An ideal dual cycle has a compression ratio of 14 and cutoff ratio of 12 The thermal efficiency amount of heat added and the maximum gas pressure and temperature are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK R 0287 kJkgK and k 14 Table A2a Analysis The specific volume of the air at the start of the compression is v P 4 1 2 3 qout x qin 1 051 m kg 80 kPa 0 287 kPa m kg K293 K 3 3 1 1 1 P RT v and the specific volume at the end of the compression is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 07508 m kg 14 2 1 051 m kg 3 3 v1 v r The pressure at the end of the compression is 3219 kPa 80 kPa 14 41 1 2 1 2 v 1 k k P r P P v nd the ma imum pressure is he temperature at the end of the compression is a x 4829 3219 kPa 51 2 3 r P P P p x kPa T 842 0 K 293 K 14 1 41 1 1 1 2 1 1 2 k k T r T T v v nd 1263 K 3219 kPa 842 0 K 4829 kPa 2 3 2 P P T Tx he remaining state temperatures are then a From the definition of cutoff ratio 0 09010 m kg 0 07508 m kg 21 3 3 2 3 v v v c c x r r T 1516 K 1263 K 007508 3 3 x Tx T v 009010 v 567 5 K 1051 1516 K 009010 1 41 1 4 3 3 4 T T v k v ng the first law and work expression to the heat addition processes gives The heat rejected is Applyi 5565 kJkg 1 005 kJkg K1516 1263K 0 718 kJkg K1263 842 0 K 3 2 in x p x T T c T T c q v 197 1 kJkg 293K 0 718 kJkg K567 5 1 4 out T T c q v 0646 5565 kJkg 1971 kJkg 1 1 in out th q q η Then preparation If you are a student using this Manual you are using it without permission 935 954 An ideal dual cycle has a compression ratio of 14 and cutoff ratio of 12 The thermal efficiency amount of heat added and the maximum gas pressure and temperature are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK R 0287 kJkgK and k 14 Table A2 Analysis The specific volume of the air at the start of the compression is 0 9076 m kg 80 kPa 0 287 kPa m kg K253 K 3 3 1 1 1 P RT v v P 4 1 2 3 qout x qin and the specific volume at the end of the compression is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 06483 m kg 14 2 0 9076 m kg 3 3 v1 v r The pressure at the end of the compression is 3219 kPa 80 kPa 14 41 1 2 1 2 v 1 k k P r P P v nd the ma imum pressure is he temperature at the end of the compression is a x 4829 3219 kPa 51 2 3 r P P P p x kPa T 727 1 K 253 K 14 1 41 1 1 1 2 1 1 2 k k T r T T v v nd 1091 K 3219 kPa 727 1 K 4829 kPa 2 3 2 P P T Tx he remaining state temperatures are then a From the definition of cutoff ratio 0 07780 m kg 0 06483 m kg 21 3 3 2 3 v v v c c x r r T 1309 K 1091 K 006483 3 3 x Tx T v 007780 v 490 0 K 09076 1309 K 007780 1 41 1 4 3 3 4 T T v k v ng the first law and work expression to the heat addition processes gives The heat rejected is Applyi 4804 kJkg 1 005 kJkg K1309 1091K 0 718 kJkg K1091 727 1 K 3 2 in x p x T T c T T c q v 170 2 kJkg 253K 0 718 kJkg K490 0 1 4 out T T c q v 0646 4804 kJkg 1702 kJkg 1 1 in out th q q η Then preparation If you are a student using this Manual you are using it without permission 936 955E An airstandard Diesel cycle with a compression ratio of 182 is considered The cutoff ratio the heat rejection per unit mass and the thermal efficiency are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable specific heats Properties The properties of air are given in Table A17E v P 4 1 2 3 qin qout 3000 R Analysis a Process 12 isentropic compression 32 144 4 Btulbm 92 0 R 540 1 1 1 r u T v Btulbm 40205 16236 R 7 93 14432 18 2 1 1 1 2 r r r r v 1 1 2 2 2 h T v v v v rocess 23 P constant heat addition P 1848 16236 R R 3000 2 3 2 3 2 2 2 3 3 3 T T T P T P v v v v b R 3000 2 3 in 3 3 h h q T v r 38863 Btulbm 40205 68 790 180 1 3 79068 Btulbm h Process 34 isentropic expansion 25091 Btulbm 11621 1 180 1 848 18 2 1 848 1 848 4 2 4 3 4 3 3 3 4 u r r r r r v v v v v v v v Process 41 v constant heat rejection c 591 Btulbm 15887 38863 Btulbm 15887 Btulbm 1 1 91 9204 250 in out th 1 4 out q q u u q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 937 956E An airstandard Diesel cycle with a compression ratio of 182 is considered The cutoff ratio the heat rejection per unit mass and the thermal efficiency are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 0240 BtulbmR cv 0171 BtulbmR and k 14 Table A2E Analysis a Process 12 isentropic compression v P 4 1 2 3 qin 1724R 540R 182 04 1 2 1 1 2 k T T v v Process 23 P constant heat addition 1741 1724 R R 3000 2 3 2 3 2 2 2 3 3 3 T T T P T P v v v v b 306 Btulbm 0240 BtulbmR 3000 1724 R 2 3 2 3 in rocess 3 isentropic expansion T T c h h q p P 4 1173 R 182 1741 3000 R 1 741 1 3 k v v 04 1 4 2 3 4 3 4 k v T T T v Process 41 v constant heat rejection c 646 Btulbm 108 306 Btulbm 108 Btulbm 1 1 540 R BtulbmR 1173 0171 in out th 1 4 1 4 out q q T T c u u q η v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 938 957 An ideal diesel engine with air as the working fluid has a compression ratio of 20 The thermal efficiency and the mean effective pressure are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK R 0287 kJkgK and k 14 Table A2 Analysis a Process 12 isentropic compression v P 4 1 2 3 qin qout K 9711 293 K 20 2 1 1 2 T T V 04 1 k V rocess 23 P constant heat addition P 2265 9711K K 2200 2 3 2 3 2 2 2 3 3 3 T T V T T P P V V V rocess 3 isentropic expansion P 4 635 1235 kJkg kJkg 7844 7844 kJkg 450 6 1235 4506 kJkg 293 K kJkg K 9206 0718 1235 kJkg 9711 K kJkg K 2200 1005 9206 K 20 2265 2200 K 2 265 265 2 in netout th out in out net 1 4 1 4 out 2 3 2 3 in 04 1 3 1 4 2 3 1 4 3 3 4 q w q q w T T c u u q T T c h h q r T T T T p k k k η v V V V V b 933 kPa kJ m kPa m kg 1 120 0885 7844 kJkg 1 1 MEP 0885 m kg 95 kPa kPa m kg K 293 K 0287 3 3 1 out net 2 1 out net max 2 min max 3 3 1 1 1 r w w r P RT v v v v v v v v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 939 958 A diesel engine with air as the working fluid has a compression ratio of 20 The thermal efficiency and the mean effective pressure are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK R 0287 kJkgK and k 14 Table A2 Analysis a Process 12 isentropic compression v P 4 1 2 3 qin qout Polytropic PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 293 K 20 04 2 1 1 2 T T V K 9711 1 k V rocess 23 P constant heat addition P 2265 9711 K K 2200 2 3 2 3 2 2 3 3 T P T P V V V V 2 3 T T Process 34 polytropic expansion 5263 kJkg 293 K kJkg K 1026 0718 1235 kJkg 9711 K kJkg K 2200 1005 1026 K 20 2265 2200 K r 1 4 1 4 out 2 3 2 3 in 035 3 4 3 4 3 4 T T c u u q T T c h h q T T T T p v V V Note that qout in this case does not represent the entire heat rejected since some heat is also rejected during the polytropic process which is determined from an energy balance on process 34 2265 2265 n 1 n 1 2 1 3 n V V kJkg 1 2200 K 0718 kJkg K 1026 kJkg 963 2200 K kJkg K 1026 0287 3 4 34out 34in 3 4 T T c w q u T T R v hich means that 1201 kJkg of heat is transferred to the com listic since the gas is at a much higher temperature than the surroundings and a hot gas loses heat during polytropic xpansion The cause of this unrealistic result is the constant specific heat assumption If we were to use u data from the ir table w would obtain 963 kJkg 1 135 1 34out in 34 system out in 3 4 out 34 u w q E E E n w 120 w bustion gases during the expansion process This is unrea e a e 128 1 kJkg 1872 4 781 3 963 3 4 34out 34in u u w q which is a heat loss as expected Then qout becomes 6544 kJkg 526 3 128 1 41out 34out out q q q and 470 1235 kJkg kJkg 5806 5806 kJkg 654 4 1235 in netout th out in out net q w q q w η 691 kPa kJ kPa m 1 m kg 1 120 0885 5806 kJkg 1 1 MEP 0885 m kg 95 kPa kPa m kg K 293 K 0287 3 3 1 out net 2 1 out net max 2 min max 3 3 1 1 1 r w w r P RT v v v v v v v v b preparation If you are a student using this Manual you are using it without permission 940 959 Problem 958 is reconsidered The effect of the compression ratio on the net work output mean effective lso Ts and Pv diagrams for the cycle are to be plotted nalysis Using EES the problem is solved as follows talq12q23q34q41 qintotalqouttotal 0 then qintotal qintotal q41 else qouttotal qouttotal q41 ssion PP2 v1T1 rocess 1 to 2 T1 heat addition nergyairTT2 ion pic process intenergyairTT3 4 to 1 pressure and thermal efficiency is to be investigated A A Procedure QTo qintotal 0 qouttotal 0 IF q12 0 THEN qintotal q12 ELSE qouttotal q12 If q23 0 then qintotal qintotal q23 else qouttotal qouttotal q23 If q34 0 then qintotal qintotal q34 else qouttotal qouttotal q34 If q41 END Input Data T1293 K P195 kPa T3 2200 K n135 rcomp 20 Process 12 is isentropic compre s1entropyairTT1PP1 s2s1 T2temperatureair ss2 P2v2T2P1 P1v1RT1 R0287 kJkgK V2 V1 rcomp Conservation of energy for p q12 w12 DELTAu12 q12 0isentropic process DELTAu12intenergyairTT2intenergyairT Process 23 is constant pressure P3P2 s3entropyair TT3 PP3 P3v3RT3 Conservation of energy for process 2 to 3 q23 w23 DELTAu23 w23 P2V3 V2constant pressure process DELTAu23intenergyairTT3inte Process 34 is polytropic expans P3P4 V4V3n s4entropyairTT4PP4 P4v4RT4 Conservation of energy for process 3 to 4 q34 w34 DELTAu34 q34 is not 0 for the ploytro DELTAu34intenergyairTT4 P3V3n Const w34P4V4P3V31n Process 41 is constant volume heat rejection V4 V1 Conservation of energy for process q41 w41 DELTAu41 w41 0 constant volume process DELTAu41intenergyairTT1intenergyairTT4 Call QTotalq12q23q34q41 qintotalqouttotal wnet w12w23w34w41 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 941 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Thermal efficiency in percent he mean effective pressure is EP wnetV1V2 r Etathwnetqintotal100 T M comp ηth MEP kPa wnet kJkg 14 4769 9708 7979 16 5014 985 8174 18 5216 9926 8298 20 5385 9954 8370 22 5529 9949 8406 24 5654 992 8415 40 45 50 55 60 65 70 75 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 s kJkgK T K 95 kPa 3401 kPa 5920 kPa 0044 01 088 m3kg Air 2 1 3 4 102 101 100 101 102 101 102 103 104 101 102 103 104 v m3kg P kPa 293 K 1049 K 2200 K 569 674 kJkgK Air preparation If you are a student using this Manual you are using it without permission 942 14 16 18 20 22 24 790 800 810 820 830 840 850 rcomp wnet kJkg 14 16 18 20 22 24 47 49 51 53 55 57 rcomp ηth 14 16 18 20 22 24 970 975 980 985 990 995 1000 rcomp MEP kPa PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 943 960 A fourcylinder ideal diesel engine with air as the working fluid has a compression ratio of 22 and a cutoff ratio of 18 The power the engine will deliver at 2300 rpm is to be determined Assumptions 1 The cold airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK R 0287 kJkgK and k 14 Table A2 Analysis Process 12 isentropic compression v P 4 1 2 3 Qin Qout PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 343 K 22 2 1 1 2 T T V K 1181 04 1 k V Process 23 P constant heat addition 2126 K 18 1181 K 81 2 2 2 3 3 2 2 2 3 3 3 T T T T T P P v v v v rocess 3 isentropic expansion P 4 480 kW 40 1 1 1 T k k k revs 1251 kJrev 230060 1 251 kJrev 0 6198 871 1 0 6198 kJ 343 K 001971 kg 0718 kJkg K 781 0 1871 kJ 1181K 0 001971 kg1005 kJkg K2216 0 001971 kg 0287 kPa m kg K343 K 97 kPa00020 m 781 K 22 81 2216 K 22 22 netout out net out in out net 1 4 1 4 out 2 3 2 3 in 3 3 1 1 1 3 4 2 3 4 3 3 4 nW W Q Q W T mc T u m u Q T T mc h m h Q RT P m r T T T v p V V V V V Discussion Note that for 2stroke engines 1 thermodynamic cycle is equivalent to 1 mechanical cycle and thus revolutions preparation If you are a student using this Manual you are using it without permission 944 961 A fourcylinder ideal diesel engine with nitrogen as the working fluid has a compression ratio of 22 and a cutoff ratio of 18 The power the engine will deliver at 2300 rpm is to be determined Assumptions 1 The airstandard assumptions are applicable with nitrogen as the working fluid 2 Kinetic and potential energy changes are negligible 3 Nitrogen is an ideal gas with constant specific heats Properties The properties of nitrogen at room temperature are cp 1039 kJkgK cv 0743 kJkgK R 02968 kJkgK and k 14 Table A2 Analysis Process 12 isentropic compression v P 4 1 2 3 Qin Qout 1181 K 343 K 22 04 1 2 1 1 2 k T T V V Process 23 P constant heat addition 2126 K 18 1181 K 81 2 2 2 3 3 2 3 T T 2 2 3 T T T P P v v v v Process 34 isentropic expansion 3 479 kW 230060 revs 1251 kJrev netout netout nW W Discussion No 1 251 kJrev 0 6202 871 1 0 6202 kJ 343 K 001906 kg 0743 kJkg K 781 0 1871 kJ 1181K 0 001906 kg1039 kJkg K2216 0 001906 kg 02968 kPa m kg K343 K 97 kPa00020 m 781 K 22 81 2216 K 22 22 out in out net 1 4 1 4 out 2 3 2 3 in 3 3 1 1 1 40 1 3 1 4 2 3 1 4 3 3 4 Q Q W T mc T u m u Q T T mc h m h Q RT P m r T T T T v p k k k V V V V V te that for 2stroke engines 1 thermodynamic cycle is equivalent to 1 mechanical cycle and thus revolutions PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 945 962 An ideal dual cycle has a compression ratio of 18 and cutoff ratio of 11 The power produced by the cycle is to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK R 0287 kJkgK and k 14 Table A2a Analysis We begin by fixing the temperatures at all states PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 924 7 K 291 K 18 1 41 1 1 1 2 1 1 2 k k T r T T v v v P 4 1 2 3 qout x qin 5148 kPa 90 kPa 18 41 1 2 1 1 2 k k P r P P v v 5663 kPa 5148 kPa 11 2 3 r P P P p x 1017 K 5148 kPa 924 7 K 5663 kPa 2 2 P P T T x x 1119 K 1017 K 11 3 rcTx T 365 8 K 18 1119 K 1 1 3 3 3 4 k cr T T T v 1 1 41 1 4 k r v pplying the first law to each of the processes gives A 1 455 0 kJkg 291K 0 718 kJkg K924 7 1 2 2 T T c w v 102 5 kJkg 1 005 kJkg K1119 1017K 3 3 x p x T T c q 3 3 3 x x T c q w v 2926 kJkg 0 718 kJkg K1119 1017K 102 5 Tx 3 540 8 kJkg 365 8 K 0 718 kJkg K1119 4 3 4 T T c w v 115 1 455 0 2926 540 8 1 2 3 3 4 net The net work of the cycle is k Jkg w w w w x The mass in the device is given by 0 003233 kg 0 287 kPa m kg K291 K 90 kPa0003 m 3 PV 3 1 1 1 RT m he net power produced by this engine is then T 248 kW kJkg400060 cycles 0 003233 kgcycle1151 net net n mw W preparation If you are a student using this Manual you are using it without permission 946 963 A dual cycle with nonisentropic compression and expansion processes is considered The power produced by the cycle is to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK R 0287 kJkgK and k 14 Table A2a Analysis We begin by fixing the temperatures at all states PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 924 7 K 291 K 18 1 41 1 1 1 2 1 1 2 k k s T r T T v v 1037 K 085 291 K 9247 291 K 1 2 1 2 1 2 1 2 T T s η η T T T T T T s v P 4 1 2 3 qout x qin 5148 kPa 90 kPa 18 41 1 2 1 1 2 k k P r P P v v 5663 kPa 5148 kPa 11 2 3 r P P P p x 1141 K 5148 kPa 1037 K 5663 kPa 2 2 P P T T x x 1255 K 1141 K 11 3 rcTx T 410 3 K 18 1255 K 11 1 41 1 3 1 4 3 3 k c k r r T T T v v 4s 494 8 K 410 3 K 0 901255 1255 K 4 3 3 4 4 3 4 3 s s T T T T T T T T η η Applying e first law to each of the processes gives th 535 6 kJkg 291K 0 718 kJkg K1037 1 2 1 2 T T c w v 114 6 kJkg 1 005 kJkg K1255 1141K 3 3 T T c q 0 718 kJkg K1255 1141 114 6 3 3 3 x p x K 3275 kJkg x x x T T c q w v 8 0 71 4 3 3 4 T T c w v 545 8 kJkg 494 8 K kJkg K1255 he net w k of the cycle is T or 32 545 8 1 2 3 3 4 net w w w w x 4295 kJkg 535 6 75 he mass in the device is given by T 0 003233 kg 0 287 kPa m kg K291 K 90 kPa0003 m 3 3 1 1 1 RT P m V The net power produced by this engine is then 926 kW 0 003233 kgcycle4295 kJkg400060 cycles net net n mw W preparation If you are a student using this Manual you are using it without permission 947 964E An ideal dual cycle has a compression ratio of 15 and cutoff ratio of 14 The net work heat addition and the thermal efficiency are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are R 03704 psiaft3lbmR Table A1E cp 0240 BtulbmR cv 0171 BtulbmR and k 14 Table A2Ea Analysis Working around the cycle the germane properties at the various states are 1580 R 535 R 15 1 41 1 1 1 2 1 1 2 k k T r T T v v v P 4 1 2 3 qout x qin 629 2 psia 14 2 psia 15 41 1 2 1 1 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2 k k P r P P v v 692 1 psia 629 2 psia 11 2 3 r P P P p x 1738 R 6292 psia 1580 R 6921 psia 2 2 P P T T x x 2433 R 1738 R14 3 3 x c x x T r T T v v 942 2 R 15 2433 R 14 1 41 1 3 1 k 4 3 3 k c r r T T T v v pplying e first law to each of the processes gives 4 A th 178 7 Btulbm 535R 0 171 Btulbm R1580 1 2 1 2 T T c w v 2 2702 Btulbm 0 171 Btulbm R1738 1580R 2 T T c q x x v 3 3 166 8 Btulbm 0 240 Btulbm R2433 1738R 0 171 Btulbm R2433 166 8 Btulbm 3 3 3 x p x T T c q 8 4796 Btulbm 173 R x x x T T c q w v 4 3 3 4 254 9 Btulbm 942 2 R 0 171 Btulbm R2433 he net w k of the cycle is T T c w v T or w w w w 1242 Btulbm 4796 178 7 254 9 1 2 3 3 4 net x nd the net heat addition is Hence the thermal efficiency is a 1938 Btulbm 2702 166 8 3 2 in x x q q q 0641 1938 Btulbm Btulbm 1242 in net th q w η preparation If you are a student using this Manual you are using it without permission 948 965 A sixcylinder compression ignition engine operates on the ideal Diesel cycle The maximum temperature in the cycle the cutoff ratio the net work output per cycle the thermal efficiency the mean effective pressure the net power output and the specific fuel consumption are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at 850 K are cp 1110 kJkgK cv 0823 kJkgK R 0287 kJkgK and k 1349 Table A2b Analysis a Process 12 Isentropic compression 1 Qin 2 3 4 Qout 5044 kPa 95 kPa 19 1349 2 1 1 2 1349 1 1 k P P v v v The clearance volume and the total volume of the engine at the beginning of compression process state 1 are 950 1 K 340 K 19 2 1 1 2 k T T v 3 3 0001778 m 0 0 0045 m 19 c c c d c r V V c V V V V 0 003378 m 0 0032 0 0001778 V V V 3 1 d c ass contained in the cylinder is The total m 0003288 kg kPa m kg K 340 K 0287 3 1 The mass of fuel burned during one cycle is 95 kPa0003378 m 3 1 1 RT P m V 0 0001134 kg 0 003288 kg 28 a m m AF f f f f f f m m m m m m rocess 23 constant pressure heat a P ddition 4 723 kJ 0 0001134 kg42500 kJkg098 HV in c m f q Q η The cutoff ratio is 2244 K 3 3 2 3 in 950 1 K 0 003288 kg0823 kJkgK 4 723 kJ T T T T mc Q v 2362 9501 K K 2244 2 3 T T β b 3 3 1 2 0 0001778 m 19 0003378 m r V V 2 3 1 4 3 3 2 3 0 0004199 m 2 362 0 0001778 m P P V V V V β PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 949 Process 34 isentropic expansion 9 kPa 302 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0003378 m 5044 kPa 3 4 3 4 P P V m 00004199 1084 K m 78 m 00004199 1349 3 3 1 1349 3 3 1 k k V V rocess 4 constant voume heat rejection 00033 2244 K 4 3 3 4 T T V P 1 2 013 kJ 340 K 0 003288 kg 0823 kJkg K 1084 1 4 out T T mc Q v The net work output and the thermal efficiency are 2 013 4 723 Q Q W 2710 kJ out in out net 57 0 5737 2710 kJ netout th W η 4 4723 kJ Qin The mean effective pressure is determined to be c 847 kPa kPa m 2710 kJ MEP 3 netout W kJ 1778m 3 The power for engine speed of 1750 rpm is 0 000 0 003378 2 1 V V d 395 kW 60 s 1min 2 revcycle 2710 kJcycle1750 revmin net 2 net n W W Note that there are two revolutions in one cycle in fourstroke engines e Finally the specific fuel consumption is gkWh 151 1kWh 3600 kJ 1kg 1000 g 2710 kJkg 0 0001134 kg sfc net W m f preparation If you are a student using this Manual you are using it without permission 950 966 An expression for cutoff ratio of an ideal diesel cycle is to be developed Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Analysis Employing the isentropic process equations PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course T r r T r T k c c v P 4 1 2 3 qin qout 1 1 2 T rk T while the ideal gas law gives 3 1 1 2 When the first law and the closed system work integral is applied to the constant pressure heat addition the result is 1 1 1 1 2 3 in T r T r r c T T c q k k c p p When this is solved for cutoff ratio the result is 1 in1 1 T c r q r k p c preparation If you are a student using this Manual you are using it without permission 951 967 An expression for the thermal efficiency of a dual cycle is to be developed and the thermal efficiency for a given case is to be calculated Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK R 0287 kJkgK and k 14 Table A2 Analysis The thermal efficiency of a dual cycle may be expressed as 1 1 3 2 1 4 in out th x p x T T c T T c T T c q q v v η By applying the isentropic process relations for ideal gases with constant specific heats to the processes 12 and 34 as well as the ideal gas equation of state the temperatures may be eliminated from the thermal efficiency expression This yields the result 1 1 1 1 1 1 th p c p k c p k r r kr r r r η v P 4 1 2 3 qout x qin where PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course P2 P r x p and x cr v v 3 hen rc rp we obtain W 1 2 1 th p k k r r η 1 1 1 1 p p k p r r r For the case r 20 and rp 2 0660 2 1 2 2 41 1 2 20 1 1 2 1 41 1 41 ηth preparation If you are a student using this Manual you are using it without permission 952 968 An expression regarding the thermal efficiency of a dual cycle for a special case is to be obtained Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Analysis The thermal efficiency of a dual cycle may be expressed as 1 1 3 2 1 4 in out th x p x T T c T T c T T c q q v v η By applying the isentropic process relations for ideal gases with constant specific heats to the processes 12 and 34 as well as the ideal gas equation of state the temperatures may be eliminated from the thermal efficiency expression This yields the result 1 1 1 1 1 1 th p c p k c p k r r kr r r r η v P 4 1 2 3 qout x qin where PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course P2 P r x p and x cr v v 3 hen rc rp we obtain W 1 2 1 th p p k r k r r η 1 1 1 1 p k p r r earrangement of this result gives R 1 th 2 1 1 1 1 k p p p k p r r r r k r η preparation If you are a student using this Manual you are using it without permission 953 969 The five processes of the dual cycle is described The Pv and Ts diagrams for this cycle is to be sketched An expression for the cycle thermal efficiency is to be obtained and the limit of the efficiency is to be evaluated for certain cases Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Analysis a The Pv and Ts diagrams for this cycle are as shown s T 5 2 1 4 3 v P 5 1 2 4 qout 3 qin b Apply first law to the closed system for processes 23 34 and 51 to show 3 2 4 3 5 1 in v p out v q C T T C T T q C T T The cycle thermal efficiency is given by 5 1 1 5 1 4 3 2 3 2 3 4 3 5 1 3 2 3 2 4 3 1 1 1 1 1 1 1 1 v out p th C T T T T T q T C T T T T T kT T T T T T T k T T η Process 12 is isentropic therefore 3 2 1 1 th in v q C T η 1 1 T T T T 1 1 2 1 k k T V 1 2 r T V Process 23 is constant volume therefore 3 3 2 2 3 3 2 2 pr T PV P T PV P Process 34 is constant pressure therefore 3 3 4 4 4 4 4 3 3 3 c PV PV T V r T T T V Process 45 is isentropic therefore 1 1 1 1 1 5 4 T r V V T V 3 2 4 4 5 1 1 1 k k k k k c c c V rV r V V V r Process 51 is constant volume however T5T1 is found from the following 1 5 5 3 4 2 1 c c k k c p p T T T T T r r r r r r 1 4 3 2 1 k T T T r T T PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 954 The ratio T3T1 is found from the following 3 3 2 1 2 1 p k 1 T T T T T T r r The efficiency becomes 1 1 1 1 1 1 k c p th k p c k p r r r r k r r r η c In the limit as rp approaches unity the cycle thermal efficiency becomes 1 1 1 1 1 1 1 lim 1 lim 1 1 1 1 lim 1 1 p p p k c p th k r r p c k c th th Diesel r c k p k r r r r k r r k r r r r η η η d In the limit as rc approaches unity the cycle thermal efficiency becomes 1 1 1 1 1 1 1 1 1 lim 1 lim 1 1 1 1 lim 1 c p c k c p p th k k r r p c th th Otto r k p k r r r r r k r r r r r r η η η 1 p PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 958 977 An ideal Ericsson cycle operates between the specified temperature limits The rate of heat addition is to be determined Analysis The thermal efficiency of this totally reversible cycle is determined from s T 3 2 qin qout 4 1 900 K 280 K 06889 900 K 280 K 1 1 th H L T T η According to the general definition of the thermal efficiency the rate of heat addition is 726 kW 06889 kW 500 th net in η W Q 978 An ideal Ericsson cycle operates between the specified temperature limits The power produced determined Analysis The power output is 500 kW when t by the cycle is to be he cycle is repeated 2000 times inute Then the work per cycle is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course per m 15 kJcycle 200060 cycles net n W 500 kJs net W hen the cycle is repeated 3000 times per minute the power output will be considered The temperature of the sourceenergy reservoir r pressure during the cycle are to be determined ature are R 03704 psiaft3lbmR Table A1E cp 0240 BtulbmR v Analysis From the thermal efficiency relation s T 3 2 qin qout 4 1 900 K 280 K W 750 kW 300060 cycles1 5 kJcycle net net nW W 979E An ideal Stirling engine with air as the working fluid is the amount of air contained in the engine and the maximum ai Assumptions Air is an ideal gas with constant specific heats Properties The properties of air at room temper c 0171 BtulbmR and k 14 Table A2Ea s T 3 2 qin qout 4 1 510 R 765 R H H H L T T T T Q W 510 R 1 6 Btu 2 Btu 1 in net ηth State 3 ma ine the mass of air in the system y be used to determ 002647 lbm 0 3704 psia ft lbm R510 R 10 psia05 ft 3 3 3 3 3 RT P m V The maximum pressure occurs at state 1 125 psia 3 3 1 1 1 ft 006 0 02647 lbm 0 3704 psia ft lbm R765 R V mRT P preparation If you are a student using this Manual you are using it without permission 959 980E An ideal Stirling engine with air as the working fluid is considered The temperature of the sourceenergy reservoir the amount of air contained in the engine and the maximum air pressure during the cycle are to be determined Assumptions Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are R 03704 psiaft3lbmR cp 0240 BtulbmR cv 0171 BtulbmR and k 14 Table A2E Analysis From the thermal efficiency relation s T 3 2 qin qout 4 1 510 R 874 R H H H T T Q 6 Btu in th L T T W 510 R 1 25 Btu 1 net η State 3 may be used to determine the mass of air in the system 002647 lbm 0 3704 psia ft lbm 3 RT3 510 R R 10 psia05 ft 3 P3 3 m V The maxim m pressure occurs at state 1 u 143 psia 3 3 1 1 1 ft 006 0 02647 lbm 0 3704 psia ft lbm R874 R V mRT P 981 An ideal Stirling engine with air as the working fluid operates between specifi imits The heat added to the cycle and the net work produced by the cycle are to be determined ed pressure l Properties The properties of air at room temperature are R 0287 kPam kgK cp 1005 kJkgK cv 0718 kJkgK nd k 14 Table A2a Analysis Applying the ideal gas equation to the isothermal process 34 gives Assumptions Air is an ideal gas with constant specific heats 3 a 600 kPa 50 kPa12 4 s T 3 2 qin qout 3 v 3 P v P 4 1 298 K 4 Since process 41 is one of constant volume 1788 K 600 kPa 298 K 3600 kPa 4 1 4 1 P P T T Adapting the first law and work integral to the heat addition process gives 12 in w q 1275 kJkg 0287 kJkg K1788 Kln12 ln 1 2 1 v v RT imilarly S 2125 kJkg 12 1 0287 kJkg K298 Kln ln 3 4 3 3 4 out v v RT w q The net work is then 1063 kJkg 212 5 1275 out in net q q w PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 960 982 An ideal Stirling engine with air as the working fluid operates between specified pressure limits The heat transfer in the regenerator is to be determined Assumptions Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are R 0287 kPam3kgK cp 1005 kJkgK cv 0718 kJkgK and k 14 Table A2a Analysis Applying the ideal gas equation to the isothermal process 34 gives s T 3 2 qin qout 4 1 298 K 600 kPa 50 kPa12 4 3 4 v 3 P v P Since process 41 is one of constant volume 1788 K 600 kPa 4 4 1 P 298 K 3600 kPa 1 P T T pplication of the first law to process 41 gives A 1070 kJkg 298K 0 718 kJkg K1788 4 1 regen T T c q v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 961 Ideal and Actual GasTurbine Brayton Cycles 983C They are 1 isentropic compression in a compressor 2 P constant heat addition 3 isentropic expansion in a turbine and 4 P constant heat rejection 984C For fixed maximum and minimum temperatures a the thermal efficiency increases with pressure ratio b the net work first increases with pressure ratio reaches a maximum and then decreases 985C Back work ratio is the ratio of the compressor or pump work input to the turbine work output It is usually between 040 and 06 for gas turbine engines 986C In gas turbine engines a gas is compressed and thus the compression work requirements are very large since the steadyflow work is proportional to the specific volume 987C As a result of turbine and compressor inefficiencies a the back work ratio increases and b the thermal efficiency decreases PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 962 988E A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10 The air temperature at the compressor exit the back work ratio and the thermal efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with variable specific heats Properties The properties of air are given in Table A17E Analysis a Noting that process 12 is isentropic s T 1 2 4 3 qin qout 2000 520 T h Pr 1 1 1 12147 520 R 12427 Btu lbm Btulbm 24011 12147 1 2147 10 2 2 1 1 2 P r r 2 h T P P P 9965 R b Process 34 is isentropic and thus 23888 Btulbm 26583 50471 4 3 Tout h h w 11584 Btulbm 11 12427 240 26583 Btulbm 17 4 10 174 0 1 0 174 50471 Btulbm R 2000 1 2 in C 4 3 4 3 3 3 4 3 h h w h P P P P P h T r r r Then the backwork ratio becomes 485 23888 Btulbm Btulbm 11584 out T Cin bw w w r 465 26460 Btulbm Btulbm 12304 12304 Btulbm 88 11584 238 26460 Btulbm 24011 71 504 in netout th Cin Tout out net 2 3 in q w w w w h h q η c PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 963 989 A simple Brayton cycle with air as the working fluid has a pressure ratio of 10 The air temperature at the etermined anges are s 17 Analysis a Noting that process 12s is isentropic turbine exit the net work output and the thermal efficiency are to be d Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy ch s T 1 2s 4s 3 qin qout 1240 K 295 K 2 4 negligible 4 Air is an ideal gas with variable specific heat Properties The properties of air are given in Table A 3068 1 29517 kJkg K 295 1 1 1 rP h T 8304 kJkg 7 70207 0 87 132493 93 1324 689 6 K 70207 kJkg and 2723 272 3 10 1 3 272 132493 kJkg K 1240 62660 kJkg 0 83 29517 57026 17 295 5649 K 57026 kJkg and 1307 1 3068 10 4 3 3 4 4 3 4 3 4 4 3 4 3 3 T 1 2 1 2 1 2 1 2 2 2 2 3 4 3 s T s T s s r r r C s s C s s r h h h h h h h h T h P P P P P h h h h h h h h h T h P P η η η η Thus T4 7644 K b c rP 1 1 2 P 2104 kJkg 487 9 3 698 4879 kJkg 29517 04 783 6983 kJkg 62660 93 1324 out in out net 1 4 out 2 3 in q q w h h q h h q 301 0 3013 6983 kJkg kJkg 2104 in netout th q w η 964 990 Problem 989 is reconsidered The mass flow rate pressure ratio turbine inlet temperature and the isentropic efficiencies of the turbine and compressor are to be varied and a general solution for the problem by taking advantage of the diagram window method for supplying data to EES is to be developed Analysis Using EES the problem is solved as follows Input data from diagram window Pratio 10 T1 295 K P1 100 kPa T3 1240 K mdot 20 kgs Etac 83100 Etat 87100 Inlet conditions h1ENTHALPYAirTT1 s1ENTROPYAirTT1PP1 Compressor anaysis ss2s1 For the ideal case the entropies are constant across the compressor PratioP2P1Definition of pressure ratio to find P2 Ts2TEMPERATUREAirsss2PP2 Ts2 is the isentropic value of T2 at compressor exit hs2ENTHALPYAirTTs2 Etac hs2h1h2h1 Compressor adiabatic efficiency Etac WdotcidealWdotcactual mdoth1 Wdotcmdoth2 SSSF First Law for the actual compressor assuming adiabatic kepe0 External heat exchanger analysis P3P2process 23 is SSSF constant pressure h3ENTHALPYAirTT3 mdoth2 Qdotin mdoth3SSSF First Law for the heat exchanger assuming W0 kepe0 Turbine analysis s3ENTROPYAirTT3PP3 ss4s3 For the ideal case the entropies are constant across the turbine Pratio P3 P4 Ts4TEMPERATUREAirsss4PP4 Ts4 is the isentropic value of T4 at turbine exit hs4ENTHALPYAirTTs4 Etat Wdott Wtsdot turbine adiabatic efficiency Wtsdot Wdott Etath3h4h3hs4 mdoth3 Wdott mdoth4 SSSF First Law for the actual compressor assuming adiabatic kepe0 Cycle analysis WdotnetWdottWdotcDefinition of the net cycle work kW EtaWdotnetQdotinCycle thermal efficiency BwrWdotcWdott Back work ratio The following state points are determined only to produce a Ts plot T2temperatureairhh2 T4temperatureairhh4 s2entropyairTT2PP2 s4entropyairTT4PP4 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 965 Bwr η Pratio Wc kW Wnet kW Wt kW Qin kW 05229 01 2 1818 1659 3477 16587 06305 01644 4 4033 2364 6396 14373 07038 01814 6 5543 2333 7876 12862 07611 01806 8 6723 2110 8833 11682 08088 01702 10 7705 1822 9527 10700 085 01533 12 8553 1510 10063 9852 08864 0131 14 9304 1192 10496 9102 09192 01041 16 9980 8772 10857 8426 09491 007272 18 10596 5679 11164 7809 09767 003675 20 11165 2661 11431 7241 2 4 6 8 10 12 14 16 18 20 012 016 02 024 028 032 036 2250 2700 3150 3600 4050 4500 Pratio η Wnet kW 45 50 55 60 65 70 75 80 85 0 500 1000 1500 s kJkgK T K 100 kPa 1000 kPa Air 1 2 3 4 2s 4s PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 966 991 A simple Brayton cycle with air as the working fluid has a pressure ratio of 10 The air temperature at the turbine exit the net work output and the thermal efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2 Analysis a Using the compressor and turbine efficiency relations 720 K 642 3 0 87 1240 1240 6258 K 0 83 295 569 6 295 6423 K 10 1 K 1240 569 6 K K 10 295 4 3 3 4 4 3 4 3 4 3 4 3 1 2 1 2 1 2 1 2 1 2 1 2 0414 1 3 4 3 4 0414 1 1 2 1 2 s T s p p s T C s p s p s s T 1 2s 4s 3 qin qout 1240 K 295 K 2 4 C η k k s k k s T T T T T T c T T c h h h h T T T T T T c T T c h h h h P P T T P P T T η η η b 1902 kJkg 427 1 3 617 4271 kJkg 295 K kJkg K 720 1005 6173 kJkg 6258 K kJkg K 1240 1005 out in out net 1 4 1 4 out 2 3 2 3 in q q w T T c h h q T T c h h q p p 308 0 3081 6173 kJkg kJkg 1902 in netout th q w η c PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 967 992E A simple ideal Brayton cycle with helium has a pressure ratio of 14 The power output is to be determined Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible 4 Helium is an ideal gas with constant specific heats Properties The properties of helium are cp 125 BtulbmR and k 1667 Table A2Ea Analysis Using the isentropic relations for an ideal gas s T 1 2 4 3 qin qout 1760 R 520 R 1495 R 520 R14 06671667 1 1 1 1 2 1 2 k k p k k T r P P T T Similarly 612 2 K 1760 R 14 3 3 4 3 4 pr T P T T 1 1 06671667 1 1 k k k k P ssure heat addition process 23 produces Applying the first law to the constantpre q 331 3 Btulbm 125 Btulbm R1760 1495R 2 3 in T c p T Similarly 115 3 Btulbm 520R 1 25 Btulbm R612 2 1 4 out T T c q p net work production is then et q q w nd The n 2160 Btulbm 115 3 331 3 out in a 5093 hp 4241 Btumin 1hp 100 lbmmin216 0 Btulbm net net mw W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 968 993E A simple Brayton cycle with helium has a pressure ratio of 14 The power output is to be determined Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible 4 Helium is an ideal gas with constant specific heats Properties The properties of helium at room temperature are cp 125 BtulbmR and k 1667 Table A2Ea Analysis For the compression process 1495 R 520 R14 06671667 1 1 1 1 2 1 2 k k p k k s T r P P T T 4 s T 1 2s 3 qin qout 1760 R 520 R 2 1546 R 0 95 520 1495 520 1 2 1 2 p C T T c h h η 1 2 1 2 1 2 1 2 C s s p s T T T T T T c h h η For the isentropic expansion process PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 612 2 R 14 1760 R 1 1 06671667 1 1 k k k k 3 3 4 pr T P P pplying the first law to the constantpressure heat addition process 23 produces 3 4 T T A 267 5 Btulbm 125 Btulbm R1760 1546R 2 3 in T T c q imilarly p S 115 3 Btulbm 520R 1 25 Btulbm R612 2 1 4 out T T c q p The net work production is then 1522 Btulbm 115 3 267 5 out in net q q w nd a 3589 hp 4241 Btumin 1hp 100 lbmmin152 2 Btulbm net net mw W preparation If you are a student using this Manual you are using it without permission 969 994 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits The effects of nonisentropic compressor and turbine on the backwork ratio is to be compared Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a Analysis For the compression process 4s s T 1 2s 4 3 qin qout 873 K 288 K 2 585 8 K 288 K12 0414 1 1 2 1 2 k k s P P T T 618 9 K 0 90 288 585 8 288 1 2 1 2 1 2 1 2 1 2 1 2 C s p s p s C T T T T T T c T T c h h h h η η F or the expa nsion process 429 2 K 12 873 K 1 0414 1 3 4 3 4 k k s P P T T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 473 6 K 429 2 0 90873 873 3 3 4 4 3 4 3 4 3 4 3 T s p p s T T T T T T c h h η η 4 T s T T c h h The isentropic and actual work of compressor and turbine are 299 3 kJkg 288K 1005 kJkg K585 8 1 2 Comp T T c W s p s 332 6 kJkg 288K 1005 kJkg K618 9 1 2 Comp T T c W 4 p 3 Turb 446 0 kJkg 429 2 K 1005 kJkg K873 T s W p s T c 401 4 kJkg 473 6 K 1005 kJkg K873 4 3 Turb T T c W p The back work ratio for 90 efficient compressor and isentropic turbine case is 07457 4460 kJkg Turb bw s W r 3326 kJkg WComp The back work ratio for 90 efficient turbine and isentropic compressor case is 07456 4014 kJkg kJkg 2993 Turb Comp bw W W r s The two results are almost identical preparation If you are a student using this Manual you are using it without permission 970 995 A gas turbine power plant that operates on the simple Brayton cycle with air as the working fluid has a specified pressure ratio The required mass flow rate of air is to be determined for two cases Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis a Using the isentropic relations 352 kgs 300 K 1 1 2 P T T s 1991 kJkg kJs 70000 1991 kJkg 31175 84 510 51084 kJkg 4917 K kJkg K 1000 1005 31175 kJkg 300 K kJkg K 6102 1005 4917 K 12 1 K 1000 6102 K 12 netout s out net sCin sTout netout s 4 3 4 3 Tout s 1 2 1 2 Cin s 0414 1 3 4 3 4 0414 1 2 w W m w w w T T c h h w T T c h h w P P T T P s s p s s p s k k s k k b The net work output is determined to be s T 1 2s 4s 3 1000 K 300 K 2 4 1037 kgs 675 kJkg kJs 70000 675 kJkg 31175 085 85 51084 0 netout a out net sCin sTout aCin aTout netout a w W m w w w w w a C T η η preparation If you are a student using this Manual you are using it without permission 971 996 An actual gasturbine power plant operates at specified conditions The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with variable specific heats Properties The properties of air are given in Table A17 Analysis a Using the isentropic relations PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 30019 kJ kg 580 K 58604 kJ kg T h T h 1 1 2 2 300 K 5420 kJkg 90583 86 153604 0 28585 kJkg 30019 04 586 90583 kJkg 6773 47411 7 1 11 474 153604kJkg 58604 950 7 100 700 4 3 Tout w 1 2 in C 4 3 4 3 2 3 in 1 2 3 4 3 s T s r r r p h h h h w h P P P P P h h h q P P r η Thus s T 1 2s 4s 3 950 kJkg 580 K 300 K 2 4 527 5420 kJkg kJkg 28585 out T Cin bw w w r 270 950 kJkg kJkg 25615 25615 kJkg 28585 0 542 in netout th Cin Tout netout q w w w w η b preparation If you are a student using this Manual you are using it without permission 972 997 A gasturbine power plant operates at specified conditions The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with constant specific heats PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2 Analysis a Using constant specific heats 5622 kJkg 8748 K 86 1005 kJkg K 15253 0 2814 kJkg 300 K kJkg K 580 1005 8748 K 7 1 K 15253 15253 K 950 kJkg 1005 kJkg K K 580 7 100 700 4 3 4 3 out T 1 2 1 2 in C 0414 1 k k 3 4 3 4s in 2 3 2 3 2 3 in 1 2 s p T s T p p p p T T c h h w T T c h h w P P T T c q T T T T c h h q P P r η η Thus s T 1 2s 4s 3 950 kJkg 580 K 300 K 2 4 501 5622 kJkg kJkg 2814 out T Cin bw w w r 296 950 kJkg kJkg 2808 2808 kJkg 2814 2 562 in netout th Cin Tout out net q w w w w η b preparation If you are a student using this Manual you are using it without permission 973 998 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid The pressure ratio and the rate of heat input are given The net power and the thermal efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a Analysis For the isentropic compression process 5271 K 273 K10 0414 1 1 2 k T rpk T s T 1 2 4 3 qin qout 273 K The heat addition is 500 kJkg 1kgs in m q 500 kW in Q Applying the first law to the heat addition process PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1025 K K 1005 kJ 2 3 cp kg 500 kJkg 1 K 527 in 2 3 in p q T T T T c he temperature at the exit of the turbine is q T 530 9 K 10 1025 K 1 1 0414 1 k k 3 4 pr T T pplying the first law to the adiabatic turbine and the compressor produce 4 3 T A 1005 kJkg K1025 T T c w 496 6 kJkg 530 9 K p 255 4 kJkg 273K 1005 kJkg K527 1 1 2 C T T c w he net po er produced by the engine is then Finally the thermal efficiency is p T w 2412 kW 255 4 kJkg 1 kgs4966 C T net w m w W 0482 500 kW kW 2412 in net th Q W η preparation If you are a student using this Manual you are using it without permission 974 999 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid The pressure ratio and the rate of heat input are given The net power and the thermal efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a Analysis For the isentropic compression process 5918 K 273 K15 0414 1 1 2 k T rpk T s T 1 2 4 3 qin qout 273 K The heat addition is 500 kJkg 1kgs in m q 500 kW in Q Applying the first law to the heat addition process PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1089 K K 1005 k 2 3 cp Jkg 500 kJkg 8 K 591 in 2 3 in p q T T T T c he temperature at the exit of the turbine is q T 502 3 K 15 1089 K 1 1 0414 1 k k 3 4 pr T T pplying the first law to the adiabatic turbine and the compressor produce 4 3 T A 1005 kJkg K1089 T T c w 589 6 kJkg K 502 3 p 320 4 kJkg 273K 1005 kJkg K591 8 1 2 C T T c w he net po er produced by the engine is then Finally the thermal efficiency is p T w 2692 kW 320 4 kJkg 1 kgs5896 C T net w m w W 0538 500 kW kW 2692 in net th Q W η preparation If you are a student using this Manual you are using it without permission 975 9100 A gasturbine plant operates on the simple Brayton cycle The net power output the back work ratio and the thermal efficiency are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable specific heats Properties The gas constant of air is R 0287 kJkgK Table A1 Analysis a For this problem we use the properties from EES software Remember that for an ideal gas enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course h s s s P 1 Combustion chamber Turbine 2 3 4 Compress 100 kPa 40C 650C 2 MPa Process 12 Compression 5 749 kJkg K kPa 100 C 40 313 6 kJkg C 40 1 1 1 1 1 s P T h T 736 7 kJkg 5 749 kJkgK 2000 kPa 2 1 2 2 811 4 kJkg 313 6 313 6 736 7 0 85 2 2 1 2 1 2 C h h s η h h h h Process 34 Expansion 959 2 kJkg 650 C 4 4 h T s s h h h h h h h 4 3 3 4 3 4 3 T 959 2 0 88 η We cannot find the enthalpy at state 3 directly However using the following lines in EES together with the isentropic efficienc 73 kJkg T3 1421ºC s3 6736 kJkgK The solution by hand would require a trial error app h4senthalpyAir PP1 ss3 ermined from y relation we find h3 18 roach h3enthalpyAir TT3 s3entropyAir TT3 PP2 The mass flow rate is det 1299 kgs 273 K kPa m kg K 40 0287 100 kPa7006 0 m s 3 3 1 1 1 RT P m V The net power output is h m h W b The back work ratio is 1 2 Cin h m h W 6464 kW 313 6 kJkg 1299 kgs8114 11868 kW 959 2 kJkg 1299 kgs1873 4 3 Tout 5404 kW 6464 11868 Cin Tout net W W W 0545 11868 kW kW 6464 out T Cin bw W W r c The rate of heat input and the thermal efficiency are 13788 kW 811 4 kJkg 1299 kgs1873 2 3 in h m h Q 392 0 392 13788 kW kW 5404 in net Q W th η preparation If you are a student using this Manual you are using it without permission 976 9101 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits The cycle is to be sketched on the Ts cycle and the isentropic efficiency of the turbine and the cycle thermal efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with constant specific heats Properties The properties of air are given as cv 0718 kJkgK cp 1005 kJkgK R 0287 kJkgK k 14 Analysis b For the compression process 4s s T 1 2s 4 3 873 K 303 K 2 60300 kW 30K 200 kgs1005 kJkg K330 1 2 Comp T T mc W p For the turbine during the isentropic process 772 9 K 800 kPa 1400 K 100 kPa 0414 1 3 4 3 4 k k s P P T T 200 kgs 4 3 Turbs s p T T mc W 126050 kW 772 9 K 1005 kJkg K1400 The actual power output from the turbine is 120 60300 60000 Turb net Turb Comp Turb net W W W The isentropic efficiency of the turbine is then W W W kW 300 954 0 954 126050 kW 300 kW 120 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course s Turb Turb rb W W η c The rate of heat input is T T mc Q p he thermal efficiency is then Tu 160200 kW 273K 330 200 kgs1005 kJkg K1400 2 3 in T 375 0 375 160200 kW 000 kW 60 in net th Q W η preparation If you are a student using this Manual you are using it without permission 977 9102 A modified Brayton cycle with air as the working fluid operates at a specified pressure ratio The Ts diagram is to be sketched and the temperature and pressure at the exit of the highpressure turbine and the mass flow rate of air are to be determined Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with constant specific heats Properties The properties of air are given as cv 0718 kJkgK cp 1005 kJkgK R 0287 kJkgK k 14 Analysis b For the compression process PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 494 5 K 273 K8 1 1 2 P T T 0414 1 2 k k P The power input to the compressor is equal to the power output from the highpressure turbine Then 494 5 273 1500 2 1 4 4 3 1 2 HP Turbout in Comp T T T T T T T T mc T T c m W W p p The pressure at this state is s T 1 4 5 3 P4 1500 K 273 K 2 P5 P3 P2 3 4 3 1 2 T T 12785 K 4573 kPa 1 k k k 40 41 1 3 4 1 4 3 4 3 4 1500 K 100 kPa 1278 5 K 8 k T T rP P T T P P c The temperature at state 5 is determined from 828 1 K 457 3 kPa 100 kPa 1278 5 K 0414 1 5 k k P T T 4 4 P he net power is that generated by the lowpressure turbine since the power output from the highpressure turbine is equal the power input to the compressor Then 5 T to 4418 kgs 828 1 K 1005 kJkg K1278 5 200000 kW 5 4 Turb LP 5 4 Turb LP T T c W m T T mc W p p preparation If you are a student using this Manual you are using it without permission 978 9103 A simple Brayton cycle with air as the working fluid operates at a specified pressure ratio and between the specified temperature and pressure limits The cycle is to be sketched on the Ts cycle and the volume flow rate of the air into the compressor is to be determined Also the effect of compressor inlet temperature on the mass flow rate and the net power output are to be investigated Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with constant specific heats Properties The properties of air are given as cv 0718 kJkgK cp 1005 kJkgK R 0287 kJkgK k 14 Analysis b For the compression process 476 0 K 273 K 7 0414 1 1 2 1 2 k k s P P T T 4s s T 1 2s 4 3 1500 K 273 K 2 526 8 K 273 273 476 0 80 0 2 2 1 2 1 2 1 2 1 2 Comp Comp W η s Comp T T T T T T T T c m T T mc W s p s p or the ex nsion process F pa PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 860 3 K 1500 K 1 0414 1 4 k k P T T 7 3 3 4 s P 924 3 K 860 3 1500 1500 90 0 4 4 4 3 4 3 4 3 4 3 s Turb Turb Turb T T T T T T T T c m T T c m W W s s p p η Given the net power the mass flow rate is determined from 463 7 kgs 273 526 8 924 3 1005 kJkg K 1500 000 kW 150 The specific volume and the volume flow rate at the inlet of the c mpressor are 1 2 4 3 net 1 2 4 3 Comp Turb net T T T T c W m T T mc T T mc W W W p p p o 1 2 4 3 net T T T T mc W p 0 7835 m kg 100 kPa 0 287 kJkg K273 K 3 1 1 1 P RT v c For a fixed compressor inlet velocity and flow area when the compressor inlet temperature increases the specific volume increases since 3632 m s 3 463 7 kgs 0 7835 m kg 3 1 1 v V m P v RT When specific volume increases the mass flow rate decreases since v V m Note that volume flow rate is the same since inlet velocity and flow area are fixed When mass flow rate decreases the net power decreases since Therefore when the inlet temperature increases both mass flow rate and the net power decrease AV V Comp Turb net w m w W preparation If you are a student using this Manual you are using it without permission 979 Brayton Cycle with Regeneration 9104C Regeneration increases the thermal efficiency of a Brayton cycle by capturing some of the waste heat from the exhaust gases and preheating the air before it enters the combustion chamber 9105C Yes At very high compression ratios the gas temperature at the turbine exit may be lower than the temperature at the compressor exit Therefore if these two streams are brought into thermal contact in a regenerator heat will flow to the exhaust gases instead of from the exhaust gases As a result the thermal efficiency will decrease 9106C The extent to which a regenerator approaches an ideal regenerator is called the effectiveness ε and is defined as ε qregen act qregen max 9107C b turbine exit 9108C The steam injected increases the mass flow rate through the turbine and thus the power output This in turn increases the thermal efficiency since η W Qin and W increases while Qin remains constant Steam can be obtained by utilizing the hot exhaust gases PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 980 9109 A Brayton cycle with regeneration produces 150 kW power The rates of heat addition and rejection are to be determined Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with constant specific heats at room temperature 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a Analysis According to the isentropic process expressions for an ideal gas 530 8 K 293 K8 0414 1 1 2 k T rpk T s T 1 2 5 4 qin 1073 K 293 K 3 6 qout 592 3 K 1073 K 8 4 5 pr T T 1 1 0414 1 k k hen the first law is applied to the heat exchanger the result is T T The simultaneous solution of these two results gives W T T T T 3 6 5 2 while the regenerator temperature specification gives 582 3 K 10 592 3 10 5 3 6 540 8 K 530 8 582 3 592 3 2 3 5 T T T T ation of the first law to the turbine and compressor gives 3 592 005 kJkg K1073 1 1 2 5 4 net T T c T T c w p p Applic K kJkg 2441 293 K 005 kJkg K530 8 1 Then 0 6145 kgs 2441 kJkg kW 150 net net w W m Applying the first law to the combustion chamber produces Similarly 3030 kW 582 3 K 1 005 kJkg K1073 0 6145 kgs 3 4 in T T mc Q p 1530 kW 293K 1 005 kJkg K540 8 0 6145 kgs 1 6 out T T mc Q p PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 981 9110 A Brayton cycle with regeneration produces 150 kW power The rates of heat addition and rejection are to be determined Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with constant specific heats at room temperature 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a Analysis For the compression and expansion processes we have PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course k T rpk T 530 8 K 293 K8 0414 1 1 2s s T 1 2 5s 4 qin 1073 K 293 K 3 6 qout 5 2s 566 3 K 0 87 293 530 8 293 1 2 1 2 1 2 1 2 C s p s p C T T T T T T c T T c η η 592 3 K 8 1073 K 1 1 0414 1 4 5 k k p s r T T 625 9 K 592 3 0 931073 1073 5 4 s p 5 4 4 5 5 4 s T p T T T T T T T c T T c η η result is T T When the first law is applied to the heat exchanger the 6 5 2 3 T T T T while the regenerator temperature specification gives 3 615 9 K 10 625 9 10 5 The simultaneous solution of these two results gives 576 3 K 566 3 615 9 625 9 2 3 5 6 T T T T Application of the first law to the turbine and compressor gives 7 kJkg 174 293 K 1 005 kJkg K566 3 625 9 K 005 kJkg K1073 1 1 2 5 4 net T T c T T c w p p Then 0 8586 kgs 1747 kJkg kW 150 net net w W m pplying the first law to the combustion chamber produces Similarly A 3944 kW 615 9 K 1 005 kJkg K1073 0 8586 kgs 3 4 in T T mc Q p 2445 kW 293K 1 005 kJkg K576 3 0 8586 kgs 1 6 out T T mc Q p preparation If you are a student using this Manual you are using it without permission 982 9111 A Brayton cycle with regeneration is considered The thermal efficiencies of the cycle for parallelflow and counter flow arrangements of the regenerator are to be compared Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with constant specific heats at room temperature 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a Analysis According to the isentropic process expressions for an ideal gas 510 9 K 293 K7 0414 1 1 2 k T rpk T s T 1 2 5 4 qin 1000 K 293 K 3 6 qout 573 5 K 1000 K 7 4 5 pr T T 1 1 0414 1 k k When the first law is applied to the heat exchanger as originally arranged the result is T T T T 3 6 5 2 while the regenerator temperature specification gives 567 5 K 6 573 5 6 5 3 T T The simultaneous solution of these two results gives 2 3 5 6 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 516 9 K 510 9 567 5 573 5 T n T T T The thermal efficiency of the cycle is the 0482 567 5 1000 293 5169 1 1 1 3 4 in 1 6 th T T T T q q η cle An energy balance on the heat exchanger gives T T T 6 T T The thermal efficiency of the cycle is then out For the rearranged version of this cy 6 6 3 T T s T 1 2 5 4 qin 1000 K 293 K 3 6 qout T6 5 2 3 The solution of these two equations is 2 K 3 539 2 K 545 0453 539 2 1000 293 5452 1 1 1 3 4 1 6 in out th T T T T q q η preparation If you are a student using this Manual you are using it without permission 983 9112E An ideal Brayton cycle with regeneration has a pressure ratio of 11 The thermal efficiency of the cycle is to be determined with and without regenerator cases Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with constant specific heats at room temperature 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 024 BtulbmR and k 14 Table A2Ea Analysis According to the isentropic process expressions for an ideal gas 1111 R 560 R11 0414 1 1 2 k T pr k T s T 1 2 5 4 qin 2400 R 560 R 3 6 qout 1210 R 2400 R 11 4 5 pr T T 1 1 0414 1 k k he regenerator is ideal i the effectiveness is 100 and thus T T T e R 1111 5 3 T T R 1210 2 6 The thermal efficiency of the cycle is then 537 0 537 1210 2400 560 1111 1 1 1 3 4 1 6 in out th T T T T q q η The solution without a regenerator is as follows s T 1 2 4 3 qin qout 2400 R 560 R 1111 R 560 R11 0414 1 1 2 k T pr k T 1210 R 11 2400 R 1 1 0414 1 3 4 k k pr T T 496 0 496 1111 2400 560 1210 1 1 1 2 3 1 4 in out th T T T T q q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 984 9113E A car is powered by a gas turbine with a pressure ratio of 4 The thermal efficiency of the car and the mass flow rate of air for a net power output of 95 hp are to be determined Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with variable specific heats 3 The ambient air is 540 R and 145 psia 4 The effectiveness of the regenerator is 09 and the isentropic efficiencies for both the compressor and the turbine are 80 5 The combustion gases can be treated as air Properties The properties of air at the compressor and turbine inlet temperatures can be obtained from Table A17E Analysis The gas turbine cycle with regeneration can be analyzed as follows PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 372 5753 23012 1 4 4 3 4 s r r h P P P 2 Btulbm 4 12 230 54935 Btulbm R 2160 192 0 Btulbm 5 544 1 386 4 386 1 12906 Btulbm 0 R 54 3 3 3 2 1 2 1 1 3 1 2 1 r s r r r P P h T h P P P P P h T and s T 1 2s 4s 3 qin 2160 R 540 R 5 4 2 40763 Btulbm 372 2 35 549 54935 4 4 3 2 1 2 h h h 080 20774 Btulbm 12906 12906 192 0 080 4 4 3 turb 2 1 2 comp h h h h h h h h h s s η η Then the thermal efficiency of the gas turbine cycle becomes 1799 Btulbm 20774 90 40763 2 4 regen h h q ε 630 Btulbm 20774 12906 40763 54935 20774 179 9 1617 Btulbm 54935 1 2 4 3 Cin Tout out net regen 2 3 in h h h h w w w q h h q 039 1617 Btulbm Btulbm 630 in netout th 39 q w η Finally the mass flow rate of air through the turbine becomes 107 lbms 1hp 0 7068 Btus 630 Btulbm hp 95 net net air w W m preparation If you are a student using this Manual you are using it without permission 985 9114 The thermal efficiency and power output of an actual gas turbine are given The isentropic efficiency of the 1 Air is an ideal gas with variable specific heats 2 Kinetic and potential energy changes are negligible 3 The same and the properties of combustion gases are the same as Analysis The properties at various states are turbine and of the compressor and the thermal efficiency of the gas turbine modified with a regenerator are to be determined Assumptions mass flow rates of air and of the combustion gases are the those of air Properties The properties of air are given in Table A17 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4847 712 5 14 7 1 4 3 4 3 4 s r r h P P P P 23 kJkg 825 5 712 17100 kJkg 1561 K C 1288 65325 kJkg 2110 1 4356 7 14 4356 1 30321 kJkg C 303 K 30 3 3 2 1 2 1 1 3 1 2 1 r s r r r P h T h P P P P P h T ass are s T 1 2s 4s 3 qin 1561 K 303 K 5 4 2 The net work output and the heat input per unit m C 658 7 931 7 K 96855 kJkg 30321 66534 66534 kJkg 37266 1038 0 net in out w q q 672 0 kJkg 1038 1710 1038 0 kJkg 0359 kJkg 37266 37266 kJkg h 1 3600 s 1536000 kgh kW 159000 4 1 out 4 1 4 t 3 2 2 3 in th net in net net T h q h h h q q h h h h q w q m W w in η ou Then the compressor and turbine efficiencies become 949 838 0 949 30321 672 30321 25 653 0 838 82523 1710 96855 1710 1 2 1 2 4 3 4 3 h h h h h h h h s C s T η η When a regenerator is added the new heat input and the thermal efficiency become 441 0 441 8452 kJkg kJkg 37266 192 8 8452 kJkg 1038 06596855 6720 1928 kJkg new in net new th regen in new in 2 4 regen q w q q q h h q η ε Discussion Note a 65 efficient regenerator would increase the thermal efficiency of this gas turbine from 359 to 441 preparation If you are a student using this Manual you are using it without permission 986 9115 Problem 9114 is reconsidered A solution that allows different isentropic efficiencies for the compressor and turbine is to be developed and the effect of the isentropic efficiencies on net work done and the heat supplied to the cycle is to be studied Also the Ts diagram for the cycle is to be plotted Analysis Using EES the problem is solved as follows Input data T3 1288 C Pratio 147 T1 30 C P1 100 kPa T4659 C Wdotnet159 MW We omit the information about the cycle net work mdot 1536000 kghConvertkghkgs Etathnoreg0359 We omit the information about the cycle efficiency Etareg 065 Etac 084 Compressor isentropic efficiency Etat 095 Turbien isentropic efficiency Isentropic Compressor anaysis s1ENTROPYAirTT1PP1 ss2s1 For the ideal case the entropies are constant across the compressor P2 PratioP1 ss2ENTROPYAirTTs2PP2 Ts2 is the isentropic value of T2 at compressor exit Etac WdotcompisenWdotcomp compressor adiabatic efficiency Wdotcomp Wdotcompisen Conservation of energy for the compressor for the isentropic case Edotin Edotout DELTAEdot0 for steadyflow mdoth1 Wdotcompisen mdoths2 h1ENTHALPYAirTT1 hs2ENTHALPYAirTTs2 Actual compressor analysis mdoth1 Wdotcomp mdoth2 h2ENTHALPYAirTT2 s2ENTROPYAirTT2 PP2 External heat exchanger analysis SSSF First Law for the heat exchanger assuming W0 kepe0 Edotin Edotout DELTAEdotcv 0 for steady flow mdoth2 Qdotinnoreg mdoth3 qinnoregQdotinnoregmdot h3ENTHALPYAirTT3 P3P2process 23 is SSSF constant pressure Turbine analysis s3ENTROPYAirTT3PP3 ss4s3 For the ideal case the entropies are constant across the turbine P4 P3 Pratio ss4ENTROPYAirTTs4PP4Ts4 is the isentropic value of T4 at turbine exit Etat Wdotturb Wdotturbisen turbine adiabatic efficiency Wdotturbisen Wdotturb SSSF First Law for the isentropic turbine assuming adiabatic kepe0 Edotin Edotout DELTAEdotcv 0 for steadyflow mdoth3 Wdotturbisen mdoths4 hs4ENTHALPYAirTTs4 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 987 Actual Turbine analysis mdoth3 Wdotturb mdoth4 h4ENTHALPYAirTT4 s4ENTROPYAirTT4 PP4 Cycle analysis Using the definition of the net cycle work and 1 MW 1000 kW Wdotnet1000WdotturbWdotcomp kJs EtathnoregWdotnet1000QdotinnoregCycle thermal efficiency BwrWdotcompWdotturbBack work ratio With the regenerator the heat added in the external heat exchanger is mdoth5 Qdotinwithreg mdoth3 qinwithregQdotinwithregmdot h5ENTHALPYAir TT5 s5ENTROPYAirTT5 PP5 P5P2 The regenerator effectiveness gives h5 and thus T5 as Etareg h5h2h4h2 Energy balance on regenerator gives h6 and thus T6 as mdoth2 mdoth4mdoth5 mdoth6 h6ENTHALPYAir TT6 s6ENTROPYAirTT6 PP6 P6P4 Cycle thermal efficiency with regenerator EtathwithregWdotnet1000Qdotinwithreg The following data is used to complete the Array Table for plotting purposes ss1s1 Ts1T1 ss3s3 Ts3T3 ss5ENTROPYAirTT5PP5 Ts5T5 ss6s6 Ts6T6 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 988 ηt ηc ηthnoreg ηthwithreg Qinnoreg kW Qinwithreg kW Wnet kW 07 075 08 085 09 095 1 084 084 084 084 084 084 084 02044 02491 02939 03386 03833 04281 04728 027 03169 03605 04011 04389 04744 05076 422152 422152 422152 422152 422152 422152 422152 319582 331856 344129 356403 368676 380950 393223 863 1052 1241 1429 1618 1807 1996 45 50 55 60 65 70 75 80 85 100 100 300 500 700 900 1100 1300 1500 1700 s kJkgK T C 100 kPa 1470 kPa Ts Diagram for Gas Turbine with Regeneration 1 2s 2 5 3 4s 4 6 07 075 08 085 09 095 1 80 100 120 140 160 180 200 ηt Wnet kW 07 075 08 085 09 095 1 310000 330000 350000 370000 390000 410000 430000 ηt Qin kW no regeneration with regeneration 07 075 08 085 09 095 1 02 025 03 035 04 045 05 055 ηt ηth no regeneration with regeneration PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 989 9116 A Brayton cycle with regeneration using air as the working fluid is considered The air temperature at the turbine exit the net work output and the thermal efficiency are to be determined Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with variable specific heats 3 Kinetic and potential energy changes are negligible PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The properties of air are given in Table A17 Analysis a The properties of air at various states are 80314 kJkg 71180 082 121925 121925 71180 kJkg 2859 20015 7 1 15 200 121925 kJkg 150 K 1 61826 kJkg 31024 075 54126 31024 54126 kJkg 1088 1 5546 7 5546 1 31024 kJkg K 310 4 3 3 4 4 3 4 3 4 3 4 3 3 T 1 2 1 2 1 2 1 2 2 1 1 3 4 3 1 2 1 s T s T s r r r C s s C s r r r h h h h h h h h h P P P P P h h h h h h h h h h P P h η η η η Thus T4 7828 K b s T 1 2s 4s 3 qin 1150 K 310 K 5 6 4 2 1 T 2 P P P 10809 kJkg 31024 61826 80314 25 1219 1 2 4 3 Cin Tout net h h h h w w w kJkg 73843 61826 065 80314 26 618 2 4 2 5 2 4 2 5 h h h h h h h h ε ε c Then 225 48082 kJkg kJkg 10809 48082 kJkg 73843 1925 12 in net th 5 3 in q w h h q η preparation If you are a student using this Manual you are using it without permission 990 9117 A stationary gasturbine power plant operating on an ideal regenerative Brayton cycle with air as the working fluid is considered The power delivered by this plant is to be determined for two cases Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas 3 Kinetic and potential energy changes are negligible Properties When assuming constant specific heats the properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a When assuming variable specific heats the properties of air are obtained from Table A17 Analysis a Assuming constant specific heats PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 39188 kW 75000 kW 05225 0 5225 607 2 1100 290 525 3 1 1 1 1 5253 K 6072 K and 100 6072 K 8 1 K 1100 K 5253 in net 5 3 1 6 5 3 1 6 in out th 2 6 4 5 0414 1 3 4 3 4 0414 1 Q W T T T T T T c T T c q q T T T T P P T T P T p p k k k k η η ε Assuming variable specific heats 290 K 8 1 2 1 2 P T T b 40283 kW 75000 kW 05371 0 5371 65137 116107 29016 52612 1 1 1 52612 kJkg 65137 kJkg and 100 65137 kJkg 2089 8 167 1 1 1 167 116107 kJkg K 1100 52612 kJkg 9 8488 1 2311 8 2311 1 29016 kJkg 0K 29 in net 5 3 1 6 in out th 2 6 4 5 4 3 4 3 3 2 1 2 1 1 3 4 3 1 2 1 Q W h h h h q q h h h h h P P P P P h T h P P P P P h T T r r r r r r η η ε s T 1 2 4 3 75000 kW 1100 K 290 K 5 qout 6 preparation If you are a student using this Manual you are using it without permission 991 9118 A regenerative gasturbine engine using air as the working fluid is considered The amount of heat transfer in the regenerator and the thermal efficiency are to be determined Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with variable specific heats 3 Kinetic and potential energy changes are negligible Properties The properties of air are given in Table A17 Analysis a The properties at various states are PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1927 kJkg 65984 80 90074 0 74 kJkg 900 83244 0 90 151542 42 1515 83244 kJkg 5006 450 5 9 1 4505 151542 kJkg 00 K 14 65984 kJkg 0 K 65 31024 kJkg K 310 9 900 100 2 4 regen 4 3 3 4 4 3 4 3 4 3 4 3 3 2 2 1 1 1 2 3 4 3 h h q h h h h h h h h h P P P P P h T h T h T P P r s T s T s r r r ε η η b p 400 0 400 66288 kJkg kJkg 26508 66288 kJkg 192 7 65984 151542 26508 kJkg 31024 65984 90074 42 1515 in net th regen 2 3 in 1 2 4 3 Cin Tout net q w q h h q h h h h w w w η s T 1 2s 4s 3 qin 1400 K 310 K 5 6 4 650 K 2 preparation If you are a student using this Manual you are using it without permission 992 9119 A regenerative gasturbine engine using air as the working fluid is considered The amount of heat transfer in the regenerator and the thermal efficiency are to be determined Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with constant specific heats 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis a Using the isentropic relations and turbine efficiency 1307 kJkg 650 K 1005 kJkg K 8126 080 8126 K 747 3 0 90 1400 1400 7473 K 9 1 K 1400 9 900 100 2 4 2 4 regen 4 3 3 4 4 3 4 3 4 3 4 3 41 40 1 3 4 3 4 1 2 T T c h h q T T T T T T c T T c h h h h P P T T P P r p s T s p p s T k k s p ε ε η η s T 1 2s 4s 3 qin 1400 K 310 K 5 6 4 650 K 2 399 0 399 6231 kJkg kJkg 2487 6231 kJkg 1307 650 K kJkg K 1400 1005 2487 kJkg 310 K 650 8126 kJkg K 1400 1005 in net th regen 2 3 regen 2 3 in 1 2 4 3 Cin Tout net q w q T T c q h h q T T c T T c w w w p p p η b preparation If you are a student using this Manual you are using it without permission 993 9120 A regenerative gasturbine engine using air as the working fluid is considered The amount of heat transfer in the regenerator and the thermal efficiency are to be determined Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with variable specific heats 3 Kinetic and potential energy changes are negligible Properties The properties of air are given in Table A17 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis a The properties at various states are 1686 kJkg 65984 0 70 90074 90074 kJkg 83244 0 90 151542 42 1515 83244 kJkg 5006 450 5 9 1 4505 151542 kJkg 00 K 14 65984 kJkg 0 K 65 31024 kJkg K 310 9 900 100 2 4 regen 4 3 3 4 4 3 4 3 4 3 4 3 3 2 2 1 1 1 2 3 4 3 h h q h h h h h h h h h P P P P P h T h T h T P P r s T s T s r r r ε η η b p s T 1 2s 4s 3 qin 1400 K 310 K 5 6 4 650 K 2 386 0 386 68718 kJkg kJkg 26508 68718 kJkg 168 6 65984 151542 26508 kJkg 31024 65984 90074 42 1515 in net th regen 2 3 in 1 2 4 3 Cin Tout net q w q h h q h h h h w w w η preparation If you are a student using this Manual you are using it without permission 994 9121 An expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator is to be developed Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with constant specific heats at room temperature 3 Kinetic and potential energy changes are negligible Analysis The expressions for the isentropic compression and expansion processes are PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2 k T rpk T 1 1 s T 1 2 4 3 qin 5 6 qout k k pr T T 1 3 4 1 For an ideal regenerator 4 5 T T 2 6 T T The thermal efficiency of the cycle is k k p k k p k k p T r T r r T T T T T T T T T T T T T T T T T T q q 1 3 1 1 1 3 1 3 4 1 2 3 1 3 5 1 6 3 1 5 3 1 6 in out th 1 1 1 1 1 1 1 1 1 1 1 1 η preparation If you are a student using this Manual you are using it without permission 995 Brayton Cycle with Intercooling Reheating and Regeneration 9122C As the number of compression and expansion stages are increased and regeneration is employed the ideal Brayton cycle will approach the Ericsson cycle 9123C Because the steadyflow work is proportional to the specific volume of the gas Intercooling decreases the average specific volume of the gas during compression and thus the compressor work Reheating increases the average specific volume of the gas and thus the turbine work output 9124C a decrease b decrease and c decrease 9125C a increase b decrease and c decrease 9126C a increase b decrease c decrease and d increase 9127C a increase b decrease c increase and d decrease 9128C c The Carnot or Ericsson cycle efficiency PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 996 9129 An ideal gasturbine cycle with two stages of compression and two stages of expansion is considered The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with variable specific heats 3 Kinetic and potential energy changes are negligible s T 3 4 1 5 qin 1200 K 300 K 8 6 7 10 9 2 Properties The properties of air are given in Table A17 Analysis a The work inputs to each stage of compressor are identical so are the work outputs of each stage of the turbine since this is an ideal cycle Then 66286 kJkg 94636 2 127779 2 22214 kJkg 30019 2 41126 2 94636 kJkg 7933 238 3 1 5 P 238 127779 kJkg 200 K 1 41126 kJkg 4 158 1 386 3 386 1 30019 kJkg K 300 6 5 out T 1 2 in C 8 6 5 6 7 5 5 4 2 1 2 1 1 5 6 1 2 1 h h w h h w h h P P P P h h T h h P P P P P h T r r r r r r Thus 335 66286 kJkg kJkg 22214 out T Cin bw w w r 368 119796 kJkg kJkg 44072 44072 kJkg 22214 86 662 119796 kJkg 94636 127779 41126 79 1277 in net th Cin Tout net 6 7 4 5 in q w w w w h h h h q η b When a regenerator is used rbw remains the same The thermal efficiency in this case becomes 553 79663 kJkg kJkg 44072 79663 kJkg 40133 96 1197 3 kJkg 401 3 41126 75 94636 0 in net th regen inold in 4 8 regen q w q q q h h q η ε PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 997 9130 A gasturbine cycle with two stages of compression and two stages of expansion is considered The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with variable specific heats 3 Kinetic and potential energy changes are negligible Properties The properties of air are given in Table A17 Analysis a The work inputs to each stage of compressor are identical so are the work outputs of each stage of the turbine Then PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 58332 kJkg 98613 2 127779 2 26446 kJkg 30019 2 43242 2 98613 kJkg 94636 0 88 127779 79 1277 94636 kJkg 7933 238 3 1 238 127779 kJkg K 1200 42 kJkg 432 41126 kJkg 4 158 1 386 3 386 1 30019 kJkg K 300 6 5 out T 1 2 in C 6 5 5 8 6 6 5 6 5 8 6 5 6 7 5 5 1 2 1 4 2 1 2 4 2 1 2 1 1 5 6 5 1 2 1 h h w h h w h h h h h h h h h h h P P P P P h h T h h h h h h h h h h h P P P P P h T s T s T s s r r r C s s C s s r r r η η η η Thus s T 3 4 1 5 qin 8 6 7 1 9 2 8 6s 0 84 30019 41126 30019 1 2 453 0 453 58332 kJkg kJkg 26446 out T Cin bw w w r 4 2s 280 0 280 113703 kJkg kJkg 31886 31886 kJkg 26446 32 583 113703 kJkg 98613 127779 43242 127779 in net th Cin Tout net 6 7 4 5 in q w w w w h h h h q η b When a regenerator is used rbw remains the same The thermal efficiency in this case becomes 442 0 442 72175 kJkg kJkg 31886 72175 kJkg 41528 03 1137 41528 kJkg 43242 75 98613 0 in net th regen inold in 4 8 regen q w q q q h h q η ε preparation If you are a student using this Manual you are using it without permission 998 9131E An ideal regenerative gasturbine cycle with two stages of compression and two stages of expansion is considered The power produced and consumed by each compression and expansion stage and the rate of heat rejected are to be determined Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with constant specific heats at room temperature 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 024 BtulbmR and k 14 Table A2Ea Analysis The pressure ratio for each stage is 12 3 464 pr s T 3 4 1 6 520 R 9 7 8 2 5 1 According to the isentropic process expressions for an ideal gas 741 6 R 520 R3464 0414 1 1 4 2 k T rpk T T Since this is an ideal cycle 741 6 50 4 9 7 5 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 50 791 6 R For the isentropic expansion processes k T rpk T T he heat i is T T T T 1129 R 791 6 R3464 0414 1 7 8 6 T nput 2 5 6 in 1620 Btulbm 791 6 R 2024 Btulbm R1129 The mass flow rate is then T T c q p 3086 lbms 0 Btulbm 162 in q 500 Btus in m Q pplication of the first law to the expansion process 67 gives A 2636 kW 7 6 67out T T mc W p 0 94782 Btus 1kW R 791 6 3 086 lbms02 4 Btulbm R1129 The same amount of power is produced in process 89 When the first law is adapted to the compression process 12 it becomes 1 2 12in T T mc W p 1732 kW 0 94782 Btus 1kW 520 R 3 086 lbms02 4 Btulbm R7416 Compression process 34 uses the same amount of power The rate of heat rejection from the cycle is 3283 Btus 520 R 3 086 lbms02 4 Btulbm R7416 2 2 3 2 out T T mc Q p preparation If you are a student using this Manual you are using it without permission 999 9132E An ideal regenerative gasturbine cycle with two stages of compression and two stages of expansion is considered The power produced and consumed by each compression and expansion stage and the rate of heat rejected are to be determined Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with constant specific heats at room temperature 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 024 BtulbmR and k 14 Table A2Ea Analysis The pressure ratio for each stage is 12 3 464 pr s T 3 4s 1 6 520 R 9 7 8 2 5 1 4 2s 7s 9s For the compression processes 2s 741 6 R 520 R3464 0414 1 1 4 k k p s T r T T 780 7 R 0 85 520 741 6 520 1 2 1 4 2 1 2 1 2 C s p s p C T T T T T T T c T T c η η Since the regenerator is ideal 830 7 R 50 780 7 50 4 9 7 5 T T T T For the expansion processes 1185 R 830 7 R3464 0414 1 7 8 6 k k p s s T r T T 1150 R 830 7 0 901185 830 7 7 6 7 8 6 7 6 7 6 c p ηT T T T T T T T c T T s T s p η The heat input is 1533 Btulbm R 2024 Btulbm R1150 830 7 2 5 6 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course in T T c q p The mass flow rate is then 3262 lbms 153 3 Btulbm Btus 500 in in q Q m Application of the first law to the expansion process 67 gives 7 6 67out T T mc W p 2637 kW 0 94782 Btus 1kW R 830 7 3 262 lbms02 4 Btulbm R1150 t becomes The same amount of power is produced in process 89 When the first law is adapted to the compression process 12 i 2153 kW 0 94782 Btus 1kW 520 R 3 262 lbms02 4 Btulbm R7807 1 2 12in T T mc W p Compression process 34 uses the same amount of power The rate of heat rejection from the cycle is 4082 Btus 520 R 3 262 lbms02 4 Btulbm R7807 2 2 3 2 out T T mc Q p preparation If you are a student using this Manual you are using it without permission 9100 9133 A regenerative gasturbine cycle with two stages of compression and two stages of expansion is considered The thermal efficiency of the cycle is to be determined Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with constant specific heats at room temperature 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a Analysis The temperatures at various states are obtained as follows PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course k T rpk T T 2 430 9 K 290 K4 0414 1 1 4 450 9 K 20 430 9 20 4 5 T T s T 3 4 1 6 290 K 9 7 8 2 5 10 749 4 K 1005 kJkg K 300 kJkg 9 K 450 in 5 6 5 6 in p p c q T T T T c q 504 3 K 4 749 4 K 1 1 0414 1 6 7 k k rp T T 802 8 K 1005 kJkg K 300 kJkg 3 K 504 in 7 8 p c q T T 540 2 K 4 802 8 K 1 1 0414 1 8 9 k k rp T T 9 e heat rejected is The thermal efficiency of the cycle is then T 20 T 520 2 K 20 540 2 10 The heat input is 600 kJkg 300 300 in q Th kJkg 3730 290 R 430 9 290 kJkg K5202 1005 3 2 1 10 out T T c T T c q p p 0378 600 373 0 1 1 in out th q q η preparation If you are a student using this Manual you are using it without permission 9101 9134 A regenerative gasturbine cycle with three stages of compression and three stages of expansion is considered The thermal efficiency of the cycle is to be determined Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with constant specific heats at room temperature 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a Analysis The temperatures at various states are obtained as follows PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course k T rpk T T T 2 430 9 K 290 K4 0414 1 1 6 4 s T 3 4 1 290 K 9 8 2 5 1 11 6 7 12 13 14 450 9 K 20 430 9 20 6 7 T T 749 4 K 1005 kJkg K 300 kJkg 9 K 450 in 7 8 7 8 in p p c q T T T T c q 504 3 K 4 749 4 K 1 1 0414 1 8 9 k k rp T T 802 8 K 1005 kJkg K 300 kJkg 3 K 504 in 9 10 p c q T T 540 2 K 4 802 8 K 1 1 0414 1 10 11 k k rp T T 838 7 K 1005 kJkg K 300 kJkg 2 K 540 in p c q 12 T11 T 564 4 K 4 838 7 K 1 1 0414 1 12 13 r T T k k p The heat input is he heat rejected is The thermal efficiency of the cycle is then 544 4 K 20 564 4 20 13 14 T T 900 kJkg 300 300 300 in q T 9 kJkg 538 290 R 430 9 290 430 9 290 kJkg K5444 1005 5 4 3 2 1 14 out T T c T T c T T c q p p p 401 0401 900 538 9 1 1 in out th q q η preparation If you are a student using this Manual you are using it without permission 9102 9135 A regenerative gasturbine cycle with three stages of compression and three stages of expansion is considered The thermal efficiency of the cycle is to be determined Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with constant specific heats at room temperature 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a Analysis Since all compressors share the same compression ratio and begin at the same temperature PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course k T rpk T T T 2 430 9 K 290 K4 0414 1 1 6 4 s T 3 4 1 290 K 9 8 2 5 1 11 6 7 12 13 14 From the problem statement 65 13 7 T T The relations for heat input and expansion processes are p p c q T T T T c q in 7 8 7 8 in k k rp T T 1 8 9 1 p c q T T in 9 10 k k rp T T 1 10 11 1 k k p cp T T in 11 12 q r T T 12 13 1 1 ing EES software gives the following results ance on the regenerator 6 14 14 13 6 14 13 6 7 T T T T T T T T The heat input is The heat rejected is 5 4 c p T T The thermal efficiency of the cycle is then The simultaneous solution of above equations us 585 7 K 870 4 K 571 9 K 8 K 849 551 3 K 819 2 K 7 K 520 13 12 11 10 9 8 7 T T T T T T T From an energy bal T 65 13 T 495 9 K 65 430 9 65 900 kJkg 300 300 300 in q 1 kJkg 490 290 R 430 9 290 430 9 290 kJkg K4959 1005 3 2 1 14 out T T c T T c q p p 455 0 455 900 490 1 1 1 in out th q q η preparation If you are a student using this Manual you are using it without permission 9103 JetPropulsion Cycles 9136C The power developed from the thrust of the engine is called the propulsive power It is equal to thrust times the aircraft velocity 9137C The ratio of the propulsive power developed and the rate of heat input is called the propulsive efficiency It is determined by calculating these two quantities separately and taking their ratio 9138C It reduces the exit velocity and thus the thrust PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9104 9139E A turboprop engine operating on an ideal cycle is considered The thrust force generated is to be determined Assumptions 1 Steady operating conditions exist 2 The air standard assumptions are applicable 3 Air is an ideal gas with constant specific heats at room temperature 4 The turbine work output is equal to the compressor work input Properties The properties of air at room temperature are R 03704 psiaft3lbmR Table A1E cp 024 BtulbmR and k 14 Table A2Ea Analysis Working across the two isentropic processes of the cycle yields s T 1 2 4 3 qin 5 qout 868 8 R 450 R10 0414 1 1 2 k T rpk T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 725 1 R 1400 R 10 3 5 rp T T 1 1 0414 1 k k ince the work produced by expansion 34 equals that used b pression 12 an energy balance gives S y com 981 2 R 450 868 8 1400 4 1 2 3 T T T T The excess enthalpy generated by expansion 45 is used to increase the kinetic energy of the flow through the propeller 2 2 2 inlet exit 5 4 V V m T T m c p e p which when solved for the velocity at which the air leaves the propeller gives 9 fts 716 600 fts 1 Btulbm 725 1 R 25037 ft s 20 0 24 Btulbm R981 2 2 1 2 2 1 2 2 2 2 1 2 inlet 5 4 exit V T T c m m V p p e The mass flow rate through the propeller is 2261 lbms 84 ft lbm 20 600 fts 4 10 ft 4 2084 ft lbm 8 psia 450 R 0 3704 psia ft 3 2 1 1 2 1 1 3 3 1 1 1 π π v v v V D AV m P RT p The thrust force generated by this propeller is then 8215 lbf 2 inlet exit lbm fts 32174 1lbf 600fts 2261 lbms7169 V V m F p preparation If you are a student using this Manual you are using it without permission 9105 9140E A turboprop engine operating on an ideal cycle is considered The thrust force generated is to be determined Assumptions 1 Steady operating conditions exist 2 The air standard assumptions are applicable 3 Air is an ideal gas with constant specific heats at room temperature 4 The turbine work output is equal to the compressor work input Properties The properties of air at room temperature are R 03704 psiaft3lbmR Table A1E cp 024 BtulbmR and k 14 Table A2Ea Analysis Working across the two isentropic processes of the cycle yields PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course k T rpk T s T 1 2 4 3 qin 5 qout 2 868 8 R 450 R10 0414 1 1 725 1 R 10 1400 R 1 1 0414 1 3 5 k k rp T T Since the work produced by expansion 34 equals that used by compression 12 an energy balance gives 981 2 R 450 868 8 1400 1 2 3 4 T T T T The mass flow rate through the propeller is 1447 lbms 84 ft lbm 20 600 fts 4 8 ft 4 2084 ft lbm 8 psia 450 R 0 3704 psia ft 3 2 1 1 1 2 1 3 3 1 π π v v v V D AV m P RT p ccording the previous problem A to 113 1 lbms 2261 lbms p e m m 20 20 he excess enthalpy generated by expansion 45 is used to increase the kinetic energy of the flow through the propeller T 2 2 inlet 2 exit 5 4 V V m T T m c p e p which when solved for the velocity at which the air leaves the propeller gives 0 fts 775 600 fts 1 Btulbm 725 1 R 25037 ft s 1447 lbms 0 24 Btulbm R981 2 113 1 lbms 2 2 2 1 2 2 2 2 1 2 inlet 5 4 exit V T T c m m V p p e The thrust force generated by this propeller is then 7870 lbf 2 inlet exit lbm fts 32174 1lbf 600fts 1447 lbms775 V V m F p preparation If you are a student using this Manual you are using it without permission 9106 9141 A turbofan engine operating on an ideal cycle produces 50000 N of thrust The air temperature at the fan outlet needed to produce this thrust is to be determined Assumptions 1 Steady operating conditions exist 2 The air standard assumptions are applicable 3 Air is an ideal gas with constant specific heats at room temperature 4 The turbine work output is equal to the compressor work input Properties The properties of air at room temperature are R 0287 kPam3kgK cp 1005 kJkgK and k 14 Table A 2a Analysis The total mass flow rate is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 676 1 kgs 452 m kg 1 200 ms 4 m 52 4 1 452 m kg 50 kPa 253 K 0 287 kPa m 3 2 1 1 2 1 1 3 3 1 π π v v v V D AV m P RT s T 1 2 4 3 qin 5 qout Now 8451 kgs 8 8 e 676 1 kgs m m The mass flow rate through the fan is In order to produce the specified thrust force the velocity at the fan exit will be 591 6 kgs 8451 676 1 e f m m m exit inlet f V V m F 2845 ms 1N 1kg ms 5916 kgs 50000 N ms 200 2 inlet exit m f F V V An energy balance on the stream passing through the fan gives 2326 K 2 2 2 2 2 inlet 2 exit 4 5 2 inlet 2 exit 5 4 m s 1000 1 kJkg 2 1 005 kJkg K 200 ms 284 5 ms K 253 2 2 p p c V V T T V V T T c preparation If you are a student using this Manual you are using it without permission 9112 9146 Problem 9145 is reconsidered The effect of compressor inlet temperature on the force that must be applied to ed nalysis Using EES the problem is solved as follows K 05 kgs PP1 SSSF First Law for the actual compressor assuming adiabatic kepe0 tant pressure in mdoth3SSSF First Law for the heat exchanger assuming W0 kepe0 SSF First Law for the actual compressor assuming adiabatic kepe0 WdotcDefinition of the net cycle work kW 5 PP5 Ts5 is the isentropic value of T5 at nozzle exit g the brakes to hold the plane stationary is to be investigat A Pratio 9 T1 7 C T1 T1273 P1 95 kPa P5P1 Vel10 ms Vdot1 181 m3s HVfuel 42700 kJkg mdotfuel Etac 10 Etat 10 EtaN 10 Inlet conditions h1ENTHALPYAirTT1 s1ENTROPYAirTT1PP1 v1volumeAirTT1 mdot Vdot1v1 Compressor anaysis ss2s1 For the ideal case the entropies are constant across the compressor PratioP2P1Definition of pressure ratio to find P2 Ts2TEMPERATUREAirsss2PP2 Ts2 is the isentropic value of T2 at compressor exit hs2ENTHALPYAirTTs2 Etac hs2h1h2h1 Compressor adiabatic efficiency Etac WdotcidealWdotcactual mdoth1 Wdotcmdoth2 External heat exchanger analysis P3P2process 23 is SSSF cons h3ENTHALPYAirTT3 Qdotin mdotfuelHVfuel mdoth2 Qdot Turbine analysis s3ENTROPYAirTT3PP3 ss4s3 For the ideal case the entropies are constant across the turbine Pratio P3 P4 Ts4TEMPERATUREAirhhs4 Ts4 is the isentropic value of T4 at turbine exit hs4ENTHALPYAirTTs4 Etat Wdott Wtsdot turbine adiabatic efficiency Wtsdot Wdott Etath3h4h3hs4 mdoth3 Wdott mdoth4 S T4TEMPERATUREAirhh4 P4pressureAirsss4hhs4 Cycle analysis WdotnetWdott Wdotnet 0 kW Exit nozzle analysis s4entropyairTT4PP4 ss5s4 For the ideal case the entropies are constant across the nozzle Ts5TEMPERATUREAirsss hs5ENTHALPYAirTTs5 EtaNh4h5h4hs5 mdoth4 mdoths5 Vels522convertm2s2kJk PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9113 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 22convertm2s2kJkg 5entropyairTT5PP5 l1 N determined only to produce a Ts plot 2entropyairTT2PP2 Force kgs K C mdoth4 mdoth5 Vel5 T5TEMPERATUREAirhh5 s Brake Force to hold the aircraft Thrust mdotVel5 Ve BrakeForce Thrust N The following state points are T2temperatureairhh2 s Brake N m T3 T1 21232 21007 20788 20576 20369 20168 19972 19782 19596 19415 19238 1977 1510 30 2368 2322 2278 2235 2194 2155 2117 208 2045 201 1284 1307 1330 1352 1375 1398 1420 1443 1466 1488 20 15 10 5 0 5 10 15 20 25 20 10 0 10 20 30 19000 19450 19900 20350 20800 21250 T1 C BrakeForce N 45 50 55 60 65 70 75 80 85 0 500 1000 1500 0x100 5x102 103 103 s kJkgK T K Air 855 kPa 95 kPa 1 2s 3 4s 5 preparation If you are a student using this Manual you are using it without permission 9115 SecondLaw Analysis of Gas Power Cycles 9148 The process with the highest exergy destruction for an ideal Otto cycle described in Prob 936 is to be determined Analysis From Prob 936 qin 5825 kJkg qout 2536 kJkg T1 288 K T2 6617 K T3 1473 K and T4 6412 K The exergy destruction during a process of the cycle is sink out source in 0 gen 0 dest T q T q s T T s x v P 4 1 3 2 s T 1 2 4 3 qin qout Application of this equation for each process of the cycle gives dest12 0 isentropic x process 0 5746 kJkg K 0 661 7 K 0 718 kJkg K ln 1473 K ln ln 2 3 2 3 1 4 2 3 v v v R T T c s s s s 5159 kJkg PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course K 1473 5825 kJkg 288 K 05746 kJkg K dest23 q T x isentropic process source in 2 3 0 T s s dest34 0 x 8812 kJkg 288 K 2536 kJkg 05746 kJkg K 288 K sink out 4 1 0 dest4 1 T q s s T x The largest exergy destruction in the cycle occurs during the heatrejection process preparation If you are a student using this Manual you are using it without permission 9116 9149E The exergy destruction associated with the heat rejection process of the Diesel cycle described in Prob 955E and the exergy at the end of the expansion stroke are to be determined Analysis From Prob 955E qout 1589 Btulbm T1 540 R P1 147 psia T4 14206 R P4 3862 psia and v 4 v 1 The entropy change during process 41 is Btulbm R 01728 0 06855 ln14 7 3862 0 83984 60078 0 ln 4 1 14206 R o 4 540R o 1 4 1 P P R s s s s Thus 656 Btulbm 540 R 01728 Btulbm R 540R 4 1 0 destroyed 41 TR s s T x 1589 Btulbm R41 q oting that state 4 is identical to the state of the surroundings the exergy at the end of the power stroke state 4 is etermined from N d 0 4 0 0 4 0 0 4 4 v v P s s T u u φ where 1 4 0 4 out 1 4 0 4 0 158 9 Btulbm R q u u u u v v v v Btulbm R 01741 1 4 0 4 s s s s Thus 656 Btulbm 0 540R 01728 Btulbm R 1589 Btulbm φ4 Discussion Note that the exergy at state 4 is identical to the exergy destruction for the process 41 since state 1 is identical to the dead state and the entire exergy at state 4 is wasted during process 41 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9117 9150 The exergy loss of each process for an ideal dual cycle described in Prob 963 is to be determined Analysis From Prob 963 qinx3 1146 kJkg T1 291 K T2 1037 K Tx 1141 K T3 1255 K and T4 4948 K Also 7467 kJkg 0 718 kJkg K1141 1037K 2 in2 T T c q x x v 146 3 kJkg 291K 0 718 kJkg K494 8 1 4 out T T c q v v P 4 1 2 3 qout x qin The exergy destruction during a process of the cycle is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course sink out source in 0 gen 0 dest T T s T T s x q q Application of this equation for each process of the cycle gives 0 1158 kJkg K 90 kPa 0 287 kJkg K ln 5148 kPa 291 K 005 kJkg K ln 1037 K 1 ln ln 1 2 1 2 1 2 P P R T T c s s p 337 kJkg 291 K01158 kJkg K 1 2 0 dest 1 2 s s T x 0 06862 kJkg K 0 1037 K 0 718 kJkg K ln 1141 K ln ln 2 2 2 v v v x x x R T T c s s 265 kJkg 1255 K 7467 kJkg 291 K 006862 kJkg K source in 2 2 0 dest 2 T q s s T x x x x 0 09571 kJkg K 0 1141 K 1 005 kJkg K ln 1255 K ln ln 3 3 3 x x p x P P R T T c s s 128 kJkg 1255 K 1146 kJkg 291 K 009571 kJkg K source in 3 3 0 dest 3 T q s s T x x x x 0 1339 kJkg K 11 0 287 kJkg K ln 18 1255 K 0 718 kJkg K ln 494 8 K ln ln ln ln 3 4 3 4 3 4 3 4 cr r R T T c R T T c s s v v v v 390 kJkg 291 K01339 kJkg K 3 4 0 dest 3 4 s s T x 0 3811 kJkg K 0 494 8 K 0 718 kJkg K ln 291 K ln ln 4 1 4 1 4 1 v v v R T T c s s 354 kJkg 291 K 1463 kJkg 03811 kJkg K 291 K sink out 4 1 0 dest4 1 T q s s T x preparation If you are a student using this Manual you are using it without permission 9118 9151 The exergy loss of each process for an airstandard Stirling cycle described in Prob 981 is to be determined Analysis From Prob 981 qin 1275 kJkg qout 2125 kJkg T1 T2 1788 K T3 T4 298 K The exergy destruction during a process of the cycle is sink out source in 0 gen 0 dest T q T q s T T s x PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Application of this equation for each process of the cycle gives 0 7132 kJkg K 0 287 kJkg K ln12 0 ln ln 1 2 1 2 1 2 v v v R T T c s s 0 0034 kJkg 1788 K 1275 kJkg 298 K 07132 kJkg K source in 1 2 0 dest 1 2 T q s s T x s T 3 2 qin qout 4 1 0 7132 kJkg K 12 1 0 287 kJkg K ln 0 ln ln 3 4 3 4 3 4 v v v R T T c s s 0 0034 kJkg out q 298 K 2125 kJkg 07132 kJkg K 298 K sink 3 4 0 dest3 4 T s s T x These results are not surprising since Stirling cycle is totally reversible Exergy destructions are not calculated for processes 23 and 41 because there is no interaction with the surroundings during these processes to alter the exergy destruction preparation If you are a student using this Manual you are using it without permission 9120 9153 Exergy analysis is to be used to answer the question in Prob 994 Analysis From Prob 994 T1 288 K T2s 5858 K T2 6189 K T3 873 K T4s 4292 K T4 4736 K rp 12 The exergy change of a flow stream between an inlet and exit state is given by 0 i e i e s s T h h ψ 4s s T 1 2s 4 3 qin qout 873 K 288 K 2 This is also the expression for reversible work Application of this equation for isentropic and actual compression processes gives 0003998 kJkg K 0 0 287 kJkg K ln12 288 K 005 kJkg K ln 585 8 K 1 ln ln 1 2 1 2 1 2 P P R T T c s s s p s 299 2 kJkg 288 K 0 0003998 kJkg K 288K 005 kJkg K585 8 1 1 2s 0 1 2 2 rev1 s s T T T c w s p s 0 05564 kJkg K 0 287 kJkg K ln12 288 K 005 kJkg K ln 618 9 K 1 ln ln 1 2 1 2 1 2 P P R T T c s s p 1 2 0 1 2 2 rev1 s s T T T c w p PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 316 5 kJkg 288 K 0 05564 kJkg K 288K 1 005 kJkg K618 9 inimum work that must be supplied to the compressor by The irreversibilities therefore increase the m 173 kJkg 299 2 316 5 2 rev1 2 rev1 revC s w w w Repeating the calculations for the turbine 0 0003944 kJkg K 0 287 kJkg K ln12 429 2 K 005 kJkg K ln 873 K 1 ln ln 4 3 4 3 4 3 P P R T T c s s s p s v s s p s s s T T T c w 4 3 0 4 3 4 3 re 445 9 kJkg 288 K 0 0003944 kJkg K 429 2 K 1 005 kJkg K873 0 09854 kJkg K 0 287 kJkg K ln12 473 6 K 005 kJkg K ln 873 K 1 ln ln 4 3 4 3 4 3 P P R T T c s s p 429 8 kJkg 0 09854 kJkg K 288 K 473 6 K 005 kJkg K873 1 4 3 0 4 3 4 rev3 s s p s s T T T c w 161 kJkg 429 8 445 9 4 rev 3 4 rev 3 revT w w w s Hence it is clear that the compressor is a little more sensitive to the irreversibilities than the turbine preparation If you are a student using this Manual you are using it without permission 9122 9155 Prob 9154 is reconsidered The effect of the cycle pressure on the total irreversibility for the cycle and the e investigated nalysis Using EES the problem is solved as follows 1 blem 9154 P6 0 estroyed53xdestroyed61 ce state 0 and state 1 are identical Problem 9116 sss2 sss4 enepsilonh4h2 exergy of the exhaust gas leaving the regenerator is to b A Given T1310 K P1100 kPa RatioP7 P2RatioPP T31150 K etaC075 etaT082 epsilon065 TH1500 K T0290 K P0100 kPa Analysis for Pro qinh3h5 qouth6h1 h5h2h4h6 s2entropyFluid PP2 hh2 s4entropyFluid hh4 PP4 s5entropyFluid hh5 PP5 P5P2 s6entropyFluid hh6 P P6P1 h0enthalpyFluid TT0 s0entropyFluid TT0 PP xdestroyed12T0s2s1 xdestroyed34T0s4s3 xdestroyedregenT0s5s2s6s4 xdestroyed53T0s3s5qinTH xdestroyed61T0s1s6qoutT0 xtotalxdestroyed12xdestroyed34xdestroyedregenxd x6h6h0T0s6s0 sin Analysis for Fluidair a h1enthalpyFluid TT1 s1entropyFluid TT1 PP1 sion ss2s1 isentropic compres hs2enthalpyFluid PP2 etaChs2h1h2h1 h3enthalpyFluid TT3 s3entropyFluid TT3 PP3 P3P2 ss4s3 isentropic expansion hs4enthalpyFluid PP4 P4P1 etaTh3h4h3hs4 qreg b wCinh2h1 wTouth3h4 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9123 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Cin inh3h2qregen tathwnetoutqin Ra P k wnetoutwToutw q e tio xtotal Jkg x6 kJkg 6 7 8 9 10 11 12 13 14 3351 3433 1771 1821 2701 280 2899 2995 3088 3178 3266 1372 1435 1496 1555 1611 1666 1719 6 7 8 9 10 11 12 13 14 130 150 170 190 210 230 250 270 290 310 330 350 RatioP x kJkg xtotaldest x6 preparation If you are a student using this Manual you are using it without permission 9124 9156 The exergy loss of each process for a regenerative Brayton cycle with three stages of reheating and intercooling described in Prob 9135 is to be determined Analysis From Prob 9135 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 551 3 K 819 2 K 7 K 520 14 13 9 8 rp 4 qin78 qin910 qin1112 300 kJkg qout141 2069 kJkg qout23 qout45 1416 kJkg T1 T3 T5 290 K T2 T4 T6 4309 K 495 9 K 7 K 585 870 4 K 571 9 K 8 K 849 12 11 10 7 T T T T T he exergy destruction during a process of a stream f m an inlet state to exit state is given by s T T T T T ro out in 0 gen 0 dest q q s s T T s x i e sink source T T Application of this equation for each process of the cycle gives 0 003 kJkg 0 287ln4 290 290 1005ln 4309 ln ln 1 2 1 2 0 dest 5 6 dest 3 4 dest 1 2 P P R T T c T x x x p 321 kJkg 870 4 300 0 5207 290 1005ln 8192 ln ln source 8 in7 7 8 7 8 0 dest 7 8 T q P P R T T c T x p 262 kJkg 870 4 300 0 5513 290 1005ln 8498 ln ln source 10 in9 7 8 9 10 0 dest9 10 T q P P R T T c T x p 225 kJkg 870 4 300 0 5719 290 1005ln 8704 ln ln source 12 in11 11 12 11 12 0 dest 1112 T q P P R T T c T x p 0 005 kJkg 4 0 287ln 1 8192 290 1005ln 5513 ln ln 8 9 8 9 0 dest 89 P P R T T c T x p 0 004 kJkg 4 0 287ln 1 8498 290 1005ln 5719 ln ln 10 11 10 11 0 dest 1011 P P R T T c T x p 0 008 kJkg 4 0 287ln 1 8704 290 1005ln 5857 ln ln 12 13 12 13 0 dest 1213 P P R T T c T x p 506 kJkg 290 206 9 0 4959 290 1005ln 290 ln ln sink 1 out14 14 1 14 1 0 dest 14 1 T q P P R T T c T x p 262 kJkg 290 141 6 0 4309 290 1005ln 290 ln ln sink 3 out2 2 3 2 3 0 dest 4 5 dest 2 3 T q P P R T T c T x x p 666 kJkg 5857 1005ln 4959 4309 290 1005ln 5207 ln ln 13 14 6 7 0 13 14 6 7 0 destregen T T c T T c T s s T x p p 3 4 1 9 8 2 5 10 11 6 7 12 13 14 preparation If you are a student using this Manual you are using it without permission 9125 9157 A gasturbine plant uses diesel fuel and operates on simple Brayton cycle The isentropic efficiency of the compressor the net power output the back work ratio the thermal efficiency and the secondlaw efficiency are to be determined PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at 500ºC 773 K are cp 1093 kJkgK cv 0806 kJkgK R 0287 kJkgK and k 1357 Table A2b Analysis a The isentropic efficiency of the compressor may be determined if we first calculate the exit temperature for the isentropic case 505 6 K 100 kPa 700 kPa K 303 135711357 1 1 2 1 2 k k s P P T T 1 Combustion chamber Turbine 2 3 4 Compress 100 kPa 30C Diesel fuel 700 kPa 260C 0881 303K 533 303K 505 6 1 2 T T T T s ηC 1 2 b The total mass flowing through the turbine and the rate of heat inp are ut 1281 kgs 0 21 kgs 12 6 kgs 60 12 6 kgs AF a a f a t m m m m 12 6 kgs m The temperature at the exit of combustion chamber is 1281 kgs109 3 8555 kJs 3 2 3 in T T T mc Q p The temperature at the turbine exit is determined using isentropic efficiency relation 8555 kW 0 21 kgs42000 kJkg097 HV in c m f q Q η 1144 K 533K kJkgK 3 T 685 7 K 700 kPa 0 kPa 11357 1357 3 3 4 s P 10 K 1144 1 4 k k P T T 754 4 K 685 7 K 1144 K 1144 0 85 4 4 4 3 4 3 T T T T T T s ηT T w he net po er and the back work ratio are 1 2 Cin T T m c W a p 3168 kW 303K 12 6 kgs109 3 kJkgK533 5455 kW 754 4 K 1281 kgs109 3 kJkgK1144 4 3 Tout T T mc W p 2287 kW 3168 5455 Cin Tout net W W W 0581 3168 kW WCin r 5455 kW Tout bw W c The thermal efficiency is 0267 8555 kW kW 2287 in net th Q W η The secondlaw efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency Carnot efficiency The maximum temperature for the cycle can be taken to be the turbine inlet temperature That is 0 735 1144 K 303 K 1 1 3 1 max T T η 0364 0 735 267 0 max th II η η η and preparation If you are a student using this Manual you are using it without permission 9126 9158 A modern compression ignition engine operates on the ideal dual cycle The maximum temperature in the cycle the net work output the thermal efficiency the mean effective pressure the net power output the secondlaw efficiency of the cycle and the rate of exergy of the exhaust gases are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at 1000 K are cp 1142 kJkgK cv 0855 kJkgK R 0287 kJkgK and k 1336 Table A2b Analysis a The clearance volume and the total volume of the engine at the beginning of compression process state 1 are x c c c c d c r V V V V V V V V 2 3 3 0 00012 m 0 0018 m 16 4 3 1 0 0018 0 00012 V V V d c 0 00192 m V Process 1 ntropic compression 1 Qin 2 3 4 P V Qout x 2 Ise 3859 kPa kPa 16 95 870 7 K 16 1336 2 1 1 2 1 1336 2 k P P v v v Process 2x and x3 Constantvolume and constant pressure heat addition processes K 343 1 1 1 2 k T T v 1692 K 3859 kPa 870 7 K 7500 kPa 2 2 P T P T x x 702 6 kJkg 870 7 K 0855 kJkgK1692 2 2 T T c q x x v T T c q q 2308 K 3 3 3 3 1692K 0855 kJkgK 702 6 kJkg T T x p x 2 x b 702 6 702 6 3 2 in x x q q q kJkg 1405 3 3 3 3 0001636 m 0 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1692 K x T 0 00012 m 2308 K x T V V Process 3 isentropic expansion 4 279 4 kPa m 000192 00001636 m kPa 7500 1009 K m 000192 00001636 m K 2308 1336 3 3 4 3 3 4 1 1336 3 3 1 4 3 3 4 k k P P T T V V V V Process 41 constant voume heat rejection 569 3 kJkg 343 K 0855 kJkg K 1009 1 4 out T T c q v The net work output and the thermal efficiency are 8358 kJkg 569 3 1405 out in netout q q w 595 0 5948 1405 kJkg kJkg 8358 in netout th q w η preparation If you are a student using this Manual you are using it without permission 9127 c The mean effective pressure is determined to be 0001853 kg kPa m kg K 343 K 0287 95 kPa000192 m 3 3 1 1 1 RT P m V 8604 kPa 0 0001 0 00192 2 1 V V kJ m kPa m 2 0001853 kg8358kJkg MEP 3 3 mwnetout d The power for engine speed of 3500 rpm is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2839 kW 60 s 0 001853 kg8358 kJkg 2 revcycle net 2 net mw W Note that there are two revolutions in one cycle in fourstroke engines e The secondlaw efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency Carnot efficiency We take the dead state temperature and pressure to be 25ºC and 100 kPa min 1 2200 revmin n 0 8709 2308 K 273 K 25 1 1 3 0 max T T η and 683 0 683 0 8709 5948 0 max th II η η η The rate of exergy of the exhaust gases is determined as follows 285 0 kJkg 100 0 287 kJkgKln 279 4 298 1 142 kJkgKln 1009 298 298 1009 0855 ln ln 0 4 0 4 0 0 4 0 4 0 0 4 4 P P R T T c T T T c s s T u u x p v 9683 kW 60s 1min 2 revcycle 0 001853 kg2850 kJkg 2200 revmin 4 2 4 mx n X preparation If you are a student using this Manual you are using it without permission 9128 Review Problems 9159 An Otto cycle with a compression ratio of 7 is considered The thermal efficiency is to be determined using constant and variable specific heats Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are R 0287 kPam3kgK cp 1005 kJkgK cv 0718 kJkgK and k 14 Table A2a Analysis a Constant specific heats 541 05408 7 1 1 1 1 14 1 1 th rk η v P 4 1 3 2 b Variable specific heats using air properties from Table A17 Process 12 isentropic compression 1 688 20548 kJkg 288 K 1 1 1 r u T v 44762 kJkg 98 3 688 1 7 2 2 1 2 r r r r v v v v 1 1 2 2 u v rocess 2 v constant heat addition 8 K 1273 2 3 3 3 u u q T in v r P 3 55089 kJkg 44762 51 998 045 12 3 99 51 kJkg u Process 34 isentropic expansion 47554 kJkg 8432 712045 4 3 3 3 4 4 u r r r r v v v v v Process 41 v constant heat rejection 27006 kJkg 20548 47554 1 4 out u u q 510 05098 55089 kJkg 27006 kJkg 1 1 in out th q q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9129 9160E An ideal diesel engine with air as the working fluid has a compression ratio of 20 The thermal efficiency is to be determined using constant and variable specific heats Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 0240 BtulbmR cv 0171 BtulbmR R 006855 BtulbmR and k 14 Table A2Ea Analysis a Constant specific heats v P 4 1 2 3 qin qout Process 12 isentropic compression 1673 8 R 505 R20 04 2 1 1 2 T T V 1 k V rocess 23 P constant heat addition P 1 350 16738 R R 2260 2 3 2 3 2 2 2 3 3 T T V 3 T T P P V V V Process 34 isentropic expansion PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 679 06794 1407 Btulbm Btulbm 9559 9559 Btulbm 11 4511 Btulbm 505 R Btulbm R 7688 0171 140 7 Btulbm 8 R 0240 768 8 R 20 2260 R 1350 1 350 350 in netout th 1 4 1 4 out 2 3 2 3 in 04 1 3 4 2 4 q w T T c u u q T T c h h q r T p k η v V V V Variable specific heats using air properties from Table A17 1 1 u T v 1 3 1 3 3 4 T T T k V 1 k 1673 Btulbm R 2260 45 140 7 out in netout q q w b Process 12 isentropic compression 82 170 8606 Btulbm R 505 1 r 01 Btulbm 391 1582 3 R 8 541 20 17082 1 1 v 2 2 1 1 1 2 h T r r r v v v v s 23 P constant heat addition 2 r Proces 1 428 15823 R R 2260 2 3 2 3 2 2 2 3 3 3 T T T P T P v v v v Process 34 isentropic expansion 18651 Btulbm 39101 52 577 922 2 57752 Btulbm R 2260 2 3 in 3 3 3 h h q h T vr 15265 Btulbm 4092 1 428 2 922 20 1 428 1 428 4 3 3 2 4 3 3 4 4 u r r r r r v v v v v v v v Process 41 v constant heat rejection Then 6659 Btulbm 8606 15265 1 4 out u u q 643 06430 18651 Btulbm 6659 Btulbm 1 1 in out th q q η preparation If you are a student using this Manual you are using it without permission 9130 9161E A simple ideal Brayton cycle with air as the working fluid operates between the specified temperature limits The net work is to be determined using constant and variable specific heats Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 0240 BtulbmR and k 14 Table A2Ea Analysis a Constant specific heats PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course k T pr k T 976 3 R 480 R12 0414 1 1 2 s T 1 2 4 3 qin qout 1460 R 480 R 717 8 R 12 1460 R 1 1 0414 1 3 4 k k pr T T 976 3 R 480 717 8 0 240 Btulbm R1460 2 1 4 3 1 2 4 3 T T T T c w w w p p p b Variable specific heats using air properties from Table A17E comp turb net T T c T T c 590 Btulbm 9182 0 11469 Btulbm 480 R 1 1 1 rP h T 23363 Btulbm 1102 12 0 9182 2 1 1 2 r2 h P P P P r 17632 Btulbm 4 12 12 5040 1 40 50 35863 Btulbm R 1460 4 3 3 4 4 3 3 3 h P P P P P h T r r r 634 Btulbm 23363 11469 35863 17632 1 2 4 3 comp turb net h h h h w w w preparation If you are a student using this Manual you are using it without permission 9131 9162 A turbocharged fourstroke V16 diesel engine produces 3500 hp at 1200 rpm The amount of power produced per cylinder per mechanical and per thermodynamic cycle is to be determined Analysis Noting that there are 16 cylinders and each thermodynamic cycle corresponds to 2 mechanical cycles revolutions we have a 816 kJcyl mech cycle 1hp 4241 Btumin 16 cylinders1200 revmin hp 3500 No of cylindersNo of mechanical cycles Total power produced mechanical Btucyl mech cycle 773 w b 1631 kJcyl therm cycle 1hp 4241 Btumin 16 cylinders12002 revmin hp 3500 No of cylindersNo of thermodynamic cycles Total power produced mic thermodyna Btucyl therm cycle 1546 w ified temperature limits is considered The pressure ratio for 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with constant specific heats Properties The specific heat ratio of air is k 14 Table A2 nalysis We treat air as an ideal gas with constant specific heats Using the isentropic relations the temperatures at the compressor and turbine exit can be expressed as 9163 A simple ideal Brayton cycle operating between the spec which the compressor and the turbine exit temperature of air are equal is to be determined Assumptions 1 Steady operating conditions exist A PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course k k p k k k k k P 1 1 3 1 Setting T2 T4 and solving for rp gives s T 1 2 4 3 q k pr T P T T 1 1 1 2 1 2 r T P P T T 3 4 3 4 1 in qout T3 T1 167 1408 1 2 1 3 300 K 1500 K k k p T T r Therefore the compressor and turbine exit temperatures will be equal when the compression ratio is 167 preparation If you are a student using this Manual you are using it without permission 9132 9164 A fourcylinder sparkignition engine with a compression ratio of 8 is considered The amount of heat supplied per cylinder the thermal efficiency and the rpm for a net power output of 60 kW are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable specific heats Properties The gas constant of air is R 0287 kJkgK Table A1 The properties of air are given in Table A17 Analysis a Process 12 isentropic compression PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 29 kJkg 564 5450 572 3 10 5 1 1 r1 572 3 v 22125 kJkg 0 K 31 2 1 2 1 1 1 2 u r u T r r r v v v v v rocess 23 v constant heat addition 1 P 05336 kJ 0287 kPa m kg 1 RT 56429 kJkg kg 17753 10 4406 kg 4406 10 310 K K kPa 00004 m 98 356 2 17753 kJkg 00 K 21 4 2 3 in 4 3 3 1 1 3 3 u m u Q P m u T r V v Process 34 isentropic expansion v P 4 1 3 2 Qin Qout 1800 K 3 b 76405 kJkg 2474 2 356 10 5 4 3 4 3 3 4 u r r r r v v v v v Process 41 v constant heat rejection 552 22125 kJkg kg 76405 10 4406 4 1 4 out u m u Q 0 5517 05336 kJ kJ 02944 02944 kJ 0 2392 5336 0 kJ 02392 in net th out in net Q W Q Q W η c 4586 rpm 1 min 60 s 02944 kJcycle 4 45 kJs 2 revcycle 2 netcyl cyl net W n W n Note that for fourstroke cycles there are two revolutions per cycle preparation If you are a student using this Manual you are using it without permission 9133 9165 Problem 9164 is reconsidered The effect of the compression ratio net work done and the efficiency of the cycle is to be investigated Also the Ts and Pv diagrams for the cycle are to be plotted Analysis Using EES the problem is solved as follows Input Data T137273 K P198 kPa T3 2100 K Vcyl04 LConvertL m3 rv105 Compression ratio Wdotnet 45 kW Ncyl4 number of cyclinders v1v2rv The first part of the solution is done per unit mass Process 12 is isentropic compression s1entropyairTT1PP1 s2s1 s2entropyair TT2 vv2 P2v2T2P1v1T1 P1v1RT1 R0287 kJkgK Conservation of energy for process 1 to 2 no heat transfer sconst with work input win DELTAu12 DELTAu12intenergyairTT2intenergyairTT1 Process 23 is constant volume heat addition s3entropyair TT3 PP3 P3v3T3P2v2T2 P3v3RT3 v3v2 Conservation of energy for process 2 to 3 the work is zero for vconst heat is added qin DELTAu23 DELTAu23intenergyairTT3intenergyairTT2 Process 34 is isentropic expansion s4entropyairTT4PP4 s4s3 P4v4T4P3v3T3 P4v4RT4 Conservation of energy for process 3 to 4 no heat transfer sconst with work output wout DELTAu34 DELTAu34intenergyairTT4intenergyairTT3 Process 41 is constant volume heat rejection v4v1 Conservation of energy for process 2 to 3 the work is zero for vconst heat is rejected qout DELTAu41 DELTAu41intenergyairTT1intenergyairTT4 wnet wout win EtathwnetqinConvert Thermal efficiency in percent The mass contained in each cylinder is found from the volume of the cylinder Vcylmv1 The net work done per cycle is WdotnetmwnetkJcylNcylNdotmechanical cyclesmin1min60s1thermal cycle2mechanical cycles PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9134 rv ηth wnet kJkg 5 6 7 8 9 10 11 42 4555 4839 5074 5273 5444 5594 5683 6019 6257 6429 6555 6646 6712 102 101 100 101 102 101 102 103 104 v m3kg P kPa 310 K 2100 K 1 2 3 4 s const Air Otto Cycle Pv Diagram 40 45 50 55 60 65 70 75 80 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 85 0 500 1000 1500 2000 2500 3000 s kJkgK T K 98 kPa 6971 kPa Air Otto Cycle Ts Diagram 1 2 3 4 v const 5 6 7 8 9 10 11 42 44 46 48 50 52 54 56 rv ηth 5 6 7 8 9 10 11 560 580 600 620 640 660 680 rv wnet kJkg preparation If you are a student using this Manual you are using it without permission 9135 9166 An ideal gas Carnot cycle with helium as the working fluid is considered The pressure ratio compression ratio and minimum temperature of the energy source are to be determined Assumptions 1 Kinetic and potential energy changes are negligible 2 Helium is an ideal gas with constant specific heats Properties The specific heat ratio of helium is k 1667 Table A2a Analysis From the definition of the thermal efficiency of a Carnot heat engine s T 3 2 qin qout 4 1 TH 288 K 576 K 050 1 1 thCarnot thCarnot η H H T An isentropic process for an i 273 K 15 1 η L L T T T deal gas is one in which Pvk remains constant Then the pressure ratio is 565 1 1 288 K T P 1 1 667 1 667 1 2 2 576 K k k T P ased on t ion the compression ratio is B he process equat 283 1 1 667 1 1 2 2 1 5 65 k P P v v 9167E An ideal gas Carnot cycle with helium as the working fluid is con and minimum temperature of the energysource reservoir are to be determ sidered The pressure ratio compression ratio ined an ideal gas with constant specific heats roperties The specific heat ratio of helium is k 1667 Table A2Ea Analysis From the definition of the thermal efficiency of a Carnot heat engine Assumptions 1 Kinetic and potential energy changes are negligible 2 Helium is P s T 3 2 qin qout 4 1 TH 520 R PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1300 R 460 R 60 1 ηthCarnot L T T 060 1 1 thCarnot η L H H T T An isentropic process for an ideal gas is one in which Pvk remains constant Then the pressure ratio is 988 1 667 1 667 1 1 1 2 1 2 520 R 1300 R k k T T P P ased on the process equation the compression ratio is B 395 1 1 667 1 1 2 2 1 9 88 k P P v v preparation If you are a student using this Manual you are using it without permission 9136 9168 The compression ratio required for an ideal Otto cycle to produce certain amount of work when consuming a given amount of fuel is to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats 4 The combustion efficiency is 100 percent Properties The properties of air at room temperature are k 14 Table A2 Analysis The heat input to the cycle for 0043 grams of fuel consumption is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course q m Q v P 4 1 3 2 qout qin 1 505 kJ kg43000 kJkg 0 035 10 3 HV fuel in The thermal efficiency is then 0 6645 1505 kJ kJ 1 in net th Q W η From the definition of thermal efficiency we obtain the required compression ratio to be 153 1 1 41 1 1 th 1 th 0 6645 1 1 1 1 1 1 k k r r η η preparation If you are a student using this Manual you are using it without permission 9137 9169 An ideal Otto cycle with air as the working fluid with a compression ratio of 92 is considered The amount of heat transferred to the air the net work output the thermal efficiency and the mean effective pressure are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable specific heats Properties The gas constant of air is R 0287 kJkgK Table A1 The properties of air are given in Table A17 Analysis a Process 12 isentropic compression v P 4 1 3 2 qin qout 2129 kPa 98 kPa 300 K 7083 K 92 9 kJkg 518 708 3 K 6752 621 2 29 1 1 2 621 21407 kJkg K 300 1 1 2 2 1 2 1 1 1 2 2 2 2 2 1 2 1 1 1 1 2 1 T P T P T P T P u T r u T r r r r v v v v v v v v v v Process 23 v constant heat addition 6098 kJkg 518 9 7 1128 593 8 11287 kJkg 14166 K 2 708 3 2 2 3 3 2 2 2 3 2 3 P T T 3 2 2 3 3 u u q u T P T T P P in r v v v b Process 34 isentropic expansion 3 48775 kJkg 7906 8 593 29 4 3 4 3 3 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4r u r r r v v v v ess 41 v constant heat rejection 1 4 t q q w u u q c v Proc 3361 kJkg 273 7 8 609 2737 kJkg 21407 75 487 out in net ou 551 6098 kJkg kJkg 3361 in net th q w η 429 kPa 1 kJ kPa m 1 m kg 1 192 0879 3361 kJkg 1 1 MEP 0879 m kg 98 kPa kPa m kg K 300 K 0287 3 3 1 net 2 1 net max 2 min 3 3 1 1 1 max r w w r P RT v v v v v v v v d preparation If you are a student using this Manual you are using it without permission 9138 9170 An ideal Otto cycle with air as the working fluid with a compression ratio of 92 is considered The amount of heat transferred to the air the net work output the thermal efficiency and the mean effective pressure are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK and k 14 Table A2a Analysis a Process 12 is isentropic compression 2190 kPa 98 kPa 300 K 7288 K 92 7288 K K 92 300 1 1 2 2 1 2 1 1 1 2 2 2 04 1 2 1 1 2 T P T P T P T P T T k v v v v v v v P 4 1 3 2 qin qout Process 23 v constant heat addition 5233 kJkg 7288 K kJkg K 14576 0718 14576 K 2 728 8 2 2 3 2 3 2 2 2 3 2 3 P T T 3 2 2 3 T T c u u q T P T T P P in v v v b Process 34 isentropic expansion 3 6000 K 92 1 K 14576 04 1 3 k v 4 3 T T v ess 41 v constant heat rejection 4 Proc 3079 kJkg 215 4 3 523 2154 kJkg 300 K kJkg K 600 0718 out in net 1 4 1 4 out q q w T T c u u q v c 588 5233 kJkg kJkg 3079 in net th q w η 393 kPa 1 kJ kPa m 1 m kg 1 192 0879 3079 kJkg 1 1 MEP 0879 m kg 98 kPa kPa m kg K 300 K 0287 3 3 1 net 2 1 net max 2 min 3 3 1 1 1 max r w w r P RT v v v v v v v v d preparation If you are a student using this Manual you are using it without permission 9139 9171E An ideal dual cycle with air as the working fluid with a compression ratio of 12 is considered The thermal efficiency of the cycle is to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 0240 BtulbmR cv 0171 BtulbmR and k 14 Table A2E Analysis The mass of air is 0 003881 lbm psia ft lbm R 580 R 03704 psia 981728 ft 147 3 3 1 1 1 RT P m V PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Process 12 isentropic compression 1667 R 580 R 14 2 1 2 T T V 04 1 1 k V rocess 2x v constant heat addition 2571 R 1667 R R Btulbm x 2 2 in 2 x x x x T T T T mc u m u Q v rocess x P constant heat addition v P 4 1 2 3 11 Btu Qout x 06 Btu P 0003881 lbm 0171 06 Btu P 3 1459 2571 R R 3752 3752 R 2571 R 0003881 lbm 0240 Btulbm R Btu 11 3 3 3 3 3 x c x x x r T T V 3 3 3 3 in 3 x x p x x T T P P T T T T V V V rocess 3 isentropic expansion mc h m h Q P 4 1519 R 14 1459 3752 R 1 459 1 459 04 1 3 1 4 1 3 1 4 3 3 4 k k k r T T T T V V V V Process 41 v constant heat rejection 634 0 6336 17 Btu 06229 Btu 1 1 06229 Btu 580 R lbm 0171 Btulbm R 1519 0003881 in out th 1 4 1 4 out Q Q T T mc u m u Q η v preparation If you are a student using this Manual you are using it without permission 9141 9173 A simple ideal Brayton cycle with air as the working fluid is considered The changes in the net work output per unit mass and the thermal efficiency are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable specific heats Properties The properties of air are given in Table A17 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course For rp 6 s T 1 2 4 3 qin qout 2 3 Analysis The properties at various states are T h P T h P r r 1 1 3 3 1 3 1 386 330 9 300 K 30019 kJ kg 1300 K 139597 kJ kg 37 9 89457 kJkg kJkg 33946 33946 kJkg 55511 89457 55511 kJkg 30019 855 3 89457 kJkg 50140 97 1395 8553 kJkg 5515 330 9 6 1 50140 kJkg 8 316 1 386 6 in net th out in net out w 1 4 2 3 in 4 3 4 2 1 2 3 4 1 2 q w q q h h q h h q h P P P P h P P P P r r r r η rp 12 For 48 5 78537 kJkg kJkg 38096 38096 kJkg 40441 78537 40441 kJkg 30019 704 6 78537 kJkg 61060 97 1395 7046 kJkg 2758 330 9 12 1 6106 kJkg 1663 1 386 12 in net th out in net 1 4 out 2 3 in 4 3 4 2 1 2 3 4 1 2 q w q q w h h q h h q h P P P P h P P P P r r r r η Thus a increase 33946 38096 net 415 kJkg w increase 37 9 48 5 th 106 η b preparation If you are a student using this Manual you are using it without permission 9142 9174 A simple ideal Brayton cycle with air as the working fluid is considered The changes in the net work output per unit mass and the thermal efficiency are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are R 0287 kJkgK cp 1005 kJkgK cv 0718 kJkgK and k 14 Table A2 Analysis Processes 12 and 34 are isentropic Therefore For rp 6 40 1 8034 kJkg kJkg 3219 3219 kJkg 481 5 4 803 4815 kJkg kJkg K 7791 300 K 1005 8034 kJkg 5006 K kJkg K 1300 1005 7791 K 6 1 K 1300 5006 K K 6 300 in net th out in net 1 4 1 4 out 2 3 2 3 in 0414 1 3 4 3 4 0414 1 1 2 1 2 q w q q w T T c h h q T T c h h q P P T T P P T T p p k k k k η s T 1 2 4 3 qin qout 2 3 For rp 12 50 8 6932 kJkg kJkg 3523 3523 kJkg 340 9 2 693 3409 kJkg 300 K kJkg K 6392 1005 6932 kJkg 6102 K kJkg K 1300 1005 6392 K 12 1 K 1300 6102 K K 12 300 in net th out in net 1 4 1 4 out 2 3 2 3 in 0414 1 3 4 3 4 0414 1 1 2 1 2 q w q q w T T c h h q T T c h h q P P T T P P T T p p k k k k η Thus a increase 321 9 352 3 net 304 kJkg w increase 40 1 50 8 th 107 η b PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9143 9175 A regenerative gasturbine engine operating with two stages of compression and two stages of expansion is considered The back work ratio and the thermal efficiency are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK and k 14 Table A2 Analysis The work inputs to each stage of compressor are identical so are the work outputs of each stage of the turbine 7915 kJkg 1006 K 2 1005 kJkg K 1400 2 2 3757 kJkg 300 K 2 1005 kJkg K 4869 2 2 8764 K 486 9 0 75 1006 9 486 1006 K 942 1 0 86 1400 1400 9421 K 4 1 K 1400 4869 K 0 78 300 445 8 300 4458 K K 4 300 7 6 7 6 out T 1 2 1 2 in C 4 9 4 5 4 9 4 5 4 9 4 5 7 6 6 7 9 7 6 7 6 7 6 7 6 0414 1 6 7 6 7 9 1 2 1 2 4 1 2 1 2 1 2 1 2 0414 1 1 2 1 2 4 T T c h h w T T c h h w T T T T T T c T T c h h h h T T T T T T T c T T c h h h h P P T T T T T T T T T T c T T c h h h h P P T T T p p p p s T s p p s T k k s s C s p s p s C k k s s ε ε η η η η s T 3 4 1 6 9 7 8 5 2 9 7s 4 2s Thus 0475 7915 kJkg kJkg 3757 out T Cin bw w w r 451 0 451 9220 kJkg kJkg 4158 4158 kJkg 375 7 5 791 9220 kJkg 1006 K 1400 8764 kJkg K 1400 1005 in net th Cin Tout net 7 8 5 6 7 8 5 6 in q w w w w T T T T c h h h h q p η preparation If you are a student using this Manual you are using it without permission 9144 9176 Problem 9175 is reconsidered The effect of the isentropic efficiencies for the compressor and turbine and regenerator effectiveness on net work done and the heat supplied to the cycle is to be investigated Also the Ts diagram for nalysis Using EES the problem is solved as follows K cy sentropic efficiency ies are constant across the compressor ssor for the isentropic case dyflow T2 PP2 ideal case the entropies are constant across the HP compressor r for the isentropic case dyflow 4 PP4 the cycle is to be plotted A Input data T6 1400 T8 T6 Pratio 4 T1 300 K P1 100 kPa T3 T1 Etareg 075 Regenerator effectiveness tac 078 Compressor isentorpic efficien E Etat 086 Turbine i LP Compressor Isentropic Compressor anaysis s1ENTROPYAirTT1PP1 ss2s1 For the ideal case the entrop P2 PratioP1 ss2ENTROPYAirTTs2PP2 Ts2 is the isentropic value of T2 at compressor exit Etac wcompisenLPwcompLP compressor adiabatic efficiency Wcomp Wcompisen Conservation of energy for the LP compre ein eout DELTAe0 for stea h1 wcompisenLP hs2 h1ENTHALPYAirTT1 hs2ENTHALPYAirTTs2 Actual compressor analysis h1 wcompLP h2 h2ENTHALPYAirTT2 s2ENTROPYAirT HP Compressor s3ENTROPYAirTT3PP3 ss4s3 For the P4 PratioP3 P3 P2 ss4ENTROPYAirTTs4PP4 Ts4 is the isentropic value of T4 at compressor exit Etac wcompisenHPwcompHP compressor adiabatic efficiency Wcomp Wcompisen Conservation of energy for the compresso ein eout DELTAe0 for stea h3 wcompisenHP hs4 h3ENTHALPYAirTT3 hs4ENTHALPYAirTTs4 Actual compressor analysis h3 wcompHP h4 h4ENTHALPYAirTT4 s4ENTROPYAirTT PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9145 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course ntercooling heat loss eat exchanger assuming W0 kepe0 in eout DELTAecv 0 for steady flow 6ENTHALPYAirTT6 s SSSF constant pressure rTT6PP6 s7ENTROPYAirTTs7PP7Ts7 is the isentropic value of T7 at HP turbine exit ic turbine assuming adiabatic kepe0 teadyflow s7 Ts7 7ENTHALPYAirTT7 AirTT7 PP7 7 qinreheat h8 T8 e analysis rTT8PP8 s9ENTROPYAirTTs9PP9Ts9 is the isentropic value of T9 at LP turbine exit n wturb ic turbine assuming adiabatic kepe0 teadyflow 9 Ts9 9ENTHALPYAirTT9 irTT9 PP9 tathnoregwnetqintotalnoregConvert Cycle thermal efficiency atio t added in the external heat exchanger is ALPYAir TT5 5ENTROPYAirTT5 PP5 I h2 qoutintercool h3 External heat exchanger analysis SSSF First Law for the h e h4 qinnoreg h6 h P6P4process 46 i HP Turbine analysis s6ENTROPYAi ss7s6 For the ideal case the entropies are constant across the turbine P7 P6 Pratio s Etat wturbHP wturbisenHP turbine adiabatic efficiency wturbisen wturb SSSF First Law for the isentrop ein eout DELTAecv 0 for s h6 wturbisenHP h hs7ENTHALPYAirT Actual Turbine analysis h6 wturbHP h7 h s7ENTROPY Reheat Qin h h8ENTHALPYAirT HL Turbin P8P7 s8ENTROPYAi ss9s8 For the ideal case the entropies are constant across the turbine P9 P8 Pratio s Etat wturbLP wturbisenLP turbine adiabatic efficiency wturbise SSSF First Law for the isentrop ein eout DELTAecv 0 for s h8 wturbisenLP hs hs9ENTHALPYAirT Actual Turbine analysis h8 wturbLP h9 h s9ENTROPYA Cycle analysis wnetwturbHPwturbLP wcompHP wcompLP qintotalnoregqinnoregqinreheat E BwrwcompHP wcompLPwturbHPwturbLPBack work r With the regenerator the hea h5 qinwithreg h6 h5ENTH s P5P4 preparation If you are a student using this Manual you are using it without permission 9146 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course h4 ives h10 and thus T10 as ALPYAir TT10 10ENTROPYAirTT10 PP10 intotalwithregqinwithregqinreheat g data is used to complete the Array Table for plotting purposes OPYAirTT5PP5 s10s10 s10T10 η qi g qin eg The regenerator effectiveness gives h5 and thus T5 as Etareg h5h4h9 Energy balance on regenerator g h4 h9h5 h10 h10ENTH s P10P9 Cycle thermal efficiency with regenerator q EtathwithregwnetqintotalwithregConvert The followin ss1s1 Ts1T1 ss3s3 Ts3T3 ss5ENTR Ts5T5 ss6s6 Ts6T6 ss8s8 Ts8T8 s T ηreg ηC ηt ηthnoreg thwithreg ntotalnore kJkg totalwithr kJkg wnet kJkg 06 065 07 075 08 085 09 095 1 078 078 086 086 3057 3057 5189 5387 1434 1434 8451 8141 4385 4385 078 078 078 078 078 078 078 086 086 086 086 086 086 086 3057 3057 3057 3057 3057 3057 3057 4129 4253 4385 4525 4675 4834 5006 1434 1434 1434 1434 1434 1434 1434 1062 1031 1000 9691 9381 9071 8761 4385 4385 4385 4385 4385 4385 4385 45 50 55 60 65 70 75 80 85 0 400 800 1200 1600 s kJkgK T K 100 kPa 400 kPa 1600 kPa Air 1 2 3 4 5 6 7 8 9 10 preparation If you are a student using this Manual you are using it without permission 9147 06 065 07 075 08 085 09 095 1 28 32 36 40 44 48 52 ηreg ηth No regeneration With regeneration 06 065 07 075 08 085 09 095 1 10 15 20 25 30 35 40 45 50 55 ηt ηth With regeneration No regeneration 06 065 07 075 08 085 09 095 1 150 200 250 300 350 400 450 500 550 600 ηt wnet kJkg PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9148 06 065 07 075 08 085 09 095 1 800 900 1000 1100 1200 1300 1400 1500 ηt qintotal kJkg With regeneration No regeneration 06 065 07 075 08 085 09 095 1 25 30 35 40 45 50 55 ηc ηth With regeneration No regeneration PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9149 9177 A regenerative gasturbine engine operating with two stages of compression and two stages of expansion is considered The back work ratio and the thermal efficiency are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Helium is an ideal gas with constant specific heats PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The properties of helium at room temperature are cp 51926 kJkgK and k 1667 Table A2 Analysis The work inputs to each stage of compressor are identical so are the work outputs of each stage of the turbine 5323 kJkg 8874 K 2 51926 kJkg K 1400 2 2 2961 kJkg 300 K 2 51926 kJkg K 5852 2 2 8118 K 585 2 0 75 887 4 2 585 8874 K 804 0 0 86 1400 1400 804 0 K 4 1 K 1400 5852 K 0 78 300 522 4 300 5224 K K 4 300 7 6 7 6 out T 1 2 1 2 in C 4 9 4 5 4 9 4 5 4 9 4 5 7 6 6 7 9 7 6 7 6 7 6 7 6 06671667 1 6 7 6 7 9 1 2 1 2 4 1 2 1 2 1 2 1 2 06671667 1 1 2 1 2 4 T T c h h w T T c h h w T T T T T T c T T c h h h h T T T T T T T c T T c h h h h P P T T T T T T T T T T c T T c h h h h P P T T T p p p p s T s p p s T k k s s C s p s p s C k k s s ε ε η η η η s T 3 4 1 6 qin 9 7 8 5 2 9 7s 4 2s Thus 0556 5323 kJkg kJkg 2961 out T Cin bw w w r 413 0 4133 5716 kJkg kJkg 2362 2362 kJkg 2961 5323 5716 kJkg 8874 K 1400 8118 kJkg K 1400 51926 in net th Cin Tout net 7 8 5 6 7 8 5 6 in q w w w w T T T T c h h h h q p η preparation If you are a student using this Manual you are using it without permission 9150 9178 An ideal gasturbine cycle with one stage of compression and two stages of expansion and regeneration is considered The thermal efficiency of the cycle as a function of the compressor pressure ratio and the highpressure turbine to compressor inlet temperature ratio is to be determined and to be compared with the efficiency of the standard regenerative cycle Analysis The Ts diagram of the cycle is as shown in the figure If the overall pressure ratio of the cycle is rp which is the pressure ratio across the compressor then the pressure ratio across each turbine stage in the ideal case becomes rp Using the isentropic relations the temperatures at the compressor and turbine exit can be expressed as k k p k k p k k p k k p k k p k k k k p k k p k k k k p k k T r r T r T r r T P T T T r r T P P T T T T r P P T T T 1 2 1 2 1 1 1 2 1 2 1 5 1 6 5 2 1 3 1 3 1 3 4 3 4 7 1 1 1 1 2 1 2 5 1 1 6 P5 Then 1 1 2 k c T r k T T c h h q 1 1 1 6 1 2 1 3 7 3 7 3 in p p p k k p p p r c T T T c h h q 6 out and thus k k p p r c T q 2 1 3 in th 1 1 1 η which simplifies to k k p p c T r q 1 2 1 out 1 k T rp k T 2 1 3 1 th 1 η The thermal efficiency of the single stage ideal regenerative cycle is given as k T rp k T 1 3 1 th 1 η Therefore the regenerative cycle with two stages of expansion has a higher thermal efficiency than the standard regenerative cycle with a single stage of expansion for any given value of the pressure ratio rp s T 1 2 7 qin 5 6 4 3 qout PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9151 9179 A gasturbine plant operates on the regenerative Brayton cycle with reheating and intercooling The back work ratio the net work output the thermal efficiency the secondlaw efficiency and the exergies at the exits of the combustion chamber and the regenerator are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable specific heats Properties The gas constant of air is R 0287 kJkgK Analysis a For this problem we use the properties from EES software Remember that for an ideal gas enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure Optimum intercooling and reheating pressure is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 346 4 kPa 1001200 4 1 2 P P P Process 12 34 Compression 5 7054 kJkg K kPa 100 K 300 30043 kJkg K 300 1 1 1 1 1 s P T h T h s s s P 42879 kJkg 7054 kJkgK 5 4 kPa 346 2 1 2 2 46088 kJkg 30043 30043 42879 0 80 2 2 1 2 1 2 C h h h h h h s η s T 3 4 1 5 qin 8 6 7 10 9 2 8s 6s 4s 2s 5 5040 kJkg K 4 kPa 346 K 350 35078 kJkg K 350 3 3 3 3 3 s P T h T 4 3 h s s s 50042 kJkg 5040 kJkgK 5 4 1200 kPa P 4 53783 kJkg 35078 35078 50042 0 80 4 4 3 4 3 4 C h h h h h h s η Process 67 89 Expansion h s s s P 6 6514 kJkg K kPa 1200 K 1400 1514 9 kJkg K 1400 6 6 6 6 6 s P T h T 1083 9 kJkg 6514 kJkgK 6 4 kPa 346 7 6 7 7 1170 1 kJkg 1083 9 1514 9 1514 9 0 80 7 7 7 ηT h6 7 6 h h h h h s 8 8 8 s P T h T 96 100 kPa 9 9 h s P 6 9196 kJkg K 4 kPa 346 K 1300 1395 6 kJkg K 1300 8 8 99600 kJkg kJkgK 6 91 8 9 s s preparation If you are a student using this Manual you are using it without permission 9152 1075 9 kJkg 99600 1395 6 1395 6 0 80 9 9 9 8 9 8 T h h h h h h s η C ycle analys is 34750 kJkg 35078 53783 30043 46088 3 4 1 2 Cin h h h h w 66450 kJkg 1075 9 1395 6 1170 1 1514 9 9 8 7 6 Tout h h h h w 0523 66450 50 347 out T Cin bw w w r 3170 kJkg 34750 66450 Cin Tout net w w w Regenerator analysis 67236 kJkg 53783 1075 9 1075 9 0 75 10 10 4 9 10 9 regen h h h h h h ε 10 10 10 s P b h h q 6 5157 kJkg K kPa 100 67236 K h 94140 kJkg 53783 67236 1075 9 5 5 4 5 10 9 regen h h h h h h q 57354 kJkg 94140 1514 9 5 6 in 0553 57354 0 317 in net th q w η c The secondlaw efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency Carnot efficiency The maximum temperature for the cycle can be taken to be the turbine inlet temperature That is 0 786 1400 K 300 K 1 1 6 1 max T T η and 0704 0 786 553 0 ηmax η η th II d The exergies at the combustion chamber exit and the regenerator exit are 9307 kJkg 5 7054kJkgK 300 K 6 6514 30043kJkg 1514 9 0 6 0 0 6 6 s s T h h x 1288 kJkg 5 7054kJkgK 300 K 6 5157 30043kJkg 67236 0 10 0 0 10 10 s s T h h x PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9153 9180 The thermal efficiency of a twostage gas turbine with regeneration reheating and intercooling to that of a three stage gas turbine is to be compared Assumptions 1 The air standard assumptions are applicable 2 Air is an ideal gas with constant specific heats at room temperature 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a Analysis Two Stages s T 3 4 1 6 873 K 283 K 9 7 8 2 5 10 The pressure ratio across each stage is 16 4 pr The temperatures at the end of compression and expansion are c 420 5 K 283 K4 0414 1 min k rpk T T 587 5 K 4 873 K 1 1 0414 1 max k k p e r T T The heat input and heat output are 573 9 kJkg 21005 kJkg K873 587 5 K 2 max in e p T T c q 276 4 kJkg 283 K 21005 kJkg K4205 2 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course min out T T c q e therm of the cycle is then c p Th al efficiency 0518 ηth 1 573 9 in q 276 4 1 qout hree Stages The pressure ratio across each stage is r The temperatures at the end of compression and expansion are T s T 3 4 1 9 8 2 5 10 11 6 7 12 13 14 p 2 520 16 3 1 368 5 K 283 K2520 0414 1 min k k p c r T T 670 4 K 2520 1 873 K 1 1 k k T T 0414 max pr he heat input and heat output are kg e T 31005 kJ 3 max in e p T T c q 610 8 kJkg K873 670 4 K 257 8 kJkg 31005 kJkg K3685 283 K 3 min out T T c q c p The thermal efficiency of the cycle is then 0578 610 8 257 8 1 1 in out th q q η preparation If you are a student using this Manual you are using it without permission 9155 9182 The electricity and the process heat requirements of a manufacturing facility are to be met by a cogeneration plant consisting of a gasturbine and a heat exchanger for steam production The mass flow rate of the air in the cycle the back work ratio the thermal efficiency the rate at which steam is produced in the heat exchanger and the utilization efficiency of the cogeneration plant are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable specific heats 1 Combustion chamber Turbine 2 3 4 Compress 100 kPa 20C 450C 1 MPa 325C 15C Sat vap 200C Heat exchanger 5 Analysis a For this problem we use the properties of air from EES software Remember that for an ideal gas enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure Process 12 Compression 5 682 kJkg K kPa 100 C 20 293 5 kJkg C 20 1 1 1 1 1 s P T h T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course h s s s P 567 2 kJkg 682 kJkgK 5 1000 kPa 2 1 2 2 611 8 kJkg 293 5 293 5 567 2 0 86 2 2 1 2 C h h η 1 2 h h h h s Process 34 Expansion 738 5 kJkg 450 C 4 4 h T s s h h h h h h h 4 3 3 4 3 4 3 T 738 5 0 88 η We cannot find the enthalpy at state 3 directly However using the following lines in EES together with the isentropic efficienc 62 kJkg T3 9132ºC s3 6507 kJkgK The solution by hand would require a trial error app 3entropyAir TT3 PP2 h4senthalpyAir PP1 ss3 The inlet water is compressed liquid at 15ºC and at the saturation pressure of steam at 200ºC 1555 kPa This is not available in the tables but we can obtain it in EES The alternative is to use saturated liquid enthalpy at the given temperature 6447 kJkg 1555 kPa C 15 2 2 1 1 1 w w w h P T The net work output is 1 4 3 out h h w y relation we find h3 12 roach h3enthalpyAir TT3 s Also 605 4 kJkg 325 C 5 5 h T 2792 kJkg 1 200 C 2 w h T x 523 4 kJkg 738 5 1262 318 2 kJkg 293 5 8 61 1 2 Cin h h w T preparation If you are a student using this Manual you are using it without permission 9156 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course C Tout net 205 2 kJkg 318 2 523 4 in The mass flow rate of air is w w w 7311 kgs 2 kJkg 205 wnet ma 1500 kJs net W b The back work ratio is 0608 523 4 2 318 out T Cin bw w w r The rate of heat input and the thermal efficiency are 4753 kW 611 8 kJkg 7 311 kgs1262 2 3 in h h m Q a 316 0 3156 4753 kW kW 1500 in net th Q W η c An energy balance on the heat exchanger gives d The heat supplied to the water in the heat exchanger process heat and the utilization efficiency are 03569 kgs w w w w w a m m h h m h h m 6447kJkg 2792 605 4 kJkg 7 311 kgs7385 1 2 5 4 973 5 kW 6447kJkg 0 3569 kgs2792 1 2 p w w w h h m Q 520 0 5204 4753 kW 973 5 1500 in p net Q Q W u ε preparation If you are a student using this Manual you are using it without permission 9157 9183 A turbojet aircraft flying is considered The pressure of the gases at the turbine exit the mass flow rate of the air through the compressor the velocity of the gases at the nozzle exit the propulsive power and the propulsive efficiency of the cycle are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Potential energy changes are negligible 3 Air is an ideal gas with variable specific heats Properties The gas constant of air is R 0287 kJkgK Table A1 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis a For this problem we use the properties from EES software Remember that for an ideal gas enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure Diffuser Process 12 23823 kJkg 35 C 1 1 h T 26937 kJkg m s 1000 1kJkg 2 ms 15 m s 1000 1kJkg 2 90036 ms 23823 kJkg 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 h h V h V h s T 1 2 4 3 qin 5 qout 6 5 7951 kJkg K kPa 50 2 26937 kJk g 2 2 s P h Compressor Process 23 3 2 h s s s 50519 kJkg 7951 kJkgK 5 3 450 kPa P 3 55350 kJkg 26937 26937 50519 0 83 3 3 2 3 2 3 C h h h h h h s η urbine P cess 34 here the ass flow rates through the compressor and the turbine are assumed equal T ro 1304 8 kJkg 950 C 4 4 h T 1020 6 kJkg 1304 8 26937 55350 5 5 5 4 2 3 h h h h h h w m 96245 kJkg 8 1304 1020 6 1304 8 5 4 h h η 0 83 5 5 5 4 T s s s h h h h 950 C 4 4 T b The mass flow rate of the air through the compressor is 450 kPa 4 s P 7725 kJkg K 6 1474 kPa 5 4 5 5 7725 kJkg K 6 96245 kJkg P s s h s 1760 kgs 26937 kJkg 55350 kJs 500 2 3 C h h W m c Nozzle Process 56 6 kJkg 1020 5 5 s h 6 8336 kJkg K 4 kPa 147 5 P preparation If you are a student using this Manual you are using it without permission 9158 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course h s s s P 70966 kJkg 8336 kJkgK 6 40 kPa 6 5 6 6 76252 kJkg 70966 1020 6 1020 6 0 83 6 6 6 5 6 5 N h h h h h h s η 7185 ms 6 2 2 2 6 2 6 6 2 5 5 m s 1000 1kJkg 2 76252 kJkg 0 1020 6 kJkg 2 2 V V V h V h where the velocity at nozzle inlet is assumed zero d The propulsive power and the propulsive efficiency are 2061 kW 2 2 1 1 6 m s 1000 1kJkg 250 ms250 ms 1 76 kgs718 5 ms V V m V Wp 1322 kW 55350kJkg 1 76 kgs1304 8 3 4 in h m h Q 0156 1322 kW 1 kW 206 in Q W p p η preparation If you are a student using this Manual you are using it without permission 9159 9184 The three processes of an air standard cycle are described The cycle is to be shown on the Pv and Ts diagrams and the expressions for back work ratio and the thermal efficiency are to be obtained Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Analysis a The Pv and Ts diagrams for this cycle are as shown b The work of compression is found by the first law for process 12 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course T 3 2 3 2 2 Pdv P v v R T T The back work ratio is 1 2 1 2 1 2 1 2 1 2 1 2 2 1 1 2 2 1 0 v comp v q w u q isentropic process w u C T w w C T T The expansion work is found by 3 exp 2 3 w w 3 1 3 1 1 2 3 2 1 1 comp v v w C T T T T C T T T T exp 3 2 w R T T R Process 12 is isentropic therefore s T 3 2 1 v P 3 2 1 1 1 2 1 2 1 k T V r 1 k T V 2 1 1 2 P and k k P V r V Process 23 is constant pressure therefore 3 3 3 3 2 2 1 3 2 2 2 2 PV T V PV V r T T T V V Process 31 is constant volume therefore 3 T T 3 3 3 1 1 2 kr T P P The back work ratio becomes CvRk1 PV T P PV P 3 1 1 1 1 1 1 exp 1 1 1 1 k comp k w 1 w k r r 2 3 1 out v T q C T T The cycle thermal efficiency is given by r c Apply first law to the closed system for processes 23 and 31 to show 3 in p q C T 3 1 1 3 1 1 1 1 1 1 v out C T T T T T q η 3 2 2 3 2 1 th in p q C T T k T T T The efficiency becomes preparation If you are a student using this Manual you are using it without permission 9160 1 1 1 1 1 1 k th k r k r r η d Determine the value of the back work ratio and efficiency as r goes to unity 1 1 exp 2 1 1 1 1 1 2 1 1 1 exp 1 lim lim lim lim 1 1 1 1 1 k k k k k r r r r w k r r k r r k kr k r 1 exp 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 lim 1 1 1 1 1 k comp k k k k comp comp r w r w k r r w k r r r w k k w k k k k 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 lim 1 lim 1 lim 1 lim 1 1 1 1 lim 1 1 0 1 1 k th k k k k th k k k k r r r r th r r k r r r r kr k r r k r r k kr k r k k k k k k η η η 2 k These results show that if there is no compression ie r 1 there can be no expansion and no net work will be done even though heat may be added to the system PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9161 9185 The three processes of an air standard cycle are described The cycle is to be shown on the Pv and Ts diagrams and the expressions for back work ratio and the thermal efficiency are to be obtained Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Analysis a The Pv and Ts diagrams for this cycle are as shown b The work of expansion is found by the first law for process 23 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 3 2 3 2 3 2 3 2 3 3 2 exp 2 3 2 3 v v q w u cess w u C T T w w C T T s found by 1 3 3 1 3 v R T T The back work ratio is 2 2 3 0 q isentropic pro The compression work i 3 1 wcomp w Pdv P v 1 3 1 comp v w C T T C 1 3 1 3 3 exp 2 3 3 2 3 2 3 1 1 1 1 v v T T T T T C w R T T R T T T R T T Process 31 is constant pressure therefore 3 3 1 1 1 1 2 T T T V V 3 1 3 3 3 1 PV PV T V V r Process 23 is isentropic therefore k1 k 1 2 2 3 2 k T V r T V and 3 2 3 2 k V P r P V The back work ratio becomes CvRk1 1 1 1 1 comp k r k w r r 1 exp 1 1 1 1 k w k r r c Apply first law to the closed system for processes 12 and 31 to show 1 3 1 t p T q C T T 2 in v q C T ou The cycle thermal efficiency is given by 3 1 1 3 1 2 1 1 2 1 1 1 1 1 1 p out th in v C T T T T T q k q C T T T T T η Process 12 is constant volume therefore 2 2 1 1 2 2 2 2 1 1 1 3 k PV PV T P P r T T T P P The efficiency becomes v P 3 2 1 s T 3 2 1 preparation If you are a student using this Manual you are using it without permission 9162 1 1 1 th k r k r η d Determine the value of the back work ratio and efficiency as r goes to unity PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1 1 exp 1 1 1 1 exp 1 1 1 1 1 1 1 1 1 lim 1 lim 1 lim 1 comp k k comp k k r r r w k r r k w r r r w r k k w r r kr 1 exp 1 lim 1 1 1 comp r w k w k 1 1 1 1 1 1 1 1 1 1 lim 1 lim 1 lim 1 1 lim 1 0 th k th k k r r r th r r k r r k k r k k k η η η r These results show that if there is no compression ie r 1 there can be no expansion and no net work will be done even though heat may be added to the system preparation If you are a student using this Manual you are using it without permission 9163 9186 The four processes of an airstandard cycle are described The cycle is to be shown on the Pv and Ts diagrams an expression for the cycle thermal efficiency is to be obtained and the limit of the efficiency as the volume ratio during heat rejection approaches unity is to be evaluated Analysis a The Pv and Ts diagrams of the cycle are shown in the figures b Apply first law to the closed system for processes 23 and 41 to show 3 2 4 1 in v out p q C T T q C T T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course The cycle thermal efficiency is given by 4 1 1 4 1 3 in v 2 2 3 2 1 1 1 1 1 p out th C T T T T T q k q C T T T T T η s T 4 2 1 3 v P 3 2 1 4 1 1 2 2 1 Process 12 is isentropic therefore 1 1 k k T V T V r Process 34 is isentropic therefore 1 1 3 4 4 3 k ek T V r T V Process 41 is constant pressure therefore 4 4 1 1 4 4 PV PV T V r 4 1 1 1 p T T T V 1 1 3 3 4 1 1 2 4 1 2 1 k k e e p p k T T r T T r r r T T T T r r tant volume and V3 V2 Since process 23 is cons 4 4 4 1 3 2 1 2 e p V V V V r r V V V V r r r k 1 T 3 2 p k p p r r T r The efficiency becomes 1 1 1 1 1 p th k k p r k r r η c In the limit as rp approaches unity the cycle thermal efficiency becomes 1 1 1 1 1 1 1 1 1 1 lim 1 lim 1 lim 1 1 1 1 lim 1 1 p p p p p th k k k k r r r p p th th Otto k k r r k k r r r kr k r k r η η η 1 1 1 preparation If you are a student using this Manual you are using it without permission 9164 9187 The four processes of an airstandard cycle are described The back work ratio and its limit as rp goes to unity are to be determined and the result is to be compared to the expression for the Otto cycle Analysis The work of compression for process 12 is found by the first law PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course T 1 4 T 1 2 1 2 1 2 1 2 1 2 1 2 2 1 1 2 1 2 2 1 0 v comp v q w u q isentropic process w u C T T w w C T The work of compression for process 41 is found by 4 1 4 1 1 4 4 wcomp w Pdv P v v R T 1 The work of expansion for process 34 is found by the first law 4 3 4 0 v v q w u q isentropic process w u C T T C T T The back work ratio is 3 4 3 4 3 4 3 4 3 4 3 4 4 3 exp3 4 3 w w 4 1 2 1 4 1 2 1 1 exp 3 4 3 4 3 1 1 1 comp v v v T T T T w R T T C T T T C w C T T T T T R Using data from the previous problem and Cv Rk1 1 1 exp 3 1 1 1 1 1 1 w T 1 k p comp k k p k r r w T r r 1 1 1 1 1 1 exp 3 3 1 1 1 1 1 1 exp 3 1 1 1 1 1 0 1 lim lim 1 1 1 1 1 lim 1 1 p p p k k p comp r r k k k p k comp r k k r r k r w T T w T T r r r w T r w T r This result is the same expression for the back work ratio for the Otto cycle preparation If you are a student using this Manual you are using it without permission 9165 9188 The effects of compression ratio on the net work output and the thermal efficiency of the Otto cycle for given operating conditions is to be investigated Analysis Using EES the problem is solved as follows Input Data T1300 K P1100 kPa T3 2000 K rcomp 12 Process 12 is isentropic compression s1entropyairTT1PP1 s2s1 T2temperatureair ss2 PP2 P2v2T2P1v1T1 P1v1RT1 R0287 kJkgK V2 V1 rcomp Conservation of energy for process 1 to 2 q12 w12 DELTAu12 q12 0isentropic process DELTAu12intenergyairTT2intenergyairTT1 Process 23 is constant volume heat addition v3v2 s3entropyair TT3 PP3 P3v3RT3 Conservation of energy for process 2 to 3 q23 w23 DELTAu23 w23 0constant volume process DELTAu23intenergyairTT3intenergyairTT2 Process 34 is isentropic expansion s4s3 s4entropyairTT4PP4 P4v4RT4 Conservation of energy for process 3 to 4 q34 w34 DELTAu34 q34 0isentropic process DELTAu34intenergyairTT4intenergyairTT3 Process 41 is constant volume heat rejection V4 V1 Conservation of energy for process 4 to 1 q41 w41 DELTAu41 w41 0 constant volume process DELTAu41intenergyairTT1intenergyairTT4 qintotalq23 qouttotal q41 wnet w12w23w34w41 EtathwnetqintotalConvert Thermal efficiency in percent PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9166 ηth rcomp wnet kJkg 4583 6 5674 4867 7 5893 5103 8 6049 5302 9 6162 5474 10 6243 5624 11 630 5757 12 6338 5875 13 6363 5983 14 6375 608 15 6379 6 7 8 9 10 11 12 13 14 15 560 570 580 590 600 610 620 630 640 rcomp wnet kJkg 6 7 8 9 10 11 12 13 14 15 45 485 52 555 59 625 rcomp ηth PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9167 9189 The effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton c be investigate ycle is to d The pressure ratios at which the net work output and the thermal efficiency are maximum are to be nalysis Using EES the problem is solved as follows P1 SSSF First Law for the actual compressor assuming adiabatic kepe0 ott mdoth4 SSSF First Law for the actual compressor assuming adiabatic kepe0 ycle work kW determined only to produce a Ts plot s4entropyairTT4PP4 determined A Pratio 8 T1 300 K P1 100 kPa T3 1800 K mdot 1 kgs Etac 100100 Etat 100100 Inlet conditions h1ENTHALPYAirTT1 s1ENTROPYAirTT1P Compressor anaysis ss2s1 For the ideal case the entropies are constant across the compressor PratioP2P1Definition of pressure ratio to find P2 Ts2TEMPERATUREAirsss2PP2 Ts2 is the isentropic value of T2 at compressor exit hs2ENTHALPYAirTTs2 Etac hs2h1h2h1 Compressor adiabatic efficiency Etac WdotcidealWdotcactual mdoth1 Wdotcmdoth2 External heat exchanger analysis P3P2process 23 is SSSF constant pressure h3ENTHALPYAirTT3 mdoth2 Qdotin mdoth3SSSF First Law for the heat exchanger assuming W0 kepe0 Turbine analysis s3ENTROPYAirTT3PP3 ss4s3 For the ideal case the entropies are constant across the turbine Pratio P3 P4 Ts4TEMPERATUREAirsss4PP4 Ts4 is the isentropic value of T4 at turbine exit hs4ENTHALPYAirTTs4 Etat Wdott Wtsdot turbine adiabatic efficiency Wtsdot Wdott Etath3h4h3hs4 mdoth3 Wd Cycle analysis WdotnetWdottWdotcDefinition of the net c EtaWdotnetQdotinCycle thermal efficiency BwrWdotcWdott Back work ratio The following state points are T2temperatureairhh2 T4temperatureairhh4 s2entropyairTT2PP2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9168 Bwr η Pratio Wc kW Wnet kW Wt kW Qin kW 0254 03383 5 1758 5163 6921 1526 02665 03689 6 2012 5537 7549 1501 02776 03938 7 2237 5822 8059 1478 02876 04146 8 2441 6045 8485 1458 02968 04324 9 2626 6224 885 1439 03052 04478 10 2797 637 9167 1422 0313 04615 11 2957 649 9447 1406 03203 04736 12 3106 6591 9696 1392 03272 04846 13 3246 6675 9921 1378 03337 04945 14 3378 6747 1013 1364 03398 05036 15 3504 6808 1031 1352 03457 0512 16 3624 6859 1048 1340 03513 05197 17 3739 6903 1064 1328 03567 05269 18 3848 6941 1079 1317 03618 05336 19 3954 6973 1093 1307 03668 05399 20 4055 700 1106 1297 03716 05458 21 4153 7023 1118 1287 03762 05513 22 4247 7043 1129 1277 03806 05566 23 4338 7059 1140 1268 0385 05615 24 4427 7072 1150 1259 03892 05663 25 4512 7083 1160 1251 03932 05707 26 4596 7092 1169 1243 03972 0575 27 4677 7098 1177 1234 0401 05791 28 4755 7103 1186 1227 04048 0583 29 4832 7106 1194 1219 04084 05867 30 4907 7107 1201 1211 0412 05903 31 498 7108 1209 1204 04155 05937 32 5051 7107 1216 1197 04189 0597 33 5121 7104 1223 1190 04222 06002 34 5189 7101 1229 1183 5 10 15 20 25 30 35 500 525 550 575 600 625 650 675 700 725 03 035 04 045 05 055 06 065 Pratio Wnet kW η th PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9169 9190 The effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton cyc is to be investigated assuming adiabatic efficiencies of 85 percent for both the turbine and the comp le ressor The mal efficiency are maximum are to be determined nalysis Using EES the problem is solved as follows w for the actual compressor assuming adiabatic kepe0 ott mdoth4 SSSF First Law for the actual compressor assuming adiabatic kepe0 ycle work kW fficiency determined only to produce a Ts plot s4entropyairTT4PP4 pressure ratios at which the net work output and the ther A Pratio 8 T1 300 K P1 100 kPa T3 1800 K mdot 1 kgs Etac 80100 Etat 80100 Inlet conditions h1ENTHALPYAirTT1 s1ENTROPYAirTT1PP1 Compressor anaysis ss2s1 For the ideal case the entropies are constant across the compressor PratioP2P1Definition of pressure ratio to find P2 Ts2TEMPERATUREAirsss2PP2 Ts2 is the isentropic value of T2 at compressor exit hs2ENTHALPYAirTTs2 Etac hs2h1h2h1 Compressor adiabatic efficiency Etac WdotcidealWdotcactual mdoth1 Wdotcmdoth2 SSSF First La External heat exchanger analysis P3P2process 23 is SSSF constant pressure h3ENTHALPYAirTT3 mdoth2 Qdotin mdoth3SSSF First Law for the heat exchanger assuming W0 kepe0 Turbine analysis s3ENTROPYAirTT3PP3 ss4s3 For the ideal case the entropies are constant across the turbine Pratio P3 P4 Ts4TEMPERATUREAirsss4PP4 Ts4 is the isentropic value of T4 at turbine exit hs4ENTHALPYAirTTs4 Etat Wdott Wtsdot turbine adiabatic efficiency Wtsdot Wdott Etath3h4h3hs4 mdoth3 Wd Cycle analysis f the net c WdotnetWdottWdotcDefinition o EtaWdotnetQdotinCycle thermal e BwrWdotcWdott Back work ratio The following state points are T2temperatureairhh2 T4temperatureairhh4 s2entropyairTT2PP2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9170 Bwr η Pratio Wc kW Wnet kW Wt kW Qin kW 03515 02551 5 2068 3815 5883 1495 03689 02764 6 2367 405 6417 1465 03843 02931 7 2632 4218 685 1439 03981 03068 8 2871 4341 7213 1415 04107 03182 9 309 4433 7522 1393 04224 03278 10 3291 4501 7792 1373 04332 03361 11 3478 4551 803 1354 04433 03432 12 3654 4588 8242 1337 04528 03495 13 3819 4614 8433 1320 04618 0355 14 3975 4632 8606 1305 04704 03599 15 4123 4642 8765 1290 04785 03643 16 4264 4647 8911 1276 04862 03682 17 4398 4647 9046 1262 04937 03717 18 4527 4644 9171 1249 05008 03748 19 4651 4636 9288 1237 05077 03777 20 4771 4626 9397 1225 05143 03802 21 4886 4614 950 1214 05207 03825 22 4997 460 9596 1202 05268 03846 23 5104 4584 9688 1192 05328 03865 24 5208 4566 9774 1181 5 9 13 17 21 25 380 390 400 410 420 430 440 450 460 470 024 026 028 03 032 034 036 038 04 Pratio Wnet kW η th Pratio for W netmax W net η th PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9171 9191 The effects of pressure ratio maximum cycle temperature and compressor and turbine inefficiencies on net work output per unit mass and the thermal efficiency of a simple Bray the ton cycle with air as the working fluid is to be e are to be used nalysis Using EES the problem is solved as follows esultT1P1rcompT3EtathConstPropEtatheasy ycle efficiency is easy 1 1rcompk1Convert ession PP2 v1T1 rocess 1 to 2 TT1 eat addition 2 to 3 nergyairTT2 ion rocess 3 to 4 TT3 n to 1 investigated Constant specific heats at room temperatur A Procedure ConstPropR For Air CV 0718 kJkgK k 14 T2 T1rcompk1 P2 P1rcompk qin23 CVT3T2 T4 T31rcompk1 qout41 CVT4T1 EtathConstProp 1qout41qin23Convert Easy Way to calculate the constant property Otto c The Etath END Input Data T1300 K P1100 kPa T3 1000 K rcomp 12 Process 12 is isentropic compr s1entropyairTT1PP1 s2s1 T2temperatureair ss2 P2v2T2P1 P1v1RT1 R0287 kJkgK V2 V1 rcomp Conservation of energy for p q12 w12 DELTAu12 q12 0isentropic process DELTAu12intenergyairTT2intenergyair Process 23 is constant volume h v3v2 s3entropyair TT3 PP3 P3v3RT3 Conservation of energy for process q23 w23 DELTAu23 w23 0constant volume process DELTAu23intenergyairTT3inte Process 34 is isentropic expans s4s3 s4entropyairTT4PP4 P4v4RT4 Conservation of energy for p q34 w34 DELTAu34 q34 0isentropic process DELTAu34intenergyairTT4intenergyair Process 41 is constant volume heat rejectio V4 V1 Conservation of energy for process 4 q41 w41 DELTAu41 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9172 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course t volume process yairTT1intenergyairTT4 all ConstPropResultT1P1rcompT3EtathConstPropEtatheasy erCentError ABSEtath EtathConstPropEtathConvert PerCentError r η w41 0 constan DELTAu41intenerg qintotalq23 qouttotal q41 wnet w12w23w34w41 EtathwnetqintotalConvert Thermal efficiency in percent C P comp ηth thConstProp ηtheasy T3 K 3604 12 608 6299 6299 1000 6681 12 5904 6299 6299 1500 9421 12 5757 6299 6299 2000 1164 12 5642 6299 6299 2500 6 7 8 9 10 11 12 36 37 38 39 4 41 42 43 rcom p PerCentError Percent Error η th η thConstProp η th Tm ax 1000 K 1000 1200 1400 1600 1800 2000 2200 2400 2600 4 62 84 106 128 15 T3 K PerCentError rcomp 6 12 preparation If you are a student using this Manual you are using it without permission 9173 9192 The effects of pressure ratio maximum cycle temperature and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with air as the working fluid is to be investigated Variable specific heats are to be used Analysis Using EES the problem is solved as follows Input data from diagram window Pratio 8 T1 300 K P1 100 kPa T3 800 K mdot 1 kgs Etac 75100 Etat 82100 Inlet conditions h1ENTHALPYAirTT1 s1ENTROPYAirTT1PP1 Compressor anaysis ss2s1 For the ideal case the entropies are constant across the compressor PratioP2P1Definition of pressure ratio to find P2 Ts2TEMPERATUREAirsss2PP2 Ts2 is the isentropic value of T2 at compressor exit hs2ENTHALPYAirTTs2 Etac hs2h1h2h1 Compressor adiabatic efficiency Etac WdotcidealWdotcactual mdoth1 Wdotcmdoth2 SSSF First Law for the actual compressor assuming adiabatic kepe0 External heat exchanger analysis P3P2process 23 is SSSF constant pressure h3ENTHALPYAirTT3 mdoth2 Qdotin mdoth3SSSF First Law for the heat exchanger assuming W0 kepe0 Turbine analysis s3ENTROPYAirTT3PP3 ss4s3 For the ideal case the entropies are constant across the turbine Pratio P3 P4 Ts4TEMPERATUREAirsss4PP4 Ts4 is the isentropic value of T4 at turbine exit hs4ENTHALPYAirTTs4 Etat Wdott Wtsdot turbine adiabatic efficiency Wtsdot Wdott Etath3h4h3hs4 mdoth3 Wdott mdoth4 SSSF First Law for the actual compressor assuming adiabatic kepe0 Cycle analysis WdotnetWdottWdotcDefinition of the net cycle work kW EtaWdotnetQdotinCycle thermal efficiency BwrWdotcWdott Back work ratio The following state points are determined only to produce a Ts plot T2temperatureairhh2 T4temperatureairhh4 s2entropyairTT2PP2 s4entropyairTT4PP4 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9174 Bwr η Pratio Wc kW Wnet kW Wt kW Qin kW 05229 01 2 1818 1659 3477 16587 06305 01644 4 4033 2364 6396 14373 07038 01814 6 5543 2333 7876 12862 07611 01806 8 6723 2110 8833 11682 08088 01702 10 7705 1822 9527 10700 085 01533 12 8553 1510 10063 9852 08864 0131 14 9304 1192 10496 9102 09192 01041 16 9980 8772 10857 8426 09491 007272 18 10596 5679 11164 7809 09767 003675 20 11165 2661 11431 7241 50 55 60 65 70 75 0 500 1000 1500 s kJkgK T K 100 kPa 800 kPa 1 2s 2 3 4 4s Air Standard Brayton Cycle Pressure ratio 8 and Tmax 1160K 2 4 6 8 10 12 14 16 18 20 000 005 010 015 020 025 0 500 1000 1500 2000 2500 Pratio Cycle efficiency W net kW η W net Tmax1160 K Note Pratio for maximum work and η η c 075 η t 082 η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9175 9193 The effects of pressure ratio maximum cycle temperature and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with helium as the working nalysis Using EES the problem is solved as follows are needed for heliums enthalpy r irTT ELSE hFunc enthalpyHeliumTTPP re the net work done and efficiency 0 Else EtaError m diagram window gs t w for the actual compressor assuming adiabatic kepe0 rst Law for the heat exchanger assuming W0 kepe0 4P4 Etat Wdott Wtsdot turbine adiabatic efficiency Wtsdot mpressor assuming adiabatic kepe0 e work kW l efficiency determined only to produce a Ts plot s4entropyairTT4PP4 fluid is to be investigated A Function hFuncWorkFluidTP The E thus T and P ES functions teat helium as a real gas IF Wo kFluid Air then hFuncenthalpyA endif END Procedure EtaCheckEtathEtaError If Etath 0 then EtaError Why a END Input data fro Pratio 8 T1 300 K P1 100 kPa T3 800 K mdot 1 k Etac 08 Etat 08 WorkFluid Helium Inlet conditions h1hFuncWorkFluidT1P1 s1ENTROPYWorkFluidTT1PP1 Compressor anaysis ss2s1 For the ideal case the entropies are constant across the compressor PratioP2P1Definition of pressure ratio to find P2 Ts2TEMPERATUREWorkFluidsss2PP2 Ts2 is the isentropic value of T2 at compressor exi hs2hFuncWorkFluidTs2P2 Etac hs2h1h2h1 Compressor adiabatic efficiency Etac WdotcidealWdotcactual mdoth1 Wdotcmdoth2 SSSF First La External heat exchanger analysis P3P2process 23 is SSSF constant pressure h3hFuncWorkFluidT3P3 mdoth2 Qdotin mdoth3SSSF Fi Turbine analysis s3ENTROPYWorkFluidTT3PP3 ss4s3 For the ideal case the entropies are constant across the turbine Pratio P3 P4 Ts4TEMPERATUREWorkFluidsss4PP4 Ts4 is the isentropic value of T4 at turbine exit hs4hFuncWorkFluidTs Wdott Etath3h4h3hs4 mdoth3 Wdott mdoth4 SSSF First Law for the actual co Cycle analysis WdotnetWdottWdotcDefinition of the net cycl EtathWdotnetQdotinCycle therma Call EtaCheckEtathEtaError BwrWdotcWdott Back work ratio The following state points are T2temperatureairhh2 T4temperatureairhh4 s2entropyairTT2PP2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9176 Bwr η Pratio Wc kW Wnet kW Wt kW Qin kW 05229 01 2 1818 1659 3477 16587 06305 01644 4 4033 2364 6396 14373 07038 01814 6 5543 2333 7876 12862 07611 01806 8 6723 2110 8833 11682 08088 01702 10 7705 1822 9527 10700 085 01533 12 8553 1510 10063 9852 08864 0131 14 9304 1192 10496 9102 09192 01041 16 9980 8772 10857 8426 09491 007272 18 10596 5679 11164 7809 09767 003675 20 11165 2661 11431 7241 50 55 60 65 70 75 0 500 1000 1500 s kJkgK T K 100 kP a 800 kP a 1 2 s 2 3 4 4 s B rayton C ycle P ressure ratio 8 and T m ax 1160K 2 4 6 8 10 12 14 16 18 20 000 005 010 015 020 025 0 500 1000 1500 2000 2500 Pratio Cycle efficiency Wnet kW η Wnet Tmax1160 K Note Pratio for maximum work and η η c 075 η t 082 η Brayton Cycle using Air m air 20 kgs PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9177 9194 The effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion and air as the working fluid is to be investigated Analysis Using EES the problem is solved as follows Input data for air CP 1005 kJkgK k 14 Nstages is the number of compression and expansion stages Nstages 1 T6 1200 K Pratio 12 T1 300 K P1 100 kPa Etareg 10 regenerator effectiveness Etac 10 Compressor isentorpic efficiency Etat 10 Turbine isentropic efficiency Rp Pratio1Nstages Isentropic Compressor anaysis T2s T1Rpk1k P2 RpP1 T2s is the isentropic value of T2 at compressor exit Etac wcompisenwcomp compressor adiabatic efficiency Wcomp Wcompisen Conservation of energy for the compressor for the isentropic case ein eout DELTAe0 for steadyflow wcompisen CPT2sT1 Actual compressor analysis wcomp CPT2 T1 Since intercooling is assumed to occur such that T3 T1 and the compressors have the same pressure ratio the work input to each compressor is the same The total compressor work is wcomptotal Nstageswcomp External heat exchanger analysis SSSF First Law for the heat exchanger assuming W0 kepe0 ein eout DELTAecv 0 for steady flow The heat added in the external heat exchanger the reheat between turbines is qintotal CPT6 T5 Nstages 1CPT8 T7 Reheat is assumed to occur until T8 T6 Turbine analysis P7 P6 Rp T7s is the isentropic value of T7 at turbine exit T7s T61Rpk1k Turbine adiabatic efficiency wturbisen wturb Etat wturb wturbisen SSSF First Law for the isentropic turbine assuming adiabatic kepe0 ein eout DELTAecv 0 for steadyflow wturbisen CPT6 T7s Actual Turbine analysis wturb CPT6 T7 wturbtotal Nstageswturb Cycle analysis wnetwturbtotalwcomptotal kJkg Bwrwcompwturb Back work ratio PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9178 P4P2 P5P4 P6P5 T4 T2 The regenerator effectiveness gives T5 as Etareg T5 T4T9 T4 T9 T7 Energy balance on regenerator gives T10 as T4 T9T5 T10 Cycle thermal efficiency with regenerator EtathregenerativewnetqintotalConvert The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min temperatures T6 and T1 for this problem EtathEricsson 1 T1T6Convert ηthEricksson ηthRegenerative Nstages 75 4915 1 75 6435 2 75 6832 3 75 7014 4 75 7233 7 75 7379 15 75 7405 19 75 7418 22 0 2 4 6 8 10 12 14 16 18 20 22 24 40 50 60 70 80 Nstages ηth Ericsson Ideal Regenerative Brayton PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9179 9195 The effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion and helium as the working fluid is to be investigated Analysis Using EES the problem is solved as follows Input data for Helium CP 51926 kJkgK k 1667 Nstages is the number of compression and expansion stages Nstages 1 T6 1200 K Pratio 12 T1 300 K P1 100 kPa Etareg 10 regenerator effectiveness Etac 10 Compressor isentorpic efficiency Etat 10 Turbine isentropic efficiency Rp Pratio1Nstages Isentropic Compressor anaysis T2s T1Rpk1k P2 RpP1 T2s is the isentropic value of T2 at compressor exit Etac wcompisenwcomp compressor adiabatic efficiency Wcomp Wcompisen Conservation of energy for the compressor for the isentropic case ein eout DELTAe0 for steadyflow wcompisen CPT2sT1 Actual compressor analysis wcomp CPT2 T1 Since intercooling is assumed to occur such that T3 T1 and the compressors have the same pressure ratio the work input to each compressor is the same The total compressor work is wcomptotal Nstageswcomp External heat exchanger analysis SSSF First Law for the heat exchanger assuming W0 kepe0 ein eout DELTAecv 0 for steady flow The heat added in the external heat exchanger the reheat between turbines is qintotal CPT6 T5 Nstages 1CPT8 T7 Reheat is assumed to occur until T8 T6 Turbine analysis P7 P6 Rp T7s is the isentropic value of T7 at turbine exit T7s T61Rpk1k Turbine adiabatic efficiency wturbisen wturb Etat wturb wturbisen SSSF First Law for the isentropic turbine assuming adiabatic kepe0 ein eout DELTAecv 0 for steadyflow wturbisen CPT6 T7s Actual Turbine analysis wturb CPT6 T7 wturbtotal Nstageswturb Cycle analysis wnetwturbtotalwcomptotal PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9180 Bwrwcompwturb Back work ratio P4P2 P5P4 P6P5 T4 T2 The regenerator effectiveness gives T5 as Etareg T5 T4T9 T4 T9 T7 Energy balance on regenerator gives T10 as T4 T9T5 T10 Cycle thermal efficiency with regenerator EtathregenerativewnetqintotalConvert The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min temperatures T6 and T1 for this problem EtathEricsson 1 T1T6Convert ηthEricksson ηthRegenerative Nstages 75 3243 1 75 589 2 75 6518 3 75 6795 4 75 7118 7 75 7329 15 75 7366 19 75 7384 22 0 2 4 6 8 10 12 14 16 18 20 22 24 30 40 50 60 70 80 Nstages ηth Ericsson Ideal Regenerative Brayton PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9181 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Fundamentals of Engineering FE Exam Problems 9196 An Otto cycle with air as the working fluid has a compression ratio of 104 Under cold air standard conditions the thermal efficiency of this cycle is a 10 b 39 c 61 d 79 e 82 Answer c 61 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values r104 k14 EtaOtto11rk1 Some Wrong Solutions with Common Mistakes W1Eta 1r Taking efficiency to be 1r W2Eta 1rk1 Using incorrect relation W3Eta 11rk11 k11667 Using wrong k value 9197 For specified limits for the maximum and minimum temperatures the ideal cycle with the lowest thermal efficiency is a Carnot b Stirling c Ericsson d Otto e All are the same Answer d Otto 9198 A Carnot cycle operates between the temperatures limits of 300 K and 2000 K and produces 600 kW of net power The rate of entropy change of the working fluid during the heat addition process is a 0 b 0300 kWK c 0353 kWK d 0261 kWK e 20 kWK Answer c 0353 kWK Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TL300 K TH2000 K Wnet600 kJs Wnet THTLDS Some Wrong Solutions with Common Mistakes W1DS WnetTH Using TH instead of THTL W2DS WnetTL Using TL instead of THTL W3DS WnetTHTL Using THTL instead of THTL 9182 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9199 Air in an ideal Diesel cycle is compressed from 2 L to 013 L and then it expands during the constant pressure heat addition process to 030 L Under cold air standard conditions the thermal efficiency of this cycle is a 41 b 59 c 66 d 70 e 78 Answer b 59 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values V12 L V2 013 L V3 030 L W1Eta 11r1k1rck1krc1 r1V1V3 Wrong r value W2Eta 1EtaDiesel Using incorrect relation W3Eta 11rk11rck11k1rc1 k11667 Using wrong k value W4Eta 11rk1 Using Otto cycle efficiency 9200 Helium gas in an ideal Otto cycle is compressed from 20C and 25 L to 025 L and its temperature increases by an additional 700C during the heat addition process The temperature of helium before the expansion process is a 1790C b 2060C c 1240C d 620C e 820C Answer a 1790C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values k1667 V125 V2025 rV1V2 T120273 K T2T1rk1 T3T2700273 C Some Wrong Solutions with Common Mistakes W1T3 T22700273 T22T1rk11 k114 Using wrong k value W2T3 T3273 Using K instead of C W3T3 T1700273 Disregarding temp rise during compression W4T3 T222700273 T222T1273rk1 Using C for T1 instead of K rV1V2 rcV3V2 k14 EtaDiesel11rk1rck1krc1 Some Wrong Solutions with Common Mistakes preparation If you are a student using this Manual you are using it without permission 9183 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9201 In an ideal Otto cycle air is compressed from 120 kgm3 and 22 L to 026 L and the net work output of the cycle is 440 kJkg The mean effective pressure MEP for this cycle is a 612 kPa b 599 kPa c 528 kPa d 416 kPa e 367 kPa Answer b 599 kPa Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values rho1120 kgm3 k14 V122 V2026 mrho1V11000 kg wnet440 kJkg Wtotalmwnet MEPWtotalV1V21000 Some Wrong Solutions with Common Mistakes W1MEP wnetV1V21000 Disregarding mass W2MEP WtotalV11000 Using V1 instead of V1V2 W3MEP rho1V21000wnetV1V21000 Finding mass using V2 instead of V1 W4MEP WtotalV1V21000 Adding V1 and V2 instead of subtracting 9202 In an ideal Brayton cycle air is compressed from 95 kPa and 25C to 1100 kPa Under cold air standard conditions the thermal efficiency of this cycle is a 45 b 50 c 62 d 73 e 86 Answer b 50 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P195 kPa P21100 kPa T125273 K rpP2P1 k14 EtaBrayton11rpk1k Some Wrong Solutions with Common Mistakes W1Eta 1rp Taking efficiency to be 1rp W2Eta 1rpk1k Using incorrect relation W3Eta 11rpk11k1 k11667 Using wrong k value 9184 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9203 Consider an ideal Brayton cycle executed between the pressure limits of 1200 kPa and 100 kPa and temperature limits of 20C and 1000C with argon as the working fluid The net work output of the cycle is a 68 kJkg b 93 kJkg c 158 kJkg d 186 kJkg e 310 kJkg Answer c 158 kJkg Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P1100 kPa P21200 kPa T120273 K T31000273 K rpP2P1 k1667 Cp05203 kJkgK Cv03122 kJkgK T2T1rpk1k qinCpT3T2 EtaBrayton11rpk1k wnetEtaBraytonqin Some Wrong Solutions with Common Mistakes W1wnet 11rpk1kqin1 qin1CvT3T2 Using Cv instead of Cp W2wnet 11rpk1kqin2 qin21005T3T2 Using Cp of air instead of argon W3wnet 11rpk11k1CpT3T22 T22T1rpk11k1 k114 Using k of air instead of argon W4wnet 11rpk1kCpT3T222 T222T1273rpk1k Using C for T1 instead of K 9204 An ideal Brayton cycle has a net work output of 150 kJkg and a backwork ratio of 04 If both the turbine and the compressor had an isentropic efficiency of 85 the net work output of the cycle would be a 74 kJkg b 95 kJkg c 109 kJkg d 128 kJkg e 177 kJkg Answer b 95 kJkg Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values wcompwturb04 wturbwcomp150 kJkg Eff085 wnetEffwturbwcompEff Some Wrong Solutions with Common Mistakes W1wnet EffwturbwcompEff Making a mistake in Wnet relation W2wnet wturbwcompEff Using a wrong relation W3wnet wturbeffwcompEff Using a wrong relation 9185 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9205 In an ideal Brayton cycle air is compressed from 100 kPa and 25C to 1 MPa and then heated to 927C before entering the turbine Under cold air standard conditions the air temperature at the turbine exit is a 349C b 426C c 622C d 733C e 825C Answer a 349C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P1100 kPa P21000 kPa T125273 K T3900273 K rpP2P1 k14 T4T31rpk1k273 Some Wrong Solutions with Common Mistakes W1T4 T3rp Using wrong relation W2T4 T3273rp Using wrong relation W3T4 T4273 Using K instead of C W4T4 T1800273 Disregarding temp rise during compression 9206 In an ideal Brayton cycle with regeneration argon gas is compressed from 100 kPa and 25C to 400 kPa and then heated to 1200C before entering the turbine The highest temperature that argon can be heated in the regenerator is a 246C b 846C c 689C d 368C e 573C Answer e 573C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values k1667 Cp05203 kJkgK P1100 kPa P2400 kPa T125273 K T31200273 K The highest temperature that argon can be heated in the regenerator is the turbine exit temperature rpP2P1 T2T1rpk1k T4T3rpk1k273 Some Wrong Solutions with Common Mistakes W1T4 T3rp Using wrong relation W2T4 T3273rpk1k Using C instead of K for T3 W3T4 T4273 Using K instead of C W4T4 T2273 Taking compressor exit temp as the answer 9186 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9207 In an ideal Brayton cycle with regeneration air is compressed from 80 kPa and 10C to 400 kPa and 175C is heated to 450C in the regenerator and then further heated to 1000C before entering the turbine Under cold air standard conditions the effectiveness of the regenerator is a 33 b 44 c 62 d 77 e 89 Answer d 77 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values k14 Cp1005 kJkgK P180 kPa P2400 kPa T110273 K T2175273 K T31000273 K T5450273 K The highest temperature that the gas can be heated in the regenerator is the turbine exit temperature rpP2P1 T2checkT1rpk1k Checking the given value of T2 It checks T4T3rpk1k EffectiveT5T2T4T2 Some Wrong Solutions with Common Mistakes W1eff T5T2T3T2 Using wrong relation W2eff T5T2T44T2 T44T3273rpk1k Using C instead of K for T3 W3eff T5T2T444T2 T444T3rp Using wrong relation for T4 9208 Consider a gas turbine that has a pressure ratio of 6 and operates on the Brayton cycle with regeneration between the temperature limits of 20C and 900C If the specific heat ratio of the working fluid is 13 the highest thermal efficiency this gas turbine can have is a 38 b 46 c 62 d 58 e 97 Answer c 62 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values k13 rp6 T120273 K T3900273 K Etaregen1T1T3rpk1k Some Wrong Solutions with Common Mistakes W1Eta 1T1273T3273rpk1k Using C for temperatures instead of K W2Eta T1T3rpk1k Using incorrect relation W3Eta 1T1T3rpk11k1 k114 Using wrong k value the one for air preparation If you are a student using this Manual you are using it without permission 101 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 10 VAPOR AND COMBINED POWER CYCLES PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 102 Carnot Vapor Cycle 101C The Carnot cycle is not a realistic model for steam power plants because 1 limiting the heat transfer processes to twophase systems to maintain isothermal conditions severely limits the maximum temperature that can be used in the cycle 2 the turbine will have to handle steam with a high moisture content which causes erosion and 3 it is not practical to design a compressor that will handle two phases 102E A steadyflow Carnot engine with water as the working fluid operates at specified conditions The thermal efficiency the quality at the end of the heat rejection process and the net work output are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a We note that and 155 0 1553 861 R 727 2 R 1 1 727 2 R F 2 267 861 R F 401 C th 40psia sat 25 0 psia sat H L L H T T T T T T η T 40 psia 250 psia 3 2 4 1 qin b Noting that s4 s1 sf 250 psia 056784 BtulbmR 0137 1 2845 0 3921 0 56784 4 4 fg f s s s x s c The enthalpies before and after the heat addition process are 1160 3 Btulbm 0 95 82547 09 376 09 Btulbm 376 2 2 250 psia 1 fg f f x h h h h h Thus and 122 Btulbm 1553 784 2 Btulbm 0 784 2 Btulbm 37609 3 1160 in th net 1 2 in q w h h q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 103 103 A steadyflow Carnot engine with water as the working fluid operates at specified conditions The thermal efficiency the amount of heat rejected and the net work output are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a Noting that TH 250C 523 K and TL Tsat 20 kPa 6006C 3331 K the thermal efficiency becomes 363 0 3632 523 K 3331 K 1 1 thC H L T T η T 20 kPa 3 2 4 1 qin qout b The heat supplied during this cycle is simply the enthalpy of vaporization 250C Thus 10923 kJkg 1715 3 kJkg 523 K K 3331 3 kJkg 1715 in out 250 in q T T q q h q H L L C fg o s c The net work output of this cycle is 6230 kJkg 0 3632 1715 3 kJkg in th net q w η 104 A steadyflow Carnot engine with water as the working fluid operates at specified conditions The thermal efficiency the amount of heat rejected and the net work output are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a Noting that TH 250C 523 K and TL Tsat 10 kPa 4581C 3188 K the thermal efficiency becomes 3904 523 K 3188 K 1 1 th C H L T T η T 10 kPa 3 2 4 1 qin qout b The heat supplied during this cycle is simply the enthalpy of vaporization Thus 10456 kJkg 1715 3 kJkg 523 K K 3188 3 kJkg 1715 in out 250 C in q T T q q h q H L L fg 250C c The net work output of this cycle is s 6697 kJkg 0 3904 1715 3 kJkg in th net q w η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 104 105 A steadyflow Carnot engine with water as the working fluid operates at specified conditions The thermal efficiency the pressure at the turbine inlet and the net work output are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a The thermal efficiency is determined from ηthC 60 273 K 350 273 K 1 1 T T L H 465 T 3 2 4 1 b Note that 350C s2 s3 sf x3sfg 08313 0891 70769 71368 kJkgK Thus 60C 140 MPa Table A6 2 2 2 1368 kJkg K 7 350 C P s T s c The net work can be determined by calculating the enclosed area on the Ts diagram Thus 1623 kJkg 1 5390 7 1368 60 350 Area 1 5390 kJkg K 7 0769 10 8313 0 4 3 net 4 4 s s T T w x s s s L H fg f PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 105 The Simple Rankine Cycle 106C The four processes that make up the simple ideal cycle are 1 Isentropic compression in a pump 2 P constant heat addition in a boiler 3 Isentropic expansion in a turbine and 4 P constant heat rejection in a condenser 107C Heat rejected decreases everything else increases 108C Heat rejected decreases everything else increases 109C The pump work remains the same the moisture content decreases everything else increases 1010C The actual vapor power cycles differ from the idealized ones in that the actual cycles involve friction and pressure drops in various components and the piping and heat loss to the surrounding medium from these components and piping 1011C The boiler exit pressure will be a lower than the boiler inlet pressure in actual cycles and b the same as the boiler inlet pressure in ideal cycles 1012C We would reject this proposal because wturb h1 h2 qout and any heat loss from the steam will adversely affect the turbine work output 1013C Yes because the saturation temperature of steam at 10 kPa is 4581C which is much higher than the temperature of the cooling water PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 106 1014 A simple ideal Rankine cycle with R134a as the working fluid operates between the specified pressure limits The mass flow rate of R134a for a given power production and the thermal efficiency of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the refrigerant tables Tables A11 A12 and A13 6489 kJkg 0 95 6394 0 95 kJkg kPa m 1 1kJ 400kPa 0 0007907 m kg1600 0007907 m kg 0 94 kJkg 63 pin 1 2 3 3 1 2 1 in p 3 04 MPa 1 MPa 40 1 w h h P P w h h f f v v v qin qout 04 MPa 1 3 2 4 16 MPa T s 27321 kJkg MPa 40 0 9875 kJkg K 30507 kJkg 80 C MPa 61 4 3 4 4 3 3 3 3 h s s P s h T P Thus 3091 kJkg 20927 18 240 20927 kJkg 6394 21 273 24018 kJkg 6489 07 305 out in net 1 4 out 2 3 in q q w h h q h h q The mass flow rate of the refrigerant and the thermal efficiency of the cycle are then 2426 kgs 3091 kJkg kJs 750 net net w W m 0129 24018 20927 1 1 in out th q q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 107 1015 A simple ideal Rankine cycle with R134a as the working fluid is considered The turbine inlet temperature the cycle thermal efficiency and the backwork ratio of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the refrigerant tables Tables A11 A12 and A13 6621 kJkg 0 78 6543 0 78 kJkg kPa m 1 1kJ 41489kPa 0 0007930 m kg1400 0007930 m kg 0 43 kJkg 65 89 kPa 414 pin 1 2 3 3 1 2 1 in p 3 10 C 1 10 C 1 sat 10 C 1 w h h P P w h h P P f f v v v qin qout 10C 1 3 2 4 14 MPa T s 0 91295 kJkg K 0 98 0 67356 25286 0 25235 kJkg 0 9819073 6543 98 0 C 10 4 4 4 4 4 4 fg f fg f x s s s x h h h x T 530C 3 3 4 3 3 27691 kJkg 91295 kJkg K 0 kPa 1400 T h s s P Thus 18692 kJkg 6543 35 252 21070 kJkg 6621 91 276 1 4 out 2 3 in h h q h h q The thermal efficiency of the cycle is 0113 21070 18692 1 1 in out th q q η The backwork ratio is determined from 00318 25235 kJkg 27691 78 kJkg 0 4 3 in P out T Pin h h w W w rbw PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 108 1016 A simple ideal Rankine cycle with water as the working fluid is considered The work output from the turbine the heat addition in the boiler and the thermal efficiency of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the steam tables Tables A4 A5 and A6 17618 kJkg 8 65 16753 8 65 kJkg kPa m 1 1kJ 7 385kPa 0 001008 m kg8588 001008 m kg 0 53 kJkg 167 kPa 8588 385 kPa 7 pin 1 2 3 3 1 2 1 in p 3 40 C 1 40 C 1 sat 300 C 2 sat 40 C 1 w h h P P w h h P P P P f f v v v qin qout 40C 1 3 2 4 300C s T 1775 1 kJkg 0 66812406 0 53 167 0 6681 7 6832 0 5724 5 7059 C 40 5 7059 kJkg K 2749 6 kJkg 1 C 300 4 4 4 4 3 4 4 3 3 3 3 fg f fg f x h h h s s s x s s T s h x T Thus 1607 6 kJkg 16753 1 1775 17618 6 2749 1775 1 6 2749 1 4 out 2 3 in 4 3 out T h h q h h q h h w kJkg 25734 kJkg 9745 The thermal efficiency of the cycle is 0375 2573 4 1607 6 1 1 in out th q q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 109 1017E A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits The rates of heat addition and rejection and the thermal efficiency of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the steam tables Tables A4E A5E and A6E 11181 Btulbm 2 40 10940 2 40 Btulbm psia ft 5404 1Btu 3psia 0 01630 ft lbm800 01630 ft lbm 0 40 Btulbm 109 pin 1 2 3 3 1 2 1 in p 3 3 psia 1 3 psia 1 w h h P P w h h f f v v v 97524 Btulbm 0 85491012 8 40 109 0 8549 1 6849 0 2009 1 6413 psia 3 1 6413 Btulbm R 1456 0 Btulbm 900 F psia 800 4 4 4 4 3 4 4 3 3 3 3 fg f fg f x h h h s s s x s s P s h T P qin qout 3 psia 1 3 2 4 800 psia T s Knowing the power output from the turbine the mass flow rate of steam in the cycle is determined from 3 450 lbms 1kJ 094782 Btu 97524Btulbm 14560 1750 kJs 4 3 Tout 4 3 Tout h h W m h m h W The rates of heat addition and rejection are Btus 2987 Btus 4637 10940Btulbm 3 450 lbms97524 11181Btulbm 3 450 lbms145 06 1 4 out 2 3 in h m h Q h m h Q and the thermal efficiency of the cycle is 356 0 3559 4637 2987 1 1 in out th Q Q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1010 1018E A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits The turbine inlet temperature and the thermal efficiency of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible qin qout 5 psia 1 3 2 4 2500 psia T s Analysis From the steam tables Tables A4E A5E and A6E 13776 Btulbm 7 58 13018 7 58 Btulbm psia ft 5404 1Btu 5psia 0 01641 ft lbm2500 01641 ft lbm 0 18 Btulbm 130 pin 1 2 3 3 1 2 1 in p 3 5 psia 1 5 psia 1 w h h P P w h h f f v v v 1 52203 Btulbm R 0 80 1 60894 23488 0 93058 Btulbm 0 801000 5 13018 80 0 psia 5 4 4 4 4 4 4 fg f fg f x s s s x h h h x P 9892F 3 3 4 3 3 1450 8 Btulbm 52203 Btulbm R 1 psia 2500 T h s s P Thus 800 4 Btulbm 58 13018 930 13130 Btulbm 13776 8 1450 1 4 out 2 3 in h h q h h q The thermal efficiency of the cycle is 0390 1313 0 800 4 1 1 in out th q q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1011 1019E A simple steam Rankine cycle operates between the specified pressure limits The mass flow rate the power produced by the turbine the rate of heat addition and the thermal efficiency of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the steam tables Tables A4E A5E and A6E 7718 Btulbm 7 46 6972 7 46 Btulbm psia ft 5404 1Btu 1 psia 0 01614 ft lbm2500 01614 ft lbm 0 72 Btulbm 69 pin 1 2 3 3 1 2 1 in p 3 6 psia 1 1 psia 1 w h h P P w h h f f v v v PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 78770 Btulbm 0 69321035 7 72 69 0 6932 1 84495 0 13262 1 4116 psia 1 1 4116 Btulbm R 1302 0 Btulbm 800 F psia 2500 4 4 4 4 3 4 4 3 3 3 3 fg s f s fg f s h x h h s s s x s s P s h T P qin qout 1 psia 1 3 2 4 2500 psia 4s s T 83913 kJkg 78770 0 901302 0 1302 0 4s 3 T 3 4 4 3 4 3 T h h h h h h h h s η η Thus 45539 Btulbm 76941 8 1224 76941 Btulbm 6972 13 839 1224 8 Btulbm 7718 13020 out in net 1 4 out 2 3 in q q w h h q h h q The mass flow rate of steam in the cycle is determined from lbms 2081 1kJ 094782 Btu 45539 Btulbm kJs 1000 net net net net w W m mw W The power output from the turbine and the rate of heat addition are Btus 2549 kW 1016 2 081 lbms12248 Btulbm 094782 Btu 1kJ 83913Btulbm 2 081 lbms13020 in in 4 3 out T mq Q h m h W and the thermal efficiency of the cycle is 03718 1kJ 094782 Btu 2549 Btus kJs 1000 in net th Q W η preparation If you are a student using this Manual you are using it without permission 1012 1020E A simple steam Rankine cycle operates between the specified pressure limits The mass flow rate the power produced by the turbine the rate of heat addition and the thermal efficiency of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the steam tables Tables A4E A5E and A6E 7718 Btulbm 7 46 6972 7 46 Btulbm psia ft 5404 1Btu 1 psia 0 01614 ft lbm2500 01614 ft lbm 0 72 Btulbm 69 pin 1 2 3 3 1 2 1 in p 3 6 psia 1 1 psia 1 w h h P P w h h f f v v v PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 78770 Btulbm 0 69321035 7 72 69 0 6932 1 84495 0 13262 1 4116 psia 1 1 4116 Btulbm R 1302 0 Btulbm 800 F psia 2500 4 4 4 4 3 4 4 3 3 3 3 fg s f s fg f s h x h h s s s x s s P s h T P qin qout 1 psia 1 3 2 4 2500 psia 4s s T 83913 kJkg 78770 0 901302 0 1302 0 4s 3 T 3 4 4 3 4 3 T h h h h h h h h s η η The mass flow rate of steam in the cycle is determined from 2048 lbms 1kJ 094782 Btu 13020 83913 Btulbm 1000 kJs 4 3 net 4 3 net h h W m h m h W The rate of heat addition is 2508 Btus 094782 Btu 1kJ 7718Btulbm 2 048 lbms13020 2 3 in h m h Q and the thermal efficiency of the cycle is 03779 1kJ 094782 Btu 2508 Btus kJs 1000 in net th Q W η The thermal efficiency in the previous problem was determined to be 03718 The error in the thermal efficiency caused by neglecting the pump work is then 164 100 0 3718 0 3718 0 3779 Error preparation If you are a student using this Manual you are using it without permission 1013 1021 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits The thermal efficiency of the cycle the mass flow rate of the steam and the temperature rise of the cooling water are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a From the steam tables Tables A4 A5 and A6 19887 kJkg 7 06 81 191 kJkg 706 kPa m 1 1 kJ m kg 7000 10 kPa 000101 00101 m kg 0 81 kJkg 191 in 1 2 3 3 1 2 1 in 3 10 kPa 1 10 kPa 1 p p f f w h h P P w h h v v v PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 21536 kJkg 0 8201 2392 1 81 191 0 8201 7 4996 0 6492 6 8000 kPa 10 8000 kJkg K 6 4114 kJkg 3 C 500 MPa 7 4 4 4 4 3 4 4 3 3 3 3 fg f fg f x h h h s s s x s s P s h T P qin qout 10 kPa 1 3 2 4 7 MPa T s Thus 1250 7 kJkg 1961 8 5 3212 1961 8 kJkg 19181 6 2153 3212 5 kJkg 19887 4 3411 out in net 1 4 out 2 3 in q q w h h q h h q and 389 32125 kJkg kJkg 12507 in net th q w η b 36 0 kg s 12507 kJkg kJs 45000 net net w W m c The rate of heat rejection to the cooling water and its temperature rise are 84C C 2000 kgs 418 kJkg 70586 kJs 70586 kJs kgs 19618 kJkg 3598 water cooling out water cooling out out mc Q T mq Q preparation If you are a student using this Manual you are using it without permission 1014 1022 A steam power plant operates on a simple nonideal Rankine cycle between the specified pressure limits The thermal efficiency of the cycle the mass flow rate of the steam and the temperature rise of the cooling water are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a From the steam tables Tables A4 A5 and A6 19992 kJkg 8 11 81 191 kJkg 811 087 kPa m 1 1 kJ 00101 m kg 7000 10 kPa 0 00101 m kg 0 9181 kJkg 1 in 1 2 3 3 1 2 1 in 3 10 kPa 1 10 kPa 1 p p p f f w h h P P w h h η v v v T 2317 1 kJkg 2153 6 0 87 3411 4 4 3411 2153 6 kJkg 0 820 2392 1 81 191 0 8201 7 4996 0 6492 6 8000 kPa 10 8000 kJkg K 6 4 kJkg 3411 C 500 MPa 7 4 3 3 4 4 3 4 3 4 4 4 4 3 4 4 3 3 3 3 s T s T fg f s fg f h h h h h h h h x h h h s s s x s s P s h T P η η 10 kPa 1 3 2 4 7 MPa s qout qin 2 4 Thus 1086 2 kJkg 2125 3 5 3211 2125 3 kJkg 19181 1 2317 3211 5 kJkg 19992 4 3411 out in net 1 4 out 2 3 in q q w h h q h h q and 338 32115 kJkg kJkg 10862 in net th q w η b 41 43 kg s 10862 kJkg kJs 45000 net net w W m c The rate of heat rejection to the cooling water and its temperature rise are 105C C 2000 kgs 418 kJkg 88051 kJs 88051 kJs kgs 21253 kJkg 4143 water cooling out water cooling out out mc Q T mq Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1015 1023 A simple Rankine cycle with water as the working fluid operates between the specified pressure limits The rate of heat addition in the boiler the power input to the pumps the net power and the thermal efficiency of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the steam tables Tables A4 A5 and A6 001026 m kg 0 31403 kJkg 75 C 36 81 3 36 kPa 50 3 75 C 1 75 C 1 sat 50 kPa 1 1 f hf h T T P v v 6 10 kJkg kPa m 1 1kJ 50kPa 0 001026 m kg6000 3 3 1 2 1 in p P P w v PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 32013 kJkg 6 10 31403 pin 1 2 w h h 2336 4 kJkg 0 86602304 7 54 340 0 8660 6 5019 1 0912 6 7219 kPa 50 6 7219 kJkg K 3302 9 kJkg 450 C kPa 6000 4 4 4 4 3 4 4 3 3 3 3 fg s f s fg f s h x h h s s s x s s P s h T P qin qout 50 kPa 1 3 2 4 6 MPa 4s s T 2394 4 kJkg 2336 4 0 943302 9 3302 9 4s 3 T 3 4 4 3 4 3 T h h h h h h h h s η η Thus kW 18050 kW 122 kW 59660 170 122 18 20 kgs610 kJkg 18170 kW 2394 4 kJkg 20 kgs33029 32013kJkg 20 kgs33029 Pin Tout net Pin in P 4 3 out T 2 3 in W W W mw W h m h W h m h Q and 03025 59660 050 18 in net th Q W η preparation If you are a student using this Manual you are using it without permission 1016 1024 The change in the thermal efficiency of the cycle in Prob 1023 due to a pressure drop in the boiler is to be determined Analysis We use the following EES routine to obtain the solution Given P26000 kPa DELTAP50 kPa P36000DELTAP kPa T3450 C P450 kPa EtaT094 DELTATsubcool63 C T1temperatureFluid PP1 xx1DELTATsubcool mdot20 kgs Analysis Fluidsteamiapws P1P4 x10 h1enthalpyFluid PP1 TT1 v1volumeFluid PP1 TT1 wpinv1P2P1 h2h1wpin h3enthalpyFluid PP3 TT3 s3entropyFluid PP3 TT3 s4s3 hs4enthalpyFluid PP4 ss4 EtaTh3h4h3hs4 qinh3h2 qouth4h1 wnetqinqout Etath1qoutqin Solution DELTAP50 kPa DELTATsubcool63 C EtaT094 Etath03022 Fluidsteamiapws h131411 kJkg h232021 kJkg h3330364 kJkg h4239601 kJkg hs423381 kJkg mdot20 kgs P150 P26000 P35950 P450 qin29834 kJkg qout20819 kJkg s367265 kJkgK s467265 kJkgK T17502 C T3450 C v10001026 m3kg wnet9015 kJkg wpin6104 kJkg x10 Discussion The thermal efficiency without a pressure drop was obtained to be 03025 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1017 1025 The net work outputs and the thermal efficiencies for a Carnot cycle and a simple ideal Rankine cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a Rankine cycle analysis From the steam tables Tables A4 A5 and A6 34564 kJkg 5 10 34054 5 10 kJkg kPa m 1 1 kJ 50 kPa 001030 m kg 5000 0 001030 m kg 0 54 kJkg 340 in 1 2 3 3 1 2 1 in 3 50 kPa 1 5 0 kPa 1 p p f f w h h P P w h h v v v T 1 2 3 4 Rankine cycle s 2 kJkg 2071 0 7509 2304 7 54 340 0 7509 6 5019 1 09120 5 9737 kPa 50 9737 kJkg K 5 2 kJkg 2794 1 MPa 5 4 4 4 4 3 4 4 3 3 3 3 fg f fg f x h h h s s s x s s P s h x P 7179 kJkg 1730 7 6 2448 1730 7 kJkg 34054 2 2071 2448 6 kJkg 34564 2 2794 out in net 1 4 out 2 3 in q q w h h q h h q 293 0 2932 2448 6 1730 7 1 1 in out th q q η b Carnot Cycle analysis T 1 2 3 4 Carnot cycle s 98905 kJkg 0 28142304 7 54 340 0 2814 6 5019 1 0912 2 9207 kPa 50 9207 kJkg K 2 5 kJkg 1154 0 C 9 263 C 9 263 2 kJkg 2794 1 MPa 5 1 1 1 1 2 1 1 2 2 2 3 2 3 3 3 3 fg f fg f x h h h s s s x s s P s h x T T T h x P 5575 kJkg 1082 2 7 1639 1082 2 kJkg 34054 2 2071 1639 7 kJkg 1154 5 2 2794 out in net 1 4 out 2 3 in q q w h h q h h q 340 0 3400 1639 7 1082 2 1 1 in out th q q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1018 1026 A 120MW coalfired steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits The overall plant efficiency and the required rate of the coal supply are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a From the steam tables Tables A4 A5 and A6 23505 kJkg 9 11 22594 911 kJkg kPa m 1 1 kJ 1 5 kPa 001014 m kg 9000 0 0010140 m kg 0 94 kJkg 225 in 1 2 3 3 1 2 1 in 3 5 kPa 1 1 15 kPa 1 p p f f w h h P P w h h v v v T 2208 8 kJkg 0 8358 2372 4 94 225 0 8358 7 2522 0 7549 6 8164 kPa 15 8164 kJkg K 6 0 kJkg 3512 C 550 MPa 9 4 4 4 4 3 4 4 3 3 3 3 fg f fg f x h h h s s s x s s P s h T P Qin Qout 15 kPa 1 3 2 4 9 MPa s The thermal efficiency is determined from 1982 9 kJkg 22594 2208 8 3276 9 kJkg 23505 0 3512 1 4 out 2 3 in h h q h h q and Thus 284 0 2843 0 96 0 75 3949 0 0 3949 3276 9 1982 9 1 1 gen comb th overall in out th η η η η η q q b Then the required rate of coal supply becomes and 519 tonsh 14404 kgs 29300 kJkg kJs 422050 422050 kJs 0 2843 kJs 120000 coal in coal overall net in C Q m W Q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1019 1027 A singleflash geothermal power plant uses hot geothermal water at 230ºC as the heat source The mass flow rate of steam through the turbine the isentropic efficiency of the turbine the power output from the turbine and the thermal efficiency of the plant are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a We use properties of water for geothermal water Tables A4 through A6 0 1661 2108 64009 99014 14 kJkg 990 kPa 500 99014 kJkg 0 C 230 2 2 1 2 2 1 1 1 fg f h h h x h h P h x T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course The mass flow rate of steam through the turbine is 3820 kgs 0 1661230 kgs 1 2 3 x m m b Turbine 2344 7 kJkg 0 902392 1 19181 90 0 kPa 10 2160 3 kJkg kPa 10 8207 kJkg K 6 1 kJkg 2748 1 kPa 500 4 4 4 4 4 3 4 4 3 3 3 3 fg f s x h h h x P h s s P s h x P production well 2 reinjection well separator steam turbine 1 Flash chamber 6 5 4 3 condenser 0686 2160 3 2748 1 2344 7 1 2748 4 3 4 3 s T h h h h η c The power output from the turbine is 15410 kW 2344 7 kJkg 3820 kJkg27481 4 3 3 Tout h h m W d We use saturated liquid state at the standard temperature for dead state enthalpy 10483 kJkg 0 25 C 0 0 0 h x T 203622 kW 230 kJkg99014 10483kJkg 0 1 1 in h m h E 76 00757 203622 410 15 in Tout th E W η preparation If you are a student using this Manual you are using it without permission 1020 1028 A doubleflash geothermal power plant uses hot geothermal water at 230ºC as the heat source The temperature of the steam at the exit of the second flash chamber the power produced from the second turbine and the thermal efficiency of the plant are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a We use properties of water for geothermal water Tables A4 through A6 0 1661 14 kJkg 990 kPa 500 99014 kJkg 0 C 230 2 1 2 2 1 1 1 x h h P h x T 19180 kgs 0 1661 230 3820 kgs 0 1661230 kgs 3 1 6 1 2 3 m m m x m m production well reinjection well separator steam turbine 1 3 condenser 4 5 6 Flash chamber Flash chamber 9 separator 8 7 2344 7 kJkg 90 0 kPa 10 2748 1 kJkg 1 kPa 500 4 4 4 3 3 3 h x P h x P 2 2693 1 kJkg 1 kPa 150 0777 0 kPa 150 64009 kJkg 0 kPa 500 8 8 8 7 7 6 7 7 6 6 6 h x P x T h h P h x P 11135 C b The mass flow rate at the lower stage of the turbine is 1490 kgs 0 077719180 kgs 6 7 8 x m m The power outputs from the high and low pressure stages of the turbine are 15410 kW 2344 7 kJkg 3820 kJkg27481 4 3 3 T1out h m h W 5191 kW 2344 7 kJkg 1490 kJkg26931 4 8 8 T2out h m h W c We use saturated liquid state at the standard temperature for the dead state enthalpy 10483 kJkg 0 25 C 0 0 0 h x T 203621 kW 10483kJkg 230 kgs99014 0 1 1 in h m h E 101 0101 203621 5193 410 15 in Tout th E W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1021 1029 A combined flashbinary geothermal power plant uses hot geothermal water at 230ºC as the heat source The mass flow rate of isobutane in the binary cycle the net power outputs from the steam turbine and the binary cycle and the thermal efficiencies for the binary cycle and the combined plant are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a We use properties of water for geothermal water Tables A4 through A6 0 1661 14 kJkg 990 kPa 500 99014 kJkg 0 C 230 2 1 2 2 1 1 1 x h h P h x T 19180 kgs 3820 230 3820 kgs 0 1661230 kgs 3 1 6 1 2 3 m m m x m m 2344 7 kJkg 90 0 kPa 10 2748 1 kJkg 1 kPa 500 4 4 4 3 3 3 h x P h x P steam turbine isobutane turbine heat exchanger pump BINARY CYCLE separator aircooled condenser condenser flash chamber 1 9 1 8 7 5 2 6 3 4 1 reinjection well production well 37704 kJkg 0 C 90 64009 kJkg 0 kPa 500 7 7 7 6 6 6 h x T h x P The isobutane properties are obtained from EES 0 001839 m kg 83 kJkg 270 0 kPa 400 69101 kJkg C 80 kPa 400 75505 kJkg C 145 kPa 3250 3 10 10 10 10 9 9 9 8 8 8 v h x P h T P h T P 27665 kJkg 5 82 27083 82 kJkg 5 0 90 kPa m 1 1 kJ 400 kPa 001819 m kg 3250 0 in 10 11 3 3 10 11 10 in p p p w h h P P w η v An energy balance on the heat exchanger gives 10546 kgs iso iso 11 8 iso 7 6 6 75505 27665kJkg 19181 kgs6400937704kJkg m m h h m h h m b The power outputs from the steam turbine and the binary cycle are 15410 kW 2344 7 kJkg 3819 kJkg27481 4 3 3 Tsteam h m h W 6139 kW 10546 kgs 5 82 kJkg 6753 6753 kW 69101kJkg 10546 kJkg75505 iso Tiso binary net 9 8 iso T in p iso w m W W h h m W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1022 c The thermal efficiencies of the binary cycle and the combined plant are 50454 kW 27665kJkg 10546 kJkg75505 11 8 iso inbinary h h m Q 122 0122 50454 6139 binary in netbinary thbinary Q W η 10483 kJkg 0 25 C 0 0 0 h x T 203622 kW 230 kJkg99014 10483kJkg 0 1 1 in h m h E 106 0106 203622 6139 410 15 in netbinary Tsteam thplant E W W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1023 The Reheat Rankine Cycle 1030C The pump work remains the same the moisture content decreases everything else increases 1031C The Ts diagram shows two reheat cases for the reheat Rankine cycle similar to the one shown in Figure 1011 In the first case there is expansion through the highpressure turbine from 6000 kPa to 4000 kPa between states 1 and 2 with reheat at 4000 kPa to state 3 and finally expansion in the lowpressure turbine to state 4 In the second case there is expansion through the highpressure turbine from 6000 kPa to 500 kPa between states 1 and 5 with reheat at 500 kPa to state 6 and finally expansion in the lowpressure turbine to state 7 Increasing the pressure for reheating increases the average temperature for heat addition makes the energy of the steam more available for doing work see the reheat process 2 to 3 versus the reheat process 5 to 6 Increasing the reheat pressure will increase the cycle efficiency However as the reheating pressure increases the amount of condensation increases during the expansion process in the lowpressure turbine state 4 versus state 7 An optimal pressure for reheating generally allows for the moisture content of the steam at the low pressure turbine exit to be in the range of 10 to 15 and this corresponds to quality in the range of 85 to 90 0 20 40 60 80 100 120 140 160 180 200 300 400 500 600 700 800 900 s kJkmolK T K 6000 kPa 4000 kPa 500 kPa 20 kPa 02 04 06 08 SteamIAPWS 1 2 3 4 5 6 7 1032C The thermal efficiency of the simple ideal Rankine cycle will probably be higher since the average temperature at which heat is added will be higher in this case PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1024 1033 An ideal reheat steam Rankine cycle produces 5000 kW power The rates of heat addition and rejection and the thermal efficiency of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the steam tables Tables A4 A5 and A6 19988 kJkg 8 07 19181 8 07 kJkg kPa m 1 1kJ 0 001010 m kg8000 10kPa 001010 m kg 0 81 kJkg 191 pin 1 2 3 3 1 2 1 in p 3 10 kPa 1 10 kPa 1 w h h P P w h h f f v v v T 2636 4 kJkg 0 94702108 0 09 640 0 9470 4 9603 1 8604 6 5579 kPa 500 6 5579 kJkg K 3273 3 kJkg 450 C kPa 8000 4 4 4 4 3 4 4 3 3 3 3 fg f fg f x h h h s s s x s s P s h T P 1 5 2 6 3 4 500 kPa 10 kPa 8 MPa s 2564 9 kJkg 0 99212392 1 81 191 0 9921 7 4996 0 6492 8 0893 kPa 10 8 0893 kJkg K 3484 5 kJkg 500 C kPa 500 6 6 6 6 5 6 6 5 5 5 5 fg f fg f x h h h s s s x s s P s h T P Thus 1548 5 kJkg 2373 1 5 3921 2373 1 kJkg 19181 9 2564 3921 5 kJkg 2636 4 3484 5 19988 3273 3 out in net 1 6 out 4 5 2 3 in q q w h h q h h h h q The mass flow rate of steam in the cycle is determined from 3229 kgs 15485 kJkg 5000 kJs net net 4 3 net w W m h m h W and the thermal efficiency of the cycle is 0395 3921 5 2373 1 1 1 in out th q q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1025 1034 An ideal reheat steam Rankine cycle produces 2000 kW power The mass flow rate of the steam the rate of heat transfer in the reheater the power used by the pumps and the thermal efficiency of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the steam tables Tables A4 A5 and A6 or EES 43305 kJkg 41751 1554 1554 kJkg kPa m 1 1kJ 0 001043 m kg15000 100kPa 001043 m kg 0 51 kJkg 417 pin 1 2 3 3 1 2 1 in p 3 100 kPa 1 100 kPa 1 w h h P P w h h f f v v v T 2703 3 kJkg 0 94971889 8 47 908 0 9497 3 8923 2 4467 6 1434 kPa 2000 6 1434 kJkg K 3157 9 kJkg 450 C 000 kPa 15 4 4 4 4 3 4 4 3 3 3 3 fg f fg f x h h h s s s x s s P s h T P 1 5 2 6 4 2 MPa 100 kPa 15 MPa 3 s 2648 0 kJkg 0 988022257 5 51 417 0 9880 6 0562 1 3028 7 2866 kPa 100 7 2866 kJkg K 3358 2 kJkg 450 C kPa 2000 6 6 6 6 5 6 6 5 5 5 5 fg f fg f x h h h s s s x s s P s h T P Thus 1149 2 kJkg 2230 5 8 379 2230 5 kJkg 41751 0 2648 3379 8 kJkg 2703 3 3358 2 43305 3157 9 out in net 1 6 out 4 5 2 3 in q q w h h q h h h h q The power produced by the cycle is 2000 kW 174 kgs11492 kJkg net net mw W The rate of heat transfer in the rehetaer is kW 27 kW 1140 740 kgs155 4 kJkg 1 kJkg 3 2703 1 740 kgs33582 Pin in P 4 5 reheater mw W h m h Q and the thermal efficiency of the cycle is 0340 3379 8 2230 5 1 1 in out th q q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1026 1035 A steam power plant that operates on the ideal reheat Rankine cycle is considered The turbine work output and the thermal efficiency of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the steam tables Tables A4 A5 and A6 25750 kJkg 6 08 42 251 608 kJkg kPa m 1 1 kJ 20 kPa 001017 m kg 6000 0 001017 m kg 0 42 kJkg 251 in 1 2 3 3 1 2 1 in 3 20 kPa 1 20 kPa 1 p p f f w h h P P w h h v v v T 2349 7 kJkg 0 8900 2357 5 42 251 0 8900 7 0752 0 8320 7 1292 kPa 20 1292 kJkg K 7 4 kJkg 3248 C 400 MPa 2 2901 0 kJkg MPa 2 5432 kJkg K 6 3 kJkg 3178 C 400 MPa 6 6 6 6 6 5 6 6 5 5 5 5 4 3 4 4 3 3 3 3 fg f fg f x h h h s s s x s s P s h T P h s s P s h T P 1 5 2 6 3 4 20 kPa 6 MPa s The turbine work output and the thermal efficiency are determined from and 3268 kJkg 2901 0 3248 4 25750 3 3178 2349 7 3248 4 2901 0 3 3178 4 5 2 3 in 6 5 4 3 out T h h h h q h h h h w kJkg 1176 Thus 358 0 358 3268 kJkg kJkg 1170 1170 kJkg 6 08 1176 in net th in net q w w w w p out T η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1027 1036 Problem 1035 is reconsidered The problem is to be solved by the diagram window data entry feature of EES by including the effects of the turbine and pump efficiencies and reheat on the steam quality at the lowpressure turbine exit Also the Ts diagram is to be plotted Analysis The problem is solved using EES and the solution is given below Input Data from diagram window P6 20 kPa P3 6000 kPa T3 400 C P4 2000 kPa T5 400 C Etat 100100 Turbine isentropic efficiency Etap 100100 Pump isentropic efficiency Pump analysis function x6x6 this function returns a string to indicate the state of steam at point 6 x6 if x61 then x6superheated if x60 then x6subcooled end FluidSteamIAPWS P1 P6 P2P3 x10 Satd liquid h1enthalpyFluidPP1xx1 v1volumeFluidPP1xx1 s1entropyFluidPP1xx1 T1temperatureFluidPP1xx1 Wpsv1P2P1SSSF isentropic pump work assuming constant specific volume WpWpsEtap h2h1Wp SSSF First Law for the pump v2volumeFluidPP2hh2 s2entropyFluidPP2hh2 T2temperatureFluidPP2hh2 High Pressure Turbine analysis h3enthalpyFluidTT3PP3 s3entropyFluidTT3PP3 v3volumeFluidTT3PP3 ss4s3 hs4enthalpyFluidsss4PP4 Ts4temperatureFluidsss4PP4 Etath3h4h3hs4Definition of turbine efficiency T4temperatureFluidPP4hh4 s4entropyFluidTT4PP4 v4volumeFluidss4PP4 h3 Wthph4SSSF First Law for the high pressure turbine Low Pressure Turbine analysis P5P4 s5entropyFluidTT5PP5 h5enthalpyFluidTT5PP5 ss6s5 hs6enthalpyFluidsss6PP6 Ts6temperatureFluidsss6PP6 vs6volumeFluidsss6PP6 Etath5h6h5hs6Definition of turbine efficiency PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course h5Wtlph6SSSF First Law for the low pressure turbine preparation If you are a student using this Manual you are using it without permission 1028 x6QUALITYFluidhh6PP6 Boiler analysis Qin h2h4h3h5SSSF First Law for the Boiler Condenser analysis h6Qouth1SSSF First Law for the Condenser T6temperatureFluidhh6PP6 s6entropyFluidhh6PP6 x6sx6x6 Cycle Statistics WnetWthpWtlpWp EffWnetQin 00 10 20 30 40 50 60 70 80 90 100 0 100 200 300 400 500 600 700 s kJkgK T C 6000 kPa 2000 kPa 20 kPa SteamIAPWS 12 3 4 5 6 Ideal Rankine cycle with reheat SOLUTION Eff0358 Etap1 Etat1 FluidSteamIAPWS Qin3268 kJkg Qout2098 kJkg Wnet1170 kJkg Wp6083 kJkg Wps6083 kJkg Wthp2772 kJkg Wtlp8987 kJkg x6s PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1029 1037E An ideal reheat steam Rankine cycle produces 5000 kW power The rates of heat addition and rejection and the thermal efficiency of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the steam tables Tables A4E A5E and A6E or EES 16306 Btulbm 1 81 16125 1 81 Btulbm psia ft 5404 1Btu 0 01659 ft lbm600 10psia 01659 ft lbm 0 25 Btulbm 161 pin 1 2 3 3 1 2 1 in p 3 10 psia 1 10 psia 1 w h h P P w h h f f v v v T 1187 5 Btulbm 0 986584333 46 355 0 9865 1 00219 0 54379 1 5325 psia 200 1 5325 Btulbm R 1289 9 Btulbm 600 F psia 600 4 4 4 4 3 4 4 3 3 3 3 fg f fg f x h h h s s s x s s P s h T P 1 5 2 6 4 200 psia 10 psia 600 psia 3 s 1071 0 Btulbm 0 926698182 25 161 0 9266 1 50391 0 28362 1 6771 psia 10 1 6771 Btulbm R 1322 3 Btulbm 600 F psia 200 6 6 4 6 5 6 6 5 5 5 5 fg f fg f x h h h s s s x s s P s h T P Thus 352 0 Btulbm 909 8 7 1261 909 7 Btulbm 16125 0 1071 1261 7 Btulbm 1187 5 12899 16306 1322 3 out in net 1 6 out 4 5 2 3 in q q w h h q h h h h q The mass flow rate of steam in the cycle is determined from 1347 lbms 1kJ 094782 Btu 3520 Btulbm kJs 5000 net net net net w W m mw W The rates of heat addition and rejection are Btus 12250 Btus 16995 1347 lbms9097 Btulbm 1347 lbms12617 Btulbm out out in in mq Q mq Q and the thermal efficiency of the cycle is 02790 1kJ 094782 Btu 16990 Btus kJs 5000 in net th Q W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1030 1038E An ideal reheat steam Rankine cycle produces 5000 kW power The rates of heat addition and rejection and the thermal efficiency of the cycle are to be determined for a reheat pressure of 100 psia Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the steam tables Tables A4E A5E and A6E or EES 16306 Btulbm 1 81 16125 1 81 Btulbm psia ft 5404 1Btu 0 01659 ft lbm600 10psia 01659 ft lbm 0 25 Btulbm 161 pin 1 2 3 3 1 2 1 in p 3 6 psia 1 10 psia 1 w h h P P w h h f f v v v T 1131 9 Btulbm 0 937488899 51 298 0 9374 1 12888 0 47427 1 5325 psia 100 1 5325 Btulbm R 1289 9 Btulbm 600 F psia 600 4 4 4 4 3 4 4 3 3 3 3 fg f fg f x h h h s s s x s s P s h T P 1 5 2 6 4 100 psia 10 psia 600 psia 3 s 1124 2 Btulbm 0 980898182 25 161 0 9808 1 50391 0 28362 1 7586 psia 10 1 7586 Btulbm R 1329 4 Btulbm 600 F psia 100 6 6 6 6 5 6 6 5 5 5 5 fg f fg f x h h h s s s x s s P s h T P Thus 361 5 Btulbm 962 9 4 1324 962 9 Btulbm 16125 2 1124 1324 4 Btulbm 1131 9 1329 4 12899 16307 out in net 1 6 out 4 5 2 3 in q q w h h q h h h h q The mass flow rate of steam in the cycle is determined from 1311 lbms 1kJ 094782 Btu 3615 Btulbm kJs 5000 net net net net w W m mw W The rates of heat addition and rejection are Btus 12620 Btus 17360 1311 lbms9629 Btulbm 1311 lbms13244 Btulbm out out in in mq Q mq Q and the thermal efficiency of the cycle is 02729 1kJ 094782 Btu 17360 Btus kJs 5000 in net th Q W η Discussion The thermal efficiency for 200 psia reheat pressure was determined in the previous problem to be 02790 Thus operating the reheater at 100 psia causes a slight decrease in the thermal efficiency PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1031 1039 An ideal reheat Rankine with water as the working fluid is considered The temperatures at the inlet of both turbines and the thermal efficiency of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis From the steam tables Tables A4 A5 and A6 19887 kJkg 7 06 19181 7 06 kJkg kPa m 1 1kJ 10kPa 0 001010 m kg7000 001010 m kg 0 81 kJkg 191 pin 1 2 3 3 1 2 1 in p 3 10 kPa 1 10 kPa 1 w h h P P w h h f f v v v 3733C 3 3 4 3 3 4 4 4 4 4 4 3085 5 kJkg kPa 7000 6 3385 kJkg K 0 93 4 6160 2 0457 2625 0 kJkg 0 932047 5 72087 0 93 kPa 800 T h s s P x s s s x h h h x P fg f fg f 1 5 2 6 3 4 800 kPa 10 kPa 7 MPa s 4162C 5 5 6 5 5 6 6 6 6 6 6 3302 0 kJkg kPa 800 7 6239 kJkg K 0 93 7 4996 0 6492 2416 4 kJkg 0 932392 1 19181 0 90 kPa 10 T h s s P x s s s x h h h x P fg f fg f Thus 2224 6 kJkg 19181 4 2416 3563 6 kJkg 2625 0 3302 0 19887 3085 5 1 6 out 4 5 2 3 in h h q h h h h q and 376 0 3757 3563 6 2224 6 1 1 in out th q q η preparation If you are a student using this Manual you are using it without permission 1032 1040 A steam power plant that operates on an ideal reheat Rankine cycle between the specified pressure limits is considered The pressure at which reheating takes place the total rate of heat input in the boiler and the thermal efficiency of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a From the steam tables Tables A4 A5 and A6 T 20695 kJkg 81 1514 191 1514 kJkg kPa m 1 1 kJ 00101 m kg 15000 10 kPa 0 00101 m kg 0 81 kJkg 191 in 1 2 3 3 1 2 1 in 3 sat 10 kPa 1 sat 10 kPa 1 p p w h h P P w h h v v v 2817 2 kJkg 15 MPa 2 61 kJkg 3466 the reheat pressure C 500 7 3988 kJkg K 7 4996 0 90 6492 0 2344 7 kJkg 0 90 2392 1 19181 kPa 10 3480 kJkg K 6 8 kJkg 3310 C 500 MPa 15 4 3 4 4 5 5 6 5 5 6 6 6 6 5 6 6 3 3 3 3 h s s P h P s s T x s s s x h h h s s P s h T P fg f fg f kPa 2150 1 5 2 6 3 4 10 kPa 15 s b The rate of heat supply is 45039 kW 2817 2 kJkg 346661 20695 kgs 3310 8 12 4 5 2 3 in h h h m h Q c The thermal efficiency is determined from Thus 426 45039 kJs 25834 kJs 1 1 25835 kJs 19181 kJkg kJs 2344 7 12 in out th 1 6 out Q Q h m h Q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1033 1041 A steam power plant that operates on a reheat Rankine cycle is considered The condenser pressure the net power output and the thermal efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a From the steam tables Tables A4 A5 and A6 Eq 3 0 85 3358 2 2 3358 Eq 2 Eq 1 95 0 2815 kJkg K 7 2 kJkg 3358 C 450 MPa 2 3027 3 kJkg 2948 1 0 85 3476 5 5 3476 2948 1 kJkg MPa 2 6317 kJkg K 6 5 kJkg 3476 C 550 5 MPa 12 6 6 5 5 6 6 5 6 5 6 5 6 6 6 6 6 5 5 5 5 4 3 3 4 4 3 4 3 4 3 4 4 3 3 3 3 s s T s T s s T s T s s h h h h h h h h h h s s P h x P s h T P h h h h h h h h h s s P s h T P η η η η 1 5 2s 6s 3 4s 125 MPa P 6 4 2 T 3 1 2 Turbine Boiler Condenser Pump 5 4 6 s The pressure at state 6 may be determined by a trialerror approach from the steam tables or by using EES from the above three equations P6 973 kPa h6 24633 kJkg b Then 20359 kJkg 1402 57 189 kJkg 1402 090 kPa m 1 1 kJ 9 73 kPa m kg 12500 000101 001010 m kg 0 57 kJkg 189 in 1 2 3 3 1 2 1 in 3 10 kPa 1 9 73 kPa 1 p p p f f w h h P P w h h η v v v Cycle analysis 10242 kW kgs36038 22737kJkg 77 2273 7 kJkg 18957 3 2463 3603 8 kJkg 2463 3 3358 2 20359 5 3476 out in net 1 6 out 4 5 2 3 in q m q W h h q h h h h q c The thermal efficiency is 369 0 369 36038 kJkg 22737 kJkg 1 1 in out th q q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1034 Regenerative Rankine Cycle 1042C Moisture content remains the same everything else decreases 1043C This is a smart idea because we waste little work potential but we save a lot from the heat input The extracted steam has little work potential left and most of its energy would be part of the heat rejected anyway Therefore by regeneration we utilize a considerable amount of heat by sacrificing little work output 1044C In open feedwater heaters the two fluids actually mix but in closed feedwater heaters there is no mixing 1045C Both cycles would have the same efficiency 1046C To have the same thermal efficiency as the Carnot cycle the cycle must receive and reject heat isothermally Thus the liquid should be brought to the saturated liquid state at the boiler pressure isothermally and the steam must be a saturated vapor at the turbine inlet This will require an infinite number of heat exchangers feedwater heaters as shown on the Ts diagram Boiler exit s T Boiler inlet qin qout PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1039 1051E The optimum bleed pressure for the open feedwater heater that maximizes the thermal efficiency of the cycle is to be determined by EES Analysis The EES program used to solve this problem as well as the solutions are given below P5500 psia T5600 F P660 psia P75 psia x30 Analysis Fluidsteamiapws pump I P1P7 x10 h1enthalpyFluid PP1 xx1 v1volumeFluid PP1 xx1 P3P6 P2P3 wpIinv1P2P1Convertpsiaft3 Btu h2h1wpIin pump II h3enthalpyFluid PP3 xx3 v3volumeFluid PP3 xx3 P4P5 wpIIinv3P4P3Convertpsiaft3 Btu h4h3wpIIin turbine h5enthalpyFluid PP5 TT5 s5entropyFluid PP5 TT5 s6s5 h6enthalpyFluid PP6 ss6 x6qualityFluid PP6 ss6 s7s5 h7enthalpyFluid PP7 ss7 x7qualityFluid PP7 ss7 open feedwater heater yh61yh2h3 ymdot6mdot3 cycle qinh5h4 qout1yh7h1 wnetqinqout Etath1qoutqin PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1040 P 6 kPa ηth 10 03011 20 03065 30 03088 40 03100 50 03107 60 03111 70 031126 80 031129 90 03112 100 03111 110 03109 120 03107 130 03104 140 03101 150 03098 160 03095 170 03091 180 03087 190 03084 200 03080 0 40 80 120 160 200 03 0302 0304 0306 0308 031 0312 Bleed pressure kPa ηth PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1041 1052 A steam power plant operates on an ideal regenerative Rankine cycle with two open feedwater heaters The net power output of the power plant and the thermal efficiency of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis P III P II P I fwh II fwh I Condenser Boiler Turbine 6 5 4 3 2 1 10 9 7 8 T 1 2 10 1 y 8 9 1 y z 06 MPa 5 kPa 02 MPa 4 y 3 10 MPa 5 6 7 s a From the steam tables Tables A4 A5 and A6 13795 kJkg 0 20 75 137 0 kJkg 20 kPa m 1 1 kJ 5 kPa m kg 200 0001005 001005 m kg 0 75 kJkg 137 in 1 2 3 3 1 2 1 in 3 5 kPa 1 5 kPa 1 pI pI f f w h h P P w h h v v v 1035 kJkg kPa m 1 1 kJ 600 kPa m kg 10000 0001101 001101 m kg 0 38 kJkg 670 liquid sat MPa 60 50513 kJkg 0 42 50471 042 kJkg kPa m 1 1 kJ 200 kPa m kg 600 0001061 001061 m kg 0 71 kJkg 504 liquid sat MPa 20 3 3 5 6 5 in 3 MPa 60 5 MPa 60 5 5 in 3 4 3 3 3 4 3 in 3 MPa 20 3 MPa 20 3 3 P P w h h P w h h P P w h h P pIII f f pII pII f f v v v v v v 2821 8 kJkg MPa 60 9045 kJkg K 6 8 kJkg 3625 C 600 MPa 10 68073 kJkg 38 1035 670 8 7 8 8 7 7 7 7 in 5 6 h s s P s h T P w h h pIII 7 kJkg 2618 0 9602 2201 6 71 504 0 9602 5 5968 1 5302 6 9045 MPa 20 9 9 9 9 7 9 9 fg f fg f x h h h s s s x s s P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1043 1053 An ideal regenerative Rankine cycle with a closed feedwater heater is considered The work produced by the turbine the work consumed by the pumps and the heat added in the boiler are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the steam tables Tables A4 A5 and A6 25445 kJkg 3 03 25142 3 03 kJkg kPa m 1 1kJ 20kPa 0 001017 m kg3000 001017 m kg 0 42 kJkg 251 pin 1 2 3 3 1 2 1 in p 3 20 kPa 1 20 kPa 1 w h h P P w h h f f v v v 2221 7 kJkg 0 83572357 5 42 251 0 8357 7 0752 0 8320 6 7450 kPa 20 2851 9 kJkg kPa 1000 6 7450 kJkg K 3116 1 kJkg 350 C kPa 3000 6 6 6 6 4 6 6 5 4 5 5 4 4 4 4 fg f fg f x h h h s s s x s s P h s s P s h T P 5 Condenser 1 2 3 4 Turbine Boiler Closed fwh 8 7 Pump 6 For an ideal closed feedwater heater the feedwater is heated to the exit temperature of the extracted steam which ideally leaves the heater as a saturated liquid at the extraction pressure 1 qin 2 6 1y 5 3 MPa 20 kPa qout y 3 1 MPa 7 8 4 s T 76353 kJkg C 9 209 kPa 3000 51 kJkg 762 C 9 179 76251 kJkg 0 kPa 1000 3 7 3 3 7 8 7 7 7 7 h T T P h h T h x P An energy balance on the heat exchanger gives the fraction of steam extracted from the turbine for closed feedwater heater m5 m4 7 3 2 5 7 7 3 3 2 2 5 5 1 1 yh h h yh m h m h m h h m m h m h e e i i Rearranging 0 2437 76251 2851 9 25445 53 763 7 5 2 3 h h h h y Then kJkg 2353 kJkg 303 kJkg 7409 76353 1 3116 2221 7 0 24372851 9 1 2851 9 3116 1 1 3 4 in in P 6 5 5 4 out T h h q w h y h h h w Also 737 8 kJkg 3 03 740 9 Pin Tout net w w w 0 3136 2353 8 737 in net th q w η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1044 1054 Problem 1053 is reconsidered The optimum bleed pressure for the open feedwater heater that maximizes the thermal efficiency of the cycle is to be determined by EES Analysis The EES program used to solve this problem as well as the solutions are given below Given P43000 kPa T4350 C P5600 kPa P620 kPa P3P4 P2P3 P7P5 P1P6 Analysis Fluidsteamiapws pump I x10 h1enthalpyFluid PP1 xx1 v1volumeFluid PP1 xx1 wpinv1P2P1 h2h1wpin turbine h4enthalpyFluid PP4 TT4 s4entropyFluid PP4 TT4 s5s4 h5enthalpyFluid PP5 ss5 T5temperatureFluid PP5 ss5 x5qualityFluid PP5 ss5 s6s4 h6enthalpyFluid PP6 ss6 x6qualityFluid PP6 ss6 closed feedwater heater x70 h7enthalpyFluid PP7 xx7 T7temperatureFluid PP7 xx7 T3T7 h3enthalpyFluid PP3 TT3 yh3h2h5h7 ymdot5mdot4 cycle qinh4h3 wTouth4h51yh5h6 wnetwToutwpin Etathwnetqin PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1045 P 6 kPa ηth 100 032380 110 032424 120 032460 130 032490 140 032514 150 032534 160 032550 170 032563 180 032573 190 032580 200 032585 210 032588 220 032590 230 032589 240 032588 250 032585 260 032581 270 032576 280 032570 290 032563 100 140 180 220 260 300 03238 0324 03243 03245 03248 0325 03253 03255 03258 0326 Bleed pressure kPa ηth PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1046 1055 A regenerative Rankine cycle with a closed feedwater heater is considered The thermal efficiency is to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the steam tables Tables A4 A5 and A6 or EES 25445 kJkg 3 03 25142 3 03 kJkg kPa m 1 1kJ 20kPa 0 001017 m kg3000 001017 m kg 0 42 kJkg 251 pin 1 2 3 3 1 2 1 in p 3 20 kPa 1 20 kPa 1 w h h P P w h h f f v v v 2221 7 kJkg 0 83572357 5 42 251 0 8357 7 0752 0 8320 6 7450 kPa 20 2851 9 kJkg kPa 1000 6 7450 kJkg K 3116 1 kJkg 350 C kPa 3000 6 6 6 6 4 6 6 5 4 5 5 4 4 4 4 fg s f s fg f s s s s s h x h h s s s x s s P h s s P s h T P 5 6 1 2 3 4 Turbine Boiler Condenser Closed fwh 7 Pump 2878 3 kJkg 2851 9 0 903116 1 3116 1 5s 4 T 4 5 5 4 5 4 T h h h h h h h h s η η 2311 1 kJkg 2221 7 0 903116 1 3116 1 6s 4 T 4 6 6 4 6 4 T h h h h h h h h s η η For an ideal closed feedwater heater the feedwater is heated to the exit temperature of the extracted steam which ideally leaves the heater as a saturated liquid at the extraction pressure 1 qin 2 6 4 5s 3 MPa 20 kPa 3 1 MPa 7 1y y qout 5 6s s T 76353 kJkg C 209 9 kPa 3000 C 179 9 76251 kJkg 0 kPa 1000 3 7 3 3 7 7 7 7 h T T P T h x P An energy balance on the heat exchanger gives the fraction of steam extracted from the turbine for closed feedwater heater m5 m4 7 3 2 5 7 7 3 3 2 2 5 5 1 1 yh h h yh m h m h m h h m m h m h e e i i Rearranging 0 2406 76251 2878 3 25445 53 763 7 5 2 3 h h h h y Then 2353 kJkg 76353 1 3116 kJkg 303 6685 kJkg 2311 1 0 24062878 3 1 2878 3 3116 1 1 3 4 in in P 6 5 5 4 out T h h q w h y h h h w Also 665 5 kJkg 3 03 668 5 Pin Tout net w w w 283 02829 2353 5 665 in net th q w η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1047 1056 A regenerative Rankine cycle with a closed feedwater heater is considered The thermal efficiency is to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the steam tables Tables A4 A5 and A6 or EES 5 Condenser 1 2 3 4 Turbine Boiler Closed fwh 7 Pump 6 1 qin 2 6 4 5s 3 MPa 20 kPa 1 MPa 1y 7 y 3 qout 5 6s s T When the liquid enters the pump 10C cooler than a saturated liquid at the condenser pressure the enthalpies become 001012 m kg 0 20934 kJkg 50 C 6006 10 10 kPa 20 3 50 C 1 50 C 1 sat 20 kPa 1 1 f hf h T T P v v 3 02 kJkg kPa m 1 1kJ 20kPa 0 001012 m kg3000 3 3 1 2 1 in p P P w v 21236 kJkg 3 02 20934 pin 1 2 w h h 2221 7 kJkg 0 83572357 5 42 251 0 8357 7 0752 0 8320 6 7450 kPa 20 2851 9 kJkg kPa 1000 6 7450 kJkg K 3116 1 kJkg 350 C kPa 3000 6 6 6 6 4 6 6 5 4 5 5 4 4 4 4 fg s f s fg f s s s s s h x h h s s s x s s P h s s P s h T P 2878 3 kJkg 2851 9 0 903116 1 3116 1 5s 4 T 4 5 5 4 5 4 T h h h h h h h h s η η 2311 1 kJkg 2221 7 0 903116 1 3116 1 6s 4 T 4 6 6 4 6 4 T h h h h h h h h s η η For an ideal closed feedwater heater the feedwater is heated to the exit temperature of the extracted steam which ideally leaves the heater as a saturated liquid at the extraction pressure 76353 kJkg C 209 9 kPa 3000 C 179 9 76251 kJkg 0 kPa 1000 3 7 3 3 7 7 7 7 h T T P T h x P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1048 An energy balance on the heat exchanger gives the fraction of steam extracted from the turbine for closed feedwater heater m5 m4 7 3 2 5 7 7 3 3 2 2 5 5 1 1 yh h h yh m h m h m h h m m h m h e e i i Rearranging 0 2605 76251 2878 3 21236 53 763 7 5 2 3 h h h h y Then 2353 kJkg 76353 1 3116 kJkg 303 6572 kJkg 2311 1 0 26052878 3 1 2878 3 3116 1 1 3 4 in in P 6 5 5 4 out T h h q w h y h h h w Also 654 2 kJkg 3 03 657 2 Pin Tout net w w w 02781 2353 2 654 in net th q w η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1049 1057 The effect of pressure drop and nonisentropic turbine on the rate of heat input is to be determined for a given power plant Analysis The EES program used to solve this problem as well as the solutions are given below Given P33000 kPa DELTAPboiler10 kPa P4P3DELTAPboiler T4350 C P51000 kPa P620 kPa etaT090 P2P3 P7P5 P1P6 Analysis Fluidsteamiapws a pump I x10 h1enthalpyFluid PP1 xx1 v1volumeFluid PP1 xx1 wpinv1P2P1 h2h1wpin turbine h4enthalpyFluid PP4 TT4 s4entropyFluid PP4 TT4 s5s4 hs5enthalpyFluid PP5 ss5 T5temperatureFluid PP5 ss5 xs5qualityFluid PP5 ss5 s6s4 hs6enthalpyFluid PP6 ss6 xs6qualityFluid PP6 ss6 h5h4etaTh4hs5 h6h4etaTh4hs6 x5qualityFluid PP5 hh5 x6qualityFluid PP6 hh6 closed feedwater heater x70 h7enthalpyFluid PP7 xx7 T7temperatureFluid PP7 xx7 T3T7 h3enthalpyFluid PP3 TT3 yh3h2h5h7 ymdot5mdot4 cycle qinh4h3 wTouth4h51yh5h6 wnetwToutwpin Etathwnetqin PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1050 Solution with 10 kPa pressure drop in the boiler DELTAPboiler10 kPa etaT09 Etath02827 Fluidsteamiapws P33000 kPa P42990 kPa qin23528 kJkg wnet6651 kJkg wpin3031 m3kPakg wTout6681 kJkg y02405 Solution without any pressure drop in the boiler DELTAPboiler0 kPa etaT1 Etath03136 Fluidsteamiapws P33000 kPa P43000 kPa qin23525 kJkg wnet7378 kJkg wpin3031 m3kPakg wTout7409 kJkg y02437 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1053 1059 Problem 1058 is reconsidered The effects of turbine and pump efficiencies on the mass flow rate and thermal efficiency are to be investigated Also the Ts diagram is to be plotted Analysis The problem is solved using EES and the solution is given below Input Data P8 10000 kPa T8 600 C P9 1200 kPa Pcfwh600 kPa P10 Pcfwh Pcond10 kPa P11 Pcond Wdotnet400 MWConvertMW kW Etaturb 100100 Turbine isentropic efficiency Etaturbhp Etaturb Turbine isentropic efficiency for high pressure stages Etaturbip Etaturb Turbine isentropic efficiency for intermediate pressure stages Etaturblp Etaturb Turbine isentropic efficiency for low pressure stages Etapump 100100 Pump isentropic efficiency Condenser exit pump or Pump 1 analysis FluidSteamIAPWS P1 P11 P2P10 h1enthalpyFluidPP1x0 Satd liquid v1volumeFluidPP1x0 s1entropyFluidPP1x0 T1temperatureFluidPP1x0 wpump1sv1P2P1SSSF isentropic pump work assuming constant specific volume wpump1wpump1sEtapump Definition of pump efficiency h1wpump1 h2 Steadyflow conservation of energy s2entropyFluidPP2hh2 T2temperatureFluidPP2hh2 Open Feedwater Heater analysis zh10 yh7 1yzh2 1h3 Steadyflow conservation of energy h3enthalpyFluidPP3x0 T3temperatureFluidPP3x0 Condensate leaves heater as sat liquid at P3 s3entropyFluidPP3x0 Boiler condensate pump or Pump 2 analysis P5P8 P4 P5 P3P10 v3volumeFluidPP3x0 wpump2sv3P4P3SSSF isentropic pump work assuming constant specific volume wpump2wpump2sEtapump Definition of pump efficiency h3wpump2 h4 Steadyflow conservation of energy s4entropyFluidPP4hh4 T4temperatureFluidPP4hh4 Closed Feedwater Heater analysis P6P9 yh9 1h4 1h5 yh6 Steadyflow conservation of energy h5enthalpyFluidPP6x0 h5 hT5 P5 where T5Tsat at P9 T5temperatureFluidPP5hh5 Condensate leaves heater as sat liquid at P6 s5entropyFluidPP6hh5 h6enthalpyFluidPP6x0 T6temperatureFluidPP6x0 Condensate leaves heater as sat liquid at P6 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1054 s6entropyFluidPP6x0 Trap analysis P7 P10 yh6 yh7 Steadyflow conservation of energy for the trap operating as a throttle T7temperatureFluidPP7hh7 s7entropyFluidPP7hh7 Boiler analysis qin h5h8SSSF conservation of energy for the Boiler h8enthalpyFluid TT8 PP8 s8entropyFluid TT8 PP8 Turbine analysis ss9s8 hs9enthalpyFluidsss9PP9 Ts9temperatureFluidsss9PP9 h9h8Etaturbhph8hs9Definition of turbine efficiency for high pressure stages T9temperatureFluidPP9hh9 s9entropyFluidPP9hh9 ss10s8 hs10enthalpyFluidsss10PP10 Ts10temperatureFluidsss10PP10 h10h9Etaturbiph9hs10Definition of turbine efficiency for Intermediate pressure stages T10temperatureFluidPP10hh10 s10entropyFluidPP10hh10 ss11s8 hs11enthalpyFluidsss11PP11 Ts11temperatureFluidsss11PP11 h11h10Etaturblph10hs11Definition of turbine efficiency for low pressure stages T11temperatureFluidPP11hh11 s11entropyFluidPP11hh11 h8 yh9 zh10 1yzh11 wturb SSSF conservation of energy for turbine Condenser analysis 1yzh11qout1yzh1SSSF First Law for the Condenser Cycle Statistics wnetwturb 1yzwpump1 wpump2 Etathwnetqin Wdotnet mdot wnet ηturb ηth m kgs 07 075 08 085 09 095 1 03834 0397 04096 04212 04321 04423 0452 369 3563 3454 3358 3274 3198 313 ηpump ηth m kgs 07 075 08 085 09 095 1 04509 04511 04513 04515 04517 04519 0452 3138 3136 3134 3133 3132 3131 313 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1055 Pcfwh kPa y z 100 200 300 400 500 600 700 800 900 1000 01702 01301 01041 008421 006794 005404 004182 003088 002094 001179 005289 009634 01226 01418 01569 01694 01801 01895 01979 02054 00 11 22 33 44 55 66 77 88 99 110 0 100 200 300 400 500 600 700 s kJkgK T C 10000 kPa 1200 kPa 600 kPa 10 kPa SteamIAPWS 12 34 56 7 8 9 10 11 07 075 08 085 09 095 1 038 039 04 041 042 043 044 045 046 ηturb ηth PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1056 07 075 08 085 09 095 1 310 320 330 340 350 360 370 ηturb m kgs 07 075 08 085 09 095 1 04508 0451 04512 04514 04516 04518 0452 04522 3129 313 3131 3132 3133 3134 3135 3136 3137 3138 ηpump ηth m kgs 100 200 300 400 500 600 700 800 900 1000 0 002 004 006 008 01 012 014 016 018 004 006 008 01 012 014 016 018 02 022 Pcfwh kPa y z PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1057 1060 A steam power plant that operates on an ideal regenerative Rankine cycle with a closed feedwater heater is considered The temperature of the steam at the inlet of the closed feedwater heater the mass flow rate of the steam extracted from the turbine for the closed feedwater heater the net power output and the thermal efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a From the steam tables Tables A4 A5 and A6 65 kJkg 260 9 223 42 251 922 kJkg 0 88 1 20 kPa 0001017 m kg8000 001017 m kg 0 42 kJkg 251 in 1 2 3 1 2 1 in 3 20 kPa 1 20 kPa 1 pI p pI f f w h h P P w h h η v v v PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 001127 m kg 0 51 kJkg 762 liquid sat MPa 1 3 1 MPa 3 1 MPa 3 3 f hf h P v v 77148 kJkg 8 97 51 762 97 kJkg 8 0 88 1000 kPa 0 001127 m kg8000 in 3 11 3 3 11 3 in pII p pII w h h P P w η v 6 5 8 1 2 3 HighP turbine Boiler Cond Closed fwh PI PII 10 11 Mixing Cham y 7 1y 9 LowP turbine 4 Also h4 h10 h11 77148 kJkg since the two fluid streams which are being mixed have the same enthalpy 3140 1 kJkg 3104 7 0 88 3399 5 5 3399 3104 7 kJkg MPa 3 7266 kJkg K 6 5 kJkg 3399 C 500 MPa 8 6 5 5 6 6 5 6 5 6 5 6 6 5 5 5 5 s T s T s h h h h h h h h h s s P s h T P η η 3499C 8 8 8 8 7 7 8 8 7 8 7 8 7 8 8 7 7 7 7 9 kJkg 3157 MPa 1 3157 9 kJkg 3117 1 0 88 3457 2 2 3457 3117 1 kJkg MPa 1 2359 kJkg K 7 2 kJkg 3457 C 500 MPa 3 T h P h h h h h h h h h s s P s h T P s T s T s η η preparation If you are a student using this Manual you are using it without permission 1058 2513 9 kJkg 2385 2 0 88 3457 2 2 3457 2385 2 kJkg kPa 20 9 7 7 9 9 7 9 7 9 7 9 9 s T s T s h h h h h h h h h s s P η η The fraction of steam extracted from the low pressure turbine for closed feedwater heater is determined from the steady flow energy balance equation applied to the feedwater heater Noting that Q W ke pe 0 0 1758 76251 3157 9 26065 77148 1 1 3 8 2 10 y y y h y h h h y The corresponding mass flow rate is 2637 kgs 0 175815 kgs 5 8 ym m c Then 1864 8 kJkg 25142 0 1758 2513 9 1 1 2945 2 kJkg 3140 1 3457 2 77148 5 3399 1 9 out 6 7 4 5 in h y h q h h h h q and 16206 kW 1864 8 kJkg 15 kgs2945 8 out in net q m q W b The thermal efficiency is determined from 367 0 3668 29458 kJkg 18648 kJkg 1 1 in out th q q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1059 1061 A Rankine steam cycle modified with two closed feedwater heaters is considered The Ts diagram for the ideal cycle is to be sketched The fraction of mass extracted for the closed feedwater heater z and the cooling water flow rate are to be determined Also the net power output and the thermal efficiency of the plant are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis b Using the data from the problem statement the enthalpies at various states are 256 1 kJkg 15 251 kJkg 15 kPa m 1 1 kJ 20 kPa m kg 5000 000102 00102 m kg 0 kJkg 251 pIin 1 2 3 3 1 2 1 in pI 3 20 kPa 1 20 kPa 1 w h h P P w h h f f v v v 1 2 9 s 5 6 245 kPa 20 kPa 14 MPa 8 4 3 1yz 1y y 5 MPa 11 12 1yz y 10 z yz 8 7 T Also throttle valve operation kJkg 533 11 12 245 kPa 11 3 h h h h h f throttle valve operation kJkg 830 9 10 1400 kPa 9 4 h h h h h f An energy balance on the closed feedwater heater gives 11 3 10 7 2 1 1 z h y h yh zh h where z is the fraction of steam extracted from the lowpressure turbine Solving for z 01017 533 2918 830 0 1153533 256 1 533 11 7 10 11 2 3 h h h y h h h z c An energy balance on the condenser gives 1 1 12 12 8 8 2 2 2 1 1 12 12 1 8 8 m h m h m h h h m m h m h m h m h h m w w w w w w w Solving for the mass flow rate of cooling water and substituting with correct units 2158 kgs 4 1810 1 251 0 1017533 0 1153 0 10172477 0 1153 50 1 1 1 1 12 8 5 w pw w T c h z h y z h y m m d The work output from the turbines is kJkg 1271 0 10172477 0 1153 1 0 10172918 0 11533406 3900 1 8 7 6 5 out T z h y zh yh h w The net work output from the cycle is 1265 9 kJkg 15 1271 Pin Tout net w w w The net power output is 633 MW 63300 kW 50 kgs12659 kJkg net net mw W The rate of heat input in the boiler is 153500 kW 830 kJkg 50 kgs3900 4 5 in h m h Q The thermal efficiency is then 412 0 412 153500 kW 300 kW 63 in net th Q W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1060 SecondLaw Analysis of Vapor Power Cycles 1062C In the simple ideal Rankine cycle irreversibilities occur during heat addition and heat rejection processes in the boiler and the condenser respectively and both are due to temperature difference Therefore the irreversibilities can be decreased and thus the 2nd law efficiency can be increased by minimizing the temperature differences during heat transfer in the boiler and the condenser One way of doing that is regeneration 1063E The exergy destructions associated with each of the processes of the Rankine cycle described in Prob 1017E are to be determined for the specified source and sink temperatures Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From Problem 1017E qin qout 3 psia 1 3 2 4 800 psia T s 865 8 Btulbm 10940 24 975 1344 2 Btulbm 11181 0 1456 6413 Btulbm R 1 2009 Btulbm R 0 1 4 out 2 3 in 4 3 3 psia 2 1 h h q h h q s s s s s f The exergy destruction during a process of a stream from an inlet state to exit state is given by sink out source in 0 gen 0 dest T q T q s s T T s x i e Application of this equation for each process of the cycle gives Btulbm 146 Btulbm 377 500 R 865 8 Btulbm 1 6413 0 2009 500 R 1960 R 1344 2 Btulbm 0 2009 1 6413 500 R sink out 4 1 0 41 destroyed source in 2 3 0 23 destroyed T q s s T x T q s s T x Processes 12 and 34 are isentropic and thus 0 0 34 destroyed destroyed 12 x x PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1062 1065 The component of the ideal reheat Rankine cycle described in Prob 1033 with the largest exergy destruction is to be identified Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From Prob 1033 1 kJkg 2373 848 1 kJkg 2636 4 4 3485 3073 4 kJkg 19988 3 3273 0893 kJkg K 8 5579 kJkg K 6 6492 kJkg K 0 out 4 5 4 5 in 2 3 2 3 in 6 5 4 3 10 kPa 2 1 q h h q h h q s s s s s s s f T 1 5 2 6 3 4 500 kPa 10 kPa 8 MPa s The exergy destruction during a process of a stream from an inlet state to exit state is given by sink out source in 0 gen 0 dest T q T q s s T T s x i e Application of this equation for each process of the cycle gives kJkg 2676 kJkg 1616 kJkg 6871 283 K 2373 1 kJkg 8 0893 0 6492 283 K 883 K 848 1 kJkg 6 5579 8 0893 283 K 883 K 3073 4 kJkg 0 6492 6 5579 283 K sink out 6 1 0 61 destroyed source in 4 5 4 5 0 45 destroyed source in 2 3 2 3 0 23 destroyed T q s s T x T q s s T x T q s s T x Processes 12 34 and 56 are isentropic and thus 0 0 0 56 destroyed 34 destroyed 12 destroyed x x x The greatest exergy destruction occurs during heat addition process 23 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1063 1066 The exergy destructions associated with each of the processes of the reheat Rankine cycle described in Prob 1035 are to be determined for the specified source and sink temperatures Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From Problem 1035 2098 3 kJkg 25142 7 2349 347 3 kJkg 2901 0 4 3248 2920 8 kJkg 25750 3 3178 1292 kJkg K 7 5432 kJkg K 6 8320 kJkg K 0 1 6 out in 45 in 23 6 5 4 3 20kPa 2 1 h h q q q s s s s s s s f Processes 12 34 and 56 are isentropic Thus i12 i34 i56 0 Also kJkg 2406 kJkg 1046 kJkg 1110 295 K 20983 kJkg 7 1292 0 8320 K 295 1500 K 3473 kJkg 6 5432 7 1292 K 295 1500 K 29208 kJkg 0 8320 6 5432 K 295 61 6 1 0 61 destroyed 45 4 5 0 45 destroyed 23 2 3 0 23 destroyed R R R R R R T q s s T x T q s s T x T q s s T x PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1064 1067 Problem 1066 is reconsidered The problem is to be solved by the diagram window data entry feature of EES by including the effects of the turbine and pump efficiencies Also the Ts diagram is to be plotted Analysis The problem is solved using EES and the solution is given below function x6x6 this function returns a string to indicate the state of steam at point 6 x6 if x61 then x6superheated if x60 then x6subcooled end Input Data from diagram window P6 20 kPa P3 6000 kPa T3 400 C P4 2000 kPa T5 400 C Etat 100100 Turbine isentropic efficiency Etap 100100 Pump isentropic efficiency Data for the irreversibility calculations To 295 K TRL 295 K TRH 1500 K Pump analysis FluidSteamIAPWS P1 P6 P2P3 x10 Satd liquid h1enthalpyFluidPP1xx1 v1volumeFluidPP1xx1 s1entropyFluidPP1xx1 T1temperatureFluidPP1xx1 Wpsv1P2P1SSSF isentropic pump work assuming constant specific volume WpWpsEtap h2h1Wp SSSF First Law for the pump v2volumeFluidPP2hh2 s2entropyFluidPP2hh2 T2temperatureFluidPP2hh2 High Pressure Turbine analysis h3enthalpyFluidTT3PP3 s3entropyFluidTT3PP3 v3volumeFluidTT3PP3 ss4s3 hs4enthalpyFluidsss4PP4 Ts4temperatureFluidsss4PP4 Etath3h4h3hs4Definition of turbine efficiency T4temperatureFluidPP4hh4 s4entropyFluidTT4PP4 v4volumeFluidss4PP4 h3 Wthph4SSSF First Law for the high pressure turbine Low Pressure Turbine analysis P5P4 s5entropyFluidTT5PP5 h5enthalpyFluidTT5PP5 ss6s5 hs6enthalpyFluidsss6PP6 Ts6temperatureFluidsss6PP6 vs6volumeFluidsss6PP6 Etath5h6h5hs6Definition of turbine efficiency PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course h5Wtlph6SSSF First Law for the low pressure turbine preparation If you are a student using this Manual you are using it without permission 1065 x6QUALITYFluidhh6PP6 Boiler analysis Qin h2h4h3h5SSSF First Law for the Boiler Condenser analysis h6Qouth1SSSF First Law for the Condenser T6temperatureFluidhh6PP6 s6entropyFluidhh6PP6 x6sx6x6 Cycle Statistics WnetWthpWtlpWp EffWnetQin The irreversibilities or exergy destruction for each of the processes are qR23 h3 h2 Heat transfer for the high temperature reservoir to process 23 i23 Tos3 s2 qR23TRH qR45 h5 h4 Heat transfer for the high temperature reservoir to process 45 i45 Tos5 s4 qR45TRH qR61 h6 h1 Heat transfer to the low temperature reservoir in process 61 i61 Tos1 s6 qR61TRL i34 Tos4 s3 i56 Tos6 s5 i12 Tos2 s1 00 10 20 30 40 50 60 70 80 90 100 0 100 200 300 400 500 600 700 s kJkgK T C 6000 kPa 2000 kPa 20 kPa SteamIAPWS 12 3 4 5 6 Ideal Rankine cycle with reheat SOLUTION Eff0358 Etap1 Etat1 FluidSteamIAPWS i120007 kJkg i231110378 kJkg i340000 kJkg i45104554 kJkg i560000 kJkg i61240601 kJkg Qin3268 kJkg Qout2098 kJkg qR232921 kJkg qR453473 kJkg qR612098 kJkg To295 K TRH1500 K TRL295 K Wnet1170 kJkg Wp6083 kJkg Wps6083 kJkg Wthp2772 kJkg Wtlp8987 kJkg PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1066 1068E The component of the ideal regenerative Rankine cycle described in Prob 1049E with the largest exergy destruction is to be identified Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From Prob 1049E 01109 1 Btulbm 732 1061 kJkg 237 6 6 1298 5590 Btulbm R 1 39213 Btulbm R 0 23488 Btulbm R 0 out 4 5 in 7 6 5 40 psia 4 3 5 psia 2 1 y q h h q s s s s s s s s s f f 1 qin 2 7 1y 6 qout 500 psia 5 psia y 40 psia 4 3 5 s T The exergy destruction during a process of a stream from an inlet state to exit state is given by sink out source in 0 gen 0 dest T q T q s s T T s x i e Application of this equation for each process of the cycle gives Btulbm 436 Btulbm 1689 520 R 732 1 Btulbm 1 5590 0 23488 520 R 1260 R 1061 Btulbm 0 39213 1 5590 520 R sink out 7 1 0 71 destroyed source in 4 5 0 45 destroyed T q s s T x T q s s T x For open feedwater heater we have 54 Btulbm 0 1109 0 23488 1 0 1109 1 5590 520 R 0 39213 1 2 6 3 0 destroyed FWH y s ys T s x Processes 12 34 56 and 67 are isentropic and thus 0 0 34 destroyed destroyed 12 x x 0 0 67 destroyed destroyed 56 x x The greatest exergy destruction occurs during heat addition process 45 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1067 1069 A singleflash geothermal power plant uses hot geothermal water at 230ºC as the heat source The power output from the turbine the thermal efficiency of the plant the exergy of the geothermal liquid at the exit of the flash chamber and the exergy destructions and exergy efficiencies for the flash chamber the turbine and the entire plant are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a We use properties of water for geothermal water Tables A4 A5 and A6 6841 kJkgK 2 1661 0 14 kJkg 990 kPa 500 6100 kJkgK 2 14 kJkg 990 0 C 230 2 2 1 2 2 1 1 1 1 s x h h P s h x T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course kgs 3819 0 1661230 kgs 1 2 3 x m m 7739 kJkg K 7 3 kJkg 2464 95 0 kPa 10 8207 kJkg K 6 1 kJkg 2748 1 kPa 500 4 4 4 4 3 3 3 3 s h x P s h x P 8604 kJkg K 1 09 kJkg 640 0 kPa 500 6 6 6 6 s h x P 2 production well reinjection well separator steam turbine 1 Flash chamber 6 5 4 3 condenser 19181 kgs 3819 230 3 1 6 m m m The power output from the turbine is 10842 kW kJkg 3819 kJkg27481 2464 3 4 3 3 T h h m W We use saturated liquid state at the standard temperature for dead state properties 3672 kJkg 0 83 kJkg 104 0 C 25 0 0 0 0 s h x T 203622 kW 230 kJkg99014 10483kJkg 0 1 1 in h m h E 53 00532 203622 842 10 in Tout th E W η b The specific exergies at various states are 21653 kJkg 0 3672kJkgK 298 K26100 99014 10483kJkg 0 1 0 0 1 1 s s T h h ψ 19444 kJkg 0 3672kJkgK 298 K26841 99014 10483kJkg 0 2 0 0 2 2 s s T h h ψ 71910 kJkg 0 3672kJkgK 298 K68207 27481 10483kJkg 0 3 0 0 3 3 s s T h h ψ 15105 kJkg 0 3672kJkgK 298 K77739 24643 10483kJkg 0 4 0 0 4 4 s s T h h ψ 8997 kJkg 0 3672kJkgK 298 K18604 64009 10483kJkg 0 6 0 0 6 6 s s T h h ψ The exergy of geothermal water at state 6 is 17257 kW 19181 kgs899 7 kJkg 6 6 6 m ψ X preparation If you are a student using this Manual you are using it without permission 1068 c Flash chamber 5080 kW 230 kgs21653 19444kJkg 2 1 1 dest FC ψ m ψ X 898 0898 21653 44 194 1 2 IIFC ψ ψ η d Turbine 10854 kW 3819 kgs71910 15105kJkg 10842 kW T 4 3 3 destT W m X ψ ψ 500 0500 3819 kgs71910 15105kJkg 10842 kW 4 3 3 T IIT ψ ψ η m W e Plant 49802 kW 230 kgs21653 kJkg 1 1 inPlant m ψ X 38960 kW 49802 10842 T inPlant destPlant W X X 218 02177 49802 kW 842 kW 10 Plant in T IIPlant X W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1069 Cogeneration 1070C The utilization factor of a cogeneration plant is the ratio of the energy utilized for a useful purpose to the total energy supplied It could be unity for a plant that does not produce any power 1071C No A cogeneration plant may involve throttling friction and heat transfer through a finite temperature difference and still have a utilization factor of unity 1072C Yes if the cycle involves no irreversibilities such as throttling friction and heat transfer through a finite temperature difference 1073C Cogeneration is the production of more than one useful form of energy from the same energy source Regeneration is the transfer of heat from the working fluid at some stage to the working fluid at some other stage PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1073 1077 A cogeneration plant modified with regeneration is to generate power and process heat The mass flow rate of steam through the boiler for a net power output of 25 MW is to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the steam tables Tables A4 A5 and A6 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 86702 kJkg 8 57 85844 8 57 kJkg kPa m 1 1 kJ 400 kPa m kg 9000 0001159 001159 m kg 0 44 kJkg 858 1 61 19341 kJkg 19181 161 kJkg kPa m 1 1 kJ 10 kPa m kg 1600 000101 00101 m kg 0 81 kJkg 191 pIIin 4 5 3 3 4 5 4 in pII 3 MPa 61 4 MPa 61 9 4 3 pIin 1 2 3 3 1 2 1 in pI 3 10 kPa 1 10 kPa 1 w h h P P w h h h h w h h P P w h h f f f f v v v v v v 1990 2 kJkg 0 7518 2392 1 81 191 0 7518 7 4996 0 6492 6 2876 kPa 10 2730 0 kJkg 0 9675 1934 4 44 858 0 9675 4 0765 2 3435 6 2876 MPa 61 6 2876 kJkg K 3118 8 kJkg 400 C MPa 9 8 8 8 8 6 8 8 7 7 7 7 6 7 7 6 6 6 6 fg f fg f fg f fg f x h h h s s s x s s P x h h h s s s x s s P s h T P 1 2 8 T 6 7 10 kPa 9 MPa 16 MPa 5 349 Boiler 6 7 Turbine 8 1 5 Condense r Process heater PI PII 9 2 3 fwh 4 s Then per kg of steam flowing through the boiler we have 869 7 kJkg 1990 2 kJkg 0 35 2730 0 1 2730 0 kJkg 3118 8 1 8 7 7 6 out T h y h h h w 860 1 kJkg 9 62 7 869 962 kJkg 857 kJkg 0 35 161 kJkg 1 1 pin Tout net pIIin pIin in p w w w w y w w Thus 291 kgs 8601 kJkg kJs 25000 net net w W m preparation If you are a student using this Manual you are using it without permission 1074 1078 Problem 1077 is reconsidered The effect of the extraction pressure for removing steam from the turbine to be used for the process heater and open feedwater heater on the required mass flow rate is to be investigated Analysis The problem is solved using EES and the solution is given below Input Data y 035 fraction of steam extracted from turbine for feedwater heater and process heater P6 9000 kPa T6 400 C Pextract1600 kPa P7 Pextract Pcond10 kPa P8 Pcond Wdotnet25 MWConvertMW kW Etaturb 100100 Turbine isentropic efficiency Etapump 100100 Pump isentropic efficiency P1 P8 P2P7 P3P7 P4 P7 P5P6 P9 P7 Condenser exit pump or Pump 1 analysis FluidSteamIAPWS h1enthalpyFluidPP1x0 Satd liquid v1volumeFluidPP1x0 s1entropyFluidPP1x0 T1temperatureFluidPP1x0 wpump1sv1P2P1SSSF isentropic pump work assuming constant specific volume wpump1wpump1sEtapump Definition of pump efficiency h1wpump1 h2 Steadyflow conservation of energy s2entropyFluidPP2hh2 T2temperatureFluidPP2hh2 Open Feedwater Heater analysis zh7 1 yh2 1 y zh3 Steadyflow conservation of energy h3enthalpyFluidPP3x0 T3temperatureFluidPP3x0 Condensate leaves heater as sat liquid at P3 s3entropyFluidPP3x0 Process heater analysis y zh7 qprocess y zh9 Steadyflow conservation of energy Qdotprocess mdoty zqprocesskW h9enthalpyFluidPP9x0 T9temperatureFluidPP9x0 Condensate leaves heater as sat liquid at P3 s9entropyFluidPP9x0 Mixing chamber at 3 4 and 9 yzh9 1yzh3 1h4 Steadyflow conservation of energy T4temperatureFluidPP4hh4 Condensate leaves heater as sat liquid at P3 s4entropyFluidPP4hh4 Boiler condensate pump or Pump 2 analysis v4volumeFluidPP4x0 wpump2sv4P5P4SSSF isentropic pump work assuming constant specific volume wpump2wpump2sEtapump Definition of pump efficiency h4wpump2 h5 Steadyflow conservation of energy PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1075 s5entropyFluidPP5hh5 T5temperatureFluidPP5hh5 Boiler analysis qin h5h6SSSF conservation of energy for the Boiler h6enthalpyFluid TT6 PP6 s6entropyFluid TT6 PP6 Turbine analysis ss7s6 hs7enthalpyFluidsss7PP7 Ts7temperatureFluidsss7PP7 h7h6Etaturbh6hs7Definition of turbine efficiency for high pressure stages T7temperatureFluidPP7hh7 s7entropyFluidPP7hh7 ss8s7 hs8enthalpyFluidsss8PP8 Ts8temperatureFluidsss8PP8 h8h7Etaturbh7hs8Definition of turbine efficiency for low pressure stages T8temperatureFluidPP8hh8 s8entropyFluidPP8hh8 h6 yh7 1 yh8 wturb SSSF conservation of energy for turbine Condenser analysis 1 yh8qout1 yh1SSSF First Law for the Condenser Cycle Statistics wnetwturb 1 ywpump1 wpump2 Etathwnetqin Wdotnet mdot wnet 00 11 22 33 44 55 66 77 88 99 110 0 100 200 300 400 500 600 700 s kJkgK T C 9000 kPa 1600 kPa 10 kPa SteamIAPWS 12 3459 6 7 8 Pextract kPa ηth m kgs Qprocess kW 200 400 600 800 1000 1200 1400 1600 1800 2000 03778 03776 03781 03787 03794 03802 03811 03819 03828 03837 254 2643 2711 2763 2807 2844 2877 2907 2934 2959 2770 2137 1745 1459 1235 1053 9007 7709 659 5618 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1076 200 400 600 800 1000 1200 1400 1600 1800 2000 0377 0378 0379 038 0381 0382 0383 0384 Pextract kPa ηth 200 400 600 800 1000 1200 1400 1600 1800 2000 25 26 27 28 29 30 500 1000 1500 2000 2500 3000 Pextract kPa m kgs Qprocess kW PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1077 1079E A cogeneration plant is to generate power while meeting the process steam requirements for a certain industrial application The net power produced the rate of process heat supply and the utilization factor of this plant are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a From the steam tables Tables A4E A5E and A6E P Boiler 2 6 1 Turbine Process heater 3 4 5 7 1229 5 Btulbm psia 120 6348 Btulbm R 1 0 Btulbm 1408 F 800 psia 600 49 Btulbm 208 7 3 7 7 6 5 4 3 7 5 3 3 3 3 1 2 240 F 1 h s s P h h h h s s s h T P h h h h f 2260 kW 2142 Btus 1229 5 Btulbm lbms 1408 0 12 7 5 5 net h h m W T 1 6 7 600 psia 120 psia 2 345 b 19450 Btus 18 20849 12 1229 5 1408 0 6 1 1 7 7 6 6 process m h m h h m m h m h Q e e i i c εu 1 since all the energy is utilized s PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1078 1080 A Rankine steam cycle modified for a closed feedwater heater and a process heater is considered The Ts diagram for the ideal cycle is to be sketched the mass flow rate of the cooling water and the utilization efficiency of the plant are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis b Using the data from the problem statement the enthalpies at various states are 201 9 kJkg 19181 10 1 10 1 kJkg kPa m 1 1 kJ m kg 10000 10 kPa 000101 00101 m kg 0 81 kJkg 191 pIin 1 2 3 3 1 2 1 in pI 3 20 kPa 1 10 kPa 1 w h h P P w h h f f v v v 1 2 8 s 4 5 700 kPa 10 kPa 2 MPa 3 1yz 1y y 10 MPa 10 11 1yz 9 z y 7 6 T 00 kJkg 697 47 kJkg 908 700 kPa 11 10 2000 kPa 9 8 3 f f h h h h h h h An energy balance on the closed feedwater heater gives 0 3495 90847 2930 201 9 47 908 8 5 2 3 2 3 8 5 h h h h y h h h h y The process heat is expressed as 603 kgs C40 C 4 18 kJkg 69700 kJkg 0 05100 kgs2714 10 6 10 6 process w p w w p w T c h zm h m T m c h zm h Q c The net power output is determined from 970 kW 94 10 1 kJkg 2089 kJkg 0 053374 0 3495 1 2714 kJkg 0 053374 2930 kJkg 0 34953374 100 kgs 1 7 4 6 4 5 4 P T net wP h z h y h z h h y h m W W W The rate of heat input in the boiler is 296550 kW 90847 kJkg 100 kgs3874 3 4 in h m h Q The rate of process heat is 10085 kW 69700 kJkg 0 05100 kgs2714 0 05 10 6 process h m h Q The utilization efficiency of this cogeneration plant is 354 0 354 296550 kW 10085 kW 94970 in process net Q Q W u ε PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1079 Combined GasVapor Power Cycles 1081C The energy source of the steam is the waste energy of the exhausted combustion gases 1082C Because the combined gassteam cycle takes advantage of the desirable characteristics of the gas cycle at high temperature and those of steam cycle at low temperature and combines them The result is a cycle that is more efficient than either cycle executed operated alone PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1080 1083 A 450MW combined gassteam power plant is considered The topping cycle is a gasturbine cycle and the bottoming cycle is an ideal Rankine cycle with an open feedwater heater The mass flow rate of air to steam the required rate of heat input in the combustion chamber and the thermal efficiency of the combined cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable specific heats Analysis a The analysis of gas cycle yields Table A17 462 02 kJkg K 460 735 8 kJkg 32 18 450 5 14 1 450 5 1515 42 kJkg K 1400 635 5 kJkg 19 40 1 386 14 1 386 300 19 kJkg K 300 12 12 11 10 11 10 10 9 8 9 8 8 10 11 10 8 9 8 h T h P P P P P h T h P P P P P h T r r r r r r PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course From the steam tables Tables A4 A5 A6 25201 kJkg 0 59 42 251 059 kJkg kPa m 1 1 kJ 20 kPa m kg 600 0001017 0 001017 m kg 42 kJkg 251 pIin 1 2 3 3 1 2 1 in pI 3 20 kPa 1 20 kPa 1 w h h P P w h h f f v v v 1 2 7 1400 K 300 K 6 8 MPa 20 kPa 06 MPa 4 9 3 Qout Qin 460 K STEAM CYCLE GAS CYCLE 10 11 12 8 5 400C T s 67853 kJkg 8 15 38 670 815 kJkg kPa m 1 1 kJ 600 kPa m kg 8000 0001101 0 001101 m kg 38 kJkg 670 pIin 3 4 3 3 3 4 3 in pII 3 MPa 60 3 MPa 60 3 w h h P P w h h f f v v v 2095 2 kJkg 0 7821 2357 5 42 251 0 7821 7 0752 0 8320 6 3658 kPa 20 2586 1 kJkg 0 9185 2085 8 38 670 0 9185 4 8285 1 9308 6 3658 MPa 60 6 3658 kJkg K 3139 4 kJkg 400 C MPa 8 7 7 7 7 5 7 7 6 6 6 6 5 6 6 5 5 5 5 fg f fg f fg f fg f x h h h s s s x s s P x h h h s s s x s s P s h T P Noting that 0 for the heat exchanger the steadyflow energy balance equation yields pe ke Q W preparation If you are a student using this Manual you are using it without permission 1082 1084 Problem 1083 is reconsidered The effect of the gas cycle pressure ratio on the ratio of gas flow rate to steam flow rate and cycle thermal efficiency is to be investigated Analysis The problem is solved using EES and the solution is given below Input data T8 300 K Gas compressor inlet P8 147 kPa Assumed air inlet pressure Pratio 14 Pressure ratio for gas compressor T10 1400 K Gas turbine inlet T12 460 K Gas exit temperature from Gastosteam heat exchanger P12 P8 Assumed air exit pressure Wdotnet450 MW Etacomp 10 Etagasturb 10 Etapump 10 Etasteamturb 10 P5 8000 kPa Steam turbine inlet T5 400273 K Steam turbine inlet P6 600 kPa Extraction pressure for steam open feedwater heater P7 20 kPa Steam condenser pressure GAS POWER CYCLE ANALYSIS Gas Compressor anaysis s8ENTROPYAirTT8PP8 ss9s8 For the ideal case the entropies are constant across the compressor P9 PratioP8 Ts9temperatureAirsss9PP9Ts9 is the isentropic value of T9 at compressor exit Etacomp wgascompisenwgascomp compressor adiabatic efficiency wcomp wcompisen h8 wgascompisen hs9SSSF conservation of energy for the isentropic compressor assuming adiabatic kepe0 per unit gas mass flow rate in kgs h8ENTHALPYAirTT8 hs9ENTHALPYAirTTs9 h8 wgascomp h9SSSF conservation of energy for the actual compressor assuming adiabatic kepe0 T9temperatureAirhh9 s9ENTROPYAirTT9PP9 Gas Cycle External heat exchanger analysis h9 qin h10SSSF conservation of energy for the external heat exchanger assuming W0 kepe0 h10ENTHALPYAirTT10 P10P9 Assume process 910 is SSSF constant pressure QdotinMW1000kWMWmdotgasqin Gas Turbine analysis s10ENTROPYAirTT10PP10 ss11s10 For the ideal case the entropies are constant across the turbine P11 P10 Pratio Ts11temperatureAirsss11PP11Ts11 is the isentropic value of T11 at gas turbine exit Etagasturb wgasturb wgasturbisen gas turbine adiabatic efficiency wgasturbisen wgasturb h10 wgasturbisen hs11SSSF conservation of energy for the isentropic gas turbine assuming adiabatic kepe0 hs11ENTHALPYAirTTs11 h10 wgasturb h11SSSF conservation of energy for the actual gas turbine assuming adiabatic kepe0 T11temperatureAirhh11 s11ENTROPYAirTT11PP11 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1083 GastoSteam Heat Exchanger SSSF conservation of energy for the gastosteam heat exchanger assuming adiabatic W0 kepe0 mdotgash11 mdotsteamh4 mdotgash12 mdotsteamh5 h12ENTHALPYAir TT12 s12ENTROPYAirTT12PP12 STEAM CYCLE ANALYSIS Steam Condenser exit pump or Pump 1 analysis FluidSteamIAPWS P1 P7 P2P6 h1enthalpyFluidPP1x0 Saturated liquid v1volumeFluidPP1x0 s1entropyFluidPP1x0 T1temperatureFluidPP1x0 wpump1sv1P2P1SSSF isentropic pump work assuming constant specific volume wpump1wpump1sEtapump Definition of pump efficiency h1wpump1 h2 Steadyflow conservation of energy s2entropyFluidPP2hh2 T2temperatureFluidPP2hh2 Open Feedwater Heater analysis yh6 1yh2 1h3 Steadyflow conservation of energy P3P6 h3enthalpyFluidPP3x0 Condensate leaves heater as sat liquid at P3 T3temperatureFluidPP3x0 s3entropyFluidPP3x0 Boiler condensate pump or Pump 2 analysis P4 P5 v3volumeFluidPP3x0 wpump2sv3P4P3SSSF isentropic pump work assuming constant specific volume wpump2wpump2sEtapump Definition of pump efficiency h3wpump2 h4 Steadyflow conservation of energy s4entropyFluidPP4hh4 T4temperatureFluidPP4hh4 wsteampumps 1ywpump1 wpump2 Total steam pump work input mass steam Steam Turbine analysis h5enthalpyFluidTT5PP5 s5entropyFluidPP5TT5 ss6s5 hs6enthalpyFluidsss6PP6 Ts6temperatureFluidsss6PP6 h6h5Etasteamturbh5hs6Definition of steam turbine efficiency T6temperatureFluidPP6hh6 s6entropyFluidPP6hh6 ss7s5 hs7enthalpyFluidsss7PP7 Ts7temperatureFluidsss7PP7 h7h5Etasteamturbh5hs7Definition of steam turbine efficiency T7temperatureFluidPP7hh7 s7entropyFluidPP7hh7 SSSF conservation of energy for the steam turbine adiabatic neglect ke and pe h5 wsteamturb yh6 1yh7 Steam Condenser analysis 1yh7qout1yh1SSSF conservation of energy for the Condenser per unit mass QdotoutConvertMW kWmdotsteamqout Cycle Statistics MassRatiogastosteam mdotgasmdotsteam WdotnetConvertMW kWmdotgaswgasturbwgascomp mdotsteamwsteamturb wsteampumpsdefinition of the net cycle work PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course EtathWdotnetQdotinConvert Cycle thermal efficiency in percent preparation If you are a student using this Manual you are using it without permission 1084 Bwrmdotgaswgascomp mdotsteamwsteampumpsmdotgaswgasturb mdotsteamwsteamturb Back work ratio Wdotnetsteam mdotsteamwsteamturb wsteampumps Wdotnetgas mdotgaswgasturb wgascomp NetWorkRatiogastosteam WdotnetgasWdotnetsteam Pratio MassRatio gastosteam Wnetgas kW Wnetsteam kW ηth NetWorkRatio gastosteam 10 7108 342944 107056 5992 3203 11 7574 349014 100986 6065 3456 12 8043 354353 95647 6129 3705 13 8519 359110 90890 6186 3951 14 9001 363394 86606 6237 4196 15 9492 367285 82715 6283 444 16 9993 370849 79151 6324 4685 17 1051 374135 75865 6362 4932 18 1103 377182 72818 6397 518 19 1157 380024 69976 6428 5431 20 1212 382687 67313 6457 5685 00 11 22 33 44 55 66 77 88 99 110 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 s kJkgK T K 8000 kPa 600 kPa 20 kPa Combined Gas and Steam Power Cycle 8 9 10 11 12 12 34 5 6 7 Steam Cycle Steam Cycle Gas Cycle Gas Cycle PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1085 5 9 13 17 21 25 55 572 594 616 638 66 P ratio ηth 5 9 14 18 23 20 25 30 35 40 45 50 55 60 65 Pratio NetWorkRatiogastosteam W dotgas W dotsteam vs Gas Pressure Ratio 5 9 14 18 23 50 60 70 80 90 100 110 120 130 140 Pratio MassRatiogastosteam Ratio of Gas Flow Rate to Steam Flow Rate vs Gas Pressure Ratio PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1086 1085 A combined gassteam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle The mass flow rate of air for a specified power output is to be determined Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable fo Brayton cycle 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a Analysis Working around the topping cycle gives the following results PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 530 8 K 293 K8 0414 1 5 6 5 6 k k s P P T T 572 8 K 0 85 293 530 8 293 5 6 5 6 5 6 5 6 5 6 5 6 C s p s p s C T T T T T T c T T c h h h h η η 758 0 K 8 1373 K 1 0414 1 7 8 7 8 k k s P P T T 5 K 819 758 0 0 901373 1373 8 7 7 8 8 7 8 7 8 7 8 7 s T s p p s T T T T T T T c T T c h h h h η η 1 2 4s 4 3 6 MPa 20 kPa 6s 1373 K 293 K Qout Qin 5 9 8s 7 GAS CYCLE STEAM CYCLE 320C 6 8 T s C 5486 K 275 6 sat 6000 kPa 9 T T Fixing the states around the bottom steam cycle yields Tables A4 A5 A6 257 5 kJkg 6 08 25142 6 08 kJkg kPa m 1 1kJ 20kPa 0 001017 m kg6000 001017 m kg 0 42 kJkg 251 pin 1 2 3 3 1 2 1 in p 3 20 kPa 1 20 kPa 1 w h h P P w h h f f v v v 2035 8 kJkg kPa 20 1871 kJkg K 6 2953 6 kJkg C 320 kPa 6000 4 3 4 4 3 3 3 3 h s s s P s h T P 6 kJkg 2127 2035 8 0 902953 6 6 2953 4 3 3 4 4 3 4 3 s T s T h h h h h h h h η η preparation If you are a student using this Manual you are using it without permission 1087 The net work outputs from each cycle are 2 kJkg 275 293K 572 7 819 5 1005 kJkg K1373 5 6 8 7 Cin Tout gas cycle net T T c T T c w w w p p 9 kJkg 819 6 08 2127 6 2953 6 Pin 4 3 Pin Tout steam cycle net w h h w w w An energy balance on the heat exchanger gives a a p w w a p m m h h T T c m h h m T T m c 0 1010 257 5 2953 6 548 6 1 005819 5 2 3 9 8 2 3 9 8 That is 1 kg of exhaust gases can heat only 01010 kg of water Then the mass flow rate of air is 2793 kgs 0 1010 819 9 kJkg air 275 2 1 000 kJs 100 net net w W ma PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1088 1086 A combined gassteam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle The mass flow rate of air for a specified power output is to be determined Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable fo Brayton cycle 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a 1 2 4s 4 3 6 MPa 20 kPa 6s Qout Qin 5 9 8s 7 GAS CYCLE STEAM CYCLE 1373 K 320C 6a 6 8 293 K T Analysis With an ideal regenerator the temperature of the air at the compressor exit will be heated to the to the temperature at the turbine exit Representing this state by 6a 5 K 819 8 6 T T a The rate of heat addition in the cycle is 370 kW 155 819 5 K C1373 279 3 kgs 1 005 kJkg 6 7 in a a p T T m c Q The thermal efficiency of the cycle is then 06436 155370 kW 000 kW 100 in net th Q W η s Without the regenerator the rate of heat addition and the thermal efficiency are 224640 kW K 572 7 C1373 1 005 kJkg 279 3 kgs 6 7 in T T m c Q a p 04452 224640 kW 000 kW 100 in net th Q W η The change in the thermal efficiency due to using the ideal regenerator is 01984 0 4452 0 6436 ηth PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1089 1087 The component of the combined cycle with the largest exergy destruction of the component of the combined cycle in Prob 1086 is to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis From Problem 1086 2821 kgs 0 1010279 3 1010 0 2 kJkg 1876 1 kJkg 2696 804 3 kJkg 4627 kJkg K 6 1871 kJkg K 6 8320 kJkg K 0 K 293 5 K 819 K 1373 1 4 out 2 3 in23 6 7 in67 4 3 20 kPa 2 1 sink 8 steam cycle source gas cycle source a w p f m m h h q h h q T T c q s s s s s T T T T 2278 kW 6 1871 2821 kgs293 K 6 4627 isentropi c process 0 3 4 0 34 destroyed 12 destroyed s s m T X X w 1 2 4s 4 3 6 MPa 20 kPa 6s Qout Qin 5 9 8s 7 GAS CYCLE STEAM CYCLE 320C 8 6 1373 K 293 K s 8665 kW 293 K 1876 2 kJkg 6 1871 0 8320 2821 kgs293 K sink out 4 1 0 41 destroyed T q s s m T X w kW 11260 0 8320 2821293 6 1871 819 5 1 005 ln 548 6 279 3 293 ln 2 3 0 8 9 0 23 0 89 0 heat exchanger destroyed s s m T T T c m T s m T s m T X a p a w a 6280 kW 0 287 ln8 293 279 3 293 1005ln 5727 ln ln 5 6 5 6 0 destroyed 56 P P R T T c m T X p a 23970 kW 1373 804 3 5727 279 3 293 1005ln 1373 ln source in 6 7 0 destroyed 67 T q T T c m T X p a 6396 kW 8 0 287 ln 1 1373 279 3 293 1005ln 8195 ln ln 7 8 7 8 0 destroyed 78 P P R T T c m T X p a The largest exergy destruction occurs during the heat addition process in the combustor of the gas cycle preparation If you are a student using this Manual you are using it without permission 1090 1088 A 280MW combined gassteam power plant is considered The topping cycle is a gasturbine cycle and the bottoming cycle is a nonideal Rankine cycle with an open feedwater heater The mass flow rate of air to steam the required rate of heat input in the combustion chamber and the thermal efficiency of the combined cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable specific heats Analysis a Using the properties of air from Table A17 the analysis of gas cycle yields T 1 2 7s 5 6s 4 9s 3 Qout Qin 8 12 11s 10 GAS CYCLE STEAM CYCLE 6 7 11 9 42126 kJkg K 420 67440 kJkg 59518 0 86 116107 07 1161 59518 kJkg 1519 11 167 1 1 167 1 116107 kJkg K 1100 66074 kJkg 0 82 30019 59584 30019 59584 kJkg 1525 1 386 11 1 386 30019 kJkg K 300 12 12 11 10 10 11 11 10 11 10 11 10 11 10 10 8 9 8 9 8 9 8 9 9 8 9 8 8 10 11 10 8 9 8 h T h h h h h h h h h P P P P P h T h h h h h h h h h P P P P P h T s T s T s r r r C s s C s r r r η η η η s From the steam tables Tables A4 A5 and A6 19260 kJkg 0 80 19181 080 kJkg kPa m 1 1 kJ 1 0 kPa m kg 800 000101 0 00101 m kg 81 kJkg 191 pIin 1 2 3 3 1 2 1 in pI 3 0 kPa 1 1 10 kPa 1 w h h P P w h h f f v v v 72555 kJkg 4 68 87 720 468 kJkg kPa m 1 1 kJ 800 kPa m kg 5000 0001115 0 001115 m kg 87 kJkg 720 pIin 3 4 3 3 3 4 3 in pII 3 MPa 80 3 MPa 80 3 w h h P P w h h f f v v v 6 4516 kJkg K 3069 3 kJkg 350 C MPa 5 5 5 5 5 s h T P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1092 1089 Problem 1088 is reconsidered The effect of the gas cycle pressure ratio on the ratio of gas flow rate to steam flow rate and cycle thermal efficiency is to be investigated Analysis The problem is solved using EES and the solution is given below Input data T8 300 K Gas compressor inlet P8 100 kPa Assumed air inlet pressure Pratio 11 Pressure ratio for gas compressor T10 1100 K Gas turbine inlet T12 420 K Gas exit temperature from Gastosteam heat exchanger P12 P8 Assumed air exit pressure Wdotnet280 MW Etacomp 082 Etagasturb 086 Etapump 10 Etasteamturb 086 P5 5000 kPa Steam turbine inlet T5 35027315 K Steam turbine inlet P6 800 kPa Extraction pressure for steam open feedwater heater P7 10 kPa Steam condenser pressure GAS POWER CYCLE ANALYSIS Gas Compressor anaysis s8ENTROPYAirTT8PP8 ss9s8 For the ideal case the entropies are constant across the compressor P9 PratioP8 Ts9temperatureAirsss9PP9Ts9 is the isentropic value of T9 at compressor exit Etacomp wgascompisenwgascomp compressor adiabatic efficiency wcomp wcompisen h8 wgascompisen hs9SSSF conservation of energy for the isentropic compressor assuming adiabatic kepe0 per unit gas mass flow rate in kgs h8ENTHALPYAirTT8 hs9ENTHALPYAirTTs9 h8 wgascomp h9SSSF conservation of energy for the actual compressor assuming adiabatic kepe0 T9temperatureAirhh9 s9ENTROPYAirTT9PP9 Gas Cycle External heat exchanger analysis h9 qin h10SSSF conservation of energy for the external heat exchanger assuming W0 kepe0 h10ENTHALPYAirTT10 P10P9 Assume process 910 is SSSF constant pressure QdotinMW1000kWMWmdotgasqin Gas Turbine analysis s10ENTROPYAirTT10PP10 ss11s10 For the ideal case the entropies are constant across the turbine P11 P10 Pratio Ts11temperatureAirsss11PP11Ts11 is the isentropic value of T11 at gas turbine exit Etagasturb wgasturb wgasturbisen gas turbine adiabatic efficiency wgasturbisen wgasturb h10 wgasturbisen hs11SSSF conservation of energy for the isentropic gas turbine assuming adiabatic kepe0 hs11ENTHALPYAirTTs11 h10 wgasturb h11SSSF conservation of energy for the actual gas turbine assuming adiabatic kepe0 T11temperatureAirhh11 s11ENTROPYAirTT11PP11 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1093 GastoSteam Heat Exchanger SSSF conservation of energy for the gastosteam heat exchanger assuming adiabatic W0 kepe0 mdotgash11 mdotsteamh4 mdotgash12 mdotsteamh5 h12ENTHALPYAir TT12 s12ENTROPYAirTT12PP12 STEAM CYCLE ANALYSIS Steam Condenser exit pump or Pump 1 analysis FluidSteamIAPWS P1 P7 P2P6 h1enthalpyFluidPP1x0 Saturated liquid v1volumeFluidPP1x0 s1entropyFluidPP1x0 T1temperatureFluidPP1x0 wpump1sv1P2P1SSSF isentropic pump work assuming constant specific volume wpump1wpump1sEtapump Definition of pump efficiency h1wpump1 h2 Steadyflow conservation of energy s2entropyFluidPP2hh2 T2temperatureFluidPP2hh2 Open Feedwater Heater analysis yh6 1yh2 1h3 Steadyflow conservation of energy P3P6 h3enthalpyFluidPP3x0 Condensate leaves heater as sat liquid at P3 T3temperatureFluidPP3x0 s3entropyFluidPP3x0 Boiler condensate pump or Pump 2 analysis P4 P5 v3volumeFluidPP3x0 wpump2sv3P4P3SSSF isentropic pump work assuming constant specific volume wpump2wpump2sEtapump Definition of pump efficiency h3wpump2 h4 Steadyflow conservation of energy s4entropyFluidPP4hh4 T4temperatureFluidPP4hh4 wsteampumps 1ywpump1 wpump2 Total steam pump work input mass steam Steam Turbine analysis h5enthalpyFluidTT5PP5 s5entropyFluidPP5TT5 ss6s5 hs6enthalpyFluidsss6PP6 Ts6temperatureFluidsss6PP6 h6h5Etasteamturbh5hs6Definition of steam turbine efficiency T6temperatureFluidPP6hh6 s6entropyFluidPP6hh6 ss7s5 hs7enthalpyFluidsss7PP7 Ts7temperatureFluidsss7PP7 h7h5Etasteamturbh5hs7Definition of steam turbine efficiency T7temperatureFluidPP7hh7 s7entropyFluidPP7hh7 SSSF conservation of energy for the steam turbine adiabatic neglect ke and pe h5 wsteamturb yh6 1yh7 Steam Condenser analysis 1yh7qout1yh1SSSF conservation of energy for the Condenser per unit mass QdotoutConvertMW kWmdotsteamqout Cycle Statistics MassRatiogastosteam mdotgasmdotsteam WdotnetConvertMW kWmdotgaswgasturbwgascomp mdotsteamwsteamturb wsteampumpsdefinition of the net cycle work PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course EtathWdotnetQdotinConvert Cycle thermal efficiency in percent preparation If you are a student using this Manual you are using it without permission 1094 Bwrmdotgaswgascomp mdotsteamwsteampumpsmdotgaswgasturb mdotsteamwsteamturb Back work ratio Wdotnetsteam mdotsteamwsteamturb wsteampumps Wdotnetgas mdotgaswgasturb wgascomp NetWorkRatiogastosteam WdotnetgasWdotnetsteam 00 11 22 33 44 55 66 77 88 99 110 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 s kJkgK T K 5000 kPa 800 kPa 10 kPa Combined Gas and Steam Power Cycle Pratio MassRatiogastosteam ηth 10 11 12 13 14 15 16 17 18 19 20 8775 9262 9743 1022 107 1117 1164 1212 1259 1307 1355 4203 4167 4122 4068 4008 394 3866 3786 3699 3607 3508 8 9 10 11 12 12 34 5 6 7 Steam Cycle Steam Cycle Gas Cycle Gas Cycle 10 12 14 16 18 20 8 9 10 11 12 13 14 Pratio MassRatiogastosteam 10 12 14 16 18 20 35 36 37 38 39 40 41 42 43 Pratio ηth PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1095 1090 A combined gassteam power plant is considered The topping cycle is a gasturbine cycle and the bottoming cycle is a nonideal reheat Rankine cycle The moisture percentage at the exit of the lowpressure turbine the steam temperature at the inlet of the highpressure turbine and the thermal efficiency of the combined cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable specific heats Analysis a We obtain the air properties from EES The analysis of gas cycle is as follows 47562 kJkg C 200 87198 kJkg 76379 0 80 1304 8 8 1304 76379 kJkg kPa 100 6 6456 kJkg kPa 700 C 950 1304 8 kJkg C 950 55721 kJkg 0 80 29016 50347 29016 50347 kJkg kPa 700 5 6648 kJkg kPa 100 C 15 28850 kJkg C 15 11 11 10 9 9 10 10 9 10 9 10 9 10 10 9 9 9 9 9 7 8 7 8 7 8 7 8 8 7 8 8 7 7 7 7 7 h T h h h h h h h h h s s P s P T h T h h h h h h h h h s s P s P T h T s T s T s C s s C s η η η η PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course From the steam tables Tables A4 A5 and A6 or from EES 19937 kJkg 7 65 81 191 56 kJkg 7 0 80 kPa m 1 1 kJ 10 kPa m kg 6000 000101 0 00101 m kg 81 kJkg 191 pIin 1 2 3 3 1 2 1 in pI 3 10 kPa 1 10 kPa 1 w h h P P w h h p f f η v v v 2366 4 kJkg 0 9091 2392 1 81 191 0 9091 7 4996 0 6492 7 4670 kPa 10 7 4670 kJkg K 3264 5 kJkg 400 C MPa 1 6 6 6 6 5 6 6 5 5 5 5 fg s f s fg f s s s h x h h s s s x s s P s h T P 1 2 6s s T 3 6 MPa 10 kPa 8s Qout Qin 7 11 10s 9 GAS CYCLE STEAM CYCLE 950C 15C 5 4 1 MPa 10 8 4s 6 3 1 2 Steam turbine Gas turbine Condenser pump 5 4 Combustion chamber 11 Compressor Heat exchanger 10 9 8 7 6 preparation If you are a student using this Manual you are using it without permission 1096 16 0 0158 0 9842 1 1 Percentage Moisture 0 9842 2546 5 kJkg kPa 10 2546 0 kJkg 2366 4 0 80 3264 5 5 3264 6 6 6 6 6 5 5 6 6 5 6 5 x x h P h h h h h h h h s T s T η η b Noting that 0 for the heat exchanger the steadyflow energy balance equation yields pe ke Q W 2965 0 kJkg 47562 1087198 3264 5 19937 15 3346 5 1 4 4 11 10 air 4 5 2 3 out in h h h h m h h m h h m m h h m E E s s e e i i Also s T s T s s h h h h h h h h h s s P s h T P 4 3 3 4 4 3 4 3 4 3 4 4 3 3 3 3 1 MPa MPa 6 η η The temperature at the inlet of the highpressure turbine may be obtained by a trialerror approach or using EES from the above relations The answer is T3 4680ºC Then the enthalpy at state 3 becomes h3 33465 kJkg c 4328 kW 87198 kJkg 10 kgs 1304 8 10 9 air Tgas h h m W 2687 kW 28850 kJkg 10 kgs 55721 7 8 air Cgas h h m W 1641 kW 2687 4328 Cgas Tgas netgas W W W 1265 kW kJkg 2546 0 3264 5 2965 0 115 kgs 3346 5 6 5 4 3 s Tsteam h h h m h W kW 78 7 564 kJkg 115 kgs s Psteam m wpump W 1256 kW 78 1265 Psteam Tsteam netsteam W W W 2897 kW 1256 1641 netsteam netgas netplant W W W d 7476 kW 55721 kJkg 10 kgs 1304 8 8 9 air in h h m Q 388 0388 7476 kW kW 2897 in netplant th Q W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1098 Review Problems 1096 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits The thermal efficiency of the cycle is to be compared when it is operated so that the liquid enters the pump as a saturated liquid against that when the liquid enters as a subcooled liquid determined power produced by the turbine and consumed by the pump are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the steam tables Tables A4 A5 and A6 34667 kJkg 6 13 34054 6 13 kJkg kPa m 1 1kJ 50kPa 0 001030 m kg6000 001030 m kg 0 54 kJkg 340 pin 1 2 3 3 1 2 1 in p 3 20 kPa 1 50 kPa 1 w h h P P w h h f f v v v 2495 0 kJkg 0 93482304 7 54 340 0 9348 6 5019 1 0912 7 1693 kPa 50 7 1693 kJkg K 3658 8 kJkg 600 C kPa 6000 4 4 4 4 3 4 4 3 3 3 3 fg f fg f x h h h s s s x s s P s h T P qin qout 50 kPa 1 3 2 4 6 MPa T s Thus 2154 5 kJkg 34054 0 2495 3312 1 kJkg 34667 8 3658 1 4 out 2 3 in h h q h h q and the thermal efficiency of the cycle is 03495 3312 1 2154 5 1 1 in out th q q η When the liquid enters the pump 113C cooler than a saturated liquid at the condenser pressure the enthalpies become 001023 m kg 0 29307 kJkg 70 C 11 3 81 3 3 11 kPa 50 3 70 C 1 70 C 1 sat 50 kPa 1 1 f hf h T T P v v 6 09 kJkg kPa m 1 1kJ 50kPa 0 001023 m kg6000 3 3 1 2 1 in p P P w v 29916 kJkg 6 09 29307 pin 1 2 w h h Then 2201 9 kJkg 29309 0 2495 3359 6 kJkg 29916 8 3658 1 4 out 2 3 in h h q h h q 03446 3359 6 2201 9 1 1 in out th q q η The thermal efficiency slightly decreases as a result of subcooling at the pump inlet PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1099 1097E A geothermal power plant operating on the simple Rankine cycle using an organic fluid as the working fluid is considered The exit temperature of the geothermal water from the vaporizer the rate of heat rejection from the working fluid in the condenser the mass flow rate of geothermal water at the preheater and the thermal efficiency of the Level I cycle of this plant are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a The exit temperature of geothermal water from the vaporizer is determined from the steadyflow energy balance on the geothermal water brine PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2674F 2 2 1 2 brine brine 325 F F 384286 lbmh 103 Btulbm 790000 Btuh 22 T T T T c m Q p b The rate of heat rejection from the working fluid to the air in the condenser is determined from the steadyflow energy balance on air 297 MBtuh 55 F F 84 5 lbmh 024 Btulbm 4195100 8 9 air air T T c m Q p c The mass flow rate of geothermal water at the preheater is determined from the steadyflow energy balance on the geothermal water 187120 lbmh geo geo in out geo geo F 211 8 F 154 0 103 Btulbm 140000 Btuh 11 m m T T c m Q p d The rate of heat input is and Q Q Q W in vaporizer reheater net Btu h kW 22 790 000 11140 000 33 930 000 1271 200 1071 Then 108 1 kWh 341214 Btu 33930000 Btuh kW 1071 in net th Q W η preparation If you are a student using this Manual you are using it without permission 10100 1098 A steam power plant operating on an ideal Rankine cycle with two stages of reheat is considered The thermal efficiency of the cycle and the mass flow rate of the steam are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a From the steam tables Tables A4 A5 and A6 29937 kJkg 1019 18 289 1019 kJkg kPa m 1 1 kJ 30 kPa m kg 10000 0001022 0 001022 m kg 18 kJkg 289 pin 1 2 3 3 1 2 1 in p 3 30 kPa 1 30 kPa 1 w h h P P w h h f f v v v T 2557 1 kJkg 0 9711 2335 3 27 289 0 9711 6 8234 0 9441 7 5706 kPa 30 7 5706 kJkg K 3578 4 kJkg 550 C MPa 2 3321 1 kJkg MPa 2 7 2335 kJkg K 3559 7 kJkg 550 C MPa 4 3204 9 kJkg MPa 4 6 7561 kJkg K 3500 9 kJkg 550 C MPa 10 8 8 8 8 7 8 8 7 7 7 7 6 5 6 6 5 5 5 5 4 3 4 4 3 3 3 3 fg f fg f x h h h s s s x s s P s h T P h s s P s h T P h s s P s h T P 1 5 2 8 3 4 7 6 2 MPa 4 MPa 30 kPa 10 MPa s Then 1545 8 kJkg 2267 9 7 3813 2267 9 kJkg 28918 2557 1 3813 7 kJkg 3321 1 3578 4 3204 9 3559 7 29937 9 3500 out in net 1 8 out 6 7 4 5 2 3 in q q w h h q h h h h h h q Thus 405 0 4053 38137 kJkg kJkg 15458 in net th q w η b The mass flow rate of the steam is then 485 kgs 15458 kJkg kJs 75000 net net w W m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10101 1099 A steam power plant operating on the ideal Rankine cycle with reheating is considered The reheat pressures of the cycle are to be determined for the cases of single and double reheat Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a Single Reheat From the steam tables Tables A4 A5 and A6 2780 kPa 5 6 5 5 6 6 6 6 6 6 C 600 7 5488 kJkg K 7 4996 0 92 0 6492 2392 5 kJkg 0 92 2392 1 19181 0 92 kPa 10 P s s T x s s s x h h h x P fg f fg f T 1 5 2 6 s 3 4 25 MPa 10 kPa SINGLE 600C 1 5 2 8 3 4 7 6 DOUBLE 10 kPa 25 MPa s T 600C b Double Reheat 600 C and 6 3637 kJkg K 600 C MPa 25 5 5 3 4 4 3 3 3 T P P s s P P s T P x x Any pressure Px selected between the limits of 25 MPa and 278 MPa will satisfy the requirements and can be used for the double reheat pressure PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10102 10100 An 150MW steam power plant operating on a regenerative Rankine cycle with an open feedwater heater is considered The mass flow rate of steam through the boiler the thermal efficiency of the cycle and the irreversibility associated with the regeneration process are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis T y P II P I Open fwh Condenser Boiler 4 3 2 1 6 Turbine 7 1y 5 1 qin 2 7s 7 5 6s 10 MPa 10 kPa 05 MPa 1y 4 y 3 qout 6 s a From the steam tables Tables A4 A5 and A6 m kg 0 001093 kJkg 09 640 liquid sat MPa 50 kJkg 33 0 52 192 81 191 kJkg 052 0 95 kPa m 1 1kJ kPa m kg 500 10 000101 m kg 00101 0 kJkg 81 191 3 MPa 50 3 MPa 50 3 3 pIin 1 2 3 3 1 2 1 in pI 3 kPa 10 1 kPa 10 1 η f f p f f h h P w h h P P w h h v v v v v 65102 kJkg 1093 64009 kJkg 1093 0 95 kPa m 1 1 kJ 500 kPa m kg 10000 0001093 pIIin 3 4 3 3 3 4 3 in pII η w h h P P w p v 2654 1 kJkg 0 9554 2108 0 09 640 0 9554 4 9603 1 8604 5995 6 MPa 50 6 5995 kJkg K 3375 1 kJkg 500 C MPa 10 6 6 6 6 5 6 6 5 5 5 5 fg s f s fg f s s s s h x h h s s s x s s P s h T P 2798 3 kJkg 2654 1 0 80 3375 1 3375 1 6 5 5 6 6 5 6 5 s T s T h h h h h h h h η η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10107 10103 A nonideal reheatregenerative Rankine cycle with one open feedwater heater is considered The fraction of steam extracted for regeneration and the thermal efficiency of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course y P II P I Open fwh Condenser 4 3 2 1 5 8 Boiler 7 6 Turbine 9 1y 1 7 2 9 5 6 8 4 1y 9s 6s 8s y 3 s a From the steam tables Tables A4 A5 and A6 0 001101 m kg 67038 kJkg sat liquid MPa 60 22654 kJkg 0 59 22594 059 kJkg kPa m 1 1 kJ m kg 600 15 kPa 0001014 001014 m kg 0 94 kJkg 225 3 MPa 60 3 MPa 60 3 3 in 1 2 3 3 1 2 1 in 3 15 kPa 1 15 kPa 1 f f pI pI f f h h P w h h P P w h h v v v v v 2783 8 kJkg MPa 01 6 5995 kJkg K 3375 1 kJkg 500 C MPa 10 68073 kJkg 1035 67038 1035 kJkg kPa m 1 1 kJ 600 kPa m kg 10000 0001101 6 5 6 6 5 5 5 5 pIIin 3 4 3 3 3 4 3 in pII s s s h s s P s h T P w h h P P w v 2878 4 kJkg 2783 8 0 84 3375 1 3375 1 6 5 5 6 6 5 6 5 η η s T s T h h h h h h h h 3337 2 kJkg 3310 2 0 84 3479 1 1 3479 3310 2 kJkg MPa 60 7 7642 kJkg K 3479 1 kJkg 500 C MPa 01 8 7 7 8 8 7 8 7 8 7 8 8 7 7 7 7 η η s T s T s s s h h h h h h h h h s s P s h T P preparation If you are a student using this Manual you are using it without permission 10109 10104 A steam power plant operating on the ideal reheatregenerative Rankine cycle with three feedwater heaters is considered Various items for this system per unit of mass flow rate through the boiler are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 16 n 13 1 LowP Turbine Boiler Condenser HighP Turbine 14 2 19 15 m z P IV P I Closed FWH I 12 11 10 9 8 6 3 4 5 7 17 Closed FWH II Open FWH P II y x P III 18 Analysis The compression processes in the pumps and the expansion processes in the turbines are isentropic Also the state of water at the inlet of pumps is saturated liquid Then from the steam tables Tables A4 A5 and A6 1011 8 kJkg 1008 3 kJkg 88586 kJkg 88446 kJkg 41928 kJkg 41751 kJkg 16884 kJkg 75 kJkg 168 10 9 6 5 4 3 2 1 h h h h h h h h 7 kJkg 2454 5 kJkg 2891 3 kJkg 3481 0 kJkg 2871 6 kJkg 3063 5 kJkg 3204 1 kJkg 3423 19 18 17 16 15 14 13 h h h h h h h For an ideal closed feedwater heater the feedwater is heated to the exit temperature of the extracted steam which ideally leaves the heater as a saturated liquid at the extraction pressure Then 1008 8 kJkg C 9 233 kPa 3000 88491 kJkg C 1 207 kPa 1800 11 9 11 11 7 5 7 7 h T T P h T T P Enthalpies at other states and the fractions of steam extracted from the turbines can be determined from mass and energy balances on cycle components as follows Mass Balances z n m z y x 1 Open feedwater heater 3 2 18 zh nh mh Closed feedwater heaterII 5 7 15 4 yh zh yh zh PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10110 Closed feedwater heaterI 9 11 14 8 xh z h y xh z h y Mixing chamber after closed feedwater heater II 8 6 7 z h y yh zh Mixing chamber after closed feedwater heater I 12 11 10 1 h z h y xh Substituting the values and solving the above equations simultaneously using EES we obtain 070882 007124 01667 005334 n m z y x h h 78000 0 0 kJkg 1009 08 kJkg 885 12 8 Note that these values may also be obtained by a hand solution by using the equations above with some rearrangements and substitutions Other results of the cycle are 439 kJkg 1620 kJkg 2890 kJkg 7696 kJkg 5023 0 4394 2890 1620 1 1 in out th 1 19 out 16 17 12 13 in 19 17 18 17 outLP T 16 13 15 13 14 13 outHP T q q h n h q h z h h h q h n h h m h w h z h h y h h x h w η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10111 10105 The optimum bleed pressure for the open feedwater heater that maximizes the thermal efficiency of the cycle is to be determined using EES Analysis The EES program used to solve this problem as well as the solutions are given below Given Pboiler6000 kPa Pcfwh13000 kPa Pcfwh21800 kPa Preheat800 kPa Pofwh100 kPa Pcondenser75 kPa Tturbine500 C Analysis Fluidsteamiapws turbines h13enthalpyFluid PPboiler TTturbine s13entropyFluid PPboiler TTturbine h14enthalpyFluid PPcfwh1 ss13 h15enthalpyFluid PPcfwh2 ss13 h16enthalpyFluid PPreheat ss13 h17enthalpyFluid PPreheat TTturbine s17entropyFluid PPreheat TTturbine h18enthalpyFluid PPofwh ss17 h19enthalpyFluid PPcondenser ss17 pump I h1enthalpyFluid PPcondenser x0 v1volumeFluid PPcondenser x0 wpIinv1PofwhPcondenser h2h1wpIin pump II h3enthalpyFluid PPofwh x0 v3volumeFluid PPofwh x0 wpIIinv3Pcfwh2Pofwh h4h3wpIIin pump III h5enthalpyFluid PPcfwh2 x0 T5temperatureFluid PPcfwh2 x0 v5volumeFluid PPcfwh2 x0 wpIIIinv5Pcfwh1Pcfwh2 h6h5wpIIIin pump IV h9enthalpyFluid PPcfwh1 x0 T9temperatureFluid PPcfwh1 x0 v9volumeFluid PPcfwh1 x0 wp4inv5PboilerPcfwh1 h10h9wp4in Mass balances xyz1 mnz PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10112 Open feedwater heater mh18nh2zh3 closed feedwater heater 2 T7T5 h7enthalpyFluid PPcfwh1 TT7 zh4yh15zh7yh5 closed feedwater heater 1 T11T9 h11enthalpyFluid PPboiler TT11 yzh8xh14yzh11xh9 Mixing chamber after closed feedwater heater 2 zh7yh6yzh8 Mixing chamber after closed feedwater heater 1 xh10yzh111h12 cycle wTouthighxh13h14yh13h15zh13h16 wToutlowmh17h18nh17h19 qinh13h12zh17h16 qoutnh19h1 Etath1qoutqin P open fwh kPa ηth 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 300 0429828 0433780 0435764 0436978 0437790 0438359 0438768 0439065 0439280 0439432 0439536 0439602 0439638 0439647 0439636 0439608 0439565 0439509 0439442 0439367 0439283 0439192 0439095 0438993 0438887 0438776 0438662 0438544 0438424 0438301 0 50 100 150 200 250 300 0428 043 0432 0434 0436 0438 044 Pofwh kPa ηth PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10113 10106 A cogeneration plant is to produce power and process heat There are two turbines in the cycle a highpressure turbine and a lowpressure turbine The temperature pressure and mass flow rate of steam at the inlet of highpressure turbine are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the steam tables Tables A4 A5 and A6 2344 1 kJkg 2047 6 0 60 2788 9 9 2788 2047 6 kJkg 0 7758 2392 1 81 191 0 7758 7 4996 0 6492 4675 6 kPa 10 6 4675 kJkg K 2788 9 kJkg sat vapor MPa 41 5 4 4 5 5 4 5 4 5 5 4 5 4 5 5 MPa 41 4 MPa 41 4 4 s T s T fg s f s fg f s s s g g h h h h h h h h x h h h s s s x s s P s s h h P η η 1 2 5 s T 5 4 4 3 and 107 9 kgmin 1 799 kgs 4448 kJkg kJs 800 444 8 kJkg 2344 1 9 2788 low turb turbII turb low 5 4 low turb w W m h h w Therefore 2843 0 kJkg 2788 9 15 54 5415 kJkg 1847 kgs kJs 1000 1108 kgmin 108 1000 4 turbhigh 3 4 3 turb high turb high turb total h w h h h m W w m I kgs 1847 6 4289 kJkg K 4 1840 0 9908 2835 2 0 9908 1958 9 82996 2770 8 MPa 41 2770 8 kJkg 0 75 2788 9 2843 0 2843 0 4 4 4 4 3 4 4 4 3 3 4 4 3 4 3 fg s f s fg f s s s s T s s T s x s s h h h x s s P h h h h h h h h η η Then from the tables or the software the turbine inlet temperature and pressure becomes 2275 C 2 MPa 3 3 3 3 6 4289 kJkg K 2843 0 kJkg T P s h PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10114 10107 A cogeneration plant is to generate power and process heat Part of the steam extracted from the turbine at a relatively high pressure is used for process heating The rate of process heat the net power produced and the utilization factor of the plant are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 1 2 8s 8 T 6 7 8 MPa 20 kPa 2 MPa 4 5 3 Qout Qin Qprocess Conden 6 Turbine Process he ater I 3 7 4 2 P I P I Boiler 5 1 8 s Analysis From the steam tables Tables A4 A5 and A6 47 kJkg 908 25371 kJkg 2 29 42 251 29 kJkg 2 0 88 kPa m 1 1 kJ 2 0 kPa m kg 2000 0001017 001017 m kg 0 42 kJkg 251 2 MPa 3 pIin 1 2 3 3 1 2 1 in pI 3 2 0 kPa 1 20 kPa 1 f p f f h h w h h P P w h h η v v v Mixing chamber 49181 kJkg 11 kgs 4 kgs25371 kJkg 11 4 kgs90847 kJkg 4 4 4 4 2 2 3 3 h h m h m h m h 49902 kJkg 7 21 81 491 21 kJkg 7 0 88 kPa m 1 1 kJ 2000 kPa m kg 8000 0001058 001058 m kg 0 IIin 4 5 3 3 4 5 4 in II 3 49181 kJkg 4 p p p h f w h h P P w f η v v v 6 7266 kJkg K 3399 5 kJkg 500 C MPa 8 6 6 6 6 s h T P 3000 4 kJkg 2 MPa 7 6 7 7 h s s s P 3048 3 kJkg 3000 4 0 88 3399 5 3399 5 7 6 6 7 7 6 7 6 s T s T h h h h h h h h η η 2215 5 kJkg 20 kPa 8 6 8 8 h s s s P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10115 2357 6 kJkg 2215 5 0 88 3399 5 3399 5 8 6 6 8 8 6 8 6 s T s T h h h h h h h h η η Then 8559 kW 90847 kJkg 4 kgs 3048 3 3 7 7 process h h m Q b Cycle analysis 8603 kW 95 8698 95 kW 11 kgs 721 kJkg kgs 229 kJkg 7 8698 kW 2357 6 kJkg 7 kgs 3399 5 3048 3 kJkg kgs 3399 5 4 pin Tout net pIIin 4 pIin 1 in p 8 6 8 7 6 7 out T W W W m w m w W h h m h h m W c Then and 538 0 538 31905 8559 8603 31905 kW 49902 kgs 3399 5 11 in process net 5 6 5 in Q Q W h h m Q u ε PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10116 10108E A combined gassteam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle The thermal efficiency of the cycle is to be determined Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable for Brayton cycle 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 0240 BtulbmR and k 14 Table A2Ea Analysis Working around the topping cycle gives the following results 1043 R 540 R10 0414 1 5 6 5 6 k k s P P T T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1099 R 0 90 540 1043 540 5 6 5 6 5 6 5 6 5 6 5 6 C s p s p s C T T T T T T c T T c h h h h η η 1326 R 10 2560 R 1 0414 1 7 8 7 8 k k s P P T T R 1449 0 902560 1326 2560 8 7 7 8 8 7 8 7 8 7 8 7 s T s p p s T T T T T T T c T T c h h h h η η 1 2 4s 4 3 800 psia 5 psia 6s 2560 R 540 R Qout Qin 5 9 8s 7 GAS CYCLE STEAM CYCLE 600F 6 8 T s 1028 R 50 978 3 R 50 sat 800 psia 9 T T Fixing the states around the bottom steam cycle yields Tables A4E A5E A6E 13259 Btulbm 2 41 13018 2 41 Btulbm psia ft 5404 1Btu 5psia 0 01641 ft lbm800 01641 ft lbm 0 18 Btulbm 130 pin 1 2 3 3 1 2 1 in p 3 5 psia 1 5 psia 1 w h h P P w h h f f v v v 908 6 Btulbm psia 5 1 4866 Btulbm R 1270 9 Btulbm 600 F psia 800 4 3 4 4 3 3 3 3 h s s s P s h T P 926 7 Btulbm 908 6 0 951270 9 9 1270 4 3 3 4 4 3 4 3 s T s T h h h h h h h h η η The net work outputs from each cycle are 5 Btulbm 132 540R 0 240 Btulbm R2560 1449 1099 5 6 8 7 Cin Tout gas cycle net T T c T T c w w w p p preparation If you are a student using this Manual you are using it without permission 10117 8 Btulbm 341 2 41 926 7 1270 9 Pin 4 3 Pin Tout steam cycle net w h h w w w An energy balance on the heat exchanger gives a a p w w a p m m h h T T c m h h m T T m c 0 08876 13259 1270 9 0 2401449 1028 2 3 9 8 2 3 9 8 That is 1 lbm of exhaust gases can heat only 008876 lbm of water Then the heat input the heat output and the thermal efficiency are 8 Btulbm 187 13018 Btulbm 0 08876 926 7 540R 0 240 Btulbm R1028 1 350 6 Btulbm 0 240 Btulbm R2560 1099R 1 4 5 9 out 6 7 in h h m m T T c m m q T T c m m q a w p a a p a a 04643 350 6 187 8 1 1 in out th q q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10118 10109E A combined gassteam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle The thermal efficiency of the cycle is to be determined Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable fo Brayton cycle 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 0240 BtulbmR and k 14 Table A2Ea 1 2 4s 4 3 800 psia 10 psia 6s 2560 R 540 R Qout Qin 5 9 8s 7 GAS CYCLE STEAM CYCLE 600F 6 8 T Analysis Working around the topping cycle gives the following results 1043 R 540 R10 0414 1 5 6 5 6 k k s P P T T 1099 R 0 90 540 1043 540 5 6 5 6 5 6 5 6 5 6 5 6 C s p s p s C T T T T T T c T T c h h h h η η 1326 R 10 2560 R 1 0414 1 7 8 7 8 k k s P P T T s R 1449 0 902560 1326 2560 8 7 7 8 8 7 8 7 8 7 8 7 s T s p p s T T T T T T T c T T c h h h h η η 1028 R 50 978 3 R 50 sat 800 psia 9 T T Fixing the states around the bottom steam cycle yields Tables A4E A5E A6E 163 7 Btulbm 2 43 16125 2 43 Btulbm psia ft 5404 1Btu 0 01659 ft lbm800 10psia 01659 ft lbm 0 25 Btulbm 161 pin 1 2 3 3 1 2 1 in p 3 10 psia 1 10 psia 1 w h h P P w h h f f v v v 946 6 Btulbm psia 10 1 4866 Btulbm R 1270 9 Btulbm 600 F psia 800 4 3 4 4 3 3 3 3 h s s s P s h T P 962 8 Btulbm 946 6 0 951270 9 9 1270 4 3 3 4 4 3 4 3 s T s T h h h h h h h h η η The net work outputs from each cycle are PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10119 5 Btulbm 132 540R 0 240 Btulbm R2560 1449 1099 5 6 8 7 Cin Tout gas cycle net T T c T T c w w w p p 305 7 Btulbm 2 43 962 8 1270 9 Pin 4 3 Pin Tout steam cycle net w h h w w w An energy balance on the heat exchanger gives a a p w w a p m m h h T T c m h h m T T m c 0 09126 163 7 1270 9 0 2401449 1028 2 3 9 8 2 3 9 8 That is 1 lbm of exhaust gases can heat only 009126 lbm of water Then the heat input the heat output and the thermal efficiency are 3 Btulbm 190 16125 Btulbm 0 09126 962 8 540R 0 240 Btulbm R1028 1 350 6 Btulbm 0 240 Btulbm R2560 1099R 1 4 5 9 out 6 7 in h h m m T T c m m q T T c m m q a w p a a p a a 04573 350 6 190 3 1 1 in out th q q η When the condenser pressure is increased from 5 psia to 10 psia the thermal efficiency is decreased from 04643 to 04573 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10120 10110E A combined gassteam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle The cycle supplies a specified rate of heat to the buildings during winter The mass flow rate of air and the net power output from the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 The air standard assumptions are applicable to Brayton cycle 3 Kinetic and potential energy changes are negligible 4 Air is an ideal gas with constant specific heats PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The properties of air at room temperature are cp 0240 BtulbmR and k 14 Table A2Ea Analysis The mass flow rate of water is 2495 lbmh 962816125 Btulbm 10 Btuh 2 6 1 4 buildings h h Q mw The mass flow rate of air is then 27340 lbmh 009126 2495 09126 0 w a m m The power outputs from each cycle are 1062 kW 341214 Btuh 1kW 540R 27340 lbmh 0 240 Btulbm R2560 1449 1099 5 6 8 7 Cin Tout gas cycle net T T m c T T c m w w m W p a p a a 1 2 4s 4 3 800 psia 10 psia 6s 2560 R 540 R Qout Qin 5 9 8s 7 GAS CYCLE STEAM CYCLE 600F 6 8 T s 224 kW 341214 Btuh 1kW 2 43 962 8 2495 lbmh1270 9 Pin 4 3 Pin Tout steam cycle net w h h m w w m W a a The net electricity production by this cycle is then 1286 kW 224 1062 net W preparation If you are a student using this Manual you are using it without permission 10121 10111 A combined gassteam power plant is considered The topping cycle is an ideal gasturbine cycle and the bottoming cycle is an ideal reheat Rankine cycle The mass flow rate of air in the gasturbine cycle the rate of total heat input and the thermal efficiency of the combined cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable specific heats Analysis a The analysis of gas cycle yields PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 52363 kJkg K 520 76838 kJkg 3754 450 5 12 1 450 5 151542 kJkg K 1400 63018 kJkg 1866 1 5546 12 1 5546 31024 kJkg K 310 11 11 10 9 10 9 9 8 7 8 7 7 9 10 9 7 8 7 h T h P P P P P h T h P P P P P h T r r r r r r From the steam tables Tables A4 A5 and A6 g w h h P P w h h f f 20442 kJk 1262 81 191 1262 kJkg kPa m 1 1 kJ 10 kPa m kg 12500 000101 00101 m kg 0 81 kJkg 191 pIin 1 2 3 3 1 2 1 in pI 3 10 kPa 1 10 kPa 1 v v v 1 2 6 3 125 MPa 10 kPa 8 Qout Qin 7 11 10 9 GAS CYCLE STEAM CYCLE 1400 K 310 K 5 4 25 MPa T s 2365 8 kJkg 0 9089 2392 1 81 191 0 9089 7 4996 0 6492 7 4653 kPa 10 7 4653 kJkg K 3574 4 kJkg 550 C MPa 52 2909 6 kJkg MPa 52 6 4651 kJkg K 3343 6 kJkg 500 C 5 MPa 12 6 6 6 6 5 6 6 5 5 5 5 4 3 4 4 3 3 3 3 fg f fg f x h h h s s s x s s P s h T P h s s P s h T P Noting that 0 for the heat exchanger the steadyflow energy balance equation yields pe ke Q W 1539 kgs 52363 12 kgs 76838 20442 6 3343 11 10 2 3 air 11 10 air 2 3 out in s s e e i i m h h h h m h h m h h m m h m h E E b The rate of total heat input is 10 kW 144 5 144200 kW kJkg 2909 6 12 kgs 3574 4 63018 kJkg 1539 kgs 151542 4 5 reheat 8 9 air reheat air in h h m h h m Q Q Q c The rate of heat rejection and the thermal efficiency are then 591 0 5913 144200 kW 58930 kW 1 1 58 930 kW 19181 kJkg 12 kgs 2365 8 31024 kJkg kgs 52363 1539 in out th 1 6 7 11 air outsteam outair out Q Q h m h h h m Q Q Q s η preparation If you are a student using this Manual you are using it without permission 10122 10112 A combined gassteam power plant is considered The topping cycle is a gasturbine cycle and the bottoming cycle is a nonideal reheat Rankine cycle The mass flow rate of air in the gasturbine cycle the rate of total heat input and the thermal efficiency of the combined cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable specific heats PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis a The analysis of gas cycle yields Table A17 52363 kJkg K 520 958 4 kJkg 86035 0 85 151542 42 1515 86035 kJkg 56 3 450 5 8 1 450 5 151542 kJkg K 1400 585 1 kJkg 0 80 29016 52612 29016 52612 kJkg 9 849 1 2311 8 1 2311 29016 kJkg K 290 11 11 10 9 9 10 10 9 10 9 10 9 10 9 9 7 8 7 8 7 8 7 8 8 7 8 7 7 9 10 9 7 8 7 η η η η h T h h h h h h h h h P P P P P h T h h h h h h h h h P P P P P h T s T s T s r s r r C s s C s r s r r s s 1 2 6s 3 15 MPa 10 kPa 8 Qout Qin 7 11 10s 9 GAS CYCLE STEAM CYCLE 1400 K 290 K 5 4 3 MPa 6 4 8 1 T s From the steam tables Tables A4 A5 and A6 20695 kJkg 1514 19181 1514 kJkg kPa m 1 1 kJ 10 kPa m kg 15000 000101 0 00101 m kg 81 kJkg 191 pIin 1 2 3 3 1 2 1 in pI 3 10 kPa 1 10 kPa 1 w h h P P w h h f f v v v 2838 1 kJkg 2781 7 0 85 3157 9 9 3157 2781 7 kJkg 0 9879 1794 9 3 1008 0 9880 3 5402 2 6454 6 1434 MPa 3 6 1428 kJkg K 3157 9 kJkg 450 C MPa 15 4 3 3 4 4 3 4 3 4 4 4 4 3 4 4 3 3 3 3 η η s T s T fg s f s fg f s s s h h h h h h h h x h h h s s s x s s P s h T P preparation If you are a student using this Manual you are using it without permission 10124 10113 A Rankine steam cycle modified with two closed feedwater heaters and one open feed water heater is considered The Ts diagram for the ideal cycle is to be sketched The fraction of mass extracted for the open feedwater heater y and the cooling water flow temperature rise are to be determined Also the rate of heat rejected in the condenser and the thermal efficiency of the plant are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis b Using the data from the problem statement the enthalpies at various states are 1 2 12 s 7 8 140 kPa 20 kPa 1910 kPa 4 3 5 6 5 MPa 9 11 14 15 1yzw 13 z w w y 10 wz kJkg 898 kJkg 676 kJkg 458 kJkg 251 1910 kPa 12 6 62 0 kPa 4 140 kPa 14 3 15 2 0 kPa 1 f f f f h h h h h h h h h h h T 620 kPa An energy balance on the open feedwater heater gives 4 3 9 1 1 h y h yh where z is the fraction of steam extracted from the lowpressure turbine Solving for z 008086 458 3154 458 676 3 9 3 4 h h h h y c An energy balance on the condenser gives w pw w w w w T m c h h m y h z h w z h y w m 1 1 2 2 1 15 11 7 Solving for the temperature rise of cooling water and substituting with correct units 995C 4200 4 18 0 08086251 1 0 0655458 0 0830 0 06552478 0 08086 0 0830 100 1 1 1 1 15 11 7 pw w w m c y h z h w z h y w m T d The rate of heat rejected in the condenser is 174700 kW C 9 95 C 4200 kgs418 kJkg out w w pw T m c Q The rate of heat input in the boiler is 300200 kW 898 kJkg 100 kgs3900 6 7 in h m h Q The thermal efficiency is then 418 0 418 300200 kW 174700 kW 1 1 in out th Q Q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10125 10114 A Rankine steam cycle modified for reheat two closed feedwater heaters and a process heater is considered The Ts diagram for the ideal cycle is to be sketched The fraction of mass w that is extracted for the closed feedwater heater is to be determined Also the mass flow rate through the boiler the rate of process heat supplied and the utilization efficiency of this cogeneration plant are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis b Using the data from the problem statement the enthalpies at various states are 256 5 kJkg 15 251 4 kJkg 15 kPa m 1 1 kJ 20 kPa m kg 5000 000102 00102 m kg 0 4 kJkg 251 pIin 1 2 3 3 1 2 1 in pI 3 20 kPa 1 20 kPa 1 w h h P P w h h f f v v v 1 2 14 s 17 11 5 6 3 245 kPa 20 kPa 14 MPa 7 4 5 MPa 9 12 13 15 z 1yzw y 8 w 16 10 12 MPa 150 kPa T kJkg 467 kJkg 830 kJkg 533 150 kPa 16 3 1400 kPa 14 15 4 245 kPa 12 13 f f f h h h h h h h h h h An energy balance on the closed feedwater heater gives 16 3 15 13 10 2 1 1 w h z y h yh zh wh h where w is the fraction of steam extracted from the lowpressure turbine Solving for z 00620 467 3023 0 1160830 0 15533 0 15467 0 1160 256 5 467 16 10 15 13 16 2 3 h h yh zh z h y h h w c The work output from the turbines is 1381 6 kJkg 0 06202620 0 15 0 1160 1 0 06203023 0 153154 0 11603692 1 0 11603349 1 0 11603400 3894 1 1 1 11 10 9 8 7 6 5 out T w h z y wh zh y h y h yh h w The net work output from the cycle is 1376 5 kJkg 15 1381 6 Pin Tout net w w w The mass flow rate through the boiler is 2179 kgs 13765 kJkg 000 kW 300 net net w W m The rate of heat input in the boiler is 700 kW 733 3349 kJkg 0 1160217 9 kgs3692 1 830 kJkg 217 9 kgs3894 1 7 8 4 5 in h y m h h m h Q The rate of process heat and the utilization efficiency of this cogeneration plant are 85670 kW 533 kJkg 0 15217 9 kgs3154 12 9 process h zm h Q 526 0 526 733700 kW 85670 kW 300000 in process net Q Q W u ε PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10126 10115 The effect of the condenser pressure on the performance a simple ideal Rankine cycle is to be investigated Analysis The problem is solved using EES and the solution is given below function x4x4 this function returns a string to indicate the state of steam at point 4 x4 if x41 then x4superheated if x40 then x4compressed end P3 10000 kPa T3 550 C P4 5 kPa Etat 10 Turbine isentropic efficiency Etap 10 Pump isentropic efficiency Pump analysis FluidSteamIAPWS P1 P4 P2P3 x10 Satd liquid h1enthalpyFluidPP1xx1 v1volumeFluidPP1xx1 s1entropyFluidPP1xx1 T1temperatureFluidPP1xx1 Wpsv1P2P1SSSF isentropic pump work assuming constant specific volume WpWpsEtap h2h1Wp SSSF First Law for the pump s2entropyFluidPP2hh2 T2temperatureFluidPP2hh2 Turbine analysis h3enthalpyFluidTT3PP3 s3entropyFluidTT3PP3 ss4s3 hs4enthalpyFluidsss4PP4 Ts4temperatureFluidsss4PP4 Etath3h4h3hs4Definition of turbine efficiency T4temperatureFluidPP4hh4 s4entropyFluidhh4PP4 x4qualityFluidhh4PP4 h3 Wth4SSSF First Law for the turbine x4sx4x4 Boiler analysis Qin h2h3SSSF First Law for the Boiler Condenser analysis h4Qouth1SSSF First Law for the Condenser Cycle Statistics WnetWtWp EtathWnetQin PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10127 P4 kPa ηth Wnet kJkg 5 15 25 35 45 55 65 75 85 100 04268 03987 03841 03739 03659 03594 03537 03488 03443 03385 1432 1302 1237 1192 1157 1129 1105 1084 1065 1040 0 2 4 6 8 10 12 0 100 200 300 400 500 600 700 s kJkgK T C 10 MPa 5 kPa SteamIAPWS 12 3 4 0 20 40 60 80 100 032 034 036 038 04 042 044 P4 kPa ηth 0 20 40 60 80 100 1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 P4 kPa Wnet kJkg PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10128 10116 The effect of superheating the steam on the performance a simple ideal Rankine cycle is to be investigated Analysis The problem is solved using EES and the solution is given below function x4x4 this function returns a string to indicate the state of steam at point 4 x4 if x41 then x4superheated if x40 then x4compressed end P3 3000 kPa T3 600 C P4 10 kPa Etat 10 Turbine isentropic efficiency Etap 10 Pump isentropic efficiency Pump analysis FluidSteamIAPWS P1 P4 P2P3 x10 Satd liquid h1enthalpyFluidPP1xx1 v1volumeFluidPP1xx1 s1entropyFluidPP1xx1 T1temperatureFluidPP1xx1 Wpsv1P2P1SSSF isentropic pump work assuming constant specific volume WpWpsEtap h2h1Wp SSSF First Law for the pump s2entropyFluidPP2hh2 T2temperatureFluidPP2hh2 Turbine analysis h3enthalpyFluidTT3PP3 s3entropyFluidTT3PP3 ss4s3 hs4enthalpyFluidsss4PP4 Ts4temperatureFluidsss4PP4 Etath3h4h3hs4Definition of turbine efficiency T4temperatureFluidPP4hh4 s4entropyFluidhh4PP4 x4qualityFluidhh4PP4 h3 Wth4SSSF First Law for the turbine x4sx4x4 Boiler analysis Qin h2h3SSSF First Law for the Boiler Condenser analysis h4Qouth1SSSF First Law for the Condenser Cycle Statistics WnetWtWp EtathWnetQin PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10129 0 2 4 6 8 10 12 0 100 200 300 400 500 600 700 s kJkgK T C 3000 kPa 10 kPa Steam 1 2 3 4 T3 C ηth Wnet kJkg x4 250 03241 8628 0752 3444 03338 9706 081 4389 03466 1083 08536 5333 03614 1206 08909 6278 03774 1340 09244 7222 03939 1485 0955 8167 04106 1639 09835 9111 04272 1803 100 1006 04424 1970 100 1100 0456 2139 100 200 300 400 500 600 700 800 900 1000 1100 032 034 036 038 04 042 044 046 T3 C ηth 200 300 400 500 600 700 800 900 1000 1100 750 1050 1350 1650 1950 2250 T3 C Wnet kJkg PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10130 10117 The effect of number of reheat stages on the performance an ideal Rankine cycle is to be investigated Analysis The problem is solved using EES and the solution is given below function x6x6 this function returns a string to indicate the state of steam at point 6 x6 if x61 then x6superheated if x60 then x6subcooled end Procedure ReheatP3T3T5h4NoRHStagesPratioEtatQinreheatWtlph6 P3P3 T5T5 h4h4 Qinreheat 0 Wtlp 0 RP1Pratio1NoRHStages1 imaxNoRHStages 1 i0 REPEAT ii1 P4 P3RP P5P4 P6P5RP FluidSteamIAPWS s5entropyFluidTT5PP5 h5enthalpyFluidTT5PP5 ss6s5 hs6enthalpyFluidsss6PP6 Ts6temperatureFluidsss6PP6 vs6volumeFluidsss6PP6 Etath5h6h5hs6Definition of turbine efficiency h6h5Etath5hs6 WtlpWtlph5h6SSSF First Law for the low pressure turbine x6QUALITYFluidhh6PP6 Qinreheat Qinreheat h5 h4 P3P4 UNTIL iimax END NoRHStages 2 P6 10kPa P3 15000kPa Pextract P6 Select a lower limit on the reheat pressure T3 500C T5 500C Etat 10 Turbine isentropic efficiency Etap 10 Pump isentropic efficiency Pratio P3Pextract P4 P31Pratio1NoRHStages1kPa FluidSteamIAPWS PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10131 Pump analysis P1 P6 P2P3 x10 Satd liquid h1enthalpyFluidPP1xx1 v1volumeFluidPP1xx1 s1entropyFluidPP1xx1 T1temperatureFluidPP1xx1 Wpsv1P2P1SSSF isentropic pump work assuming constant specific volume WpWpsEtap h2h1Wp SSSF First Law for the pump v2volumeFluidPP2hh2 s2entropyFluidPP2hh2 T2temperatureFluidPP2hh2 High Pressure Turbine analysis h3enthalpyFluidTT3PP3 s3entropyFluidTT3PP3 v3volumeFluidTT3PP3 ss4s3 hs4enthalpyFluidsss4PP4 Ts4temperatureFluidsss4PP4 Etath3h4h3hs4Definition of turbine efficiency T4temperatureFluidPP4hh4 s4entropyFluidhh4PP4 v4volumeFluidss4PP4 h3 Wthph4SSSF First Law for the high pressure turbine Low Pressure Turbine analysis Call ReheatP3T3T5h4NoRHStagesPratioEtatQinreheatWtlph6 h6h6 P5P4 s5entropyFluidTT5PP5 h5enthalpyFluidTT5PP5 ss6s5 hs6enthalpyFluidsss6PP6 Ts6temperatureFluidsss6PP6 vs6volumeFluidsss6PP6 Etath5h6h5hs6Definition of turbine efficiency h5Wtlph6SSSF First Law for the low pressure turbine x6QUALITYFluidhh6PP6 Wtlptotal NoRHStagesWtlp Qinreheat NoRHStagesh5 h4 Boiler analysis Qinboiler h2h3SSSF First Law for the Boiler Qin QinboilerQinreheat Condenser analysis h6Qouth1SSSF First Law for the Condenser T6temperatureFluidhh6PP6 s6entropyFluidhh6PP6 x6QUALITYFluidhh6PP6 x6sx6x6 Cycle Statistics WnetWthpWtlp Wp EtathWnetQin PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10132 ηth NoRH Stages Qin kJkg Wnet kJkg 04097 1 4085 1674 04122 2 4628 1908 04085 3 5020 2051 04018 4 5333 2143 03941 5 5600 2207 0386 6 5838 2253 03779 7 6058 2289 03699 8 6264 2317 03621 9 6461 2340 03546 10 6651 2358 1 2 3 4 5 6 7 8 9 10 1600 1700 1800 1900 2000 2100 2200 2300 2400 NoRHStages Wnet kJkg 1 2 3 4 5 6 7 8 9 10 035 036 037 038 039 04 041 042 NoRHStages ηth 1 2 3 4 5 6 7 8 9 10 4000 4500 5000 5500 6000 6500 7000 NoRHStages Qin kJkg PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10133 10118 The effect of number of regeneration stages on the performance an ideal regenerative Rankine cycle with one open feedwater heater is to be investigated Analysis The problem is solved using EES and the solution is given below Procedure ReheatNoFwhT5P5PcondEtaturbEtapumpqinwnet FluidSteamIAPWS Tcond temperatureFluidPPcondx0 Tboiler temperatureFluidPP5x0 P7 Pcond s5entropyFluid TT5 PP5 h5enthalpyFluid TT5 PP5 h1enthalpyFluid PP7x0 P41 P5 NOTICE THIS IS P4i WITH i 1 DELTATcondboiler Tboiler Tcond If NoFWH 0 Then the following are h7 h2 wnet and qin for zero feedwater heaters NoFWH 0 h7enthalpyFluid ss5PP7 h2h1volumeFluid PP7x0P5 P7Etapump wnet Etaturbh5h7h2h1 qin h5 h2 else i0 REPEAT ii1 The following maintains the same temperature difference between any two regeneration stages TFWHi NoFWH 1 iDELTATcondboilerNoFWH 1TcondC Pextracti pressureFluidTTFWHix0kPa P3iPextracti P6iPextracti If i 1 then P4i P6i 1 UNTIL iNoFWH P4NoFWH1P6NoFWH h4NoFWH1h1volumeFluid PP7x0P4NoFWH1 P7Etapump i0 REPEAT ii1 Boiler condensate pump or the Pumps 2 between feedwater heaters analysis h3ienthalpyFluidPP3ix0 v3ivolumeFluidPP3ix0 wpump2sv3iP4iP3iSSSF isentropic pump work assuming constant specific volume wpump2iwpump2sEtapump Definition of pump efficiency h4i wpump2i h3i Steadyflow conservation of energy s4ientropyFluidPP4ihh4i T4itemperatureFluidPP4ihh4i Until i NoFWH i0 REPEAT PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10134 ii1 Open Feedwater Heater analysis h2i h6i s5i s5 ss6is5i hs6ienthalpyFluidsss6iPP6i Ts6itemperatureFluidsss6iPP6i h6ih5Etaturbh5hs6iDefinition of turbine efficiency for high pressure stages If i1 then y1h31 h42h61 h42 Steadyflow conservation of energy for the FWH If i 1 then js i 1 j 0 sumyj 0 REPEAT j j1 sumyj sumyj y j UNTIL j js yi 1 sumyjh3i h4i1h6i h4i1 ENDIF T3itemperatureFluidPP3ix0 Condensate leaves heater as sat liquid at P3 s3ientropyFluidPP3ix0 Turbine analysis T6itemperatureFluidPP6ihh6i s6ientropyFluidPP6ihh6i yh6i yih6i UNTIL iNoFWH ss7s6i hs7enthalpyFluidsss7PP7 Ts7temperatureFluidsss7PP7 h7h6iEtaturbh6ihs7Definition of turbine efficiency for low pressure stages T7temperatureFluidPP7hh7 s7entropyFluidPP7hh7 sumyi 0 sumyh6i 0 wp2i Wpump21 i0 REPEAT ii1 sumyi sumyi yi sumyh6i sumyh6i yh6i If NoFWH 1 then wp2i wp2i 1 sumyiWpump2i UNTIL i NoFWH Condenser PumpPump1 Analysis P2 P6 NoFWH P1 Pcond h1enthalpyFluidPP1x0 Satd liquid v1volumeFluidPP1x0 s1entropyFluidPP1x0 T1temperatureFluidPP1x0 wpump1sv1P2P1SSSF isentropic pump work assuming constant specific volume wpump1wpump1sEtapump Definition of pump efficiency h2wpump1 h1 Steadyflow conservation of energy s2entropyFluidPP2hh2 T2temperatureFluidPP2hh2 Boiler analysis qin h5 h41SSSF conservation of energy for the Boiler wturb h5 sumyh6i 1 sumyih7 SSSF conservation of energy for turbine PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10135 Condenser analysis qout1 sumyih7 h1SSSF First Law for the Condenser Cycle Statistics wnetwturb 1 sumyiwpump1 wp2i endif END Input Data NoFWH 2 P5 10000 kPa T5 500 C Pcond10 kPa Etaturb 10 Turbine isentropic efficiency Etapump 10 Pump isentropic efficiency P1 Pcond P4 P5 Condenser exit pump or Pump 1 analysis Call ReheatNoFwhT5P5PcondEtaturbEtapumpqinwnet Etathwnetqin No FWH ηth wnet kJkg qin kJkg 0 1 2 3 4 5 6 7 8 9 10 04019 04311 04401 04469 04513 04544 04567 04585 04599 04611 0462 1275 1125 1061 1031 1013 1000 9905 9833 9777 9731 9694 3173 2609 2411 2307 2243 2200 2169 2145 2126 2111 2098 0 1 2 3 4 5 6 7 8 9 10 04 041 042 043 044 045 046 047 NoFwh ηth 0 1 2 3 4 5 6 7 8 9 10 950 1000 1050 1100 1150 1200 1250 1300 NoFwh wnet kJkg 0 1 2 3 4 5 6 7 8 9 10 2000 2200 2400 2600 2800 3000 3200 NoFwh qin kJkg PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10136 10119 It is to be demonstrated that the thermal efficiency of a combined gassteam power plant ηcc can be expressed as η η η η η cc g s g s where ηg g i W Q n and ηs s gout W Q are the thermal efficiencies of the gas and steam cycles respectively and the efficiency of a combined cycle is to be obtained Analysis The thermal efficiencies of gas steam and combined cycles can be expressed as η η η cc total in out in g g in gout in s s gout out gout W Q Q Q W Q Q Q W Q Q Q 1 1 1 where Qin is the heat supplied to the gas cycle where Qout is the heat rejected by the steam cycle and where Qgout is the heat rejected from the gas cycle and supplied to the steam cycle Using the relations above the expression η η η η g s g s can be expressed as cc Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q η η η η η in out in out out g out in out g out g out in out g out g out in out g out g out in gout s g s g 1 1 1 1 1 1 1 1 Therefore the proof is complete Using the relation above the thermal efficiency of the given combined cycle is determined to be η η η η η cc g s g s 0 4 0 30 0 40 0 30 058 10120 The thermal efficiency of a combined gassteam power plant ηcc can be expressed in terms of the thermal efficiencies of the gas and the steam turbine cycles as η η η η η cc g s g s It is to be shown that the value of ηcc is greater than either of η η g or s Analysis By factoring out terms the relation η η η η η cc g s g s can be expressed as η η η η η η η η η η cc g s g s g s g Positive since 1 g g 1 1 2 4 3 4 or η η η η η η η η η η cc g s g s s g s Positive since 1 s s 1 1 2 4 3 4 Thus we conclude that the combined cycle is more efficient than either of the gas turbine or steam turbine cycles alone PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10137 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course b 10121 It is to e shown that the exergy destruction associated with a simple ideal Rankine cycle can be expressed as th qin x η η thCarnot destroyed where ηth is efficiency of the Rankine cycle and ηth Carnot is the efficiency of the Carnot cycle operating between the same temperature limits Analysis The exergy destruction associated with a cycle is given on a unit mass basis as R R T q T x 0 destroyed where the direction of qin is determined with respect to the reservoir positive if to the reservoir and negative if from the reservoir For a cycle that involves heat transfer only with a source at TH and a sink at T0 the irreversibility becomes th th C th C th H H H q q T T q q q q T T q T q T q T x η η η η in in 0 in out in in 0 out in 0 out 0 destroyed 1 1 preparation If you are a student using this Manual you are using it without permission 10138 Fundamentals of Engineering FE Exam Problems 10122 Consider a simple ideal Rankine cycle If the condenser pressure is lowered while keeping turbine inlet state the same select the correct statement a the turbine work output will decrease b the amount of heat rejected will decrease c the cycle efficiency will decrease d the moisture content at turbine exit will decrease e the pump work input will decrease Answer b the amount of heat rejected will decrease 10123 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures If the steam is superheated to a higher temperature select the correct statement a the turbine work output will decrease b the amount of heat rejected will decrease c the cycle efficiency will decrease d the moisture content at turbine exit will decrease e the amount of heat input will decrease Answer d the moisture content at turbine exit will decrease 10124 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures If the cycle is modified with reheating select the correct statement a the turbine work output will decrease b the amount of heat rejected will decrease c the pump work input will decrease d the moisture content at turbine exit will decrease e the amount of heat input will decrease Answer d the moisture content at turbine exit will decrease 10125 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures If the cycle is modified with regeneration that involves one open feed water heater select the correct statement per unit mass of steam flowing through the boiler a the turbine work output will decrease b the amount of heat rejected will increase c the cycle thermal efficiency will decrease d the quality of steam at turbine exit will decrease e the amount of heat input will increase Answer a the turbine work output will decrease PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10139 10126 Consider a steadyflow Carnot cycle with water as the working fluid executed under the saturation dome between the pressure limits of 3 MPa and 10 kPa Water changes from saturated liquid to saturated vapor during the heat addition process The net work output of this cycle is a 666 kJkg b 888 kJkg c 1040 kJkg d 1130 kJkg e 1440 kJkg Answer a 666 kJkg Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P133000 kPa P210 kPa hfgENTHALPYSteamIAPWSx1PP1ENTHALPYSteamIAPWSx0PP1 T1TEMPERATURESteamIAPWSx0PP1273 T2TEMPERATURESteamIAPWSx0PP2273 qinhfg EtaCarnot1T2T1 wnetEtaCarnotqin Some Wrong Solutions with Common Mistakes W1work Eta1qin Eta1T2T1 Taking Carnot efficiency to be T2T1 W2work Eta2qin Eta21T2273T1273 Using C instead of K W3work EtaCarnotENTHALPYSteamIAPWSx1PP1 Using hg instead of hfg W4work EtaCarnotq2 q2ENTHALPYSteamIAPWSx1PP2ENTHALPYSteamIAPWSx0PP2 Using hfg at P2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10140 10127 A simple ideal Rankine cycle operates between the pressure limits of 10 kPa and 3 MPa with a turbine inlet temperature of 600C Disregarding the pump work the cycle efficiency is a 24 b 37 c 52 d 63 e 71 Answer b 37 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P110 kPa P23000 kPa P3P2 P4P1 T3600 C s4s3 h1ENTHALPYSteamIAPWSx0PP1 v1VOLUMESteamIAPWSx0PP1 wpumpv1P2P1 kJkg h2h1wpump h3ENTHALPYSteamIAPWSTT3PP3 s3ENTROPYSteamIAPWSTT3PP3 h4ENTHALPYSteamIAPWSss4PP4 qinh3h2 qouth4h1 Etath1qoutqin Some Wrong Solutions with Common Mistakes W1Eff qoutqin Using wrong relation W2Eff 1h44h1h3h2 h44 ENTHALPYSteamIAPWSx1PP4 Using hg for h4 W3Eff 1T1273T3273 T1TEMPERATURESteamIAPWSx0PP1 Using Carnot efficiency W4Eff h3h4qin Disregarding pump work PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10141 10128 A simple ideal Rankine cycle operates between the pressure limits of 10 kPa and 5 MPa with a turbine inlet temperature of 600C The mass fraction of steam that condenses at the turbine exit is a 6 b 9 c 12 d 15 e 18 Answer c 12 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P110 kPa P25000 kPa P3P2 P4P1 T3600 C s4s3 h3ENTHALPYSteamIAPWSTT3PP3 s3ENTROPYSteamIAPWSTT3PP3 h4ENTHALPYSteamIAPWSss4PP4 x4QUALITYSteamIAPWSss4PP4 moisture1x4 Some Wrong Solutions with Common Mistakes W1moisture x4 Taking quality as moisture W2moisture 0 Assuming superheated vapor PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10142 10129 A steam power plant operates on the simple ideal Rankine cycle between the pressure limits of 5 kPa and 10 MPa with a turbine inlet temperature of 600C The rate of heat transfer in the boiler is 300 kJs Disregarding the pump work the power output of this plant is a 93 kW b 118 kW c 190 kW d 216 kW e 300 kW Answer b 118 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P110 kPa P25000 kPa P3P2 P4P1 T3600 C s4s3 Qrate300 kJs mQrateqin h1ENTHALPYSteamIAPWSx0PP1 h2h1 pump work is neglected v1VOLUMESteamIAPWSx0PP1 wpumpv1P2P1 h2h1wpump h3ENTHALPYSteamIAPWSTT3PP3 s3ENTROPYSteamIAPWSTT3PP3 h4ENTHALPYSteamIAPWSss4PP4 qinh3h2 Wturbmh3h4 Some Wrong Solutions with Common Mistakes W1power Qrate Assuming all heat is converted to power W3power QrateCarnot Carnot 1T1273T3273 T1TEMPERATURESteamIAPWSx0PP1 Using Carnot efficiency W4power mh3h44 h44 ENTHALPYSteamIAPWSx1PP4 Taking h4hg PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10143 10130 Consider a combined gassteam power plant Water for the steam cycle is heated in a wellinsulated heat exchanger by the exhaust gases that enter at 800 K at a rate of 60 kgs and leave at 400 K Water enters the heat exchanger at 200C and 8 MPa and leaves at 350C and 8 MPa If the exhaust gases are treated as air with constant specific heats at room temperature the mass flow rate of water through the heat exchanger becomes a 11 kgs b 24 kgs c 46 kgs d 53 kgs e 60 kgs Answer a 11 kgs Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values mgas60 kgs Cp1005 kJkgK T3800 K T4400 K QgasmgasCpT3T4 P18000 kPa T1200 C P28000 kPa T2350 C h1ENTHALPYSteamIAPWSTT1PP1 h2ENTHALPYSteamIAPWSTT2PP2 Qsteammsteamh2h1 QgasQsteam Some Wrong Solutions with Common Mistakes mgasCpT3 T4W1msteam418T2T1 Assuming no evaporation of liquid water mgasCvT3 T4W2msteamh2h1 Cv0718 Using Cv for air instead of Cp W3msteam mgas Taking the mass flow rates of two fluids to be equal mgasCpT3 T4W4msteamh2h11 h11ENTHALPYSteamIAPWSx0PP1 Taking h1hfP1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10144 10131 An ideal reheat Rankine cycle operates between the pressure limits of 10 kPa and 8 MPa with reheat occurring at 4 MPa The temperature of steam at the inlets of both turbines is 500C and the enthalpy of steam is 3185 kJkg at the exit of the highpressure turbine and 2247 kJkg at the exit of the lowpressure turbine Disregarding the pump work the cycle efficiency is a 29 b 32 c 36 d 41 e 49 Answer d 41 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P110 kPa P28000 kPa P3P2 P44000 kPa P5P4 P6P1 T3500 C T5500 C s4s3 s6s5 h1ENTHALPYSteamIAPWSx0PP1 h2h1 h443185 kJkg for checking given data h662247 kJkg for checking given data h3ENTHALPYSteamIAPWSTT3PP3 s3ENTROPYSteamIAPWSTT3PP3 h4ENTHALPYSteamIAPWSss4PP4 h5ENTHALPYSteamIAPWSTT5PP5 s5ENTROPYSteamIAPWSTT5PP5 h6ENTHALPYSteamIAPWSss6PP6 qinh3h2h5h4 qouth6h1 Etath1qoutqin Some Wrong Solutions with Common Mistakes W1Eff qoutqin Using wrong relation W2Eff 1qouth3h2 Disregarding heat input during reheat W3Eff 1T1273T3273 T1TEMPERATURESteamIAPWSx0PP1 Using Carnot efficiency W4Eff 1qouth5h2 Using wrong relation for qin PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10145 10132 Pressurized feedwater in a steam power plant is to be heated in an ideal open feedwater heater that operates at a pressure of 2 MPa with steam extracted from the turbine If the enthalpy of feedwater is 252 kJkg and the enthalpy of extracted steam is 2810 kJkg the mass fraction of steam extracted from the turbine is a 10 b 14 c 26 d 36 e 50 Answer c 26 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values hfeed252 kJkg hextracted2810 kJkg P32000 kPa h3ENTHALPYSteamIAPWSx0PP3 Energy balance on the FWH h3xexthextracted1xexthfeed Some Wrong Solutions with Common Mistakes W1ext hfeedhextracted Using wrong relation W2ext h3hextractedhfeed Using wrong relation W3ext hfeedhextractedhfeed Using wrong relation 10133 Consider a steam power plant that operates on the regenerative Rankine cycle with one open feedwater heater The enthalpy of the steam is 3374 kJkg at the turbine inlet 2797 kJkg at the location of bleeding and 2346 kJkg at the turbine exit The net power output of the plant is 120 MW and the fraction of steam bled off the turbine for regeneration is 0172 If the pump work is negligible the mass flow rate of steam at the turbine inlet is a 117 kgs b 126 kgs c 219 kgs d 288 kgs e 679 kgs Answer b 126 kgs Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values hin3374 kJkg hout2346 kJkg hextracted2797 kJkg Wnetout120000 kW xbleed0172 wturbhinhextracted1xbleedhextractedhout mWnetoutwturb Some Wrong Solutions with Common Mistakes W1mass Wnetouthinhout Disregarding extraction of steam W2mass Wnetoutxbleedhinhout Assuming steam is extracted at trubine inlet W3mass Wnetouthinhoutxbleedhextracted Using wrong relation PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10146 10134 Consider a cogeneration power plant modified with regeneration Steam enters the turbine at 6 MPa and 450C at a rate of 20 kgs and expands to a pressure of 04 MPa At this pressure 60 of the steam is extracted from the turbine and the remainder expands to a pressure of 10 kPa Part of the extracted steam is used to heat feedwater in an open feedwater heater The rest of the extracted steam is used for process heating and leaves the process heater as a saturated liquid at 04 MPa It is subsequently mixed with the feedwater leaving the feedwater heater and the mixture is pumped to the boiler pressure The steam in the condenser is cooled and condensed by the cooling water from a nearby river which enters the adiabatic condenser at a rate of 463 kgs Condenser Boiler 6 7 10 8 Turbine 5 Process heater P I P II 9 2 3 fwh 4 1 11 h1 19181 h2 19220 h3 h4 h9 60466 h5 61073 h6 33029 h7 h8 h10 26656 h11 21288 1 The total power output of the turbine is a 170 MW b 84 MW c 122 MW d 200 MW e 34 MW Answer a 170 MW 2 The temperature rise of the cooling water from the river in the condenser is a 80C b 52C c 96C d 129C e 162C Answer a 80C 3 The mass flow rate of steam through the process heater is a 16 kgs b 38 kgs c 52 kgs d 76 kgs e 104 kgs Answer e 104 kgs 4 The rate of heat supply from the process heater per unit mass of steam passing through it is a 246 kJkg b 893 kJkg c 1344 kJkg d 1891 kJkg e 2060 kJkg Answer e 2060 kJkg 5 The rate of heat transfer to the steam in the boiler is a 260 MJs b 538 MJs c 395 MJs d 628 MJs e 1254 MJs Answer b 538 MJs Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Note The solution given below also evaluates all enthalpies given on the figure PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 111 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 11 REFRIGERATION CYCLES PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 112 The Reversed Carnot Cycle 111C The reversed Carnot cycle serves as a standard against which actual refrigeration cycles can be compared Also the COP of the reversed Carnot cycle provides the upper limit for the COP of a refrigeration cycle operating between the specified temperature limits 112C Because the compression process involves the compression of a liquidvapor mixture which requires a compressor that will handle two phases and the expansion process involves the expansion of highmoisture content refrigerant 113 A steadyflow Carnot refrigeration cycle with refrigerant134a as the working fluid is considered The coefficient of performance the amount of heat absorbed from the refrigerated space and the net work input are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a Noting that TH 40C 313 K and TL Tsat 100 kPa 2637C 2466 K the COP of this Carnot refrigerator is determined from T 372 1 313 K 246 6 K 1 1 1 COP RC L H T T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course b From the refrigerant tables Table A11 10826 kJkg 27 kJkg 271 40 C 4 40 C 3 f g h h h h Thus and 1284 kJkg 1630 kJkg 313 K K 2466 163 0 kJkg 10826 27127 4 3 H H L L L H L H H q T T q T T q q h h q QH QL 40C 4 3 2 1 100 kPa s c The net work input is determined from 346 kJkg 128 4 163 0 net L H q q w preparation If you are a student using this Manual you are using it without permission 113 114E A steadyflow Carnot refrigeration cycle with refrigerant134a as the working fluid is considered The coefficient of performance the quality at the beginning of the heatabsorption process and the net work input are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a Noting that TH Tsat 90 psia 7278F 5328 R and TL Tsat 30 psia 1537F 4754 R 828 1 5328 R 4754 R 1 1 1 COP RC L H T T T QH QL 4 3 2 1 b Process 41 is isentropic and thus 02374 0 18589 0 03793 08207 0 Btulbm R 008207 0 14525 0 05 07481 0 30 psia 1 1 90 psia 4 4 1 fg f fg f s s s x x s s s s s c Remembering that on a Ts diagram the area enclosed represents the net work and s3 sg 90 psia 022006 BtulbmR 792 Btulbm 0 08207 Btulbm R 7278 1537 0 22006 4 3 netin s s T T w L H PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 114 Ideal and Actual VaporCompression Refrigeration Cycles 115C Yes the throttling process is an internally irreversible process 116C To make the ideal vaporcompression refrigeration cycle more closely approximate the actual cycle 117C No Assuming the water is maintained at 10C in the evaporator the evaporator pressure will be the saturation pressure corresponding to this pressure which is 12 kPa It is not practical to design refrigeration or airconditioning devices that involve such extremely low pressures 118C Allowing a temperature difference of 10C for effective heat transfer the condensation temperature of the refrigerant should be 25C The saturation pressure corresponding to 25C is 067 MPa Therefore the recommended pressure would be 07 MPa 119C The area enclosed by the cyclic curve on a Ts diagram represents the net work input for the reversed Carnot cycle but not so for the ideal vaporcompression refrigeration cycle This is because the latter cycle involves an irreversible process for which the process path is not known 1110C The cycle that involves saturated liquid at 30C will have a higher COP because judging from the Ts diagram it will require a smaller work input for the same refrigeration capacity 1111C The minimum temperature that the refrigerant can be cooled to before throttling is the temperature of the sink the cooling medium since heat is transferred from the refrigerant to the cooling medium PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 115 1112E A refrigerator operating on the ideal vaporcompression refrigeration cycle with refrigerant134a as the working fluid is considered The increase in the COP if the throttling process were replaced by an isentropic expansion is to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis In an ideal vaporcompression refrigeration cycle the compression process is isentropic the refrigerant enters the compressor as a saturated vapor at the evaporator pressure and leaves the condenser as saturated liquid at the condenser pressure From the refrigerant tables Tables A11E A12E and A13E isentropi c expansion 0 4723 5980 Btulbm F 20 339 Btulbm throttling 66 0 12715 Btulbm R 66339 Btulbm sat liquid psia 300 12568 Btulbm psia 300 0 22341 Btulbm R 10598 Btulbm sat vapor F 20 4 4 3 4 4 3 4 300 psia 3 300 psia 3 3 2 1 2 2 20 F 1 20 F 1 1 s s f f g g x h s s T h h s s h h P h s s P s s h h T QH QL 20F 1 2 3 4 300 psia Win 4s s T The COP of the refrigerator for the throttling case is 2012 12568 10598 66339 10598 COP 1 2 4 1 in R h h h h w qL The COP of the refrigerator for the isentropic expansion case is 2344 12568 10598 5980 10598 COP 1 2 4 1 in R h h h h w q s L The increase in the COP by isentropic expansion is 165 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 116 1113 An ideal vaporcompression refrigeration cycle with refrigerant134a as the working fluid is considered The COP and the power requirement are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis In an ideal vaporcompression refrigeration cycle the compression process is isentropic the refrigerant enters the compressor as a saturated vapor at the evaporator pressure and leaves the condenser as saturated liquid at the condenser pressure From the refrigerant tables Tables A11 A12 and A13 32 kJkg throttling 107 10732 kJkg sat liquid MPa 1 27529 kJkg MPa 1 0 92927 kJkg K 25277 kJkg sat vapor C 4 3 4 1 MPa 3 3 2 1 2 2 4 C 1 4 C 1 1 h h h h P h s s P s s h h T f g g PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course The mass flow rate of the refrigerant is 2 750 kgs 25277 10732 kJkg 400 kJs 4 1 4 1 h h Q m h m h Q L L QH QL 4C 1 2 3 4 1 MPa Win 4s s T The power requirement is 6193 kW 25277 kJkg 2 750 kgs27529 1 2 in h m h W The COP of the refrigerator is determined from its definition 646 6193 kW 400 kW COP in R W QL preparation If you are a student using this Manual you are using it without permission 117 1114 An ideal vaporcompression refrigeration cycle with refrigerant134a as the working fluid is considered The rate of heat removal from the refrigerated space the power input to the compressor the rate of heat rejection to the environment and the COP are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a In an ideal vaporcompression refrigeration cycle the compression process is isentropic the refrigerant enters the compressor as a saturated vapor at the evaporator pressure and leaves the condenser as saturated liquid at the condenser pressure From the refrigerant tables Tables A12 and A13 82 kJkg throttling 88 8882 kJkg sat liquid MPa 70 3495 C 27350 kJkg MPa 70 0 94779 kJkg K 23697 kJkg sat vapor kPa 120 3 4 MPa 70 3 3 2 2 1 2 2 120 kPa 1 120 kPa 1 1 h h h h P T h s s P s s h h P f g g T QH QL 012 1 2 3 4 07 MPa Win 4s Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from s and kW 183 kW 741 23697 kJkg kgs 27350 005 8882 kJkg kgs 23697 005 1 2 in 4 1 h m h W h m h QL b The rate of heat rejection to the environment is determined from 923 kW 1 83 7 41 Win Q Q L H c The COP of the refrigerator is determined from its definition 406 183 kW 741 kW COP in R W QL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 118 1115 An ideal vaporcompression refrigeration cycle with refrigerant134a as the working fluid is considered The rate of heat removal from the refrigerated space the power input to the compressor the rate of heat rejection to the environment and the COP are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a In an ideal vaporcompression refrigeration cycle the compression process is isentropic the refrigerant enters the compressor as a saturated vapor at the evaporator pressure and leaves the condenser as saturated liquid at the condenser pressure From the refrigerant tables Tables A12 and A13 61 kJkg throttling 101 10161 kJkg sat liquid MPa 90 4445 C 27893 kJkg MPa 90 0 94779 kJkg K 23697 kJkg sat vapor kPa 120 3 4 MPa 90 3 3 2 2 1 2 2 120 kPa 1 120 kPa 1 1 h h h h P T h s s P s s h h P f g g T QH QL 012 MPa 1 2 3 4 09 MPa Win 4s Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from s and kW 210 kW 677 23697 kJkg kgs 27893 005 10161 kJkg kgs 23697 005 1 2 in 4 1 h m h W h m h QL b The rate of heat rejection to the environment is determined from 887 kW 2 10 6 77 Win Q Q L H c The COP of the refrigerator is determined from its definition 323 210 kW 677 kW COP in R W QL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 119 1116 An ideal vaporcompression refrigeration cycle with refrigerant134a as the working fluid is considered The throttling valve in the cycle is replaced by an isentropic turbine The percentage increase in the COP and in the rate of heat removal from the refrigerated space due to this replacement are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible T QH QL 012 MPa 1 2 3 4 07 MPa Win 4s Analysis If the throttling valve in the previous problem is replaced by an isentropic turbine we would have s4s s3 sf 07 MPa 033230 kJkgK and the enthalpy at the turbine exit would be 8258 kJkg 0 2802 21448 49 22 0 2802 0 85503 0 09275 33230 0 120 kPa 4 4 120 kPa 3 4 fg s f s fg f s h x h h s s s x s Then 772 kW 005 kgs 23697 8258 kJkg 4 1 s L h m h Q and 4 23 183 kW 772 kW COP in R W QL Then the percentage increase in and COP becomes Q 42 42 4 06 4 06 4 23 COP COP in COP Increase 7 41 7 41 7 72 in Increase R R R L L L Q Q Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1110 1117 A refrigerator with refrigerant134a as the working fluid is considered The rate of heat removal from the refrigerated space the power input to the compressor the isentropic efficiency of the compressor and the COP of the refrigerator are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a From the refrigerant tables Tables A12 and A13 T QH QL 021 MPa 1 2s 3 4 115 MPa s 2 Win 020 MPa 5C 12 MPa 70C 28 kJkg throttling 114 11428 kJkg 44 C 15 MPa 1 28721 kJkg MPa 21 30061 kJkg 70 C MPa 21 0 95407 kJkg K 24880 kJkg 5 C 20 MPa 0 3 4 4 4 C 3 3 3 2 1 2 2 2 2 2 1 1 1 1 h h h h T P h s s P h T P s h T P f s s s Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from and kW 363 kW 942 24880 kJkg kgs 30061 007 11428 kJkg kgs 24880 007 1 2 in 4 1 h m h W h m h QL b The isentropic efficiency of the compressor is determined from 741 0 741 24880 30061 24880 21 287 1 2 1 2 h h h h s ηC c The COP of the refrigerator is determined from its definition 260 363 kW 942 kW COP in R W QL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1111 1118E An ideal vaporcompression refrigeration cycle with refrigerant134a as the working fluid is considered The mass flow rate of the refrigerant and the power requirement are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis In an ideal vaporcompression refrigeration cycle the compression process is isentropic the refrigerant enters the compressor as a saturated vapor at the evaporator pressure and leaves the condenser as saturated liquid at the condenser pressure From the refrigerant tables Tables A11E A12E and A13E 50 Btulbm throttling 51 5150 Btulbm sat liquid psia 180 12199 Btulbm psia 180 0 22485 Btulbm R 10382 Btulbm sat vapor F 5 3 4 180 psia 3 3 2 1 2 2 5 F 1 5 0 F 1 1 h h h h P h s s P s s h h T f g g PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course The mass flow rate of the refrigerant is 8601 lbmh 5150 Btulbm 10382 45000 Btuh 4 1 4 1 h h Q m h m h Q L L QH QL 5F 1 2 3 4 180 psia Win 4s s T The power requirement is 4582 kW 341214 Btuh 1kW 10382 Btulbm 860 1 lbmh12199 1 2 in h m h W preparation If you are a student using this Manual you are using it without permission 1112 1119E Problem 1118E is to be repeated if ammonia is used as the refrigerant Analysis The problem is solved using EES and the solution is given below Given x11 T15 F x30 P3180 psia QdotL45000 Btuh Analysis Fluidammonia compressor h1enthalpyFluid TT1 xx1 s1entropyFluid TT1 xx1 s2s1 P2P3 h2enthalpyFluid PP2 ss2 expansion valve h3enthalpyFluid PP3 xx3 h4h3 cycle mdotRQdotLh1h4 WdotinmdotRh2h1ConvertBtuh kW Solution for ammonia COPR4515 Fluidammonia mdotR958 lbmh QdotL45000 Btuh Wdotin2921 kW Solution for R134a COPR2878 FluidR134a mdotR8601 lbmh QdotL45000 Btuh Wdotin4582 kW PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1113 1120 A commercial refrigerator with refrigerant134a as the working fluid is considered The quality of the refrigerant at the evaporator inlet the refrigeration load the COP of the refrigerator and the theoretical maximum refrigeration load for the same power input to the compressor are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a From refrigerant134a tables Tables A11 through A13 04795 4 4 4 3 4 3 3 3 2 2 2 1 1 1 23 kJkg 111 kPa 60 23 kJkg 111 11123 kJkg C 42 kPa 1200 29516 kJkg C 65 kPa 1200 23003 kJkg C 34 kPa 60 x h P h h h T P h T P h T P 60 kPa 34C 1 2 3 4 QH 42C Win Condenser Evaporator Compressor Expansion valve QL 12 MPa 65C Qin 26C Water 18C Using saturated liquid enthalpy at the given temperature for water we have Table A4 94 kJkg 108 47 kJkg 75 26 C 2 18 C 1 f w f w h h h h b The mass flow rate of the refrigerant may be determined from an energy balance on the compressor 0455 kgs 0 7547kJkg 025 kgs10894 11123kJkg 29516 1 2 3 2 R R w w w R m m h h m h h m The waste heat transferred from the refrigerant the compressor power input and the refrigeration load are 8 367 kW 0 0455 kgs29516 11123kJkg 3 2 h h m Q R H 2 513 kW 0450 kW 23003kJkg 0 0455 kgs29516 in 1 2 in Q h h m W R 5404 kW 0 450 2 513 8 367 in in Q W Q Q H L c The COP of the refrigerator is determined from its definition T QH QL 1 2 3 4 2 Win 215 2 513 5 404 COP in L W Q d The reversible COP of the refrigerator for the same temperature limits is 5 063 273 1 273 30 18 1 1 1 COPmax L H T T Then the maximum refrigeration load becomes s 1272 kW 5 063 2 513 kW in max Lmax W COP Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1114 1121 A refrigerator with refrigerant134a as the working fluid is considered The power input to the compressor the rate of heat removal from the refrigerated space and the pressure drop and the rate of heat gain in the line between the evaporator and the compressor are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a From the refrigerant tables Tables A12 and A13 23468 kJkg 0 10173 MPa sat vapor C 26 83 kJkg throttling 87 8783 kJkg 26 C 75 MPa 0 28407 kJkg MPa 80 0 19841 m kg 0 97207 kJkg K 50 kJkg 239 20 C kPa 100 5 5 5 3 4 26 C 3 3 3 2 1 2 2 3 1 1 1 1 1 h P T h h h h T P h s s P s h T P f s s v T QH QL 010 MPa 1 2s 3 4 075 MPa s 08 MPa 2 Win 010 M 20C 26C Pa Then the mass flow rate of the refrigerant and the power input becomes 240 kW 0 78 23950 kJkg 28407 00420 kgs 00420 kgs m kg 019841 m s 0560 1 2 in 3 3 1 1 C s h m h W m η v V b The rate of heat removal from the refrigerated space is 617 kW 8783 kJkg 00420 kgs 23468 4 5 h m h QL c The pressure drop and the heat gain in the line between the evaporator and the compressor are and kW 0203 173 23468 kJkg kgs 23950 00420 100 73 101 5 1 gain 1 5 h m h Q P P P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1115 1122 Problem 1121 is reconsidered The effects of the compressor isentropic efficiency and the compressor inlet volume flow rate on the power input and the rate of refrigeration are to be investigated Analysis The problem is solved using EES and the solution is given below Given P1100 kPa T120 C Vdot05 m3min P2800 kPa EtaC078 P3750 kPa T326 C T526 C x51 Analysis FluidR134a compressor h1enthalpyFluid PP1 TT1 s1entropyFluid PP1 TT1 06 065 07 075 08 085 09 095 1 0 1 2 3 4 5 6 7 ηC Win kW 01 m3min 05 m3min 1 m3min v1volumeFluid PP1 TT1 ss2s1 hs2enthalpyFluid PP2 sss2 expansion valve x30 assumed saturated liquid h3enthalpyFluid TT3 xx3 h4h3 evaporator exit h5enthalpyFluid TT5 xx5 P5pressureFluid TT5 xx5 cycle mdotVdotv1Convertkgmin kgs Wdotinmdoths2h1EtaC QdotLmdoth5h4 DELTAPP5P1 Qdotgainmdoth1h5 06 065 07 075 08 085 09 095 1 0 2 4 6 8 10 12 14 ηC QL kW 05 m3min 1 m3min 01 m3min ηc Win kW QL kW 06 065 07 075 08 085 09 095 1 312 288 2674 2496 234 2202 208 1971 1872 6168 6168 6168 6168 6168 6168 6168 6168 6168 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1116 1123 A refrigerator uses refrigerant134a as the working fluid and operates on the ideal vaporcompression refrigeration cycle except for the compression process The mass flow rate of the refrigerant the condenser pressure and the COP of the refrigerator are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a b From the refrigerant134a tables Tables A11 through A13 QH 60C Win Condenser Evaporator Compressor Expansion valve 1 2 3 4 QL 120 kPa x03 23697 kJkg sat vap 1 kPa 120 29887 kJkg C 60 8 kPa 671 sat liq 0 83 kJkg 86 8683 kJkg 30 0 kPa 120 1 1 4 1 2 2 2 3 2 3 3 3 4 3 4 4 4 h x P P h T P P P P x h h h h x P kPa 6718 The mass flow rate of the refrigerant is determined from T QH QL 120 kPa 1 2 3 4 Win 4s 000727 kgs 23697kJkg 29887 45 kW 0 1 2 in h h W m c The refrigeration load and the COP are 1 091 kW 8683kJkg 0 0727 kgs23697 4 1 h m h QL 243 045 kW 1 091 kW COP in L W Q s PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1117 1124 A vaporcompression refrigeration cycle with refrigerant22 as the working fluid is considered The hardware and the Ts diagram for this air conditioner are to be sketched The heat absorbed by the refrigerant the work input to the compressor and the heat rejected in the condenser are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a In this normal vaporcompression refrigeration cycle the refrigerant enters the compressor as a saturated vapor at the evaporator pressure and leaves the condenser as saturated liquid at the condenser pressure 1 2 3 4 qH 5C win Condenser Evaporator Compressor Expansion valve 45C qL sat vap QH QL 5C 1 2 3 4 45C Win 2s 4s s T b The properties as given in the problem statement are h4 h3 hf 45C 101 kJkg h1 hg 5C 2481 kJkg The heat absorbed by the refrigerant in the evaporator is 1471 kJkg 101 248 1 4 1 h h qL c The COP of the air conditioner is 4 689 3412 Btuh 1 W W 16 Btuh 3412 Btuh 1 W SEER COPR The work input to the compressor is 314 kJkg 4 689 147 1 kJkg COP COP R in in R L L q w w q The enthalpy at the compressor exit is 279 5 kJkg 31 4 kJkg 1 kJkg 248 in 1 2 1 2 in w h h h h w The heat rejected from the refrigerant in the condenser is then 1785 kJkg 101 279 5 3 2 h h qH PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1118 1125 A vaporcompression refrigeration cycle with refrigerant134a as the working fluid is considered The amount of cooling the work input and the COP are to be determined Also the same parameters are to be determined if the cycle operated on the ideal vaporcompression refrigeration cycle between the same temperature limits Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a The expansion process through the expansion valve is isenthalpic h4 h3 Then T qH qL 1 2s 3 4 2 win 40C 15C 1593 kJkg 24319 40249 4 1 h h qL 210 8 kJkg 24319 45400 3 2 h h qH 5151 kJkg 40249 45400 1 2 in h h w 3093 5151 kJkg 159 3 kJkg COP win qL s c Ideal vaporcompression refrigeration cycle solution T qH qL 1 2 3 4 40C 15C win 1492 kJkg 24980 39904 4 1 h h qL 1909 kJkg 24980 44071 3 2 h h qH 4167 kJkg 39904 44071 1 2 in h h w 3582 4167 kJkg 149 2 kJkg COP win qL Discussion In the ideal operation the refrigeration load decreases by 63 and the work input by 191 while the COP increases by 158 s PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1119 1126 A vaporcompression refrigeration cycle with refrigerant134a as the working fluid is considered The rate of cooling the power input and the COP are to be determined Also the same parameters are to be determined if the cycle operated on the ideal vaporcompression refrigeration cycle between the same pressure limits Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a From the refrigerant134a tables Tables A11 through A13 39 kJkg 120 12039 kJkg 48 C 44 4 52 kPa 1400 C 4 52 29590 kJkg kPa 1400 0 9698 kJkg K 05 kJkg 253 0 C 10 1 1 10 kPa 200 C 1 10 3 4 48 C 3 3 3 kPa sat1400 2 1 1 2 1 1 1 1 kPa sat200 h h h h T P T h s s P s h T P T f s T QH QL 200 kPa 1 2s 3 4 14 MPa 2 Win s 30174 kJkg 25305 25305 29590 0 88 2 2 1 2 1 2 h h h h h h s C η 3317 kW 12039 0 025 kgs25305 4 1 h m h QL 4534 kW 12039 0 025 kgs30174 3 2 h m h QH 1217 kW 25305 0 025 kgs30174 1 2 in h m h W 2725 1 217 kW 3 317 kW COP in W QL b Ideal vaporcompression refrigeration cycle solution From the refrigerant134a tables Tables A11 through A13 T QH QL 200 kPa 1 2 3 4 14 MPa Win 4s 22 kJkg 127 12722 kJkg 0 kPa 1400 28508 kJkg kPa 1400 9377 kJkg K 0 46 kJkg 244 1 kPa 200 3 4 3 3 3 2 1 1 2 1 1 1 1 h h h x P h s s P s h x P s kW 3931 12722 0 025 kgs24446 4 1 h m h QL 3947 kW 12722 0 025 kgs28508 3 2 h m h QH 1016 kW 24446 0 025 kgs28508 1 2 in h m h W 2886 1 016 kW 3 931 kW COP in W QL Discussion The cooling load increases by 185 while the COP increases by 59 when the cycle operates on the ideal vaporcompression cycle PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1120 SecondLaw Analysis of VaporCompression Refrigeration Cycles 1127C The secondlaw efficiency of a refrigerator operating on the vaporcompression refrigeration cycle is defined as W X W W W X QL desttotal min IIR 1 η where is the exergy of the heat transferred from the lowtemperature medium and it is expressed as QL X L L Q T T Q X L 0 1 X desttotal is the total exergy destruction in the cycle and W is the actual power input to the cycle The secondlaw efficiency can also be expressed as the ratio of the actual COP to the Carnot COP Carnot R IIR COP COP η 1128C The secondlaw efficiency of a heat pump operating on the a vaporcompression refrigeration cycle is defined as W x E W W W Ex QH desttotal min IIHP 1 η Substituting HP COP QH W and H H Q T T Q Ex H 0 1 into the secondlaw efficiency equation Carnot HP HP HP 0 HP 0 IIHP COP COP COP COP 1 COP 1 L H H H H H H H H Q T T T Q T T Q Q T T Q W Ex H η since T0 TL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1121 1129C In an isentropic compressor s2 s1 and h2s h2 Applying these to the two the efficiency definitions we obtain 100 1 1 2 1 2 1 2 1 2 isen sComp h h h h h h h h w w s η 1 100 1 2 1 2 1 2 1 2 0 1 2 rev IIComp h h h h h h s s T h h w w η Thus the isentropic efficiency and the exergy efficiency of an isentropic compressor are both 100 The exergy efficiency of a compressor is not necessarily equal to its isentropic efficiency The two definitions are different as shown in the above equations In the calculation of isentropic efficiency the exit enthalpy is found at the hypothetical exit state at the exit pressure and the inlet entropy while the exergy efficiency involves the actual exit state The two efficiencies are usually close but different In the special case of an isentropic compressor the two efficiencies become equal to each other as proven above 1130 A vaporcompression refrigeration system is used to keep a space at a low temperature The power input the COP and the secondlaw efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis The power input is 06944 kW 3600 kJh 1kW 2500 kJh 2500 kJh 3500 6000 in L H Q Q W The COP is 14 0 6944 kW 3500 3600 kW COP in R W QL The COP of the Carnot cycle operating between the space and the ambient is 5 208 250 K 298 250 K COPCarnot L H L T T T The secondlaw efficiency is then 269 0 2688 5 208 41 COP COP Carnot R ηII PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1122 1131 A refrigerator is used to cool bananas at a specified rate The rate of heat absorbed from the bananas the COP The minimum power input the secondlaw efficiency and the exergy destruction are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a The rate of heat absorbed from the bananas is 61100 kJh 12 C C28 1140 kgh 3 35 kJkg 2 1 T T mc Q p L The COP is 197 kW 68 1697 kW kW 68 61100 3600 kW COP in W QL b Theminimum power input is equal to the exergy of the heat transferred from the lowtemperature medium 0463 kW 273 20 273 28 1697 kW 1 1 0 L L Q T T Q Ex L where the dead state temperature is taken as the inlet temperature of the eggplants T0 28C and the temperature of the lowtemperature medium is taken as the average temperature of bananas T 12282 20C c The secondlaw efficiency of the cycle is 539 00539 68 463 0 in II W Ex QL η The exergy destruction is the difference between the exergy expended power input and the exergy recovered the exergy of the heat transferred from the lowtemperature medium 814 kW 0 463 68 in dest ExQL W Ex PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1123 1132 A vaporcompression refrigeration cycle is used to keep a space at a low temperature The power input the mass flow rate of water in the condenser the secondlaw efficiency and the exergy destruction are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a The power input is 3431 kW 2 05 7 034 kW 2 05 3412 Btuh 1kW 24000 Btuh COP in QL W b From an energy balance on the cycle 3 431 1046 kW 7 034 in W Q Q L H The mass flow rate of the water is then determined from 02086 kgs C12 C 4 18 kJkg 46 kW 10 w pw H w pw H T c Q m T mc Q c The exergy of the heat transferred from the lowtemperature medium is 05153 kW 273 0 273 20 7 034 kW 1 1 0 L L Q T T Q Ex L The secondlaw efficiency of the cycle is 150 01502 3 431 5153 0 in II W Ex QL η The exergy destruction is the difference between the exergy supplied power input and the exergy recovered the exergy of the heat transferred from the lowtemperature medium 2916 kW 0 5153 3 431 in dest ExQL W Ex Alternative Solution The exergy efficiency can also be determined as follows 1365 0 20 273 0 COP RCarnot L H L T T T 150 01502 1365 2 05 COP COP RCarnot II η The result is identical as expected PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1124 1133E A vaporcompression refrigeration cycle is used to keep a space at a low temperature The mass flow rate of R 134a the COP The exergy destruction in each component and the exergy efficiency of the compressor the secondlaw efficiency and the exergy destruction are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a The properties of R134a are Tables A11E through A13E 0 1001 Btulbm R 30 Btulbm 45 psia 20 30 Btulbm 45 0 0921 Btulbm R 30 Btulbm 45 0 psia 140 0 2444 Btulbm R 36 Btulbm 131 F 160 psia 140 0 2257 Btulbm R 73 Btulbm 102 1 psia 20 4 4 4 3 4 3 3 3 3 2 2 2 2 1 1 1 1 s h P h h s h x P s h T P s h x P PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course The energy interactions in each component and the mass flow rate of R134a are QH QL 20 psia 1 2 3 4 140 psia Win 2s 4s s T 2863 Btulbm 10273 13136 1 2 in h h w 8606 Btulbm 4530 13136 3 2 h h qH 5743 Btulbm 2863 8606 in w q q H L 02177 lbms 5743 Btulbm 45000 3600 Btus L L q Q m The COP is 2006 2863 Btulbm 5743 Btulbm COP win qL b The exergy destruction in each component of the cycle is determined as follows Compressor 0 01874 Btulbm R 0 2257 0 2444 1 2 gen1 2 s s s 2203 Btus 02177 lbms540 R001874 Btulbm R 0 gen12 dest12 mT s Ex Condenser 0 007073 Btulbm R 540 R 8606 Btulbm 0 2444 Btulbm R 0 0921 2 3 gen2 3 H H T q s s s 08313 Btus 02177 lbms540 R000707 3 Btulbm R 0 gen23 dest23 mT s Ex Expansion valve 0 007962 Btulbm R 0 0921 0 1001 3 4 gen3 4 s s s 09359 Btus 02177 lbms540 R000796 2 Btulbm R 0 gen34 dest34 mT s Ex Evaporator 0 003400 Btulbm R 470 R 5743 Btulbm 0 1001 Btulbm R 0 2257 4 1 gen4 1 L L T q s s s 03996 Btus 02177 lbms540 R000340 0 Btulbm R 0 gen4 1 dest4 1 mT s Ex preparation If you are a student using this Manual you are using it without permission 1125 The power input and the exergy efficiency of the compressor is determined from 6 232 Btus 02177 lbms2863 Btulbm in in mw W 647 06465 6 232 Btus 2 203 Btus 1 1 in dest1 2 II W Ex η c The exergy of the heat transferred from the lowtemperature medium is 1 862 Btus 470 540 45000 3600 Btus 1 1 0 L L Q T T Q Ex L The secondlaw efficiency of the cycle is 299 02987 6 232 Btus 862 Btus 1 in II W Ex QL η The total exergy destruction in the cycle is the difference between the exergy supplied power input and the exergy recovered the exergy of the heat transferred from the lowtemperature medium 4370 Btus 1 862 6 232 in desttotal ExQL W Ex The total exergy destruction can also be determined by adding exergy destructions in each component 370 Btus 4 0 3996 0 9359 0 8313 203 2 dest4 1 dest34 dest23 dest12 total dest Ex Ex Ex Ex Ex The result is the same as expected PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1126 1134 A vaporcompression refrigeration cycle is used to keep a space at a low temperature The mass flow rate of R134a the COP The exergy destruction in each component and the exergy efficiency of the compressor the secondlaw efficiency and the exergy destruction are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis a The properties of R134a are Tables A11 through A13 0 4244 kJkg K 77 kJkg 117 0 MPa 21 9267 kJkg K 0 27 kJkg 278 C 50 MPa 21 3 3 3 3 2 2 2 2 s h x P s h T P The rate of heat transferred to the water is the energy change of the water from inlet to exit 5 016 kW 20 C C28 0 15 kgs 4 18 kJkg 1 2 w w p w H T T m c Q QH QL 1 2 3 4 12 MPa Win 4s 2s s T The energy decrease of the refrigerant is equal to the energy increase of the water in the condenser That is 003125 kgs 27827 11777 kJkg 5 016 kW 3 2 3 2 h h Q m h h m Q H R R H The refrigeration load is 9610 Btuh 1kW 2816 kW 3412 Btuh 2816 kW 22 5 016 Win Q Q H L The COP of the refrigerator is determined from its definition 128 kW 22 2 816 kW COP in W QL b The COP of a reversible refrigerator operating between the same temperature limits is 8 156 273 12 273 20 273 12 COPCarnot L H L T T T The minimum power input to the compressor for the same refrigeration load would be 03453 kW 8 156 2 816 kW COPCarnot inmin QL W The secondlaw efficiency of the cycle is 157 01569 22 3453 0 in inmin II W W η The total exergy destruction in the cycle is the difference between the actual and the minimum power inputs 185 kW 0 3453 22 inmin in desttotal W W Ex c The entropy generation in the condenser is 001191 kWK 0 0 9267 kJkg K 0 03125 kgs 0 4004 273 20 273 Cln 28 0 15 kgs 4 18 kJkg ln 2 3 1 2 cond gen s s m T T m c S R w w w p The exergy destruction in the condenser is 0349 kW 293 K 0 001191 kWK gencond 0 destcond T S Ex preparation If you are a student using this Manual you are using it without permission 1127 1135 An ideal vaporcompression refrigeration cycle is used to keep a space at a low temperature The cooling load the COP the exergy destruction in each component the total exergy destruction and the secondlaw efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a The properties of R134a are Tables A11 through A13 0 5251 kJkg K 93 kJkg 135 C 10 93 kJkg 135 0 4791 kJkg K 93 kJkg 135 0 kPa 1600 28785 kJkg kPa 1600 0 9377 kJkg K 51 kJkg 244 1 C 10 4 4 4 3 4 3 3 3 3 2 1 2 sat579 C 2 1 1 1 1 s h T h h s h x P h s s P P s h x T T QH QL 10C 1 2 3 4 579C Win 4s s The energy interactions in the components and the COP are 1086 kJkg 24451 13593 4 1 h h qL 1519 kJkg 13593 28785 3 2 h h qH 4333 kJkg 24451 28785 1 2 in h h w 2506 4333 kJkg 108 6 kJkg COP win qL b The exergy destruction in each component of the cycle is determined as follows Compressor 0 1 2 gen1 2 s s s 0 0 gen12 dest12 T s Ex Condenser 0 05124 kJkg K 298 K 151 9 kJkg 0 9377 kJkg K 0 4791 2 3 gen2 3 H H T q s s s 1527 kJkg 298 K005124 kJkg K 0 gen23 dest23 T s Ex Expansion valve 0 04595 kJkg K 0 4791 0 5251 3 4 gen3 4 s s s 1369 kJkg 298 K004595 kJkg K 0 gen34 dest34 T s Ex Evaporator 0 02201 kJkg K 278 K 108 6 kJkg 0 5251 kJkg K 0 9377 4 1 gen4 1 L L T q s s s 656 kJkg kJkg K 298 K002201 0 gen4 1 dest4 1 T s Ex The total exergy destruction can be determined by adding exergy destructions in each component 3552 kJkg 6 56 1369 1527 0 dest4 1 dest34 dest23 dest12 desttotal Ex Ex Ex Ex Ex PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1128 c The exergy of the heat transferred from the lowtemperature medium is 7 812 kJkg 278 298 108 6 kJkg 1 1 0 L L q T T q Ex L The secondlaw efficiency of the cycle is 180 01803 4333 812 7 in II w Ex qL η The total exergy destruction in the cycle can also be determined from 3552 kJkg 7 812 4333 in desttotal ExqL w Ex The result is identical as expected The secondlaw efficiency of the compressor is determined from 1 2 1 2 0 1 2 in act rev expended recovered IIComp h m h s s T h h m W W X X η since the compression through the compressor is isentropic s2 s1 the secondlaw efficiency is ηIIComp 1100 The secondlaw efficiency of the evaporator is determined from 1 4 1 dest4 1 4 0 1 4 0 expended recovered II Evap 1 X X X s s T h m h T T T Q X X L L L η where 1437 kJkg 0 9377 kJkg K 298 K 0 5251 24451 kJkg 13593 1 4 0 1 4 1 4 s s T h h x x Substituting 544 0 544 1437 kJkg 6 56 kJkg 1 1 1 4 dest4 1 II Evap x x x η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1129 1136 An ideal vaporcompression refrigeration cycle uses ammonia as the refrigerant The volume flow rate at the compressor inlet the power input the COP the secondlaw efficiency and the total exergy destruction are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a The properties of ammonia are given in problem statement An energy balance on the cındenser gives 1361 kJkg 437 4 1439 3 4 1 h h qH T QH QL 200 kPa 1 2 3 4 2 MPa Win 4s 0 01323 kgs 1361 kJkg 18 kW H H q Q m The volume flow rate is determined from 787 Ls 007865 m s 0 0 01323 kgs 0 5946 m kg 3 3 1 1 mv V b The power input and the COP are s 475 kW kJkg 1439 3 0 01323 kgs17983 1 2 in h m h W 1325 kW 437 4 kJkg 0 01323 kgs14393 4 1 h m h QL 279 4 75 kW 1325 kW COP in W QL c The exergy of the heat transferred from the lowtemperature medium is 1 81 kW 264 300 1325 kW 1 1 0 L L Q T T Q Ex L The secondlaw efficiency of the cycle is 381 0381 4 75 81 1 in II W Ex QL η The total exergy destruction in the cycle is the difference between the exergy supplied power input and the exergy recovered the exergy of the heat transferred from the lowtemperature medium 294 kW 1 81 4 75 in desttotal ExQL W Ex PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1130 1137 Prob 1136 is reconsidered Using EES software the problem is to be repeated ammonia R134a and R22 is used as a refrigerant and the effects of evaporator and condenser pressures on the COP the secondlaw efficiency and the total exergy destruction are to be investigated Analysis The equations as written in EES are GIVEN P1200 kPa P22000 kPa QdotH18 kW TL9273 K TH27273 K PROPERTIES Fluidammonia x11 x30 h1enthalpyFluid PP1 xx1 s1entropyFluid PP1 xx1 v1volumeFluid PP1 xx1 h2enthalpyFluid PP2 ss1 s2s1 h3enthalpyFluid PP2 xx3 s3entropyFluid PP2 xx3 h4h3 s4entropyFluid PP1 hh4 qHh2h3 mdotQdotHqH Voldot1mdotv1 QdotLmdoth1h4 Wdotinmdoth2h1 COPQdotLWdotin ExdotQLQdotL1THTL etaIIExdotQLWdotin ExdotdestWdotinExdotQL The solutions in the case of ammonia R134a and R22 are PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 1131 Now we investigate the effects of evaporating and condenser pressures on the COP the secondlaw efficiency and the total exergy destruction The results are given by tables and figures 100 150 200 250 300 350 400 2 25 3 35 4 45 025 03 035 04 045 05 055 06 P1 kPa COP ηII PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1132 100 150 200 250 300 350 400 1 15 2 25 3 35 4 45 P1 kPa Exdest kW 1000 1200 1400 1600 1800 2000 275 32 365 41 455 5 035 04 045 05 055 06 065 07 P2 kPa COP ηII 1000 1200 1400 1600 1800 2000 1 14 18 22 26 3 P2 kPa Exdest kW PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1133 Selecting the Right Refrigerant 1138C The desirable characteristics of a refrigerant are to have an evaporator pressure which is above the atmospheric pressure and a condenser pressure which corresponds to a saturation temperature above the temperature of the cooling medium Other desirable characteristics of a refrigerant include being nontoxic noncorrosive nonflammable chemically stable having a high enthalpy of vaporization minimizes the mass flow rate and of course being available at low cost 1139C The minimum pressure that the refrigerant needs to be compressed to is the saturation pressure of the refrigerant at 30C which is 0771 MPa At lower pressures the refrigerant will have to condense at temperatures lower than the temperature of the surroundings which cannot happen 1140C Allowing a temperature difference of 10C for effective heat transfer the evaporation temperature of the refrigerant should be 20C The saturation pressure corresponding to 20C is 0133 MPa Therefore the recommended pressure would be 012 MPa 1141 A refrigerator that operates on the ideal vaporcompression cycle with refrigerant134a is considered Reasonable pressures for the evaporator and the condenser are to be selected Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis Allowing a temperature difference of 10C for effective heat transfer the evaporation and condensation temperatures of the refrigerant should be 20C and 35C respectively The saturation pressures corresponding to these temperatures are 0133 MPa and 0888 MPa Therefore the recommended evaporator and condenser pressures are 0133 MPa and 0888 MPa respectively 1142 A heat pump that operates on the ideal vaporcompression cycle with refrigerant134a is considered Reasonable pressures for the evaporator and the condenser are to be selected Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis Allowing a temperature difference of 10C for effective heat transfer the evaporation and condensation temperatures of the refrigerant should be 4C and 36C respectively The saturation pressures corresponding to these temperatures are 338 kPa and 912 kPa Therefore the recommended evaporator and condenser pressures are 338 kPa and 912 kPa respectively PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1134 Heat Pump Systems 1143C A heat pump system is more cost effective in Miami because of the low heating loads and high cooling loads at that location 1144C A watersource heat pump extracts heat from water instead of air Watersource heat pumps have higher COPs than the airsource systems because the temperature of water is higher than the temperature of air in winter 1145E A heat pump operating on the ideal vaporcompression refrigeration cycle with refrigerant134a as the working fluid is considered The COP of the heat pump is to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis In an ideal vaporcompression refrigeration cycle the compression process is isentropic the refrigerant enters the compressor as a saturated vapor at the evaporator pressure and leaves the condenser as saturated liquid at the condenser pressure From the refrigerant tables Tables A11E A12E and A13E 869 Btulbm throttling 37 37869 Btulbm sat liquid psia 100 11498 Btulbm psia 100 0 22189 kJkg K 10878 Btulbm sat vapor F 40 3 4 100 psia 3 3 2 1 2 2 40 F 1 40 F 1 1 h h h h P h s s P s s h h T f g g QH QL 40F 1 2 3 4 100 psia Win 4s s T The COP of the heat pump is determined from its definition 1243 11498 10878 37869 11498 COP 1 2 3 2 in HP h h h h w qH PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1135 1146 A heat pump operating on the ideal vaporcompression refrigeration cycle with refrigerant134a as the working fluid is considered The COP and the rate of heat supplied to the evaporator are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis In an ideal vaporcompression refrigeration cycle the compression process is isentropic the refrigerant enters the compressor as a saturated vapor at the evaporator pressure and leaves the condenser as saturated liquid at the condenser pressure From the refrigerant tables Tables A11 A12 and A13 32 kJkg throttling 107 10732 kJkg sat liquid kPa 1000 27798 kJkg kPa 1000 0 93773 kJkg K 24446 kJkg sat vapor kPa 200 3 4 1000 kPa 3 3 2 1 2 2 200 kPa 1 200 kPa 1 1 h h h h P h s s P s s h h P f g g PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course The mass flow rate of the refrigerant is determined from 0179 kgs 24446 kJkg 27798 6 kJs 1 2 in 1 2 in h h W m h m h W QH QL 200 kPa 1 2 3 4 10 MPa Win 4s s T Then the rate of heat supplied to the evaporator is 245 kW 10732 kJkg 0 179 kgs24446 4 1 h m h QL The COP of the heat pump is determined from its definition 509 24446 27798 10732 27798 COP 1 2 3 2 in HP h h h h w qH preparation If you are a student using this Manual you are using it without permission 1136 1147 A heat pump operating on the ideal vaporcompression refrigeration cycle with refrigerant134a as the working fluid is considered The rate of heat transfer to the heated space and the COP are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a In an ideal vaporcompression refrigeration cycle the compression process is isentropic the refrigerant enters the compressor as a saturated vapor at the evaporator pressure and leaves the condenser as saturated liquid at the condenser pressure 1 2 3 4 20C Condenser Evaporator Compressor Expansion valve QH 14 MPa sat vap QH QL 20C 1 3 4 14 MPa W in 2 4s s T in W L Q b The properties as given in the problem statement are h4 h3 hf 1400 kPa 1272 kJkg h1 hg 20C 2616 kJkg The enthalpy at the compressor exit is 281 6 kJkg 20 kJkg 6 kJkg 261 in 1 2 1 2 in w h h h h w The mass flow rate through the cycle is 0 02009 kgs kJkg 127 2 2616 kJs 72 4 1 4 1 h h Q m h m h Q L L The rate of heat transfer to the heated space is 310 kW kJkg 127 2 0 02009 kgs2816 3 2 h m h QH c The COP of the heat pump is 772 002009 kgs20 kJkg 310 kW COP in in HP mw Q W Q H H PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1137 1148 A heat pump vaporcompression refrigeration cycle with refrigerant134a as the working fluid is considered The hardware and the Ts diagram for this heat pump are to be sketched The power input and the COP are to be determined Analysis a In a normal vaporcompression refrigeration cycle the refrigerant enters the compressor as a saturated vapor at the evaporator pressure and leaves the condenser as saturated liquid at the condenser pressure 1 4 240 kPa sat vap Evaporator Compressor Expansion valve 2 3 Condenser QH 16 MPa QH QL 240 kPa 1 2 3 4 16 MPa Win 2s 4s s T in W L Q b The properties as given in the problem statement are h4 h3 hf 1600 kPa 134 kJkg h1 hg 240 kPa 244 kJkg h2s 285 kJkg From the definition of isentropic efficiency for a compressor 292 2 kJkg 0 85 244 285 244 Comp 1 2 1 2 1 2 1 2 Comp η η h h h h h h h h s s Then the work input to the compressor is 48 2 kJkg 244 292 2 1 2 in h h w The mass flow rate through the cycle is 0 04446 kgs 134 kJkg 2922 1 ton 2 ton 21160 kJs 3 2 3 2 h h Q m h m h Q H H Then the power input to the compressor is 214 kW 004446 kgs482 kJkg in in mw W The COP of the heat pump is 329 214 kW 1 ton 2 ton 21160 kJs COP in HP W QH PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1138 1149 A geothermal heat pump is considered The degrees of subcooling done on the refrigerant in the condenser the mass flow rate of the refrigerant the heating load the COP of the heat pump the minimum power input are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a From the refrigerant134a tables Tables A11 through A13 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 28000 kJkg kPa 1400 9223 kJkg 0 59 kJkg 261 sat vap 1 1 kPa 572 12124 kJkg 1 kPa 572 23 0 C 20 2 1 2 2 1 1 1 1 4 3 4 4 4 4 h s s P s h x P h h h P x T From the steam tables Table A4 53 kJkg 167 34 kJkg 209 40 C 2 50 C 1 f w f w h h h h The saturation temperature at the condenser pressure of 1400 kPa and the actual temperature at the condenser outlet are 5240 C sat 1400 kPa T 4859 C 24 kJkg 121 1400 kPa 3 3 3 T h P from EES Then the degrees of subcooling is 381C 4859 5240 3 sat subcool T T T b The rate of heat absorbed from the geothermal water in the evaporator is QH QL 1 2 3 4 14 MPa s T Win 40C 14 MPa s2 s1 1 2 3 4 QH 20C x023 Condenser Evaporator Compressor Expansion valve Water 50C Win sat vap 4s 2 718 kW 0 065 kgs20934 16753kJkg 2 1 w w w L h h m Q This heat is absorbed by the refrigerant in the evaporator 001936 kgs 26159 12124kJkg 718 kW 2 4 1 h h Q m L R c The power input to the compressor the heating load and the COP are 0 6564 kW 26159kJkg 0 01936 kgs28000 out 1 2 in Q h h m W R 3074 kW 0 01936 kgs28000 12124kJkg 3 2 h h m Q R H 468 06564 kW 3 074 kW COP in H W Q d The reversible COP of the cycle is 1292 273 273 50 1 25 1 1 1 COPrev TL TH The corresponding minimum power input is 0238 kW 1292 3 074 kW COPrev inmin QH W preparation If you are a student using this Manual you are using it without permission 1139 1150 An actual heat pump cycle with R134a as the refrigerant is considered The isentropic efficiency of the compressor the rate of heat supplied to the heated room the COP of the heat pump and the COP and the rate of heat supplied to the heated room if this heat pump operated on the ideal vaporcompression cycle between the same pressure limits are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a The properties of refrigerant134a are Tables A11 through A13 27726 kPa 800 0 9506 kJkg 87 kJkg 247 4 C 1009 kPa 200 09 C 10 91 kJkg 87 8791 kJkg 3 C 2906 kPa 750 06 C 29 29176 kJkg C 55 kPa 800 2 1 2 2 1 1 1 1 kPa sat200 3 4 3 3 3 sat750 kPa 3 2 2 2 h s s s P s h T P T h h h T P T T h T P QH 750 kPa Win Condenser Evaporator Compressor Expansion valve 1 2 3 4 QL 800 kPa 55C The isentropic efficiency of the compressor is T Q QL 1 2 3 4 2 Win 0670 24787 29176 24787 26 277 1 2 1 2 h h h h s ηC b The rate of heat supplied to the room is 367 kW 8791kJkg 0 018 kgs29176 3 2 h m h QH c The power input and the COP are 0 790 kW 24787kJkg 0 018 kgs29176 1 2 in h m h W s 464 0 790 3 67 COP in W QH d The ideal vaporcompression cycle analysis of the cycle is as follows 9377 kJkgK 0 46 kJkg 244 200 kPa 1 200 kPa 1 g g s s h h T QH QL 02 MPa 1 2 3 4 08 MPa Win 4s 27325 kJkg 800 kPa 2 1 2 2 h s s P 3 4 800 kPa 3 47 kJkg 95 h h h h f 618 24446 27325 9547 27325 COP 1 2 3 2 h h h h s 320 kW 9547kJkg 0 018 kgs27325 3 2 h m h QH PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1140 Innovative Refrigeration Systems 1151C Performing the refrigeration in stages is called cascade refrigeration In cascade refrigeration two or more refrigeration cycles operate in series Cascade refrigerators are more complex and expensive but they have higher COPs they can incorporate two or more different refrigerants and they can achieve much lower temperatures 1152C Cascade refrigeration systems have higher COPs than the ordinary refrigeration systems operating between the same pressure limits 1153C The saturation pressure of refrigerant134a at 32C is 77 kPa which is below the atmospheric pressure In reality a pressure below this value should be used Therefore a cascade refrigeration system with a different refrigerant at the bottoming cycle is recommended in this case 1154C We would favor the twostage compression refrigeration system with a flash chamber since it is simpler cheaper and has better heat transfer characteristics 1155C Yes by expanding the refrigerant in stages in several throttling devices 1156C To take advantage of the cooling effect by throttling from high pressures to low pressures PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1143 1159 Problem 1157 is reconsidered The effects of the various refrigerants in EES data bank for compressor efficiencies of 80 90 and 100 percent is to be investigated Analysis The problem is solved using EES and the results are tabulated and plotted below FluidR134a Input Data P1100 kPa P4 1400 kPa P6400 kPa Etacomp 10 mdotA025 kgs High Pressure Compressor A P9P6 h4senthalpyFluidPP4ss9 State 4s is the isentropic value of state 4 h9wcompAsh4s energy balance on isentropic compressor wcompAwcompAsEtacompdefinition of compressor isentropic efficiency h9wcompAh4 energy balance on real compressorassumed adiabatic s4entropyFluidhh4PP4 properties for state 4 T4temperatureFluidhh4PP4 WdotcompAmdotAwcompA Condenser P5P4 neglect pressure drops across condenser T5temperatureFluidPP5x0 properties for state 5 assumes sat liq at cond exit h5enthalpyFluidTT5x0 properties for state 5 s5entropyFluidTT5x0 h4qHh5 energy balance on condenser QdotH mdotAqH Throttle Valve A h6h5 energy balance on throttle isenthalpic x6qualityFluidhh6PP6 properties for state 6 s6entropyFluidhh6PP6 T6temperatureFluidhh6PP6 Flash Chamber mdotB 1x6 mdotA P7 P6 h7enthalpyFluid PP7 x0 s7entropyFluidhh7PP7 T7temperatureFluidhh7PP7 Mixing Chamber x6mdotAh3 mdotBh2 x6 mdotA mdotBh9 P3 P6 h3enthalpyFluid PP3 x1 properties for state 3 s3entropyFluidPP3x1 T3temperatureFluidPP3xx1 s9entropyFluidhh9PP9 properties for state 9 T9temperatureFluidhh9PP9 Low Pressure Compressor B x11 assume flow to compressor inlet to be saturated vapor h1enthalpyFluidPP1xx1 properties for state 1 T1temperatureFluidPP1 xx1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1144 s1entropyFluidPP1xx1 P2P6 h2senthalpyFluidPP2ss1 state 2s is isentropic state at comp exit h1wcompBsh2s energy balance on isentropic compressor wcompBwcompBsEtacompdefinition of compressor isentropic efficiency h1wcompBh2 energy balance on real compressorassumed adiabatic s2entropyFluidhh2PP2 properties for state 2 T2temperatureFluidhh2PP2 WdotcompBmdotBwcompB Throttle Valve B h8h7 energy balance on throttle isenthalpic x8qualityFluidhh8PP8 properties for state 8 s8entropyFluidhh8PP8 T8temperatureFluidhh8PP8 Evaporator P8P1 neglect pressure drop across evaporator qL h8h1 energy balance on evaporator QdotLmdotBqL Cycle Statistics Wdotintotal WdotcompA WdotcompB COPQdotLWdotintotal definition of COP ηcomp QL kW COP 06 065 07 075 08 085 09 095 1 2855 2855 2855 2855 2855 2855 2855 2855 2855 1438 157 1702 1835 1968 2101 2234 2368 2501 025 000 025 050 075 100 125 150 175 100 50 0 50 100 150 200 250 s kJkgK T C 1400 kPa 400 kPa 100 kPa R134a 1 239 4 5 6 7 8 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1145 06 065 07 075 08 085 09 095 1 14 16 18 2 22 24 26 28 ηcomp COP R134a R22 ammonia 06 065 07 075 08 085 09 095 1 0 50 100 150 200 250 300 ηcomp QL kW ammonia R22 R134a 100 200 300 400 500 600 700 800 900 1000 19 2 21 22 23 24 25 26 P6 kPa COP R134a PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1146 1160 A twostage cascade refrigeration cycle is considered The mass flow rate of the refrigerant through the upper cycle the rate of heat removal from the refrigerated space and the COP of the refrigerator are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a The properties are to be obtained from the refrigerant tables Tables A11 through A13 9377 kJkgK 0 46 kJkg 244 200 kPa 1 200 kPa 1 g g s s h h 5 6 7 8 QH Condenser Evaporator Compressor Expansion valve Win 1 2 3 4 Win Condenser Evaporator Compressor Expansion valve QL 26330 kJkg 500 kPa 2 1 2 2 h s s s P 26801 kJkg 24446 24446 26330 0 80 2 2 1 2 1 2 h h h h h h s C η 33 kJkg 73 33 kJkg 73 3 4 500 kPa 3 h h h h f 9269 kJkgK 0 55 kJkg 255 400 kPa 5 400 kPa 5 g g s s h h 27833 kJkg 1200 kPa 6 5 6 6 h s s s P 28402 kJkg 25555 25555 27833 0 80 6 6 5 6 5 6 h h h h h h s C η 77 kJkg 117 77 kJkg 117 7 8 1200 kPa 7 h h h h f The mass flow rate of the refrigerant through the upper cycle is determined from an energy balance on the heat exchanger 0212 kgs A A B A m m h h m h h m 0 15 kgs26801 7333kJkg 25555 11777kJkg 3 2 8 5 b The rate of heat removal from the refrigerated space is 2567 kW 7333kJkg 0 15 kgs24446 4 1 h h m Q B L c The power input and the COP are 9 566 kW 24446kJkg 0 212 kgs26801 25555kJkg 0 15 kgs28402 1 2 5 6 in h h m h h m W B A 268 9 566 2567 COP in L W Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1147 1161 A twoevaporator compression refrigeration cycle with refrigerant134a as the working fluid is considered The cooling rate of the hightemperature evaporator the power required by the compressor and the COP of the system are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 4 5 6 Compressor Expansion valve 1 2 Evaporator 2 Expansion valve Expansion valve Condenser Evaporator 1 7 QH QL 264C 1 2 3 6 800 kPa Win 4 5 7 0C s T 3 Analysis From the refrigerant tables Tables A11 A12 and A13 23444 kJkg sat vapor C 4 26 25045 kJkg sat vapor C 0 47 kJkg throttling 95 9547 kJkg sat liquid kPa 800 C 26 4 7 7 0 C 5 5 3 6 4 800 kPa 3 3 g g f h h T h h T h h h h h P The mass flow rate through the lowtemperature evaporator is found by 0 05757 kgs 9547 kJkg 23444 8 kJs 6 7 2 6 7 2 h h Q m h h m Q L L The mass flow rate through the warmer evaporator is then 0 04243 kgs 0 05757 10 2 1 m m m Applying an energy balance to the point in the system where the two evaporator streams are recombined gives 24123 kJkg 10 0 0575723444 0 0424325045 7 2 1 5 1 1 7 2 1 5 m m h m h h mh m h m h Then 28626 kJkg kPa 800 0 9789 kJkg K 23 kJkg 241 kPa 100 2 1 2 2 1 1 sat 264 C 1 h s s P s h P P The cooling rate of the hightemperature evaporator is 658 kW 9547 kJkg 0 04243 kgs25045 4 5 1 h m h QL The power input to the compressor is 450 kW 24123 kJkg kgs28626 10 1 2 in h m h W The COP of this refrigeration system is determined from its definition 324 450 kW 658 kW 8 COP in R W QL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1148 1162E A twoevaporator compression refrigeration cycle with refrigerant134a as the working fluid is considered The power required by the compressor and the COP of the system are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 4 5 6 Compressor Expansion valve 1 2 Evaporator 2 Expansion valve Expansion valve Condenser Evaporator 1 7 QH QL 10 psia 1 2 3 6 180 psia Win 4 5 7 30 psia s T 3 Analysis From the refrigerant tables Tables A11E A12E and A13E 9868 Btulbm sat vapor psia 10 10532 Btulbm sat vapor psia 30 50 Btulbm throttling 51 5150 Btulbm sat liquid psia 180 10 psia 7 7 0 psia 3 5 5 3 6 4 180 psia 3 3 g g f h h P h h P h h h h h P The mass flow rates through the hightemperature and lowtemperature evaporators are found by 167 2 lbmh 5150 Btulbm 10532 9000 Btuh 4 5 1 1 4 5 1 1 h h Q m h m h Q L L 508 6 lbmh 5150 Btulbm 9868 24000 Btuh 6 7 2 2 6 7 2 2 h h Q m h h m Q L L Applying an energy balance to the point in the system where the two evaporator streams are recombined gives 10033 Btulbm 508 6 167 2 508 6 9868 167 2 10532 2 1 7 2 1 5 1 1 2 1 7 2 1 5 m m m h m h h h m m m h m h Then 12705 Btulbm psia 180 0 2333 Btulbm R 10033 Btulbm psia 10 2 1 2 2 1 1 1 h s s P s h P The power input to the compressor is 529 kW 341214 Btuh 1kW 10033 Btulbm 508 6 lbmh12705 167 2 1 2 2 1 in h h m m W The COP of this refrigeration system is determined from its definition 183 341214 Btuh 1kW 529 kW 9000 Btuh 24000 COP in R W QL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1149 1163E A twoevaporator compression refrigeration cycle with refrigerant134a as the working fluid is considered The power required by the compressor and the COP of the system are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 4 5 6 Compressor Expansion valve 1 2 Evaporator 2 Expansion valve Expansion valve Condenser Evaporator 1 7 QH QL 10 psia 1 2 3 6 180 psia Win 4 5 7 60 psia s T 3 Analysis From the refrigerant tables Tables A11E A12E and A13E 9868 Btulbm sat vapor psia 10 11011 Btulbm sat vapor psia 60 50 Btulbm throttling 51 5150 Btulbm sat liquid psia 180 10 psia 7 7 0 psia 6 5 5 3 6 4 180 psia 3 3 g g f h h P h h P h h h h h P The mass flow rates through the hightemperature and lowtemperature evaporators are found by 511 8 lbmh 5150 Btulbm 11011 30000 Btuh 4 5 1 1 4 5 1 1 h h Q m h m h Q L L 508 6 lbmh 5150 Btulbm 9868 24000 Btuh 6 7 2 2 6 7 2 2 h h Q m h h m Q L L Applying an energy balance to the point in the system where the two evaporator streams are recombined gives 10441 Btulbm 508 6 511 8 508 6 9868 511 8 11011 2 1 7 2 1 5 1 1 2 1 7 2 1 5 m m m h m h h h m m m h m h Then 13269 Btulbm psia 180 0 2423 Btulbm R 10441 Btulbm psia 10 2 1 2 2 1 1 1 h s s P s h P The power input to the compressor is 846 kW 341214 Btuh 1kW 10441 Btulbm 508 6 lbmh13269 511 8 1 2 2 1 in h h m m W The COP of this refrigeration system is determined from its definition 187 341214 Btuh 1kW 846 kW 30000 Btuh 24000 COP in R W QL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1150 1164 A twostage cascade refrigeration system is considered Each stage operates on the ideal vaporcompression cycle with upper cycle using water and lower cycle using refrigerant134a as the working fluids The mass flow rate of R134a and water in their respective cycles and the overall COP of this system are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 The heat exchanger is adiabatic Analysis From the water and refrigerant tables Tables A4 A5 A6 A11 A12 and A13 94 kJkg throttling 63 6394 kJkg sat liquid kPa 400 26759 kJkg kPa 400 0 96866 kJkg K 22586 kJkg sat vapor C 40 44 kJkg throttling 858 85844 kJkg sat liquid MPa 61 5083 4 kJkg MPa 61 9 0249 kJkg K 2510 1 kJkg sat vapor C 5 7 8 40 0 kPa 7 7 6 5 6 6 40 C 5 40 C 5 5 3 4 MPa 61 3 3 2 1 2 2 5 C 1 5 C 1 1 h h h h P h s s P s s h h T h h h h P h s s P s s h h T f g g f g g 40C QL 5 6 3 8 4 400 kPa 5C 7 1 16 MPa 2 s T The mass flow rate of R134a is determined from 01235 kgs 6394 kJkg 22586 20 kJs 8 5 8 5 h h Q m h h m Q L R R L An energy balance on the heat exchanger gives the mass flow rate of water 001523 kgs 85844 2510 1 6394 0 1235 kgs 26759 4 1 7 6 4 1 7 6 h h h h m m h h m h h m R w w R The total power input to the compressors is 4435 kJs kJkg 1 2510 0 01523 kgs50834 22586 kJkg 0 1235 kgs26759 1 2 5 6 in h h m h h m W w R The COP of this refrigeration system is determined from its definition 0451 4435 kJs 20 kJs COP in R W QL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1151 1165 A twostage vaporcompression refrigeration system with refrigerant134a as the working fluid is considered The process with the greatest exergy destruction is to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From Prob 1155 and the water and refrigerant tables Tables A4 A5 A6 A11 A12 and A13 303 K 30 C 303 K 30 C 243 K 30 C 4225 0 kJkg 16192 kJkg 0 01523 kgs 0 1235 kgs 0 27423 kJkg K 0 24757 kJkg K 0 96866 kJkg K 3 0869 kJkg K 2 3435 kJkg K 0249 kJkg K 9 0 3 2 8 5 8 7 6 5 4 3 2 1 T T T h h q h h q m m s s s s s s s s H L H L w R 40C QL 5 6 3 8 4 400 kPa 5C 7 1 16 MPa 2 s T The exergy destruction during a process of a stream from an inlet state to exit state is given by sink out source in 0 gen 0 dest T q T q s s T T s x i e Application of this equation for each process of the cycle gives 0 417 kJs 0 96866 0 1235 0 24757 3 0869 303 0 01523 9 0249 1 05 kJs 243 16192 0 27423 0 96866 0 1235303 0 996 kJs 0 24757 0 1235303 0 27423 3 43 kJs 2 3435 0 01523303 3 0869 3352 kJs 303 4225 0 9 0249 2 3435 0 01523303 K 6 7 R 4 1 w 0 heat exch destroyed 8 5 0 R 85 destroyed 7 8 0 R 78 destroyed 3 4 0 w 34 destroyed 2 3 0 23 destroyed s s m s s T m X T q s s m T X s s m T X s s m T X T q s s m T X L L H H w For isentropic processes the exergy destruction is zero 0 0 56 destroyed 12 destroyed X X Note that heat is absorbed from a reservoir at 30C 243 K and rejected to a reservoir at 30C 303 K which is also taken as the dead state temperature Alternatively one may use the standard 25C 298 K as the dead state temperature and perform the calculations accordingly The greatest exergy destruction occurs in the condenser PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1152 1166 A twostage cascade refrigeration cycle with a flash chamber with refrigerant134a as the working fluid is considered The mass flow rate of the refrigerant through the highpressure compressor the rate of refrigeration the COP are to be determined Also the rate of refrigeration and the COP are to be determined if this refrigerator operated on a singlestage vaporcompression cycle under similar conditions Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a From the refrigerant134a tables Tables A11 through A13 9377 kJkgK 0 51 kJkg 244 10 C 1 10 C 1 g g s s h h T 1 2 5 8 045 MPa 4s 9 2s 6 7 3 B 10C A 16 MPa 4 26107 kJkg 450 kPa 2 1 2 2 h s s s P 26376 kJkg 24451 24451 26107 0 86 2 2 1 2 1 2 h h h h h h s C η s 81 kJkg 68 81 kJkg 68 93 kJkg 135 93 kJkg 135 53 kJkg 257 7 8 450 kPa 7 5 6 1600 kPa 5 450 kPa 3 h h h h h h h h h h f f g 0 3557 kPa 450 13593 kJkg 6 6 6 x P h The mass flow rate of the refrigerant through the high pressure compressor is determined from a mass balance on the flash chamber 01707 kgs 03557 1 0 11 kgs 1 6 7 x m m Also 0 06072 kgs 0 11 0 1707 7 3 m m m b The enthalpy at state 9 is determined from an energy balance on the mixing chamber 26154 kJkg 006072 kgs25753 kJkg 011 kgs26376 kJkg 0 1707 kgs 9 9 3 3 2 7 9 h h m h m h mh Then 0 9393 kJkg 54 kJkg 261 450 kPa 9 9 9 s h P 28841 kJkg 1600 kPa 4 9 4 4 h s s s P 29278 kJkg 26154 26154 28841 0 86 4 4 9 4 9 4 h h h h h h s C η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1153 The rate of heat removal from the refrigerated space is 1933 kW 6881kJkg 0 11 kgs24451 8 1 7 h h m QL c The power input and the COP are 7 45 kW 26154kJkg 0 1707 kgs29278 24451kJkg 0 11 kgs26376 9 4 1 2 7 in h m h h h m W 259 7 45 1933 COP in L W Q d If this refrigerator operated on a singlestage cycle between the same pressure limits we would have 9377 kJkgK 0 51 kJkg 244 10 C 1 10 C 1 g g s s h h T QH QL 1 2s 3 4 16 MPa 2 Win 10C 28785 kJkg 1600 kPa 2 1 2 2 h s s s P 29490 kJkg 24451 24451 28785 0 86 2 2 1 2 1 2 h h h h h h s C η s 93 kJkg 135 93 kJkg 135 3 4 1600 kPa 3 h h h h f 1854 kW 0 1707 kgs24451 13593kJkg 4 1 h m h QL 8 60 kW 24451kJkg 0 1707 kgs29490 1 2 in h m h W 216 8 60 1854 COP in L W Q Discussion The cooling load decreases by 41 while the COP decreases by 166 when the cycle operates on the single stage vaporcompression cycle PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1154 Gas Refrigeration Cycles 1167C The ideal gas refrigeration cycle is identical to the Brayton cycle except it operates in the reversed direction 1168C In the ideal gas refrigeration cycle the heat absorption and the heat rejection processes occur at constant pressure instead of at constant temperature 1169C The reversed Stirling cycle is identical to the Stirling cycle except it operates in the reversed direction Remembering that the Stirling cycle is a totally reversible cycle the reversed Stirling cycle is also totally reversible and thus its COP is COPRStirling 1 1 T T H L 1170C In aircraft cooling the atmospheric air is compressed by a compressor cooled by the surrounding air and expanded in a turbine The cool air leaving the turbine is then directly routed to the cabin 1171C No because h hT for ideal gases and the temperature of air will not drop during a throttling h1 h2 process 1172C By regeneration PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1155 1173 An idealgas refrigeration cycle with air as the working fluid is considered The rate of refrigeration the net power input and the COP are to be determined Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with variable specific heats 3 Kinetic and potential energy changes are negligible Analysis a We assume both the turbine and the compressor to be isentropic the turbine inlet temperature to be the temperature of the surroundings and the compressor inlet temperature to be the temperature of the refrigerated space From the air table Table A17 1 5546 31024 kJkg K 310 1 0889 28013 kJkg K 280 3 1 3 3 1 1 r r P h T P h T T 1 2 QH 3 4 QL 37C 7C Thus 20057 kJkg 200 6 K 0 3401 1 5546 160 35 43296 kJkg 431 5 K 4 978 1 0889 35 160 4 4 3 4 2 2 1 2 3 4 1 2 h T P P P P h T P P P P r r r r s Then the rate of refrigeration is 159 kW 20057 kJkg 02 kgs 28013 4 1 h m h m q Q L L b The net power input is determined from W W W net in comp in turb out where 2193 kW 20057 kJkg kgs 31024 02 3057 kW 28013 kJkg kgs 43296 02 4 3 out turb 1 2 compin h m h W h m h W Thus 864 kW 2193 3057 Wnetin c The COP of this ideal gas refrigeration cycle is determined from 184 864 kW 159 kW COP netin R W QL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1156 1174 An idealgas refrigeration cycle with air as the working fluid is considered The rate of refrigeration the net power input and the COP are to be determined Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with variable specific heats 3 Kinetic and potential energy changes are negligible Analysis a We assume the turbine inlet temperature to be the temperature of the surroundings and the compressor inlet temperature to be the temperature of the refrigerated space From the air table Table A17 1 5546 31024 kJkg K 310 1 0889 28013 kJkg K 280 3 1 3 3 1 1 r r P h T P h T 1 2 QH 3 4s QL 2 4 1 T 37C Thus 7C 20057 kJkg 200 6 K 0 3401 1 5546 160 35 43296 kJkg 431 5 K 4 978 1 0889 35 160 4 4 3 4 2 2 1 2 3 4 1 2 s s r r s s r r h T P P P P h T P P P P s Also 21702 kJkg 20057 0 85 31024 31024 4 3 3 4 4 3 4 3 s T s T h h h h h h h h η η Then the rate of refrigeration is 126 kW 21702 kJkg 02 kgs 28013 4 1 h m h m q Q L L b The net power input is determined from W W W netin compin turbout where 1864 kW 21702 kJkg kgs 31024 02 3821 kW 28013 kJkg 080 43296 02 kgs 4 3 out turb 1 2 1 2 in comp h m h W h m h h m h W C s η Thus 196 kW 1864 3821 Wnetin c The COP of this ideal gas refrigeration cycle is determined from 0643 196 kW 126 kW COP netin R W QL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1157 1175 Problem 1174 is reconsidered The effects of compressor and turbine isentropic efficiencies on the rate of refrigeration the net power input and the COP are to be investigated Analysis The problem is solved using EES and the solution is given below Input data T1 7 C P1 35 kPa T3 37 C P3160 kPa mdot02 kgs Etacomp 100 Etaturb 10 Compressor anaysis s1ENTROPYAirTT1PP1 s2ss1 For the ideal case the entropies are constant across the compressor P2 P3 s2sENTROPYAirTTs2PP2Ts2 is the isentropic value of T2 at compressor exit Etacomp WdotcompisenWdotcomp compressor adiabatic efficiency Wdotcomp Wdotcompisen mdoth1 Wdotcompisen mdoths2SSSF First Law for the isentropic compressor assuming adiabatic kepe0 mdot is the mass flow rate in kgs h1ENTHALPYAirTT1 hs2ENTHALPYAirTTs2 mdoth1 Wdotcomp mdoth2SSSF First Law for the actual compressor assuming adiabatic kepe0 h2ENTHALPYAirTT2 s2ENTROPYAirhh2PP2 Heat Rejection Process 23 assumed SSSF constant pressure process mdoth2 Qdotout mdoth3SSSF First Law for the heat exchanger assuming W0 kepe0 h3ENTHALPYAirTT3 Turbine analysis s3ENTROPYAirTT3PP3 s4ss3 For the ideal case the entropies are constant across the turbine P4 P1 s4sENTROPYAirTTs4PP4Ts4 is the isentropic value of T4 at turbine exit Etaturb Wdotturb Wdotturbisen turbine adiabatic efficiency Wdotturbisen Wdotturb mdoth3 Wdotturbisen mdoths4SSSF First Law for the isentropic turbine assuming adiabatic kepe0 hs4ENTHALPYAirTTs4 mdoth3 Wdotturb mdoth4SSSF First Law for the actual compressor assuming adiabatic kepe0 h4ENTHALPYAirTT4 s4ENTROPYAirhh4PP4 Refrigeration effect mdoth4 QdotRefrig mdoth1 Cycle analysis WdotinnetWdotcompWdotturbExternal work supplied to compressor COP QdotRefrigWdotinnet The following is for plotting data only Ts1Ts2 ss1s2s Ts2Ts4 ss2s4s PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1158 ηcomp COP QRefrig kW Winnet kW 07 075 08 085 09 095 1 03291 03668 04077 04521 05006 05538 06123 9334 9334 9334 9334 9334 9334 9334 2836 2545 229 2065 1865 1686 1524 50 55 60 65 70 100 50 0 50 100 150 200 250 s kJkgK T C 160 kPa 35 kPa Air 1 2 3 4 Wcomp Wturb Qrefrig QH 07 075 08 085 09 095 1 0 02 04 06 08 1 12 14 16 18 2 ηcomp COP ηturb10 ηturb085 ηturb070 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1159 07 075 08 085 09 095 1 9 12 15 18 21 24 27 30 ηcomp Winnet kW ηturb10 ηturb085 ηturb070 07 075 08 085 09 095 1 8 12 16 20 ηcomp QRefrig kW ηturb085 ηturb10 ηturb070 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1160 1176 A gas refrigeration cycle with helium as the working fluid is considered The minimum temperature in the cycle the COP and the mass flow rate of the helium are to be determined Assumptions 1 Steady operating conditions exist 2 Helium is an ideal gas with constant specific heats 3 Kinetic and potential energy changes are negligible Properties The properties of helium are cp 51926 kJkgK and k 1667 Table A2 1 2 QH 3 4s QRefrig 2 4 1 T Analysis a From the isentropic relations 208 1 K 3 1 K 323 408 2 K K 3 263 0 667 1 667 1 k k 3 4 3 4 0 667 1 667 1 k k 1 2 1 2 P P T T P P T T s s 50C 10C s and 444 5 K 0 80 263 408 2 263 208 1 0 80 323 323 1 2 1 2 1 2 1 2 1 2 1 2 min 4 3 3 4 4 3 4 3 4 3 4 3 C s s s C s T s s T T T T T T T T T h h h h T T T T T T T T T h h h h η η η η K 2311 b The COP of this gas refrigeration cycle is determined from 0356 231 1 323 263 444 5 231 1 263 COP 4 3 1 2 4 1 4 3 1 2 4 1 turbout compin netin R T T T T T T h h h h h h w w q w q L L c The mass flow rate of helium is determined from 0109 kgs 2311 K 51926 kJkg K 263 kJs 18 4 1 refrig 4 1 refrig refrig T T c Q h h Q q Q m p L PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1161 1177E An ideal gas refrigeration cycle with air as the working fluid has a compression ratio of 4 The COP of the cycle is to be determined Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant specific heats 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 0240 BtulbmR and k 14 Table A2Ea Analysis From the isentropic relations 376 8 R 4 560 R 1 668 7 R 450 R4 41 40 1 3 4 3 4 41 40 1 1 2 1 2 k k k k P P T T P P T T 1 2 QH 3 4 QRefrig s T 100F 10F The COP of this ideal gas refrigeration cycle is determined from 206 376 8 560 450 668 7 376 8 450 COP 4 3 1 2 4 1 4 3 1 2 4 1 turbout compin netin R T T T T T T h h h h h h w w q w q L L PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1162 1178E An gas refrigeration cycle with air as the working fluid has a compression ratio of 4 The COP of the cycle is to be determined Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant specific heats 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 0240 BtulbmR and k 14 Table A2Ea T 1 2s QH 3 4s QRefrig 2 4 1 100F s Analysis From the isentropic relations 402 9 R 19 psia 6 psia 560 R 668 7 R 450 R4 41 40 1 3 4 3 4 41 40 1 1 2 1 2 k k s k k s P P T T P P T T 10F and 7014 R 450 0 87 668 7 450 412 3 R 402 9 0 94560 560 1 2 1 2 1 2 1 2 1 2 1 2 4 3 3 4 4 3 4 3 4 3 4 3 C s s s C s T s s T T T T T T T T T h h h h T T T T T T T T h h h h η η η η The COP of this gas refrigeration cycle is determined from 0364 412 3 560 450 701 4 412 3 450 COP 4 3 1 2 4 1 4 3 1 2 4 1 turbout compin netin R T T T T T T h h h h h h w w q w q L L PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1163 1179 An ideal gas refrigeration cycle with air as the working fluid is considered The minimum pressure ratio for this system to operate properly is to be determined Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant specific heats 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a 1 2 QH 3 4 QRefrig s T Analysis An energy balance on process 41 gives 224 2 K 1005 kJkg K 36 kJkg K 260 Refrig 1 4 4 1 Refrig p p c q T T T T c q 25C 13C The minimum temperature at the turbine inlet would be the same as that to which the heat is rejected That is T3 298 K Then the minimum pressure ratio is determined from the isentropic relation to be 271 40 41 1 4 3 4 3 2242 K 298 K k k T T P P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1164 1180 A regenerative gas refrigeration cycle using air as the working fluid is considered The effectiveness of the regenerator the rate of heat removal from the refrigerated space the COP of the cycle and the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2 Analysis a From the isentropic relations 3 4 5 6 QH Compressor 1 2 QL Turbine Heat Exch Heat Exch Regenerator 432 4 K 2 K 5 273 41 40 1 k k 1 2 1 2 P P T T s 472 5 K 273 2 273 2 432 4 0 80 2 2 1 2 1 2 1 2 1 2 T T T T T T h h h h s s C η The temperature at state 4 can be determined by solving the following two equations simultaneously 41 40 4 1 k k 4 5 4 5 5 1 T P P T T s s s T T T T h h h h 5 4 4 5 4 5 4 193 2 0 85 η Using EES we obtain T4 2813 K s 1 3 5 6 4 2s QH QRefrig Qrege 5 2 T An energy balance on the regenerator may be written as or 246 3 K 281 3 308 2 273 2 4 3 1 6 6 1 4 3 6 1 4 3 T T T T T T T T T T mc T T mc p p 35C 0C The effectiveness of the regenerator is 0434 246 3 308 2 281 3 2 308 6 3 4 3 6 3 4 3 regen T T T T h h h h ε 80C b The refrigeration load is 2136 kW kgs100 5 kJkgK2463 193 2 K 40 5 6 T T mc Q p L c The turbine and compressor powers and the COP of the cycle are 8013 kW 273 2 K kgs100 5 kJkgK4725 40 1 2 Cin T T mc W p 3543 kW kgs100 5 kJkgK2813 193 2 kJkg 40 5 4 Tout T T mc W p 0478 3543 8013 2136 COP Tout Cin netin W W Q W Q L L PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1165 d The simple gas refrigeration cycle analysis is as follows 194 6 K 5 1 308 2 K 1 41 40 1 k k 3 4 r T T s 1 2 QH 3 4s QRefrig 2 4 1 T 211 6 K 194 6 308 2 308 2 0 85 4 4 4 3 4 3 T T T T T T s ηT 35C 0C 2474 kW 211 6 kJkg kgs100 5 kJkgK2732 40 4 1 T T mc Q p L s 32 kW 41 211 6 kJkg 3082 273 2 kgs100 5 kJkgK 4725 40 4 3 1 2 in net T T mc T T mc W p p 0599 4132 2474 COP netin W QL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1166 1181 An ideal gas refrigeration cycle with with two stages of compression with intercooling using air as the working fluid is considered The COP of this system and the mass flow rate of air are to be determined Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant specific heats 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a 3 4 5 6 Q 1 2 Turbine Q Heat exch xch Heat e Heat exch Q 10C 4 3 6 5 2 1 s T 18C Analysis From the isentropic relations 128 2 K 16 283 K 1 420 5 K 283 K4 378 9 K 255 K4 41 40 1 5 6 5 6 41 40 1 3 4 3 4 41 40 1 1 2 1 2 k k k k k k P P T T P P T T P P T T The COP of this ideal gas refrigeration cycle is determined from 119 283 128 2 283 420 5 255 378 9 128 2 255 COP 6 5 3 4 1 2 6 1 6 5 3 4 1 2 6 1 turbout compin netin R T T T T T T T T h h h h h h h h w w q w q L L The mass flow rate of the air is determined from 0163 kgs K 1005 kJkg K255 128 2 75000 3600 kJs 6 1 Refrig 6 1 Refrig T T c Q m T T mc Q p p PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1167 1182 A gas refrigeration cycle with with two stages of compression with intercooling using air as the working fluid is considered The COP of this system and the mass flow rate of air are to be determined Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant specific heats 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis From the isentropic relations 3 4 5 6 Q 1 2 Turbine Q Heat exch xch Heat e Heat exch Q 10C 6s 2 T 1 3 4 6 5 4s 2s 18C s 128 2 K 16 283 K 1 420 5 K 283 K4 378 9 K 255 K4 41 40 1 5 6 5 6 41 40 1 3 4 3 4 41 40 1 1 2 1 2 k k s k k s k k s P P T T P P T T P P T T and 135 9 K 0 95283 128 2 283 444 8 K 283 0 85 420 5 283 400 8 K 255 0 85 378 9 255 6 5 5 6 6 5 6 5 6 5 6 5 3 4 3 4 3 4 3 4 3 4 3 4 1 2 1 2 1 2 1 2 1 2 1 2 s T s s T C s s s C C s s s C T T T T T T T T h h h h T T T T T T T T h h h h T T T T T T T T h h h h η η η η η η The COP of this ideal gas refrigeration cycle is determined from 0742 283 135 9 283 444 8 255 400 8 135 9 255 COP 6 5 3 4 1 2 6 1 6 5 3 4 1 2 6 1 turbout compin netin R T T T T T T T T h h h h h h h h w w q w q L L The mass flow rate of the air is determined from 0174 kgs K 1005 kJkg K255 135 9 75000 3600 kJs 6 1 Refrig 6 1 Refrig T T c Q m T T mc Q p p preparation If you are a student using this Manual you are using it without permission 1168 1183 A regenerative gas refrigeration cycle with argon as the working fluid is considered Te refrigeration load the COP the minimum power input the secondlaw efficiency and the total exergy destruction in the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Properties The properties of argon are cp 05203 kJkgK and k 1667 Analysis a From the isentropic relations PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 123 4 K 5 1 K 235 462 7 K K 5 243 0 667 1 667 1 k k 4 5 4 5 0 667 1 667 1 k k 1 2 1 2 P P T T P P T T s s 510 9 K 243 0 82 462 7 243 143 5 K 123 4 0 82 235 235 1 2 1 2 1 2 1 2 1 2 1 2 5 4 4 5 5 4 5 4 5 4 5 4 C s s s C s T s s T T T T T T T T T h h h h T T T T T T T T h h h h η η η η s T 1 3 30C 38C 5s 6 4 2s QH QRefrig Qrege 5 2 15C From an energy balance on the regenerator or 190 K 235 288 243 4 3 1 6 6 1 4 3 6 1 4 3 T T T T T T T T T T mc T T mc p p 1935 kW 143 5 K 0 08 kgs 0 5203 kJkg K190 5 6 T T mc Q p L 7 343 kW 143 5 K 235 243 0 08 kgs 0 5203 kJkg K510 9 5 4 1 2 T T T T mc W p net 02636 7 343 1 935 COP net W QL b The exergy of the heat transferred from the lowtemperature medium is 0 382 kW 228 273 1 935 kW 1 1 0 L L Q T T Q Ex L This is the minimum power input 0382 kW ExQL W min The secondlaw efficiency of the cycle is 52 005202 7 343 382 0 net II W Ex QL η The total exergy destruction in the cycle can be determined from 6961kW 0 382 7 343 net desttotal ExQL W Ex preparation If you are a student using this Manual you are using it without permission 1169 Absorption Refrigeration Systems 1184C In absorption refrigeration water can be used as the refrigerant in air conditioning applications since the temperature of water never needs to fall below the freezing point 1185C Absorption refrigeration is the kind of refrigeration that involves the absorption of the refrigerant during part of the cycle In absorption refrigeration cycles the refrigerant is compressed in the liquid phase instead of in the vapor form 1186C The main advantage of absorption refrigeration is its being economical in the presence of an inexpensive heat source Its disadvantages include being expensive complex and requiring an external heat source 1187C The fluid in the absorber is cooled to maximize the refrigerant content of the liquid the fluid in the generator is heated to maximize the refrigerant content of the vapor 1188C The coefficient of performance of absorption refrigeration systems is defined as gen pumpin gen R requiredinput desiredoutput COP Q Q W Q Q L L 1189C The rectifier separates the water from NH3 and returns it to the generator The regenerator transfers some heat from the waterrich solution leaving the generator to the NH3rich solution leaving the pump 1190 The COP of an absorption refrigeration system that operates at specified conditions is given It is to be determined whether the given COP value is possible Analysis The maximum COP that this refrigeration system can have is 2 97 273 292 273 368 K 292 K 1 1 COP 0 0 Rmax L L s T T T T T which is smaller than 31 Thus the claim is not possible PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1170 1191 The conditions at which an absorption refrigeration system operates are specified The maximum COP this absorption refrigeration system can have is to be determined Analysis The maximum COP that this refrigeration system can have is 264 273 298 273 393 K 298 K 1 1 COP 0 0 Rmax L L s T T T T T 1192 The conditions at which an absorption refrigeration system operates are specified The maximum rate at which this system can remove heat from the refrigerated space is to be determined Analysis The maximum COP that this refrigeration system can have is 1 15 243 298 243 403 K 298 K 1 1 COP 0 0 Rmax T T T T T L L s Thus 10 kJh 575 5 1 15 5 10 kJh COP 5 gen Rmax Lmax Q Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1171 1193 A reversible absorption refrigerator consists of a reversible heat engine and a reversible refrigerator The rate at which the steam condenses the power input to the reversible refrigerator and the second law efficiency of an actual chiller are to be determined Properties The enthalpy of vaporization of water at 150C is hfg 21138 kJkg Table A4 Analysis a The thermal efficiency of the reversible heat engine is 0 2954 27315 K 150 27315 K 25 1 1 0 threv Ts T η PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course The COP of the reversible refrigerator is 6 454 27315 K 15 27315 25 27315 K 15 COP 0 Rrev L L T T T Rev Ref T0 TL Rev HE Ts T0 The COP of the reversible absorption refrigerator is 1 906 0 2954 6 454 COP COP Rrev threv absrev η The heat input to the reversible heat engine is 3672 kW 1906 70 kW COP absrev in QL Q Then the rate at which the steam condenses becomes 00174 kgs 21138 kJkg 3672 kJs in fg s h Q m b The power input to the refrigerator is equal to the power output from the heat engine 109 kW 0 29543672 kW in threv outHE inR Q W W η c The secondlaw efficiency of an actual absorption chiller with a COP of 08 is 420 0 420 1906 80 COP COP rev abs actual ηII preparation If you are a student using this Manual you are using it without permission 1172 1194E An ammoniawater absorption refrigeration cycle is considered The rate of cooling the COP and the secondlaw efficiency of the system are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Properties The properties of ammonia are as given in the problem statement The specific heat of geothermal water is given to be 10 BtulbmF Analysis a The rate of cooling provided by the system is 61700 Btuh 13 Btus 17 Btulbm 190 9 0 04 lbms6192 4 1 h h m Q R L b The rate of heat input to the generator is 22 0 Btus F240 200 F Btulbm 01 0 55 lbms geoout geoin geo gen T T c m Q p Then the COP becomes 0779 22 0 Btus 1713 Btus COP gen Q QL c The reversible COP of the system is 2 38 25 70 460 25 460 220 460 70 1 1 COP 0 0 absrev L L s T T T T T The temperature of the heat source is taken as the average temperature of the geothermal water 2402002220F Then the secondlaw efficiency becomes 328 0328 238 0 779 COP COP absrev ηII PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1173 Special Topic Thermoelectric Power Generation and Refrigeration Systems 1195C The circuit that incorporates both thermal and electrical effects is called a thermoelectric circuit 1196C When two wires made from different metals joined at both ends junctions forming a closed circuit and one of the joints is heated a current flows continuously in the circuit This is called the Seebeck effect When a small current is passed through the junction of two dissimilar wires the junction is cooled This is called the Peltier effect 1197C No 1198C No 1199C Yes 11100C When a thermoelectric circuit is broken the current will cease to flow and we can measure the voltage generated in the circuit by a voltmeter The voltage generated is a function of the temperature difference and the temperature can be measured by simply measuring voltages 11101C The performance of thermoelectric refrigerators improves considerably when semiconductors are used instead of metals 11102C The efficiency of a thermoelectric generator is limited by the Carnot efficiency because a thermoelectric generator fits into the definition of a heat engine with electrons serving as the working fluid 11103E A thermoelectric generator that operates at specified conditions is considered The maximum thermal efficiency this thermoelectric generator can have is to be determined Analysis The maximum thermal efficiency of this thermoelectric generator is the Carnot efficiency 31 3 800R 550R 1 1 thCarnot thmax H L T T η η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1174 11104 A thermoelectric refrigerator that operates at specified conditions is considered The maximum COP this thermoelectric refrigerator can have and the minimum required power input are to be determined Analysis The maximum COP of this thermoelectric refrigerator is the COP of a Carnot refrigerator operating between the same temperature limits Thus W 121 1072 1072 130 W COP 1 293 K 268 K 1 1 1 COP COP max min in RCarnot max L L H Q W T T 11105 A thermoelectric cooler that operates at specified conditions with a given COP is considered The required power input to the thermoelectric cooler is to be determined Analysis The required power input is determined from the definition of COPR COP COP 180 W 015 R in in R Q W W Q L L 1200 W 11106E A thermoelectric cooler that operates at specified conditions with a given COP is considered The rate of heat removal is to be determined Analysis The required power input is determined from the definition of COPR 137 Btumin 1hp hp 4241 Btumin 81 0 18 COP COP in R in R W Q W Q L L PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1175 11107 A thermoelectric refrigerator powered by a car battery cools 9 canned drinks in 12 h The average COP of this refrigerator is to be determined Assumptions Heat transfer through the walls of the refrigerator is negligible Properties The properties of canned drinks are the same as those of water at room temperature ρ 1 kgL and cp 418 kJkgC Table A3 Analysis The cooling rate of the refrigerator is simply the rate of decrease of the energy of the canned drinks 0 00671 kW 671 W 12 3600 s kJ 290 C253 C 290 kJ 3 15 kg418 kJkg 1 kgL035 0 L 315 kg 9 cooling cooling cooling t Q Q mc T Q m V ρ The electric power consumed by the refrigerator is 12 V3 A 36 W in I W V Then the COP of the refrigerator becomes COP W 36 W cooling in Q W 6 71 0 20 0186 11108E A thermoelectric cooler is said to cool a 12oz drink or to heat a cup of coffee in about 15 min The average rate of heat removal from the drink the average rate of heat supply to the coffee and the electric power drawn from the battery of the car are to be determined Assumptions Heat transfer through the walls of the refrigerator is negligible Properties The properties of canned drinks are the same as those of water at room temperature cp 10 BtulbmF Table A3E Analysis a The average cooling rate of the refrigerator is simply the rate of decrease of the energy content of the canned drinks 362 W 1Btu 1055 J 15 60 s 84 Btu 30 F7838 F 3084 Btu 0 771 lbm10 Btulbm cooling cooling cooling t Q Q T mc Q p b The average heating rate of the refrigerator is simply the rate of increase of the energy content of the canned drinks 497 W 1Btu 1055 J 15 60 s 4 Btu 42 F13075 F 424 Btu 0 771 lbm10 Btulbm heating heating heating t Q Q T mc Q p c The electric power drawn from the car battery during cooling and heating is 4 W 41 W 181 12 49 7 W COP 21 1 20 1 COP COP 02 36 2 W COP heating heating heating in cooling heating cooling cooling cooling in Q W Q W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1176 11109 The maximum power a thermoelectric generator can produce is to be determined Analysis The maximum thermal efficiency this thermoelectric generator can have is 0 1873 363 K 295 K 1 1 thmax H L T T η Thus 364 kW 1 31 10 kJh 10 kJh 018737 6 6 in thmax outmax Q W η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1177 Review Problems 11110 A steadyflow Carnot refrigeration cycle with refrigerant134a as the working fluid is considered The COP the condenser and evaporator pressures and the net work input are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a The COP of this refrigeration cycle is determined from T qL 4 3 2 1 506 1 303 K 253 K 1 1 1 COP RC L H T T b The condenser and evaporative pressures are Table A11 30C 77064 kPa 13282 kPa sat30 C cond sat 20 C evap P P P P 20C c The net work input is determined from s 19582 kJkg 0 80 21291 49 25 5743 kJkg 0 15 21291 49 25 20 C 2 2 20 C 1 1 fg f fg f x h h h x h h h 2735 kJkg 5 06 1384 kJkg COP 138 4 kJkg 5743 82 195 R in net 1 2 L L q w h h q 11111 A room is cooled adequately by a 5000 Btuh window airconditioning unit The rate of heat gain of the room when the airconditioner is running continuously is to be determined Assumptions 1 The heat gain includes heat transfer through the walls and the roof infiltration heat gain solar heat gain internal heat gain etc 2 Steady operating conditions exist Analysis The rate of heat gain of the room in steady operation is simply equal to the cooling rate of the airconditioning system Q Q heat gain cooling 5000 Btu h 11112 A heat pump water heater has a COP of 34 and consumes 6 kW when running It is to be determined if this heat pump can be used to meet the cooling needs of a room by absorbing heat from it Assumptions The COP of the heat pump remains constant whether heat is absorbed from the outdoor air or room air Analysis The COP of the heat pump is given to be 34 Then the COP of the airconditioning system becomes 42 1 43 1 COP COP heat pump aircond Then the rate of cooling heat absorption from the air becomes 20 4 kW 51840 kJh 43 6 kW COP aircond cooling Win Q since 1 kW 3600 kJh We conclude that this heat pump can meet the cooling needs of the room since its cooling rate is greater than the rate of heat gain of the room PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1178 11113 A heat pump that operates on the ideal vaporcompression cycle with refrigerant134a as the working fluid is used to heat a house The rate of heat supply to the house the volume flow rate of the refrigerant at the compressor inlet and the COP of this heat pump are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a In an ideal vaporcompression refrigeration cycle the compression process is isentropic the refrigerant enters the compressor as a saturated vapor at the evaporator pressure and leaves the condenser as saturated liquid at the condenser pressure From the refrigerant tables Tables A12 and A13 QH QL 200 kPa 1 2 3 4 09 MPa Win House 61 kJkg throttling 101 10161 kJkg sat liquid MPa 90 27575 kJkg MPa 90 099867 m kg 0 0 93773 kJkg K 46 kJkg 244 sat vapor kPa 200 3 4 09 MPa 3 3 2 1 2 2 3 200 kPa 1 200 kPa 1 200 kPa 1 1 h h h h P h s s P s s h h P f g g g v v T s The rate of heat supply to the house is determined from 5573 kW 10161 kJkg 032 kgs 27575 3 2 h m h QH b The volume flow rate of the refrigerant at the compressor inlet is 00320 m s 3 032 kgs 0099867 m kg 3 1 1 v V m c The COP of t his heat pump is determined from 557 24446 27575 10161 27575 COP 1 2 3 2 in R h h h h w qL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1179 11114 A groundcoupled heat pump that operates on the vaporcompression refrigeration cycle with refrigerant134a as the working fluid is considered The hardware and the Ts diagram for this air conditioner are to be sketched The exit temperature of the water in the condenser and the COP are to be determined Analysis a In an ideal vaporcompression refrigeration cycle the compression process is isentropic the refrigerant enters the compressor as a saturated vapor at the evaporator pressure and leaves the condenser as saturated liquid at the condenser pressure 1 2 3 4 20C Condenser Evaporator Compressor Expansion valve 14 MPa in W Tw2 Water 10C 032 kgs sat vap QH QL 20C 1 3 4 14 MPa W n i 2 4s s T L Q b The properties as given in the problem statement are h4 h3 hf 1400 kPa 1272 kJkg h1 hg 20C 2616 kJkg The rate of heat transfer in the condenser is determined from 21 kW 6 1 18 kW 1 COP 1 1 COP R in R L H L L H Q Q W Q Q Q An energy balance on the condenser gives 257C 0 32 kgs418 kJkg C 21 kW C 10 1 2 1 2 3 2 pw w H w w w w pw w H m c Q T T T T m c h m h Q c The COP of the heat pump is 7 18 kW 21 kW 21 kW COPHP L H H Q Q Q It may also be determined from 7 1 6 1 COP COP R HP PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1180 11115 An ideal vaporcompression refrigeration cycle with refrigerant22 as the working fluid is considered The evaporator is located inside the air handler of building The hardware and the Ts diagram for this heat pump application are to be sketched The COP of the unit and the ratio of volume flow rate of air entering the air handler to mass flow rate of R 22 through the air handler are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a In an ideal vaporcompression refrigeration cycle the compression process is isentropic the refrigerant enters the compressor as a saturated vapor at the evaporator pressure and leaves the condenser as saturated liquid at the condenser pressure From the refrigerant22 data from the problem statement 1 2 3 4 QH 5C Win Condenser Evaporator Compressor Expansion valve 45C QL Air 27C sat vap QH QL 5C 1 2 3 4 45C Win 4s s T 7C kJkg throttling 101 101 kJkg sat liquid kPa 1728 283 7 kJkg kPa 1728 0 9344 kJkg K 248 1 kJkg sat vapor C 5 3 4 1728 kPa 3 3 2 1 2 2 5 C 1 5 C 1 1 h h h h P h s s P s s h h T f g g b The COP of the refrigerator is determined from its definition 413 248 1 283 7 101 248 1 COP 1 2 4 1 in R h h h h w qL c An energy balance on the evaporator gives T c T m c h h m Q p a a p a R L v V 4 1 Rearranging we obtain the ratio of volume flow rate of air entering the air handler to mass flow rate of R22 through the air handler 365 m airminkg R22s 3 091 m airskg R22s 6 0 8323 m kg 1 005 kJkg K20 K 1 101 kJkg 248 1 1 3 3 4 1 T c h h m p R a v V Note that the specific volume of air is obtained from ideal gas equation taking the pressure of air to be 100 kPa given and using the average temperature of air 17C 290 K to be 08323 m3kg PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1181 11116 An air conditioner operates on the vaporcompression refrigeration cycle The rate of cooling provided to the space the COP the isentropic efficiency and the exergetic efficiency of the compressor the exergy destruction in each component of the cycle the total exergy destruction the minimum power input and the secondlaw efficiency of the cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a The properties of R134a are Tables A11 through A13 0 4228 kJkg K 26 kJkg 108 kPa 180 26 kJkg 108 0 3948 kJkg K 26 kJkg 108 40 C 36 3 46 kPa 1200 C 3 46 9614 kJkg K 0 64 kJkg 289 C 60 kPa 1200 28532 kJkg kPa 1200 9483 kJkg K 0 14 kJkg 245 10 C 72 7 12 kPa 180 C 7 12 4 4 4 3 4 40 C 3 40 C 3 3 3 kPa sat1200 2 2 2 2 2 1 1 2 1 1 1 1 kPa sat180 s h P h h s s h h T P T s h T P h s s P s h T P T f f s QH QL 180 kPa 1 2s 3 4 s T 12 MPa 2 Win The cooling load and the COP are 28020 Btuh 1kW kW 3412 Btuh 8213 8213 kW 10826kJkg 0 06 kgs24514 4 1 h m h QL 1088 kW 10826kJkg 0 06 kgs28964 3 2 h m h QH 2 670 kW 24514kJkg 0 06 kgs28964 1 2 in h m h W 3076 2 670 kW 8 213 kW COP in W QL b The isentropic efficiency of the compressor is 903 0 9029 24514 28964 24514 32 285 1 2 1 2 h h h h s ηC The reversible power and the exergy efficiency for the compressor are 428 kW 2 0 9483kJkg K 310 K09614 24514kJkg 0 06 kgs 28964 1 2 0 1 2 s s T h m h Wrev 909 0 9091 2 670 kW 428 kW 2 in rev W W ex C η c The exergy destruction in each component of the cycle is determined as follows Compressor 0 0007827 kWK 0 9483 kJkg K 0 06 kgs 0 9614 1 2 gen1 2 s m s S 02426 kW 310 K 0 0007827 kWK gen12 0 dest12 T S Ex PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1182 Condenser 0 001114 kWK 310 K 1088 kW 0 9614 kJkg K 0 06 kgs 0 3948 2 3 gen2 3 H H T Q s m s S 03452 kW 310 K000111 4 kJkg K gen23 0 dest23 T S Ex Expansion valve 0 001678 kWK 003948 kJkg K 0 06 kgs 0 4228 3 4 gen3 4 s m s S 05203 kJkg 310 K000167 8 kJkg K gen34 0 dest34 T S Ex Evaporator 0 003597 kWK 294 K 8 213 kW 0 4228 kJkg K 0 06 kgs 0 9483 4 1 gen4 1 L L T Q s m s S 1115 kW 310 K000359 7 kJkg K gen4 1 0 dest4 1 T S Ex The total exergy destruction can be determined by adding exergy destructions in each component 2223 kW 1 115 0 5203 0 3452 2426 0 dest4 1 dest34 dest23 dest12 desttotal Ex Ex Ex Ex Ex d The exergy of the heat transferred from the lowtemperature medium is 0 4470 kW 294 310 8 213 kW 1 1 0 L L Q T T Q Ex L This is the minimum power input to the cycle 04470 kW ExQL W min in The secondlaw efficiency of the cycle is 167 01674 2 670 4470 0 in inmin II W W η The total exergy destruction in the cycle can also be determined from 2 223 kW 0 4470 2 670 in desttotal ExQL W Ex The result is the same as expected PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1183 11117 An ideal vaporcompression refrigeration cycle with refrigerant134a as the working fluid is considered Cooling water flows through the water jacket surrounding the condenser To produce ice potable water is supplied to the chiller section of the refrigeration cycle The hardware and the Ts diagram for this refrigerantice making system are to be sketched The mass flow rates of the refrigerant and the potable water are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a In an ideal vaporcompression refrigeration cycle the compression process is isentropic the refrigerant enters the compressor as a saturated vapor at the evaporator pressure and leaves the condenser as saturated liquid at the condenser pressure From the refrigerant134a tables 1 2 3 4 140 kPa Win Condenser Evaporator Compressor Expansion valve Water 200 kgs 12 MPa Potable water sat vap QH QL 140 kPa 1 2 3 4 12 MPa Win 4s s T 77 kJkg throttling 117 11777 kJkg sat liquid kPa 1200 28407 kJkg kPa 1200 0 94456 kJkg K 23916 kJkg sat vapor kPa 140 3 4 1200 kPa 3 3 2 1 2 2 140 kPa 1 140 kPa 1 1 h h h h P h s s P s s h h T f g g b An energy balance on the condenser gives T m c h h m Q p w R H 3 2 Solving for the mass flow rate of the refrigerant 503 kgs 28407 11777kJkg 200 kgs 4 18 kJkg K10 K 3 2 h h T m c m p w R c An energy balance on the evaporator gives if w R L m h h h m Q 4 1 Solving for the mass flow rate of the potable water 183 kgs 333 kJkg 50 3 kgs23916 11777kJkg 4 1 if R w h h h m m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1184 11118 A refrigerator operating on a vaporcompression refrigeration cycle with refrigerant134a as the working fluid is considered The process with the greatest exergy loss is to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis In this cycle the refrigerant leaves the condenser as saturated liquid at the condenser pressure The compression process is not isentropic From the refrigerant tables Tables A11 A12 and A13 0 4988 kJkg K 0 5089 11777 kJkg C 37 77 kJkg throttling 117 0 42441 kJkg K 11777 kJkg sat liquid MPa 21 29811 kJkg MPa 21 0 9867 kJkg K 23309 kJkg 30 C 7 37 kPa 60 4 4 4 4 3 4 12 MPa 3 12 MPa 3 3 2 1 2 2 1 1 1 sat 37 C 1 s x h T h h s s h h P h s s P s h T P P f f s QH QL 37C 1 2s 3 4 12 MPa 2 Win 4s s T The actual enthalpy at the compressor exit is determined by using the compressor efficiency 30533 kJkg 0 90 23309 29811 23309 C 1 2 1 2 1 2 1 2 C η η h h h h h h h h s s and 1 0075 kJkg K 30533 Btulbm MPa 21 2 2 2 s h P The heat added in the evaporator and that rejected in the condenser are 18756 kJkg 30533 11777 kJkg 11532 kJkg 23309 11777 kJkg 3 2 4 1 h h q h h q H L The exergy destruction during a process of a stream from an inlet state to exit state is given by sink out source in 0 gen 0 dest T q T q s s T T s x i e Application of this equation for each process of the cycle gives 1 66 kJkg 273 K 34 115 3 kJkg 0 4988 0 9867 303 K 0 42441 kJkg K 303 K 0 4988 1088 kJkg 303 K 18756 kJkg 1 0075 0 42441 303 K 6 37 kJkg 0 9867 kJkg K 303 K 1 0075 4 1 0 41 destroyed 3 4 0 34 destroyed 2 3 0 23 destroyed 1 2 0 12 destroyed L L H H T q s s T x s s T x T q s s T x s s T x 2254 kJkg The greatest exergy destruction occurs in the expansion valve Note that heat is absorbed from fruits at 34C 239 K and rejected to the ambient air at 30C 303 K which is also taken as the dead state temperature Alternatively one may use the standard 25C 298 K as the dead state temperature and perform the calculations accordingly PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1185 11119 A refrigerator operating on a vaporcompression refrigeration cycle with refrigerant134a as the working fluid is considered The process with the greatest exergy loss is to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From the refrigerant tables Tables A11 A12 and A13 0 4585 kJkg K 0 4665 10826 kJkg C 37 26 kJkg throttling 108 0 39486 kJkg K 10826 kJkg 40 C 36 3 46 36 MPa 21 29811 kJkg MPa 21 0 9867 kJkg K 23309 kJkg 30 C 7 37 kPa 60 4 4 4 4 3 4 40 C 3 40 C 3 sat 12 MPa 3 3 2 1 2 2 1 1 1 sat 37 C 1 s x h T h h s s h h T T P h s s P s h T P P f f s QH QL 37C 1 2s 3 4 12 MPa 2 Win s T The actual enthalpy at the compressor exit is determined by using the compressor efficiency 30533 kJkg 0 90 23309 29811 23309 C 1 2 1 2 1 2 1 2 C η η h h h h h h h h s s and 1 0075 kJkg K 30533 Btulbm MPa 21 2 2 2 s h P The heat added in the evaporator and that rejected in the condenser are 19707 kJkg 30533 10826 kJkg 12483 kJkg 23309 10826 kJkg 3 2 4 1 h h q h h q H L The exergy destruction during a process of a stream from an inlet state to exit state is given by sink out source in 0 gen 0 dest T q T q s s T T s x i e Application of this equation for each process of the cycle gives 1 80 kJkg 273 K 34 12483 kJkg 0 4585 0 9867 303 K 0 39486 kJkg K 303 K 0 4585 1144 kJkg 303 K 19707 kJkg 1 0075 0 39486 303 K 6 37 kJkg 0 9867 kJkg K 303 K 1 0075 4 1 0 41 destroyed 3 4 0 34 destroyed 2 3 0 23 destroyed 1 2 0 12 destroyed L L H H T q s s T x s s T x T q s s T x s s T x 1928 kJkg The greatest exergy destruction occurs in the expansion valve Note that heat is absorbed from fruits at 34C 239 K and rejected to the ambient air at 30C 303 K which is also taken as the dead state temperature Alternatively one may use the standard 25C 298 K as the dead state temperature and perform the calculations accordingly PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1187 11121E A twoevaporator compression refrigeration cycle with refrigerant134a as the working fluid is considered The cooling load of both evaporators per unit of flow through the compressor and the COP of the system are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 4 5 6 Compressor Expansion valve 1 2 Evaporator 2 Expansion valve Expansion valve Condenser Evaporator 1 7 Analysis From the refrigerant tables Tables A11E A 12E and A13E 3 9868 Btulbm sat vapor F 5 29 10740 Btulbm sat vapor F 30 519 Btulbm throttling 48 48519 Btulbm sat liquid psia 160 F 29 5 7 7 30 F 5 5 3 6 4 160 psia 3 3 g g f h h T h h T h h h h h P PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course For a unit mass flowing through the compressor the fraction of mass flowing through Evaporator II is denoted by x and that through Evaporator I is y y 1x From the cooling loads specification 2 2 6 7 4 5 e vap 2 e vap1 h y h h h x Q Q L L where y x 1 Combining these results and solving for y gives 0 3698 48519 10740 48519 29868 48519 10740 2 4 5 6 7 4 5 h h h h h h y QH QL 295F 1 2 3 6 160 psia Win 4 5 7 30F s T Then 0 6302 0 3698 1 1 y x Applying an energy balance to the point in the system where the two evaporator streams are recombined gives 10418 Btulbm 1 0 36989868 0 630210740 1 7 5 1 1 7 5 yh xh h h yh xh Then 13114 Btulbm psia 160 0 2418 Btulbm R 18 Btulbm 104 psia 10 2 1 2 2 1 1 sat 295 F 1 h s s P s h P P The cooling load of both evaporators per unit mass through the compressor is 557 Btulbm 48519 Btulbm 0 36989868 48519 Btulbm 0 630210740 6 7 4 5 h y h h x h qL The work input to the compressor is 270 Btulbm 10418 Btulbm 13114 1 2 in h h w The COP of this refrigeration system is determined from its definition 206 270 Btulbm 557 Btulbm COP in R w qL preparation If you are a student using this Manual you are using it without permission 1188 11122E A twoevaporator compression refrigeration cycle with refrigerant134a as the working fluid is considered The process with the greatest exergy destruction is to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From Prob 11121E and the refrigerant tables Tables A11E A12E and A13E 8261 Btulbm 5017 Btulbm 5888 Btulbm 0 3698 1 0 6302 0 2295 Btulbm R 0 1129 Btulbm R 0 2226 Btulbm R 0 1024 Btulbm R 0 09774 Btulbm R 2418 Btulbm R 0 6 7 67 4 5 45 7 6 5 4 3 2 1 H L L q h h q h h q x y x s s s s s s s QH QL 295F 1 2 3 6 160 psia Win 4 5 7 30F s T The exergy destruction during a process of a stream from an inlet state to exit state is given by sink out source in 0 gen 0 dest T q T q s s T T s x i e Application of this equation for each process of the cycle gives the exergy destructions per unit mass flowing through the compressor 900 Btulbm 0 3698 0 2295 0 6302 0 2226 540 R 0 2418 0 77 Btulbm 445 R 5017 Btulbm 0 1129 0 2295 0 3698540 R 0 84 Btulbm 500 R 5888 Btulbm 0 1024 0 2226 0 6302540 R 4 60 Btulbm 0 09774 Btulbm R 0 1129 0 3698 0 1024 540 R 0 6302 482 Btulbm 540 R 8261 Btulbm 0 2418 0 09774 540 R 7 5 1 0 mixing destroyed 67 6 7 0 67 destroyed 45 4 5 0 45 destroyed 3 6 4 0 346 destroyed 2 3 0 23 destroyed ys xs s T X T q s s yT x T q s s xT x s ys xs T x T q s s T x L L L L H H For isentropic processes the exergy destruction is zero 0 destroyed 12 X The greatest exergy destruction occurs during the mixing process Note that heat is absorbed in evaporator 2 from a reservoir at 15F 445 R in evaporator 1 from a reservoir at 40F 500 R and rejected to a reservoir at 80F 540 R which is also taken as the dead state temperature PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1189 11123 A twostage compression refrigeration system with a separation unit is considered The rate of cooling and the power requirement are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 4 5 6 Compressor Expansion valve 1 2 Evaporator 7 Separator Condenser Expansion valve Compressor 8 QL 32C 7 8 3 6 4 89C 5 1 14 MPa 2 s T 3 Analysis From the refrigerant tables Tables A11 A12 and A13 26451 kJkg kPa 400 0 95813 kJkg K 23091 kJkg sat vapor C 32 94 kJkg throttling 63 6394 kJkg sat liquid C 98 22 kJkg throttling 127 12722 kJkg sat liquid kPa 1400 28149 kJkg kPa 1400 0 92691 kJkg K 25555 kJkg sat vapor C 98 8 7 8 sat 89 C 8 32 C 7 32 C 7 7 5 6 C 98 5 5 3 4 1400 kPa 3 3 2 1 2 2 C 98 1 C 98 1 1 h s s P P s s h h T h h h h T h h h h P h s s P s s h h T g g f f g g An energy balance on the separator gives 1 280 kgs 26451 6394 2 kgs 25555 12722 5 8 4 1 2 6 4 1 2 5 8 6 h h h h m m h h m h h m The rate of cooling produced by this system is then 2137 kJs 1 280 kgs23091 6394 kJkg 6 7 6 h h m QL The total power input to the compressors is 9489 kW 25555 kJkg 2 kgs28149 23091 kJkg 280 kgs26451 1 1 2 2 7 8 6 in h h m h h m W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1190 11124 A twostage vaporcompression refrigeration system with refrigerant134a as the working fluid is considered The process with the greatest exergy destruction is to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis From Prob 11109 and the refrigerant tables Tables A11 A12 and A13 298 K 273 25 255 K 273 18 15427 kJkg 16697 kJkg 1 280 kgs 2 kgs 0 95813 kJkg K 0 2658 kJkg K 0 24761 kJkg K 0 4720 kJkg K 0 45315 kJkg K 92691 kJkg K 0 0 3 2 6 7 lower upper 8 7 6 5 4 3 2 1 T T T h h q h h q m m s s s s s s s s H L H L QL 32C 7 8 3 6 4 89C 5 1 14 MPa 2 s T The exergy destruction during a process of a stream from an inlet state to exit state is given by sink out source in 0 gen 0 dest T q T q s s T T s x i e Application of this equation for each process of the cycle gives 0 11 kW 0 4720 2 kgs 0 92691 0 95813 1 280 kgs 0 24761 298 K 1432 kW 255 K 16697 kJkg 0 2658 0 95813 280 kgs298 K 1 6 94 kW 0 24761 kJkg K 1 280 kgs298 K 0 2658 1123 kW 0 45315 kJkg K 2 kgs298 K 0 4720 298 K 15427 kJkg 0 92691 0 45315 2 kgs298 K 4 1 upper 8 5 lower 0 separator destroyed 6 7 0 lower 67 destroyed 5 6 0 lower 56 destroyed 3 4 0 upper 34 destroyed 2 3 0 upper 23 destroyed s s m s s T m X T q s s T m X s s T m X s s T m X T q s s T m X L L H H kW 2618 For isentropic processes the exergy destruction is zero 0 0 78 destroyed 12 destroyed X X Note that heat is absorbed from a reservoir at 0F 460 R and rejected to the standard ambient air at 77F 537 R which is also taken as the dead state temperature The greatest exergy destruction occurs during the condensation process PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1192 11126 An absorption refrigeration system operating at specified conditions is considered The minimum rate of heat supply required is to be determined Analysis The maximum COP that this refrigeration system can have is 2 274 275 298 275 368 K 298 K 1 1 COP 0 0 Rmax L L s T T T T T Thus 3 kW 12 2274 28 kW COP Rmax genmin QL Q 11127 Problem 11126 is reconsidered The effect of the source temperature on the minimum rate of heat supply is to be investigated Analysis The problem is solved using EES and the solution is given below Input Data TL 2 C T0 25 C Ts 95 C QdotL 28 kW The maximum COP that this refrigeration system can have is COPRmax 1T0273Ts273TL273T0 TL The minimum rate of heat supply is Qdotgenmin QdotLCOPRmax 50 90 130 170 210 250 5 10 15 20 25 30 35 Ts C Qgenmin kW Ts C Qgenmin kW 50 75 100 125 150 175 200 225 250 3026 163 1165 932 7925 6994 633 5831 5443 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1193 11128 A regenerative gas refrigeration cycle using air as the working fluid is considered The effectiveness of the regenerator the rate of heat removal from the refrigerated space the COP of the cycle and the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable specific heats Analysis a For this problem we use the properties of air from EES Note that for an ideal gas enthalpy is a function of temperature only while entropy is functions of both temperature and pressure 3 4 5 6 QH Compressor 1 2 QL Turbine Heat Exch Heat Exch Regenerator 43350 kJkg kPa 500 5 6110 kJkgK C 0 kPa 100 27340 kJkg C 0 2 1 2 2 1 1 1 1 1 h s s s P s T P h T 30863 kJkg C 35 52 kJkg 473 27340 27340 43350 80 0 3 3 2 2 1 2 1 2 h T h h h h h h s C η For the turbine inlet and exit we have 19345 kJkg 80 C 5 5 h T s 1 3 5s 6 4 2s QH QRefrig Qregen 5 2 T s T h h h h h T 5 4 5 4 4 4 η 35C 0C h s s s P s T P s T P 5 4 5 5 4 4 4 1 1 1 kPa 500 kPa 500 5 6110 kJkgK C 0 kPa 100 80C We can determine the temperature at the turbine inlet from EES using the above relations A hand solution would require a trialerror approach T4 2818 K h4 28208 kJkg An energy balance on the regenerator gives 24685 kJkg 28208 30863 27340 4 3 1 6 h h h h The effectiveness of the regenerator is determined from 0430 24685 30863 28208 63 308 6 3 4 3 regen h h h h ε b The refrigeration load is 2136 kW kgs24685 19345kJkg 40 5 6 h m h QL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1194 c The turbine and compressor powers and the COP of the cycle are 8005 kW 27340kJkg kgs47352 40 1 2 Cin h m h W 3545 kW kgs28208 19345kJkg 40 5 4 Tout h m h W 0479 3545 8005 2136 COP Tout Cin netin W W Q W Q L L d The simple gas refrigeration cycle analysis is as follows T 1 2 QH 3 4s QRefrig 2 4 1 0C 35C 63 kJkg 308 52 kJkg 473 40 kJkg 273 3 2 1 h h h 5 2704 kJkg C 35 500 kPa 3 3 3 s T P 19452 kJkgK 100 kPa 4 3 4 1 h s s s P s 21164 kJkg 30863 19452 30863 0 85 4 4 4 3 4 3 h h h h h h s ηT 2470 kW 21164kJkg kgs27340 40 4 1 h m h QL 4125 kW 21164kJkg 30863 27340 kgs 47352 40 4 3 1 2 netin h m h h m h W 0599 4125 2470 COP netin W QL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1195 11129 An innovative vaporcompression refrigeration system with a heat exchanger is considered The systems COP is to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 4 5 6 Compressor 1 2 Heat exchanger Condenser Evaporator QH QL 101C 1 2 3 5 800 kPa 4 Win 6 s T Throttle valve Analysis From the refrigerant tables Tables A11 A12 and A13 kPa 200 24446 kJkg sat vapor C 1 10 32 kJkg throttling 79 7932 kJkg 800 kPa 20 C 11 3 3 31 3 11 9547 kJkg sat liquid kPa 800 sat 101 C 6 C 10 1 6 6 4 5 20 C 4 4 sat 800 kPa 4 800 kPa 3 3 P P h h T h h h h P T T h h P g f f An energy balance on the heat exchanger gives 26061 kJkg 24446 7932 9547 6 4 3 1 4 3 6 1 h h h h h m h h m h Then 29217 kJkg kPa 800 0 9970 kJkg K 26061 kJkg kPa 200 2 1 2 2 1 1 1 h s s P s h P The COP of this refrigeration system is determined from its definition 523 26061 29217 7932 24446 COP 1 2 5 6 in R h h h h w qL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1196 11130 An innovative vaporcompression refrigeration system with a heat exchanger is considered The systems COP is to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 4 5 6 Compressor 1 2 Heat exchanger Condenser Evaporator QH QL 200 kPa 1 2 3 5 800 kPa 4 Win 6 s T Throttle valve Analysis From the refrigerant tables Tables A11 A12 and A13 kPa 200 24446 kJkg sat vapor C 1 10 43 kJkg throttling 65 6543 kJkg 800 kPa 10 C 21 3 3 31 3 21 9547 kJkg sat liquid kPa 800 sat 101 C 6 C 10 1 6 6 4 5 10 C 4 4 sat 800 kPa 4 800 kPa 3 3 P P h h T h h h h P T T h h P g f f An energy balance on the heat exchanger gives 27450 kJkg 24446 6543 9547 6 4 3 1 4 3 6 1 h h h h h m h h m h Then 30828 kJkg kPa 800 1 0449 kJkg K 27450 kJkg kPa 200 2 1 2 2 1 1 1 h s s P s h P The COP of this refrigeration system is determined from its definition 530 27450 30828 6543 24446 COP 1 2 5 6 in R h h h h w qL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1197 11131 An ideal gas refrigeration cycle with with three stages of compression with intercooling using air as the working fluid is considered The COP of this system is to be determined Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with constant specific heats 3 Kinetic and potential energy changes are negligible Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Table A2a Analysis From the isentropic relations 3 4 8 7 5 6 2 1 s 54 3 K 7 7 7 1 288 K 502 2 K 288 K7 423 7 K 243 K7 41 40 1 7 8 7 8 41 40 1 3 4 3 6 4 41 40 1 1 2 1 2 k k k k k k P P T T P P T T T P P T T T 15C 30C The COP of this ideal gas refrigeration cycle is determined from 0503 54 3 288 288 2502 2 243 423 7 54 3 243 2 COP 8 7 3 4 1 2 8 1 8 7 5 6 3 4 1 2 8 1 turbout compin netin R T T T T T T T T h h h h h h h h h h w w q w q L L PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1198 11132 A vortex tube receives compressed air at 500 kPa and 300 K and supplies 25 percent of it as cold air and the rest as hot air The COP of the vortex tube is to be compared to that of a reversed Brayton cycle for the same pressure ratio the exit temperature of the hot fluid stream and the COP are to be determined and it is to be shown if this process violates the second law Assumptions 1 The vortex tube is adiabatic 2 Air is an ideal gas with constant specific heats at room temperature 3 Steady operating conditions exist Properties The gas constant of air is 0287 kJkgK Table A1 The specific heat of air at room temperature is cp 1005 kJkgK Table A2 The enthalpy of air at absolute temperature T can be expressed in terms of specific heats as h cpT Analysis a The COP of the vortex tube is much lower than the COP of a reversed Brayton cycle of the same pressure ratio since the vortex tube involves vortices which are highly irreversible Owing to this irreversibility the minimum temperature that can be obtained by the vortex tube is not as low as the one that can be obtained by the revered Brayton cycle b We take the vortex tube as the system This is a steady flow system with one inlet and two exits and it involves no heat or work interactions Then the steadyflow energy balance equation for this system for a unit mass flow rate at the inlet can be expressed as E E in out m1 kg s 1 Warm air Cold air Compressed air 2 3 1 3 2 1 3 3 2 2 1 1 3 3 2 2 1 1 0 75 0 25 1 c T c T T c m c T m c T c T m m h m h h m p p p p p p Canceling cp and solving for T3 gives 3073 K 0 75 0 25 278 300 0 75 0 25 2 1 3 T T T Therefore the hot air stream will leave the vortex tube at an average temperature of 3073 K c The entropy balance for this steady flow system can be expressed as with one inlet and two exits and it involves no heat or work interactions Then the steadyflow entropy balance equation for this system for a unit mass flow rate at the inlet can be expressed S S S in out gen 0 m1 kg s 1 1 3 1 3 1 2 1 2 1 3 1 2 1 3 3 1 2 2 1 3 2 3 3 2 2 1 1 3 3 2 2 in out gen ln ln 0 75 ln ln 25 0 0 75 25 0 P P R T T c P P R T T c s s s s s s m s s m s m m m s m s m s m s s m S S S p p Substituting the known quantities the rate of entropy generation is determined to be 0 461 kWK 0 500 kPa 0 287 kJkgK ln 100 kPa 300 K 1 005 kJkgK ln 307 3 K 75 0 500 kPa 0 287 kJkgK ln 100 kPa 300 K 1 005 kJkgK ln 278 K 0 25 gen S which is a positive quantity Therefore this process satisfies the 2nd law of thermodynamics d For a unit mass flow rate at the inlet m1 1 kg s the cooling rate and the power input to the compressor are determined to 0 25 kgs100 5 kJkgK300 278K 553 kW c 1 c 1 cooling T T m c h h m Q p c c PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1199 157 1 kW 1 100 kPa 500 kPa 1 0 80 41 kgs028 7 kJkgK300 K 1 1 1 41 1 41 1 0 1 comp 0 0 in comp k k P P k m RT W η Then the COP of the vortex refrigerator becomes 0035 157 1 kW 553 kW COP in comp cooling W Q The COP of a Carnot refrigerator operating between the same temperature limits of 300 K and 278 K is COP 278 K K Carnot T T T L H L 300 278 126 Discussion Note that the COP of the vortex refrigerator is a small fraction of the COP of a Carnot refrigerator operating between the same temperature limits PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 11100 11133 A vortex tube receives compressed air at 600 kPa and 300 K and supplies 25 percent of it as cold air and the rest as hot air The COP of the vortex tube is to be compared to that of a reversed Brayton cycle for the same pressure ratio the exit temperature of the hot fluid stream and the COP are to be determined and it is to be shown if this process violates the second law Assumptions 1 The vortex tube is adiabatic 2 Air is an ideal gas with constant specific heats at room temperature 3 Steady operating conditions exist Properties The gas constant of air is 0287 kJkgK Table A1 The specific heat of air at room temperature is cp 1005 kJkgK Table A2 The enthalpy of air at absolute temperature T can be expressed in terms of specific heats as h cp T Analysis a The COP of the vortex tube is much lower than the COP of a reversed Brayton cycle of the same pressure ratio since the vortex tube involves vortices which are highly irreversible Owing to this irreversibility the minimum temperature that can be obtained by the vortex tube is not as low as the one that can be obtained by the revered Brayton cycle b We take the vortex tube as the system This is a steady flow system with one inlet and two exits and it involves no heat or work interactions Then the steadyflow entropy balance equation for this system for a unit mass flow rate at the inlet can be expressed as E E in out m1 kg s 1 Warm air Cold air Compressed air 2 3 1 3 2 1 3 3 2 2 1 1 3 3 2 2 1 1 0 75 0 25 1 c T c T T c m c T m c T c T m m h m h h m p p p p p p Canceling cp and solving for T3 gives 3073 K 0 75 0 25 278 300 0 75 0 25 2 1 3 T T T Therefore the hot air stream will leave the vortex tube at an average temperature of 3073 K c The entropy balance for this steady flow system can be expressed as with one inlet and two exits and it involves no heat or work interactions Then the steadyflow energy balance equation for this system for a unit mass flow rate at the inlet can be expressed S S S in out gen 0 m1 kg s 1 1 3 1 3 1 2 1 2 1 3 1 2 1 3 3 1 2 2 1 3 2 3 3 2 2 1 1 3 3 2 2 in out gen ln ln 0 75 ln ln 25 0 0 75 25 0 P P R T T c P P R T T c s s s s s s m s s m s m m m s m s m s m s s m S S S p p Substituting the known quantities the rate of entropy generation is determined to be 0 513 kWK 0 600 kPa 0 287 kJkgKln 100 kPa 300 K 75 1 005 kJkgKln 307 3 K 0 600 kPa 0 287 kJkgKln 100 kPa 300 K 0 25 1 005 kJkgKln 278 K gen S which is a positive quantity Therefore this process satisfies the 2nd law of thermodynamics d For a unit mass flow rate at the inlet m1 1 kg s the cooling rate and the power input to the compressor are determined to 0 25 kgs100 5 kJkgK300 278K 553 kW c 1 c 1 cooling T T m c h h m Q p c c PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 11101 179 9 kW 1 100 kPa 600 kPa 1 0 80 41 kgs028 7 kJkgK300 K 1 1 1 41 1 41 1 0 1 comp 0 0 in comp k k P P k m RT W η Then the COP of the vortex refrigerator becomes 0031 179 9 kW 553 kW COP in comp cooling W Q The COP of a Carnot refrigerator operating between the same temperature limits of 300 K and 278 K is 126 278 K 300 278 K COPCarnot L H L T T T Discussion Note that the COP of the vortex refrigerator is a small fraction of the COP of a Carnot refrigerator operating between the same temperature limits PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 11102 11134 The effect of the evaporator pressure on the COP of an ideal vaporcompression refrigeration cycle with R 134a as the working fluid is to be investigated Analysis The problem is solved using EES and the solution is given below Input Data P1100 kPa P2 1400 kPa FluidR134a Etac10 Compressor isentropic efficiency Compressor h1enthalpyFluidPP1x1 properties for state 1 s1entropyFluidPP1x1 T1temperatureFluidhh1PP1 h2senthalpyFluidPP2ss1 Identifies state 2s as isentropic h1Wcsh2s energy balance on isentropic compressor WcWcsEtacdefinition of compressor isentropic efficiency h1Wch2 energy balance on real compressorassumed adiabatic s2entropyFluidhh2PP2 properties for state 2 T2temperatureFluidhh2PP2 Condenser P3 P2 h3enthalpyFluidPP3x0 properties for state 3 s3entropyFluidPP3x0 h2Qouth3 energy balance on condenser Throttle Valve h4h3 energy balance on throttle isenthalpic x4qualityFluidhh4PP4 properties for state 4 s4entropyFluidhh4PP4 T4temperatureFluidhh4PP4 Evaporator P4 P1 Qin h4h1 energy balance on evaporator Coefficient of Performance COPQinWc definition of COP 100 150 200 250 300 350 400 450 500 1 2 3 4 5 6 7 P1 kPa COP η comp 07 η comp 10 P1 kPa COP ηc 100 150 200 250 300 350 400 450 500 1937 2417 2886 3363 3859 4384 4946 5555 622 1 1 1 1 1 1 1 1 1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 11103 11135 The effect of the condenser pressure on the COP of an ideal vaporcompression refrigeration cycle with R 134a as the working fluid is to be investigated Analysis The problem is solved using EES and the solution is given below Input Data P1150 kPa P2 400 kPa FluidR134a Etac07 Compressor isentropic efficiency Compressor h1enthalpyFluidPP1x1 properties for state 1 s1entropyFluidPP1x1 T1temperatureFluidhh1PP1 h2senthalpyFluidPP2ss1 Identifies state 2s as isentropic h1Wcsh2s energy balance on isentropic compressor WcWcsEtacdefinition of compressor isentropic efficiency h1Wch2 energy balance on real compressorassumed adiabatic s2entropyFluidhh2PP2 properties for state 2 T2temperatureFluidhh2PP2 Condenser P3 P2 h3enthalpyFluidPP3x0 properties for state 3 s3entropyFluidPP3x0 h2Qouth3 energy balance on condenser Throttle Valve h4h3 energy balance on throttle isenthalpic x4qualityFluidhh4PP4 properties for state 4 s4entropyFluidhh4PP4 T4temperatureFluidhh4PP4 Evaporator P4 P1 Qin h4h1 energy balance on evaporator Coefficient of Performance COPQinWc definition of COP 400 600 800 1000 1200 1400 1 2 3 4 5 6 7 8 9 P2 kPa COP ηcomp07 ηcomp10 P2 kPa COP ηc 400 500 600 700 800 900 1000 1100 1200 1300 1400 6162 4722 3881 332 2913 26 2351 2145 1971 1822 1692 07 07 07 07 07 07 07 07 07 07 07 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 11104 11136 A relation for the COP of the twostage refrigeration system with a flash chamber shown in Fig 1112 is to be derived Analysis The coefficient of performance is determined from COPR in q w L where 9 4 1 2 6 compIIin compIin in 6 6 8 1 6 1 1 with 1 h h h h x w w w h h h x h h x q fg f L PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 11105 Fundamentals of Engineering FE Exam Problems 11137 Consider a heat pump that operates on the reversed Carnot cycle with R134a as the working fluid executed under the saturation dome between the pressure limits of 140 kPa and 800 kPa R134a changes from saturated vapor to saturated liquid during the heat rejection process The net work input for this cycle is a 28 kJkg b 34 kJkg c 49 kJkg d 144 kJkg e 275 kJkg Answer a 28 kJkg Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P1800 kPa P2140 kPa hfgENTHALPYR134ax1PP1ENTHALPYR134ax0PP1 THTEMPERATURER134ax0PP1273 TLTEMPERATURER134ax0PP2273 qHhfg COPTHTHTL wnetqHCOP Some Wrong Solutions with Common Mistakes W1work qHCOP1 COP1TLTHTL Using COP of regrigerator W2work qHCOP2 COP2TH273THTL Using C instead of K W3work hfg3COP hfg3 ENTHALPYR134ax1PP2ENTHALPYR134ax0PP2 Using hfg at P2 W4work qHTLTH Using the wrong relation 11138 A refrigerator removes heat from a refrigerated space at 0C at a rate of 22 kJs and rejects it to an environment at 20C The minimum required power input is a 89 W b 150 W c 161 W d 557 W e 2200 W Answer c 161 W Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TH20273 TL0273 QL22 kJs COPmaxTLTHTL wminQLCOPmax Some Wrong Solutions with Common Mistakes W1work QLCOP1 COP1THTHTL Using COP of heat pump W2work QLCOP2 COP2TH273THTL Using C instead of K W3work QLTLTH Using the wrong relation W4work QL Taking the rate of refrigeration as power input PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 11106 11139 A refrigerator operates on the ideal vapor compression refrigeration cycle with R134a as the working fluid between the pressure limits of 120 kPa and 800 kPa If the rate of heat removal from the refrigerated space is 32 kJs the mass flow rate of the refrigerant is a 019 kgs b 015 kgs c 023 kgs d 028 kgs e 081 kgs Answer c 023 kgs Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P1120 kPa P2800 kPa P3P2 P4P1 s2s1 Qrefrig32 kJs mQrefrigh1h4 h1ENTHALPYR134ax1PP1 s1ENTROPYR134ax1PP1 h2ENTHALPYR134ass2PP2 h3ENTHALPYR134ax0PP3 h4h3 Some Wrong Solutions with Common Mistakes W1mass Qrefrigh2h1 Using wrong enthalpies for Win W2mass Qrefrigh2h3 Using wrong enthalpies for QH W3mass Qrefrigh1h44 h44ENTHALPYR134ax0PP4 Using wrong enthalpy h4 at P4 W4mass Qrefrighfg hfgENTHALPYR134ax1PP2 ENTHALPYR134ax0PP2 Using hfg at P2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 11107 11140 A heat pump operates on the ideal vapor compression refrigeration cycle with R134a as the working fluid between the pressure limits of 032 MPa and 12 MPa If the mass flow rate of the refrigerant is 0193 kgs the rate of heat supply by the heat pump to the heated space is a 33 kW b 23 kW c 26 kW d 31 kW e 45 kW Answer d 31 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P1320 kPa P21200 kPa P3P2 P4P1 s2s1 m0193 kgs Qsupplymh2h3 kJs h1ENTHALPYR134ax1PP1 s1ENTROPYR134ax1PP1 h2ENTHALPYR134ass2PP2 h3ENTHALPYR134ax0PP3 h4h3 Some Wrong Solutions with Common Mistakes W1Qh mh2h1 Using wrong enthalpies for Win W2Qh mh1h4 Using wrong enthalpies for QL W3Qh mh22h4 h22ENTHALPYR134ax1PP2 Using wrong enthalpy h2 hg at P2 W4Qh mhfg hfgENTHALPYR134ax1PP1 ENTHALPYR134ax0PP1 Using hfg at P1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 11108 11141 An ideal vapor compression refrigeration cycle with R134a as the working fluid operates between the pressure limits of 120 kPa and 700 kPa The mass fraction of the refrigerant that is in the liquid phase at the inlet of the evaporator is a 069 b 063 c 058 d 043 e 035 Answer a 069 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P1120 kPa P2700 kPa P3P2 P4P1 h1ENTHALPYR134ax1PP1 h3ENTHALPYR134ax0PP3 h4h3 x4QUALITYR134ahh4PP4 liquid1x4 Some Wrong Solutions with Common Mistakes W1liquid x4 Taking quality as liquid content W2liquid 0 Assuming superheated vapor W3liquid 1x4s x4sQUALITYR134ass3PP4 Assuming isentropic expansion s3ENTROPYR134ax0PP3 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 11109 11142 Consider a heat pump that operates on the ideal vapor compression refrigeration cycle with R134a as the working fluid between the pressure limits of 032 MPa and 12 MPa The coefficient of performance of this heat pump is a 017 b 12 c 31 d 49 e 59 Answer e 59 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P1320 kPa P21200 kPa P3P2 P4P1 s2s1 h1ENTHALPYR134ax1PP1 s1ENTROPYR134ax1PP1 h2ENTHALPYR134ass2PP2 h3ENTHALPYR134ax0PP3 h4h3 COPHPqHWin Winh2h1 qHh2h3 Some Wrong Solutions with Common Mistakes W1COP h1h4h2h1 COP of refrigerator W2COP h1h4h2h3 Using wrong enthalpies QLQH W3COP h22h3h22h1 h22ENTHALPYR134ax1PP2 Using wrong enthalpy h2 hg at P2 11143 An ideal gas refrigeration cycle using air as the working fluid operates between the pressure limits of 80 kPa and 280 kPa Air is cooled to 35C before entering the turbine The lowest temperature of this cycle is a 58C b 26C c 0C d 11C e 24C Answer a 58C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values k14 P1 80 kPa P2280 kPa T335273 K Mimimum temperature is the turbine exit temperature T4T3P1P2k1k 273 Some Wrong Solutions with Common Mistakes W1Tmin T3273P1P2k1k Using C instead of K W2Tmin T3P1P2k1 273 Using wrong exponent W3Tmin T3P1P2k 273 Using wrong exponent PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 11110 11144 Consider an ideal gas refrigeration cycle using helium as the working fluid Helium enters the compressor at 100 kPa and 17C and is compressed to 400 kPa Helium is then cooled to 20C before it enters the turbine For a mass flow rate of 02 kgs the net power input required is a 283 kW b 405 kW c 647 kW d 937 kW e 113 kW Answer d 937 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values k1667 Cp51926 kJkgK P1 100 kPa T117273 K P2400 kPa T320273 K m02 kgs Mimimum temperature is the turbine exit temperature T2T1P2P1k1k T4T3P1P2k1k WnetinmCpT2T1T3T4 Some Wrong Solutions with Common Mistakes W1Win mCpT22T1T3T44 T22T1P2P1 T44T3P1P2 Using wrong relations for temps W2Win mCpT2T1 Ignoring turbine work W3Winm1005T2BT1T3T4B T2BT1P2P1kB1kB T4BT3P1P2kB1kB kB14 Using air properties W4WinmCpT2AT1273T3273T4A T2AT1273P2P1k1k T4AT3273P1P2k 1k Using C instead of K PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 11111 11145 An absorption airconditioning system is to remove heat from the conditioned space at 20C at a rate of 150 kJs while operating in an environment at 35C Heat is to be supplied from a geothermal source at 140C The minimum rate of heat supply required is a 86 kJs b 21 kJs c 30 kJs d 61 kJs e 150 kJs Answer c 30 kJs Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TL20273 K Qrefrig150 kJs To35273 K Ts140273 K COPmax1ToTsTLToTL QinQrefrigCOPmax Some Wrong Solutions with Common Mistakes W1Qin Qrefrig Taking COP 1 W2Qin QrefrigCOP2 COP2TLTsTL Wrong COP expression W3Qin QrefrigCOP3 COP31ToTsTsToTL Wrong COP expression COPHP W4Qin QrefrigCOPmax Multiplying by COP instead of dividing PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 121 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 12 THERMODYNAMIC PROPERTY RELATIONS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 122 Partial Derivatives and Associated Relations 121C For functions that depend on one variable they are identical For functions that depend on two or more variable the partial differential represents the change in the function with one of the variables as the other variables are held constant The ordinary differential for such functions represents the total change as a result of differential changes in all variables 122C a xy dx b z y dz and c dz zx z y 123C Yes 124C Yes 125 Air at a specified temperature and specific volume is considered The changes in pressure corresponding to a certain increase of different properties are to be determined Assumptions Air is an ideal gas Properties The gas constant of air is R 0287 kPam3kgK Table A1 Analysis An ideal gas equation can be expressed as P RTv Noting that R is a constant and P PT v 2 v v v v v RTd RdT dv T P dT T P dP a The change in T can be expressed as dT T 300 001 30 K At v constant 07175 kPa m kg 12 kPa m kg K30 K 0287 3 3 v v RdT dP b The change in v can be expressed as dv v 12 001 0012 m3kg At T constant 07175 kPa 2 3 3 3 2 m kg 12 kPa m kg K300 K0012 m kg 0287 v RTdv dP T c When both v and T increases by 1 the change in P becomes 0 0 7175 0 7175 dP T dP dP v Thus the changes in T and v balance each other PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 123 126 Helium at a specified temperature and specific volume is considered The changes in pressure corresponding to a certain increase of different properties are to be determined Assumptions Helium is an ideal gas Properties The gas constant of helium is R 20769 kPam3kgK Table A1 Analysis An ideal gas equation can be expressed as P RTv Noting that R is a constant and P PT v 2 v v v v v v RTd RdT d T P dT T P dP a The change in T can be expressed as dT T 300 001 30 K At constant v 5192 kPa m kg 12 kPa m kg K30 K 20769 3 3 v v RdT dP b The change in v can be expressed as dv v 12 001 0012 m3kg At T constant 5192 kPa 2 3 3 3 2 m kg 12 kPa m kg K300 K0012 m 20769 v RTdv dP T c When both v and T increases by 1 the change in P becomes 0 5 192 5 192 dP T dP dP v Thus the changes in T and v balance each other 127 Nitrogen gas at a specified state is considered The cp and cv of the nitrogen are to be determined using Table A18 and to be compared to the values listed in Table A2b Analysis The cp and cv of ideal gases depends on temperature only and are expressed as cpT dhTdT and cvT duTdT Approximating the differentials as differences about 400 K the cp and cv values are determined to be 1045 kJkg K 390K 410 11347280 kJkg 11932 390 K 410 390 K K 410 400 K 400 K K 400 h h T h T dT dh T c T T p cp h Compare Table A2b at 400 K cp 1044 kJkgK T 0748 kJkg K 390K 410 810428 0 kJkg 8523 390 K 410 390 K K 410 400K 400 K K 400 u u T u T dT du T c T T v Compare Table A2b at 400 K cv 0747 kJkgK PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 124 128E Nitrogen gas at a specified state is considered The cp and cv of the nitrogen are to be determined using Table A18E and to be compared to the values listed in Table A2Eb Analysis The cp and cv of ideal gases depends on temperature only and are expressed as cpT dhTdT and cvT duTdT Approximating the differentials as differences about 600 R the cp and cv values are determined to be 0250 Btulbm R 780R 820 5424 2 28013 Btulbm 57047 780 R 820 780 R R 820 800 R 800 R R 800 h h T h T dT dh T c T T p Compare Table A2Eb at 800 R 340F cp 0250 BtulbmR 0179 Btulbm R 780R 820 3875 2 28013 Btulbm 40763 780 R 820 780 R R 820 800 R 800 R R 800 u u T u T dT du T c T T v Compare Table A2Eb at 800 R 340F cv 0179 BtulbmR PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 125 129 The state of an ideal gas is altered slightly The change in the specific volume of the gas is to be determined using differential relations and the idealgas relation at each state Assumptions The gas is air and air is an ideal gas Properties The gas constant of air is R 0287 kPam3kgK Table A1 Analysis a The changes in T and P can be expressed as 4 kPa 100kPa 96 4 K 400K 404 P dP T dT The ideal gas relation Pv RT can be expressed as v RTP Note that R is a constant and v v T P Applying the total differential relation and using average values for T and P 00598 m kg 3 004805 m kg m kg 00117 kPa 98 4 kPa 402 K 98 kPa 4 K kPa m kg K 0287 3 3 2 3 2 P RTdP P RdT dP P dT T d T P v v v b Using the ideal gas relation at each state 12078 m kg 96 kPa kPa m kg K404 K 0287 11480 m kg 100 kPa kPa m kg K400 K 0287 3 3 2 2 2 3 3 1 1 1 P RT P RT v v Thus 00598 m kg 3 1 1480 1 2078 1 2 v v v The two results are identical PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 126 1210 Using the equation of state Pv a RT the cyclic relation and the reciprocity relation at constant v are to be verified Analysis a This equation of state involves three variables P v and T Any two of these can be taken as the independent variables with the remaining one being the dependent variable Replacing x y and z by P v and T the cyclic relation can be expressed as 1 v v v P T T P P T where R a P T R a P T P R T a P RT a P a RT P a RT P P T v v v v v v v v v 2 Substituting 1 R a P R a P P T T P P T v v v v v which is the desired result b The reciprocity rule for this gas at v constant can be expressed as a R T P a RT P R a P T R a P T P T T P v v v v v v v v 1 We observe that the first differential is the inverse of the second one Thus the proof is complete PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 127 1211 It is to be proven for an ideal gas that the P constant lines on a T diagram are straight lines and that the high pressure lines are steeper than the lowpressure lines v Analysis a For an ideal gas Pv RT or T PvR Taking the partial derivative of T with respect to v holding P constant yields T P const v R P T P v which remains constant at P constant Thus the derivative TvP which represents the slope of the P const lines on a Tv diagram remains constant That is the P const lines are straight lines on a Tv diagram b The slope of the P const lines on a Tv diagram is equal to PR which is proportional to P Therefore the high pressure lines are steeper than low pressure lines on the Tv diagram 1212 A relation is to be derived for the slope of the v constant lines on a TP diagram for a gas that obeys the van der Waals equation of state Analysis The van der Waals equation of state can be expressed as b a R P T v v 2 1 Taking the derivative of T with respect to P holding v constant R b b R P T v v v 1 1 0 which is the slope of the v constant lines on a TP diagram PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 128 The Maxwell Relations 1213 The validity of the last Maxwell relation for refrigerant134a at a specified state is to be verified Analysis We do not have exact analytical property relations for refrigerant134a and thus we need to replace the differential quantities in the last Maxwell relation with the corresponding finite quantities Using property values from the tables about the specified state m kg K 1602 10 m kg K 10 1621 30 C 70 0029966m kg 0036373 500kPa 900 10309kJkg K 09660 30 C 70 500 kPa 900 3 4 3 4 3 kPa 700 30 C 70 C C 50 500 kPa kPa 900 kPa 700 C 50 P T P T P T s s T P s T P s v v v v since kJ kPam³ and K C for temperature differences Thus the last Maxwell relation is satisfied PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 129 1214 Problem 1213 is reconsidered The validity of the last Maxwell relation for refrigerant 134a at the specified state is to be verified Analysis The problem is solved using EES and the solution is given below Input Data T50 C P700 kPa Pincrement 200 kPa Tincrement 20 C P2PPincrement P1PPincrement T2TTincrement T1TTincrement DELTAP P2P1 DELTAT T2T1 v1volumeR134aTT1PP v2volumeR134aTT2PP s1entropyR134aTTPP1 s2entropyR134aTTPP2 DELTAss2 s1 DELTAvv2 v1 The partial derivatives in the last Maxwell relation Eq 1219 is associated with the Gibbs function and are approximated by the ratio of ordinary differentials LeftSide DELTAsDELTAPConvertkJm3kPa m3kgK at T Const RightSideDELTAvDELTAT m3kgK at P Const SOLUTION DELTAP400 kPa DELTAs006484 kJkgK DELTAT40 C DELTAv0006407 m3kg LeftSide00001621 m3kgK P700 kPa P1500 kPa P2900 kPa Pincrement200 kPa RightSide00001602 m3kgK s110309 kJkgK s209660 kJkgK T50 C T130 C T270 C Tincrement20 C v1002997 m3kg v2003637 m3kg PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1210 1215E The validity of the last Maxwell relation for steam at a specified state is to be verified Analysis We do not have exact analytical property relations for steam and thus we need to replace the differential quantities in the last Maxwell relation with the corresponding finite quantities Using property values from the tables about the specified state ft lbm R 1635 10 ft lbm R 10 1639 700 F 900 19777 16507ft lbm 350psia 450 17009Btulbm R 16706 700 F 900 350 psia 450 3 3 3 3 3 psia 400 700 F 900 F F 800 350 psia psia 450 psia 400 F 800 P T P T P T s s T P s T P s v v v v since 1 Btu 54039 psiaft3 and R F for temperature differences Thus the fourth Maxwell relation is satisfied 1216 Using the Maxwell relations a relation for sPT for a gas whose equation of state is Pvb RT is to be obtained Analysis This equation of state can be expressed as b P v RT Then P R T P v From the fourth Maxwell relation P R P T T P s v 1217 Using the Maxwell relations a relation for svT for a gas whose equation of state is Pav2vb RT is to be obtained Analysis This equation of state can be expressed as v 2 v a b RT P Then b R T P v v From the third Maxwell relation b R v v T v P s T PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1211 1218 Using the Maxwell relations and the idealgas equation of state a relation for svT for an ideal gas is to be obtained Analysis The ideal gas equation of state can be expressed as v P RT Then v v R T P From the third Maxwell relation v v v R T P s T PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1212 1219 It is to be proven that v T P k k T P s 1 Analysis Using the definition of cv v v v v T P P s T T s T c Substituting the first Maxwell relation T s P s v v v v v T P T T c s Using the definition of cp P P P p T s T T s T c v v Substituting the second Maxwell relation s P T P s v P s p T T P T c v From Eq 1246 T P p P T T c c v v v 2 Also cv c c k k p p 1 Then T P s T P P s P T T P P T T T P k k v v v v v 2 1 Substituting this into the original equation in the problem statement produces v v v T P P T T P T P T P s s But according to the cyclic relation the last three terms are equal to 1 Then s s T P T P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1213 1220 It is to be shown how T v u a and g could be evaluated from the thermodynamic function h hs P Analysis Forming the differential of the given expression for h produces dP P h ds s h dh s P Solving the dh Gibbs equation gives dP Tds dh v Comparing the coefficient of these two expressions s P P h s h T v both of which can be evaluated for a given P and s From the definition of the enthalpy P s h hP P h u v Similarly the definition of the Helmholtz function P s s h s P h P h Ts u a while the definition of the Gibbs function gives s P h s h Ts h q All of these can be evaluated for a given P and s and the fundamental hsP equation PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1214 The Clapeyron Equation 1221C It enables us to determine the enthalpy of vaporization from hfg at a given temperature from the P v T data alone 1222C It is assumed that vfg vg RTP and hfg constant for small temperature intervals 1223 Using the Clapeyron equation the enthalpy of vaporization of steam at a specified pressure is to be estimated and to be compared to the tabulated data Analysis From the Clapeyron equation 21599 kJkg 13627 13058 C 50 kPa 0001073 m kg 27315 K060582 13352 275kPa 325 3 sat275 kPa sat325 kPa 300 kPa 300 kPa sat sat 300 kPa 300 kPa sat T T T T P T dT dP T h f g f g fg fg v v v v v The tabulated value of hfg at 300 kPa is 21635 kJkg 1224 The hfg and sfg of steam at a specified temperature are to be calculated using the Clapeyron equation and to be compared to the tabulated data Analysis From the Clapeyron equation 2206 8 kJ kg 10 K 0001060 m kg 23223 16918kPa 27315 K089133 120 125 C 115 C 3 sat115 C sat125 C 120 C sat120 C 120 C sat P P T T P T dT dP T h f g f g fg fg v v v v v Also 5 6131 kJkg K 27315K 120 22068 kJkg T h s fg fg The tabulated values at 120C are hfg 22021 kJkg and sfg 56013 kJkgK PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1215 1225E The hfg of refrigerant134a at a specified temperature is to be calculated using the Clapeyron equation and ClapeyronClausius equation and to be compared to the tabulated data Analysis a From the Clapeyron equation 01 error psia ft lbm 4826 10 R 23793 psia 0 01201 ft lbm 29759 45967 R17345 10 5 F 15 F 3 3 sat5 F sat 15 F 1 0 F sat 10 F 1 0 F sat 8931 Btulbm P P T T P T dT dP T h f g f g fg fg v v v v v since 1 Btu 54039 psiaft3 b From the ClapeyronClausius equation error 76 45967 R 5 1 45967 R 15 1 001946 Btulbm R 29759 psia 23793 psia ln 1 1 ln sat 2 1 sat 1 2 9604 Btulbm fg fg fg h h T T R h P P The tabulated value of hfg at 10F is 8923 Btulbm PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1216 1226 The enthalpy of vaporization of steam as a function of temperature using Clapeyron equation and steam data in EES is to be plotted Analysis The enthalpy of vaporization is determined using Clapeyron equation from T P T h fg fg Clapeyron v At 100ºC for an increment of 5ºC we obtain 1 6710 m kg 0 001043 6720 1 6720 m kg 1 001043 m kg 0 3629 kPa 8461 12090 10 C 95 105 90 kPa 120 8461 kPa 105 C 5 100 95 C 5 100 3 3 100 C 3 100 C 1 2 1 2 sat 105 C 2 sat 95 C 1 increment 2 increment 1 f g fg g f P P P T T T P P P P T T T T T T v v v v v Substituting 22628 kJkg 10 K 27315 K16710 m kg 3629 kPa 100 3 Clapeyron T P T h fg fg v The enthalpy of vaporization from steam table is 22564 m kg 3 fg 100C h The percent error in using Clapeyron equation is 028 100 2256 4 2256 4 2262 8 PercentError We repeat the analysis over the temperature range 10 to 200ºC using EES Below the copy of EES solution is provided Input Data T100 C Tincrement 5C T2TTincrementC T1TTincrementC P1 pressureSteamiapwsTT1x0kPa P2 pressureSteamiapwsTT2x0kPa DELTAP P2P1kPa DELTAT T2T1C vfvolumeSteamiapwsTTx0m3kg vgvolumeSteamiapwsTTx1m3kg hfenthalpySteamiapwsTTx0kJkg hgenthalpySteamiapwsTTx1kJkg hfghg hfkJkgK vfgvg vfm3kg The Clapeyron equation Eq 1122 provides a means to calculate the enthalpy of vaporization hfg at a given temperature by determining the slope of the saturation curve on a PT diagram and the specific volume of the saturated liquid and satruated vapor at the temperature PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1217 hfgClapeyronT27315vfgDELTAPDELTATConvertm3kPakJkJkg PercentErrorABShfgClapeyronhfghfg100 hfg kJkg hfgClapeyron kJkg PercentError T C 247720 250809 1247 10 242982 245109 08756 30 238195 239669 06188 50 233304 234347 04469 70 228251 229007 03311 90 222968 223525 025 110 217373 217786 01903 130 211377 211684 01454 150 201417 201615 009829 180 189967 190098 006915 210 176550 176638 005015 240 0 50 100 150 200 250 1700 1800 1900 2000 2100 2200 2300 2400 2500 2600 T C hfg kJkg hfg calculated by EES hfg calculated by Clapeyron equation PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1218 1227E A substance is cooled in a pistoncylinder device until it turns from saturated vapor to saturated liquid at a constant pressure and temperature The boiling temperature of this substance at a different pressure is to be estimated Analysis From the Clapeyron equation 1 896 psiaR R15 ft 05 lbm 475 05 lbm 1Btu Btu 5404 psia ft 250 3 3 sat fg fg T h dT dP v Weight 50 psia 15F 05 lbm Sat vapor Q Using the finite difference approximation sat 1 2 1 2 sat T T P P dT dP Solving for T2 4803 R 1896 psiaR 50psia 60 475 R 1 2 1 2 dP dT P P T T 1228E A substance is cooled in a pistoncylinder device until it turns from saturated vapor to saturated liquid at a constant pressure and temperature The saturation pressure of this substance at a different temperature is to be estimated Analysis From the Clapeyron equation 1 896 psiaR R15 ft 05 lbm 475 05 lbm 1Btu Btu 5404 psia ft 250 3 3 sat fg fg T h dT dP v Weight 50 psia 10F 05 lbm Sat vapor Q Using the finite difference approximation sat 1 2 1 2 sat T T P P dT dP Solving for P2 4052 psia 475R 1896 psiaR470 50 psia 1 2 1 2 T dT T dP P P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1219 1229E A substance is cooled in a pistoncylinder device until it turns from saturated vapor to saturated liquid at a constant pressure and temperature The sfg of this substance at the given temperature is to be estimated Analysis From the Clapeyron equation Weight 50 psia 15F 05 lbm Sat vapor Q fg fg fg fg s T h dT dP v v sat Solving for sfg 1053 Btulbm R 475 R 250 Btu05 lbm T h s fg fg Alternatively 1 053 Btulbm R psia ft 5404 1Btu 05 lbm 896 psiaR 15 ft 1 3 3 sat fg fg dT dP s v 1230E Saturation properties for R134a at a specified temperature are given The saturation pressure is to be estimated at two different temperatures Analysis From the Clapeyron equation 0 4979 psiaR 1Btu psia ft 5404 R21446 ft lbm 460 90886 Btulbm 3 3 sat fg fg T h dT dP v Using the finite difference approximation sat 1 2 1 2 sat T T P P dT dP Solving for P2 at 15F 1372 psia 460R 0 4979 psiaR445 21185 psia 1 2 1 2 T dT T dP P P Solving for P2 at 30F 625 psia 460R 0 4979 psiaR430 21185 psia 1 2 1 2 T dT T dP P P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1220 1231E A table of properties for methyl chloride is given The saturation pressure is to be estimated at two different temperatures Analysis The Clapeyron equation is fg fg T h dT dP v sat Using the finite difference approximation fg fg T h T T P P dT dP v sat 1 2 1 2 sat Solving this for the second pressure gives for T2 110F 1340 psia 110 100R 1Btu 404 psia ft 5 R086332 ft lbm 560 15485 Btulbm 7 psia 116 3 3 1 2 1 2 T T T h P P fg fg v When T2 90F 994 psia 90 100R 1Btu 404 psia ft 5 R086332 ft lbm 560 15485 Btulbm 7 psia 116 3 3 1 2 1 2 T T T h P P fg fg v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1221 1232 It is to be shown that sat T P T T h T c c fg P fg p f p g v Analysis The definition of specific heat and Clapeyron equation are fg fg P p T h dT dP T h c v sat According to the definition of the enthalpy of vaporization T h T h T h f g fg Differentiating this expression gives 2 2 2 1 1 T h h T c T c T h T h T T h T h T T T h T T h T T h f g p f g p f P f g P g P f P g P fg Using ClasiusClapeyron to replace the last term of this expression and solving for the specific heat difference gives sat T P T T h T c c fg P fg p f p g v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1222 General Relations for du dh ds cv and cp 1233C Yes through the relation P T p T T P c 2 2v 1234E The specific heat difference cp cv for liquid water at 1000 psia and 300F is to be estimated Analysis The specific heat difference cp cv is given as T P p P T T c c v v v 2 Approximating differentials by differences about the specified state 54039 psia ft Btu 1 986 psia ft lbm R 0 0017417ft lbm 0017345 1000 psia 50 R 0017151ft lbm 60967 R 0017633 500psia 1500 275 F 45967 R 325 300 3 3 3 2 3 300 F 500psia psia 1500 2 psia 1000 275 F F 325 F 300 2 psia 1000 Btulbm R 0183 T P T P p P T T c c v v v v v v v Properties are obtained from Table A7E PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1223 1235 The volume expansivity β and the isothermal compressibility α of refrigerant134a at 200 kPa and 30C are to be estimated Analysis The volume expansivity and isothermal compressibility are expressed as T P P T α β v v v v 1 and 1 Approximating differentials by differences about the specified state 0 00381 K 1 20 K 011418m kg 012322 m kg 011874 1 20 C 40 1 1 3 3 kPa 200 20 C C 40 200kPa P P T v v v v v β and 0 00482 kPa 1 60 kPa 013248m kg 009812 m kg 011874 1 240 180kPa 1 1 3 3 C 30 180kPa kPa 240 30 C T T P v v v v v α PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1224 1236 The internal energy change of air between two specified states is to be compared for two equations of states Assumptions Constant specific heats for air can be used Properties For air at the average temperature 203002160C433 K cv 0731 kJkgK Table A2b Analysis Solving the equation of state for P gives a RT P v Then a R T P v v Using equation 1229 v v v P d T P T c dT du Substituting dT c a d RT a RT c dT du v v v v v Integrating this result between the two states with constant specific heats gives 205 kJkg 20K 0 731 kJkg K300 1 2 1 2 T T c u u v The ideal gas model for the air gives c dT du v which gives the same answer PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1225 1237 The enthalpy change of air between two specified states is to be compared for two equations of states Assumptions Constant specific heats for air can be used Properties For air at the average temperature 203002160C433 K cp 1018 kJkgK Table A2b Analysis Solving the equation of state for v gives a P v RT Then P R T P v Using equation 1235 dP T T c dT dh P p v v Substituting adP dT c dP P RT a P RT c dT dh p p Integrating this result between the two states with constant specific heats gives 2900 kJkg 100kPa 001 m kg600 20K 018 kJkg K300 1 3 1 2 1 2 1 2 P a P T T c h h p For an ideal gas c dT dh p which when integrated gives 2850 kJkg 20K 1 018 kJkg K300 1 2 1 2 T T c h h p PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1226 1238 The entropy change of air between two specified states is to be compared for two equations of states Assumptions Constant specific heats for air can be used Properties For air at the average temperature 203002160C433 K cp 1018 kJkgK Table A2b and R 0287 kJkgK Table A1 Analysis Solving the equation of state for v gives a P v RT Then P R T P v The entropy differential is P R dP T dT c dP T T dT c ds p P p v which is the same as that of an ideal gas Integrating this result between the two states with constant specific heats gives 01686 kJkg K 100 kPa 0 287 kJkg Kln 600 kPa 293 K 018 kJkg Kln 573 K 1 ln ln 1 2 1 2 1 2 P P R T T c s s p PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1227 1239 The internal energy change of helium between two specified states is to be compared for two equations of states Properties For helium cv 31156 kJkgK Table A2a Analysis Solving the equation of state for P gives a RT P v Then a R T P v v Using equation 1229 v v v P d T P T c dT du Substituting dT c a d RT a RT c dT du v v v v v Integrating this result between the two states gives 8724 kJkg 20K 3 1156 kJkg K300 1 2 1 2 T T c u u v The ideal gas model for the helium gives c dT du v which gives the same answer PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1228 1240 The enthalpy change of helium between two specified states is to be compared for two equations of states Properties For helium cp 51926 kJkgK Table A2a Analysis Solving the equation of state for v gives a P v RT Then P R T P v Using equation 1235 dP T T c dT dh P p v v Substituting adP dT c dP P RT a P RT c dT dh p p Integrating this result between the two states gives 1459 kJkg 100kPa 001 m kg600 20K 5 1926 kJkg K300 3 1 2 1 2 1 2 P a P T T c h h p For an ideal gas c dT dh p which when integrated gives 1454 kJkg 20K 5 1926 kJkg K300 1 2 1 2 T T c h h p PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1229 1241 The entropy change of helium between two specified states is to be compared for two equations of states Properties For helium cp 51926 kJkgK and R 20769 kJkgK Table A2a Analysis Solving the equation of state for v gives a P v RT Then P R T P v The entropy differential is P R dP T dT c dP T T dT c ds p P p v which is the same as that of an ideal gas Integrating this result between the two states gives 02386 kJkg K 100 kPa 2 0769 kJkg Kln 600 kPa 293 K 5 1926 kJkg Kln 573 K ln ln 1 2 1 2 1 2 P P R T T c s s p PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1230 1242 General expressions for u h and s for a gas whose equation of state is Pva RT for an isothermal process are to be derived Analysis a A relation for u is obtained from the general relation 2 1 2 1 1 2 v v v v P dv T P T c dT u u u T T The equation of state for the specified gas can be expressed as Thus 0 P P P a RT P T P T a R T P a RT P v v v v v Substituting 2 1 T T c dT u v b A relation for h is obtained from the general relation 2 1 2 1 1 2 P P P T T P dP T v T c dT h h h v The equation of state for the specified gas can be expressed as Thus a a P T R T v T P R T a P RT P P v v v v v v Substituting 1 2 2 1 2 1 2 1 P a P c dT adP c dT h T T p P P T T p c A relation for s is obtained from the general relation 2 1 2 1 1 2 P P P T T p dP T dT T c s s s v Substituting vTP RT 1 2 ln 2 1 2 1 2 1 P P R dT T c dP P R dT T c s T T p P P P T T p For an isothermal process dT 0 and these relations reduce to 1 2 1 2 ln and 0 P P R s P a P h u PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1231 1243 General expressions for uPT and hvT in terms of P v and T only are to be derived Analysis The general relation for du is v v v P d T P T c dT du Differentiating each term in this equation with respect to P at T constant yields T T T T P P P T P T P P T P T P u v v v v v 0 Using the properties P T v the cyclic relation can be expressed as P T T P T P T P P T T P v v v v v v 1 Substituting we get T P T P P T T P u v v The general relation for dh is dP T T c dT dh P p v v Differentiating each term in this equation with respect to v at T constant yields T P T T P T P T T P P T T h v v v v v v v v 0 Using the properties v T P the cyclic relation can be expressed as v v v v v v P T P T P P T T T P T P 1 Substituting we get v v v v P T T P h T T PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1232 1244 It is to be shown that P p T T P T c c v v v Analysis We begin by taking the entropy to be a function of specific volume and temperature The differential of the entropy is then v v v d s dT T s ds T Substituting T c T s v v from Eq 1228 and the third Maxwell equation changes this to v v v d T P T dT c ds Taking the entropy to be a function of pressure and temperature dP P s dT T s ds T P Combining this result with T c T s p P from Eq 1234 and the fourth Maxwell equation produces dP T dT T c ds P p v Equating the two previous ds expressions and solving the result for the specific heat difference v v v v d T P dP T T dT c c P p Taking the pressure to be a function of temperature and volume v v v d P dT T P dP T When this is substituted into the previous expression the result is v v v v v v v d T P P T T dT T P T T dT c c T P P p According to the cyclic relation the term in the bracket is zero Then canceling the common dT term P p T T P T c c v v v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1233 1245 It is to be proven that the definition for temperature v s u T reduces the net entropy change of two constant volume systems filled with simple compressible substances to zero as the two systems approach thermal equilibrium Analysis The two constantvolume systems form an isolated system shown here For the isolated system 0 B A tot dS dS dS VA const TA VB const TB Isolated system boundary Assume S u v S Then v v v d s du u s ds u Since 0 const and v v d du u s ds v and from the definition of temperature from the problem statement T du s u du v Then B B B A A A T du m T du m dS tot The first law applied to the isolated system yields A A B B B B A A m du m du m du m du dU dU E E 0 0 out in Now the entropy change may be expressed as B A A B A A B A A A T T T T m du T T m du dS 1 1 tot As the two systems approach thermal equilibrium 0 lim tot B A T T dS PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1234 1246 An expression for the volume expansivity of a substance whose equation of state is RT is to be derived a P v Analysis Solving the equation of state for v gives a P v RT The specific volume derivative is then P R T P v The definition for volume expansivity is P T β v v 1 Combining these two equations gives aP RT R β 1247 An expression for the specific heat difference of a substance whose equation of state is is to be derived RT a P v Analysis The specific heat difference is expressed by T P p P T T c c v v v 2 Solving the equation of state for specific volume a P v RT The specific volume derivatives are then RT P P P RT P T T 2 2 v v P R T P v Substituting R RT P P T R RT P P T R c c p 2 2 2 2 2 v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1235 1248 An expression for the isothermal compressibility of a substance whose equation of state is RT a P v is to be derived Analysis The definition for the isothermal compressibility is T P α v v 1 Solving the equation of state for specific volume a P v RT The specific volume derivative is then 2 P RT P T v Substituting these into the isothermal compressibility equation gives 2 aP P RT RT aP RT P P RT α 1249 An expression for the isothermal compressibility of a substance whose equation of state is 1 2 T b a b RT P v v v is to be derived Analysis The definition for the isothermal compressibility is T P α v v 1 The derivative is 2 2 1 2 2 2 b b T a b RT P T v v v v v Substituting 2 1 2 2 2 2 1 2 2 2 1 2 1 1 1 b b T a b RT b b T a b RT P T v v v v v v v v v v v α PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1236 1250 An expression for the volume expansivity of a substance whose equation of state is 1 2 T b a b RT P v v v is to be derived Analysis The definition for volume expansivity is P T β v v 1 According to the cyclic relation 1 v v v P T P T T P which on rearrangement becomes T P P T P T v v v Proceeding to perform the differentiations gives 3 2 2 T b a b R T P v v v v and 2 2 1 2 2 2 2 1 2 2 2 1 1 b b T a b RT b bT a b RT P T v v v v v v v v Substituting these results into the definition of the volume expansivity produces 2 2 1 2 2 2 3 2 2 1 b b T a b RT T b a b R β v v v v v v v v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1237 1251 An expression for the volume expansivity of a substance whose equation of state is T a b RT P v 2 v is to be derived Analysis The definition for volume expansivity is P T β v v 1 According to the cyclic relation 1 v v v P T P T T P which on rearrangement becomes T P P T P T v v v Proceeding to perform the differentiations gives 2 2T a b R T P v v v and T a b RT P T 3 2 2 v v v Substituting these results into the definition of the volume expansivity produces T a b RT T a b R 3 2 2 2 2 1 v v v v v β PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1238 1252 It is to be shown that v T P α β Analysis The definition for the volume expansivity is P T β v v 1 The definition for the isothermal compressibility is T P α v v 1 According to the cyclic relation 1 v v v P T P T T P which on rearrangement becomes v v v T P P T T P When this is substituted into the definition of the volume expansivity the result is v v v v α β T P T P P T 1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1239 1253 It is to be demonstrated that s p P c c k α v v v Analysis The relations for entropy differential are dP T T dT c ds d T P T dT c ds P p v v v v For fixed s these basic equations reduce to dP T T dT c d T P T dT c P p v v v v Also when s is fixed P s P v v Forming the specific heat ratio from these expressions gives s P P P T T k v v v The cyclic relation is 1 v v v P T P T T P Solving this for the numerator of the specific heat ratio expression and substituting the result into this numerator produces s s T P P P k v v v v α PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1240 1254 The Helmholtz function of a substance has the form 0 0 0 0 0 ln 1 ln T T T T T T cT RT a v v It is to be shown how to obtain P h s cv and cp from this expression Analysis Taking the Helmholtz function to be a function of temperature and specific volume yields v v v d a dT T a da T while the applicable Helmholtz equation is sdT Pd da v Equating the coefficients of the two results produces v v T a s a P T Taking the indicated partial derivatives of the Helmholtz function given in the problem statement reduces these expressions to 0 0 ln ln T T c R s RT P v v v The definition of the enthalpy h u Pv and Helmholtz function a uTs may be combined to give RT cT cT RT T T cT RT T T T T T T cT RT a T a T a P Ts a P u h T 0 0 0 0 0 0 0 0 ln ln ln 1 ln v v v v v v v v v According to T c T s v v given in the text Eq 1228 c T T c T s T c v v The preceding expression for the temperature indicates that the equation of state for the substance is the same as that of an ideal gas Then c R c R c p v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1241 The JouleThomson Coefficient 1255C It represents the variation of temperature with pressure during a throttling process 1256C The line that passes through the peak points of the constant enthalpy lines on a TP diagram is called the inversion line The maximum inversion temperature is the highest temperature a fluid can be cooled by throttling 1257C No The temperature may even increase as a result of throttling 1258C Yes 1259C No Helium is an ideal gas and h hT for ideal gases Therefore the temperature of an ideal gas remains constant during a throttling h constant process 1260E The JouleThompson coefficient of nitrogen at two states is to be estimated Analysis a The enthalpy of nitrogen at 120 psia and 350 R is from EES h 8488 Btulbm Approximating differentials by differences about the specified state the JouleThomson coefficient is expressed as 8488 Btulbm h h P T P T µ Considering a throttling process from 130 psia to 110 psia at h 8488 Btulbm the JouleThomson coefficient is determined to be 00599 Rpsia 130 psia 110 35060 R 34940 130 psia 110 88 Btulbm 84 130 psia psia 110 h T T µ b The enthalpy of nitrogen at 1200 psia and 700 R is from EES h 17014 Btulbm Approximating differentials by differences about the specified state the JouleThomson coefficient is expressed as 17014Btulbm h h P T P T µ Considering a throttling process from 1210 psia to 1190 psia at h 17014 Btulbm the JouleThomson coefficient is determined to be 000929 Rpsia 1210 psia 1190 70009 R 69991 1210 psia 1190 14 Btulbm 170 1210 psia psia 1190 h T T µ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1242 1261E Problem 1260E is reconsidered The JouleThompson coefficient for nitrogen over the pressure range 100 to 1500 psia at the enthalpy values 100 175 and 225 Btulbm is to be plotted Analysis The problem is solved using EES and the results are tabulated and plotted below Gas Nitrogen Pref120 psia Tref350 R P Pref h100 Btulbm henthalpyGas TTref PPref dP 10 psia T temperatureGas PP hh P1 P dP P2 P dP T1 temperatureGas PP1 hh T2 temperatureGas PP2 hh Mu DELTATDELTAP Approximate the differential by differences about the state at hconst DELTATT2T1 DELTAPP2P1 h 225 Btulbm P psia µ Rpsia 100 275 450 625 800 975 1150 1325 1500 0004573 000417 0003781 0003405 0003041 0002688 0002347 0002015 0001694 0 200 400 600 800 1000 1200 1400 1600 0 0005 001 0015 002 0025 003 0035 004 0045 005 P psia µ Rpsia h100 Btulbm h175 Btulbm h225 Btulbm 1262 Steam is throttled slightly from 1 MPa and 300C It is to be determined if the temperature of the steam will increase decrease or remain the same during this process Analysis The enthalpy of steam at 1 MPa and T 300C is h 30516 kJkg Now consider a throttling process from this state to 08 MPa which is the next lowest pressure listed in the tables The temperature of the steam at the end of this throttling process will be 29752 C 3051 6 kJkg MPa 80 2 T h P Therefore the temperature will decrease PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1243 1263E The JouleThomson coefficient of refrigerant134a at a given state is to be estimated Analysis The JouleThomson coefficient is defined as P h T µ We use a finite difference approximation as 1 2 1 2 P P T T µ at constant enthalpy At the given state we call it state 1 the enthalpy of R134a is 11379 Btulbm Table A 13E 0 F 6 40 psia 1 1 1 h T P The second state will be selected for a pressure of 30 psia At this pressure and the same enthalpy we have 5678 F Table A 13E 11379 kJkg 30 psia 2 1 2 2 T h h P Substituting 0322 Rpsia 40psia 30 60R 5678 1 2 1 2 P P T T µ 1264 The JouleThomson coefficient of refrigerant134a at a given state is to be estimated Analysis The JouleThomson coefficient is defined as P h T µ We use a finite difference approximation as 1 2 1 2 P P T T µ at constant enthalpy At the given state we call it state 1 the enthalpy of R134a is 33393 kJkg Table A 13 0 C 9 200 kPa 1 1 1 h T P The second state will be selected for a pressure of 180 kPa At this pressure and the same enthalpy we have 8978 C Table A 13 33393 kJkg 180 kPa 2 1 2 2 T h h P Substituting 00110 KkPa 200kPa 180 90K 8978 1 2 1 2 P P T T µ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1244 1265 The equation of state of a gas is given by 2 T bP P v RT An equation for the JouleThomson coefficient inversion line using this equation is to be derived Analysis From Eq 1252 of the text µ v v P p T T c 1 When µ 0 as it does on the inversion line this equation becomes v v P T T Using the equation of state to evaluate the partial derivative 3 2 T bP P R T P v Substituting this result into the previous expression produces 0 3 2 2 2 2 T bP T bP P RT T bP P RT The condition along the inversion line is then 0 P 1266 It is to be demonstrated that the JouleThomson coefficient is given by P p T T c T 2 v µ Analysis From Eq 1252 of the text µ v v P p T T c 1 Expanding the partial derivative of vT produces 2 1 T T T T T P P v v v When this is multiplied by T2 the righthand side becomes the same as the bracketed quantity above Then P p T T c T 2 v µ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1245 1267 The most general equation of state for which the JouleThomson coefficient is always zero is to be determined Analysis From Eq 1252 of the text µ v v P p T T c 1 When µ 0 this equation becomes T T P v v This can only be satisfied by an equation of state of the form T f P v where fP is an arbitrary function of the pressure PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1246 The dh du and ds of Real Gases 1268C It is the variation of enthalpy with pressure at a fixed temperature 1269C As PR approaches zero the gas approaches ideal gas behavior As a result the deviation from ideal gas behavior diminishes 1270C So that a single chart can be used for all gases instead of a single particular gas 1271 The errors involved in the enthalpy and internal energy of CO2 at 350 K and 10 MPa if it is assumed to be an ideal gas are to be determined Analysis a The enthalpy departure of CO2 at the specified state is determined from the generalized chart to be Fig A29 and 51 1 353 7 39 10 1 151 304 2 350 cr ideal cr cr T R h h Z P P P T T T u T P h R R CO2 350 K 10 MPa Thus and 7 557 7 557 11351 Error 7 557kJkmol 8 314 304 2 51 351 11 ideal cr ideal 50 2 h h h Z R T h h P T u h b At the calculated TR and PR the compressibility factor is determined from the compressibility chart to be Z 065 Then using the definition of enthalpy the internal energy is determined to be and 48 9 5 666 5 666 8 439 Error 5 666kJkmol 8 314 350 0 65 7557 ideal u u u ZR T h P h u u v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1247 1272 The enthalpy and entropy changes of nitrogen during a process are to be determined assuming ideal gas behavior and using generalized charts Analysis a Using data from the ideal gas property table of nitrogen Table A18 and K 6 8 314 ln 12 193562 183289 ln 6 537 9306 1 2 1 2 1 ideal 2 1ideal 2ideal 1 ideal 2 510 kJ kmol 4 kJ kmol 2769 P P R s s s s h h h h u o o b The enthalpy and entropy departures of nitrogen at the specified states are determined from the generalized charts to be Figs A29 A30 0 25 and 60 1 770 3 39 6 1 783 126 2 225 1 1 cr 1 1 cr 1 1 s h R R Z Z P P P T T T and 0 15 and 40 2 540 3 39 12 2 536 126 2 320 2 2 cr 2 2 cr 2 2 s h R R Z Z P P P T T T Substituting 341 kJkmol K 5 kJkmol 2979 4 510 0 15 0 25 314 8 2769 40 60 314 126 2 8 1 ideal 2 2 1 1 2 1 ideal 2 2 1 1 2 s s Z Z R s s h h Z Z R T h h s s u h h cr u PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1248 1273E The enthalpy and entropy changes of water vapor during a change of state are to be determined using the departure charts and the property tables Properties The properties of water are Table A1E 3200 psia 1164 8 R 18015 lbmlbmol cr cr P T M Analysis a The pressure of water vapor during this process is 68056 psia sat 500 F 2 1 P P P Using data from the ideal gas property table of water vapor Table A23 and 3 713 Btulbmol R 0 49843 53556 ln 4440 8 Btulbmol 7738 0 12178 8 1 2 1 2 1 ideal 2 1ideal 2ideal 1 ideal 2 P P R s s s s h h h h u o o The enthalpy and entropy departures of water vapor at the specified states are determined from the generalized charts to be Figs A29 A30 or from EES 0 277 0 340 and 0 213 3200 56 680 0 824 1164 8 960 1 1 cr 1 1 cr 1 1 s h R R Z Z P P P T T T and 0 0903 0 157 and 0 213 3200 56 680 1 253 1164 8 1460 2 2 cr 2 2 cr 2 2 s h R R Z Z P P P T T T The enthalpy and entropy changes per mole basis are 4 084 Btulbmol R 0 277 1 9858 0 0903 713 3 4864 kJkmol 0 340 1 98581164 8 0 157 8 4440 1 2 1 ideal 2 1 2 1 2 cr 1 ideal 2 1 2 s s u h h u Z Z R s s s s Z Z R T h h h h The enthalpy and entropy changes per mass basis are Btulbm R 02267 Btulbm 2700 18015 lbmlbmol 4 084 Btulbmol K 18015 lbmlbmol Btulbmol 4864 1 2 1 2 1 2 1 2 M s s s s M h h h h b The inlet and exit state properties of water are 4334 Btulbm R Table A4E 1 1202 3 Btulbm 1 F 500 1 1 1 1 s h x T 7008 Btulbm R from EES 1 1515 7 Btulbm F 1000 56 psia 680 2 2 2 2 s h T P The enthalpy and entropy changes are Btulbm R 02674 Btulbm 3134 1 4334 7008 1 1202 3 7 1515 1 2 1 2 s s h h PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1249 1274E The enthalpy and entropy changes of water vapor during a change of state are to be determined using the departure charts and the property tables Properties The properties of water are Table A1E 3200 psia 1164 8 R 18015 lbmlbmol cr cr P T M Analysis a Using data from the ideal gas property table of water vapor Table A23E and 06734 Btulbmol R 3000 1 9858 ln 1000 56411 53556 ln 4853 7 Btulbmol 17032 5 12178 8 1 2 1 2 1 ideal 2 1ideal 2ideal 1 ideal 2 P P R s s s s h h h h u o o The enthalpy and entropy departures of water vapor at the specified states are determined from the generalized charts to be Figs A29 A30 or from EES 0 188 0 387 and 0 9375 3200 3000 1 683 1164 8 1960 1 1 cr 1 1 cr 1 1 s h R R Z Z P P P T T T and 0 134 0 233 and 0 3125 3200 1000 1 253 1164 8 1460 2 2 cr 2 2 cr 2 2 s h R R Z Z P P P T T T The enthalpy and entropy changes per mole basis are 05662 Btulbmol R 0 188 1 9858 0 134 6734 0 44975 Btulbmol 0 387 1 98581164 8 0 233 7 4853 1 2 1 ideal 2 1 2 1 2 cr 1 ideal 2 1 2 s s u h h u Z Z R s s s s Z Z R T h h h h The enthalpy and entropy changes per mass basis are Btulbm R 00314 Btulbm 2497 18015 lbmlbmol 05662 Btulbmol R 18015 lbmlbmol Btulbmol 44975 1 2 1 2 1 2 1 2 M s s s s M h h h h b Using water tables Table A6E 6883 Btulbm R 1 1764 6 Btulbm F 1500 psia 3000 1 1 1 1 s h T P 1 6535 Btulbm R 1506 2 Btulbm 1000 F psia 1000 2 2 2 2 s h T P The enthalpy and entropy changes are Btulbm R 00348 Btulbm 2584 1 6883 6535 1 1764 6 2 1506 1 2 1 2 s s h h PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1250 1275 The enthalpy and entropy changes of water vapor during a change of state are to be determined using the departure charts and the property tables Properties The properties of water are Table A1 2206 MPa 647 1 K 18015 kgkmol cr cr P T M Analysis Using data from the ideal gas property table of water vapor Table A23 and 4 2052 kJkmol K 1000 8 314 ln 500 227109 217141 ln 7672 kJkmol 30754 23082 1 2 1 2 1 ideal 2 1ideal 2ideal 1 ideal 2 P P R s s s s h h h h u o o The enthalpy and entropy departures of water vapor at the specified states are determined from the generalized charts to be Figs A29 A30 or from EES 0 0157 0 0288 and 0 0453 2206 1 1 349 647 1 873 1 1 cr 1 1 cr 1 1 s h R R Z Z P P P T T T and 0 0146 0 0223 and 0 0227 2206 50 1 040 647 1 673 2 2 cr 2 2 cr 2 2 s h R R Z Z P P P T T T The enthalpy and entropy changes per mole basis are 4 1961 kJkmol K 0 0157 8 314 0 0146 2052 4 7637 kJkmol 0 0288 8 314647 1 0 0223 7672 1 2 1 ideal 2 1 2 1 2 cr 1 ideal 2 1 2 s s u h h u Z Z R s s s s Z Z R T h h h h The enthalpy and entropy changes per mass basis are kJkg K 02329 kJkg 4239 18015 kgkmol 1961 kJkmol K 4 18015 kgkmol kJkmol 7637 1 2 1 2 1 2 1 2 M s s s s M h h h h The inlet and exit state properties of water vapor from Table A6 are 0311 kJkg K 8 3698 6 kJkg C 600 kPa 1000 1 1 1 1 s h T P 7 7956 kJkg K 3272 4 kJkg 400 C kPa 500 2 2 2 2 s h T P The enthalpy and entropy changes are kJkg K 02355 kJkg 4262 8 0311 7956 7 3698 6 4 3272 1 2 1 2 s s h h PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1252 1277 Propane is to be adiabatically and reversibly compressed in a steadyflow device The specific work required for this compression is to be determined using the departure charts and treating the propane as an ideal gas with temperature variable specific heats Properties The properties of propane are Table A1 4 26 MPa 370 K 0 1885 kJkg K 44097 kgkmol cr cr P T R M Analysis The temperature at the exit state may be determined by the fact that the process is isentropic and the molar entropy change between the inlet and exit is zero When the entropy change equation is integrated with variable specific heats it becomes PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1 2 2 1 1 ideal 2 ln P P R dT T c s s u p When the expression of Table A2c is substituted for cp and the integration performed we obtain 750 8 314ln 7000 4 50 100 0 01058 4 50 100 0 786 4 50 3048 100 4 04ln 450 0 ln 3 2 ln ln ln 3 3 2 2 2 2 2 2 1 2 3 1 3 2 2 1 2 2 1 2 1 2 1 2 2 1 2 1 2 2 1 1 ideal 2 T T T T P P R T d T T c T T b T T T a P P R dT dT cT b T a P P R dT T c s s u u u p 7 MPa Propane 750 kPa 177C Solving this equation by EES or an iterative solution by hand gives 2 532 K T When en energy balance is applied to the compressor it becomes 9111 kJkmol 4 50 0 7935 5 32 4 50 52 4 5 32 450 0 1524532 450 04532 4 4 3 2 4 4 3 3 2 2 4 1 4 2 3 1 3 2 2 1 2 2 1 2 2 1 3 2 2 1 1 ideal 2 in T d T T c T T b T T T a dT dT cT bT a c dT h h w p The work input per unit mass basis is 207 kJkg 44097 kgkmol kJkmol 9111 in in M w w The enthalpy departures of propane at the specified states are determined from the generalized charts to be Fig A29 or from EES 0 136 0 176 4 26 50 1 22 370 450 1 cr 1 1 cr 1 1 h R R Z P P P T T T and 0 971 1 64 4 26 7 1 44 370 532 2 cr 2 2 cr 2 2 h R R Z P P P T T T The work input ie enthalpy change is determined to be 148 kJkg 0 136 0 1885370 0 971 207 1 2 cr 1 ideal 2 1 2 in h h Z Z RT h h h h w preparation If you are a student using this Manual you are using it without permission 1253 1278E Oxygen is to be adiabatically and reversibly expanded in a nozzle The exit velocity is to be determined using the departure charts and treating the oxygen as an ideal gas with temperature variable specific heats Properties The properties of oxygen are Table A1 736 psia 278 6 R 0 06206 Btulbm R 31999 lbmlbmol cr cr P T R M Analysis The temperature at the exit state may be determined by the fact that the process is isentropic and the molar entropy change between the inlet and exit is zero From the entropy change equation for an ideal gas with variable specific heats 2 085 Btulbmol R 200 1 9858 ln 70 ln 0 1 2 1 2 1 ideal 2 P P R s s s s u o o 200 psia 600F 0 fts O2 70 psia Then from Table A19E 5614 1 Btulbmol 802 R 836 Btulbmol R 51 51836 Btulbmol R 2 085 53921 085 2 53921 Btulbmol R 7543 6 Btulbmol R 1060 2ideal 2 2 1 2 1 1ideal 1 h T s s s s h T o o o o The enthalpy change per mole basis is 1929 5 Btulbmol 7543 6 5614 1 1ideal 2ideal 1 ideal 2 h h h h The enthalpy change per mass basis is 6030 Btulbm 31999 lbmlbmol 1929 5 Btulbmol 1 ideal 2 1 ideal 2 M h h h h An energy balance on the nozzle gives 2 2 2 2 2 2 2 2 1 1 2 2 2 2 1 1 out in V h V h V m h V h m E E Solving for the exit velocity 1738 fts 50 2 2 2 50 2 1 2 1 2 1Btulbm 26030 Btulbm 25037 ft s 0 fts 2 h h V V The enthalpy departures of oxygen at the specified states are determined from the generalized charts to be Fig A29 or from EES 0 000759 0 272 736 200 3 805 278 6 1060 1 cr 1 1 cr 1 1 h R R Z P P P T T T 0 00894 0 0951 736 70 2 879 278 6 802 2 cr 2 2 cr 2 2 h R R Z P P P T T T The enthalpy change is 6044 Btulbm 0 000759 0 06206 Btulbm R278 6 R 0 00894 30 Btulbm 60 1 2 cr 1 ideal 2 1 2 h h Z Z RT h h h h The exit velocity is 1740 fts 50 2 2 2 50 2 1 2 1 2 1Btulbm 26044 Btulbm 25037 ft s 0 fts 2 h h V V PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1254 1279 Propane is compressed isothermally by a pistoncylinder device The work done and the heat transfer are to be determined using the generalized charts Assumptions 1 The compression process is quasiequilibrium 2 Kinetic and potential energy changes are negligible Analysis a The enthalpy departure and the compressibility factors of propane at the initial and the final states are determined from the generalized charts to be Figs A29 A15 and 0 50 and 81 0 939 4 26 4 1 008 370 373 0 92 0 28and 0 235 4 26 1 1 008 370 373 2 2 cr 2 2 cr 2 2 1 1 cr 1 1 cr 1 1 Z Z P P P T T T Z Z P P P T T T h R R h R R Propane 1 MPa 100 C Q Treating propane as a real gas with Zavg Z1Z22 092 0502 071 constant avg C RT Z ZRT Pv Then the boundary work becomes 996 kJkg 092 4 71 01885 kJkg K 373 K ln 050 1 0 ln ln ln 2 1 1 2 ave 1 1 2 2 avg 1 2 2 1 2 1 in Z P Z P RT Z P Z RT P Z RT RT Z C C d Pd wb v v v v v Also 76 5 kJkg 0 92 373 373 50 0 1885 106 106 kJkg 0 81 0 28 0 1885 370 1 1 2 2 1 2 1 2 1 ideal 2 2 1 1 2 Z T R Z T h h u u h h Z Z RT h h h h cr Then the heat transfer for this process is determined from the closed system energy balance to be 1761 kJkg 176 1 kJkg 99 6 76 5 out in 1 2 in 1 2 in in system out in q w u u q u u u w q E E E b b PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1255 1280 Problem 1279 is reconsidered This problem is to be extended to compare the solutions based on the ideal gas assumption generalized chart data and real fluid EES data Also the solution is to be extended to carbon dioxide nitrogen and methane Analysis The problem is solved using EES and the solution is given below Procedure INFOName T1 Fluid Tcritical pcritical If NamePropane then Tcritical370 pcritical4620 FluidC3H8 goto 10 endif If NameMethane then Tcritical1911 pcritical4640 FluidCH4 goto 10 endif If NameNitrogen then Tcritical1262 pcritical3390 FluidN2 goto 10 endif If NameOxygen then Tcritical1548 pcritical5080 FluidO2 goto 10 endif If NameCarbonDioxide then Tcritical3042 pcritical7390 FluidCO2 goto 10 endif If NamenButane then Tcritical4252 pcritical3800 FluidC4H10 goto 10 endif 10 If T1Tcritical then CALL ERRORThe supplied temperature must be greater than the critical temperature for the fluid A value of XXXF1 K was suppliedT1 endif end Data from the Diagram Window T110027315 p11000 p24000 NamePropane FluidC3H8 Call INFOName T1 Fluid Tcritical pcritical Ru8314 MmolarmassFluid RRuM IDEAL GAS SOLUTION State 1 hideal1enthalpyFluid TT1 Enthalpy of ideal gas sideal1entropyFluid TT1 pp1 Entropy of ideal gas uideal1hideal1RT1 Internal energy of ideal gas State 2 hideal2enthalpyFluid TT2 Enthalpy of ideal gas sideal2entropyFluid TT2 pp2 Entropy of ideal gas uideal2hideal2RT2 Internal energy of ideal gas Work is the integral of p dv which can be done analytically widealRT1Lnp1p2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1256 First Law note that uideal2 is equal to uideal1 qidealwidealuideal2uideal1 Entropy change DELTAsidealsideal2sideal1 COMPRESSABILITY CHART SOLUTION State 1 Tr1T1Tcritical pr1p1pcritical Z1COMPRESSTr1 Pr1 DELTAh1ENTHDEPTr1 Pr1RTcritical Enthalpy departure h1hideal1DELTAh1 Enthalpy of real gas using charts u1h1Z1RT1 Internal energy of gas using charts DELTAs1ENTRDEPTr1 Pr1R Entropy departure s1sideal1DELTAs1 Entropy of real gas using charts State 2 T2T1 Tr2Tr1 pr2p2pcritical Z2COMPRESSTr2 Pr2 DELTAh2ENTHDEPTr2 Pr2RTcritical Enthalpy departure DELTAs2ENTRDEPTr2 Pr2R Entropy departure h2hideal2DELTAh2 Enthalpy of real gas using charts s2sideal2DELTAs2 Entropy of real gas using charts u2h2Z2RT2 Internal energy of gas using charts Work using charts note use of EES integral function to evaluate the integral of p dv wchartIntegralpvv1v2 We need an equation to relate p and v in the above INTEGRAL function pvCOMPRESSTr2ppcriticalRT1 To specify relationship between p and v Find the limits of integration p1v1Z1RT1 to get v1 the lower bound p2v2Z2RT2 to get v2 the upper bound First Law note that u2 is not equal to u1 qchartwchartu2u1 Entropy Change DELTAscharts2s1 SOLUTION USING EES BUILTIN PROPERTY DATA At state 1 uees1intEnergyNameTT1pp1 sees1entropyNameTT1pp1 At state 2 uees2IntEnergyNameTT2pp2 sees2entropyNameTT2pp2 Work using EES builtin properties note use of EES Integral funcion to evaluate the integral of pdv weesintegralpees vees vees1vees2 The following equation relates p and v in the above INTEGRAL peespressureNameTT1 vvees To specify relationship between p and v Find the limits of integration vees1volumeName TT1pp1 to get lower bound vees2volumeName TT2pp2 to get upper bound First law note that uees2 is not equal to uees1 qeesweesuees2uees1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1257 Entropy change DELTAseessees2sees1 Note In all three solutions to this problem we could have calculated the heat transfer by qTDELTAs since T is constant Then the first law could have been used to find the work The use of integral of p dv to find the work is a more fundamental approach and can be used if T is not constant SOLUTION DELTAh11648 kJkg DELTAh29196 kJkg DELTAs1003029 kJkgK DELTAs201851 kJkgK DELTAschart04162 kJkgK DELTAsees04711 kJkgK DELTAsideal02614 kJkgK FluidC3H8 h12232 kJkg h22308 kJkg hideal12216 kJkg hideal22216 kJkg M441 NamePropane p4000 p11000 kPa p24000 kPa pr102165 pr208658 pcritical4620 kPa pees4000 qchart1553 kJkg qees1758 kJkg qideal9754 kJkg R01885 kJkgK Ru8314 kJmoleK s16073 kJkgK s25657 kJkgK sees12797 kJkgK sees22326 kJkgK sideal16103 kJkgK sideal25842 kJkgK T13732 K T23732 K Tr11009 Tr21009 Tcritical370 K u12298 kJkg u22351 kJkg uees16884 kJkg uees26171 kJkg uideal12286 kJkg uideal22286 kJkg v001074 v1006506 m3kg v2001074 m3kg vees0009426 vees100646 m3kg vees20009426 m3kg wchart1019 kJkg wees1045 kJkg wideal9754 kJkg Z109246 Z206104 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1259 1282 A paddlewheel placed in a wellinsulated rigid tank containing oxygen is turned on The final pressure in the tank and the paddlewheel work done during this process are to be determined Assumptions 1The tank is wellinsulated and thus heat transfer is negligible 2 Kinetic and potential energy changes are negligible Properties The gas constant of O2 is R 02598 kJkgK Table A1 Analysis a For this problem we use critical properties compressibility factor and enthalpy departure factors in EES The compressibility factor of oxygen at the initial state is determined from EES to be O2 175 K 6 MPa 1 33 0 682 and 1 19 5 043 6 1 13 154 6 175 1 1 cr 1 1 cr 1 1 h R R Z Z P P P T T T Then 9 68 kg m kg 000516 m 005 000516 m kg 6000 kPa 2598 kPa m kg K175 K 06820 3 3 1 3 3 1 v V v v m ZRT P The specific volume of oxygen remains constant during this process v2 v1 Thus 9652 kPa 91 5043 1 91 1 09 1 853 0 0 649 kPa m kg K1546 K 5043 kPa 02598 000516 m kg 1 46 154 8 225 cr 2 2 2 2 2 3 3 cr cr 2 2 cr 2 2 P P P P Z Z P RT T T T R R h R R v v b The energy balance relation for this closed system can be expressed as 1 1 2 2 1 2 1 1 2 2 1 2 in 1 2 in system out in Z T R Z T h m h P P h m h W u m u U W E E E v v where 51 kJkg 62 5296 1 09 0 2598154 6 1 33 1 ideal 2 2 1 cr 1 2 h h Z Z RT h h h h Substituting 423 kJ 0682 175 K 02598 kJkg K 0853 225 9 68 kg 6251 Win Discussion The following routine in EES is used to get the solution above Reading values from Fig A15 and A29 together with properties in the book could yield different results Given V005 m3 T1175 K P16000 kPa T2225 K Properties FluidO2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1260 Ru8314 kJkmolK TcrTCRITFluid PcrPCRITFluid MMmolarmassFluid RRuMM Analysis a TR1T1Tcr PR1P1Pcr Zh1ENTHDEPTR1 PR1 the function that returns enthalpy departure factor at TR1 and PR1 Z1COMPRESSTR1 PR1 the function that returns compressibility factor at TR1 and PR1 TR2T2Tcr vR2v2PcrRTcr v1Z1RT1P1 mVv1 v2v1 Zh2ENTHDEPTR2 PR2 the function that returns enthalpy departure factor at TR2 and PR2 Z2COMPRESSTR2 PR2 the function that returns compressibility factor at TR2 and PR2 P2Z2RT2v2 P2PR2Pcr b h1idealenthalpyFluid TT1 h2idealenthalpyFluid TT2 DELTAhidealh2idealh1ideal DELTAhRTcrZh1Zh2DELTAhideal DELTAuDELTAhRZ2T2Z1T1 WinmDELTAu Solution DELTAh6251 kJkg DELTAhideal5296 kJkg DELTAu4365 kJkg FluidO2 h1ideal1218 kJkg h2ideal688 kJkg m9682 kg MM32 kgkmol P16000 kPa P29652 kPa Pcr5043 kPa PR1119 PR21914 R02598 kJkgK Ru8314 kJkmolK T1175 K T2225 K Tcr1546 K TR11132 TR21456 V005 m3 v10005164 m3kg v20005164 m3kg vR206485 Win4226 kJ Z106815 Z208527 Zh11331 Zh21094 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1261 1283 The heat transfer and entropy changes of CO2 during a process are to be determined assuming ideal gas behavior using generalized charts and real fluid EES data Analysis The temperature at the final state is 2984 K 1MPa 273 K 8 MPa 100 1 2 1 2 P T P T Using data from the ideal gas property table of CO2 Table A20 3386 9 kJkg 44 kgkmol 149024 kJkmol 94115 kJkmol K 1 8 314 ln 8 222367 333770 ln 149024 kJkmol 161293 12 269 1 ideal 2 1 ideal 2 1 2 1 2 1 ideal 2 1ideal 2ideal 1 ideal 2 M h h h h P P R s s s s h h h h u o o The heat transfer is determined from an energy balance noting that there is no work interaction 28937 kJkg 9 kJkg 01889 kJkgK2984373 3386 1 2 1 ideal 2 1 ideal 2 ideal T R T h h u u q The entropy change is 21390 kJkgK 44 kgkmol 94115 kJkmol 1 ideal 2 1 ideal 2 ideal M s s s s s The compressibility factor and the enthalpy and entropy departures of CO2 at the specified states are determined from the generalized charts to be we used EES and 0 002685 0 1144and 1 009 1 083 7 39 8 9 813 304 2 2985 0 05987 0 1028and 0 976 0 135 7 39 1 1 226 304 2 373 2 2 2 cr 2 2 cr 2 2 1 1 1 cr 1 1 cr 1 1 s h R R s h R R Z Z Z P P P T T T Z Z Z P P P T T T Thus kJkgK 2151 kJkg 29359 2 1390 0 002685 0 05987 1889 0 373 0 976 0 18892887 0 1028 0 1144 0 1889304 2 9 3386 1 ideal 2 2 1 1 chart 2 chart 1 2 1 1 2 cr 1 ideal 2 1 2 chart s s Z R Z s s s T Z R T Z Z RT h h u u q s s h h Note that the temperature at the final state in this case was determined from 2888 K 1 009 0 976 1MPa 273 K 8 MPa 100 2 1 1 2 1 2 Z Z P T P T The solution using EES builtin property data is as follows 2464 kJkgK 0 614 kJkg 8 06885 m kg 0 MPa 1 K 373 1 1 3 1 1 1 s u P T v 85 kJkgK 1 kJkg 2754 K 2879 06885 m kg 0 MPa 8 2 2 2 3 1 2 2 s u T P v v Then kJkgK 2097 kJkg 2763 0 2464 1 85 8 614 2754 1 2 1 EES 2 EES 1 2 EES s s s s s u u q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1262 Review Problems 1284 It is to be shown that the slope of a constantpressure line on an hs diagram is constant in the saturation region and increases with temperature in the superheated region Analysis For P constant dP 0 and the given relation reduces to dh Tds which can also be expressed as T s h P h P const s Thus the slope of the P constant lines on an hs diagram is equal to the temperature a In the saturation region T constant for P constant lines and the slope remains constant b In the superheat region the slope increases with increasing temperature since the slope is equal temperature 1285 Using the cyclic relation and the first Maxwell relation the other three Maxwell relations are to be obtained Analysis 1 Using the properties P s v the cyclic relation can be expressed as 1 s P P s s P v v v Substituting the first Maxwell relation v v s P T s P s P s s P s s P T s P T P s T v v v v v 1 1 2 Using the properties T v s the cyclic relation can be expressed as 1 v v v T s s T T s Substituting the first Maxwell relation v v s P T s v v v v v v v T P s s T P T s s s P T T T 1 1 3 Using the properties P T v the cyclic relation can be expressed as 1 T P P T T P v v v Substituting the third Maxwell relation v v T P s T P T P T T P T T P s T P s P T s v v v v v 1 1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1263 1286 For β 0 it is to be shown that at every point of a singlephase region of an hs diagram the slope of a constant pressure line is greater than the slope of a constanttemperature line but less than the slope of a constantvolume line Analysis It is given that β 0 Using the Tds relation ds dP T ds dh dP Tds dh v v 1 P constant T s h P 2 T constant T T s P T s h v But the 4th Maxwell relation P T T s P v Substituting β 1 T T T s h P T v v Therefore the slope of P constant lines is greater than the slope of T constant lines 3 v constant a s P T s h v v v From the ds relation v v v d T P T dT c ds Divide by dP holding v constant or b T P c T s P P T T c P s v v v v v v Using the properties P T v the cyclic relation can be expressed as 1 1 c P T T P P T T P T P T P α β α β v v v v v v v v where we used the definitions of α and β Substituting b and c into a T c T T s P T s h α β v v v v v Here α is positive for all phases of all substances T is the absolute temperature that is also positive so is cv Therefore the second term on the right is always a positive quantity since β is given to be positive Then we conclude that the slope of P constant lines is less than the slope of v constant lines PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1264 1287 It is to be shown that v v v T P T T c s and P s p T T P T c v Analysis Using the definition of cv v v v v T P P s T T s T c Substituting the first Maxwell relation T s P s v v v v v T P T T c s Using the definition of cp P P P p T s T T s T c v v Substituting the second Maxwell relation s P T P s v P s p T T P T c v 1288 It is to be proven that for a simple compressible substance T P s u v Analysis The proof is simply obtained as T P T P s u u s s u v v v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1265 1289 It is to be proven by using the definitions of pressure and temperature v s u T and s u P v that for ideal gases the development of the constantpressure specific heat yields 0 P T h Analysis The definition for enthalpy is Pv u h Then T T T T P P P P P u P h v v Assume usv u Then v v v d u ds s u du s T s T T P u P s s u P u v v v T T P T P P T P P T T P u v v v v v v v v T T T T P T P P P P T P h For ideal gases P R P P RT T v v and Then 0 v v v P TR P h T PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1266 1290 It is to be proven by using the definitions of pressure and temperature v s u T and s u P v that for ideal gases the development of the constantvolume specific heat yields 0 T u v Analysis Assume usv u Then v v v d u ds s u du s P s T u s s u u T T s T T v v v v v v v From Maxwell equation P T P T P s T T v v For ideal gases v v v R T P RT P and Then 0 P P P T R u T v v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1267 1291 Expressions for h u so Pr and vr for an ideal gas whose cp o is given by n i r i p T T a c ln o are to be developed Analysis By making the change in variable x lnTTr the enthalpy of this substance relative to a reference state is given by 1 1 1 1 2 1 2 1 ref ref n i x n n i i n x i x n i x n n i i n x i x e a e dx x a c dT h i n i n ref i n ref i n ref i n i n i n i n x i x x x i n i T T p Similarly s0 is given by 1 12 2 2 1 12 2 2 2 2 1 2 1 2 2 ref ref n i x n n i i x n i x n i x n n i i x n i x e a dx e x a T dT T c s i n i n ref i n ref i n ref i n i n i n n i n i x i x x x i n r i T T p o With these two results R s r e P P h u o v According to the du form of Gibbs equations v R dv T du Noting that for ideal gases R and c c p v c dT du v this expression reduces to v R dv T R dT c p When this is integrated between the reference and actual states the result is ref ref ln ln v v R T T R T dT c p Solving this for the specific volume ratio gives R T R T s ref o exp ref v v The ratio of the specific volumes at two states which have the same entropy is then R T R T s s exp 1 2 o 1 o 2 ref v v Inspection of this result gives R RT so r exp v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1268 1292 It is to be shown that the position of the JouleThompson coefficient inversion curve on the TP plane is given by ZTP 0 Analysis The inversion curve is the locus of the points at which the JouleThompson coefficient µ is zero 0 1 µ v v P p T T c which can also be written as 0 P ZRT T T P v a since it is given that P v ZRT b Taking the derivative of b with respect to T holding P constant gives Z T P T R T P ZRT T P P P Z v Substituting in a 0 0 0 P P P T Z Z Z T Z T P ZRT Z T Z T P TR which is the desired relation PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1269 1293 It is to be shown that for an isentropic expansion or compression process Pv k constant It is also to be shown that the isentropic expansion exponent k reduces to the specific heat ratio cpcv for an ideal gas Analysis We note that ds 0 for an isentropic process Taking s sP v the total differential ds can be expressed as 0 v v v d s dP P s ds P a We now substitute the Maxwell relations below into a to get 0 and v v v v v d T P dP T T P s T P s s s s P s Rearranging 0 0 v v v v d P dP d T P T dP s s s Dividing by P 0 1 v v d P P P dP s b We now define isentropic expansion exponent k as s P P k v v Substituting in b 0 v k dv P dP Taking k to be a constant and integrating Thus constant constant ln constant ln ln k k P P k P v v v To show that k cpcv for an ideal gas we write the cyclic relations for the following two groups of variables 1 1 1 1 d P T s P T c P T s P T s T P s c T s T c T s T s T s s T p s T P s T s T v v v v v v v where we used the relations P p T s T c T s T c and v v Setting Eqs c and d equal to each other s T s T p T s T c P T s P T c v v v or s T s T s T s T p P P T T P s P s T s T P P s c c v v v v v v v but P P P RT P T T v v Substituting k P P c c s p v v v which is the desired relation PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1270 1294 The cp of nitrogen at 300 kPa and 400 K is to be estimated using the relation given and its definition and the results are to be compared to the value listed in Table A2b Analysis a We treat nitrogen as an ideal gas with R 0297 kJkgK and k 1397 Note that PTkk1 C constant for the isentropic processes of ideal gases The cp relation is given as 1 1 1 1 1 1 1 1 1 T k kP T PT k k CT k k T P CT P P R T P RT T T P T c k k k k k k s k k P P s p v v v Substituting 1045 kJkg K 1 1 397 1397029 7 kJkg K 1 1 k kR P R T k kP T c p b The cp is defined as cp T P h Replacing the differentials by differences 1045 kJkg K 390 K 410 11347 280 kJkg 11932 390 K 410 390 K 410 kPa 300 h K h T h c P p Compare Table A2b at 400 K cp 1044 kJkgK 1295 The temperature change of steam and the average JouleThompson coefficient during a throttling process are to be estimated Analysis The enthalpy of steam at 45 MPa and T 300C is h 29442 kJkg Now consider a throttling process from this state to 25 MPa The temperature of the steam at the end of this throttling process will be 27372 C 2 kJkg 2944 MPa 52 2 T h P Thus the temperature drop during this throttling process is 2628C 300 27372 1 2 T T T The average JouleThomson coefficient for this process is determined from 1314CMPa µ 45 MPa 25 300 C 27372 h 3204 7 kJkg h P T P T PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1272 1297 Problem 1296 is reconsidered The problem is to be solved assuming steam is the working fluid by using the generalized chart method and EES data for steam The power output and the exergy destruction rate for these two calculation methods against the turbine exit pressure are to be plotted Analysis The problem is solved using EES and the results are tabulated and plotted below Input Data T1800 K P19000 kPa Vel1100 ms T2450 K P21500 kPa Vel2150 ms Qdotout20 kW To25273 K mdot3 kgs NameSteamiapws Tcritical6473 K Pcritical22090 kPa FluidH2O Ru8314 MmolarmassFluid RRuM IDEAL GAS SOLUTION State 1 hideal1enthalpyFluidTT1 Enthalpy of ideal gas sideal1entropyFluid TT1 PP1 Entropy of ideal gas State 2 hideal2enthalpyFluidTT2 Enthalpy of ideal gas sideal2entropyFluid TT2 PP2 Entropy of ideal gas Conservation of Energy Steadyflow EdotinEdotout mdothideal1Vel122convertm2s2kJkgmdothideal2Vel222convertm2s2kJkgQ dotoutWdotoutideal Second Law analysis SdotinSdotoutSdotgen 0 mdotsideal1 mdotsideal2 QdotoutTo Sdotgenideal 0 Exergy Destroyed Xdotdestroyedideal ToSdotgenideal COMPRESSABILITY CHART SOLUTION State 1 Tr1T1Tcritical Pr1P1Pcritical Z1COMPRESSTr1 Pr1 DELTAh1ENTHDEPTr1 Pr1RTcritical Enthalpy departure hchart1hideal1DELTAh1 Enthalpy of real gas using charts DELTAs1ENTRDEPTr1 Pr1R Entropy departure schart1sideal1DELTAs1 Entropy of real gas using charts State 2 Tr2T2Tcritical Pr2P2Pcritical Z2COMPRESSTr2 Pr2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1273 DELTAh2ENTHDEPTr2 Pr2RTcritical Enthalpy departure DELTAs2ENTRDEPTr2 Pr2R Entropy departure hchart2hideal2DELTAh2 Enthalpy of real gas using charts schart2sideal2DELTAs2 Entropy of real gas using charts Conservation of Energy Steadyflow EdotinEdotout mdothchart1Vel122convertm2s2kJkgmdothchart2Vel222convertm2s2kJkgQ dotoutWdotoutchart Second Law analysis SdotinSdotoutSdotgen 0 mdotschart1 mdotschart2 QdotoutTo Sdotgenchart 0 Exergy Destroyed Xdotdestroyedchart ToSdotgenchartkW SOLUTION USING EES BUILTIN PROPERTY DATA At state 1 hees1enthalpyNameTT1PP1 sees1entropyNameTT1PP1 At state 2 hees2enthalpyNameTT2PP2 sees2entropyNameTT2PP2 Conservation of Energy Steadyflow EdotinEdotout mdothees1Vel122convertm2s2kJkgmdothees2Vel222convertm2s2kJkgQd otoutWdotoutees Second Law analysis SdotinSdotoutSdotgen 0 mdotsees1 mdotsees2 QdotoutTo Sdotgenees 0 Exergy Destroyed Xdotdestroyedees ToSdotgenees PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1274 P2 kPa T2 K Woutchart kW Woutees kW Woutideal kW Xdestroyedchart kW Xdestroyedees kW Xdestroyedideal kW 1100 200 300 400 500 600 700 800 900 450 450 450 450 450 450 450 450 450 1822 1829 1837 1844 1852 1859 1867 1874 1882 1836 1853 1871 1890 1909 1929 1950 1971 1994 2097 2097 2097 2097 2097 2097 2097 2097 2097 9091 620 4497 3279 2327 1543 8751 2918 227 9057 6109 4344 3062 2042 1186 4406 2256 8347 836 5501 3828 2641 172 9679 3319 219 705 100 200 300 400 500 600 700 800 900 1800 1850 1900 1950 2000 2050 2100 2150 2200 P2 kPa Wout kW chart ees ideal 100 200 300 400 500 600 700 800 900 400 200 0 200 400 600 800 1000 P2 kPa Xdestroyed kW Xdestroyedideal Xdestroyedideal Xdestroyedees Xdestroyedees Xdestroyedchart Xdestroyedchart PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1275 1298 An adiabatic storage tank that is initially evacuated is connected to a supply line that carries nitrogen A valve is opened and nitrogen flows into the tank The final temperature in the tank is to be determined by treating nitrogen as an ideal gas and using the generalized charts and the results are to be compared to the given actual value Assumptions 1 Uniform flow conditions exist 2 Kinetic and potential energies are negligible Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance m m m m m m m i in out system out initial since 2 0 Energy balance E E E m h m u i i in out system 0 2 2 10 MPa V1 02 m3 Initially evacuated N2 Combining the two balances u2 hi a From the ideal gas property table of nitrogen at 225 K we read u h h i 2 6 537 225 K kJ kmol The temperature that corresponds to this u2 value is 74 error T2 3148 K b Using the generalized enthalpy departure chart hi is determined to be 90 2 95 3 39 10 1 78 126 2 225 ideal cr cr cr u i i h i i i R i i R T R h h Z P P P T T T Fig A29 Thus and 593 kJkmol 5 5 593 kJkmol 8 314 126 2 90 6 537 90 2 cr ideal i u i i h u R T h h Try T2 280 K Then at PR2 295 and TR2 222 we read Z2 098 and cr 2 2 R T h h u ideal 055 Thus 5 283 kJkmol 8 314 280 0 98 564 7 7 564 kJkmol 8 314 126 2 0 55 8141 55 0 2 2 2 cr 2ideal 2 ZR T h u R T h h u u Try T2 300 K Then at PR2 295 and TR2 238 we read Z2 10 and cr 2 2 R T h h u ideal 050 Thus 5 704 kJkmol 8 314 300 01 198 8 8198 kJkmol 8 314 126 2 0 50 8 723 50 0 2 2 2 cr 2ideal 2 ZR T h u R T h h u u By linear interpolation 06 error 2 2947 K T PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1276 1299E Methane is to be adiabatically and reversibly compressed in a steadyflow device The specific work required for this compression is to be determined using the departure charts and treating the methane as an ideal gas with temperature variable specific heats Properties The properties of methane are Table A1E 673 psia 343 9 R 0 1238 Btulbm R 16043 lbmlbmol cr cr P T R M Analysis The temperature at the exit state may be determined by the fact that the process is isentropic and the molar entropy change between the inlet and exit is zero When the expression of Table A2Ec is substituted for cp and the integration performed we obtain 1 2 3 1 3 2 2 1 2 2 1 2 1 2 1 2 2 1 2 1 2 2 1 1 ideal 2 ln 3 2 ln ln ln P P R T d T T c T T b T T T a P P R dT dT cT b T a P P R dT T c s s u u u p 500 psia Methane Substituting 50 1 9858 ln 500 560 3 4510 10 0 560 2 0 09352 10 560 0 006666 4 75ln 560 0 3 3 2 9 2 2 2 5 2 2 T T T T 50 psia 100F Solving this equation by EES or an iterative solution gives 2 892 R T When en energy balance is applied to the compressor it becomes Btulbmol 3290 560 892 4 4510 10 0 560 892 3 0 09352 10 560 892 2 0 006666 560 75892 4 4 3 2 4 4 9 3 3 5 2 2 4 1 4 2 3 1 3 2 2 1 2 2 1 2 2 1 3 2 2 1 1 ideal 2 in T d T T c T T b T T T a dT dT cT bT a c dT h h w p The work input per unit mass basis is 2051 Btulbm 16043 lbmlbmol 3290 Btulbmol in in M w w The enthalpy departures of propane at the specified states are determined from the generalized charts to be Fig A29 or from EES 0 0332 0 0743 673 50 1 628 343 9 560 1 cr 1 1 cr 1 1 h R R Z P P P T T T and 0 0990 0 743 673 500 2 594 343 9 892 2 cr 2 2 cr 2 2 h R R Z P P P T T T The work input is determined to be 2023 Btulbm 0 0332 0 1238 Btulbm R343 9 R 0 0990 1 Btulbm 205 1 2 cr 1 ideal 2 1 2 in h h Z Z RT h h h h w PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1277 12100 The volume expansivity of water is given The change in volume of water when it is heated at constant pressure is to be determined Properties The volume expansivity of water is given to be 0207106 K1 at 20C Analysis We take v v P T Its total differential is dP P dT T d T P v v v which for a constant pressure process reduces to dT T d P v v Dividing by v and using the definition of β dT dT T d P β v v v v 1 Taking β to be a constant integration from 1 to 2 yields or 1 2 1 2 1 2 1 2 exp ln T T T T β β v v v v Substituting the given values and noting that for a fixed mass V2V1 v2v1 3 1 6 3 1 2 1 2 50000414 m 0 10 C 50 K 10 0 207 exp m 50 exp T β T V V Therefore 414 cm3 3 1 2 0 00000414 m 50 0 50000414 V V V PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1278 12101 The work done by the refrigerant 134a as it undergoes an isothermal process in a closed system is to be determined using the tabular EES data and the generalized charts Analysis The solution using EES builtin property data is as follows 1 1484 kJkgK 25 kJkg 264 MPa 10 C 40 3916 kJkgK 0 37 kJkg 106 MPa 2 C 40 2 2 2 2 1 1 1 1 s u P T s u P T 791 kJkg 10637 26425 23700 23700 kJkg 27315 K 0 7568 kJkgK 40 07568 kJkgK 0 3916 1484 1 1 2 EES EES EES 1 EES 1 2 EES u u q w s T q s s s For the generalized chart solution we first determine the following factors using EES as 0 02635 0 03396 and 0 9857 0 02464 4 059 10 0 8369 374 2 2 313 5 147 4 82 and 0 08357 0 4927 4 059 2 0 8369 374 2 2 313 2 2 2 2 2 2 2 1 1 1 1 1 1 1 s h cr R cr R s h cr R cr R Z Z Z P P P T T T Z Z Z P P P T T T Then 0 4194 kJkgK 5 147 0 08148 kJkgK 14697 kJkg 4 82 0 08148 kJkgK3742 K 1 1 cr 1 1 Z R s RT Z h s h 0 002147 kJkgK 0 02635 0 08148 kJkgK 1 04 kJkg 0 03396 0 08148 kJkgK3742 K 2 2 cr 2 2 R Z s RT Z h s h 0 2441 kJ kg K 2 0 08148 kJkgKln 01 ln 1 2 ideal P P R s 06613 kJkg K 0 4194 0 002147 0 2441 1 2 ideal chart s s s s 20709 kJkg 27315 K 0 6613 kJkgK 40 chart 1 chart s T q 842 kJkg 12292 09 207 12292 kJkg 0 08357 0 08148313 2 0 9857 0 08148313 2 14697 1 04 0 chart chart chart 1 1 2 2 1 2 ideal chart u q w Z RT Z RT h h h u The copy of the EES solution of this problem is given next Input data TcriticalTCRITR134a K PcriticalPCRITR134a kpa T14027315K T2T1K P12000kPa PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1279 P2100kPa Ru8314kJkmolK MmolarmassR134a RRuMkJkgK SOLUTION USING EES BUILTIN PROPERTY DATA For the isothermal process the heat transfer is Ts2 s1 DELTAsEESentropyR134aTT2PP2entropyR134aTT1PP1 qEEST1DELTAsEES s2entropyR134aTT2PP2 s1entropyR134aTT1PP1 Conservation of energy for the closed system DELTAuEESintEnergyR134aTT2pP2intEnergyR134aTT1PP1 qEESwEESDELTAuEES u1intEnergyR134aTT1PP1 u2intEnergyR134aTT2pP2 COMPRESSABILITY CHART SOLUTION State 1 Tr1T1Tcritical pr1p1pcritical Z1COMPRESSTr1 Pr1 DELTAh1ENTHDEPTr1 Pr1RTcriticalEnthalpy departure Zh1ENTHDEPTr1 Pr1 DELTAs1ENTRDEPTr1 Pr1R Entropy departure Zs1ENTRDEPTr1 Pr1 State 2 Tr2T2Tcritical Pr2P2Pcritical Z2COMPRESSTr2 Pr2 DELTAh2ENTHDEPTr2 Pr2RTcriticalEnthalpy departure Zh2ENTHDEPTr2 Pr2 DELTAs2ENTRDEPTr2 Pr2R Entropy departure Zs2ENTRDEPTr2 Pr2 Entropy Change DELTAsideal RlnP2P1 DELTAschartDELTAsidealDELTAs2DELTAs1 For the isothermal process the heat transfer is Ts2 s1 qchartT1DELTAschart Conservation of energy for the closed system DELTAhideal0 DELTAuchartDELTAhidealDELTAh2DELTAh1Z2RT2Z1RT1 qchartwchartDELTAuchart PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1280 SOLUTION DELTAh114697 kJkg DELTAh2104 kJkg DELTAhideal0 kJkg DELTAs104194 kJkgK DELTAs20002147 kJkgK DELTAschart06613 kJkgK DELTAsEES07568 kJkgK DELTAsideal02441 kJkgK DELTAuchart12292 kJkg DELTAuEES1579 kJkg M102 P12000 kPa P2100 kPa pr104927 Pr2002464 Pcritical4059 kpa qchart20709 kJkg qEES23700 kJkg R008148 kJkgK Ru8314 kJkmolK s103916 s211484 kJkgK T13132 K T23132 K Tr108369 Tr208369 Tcritical3742 K u110637 u226425 kJkg wchart8418 kJkg wEES7912 kJkg Z1008357 Z209857 Zh1482 Zh2003396 Zs15147 Zs2002635 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1281 12102 The heat transfer work and entropy changes of methane during a process in a pistoncylinder device are to be determined assuming ideal gas behavior using generalized charts and real fluid EES data Analysis The ideal gas solution Properties are obtained from EES State 1 0 04834 m kg 4000 kPa 27315 K 0 5182 kJkgK 100 4685 kJkg 27315 0 5182100 4492 1022 kJkgK 4 MPa C 100 4492 kJkg C 100 3 1 1 1 1 1 1 1 1 1 1 1 P T R RT h u s P T h T v State 2 0 08073 m kg 4000 kPa 27315 K 0 5182 kJkgK 350 4093 kJkg 27315 0 5182350 3770 1168 kJkgK 4 MPa C 350 3770 kJkg C 350 3 2 2 2 2 2 2 2 2 2 2 2 P T R RT h u s P T h T v 12956 kJkg 4000 kPa008073004834m kg 3 1 2 ideal v P v w 72170 kJkg 4685 4093 12956 1 2 ideal ideal u u w q 146 kJkg 1022 1168 1 2 ideal s s s For the generalized chart solution we first determine the following factors using EES as 0 2555 0 4318 and 0 9023 0 5413 7 39 4 1 227 304 2 373 1 1 1 cr 1 1 cr 1 1 s h R R Z Z Z P P P T T T 0 06446 0 1435 and 0 995 0 5413 7 39 4 2 048 304 2 623 2 2 2 cr 2 2 cr 2 2 s h R R Z Z Z P P P T T T State 1 4734 kJkg 0 9023 0 518237315 4560 4560 kJkg 6807 4492 6807 kJkg 0 4318 0 5182 kJkgK3042 K 1 1 1 1 1 1ideal 1 cr 1 1 Z RT h u h h h RT Z h h 0 04362 m kg 4000 0 9023 0 5182 37315 3 1 1 1 1 P Z R T v 1009 kJkgK 0 1324 22 10 0 1324 kJkgK 0 2555 0 5182 kJkgK 1 1ideal 1 1 1 s s s Z R s s State 2 4114 kJkg 0 995 0 518262315 3793 3793 kJkg 2262 3770 2262 kJkg 0 1435 0 5182 kJkgK3042 K 2 2 2 2 2 2ideal 2 cr 2 2 Z RT h u h h h RT Z h h PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1282 0 08033 m kg 4000 0 995 0 5182 62315 3 2 2 2 2 P Z R T v 1165 kJkgK 0 03341 68 11 0 03341 kJkgK 0 06446 0 5182 kJkgK 2 2ideal 2 2 2 s s s R Z s s Then 14684 kJkg 4000 kPa008033004362m kg 3 1 2 chart v P v w 76684 kJkg 4734 4114 14684 1 2 chart chart u u w q 156 kJkg 1009 1165 1 2 chart s s s The solution using EES builtin property data is as follows 439 kJkgK 1 82 kJkg 39 04717 m kg 0 MPa 4 C 100 1 1 3 1 1 1 s u P T v 06329 kJkgK 0 52 kJkg 564 08141 m kg 0 MPa 4 C 350 2 2 3 2 2 2 s u P T v 13696 kJkg 4000 kPa008141004717m kg 3 1 2 EES v P v w 74131 kJkg 3982 56452 13697 1 2 EES EES u u w q 150 kJkg 1 439 0 06329 1 2 EES s s s PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1283 12103E Methane is compressed steadily The entropy change and the specific work required are to be determined using the departure charts and the property tables Properties The properties of methane are Table A1E 0 1238 Btulbm R 0 532 Btulbm R 673 psia 343 9 R 16043 lbmlbmol cr cr R c P T M p Analysis a Using empirical correlation for the cp of methane as given in Table A2Ec gives 288 Btulbmol R 12 560 1560 4 4510 10 0 560 1560 3 0 09352 10 560 1560 2 0 6666 10 7501000 4 4 3 2 4 4 9 3 3 5 2 2 2 4 1 4 2 3 1 3 2 2 1 2 2 1 2 3 2 1 2 T d T T c T T b T T T a dT dT cT bT a c dT h h p 500 psia 1100F Methane 50 psia 100F The work input is equal to the enthalpy change The enthalpy change per unit mass is 7659 Btulbm 16043 lbmlbmol 12288 Btulbmol 1 2 1 2 M h h h h win Similarly the entropy change is given by 7 407 Btulbmol R 50 1 9858 ln 500 560 1560 3 4510 10 0 560 1560 2 0 09352 10 560 1560 0 6666 10 560 750ln 1560 4 ln 3 2 ln ln ln 3 3 9 2 2 5 2 1 2 3 1 3 2 2 1 2 2 1 2 1 2 1 2 2 1 2 1 2 2 1 1 ideal 2 P P R T d T T c T T b T T T a P P R dT dT cT b T a P P R dT T c s s u u u p The entropy change per unit mass is 04617 Btulbm R 16043 lbmlbmol 7 407 Btulbmol R 1 ideal 2 1 ideal 2 M s s s s b The enthalpy and entropy departures of water vapor at the specified states are determined from the generalized charts to be Figs A29 A30 or from EES We used EES 0 01617 0 03313 and 0 0743 673 50 1 63 343 9 560 1 1 cr 1 1 cr 1 1 s h R R Z Z P P P T T T and 0 00695 0 and 0 743 673 500 4 54 343 9 1560 2 2 cr 2 2 cr 2 2 s h R R Z Z P P P T T T The work input and entropy changes are Btulbm R 04628 Btulbm 7674 0 01617 0 1238 0 00695 4617 0 0 03313 0 1238343 9 0 9 765 1 2 1 ideal 2 1 2 1 2 cr 1 ideal 2 1 2 s s h h in Z R Z s s s s Z Z RT h h h h w PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1284 12104E Methane is compressed in a steadyflow device The secondlaw efficiency of the compression process is to be determined Analysis The reversible work input to the compressor is determined from 518 8 Btulbm 537 R 0 4628 Btulbm R 767 4 Btulbm 1 2 0 1 2 rev s s T h h w The secondlaw efficiency of the compressor is 676 0 676 767 4 8 518 actual rev II w w η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1285 Fundamentals of Engineering FE Exam Problems 12105 A substance whose JouleThomson coefficient is negative is throttled to a lower pressure During this process select the correct statement a the temperature of the substance will increase b the temperature of the substance will decrease c the entropy of the substance will remain constant d the entropy of the substance will decrease e the enthalpy of the substance will decrease Answer a the temperature of the substance will increase 12106 Consider the liquidvapor saturation curve of a pure substance on the PT diagram The magnitude of the slope of the tangent line to this curve at a temperature T in Kelvin is a proportional to the enthalpy of vaporization hfg at that temperature b proportional to the temperature T c proportional to the square of the temperature T d proportional to the volume change vfg at that temperature e inversely proportional to the entropy change sfg at that temperature Answer a proportional to the enthalpy of vaporization hfg at that temperature 12107 Based on the generalized charts the error involved in the enthalpy of CO2 at 300 K and 5 MPa if it is assumed to be an ideal gas is a 0 b 9 c 16 d 22 e 27 Answer e 27 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values T300 K P5000 kPa PcrPCRITCarbonDioxide TcrTCRITCarbonDioxide TrTTcr PrPPcr hRENTHDEPTr Pr hideal11351MolarmassCO2 Table A20 of the text hcharthidealRTcrhR R01889 ErrorhcharthidealhchartConvert PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 131 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 13 GAS MIXTURES PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 132 Composition of Gas Mixtures 131C The ratio of the mass of a component to the mass of the mixture is called the mass fraction mf and the ratio of the mole number of a component to the mole number of the mixture is called the mole fraction y 132C The mass fractions will be identical but the mole fractions will not 133C Yes 134C Yes because both CO2 and N2O has the same molar mass M 44 kgkmol 135C No We can do this only when each gas has the same mole fraction 136C It is the average or the equivalent molar mass of the gas mixture No 137 From the definition of mass fraction m i i m m i i m i i M M y N M N M m m mf PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 133 138 A mixture consists of two gases Relations for mole fractions when mass fractions are known are to be obtained Analysis The mass fractions of A and B are expressed as m B B B m A A m m A A m A A M M y M M y N M N M m m and mf mf Where m is mass M is the molar mass N is the number of moles and y is the mole fraction The apparent molar mass of the mixture is B B A A m B B A A m m m y M y M N N M N M N m M Combining the two equation above and noting that 1 B A y y gives the following convenient relations for converting mass fractions to mole fractions 1 1 mf B A A B A M M M y and 1 A B y y which are the desired relations 139 The definitions for the mass fraction weight and the weight fractions are total total wf mf W W mg W m m i i i i Since the total system consists of one mass unit the mass of the ith component in this mixture is mi The weight of this one component is then i i g W mf Hence the weight fraction for this one component is i i i i g g mf mf mf wf PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 134 1310E The moles of components of a gas mixture are given The mole fractions and the apparent molecular weight are to be determined Properties The molar masses of He O2 N2 and H2O are 40 320 280 and 180 lbmlbmol respectively Table A1 Analysis The total mole number of the mixture is lbmol 37 52 30 51 3 N2 H2O O2 He N N N N N m and the mole fractions are 0343 00411 0206 0411 73 lbmol lbmol 25 73 lbmol lbmol 03 73 lbmol lbmol 15 73 lbmol lbmol 3 N2 N2 H2O H2O O2 O2 He He m m m m N N y N N y N N y N N y 3 lbmol He 15 lbmol O2 03 lbmol H2O 25 lbmol N2 The total mass of the mixture is 135 4 lbm lbm28 lbmlbmol 52 lbm18 lbmlbmol 30 lbm32 lbmlbmol 51 3lbm4 lbmlbmol N2 N2 H2O H2O O2 O2 He He N2 H2O O2 He M N M N M N M N m m m m mm Then the apparent molecular weight of the mixture becomes 186 lbmlbmol 73 lbmol lbm 1354 m m m N m M PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 135 1311 The masses of the constituents of a gas mixture are given The mass fractions the mole fractions the average molar mass and gas constant are to be determined Properties The molar masses of O2 N2 and CO2 are 320 280 and 440 kgkmol respectively Table A1 Analysis a The total mass of the mixture is 23 kg 10 kg 8 kg 5 kg 2 2 2 CO N O m m m mm 5 kg O2 8 kg N2 10 kg CO2 Then the mass fraction of each component becomes 0435 0348 0217 23 kg 10 kg mf 23 kg 8 kg mf 23 kg 5 kg mf 2 2 2 2 2 2 CO CO N N O O m m m m m m m m m b To find the mole fractions we need to determine the mole numbers of each component first 0227 kmol 44 kgkmol kg 10 0286 kmol 28 kgkmol kg 8 0156 kmol 32 kgkmol kg 5 2 2 2 2 2 2 2 2 2 CO CO CO N N N O O O M m N M m N M m N Thus 0669 kmol 0227 kmol 0286 kmol 0 15 6 kmol 2 2 2 CO N O N N N N m and 0339 0428 0233 0669 kmol kmol 0227 0669 kmol kmol 0286 0699 kmol kmol 0156 2 2 2 2 2 2 CO CO N N O O m m m N N y N N y N N y c The average molar mass and gas constant of the mixture are determined from their definitions and kJkg K 0242 kgkmol 344 344 kgkmol kJkmol K 8314 0669 kmol kg 23 m u m m m m M R R N m M PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 136 1312 The mass fractions of the constituents of a gas mixture are given The mole fractions of the gas and gas constant are to be determined Properties The molar masses of CH4 and CO2 are 160 and 440 kgkmol respectively Table A1 Analysis For convenience consider 100 kg of the mixture Then the number of moles of each component and the total number of moles are 0568 kmol 44 kgkmol 25 kg kg 25 4688 kmol 16 kgkmol 75 kg kg 75 2 2 2 2 4 4 4 4 CO CO CO CO CH CH CH CH M m N m M m N m mass 75 CH4 25 CO2 5256 kmol 0568 kmol 4 688 kmol 2 4 CO CH N N N m Then the mole fraction of each component becomes 108 892 0108 or 5256 kmol kmol 0568 0892 or 5256 kmol kmol 4688 2 2 4 4 CO CO CH CH m m N N y N N y The molar mass and the gas constant of the mixture are determined from their definitions and kJkg K 0437 1903 kgkmol kJkmol K 8314 1903 kg kmol 5256 kmol kg 100 m u m m m m M R R N m M 1313 The mole numbers of the constituents of a gas mixture are given The mass of each gas and the apparent gas constant are to be determined Properties The molar masses of H2 and N2 are 20 and 280 kgkmol respectively Table A1 Analysis The mass of each component is determined from kg 112 kg 10 4 kmol 28 kgkmol kmol 4 5 kmol 20 kgkmol kmol 5 2 2 2 2 2 2 2 2 N N N N H H H H M N m N M N m N 5 kmol H2 4 kmol N2 The total mass and the total number of moles are 9 kmol 4 kmol kmol 5 122 kg 112 kg kg 10 2 2 2 2 N H N H N N N m m m m m The molar mass and the gas constant of the mixture are determined from their definitions and 0613 kJkg K 1356 kgkmol kJkmol K 8314 1356 kg kmol 9 kmol kg 122 m u m m m m M R R N m M PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 137 1314 The mass fractions of the constituents of a gas mixture are given The volumetric analysis of the mixture and the apparent gas constant are to be determined Properties The molar masses of O2 N2 and CO2 are 320 28 and 440 kgkmol respectively Table A1 Analysis For convenience consider 100 kg of the mixture Then the number of moles of each component and the total number of moles are 1 136 kmol 44 kgkmol 50 kg kg 50 1 071 kmol 28 kgkmol 30 kg kg 20 0 625 kmol 32 kgkmol 20 kg kg 20 2 2 2 2 2 2 2 2 2 2 2 2 CO CO CO CO N N N N O O O O M m N m M m N m M m N m mass 20 O2 30 N2 50 CO2 2 832 kmol 1 136 1 071 0 625 2 2 2 CO N O N N N N m Noting that the volume fractions are same as the mole fractions the volume fraction of each component becomes 401 378 221 0401 or 2832 kmol kmol 1136 0378 or 2832 kmol kmol 1071 0221 or 2832 kmol kmol 0625 2 2 2 2 2 2 CO CO N N O O m m m N N y N N y N N y The molar mass and the gas constant of the mixture are determined from their definitions 3531 kg kmol 2832 kmol 100 kg m m m N m M and 0235 kJkg K 3531 kgkmol kJkmol K 8314 m u m M R R PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 138 PvT Behavior of Gas Mixtures 1315C Normally yes Air for example behaves as an ideal gas in the range of temperatures and pressures at which oxygen and nitrogen behave as ideal gases 1316C The pressure of a gas mixture is equal to the sum of the pressures each gas would exert if existed alone at the mixture temperature and volume This law holds exactly for ideal gas mixtures but only approximately for real gas mixtures 1317C The volume of a gas mixture is equal to the sum of the volumes each gas would occupy if existed alone at the mixture temperature and pressure This law holds exactly for ideal gas mixtures but only approximately for real gas mixtures 1318C The PvT behavior of a component in an ideal gas mixture is expressed by the ideal gas equation of state using the properties of the individual component instead of the mixture Pivi RiTi The PvT behavior of a component in a real gas mixture is expressed by more complex equations of state or by Pivi ZiRiTi where Zi is the compressibility factor 1319C Component pressure is the pressure a component would exert if existed alone at the mixture temperature and volume Partial pressure is the quantity yiPm where yi is the mole fraction of component i These two are identical for ideal gases 1320C Component volume is the volume a component would occupy if existed alone at the mixture temperature and pressure Partial volume is the quantity yiVm where yi is the mole fraction of component i These two are identical for ideal gases 1321C The one with the highest mole number 1322C The partial pressures will decrease but the pressure fractions will remain the same 1323C The partial pressures will increase but the pressure fractions will remain the same 1324C No The correct expression is the volume of a gas mixture is equal to the sum of the volumes each gas would occupy if existed alone at the mixture temperature and pressure PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 139 1325C No The correct expression is the temperature of a gas mixture is equal to the temperature of the individual gas components 1326C Yes it is correct 1327C With Kays rule a realgas mixture is treated as a pure substance whose critical pressure and temperature are defined in terms of the critical pressures and temperatures of the mixture components as i i m i i m y T T y P P cr cr cr cr and The compressibility factor of the mixture Zm is then easily determined using these pseudocritical point values 1328 The partial pressure of R134a in atmospheric air to form a 100ppm contaminant is to be determined Analysis Noting that volume fractions and mole fractions are equal the molar fraction of R134a in air is 0 0001 10 100 6 R134a y The partial pressure of R134a in air is then 001 kPa 0 0001100 kPa R134a R134a Pm y P 1329 A tank contains a mixture of two gases of known masses at a specified pressure and temperature The mixture is now heated to a specified temperature The volume of the tank and the final pressure of the mixture are to be determined Assumptions Under specified conditions both Ar and N2 can be treated as ideal gases and the mixture as an ideal gas mixture Analysis The total number of moles is and 233 m3 250 kPa kmol831 4 kPa m kmol K280 K 25 25 kmol 2 kmol kmol 50 3 N Ar 2 m m u m m m P R T N N N N V 05 kmol Ar 2 kmol N2 280 K 250 kPa Q Also 3571 kPa 280 K 250 kPa 400 K 1 1 2 2 1 1 1 2 2 2 T P T P T P T P V V PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1310 1330 The volume fractions of components of a gas mixture are given The mass fractions and apparent molecular weight of the mixture are to be determined Properties The molar masses of H2 He and N2 are 20 40 and 280 kgkmol respectively Table A1 Analysis We consider 100 kmol of this mixture Noting that volume fractions are equal to the mole fractions mass of each component are 840 kg 30 kmol28 kgkmol 16 0 kg 40 kmol4 kgkmol 6 0 kg 30 kmol2 kgkmol N2 N2 N2 He He He H2 H2 H2 M N m M N m M N m 30 H2 40 He 30 N2 by volume The total mass is 1060 kg 840 160 60 N2 He H2 N m m mm Then the mass fractions are 07925 01509 005660 1060 kg 840 kg mf 1060 kg 160 kg mf 1060 kg 60 kg mf N2 N2 He He H2 H2 m m m m m m m m m The apparent molecular weight of the mixture is 1060 kgkmol 100 kmol kg 1060 m m m N m M PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1311 1331 The partial pressures of a gas mixture are given The mole fractions the mass fractions the mixture molar mass the apparent gas constant the constantvolume specific heat and the specific heat ratio are to be determined Properties The molar masses of CO2 O2 and N2 are 440 320 and 280 kgkmol respectively Table A1 The constant volume specific heats of these gases at 300 K are 0657 0658 and 0743 kJkgK respectively Table A2a Analysis The total pressure is 100 kPa 50 37 5 12 5 N2 O2 CO2 total P P P P Partial pressures CO2 125 kPa O2 375 kPa N2 50 kPa The volume fractions are equal to the pressure fractions Then 050 0375 0125 100 50 100 5 37 100 5 12 total N2 N2 total O2 O2 total CO2 CO2 P P y P P y P P y We consider 100 kmol of this mixture Then the mass of each component are 1400 kg 50 kmol28 kgkmol 1200 kg 37 5 kmol32 kgkmol 550 kg 12 5 kmol44 kgkmol N2 N2 N2 O2 O2 O2 CO2 CO2 CO2 M N m M N m M N m The total mass is 3150 kg 1400 1200 550 Ar O2 N2 m m m mm Then the mass fractions are 04444 03810 01746 3150 kg 1400 kg mf 3150 kg 1200 kg mf 3150 kg 550 kg mf N2 N2 O2 O2 CO2 CO2 m m m m m m m m m The apparent molecular weight of the mixture is 3150 kgkmol 100 kmol kg 3150 m m m N m M The constantvolume specific heat of the mixture is determined from 06956 kJkg K 0 743 0 4444 0 658 0 3810 0 657 1746 0 mf mf mf N2 N2 O2 O2 CO2 Co2 v v v v c c c c The apparent gas constant of the mixture is 02639 kJkg K 3150 kgkmol kJkmol K 8314 m u M R R The constantpressure specific heat of the mixture and the specific heat ratio are 09595 kJkg K 0 2639 0 6956 R c c v p 1379 06956 kJkg K kJkg K 09595 v p c c k PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1312 1332 The mole numbers of combustion gases are given The partial pressure of water vapor is to be determined Analysis The total mole of the mixture and the mole fraction of water vapor are 8 06 kmol 5 65 1 66 0 75 total N 0 2060 8 06 66 1 total H2O H2O N N y Noting that molar fraction is equal to pressure fraction the partial pressure of water vapor is 2086 kPa total 0 2060101 3 kPa H2O H2O P y P 1333 An additional 5 of oxygen is mixed with standard atmospheric air The molecular weight of this mixture is to be determined Properties The molar masses of N2 and O2 are 280 and 320 kgkmol respectively Table A1 Analysis Standard air is taken as 79 nitrogen and 21 oxygen by mole That is 79 0 21 0 N2 O2 y y Adding another 005 moles of O2 to 1 kmol of standard air gives 07524 1 05 79 0 0 2476 1 05 26 0 N2 O2 y y Then 2899 kgkmol 0 7524 28 0 2476 32 N2 N2 O2 O2 M y M y M m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1313 1334 The masses temperatures and pressures of two gases contained in two tanks connected to each other are given The valve connecting the tanks is opened and the final temperature is measured The volume of each tank and the final pressure are to be determined Assumptions Under specified conditions both N2 and O2 can be treated as ideal gases and the mixture as an ideal gas mixture Properties The molar masses of N2 and O2 are 280 and 320 kgkmol respectively The gas constants of N2 and O2 are 02968 and 02598 kPam3kgK respectively Table A1 Analysis The volumes of the tanks are 3 3 m 2065 m 0322 150 kPa kg02598 kPa m kg K298 K 4 550 kPa kg02968 kPa m kg K298 K 2 3 O O 3 N N 2 2 2 2 P mRT P mRT V V 4 kg O2 25C 150 kPa 2 kg N2 25C 550 kPa 3 3 3 O N total 2386 m 2 065 m 0 322 m 2 2 V V V Also 0125 kmol 32 kgkmol kg 4 007143 kmol 28 kgkmol kg 2 2 2 2 2 2 2 O O O N N N M m N M m N 01964 kmol 0125 kmol 0 07143 kmol 2 2 O N N N Nm Thus 204 kPa 3 3 m 2386 kmol831 4 kPa m kmol K298 K 01964 m u m NR T P V PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1314 1335 The masses of components of a gas mixture are given The apparent molecular weight of this mixture the volume it occupies the partial volume of the oxygen and the partial pressure of the helium are to be determined Properties The molar masses of O2 CO2 and He are 320 440 and 40 kgkmol respectively Table A1 Analysis The total mass of the mixture is kg 61 50 1 10 He CO2 O2 m m m mm The mole numbers of each component are 01 kg O2 1 kg CO2 05 kg He 0 125 kmol 4 kgkmol kg 05 0 02273 kmol 44 kgkmol kg 1 0 003125 kmol 32 kgkmol kg 01 He He He CO2 CO2 CO2 O2 O2 O2 M m N M m N M m N The mole number of the mixture is 0 15086 kmol 0 125 0 02273 0 003125 He CO2 O2 N N N N m Then the apparent molecular weight of the mixture becomes 1061 kgkmol 015086 kmol kg 16 m m m N m M The volume of this ideal gas mixture is 3764 m3 100 kPa 01509 kmol831 4 kPa m kmol K300 K 3 P R T N u m Vm The partial volume of oxygen in the mixture is 007795 m3 01509 kmol 3764 m 0003125 kmol 3 O2 O2 O2 m m m N N y V V V The partial pressure of helium in the mixture is 8284 kPa 01509 kmol 100 kPa 0125 kmol He He He m m m P N N P y P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1315 1336 The mass fractions of components of a gas mixture are given The volume occupied by 100 kg of this mixture is to be determined Properties The molar masses of CH4 C3H8 and C4H10 are 160 440 and 580 kgkmol respectively Table A1 Analysis The mole numbers of each component are 0 2586 kmol 58 kgkmol kg 15 0 5682 kmol 44 kgkmol kg 25 3 75 kmol 16 kgkmol kg 60 C4H10 C4H10 C4H10 C3H8 C3H8 C3H8 CH4 CH4 CH4 M m N M m N M m N 60 CH4 25 C3H8 15 C4H10 by mass The mole number of the mixture is 4 5768 kmol 0 2586 0 5682 3 75 C4H10 C3H8 CH4 N N N N m The apparent molecular weight of the mixture is 2185 kgkmol 45768 kmol 100 kg m m m N m M Then the volume of this ideal gas mixture is 393 m3 3000 kPa 45768 kmol831 4 kPa m kmol K310 K 3 P R T N u m Vm PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1316 1337E The mass fractions of components of a gas mixture are given The mass of 7 ft3 of this mixture and the partial volumes of the components are to be determined Properties The molar masses of N2 O2 and He are 280 320 and 40 lbmlbmol respectively Table A1E Analysis We consider 100 lbm of this mixture for calculating the molar mass of the mixture The mole numbers of each component are 5 lbmol 4 lbmlbmol lbm 20 1 094 lbmol 32 lbmlbmol lbm 35 1 607 lbmol 28 lbmlbmol lbm 45 He He He O2 O2 O2 N2 N2 N2 M m N M m N M m N 7 ft3 45 N2 35 O2 20 He by mass The mole number of the mixture is 7 701 lbmol 5 1 094 1 607 He O2 N2 N N N N m The apparent molecular weight of the mixture is 1299 lbmlbmol 7701 lbmol 100 lbm m m m N m M Then the mass of this ideal gas mixture is 4887 lbm psia ft lbmol R520 R 1073 psia7 ft 1299 lbmlbmol 300 3 3 R T P M m u m V The mole fractions are 06493 7701 lbmol lbmol 5 0142 7701 lbmol lbmol 1094 02087 7701 lbmol lbmol 1607 He He O2 O2 N2 N2 m m m N N y N N y N N y Noting that volume fractions are equal to mole fractions the partial volumes are determined from 3 3 3 ft 4545 ft 0994 ft 1461 0 64937 ft 0 1427 ft 0 20877 ft 3 He He 3 O2 O2 3 N2 N2 m m m y y y V V V V V V PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1317 1338 The mass fractions of components of a gas mixture are given The partial pressure of ethane is to be determined Properties The molar masses of CH4 and C2H6 are 160 and 300 kgkmol respectively Table A1 Analysis We consider 100 kg of this mixture The mole numbers of each component are kmol 01 30 kgkmol kg 30 4 375 kmol 16 kgkmol kg 70 C2H6 C2H6 C2H6 CH4 CH4 CH4 M m N M m N 70 CH4 30 C2H6 by mass 100 m3 130 kPa 25C The mole number of the mixture is 5 375 kmol 01 4 375 C2H6 CH4 N N N m The mole fractions are 01861 5375 kmol kmol 10 08139 5375 kmol kmol 4375 C2H6 C2H6 CH4 CH4 m m N N y N N y The final pressure of ethane in the final mixture is 2419 kPa 0 1861130 kPa C2H6 C2H6 Pm y P 1339 A container contains a mixture of two fluids The volume of the container and the total weight of its contents are to be determined Assumptions The volume of the mixture is the sum of the volumes of the two constituents Properties The specific volumes of the two fluids are given to be 0001 m3kg and 0008 m3kg Analysis The volumes of the two fluids are given by 3 3 3 3 0016 m 2 kg0008 m kg 0001 m m kg kg0001 1 B B B A A A m m v V v V 1 kg fluid A 2 k g fluid B The volume of the container is then 0017 m3 0 016 0 001 B A V V V The total mass is 3 kg 2 1 B A m m m and the weight of this mass will be 288 N 2 2 28 8 kg ms 3kg96 ms mg W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1318 1340E A mixture consists of liquid water and another fluid The specific weight of this mixture is to be determined Properties The densities of water and the fluid are given to be 624 lbmft3 and 500 lbmft3 respectively Analysis We consider 1 ft3 of this mixture The volume of the water in the mixture is 07 ft3 which has a mass of 4368 lbm 62 4 lbmft 07 ft 3 3 w w mw V ρ 07 ft3 water 03 ft3 fluid The weight of this water is 4331 lbf lbm fts 32174 1lbf 4368 lbm319 fts 2 2 m g W w w Similarly the volume of the second fluid is 03 ft3 and the mass of this fluid is 15 lbm 50 lbmft 03 ft 3 3 f f m f V ρ The weight of the fluid is 1487 lbf lbm fts 32174 1lbf 15 lbm319 fts 2 2 m g W f f The specific weight of this mixture is then 582 lbfft3 03 ft3 07 4331 1487 lbf f w f w W W V V γ 1341 The mole fractions of components of a gas mixture are given The mass flow rate of the mixture is to be determined Properties The molar masses of air and CH4 are 2897 and 160 kgkmol respectively Table A1 Analysis The molar fraction of air is 15 CH4 85 air by mole 0 85 0 15 1 1 CH4 air y y The molar mass of the mixture is determined from 02 kgkmol 27 0 85 2897 15 16 0 air air CH4 CH4 M y M y M m Given the engine displacement and speed and assuming that this is a 4stroke engine 2 revolutions per cycle the volume flow rate is determined from 75 m min 2 revcycle 3000 revmin0005 m 2 3 3 nVd V The specific volume of the mixture is 1 127 m kg 2702 kgkmol8 0 kPa 8314 kPa m kmol K293 K 3 3 P M T R m u v Hence the mass flow rate is 665 kgmin m kg 1127 m min 75 3 3 v V m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1319 1342E The volumetric fractions of components of a natural gas mixture are given The mass and volume flow rates of the mixture are to be determined Properties The molar masses of CH4 and C2H6 are 160 and 300 lbmlbmol respectively Table A1E Analysis The molar mass of the mixture is determined from 1670 lbmlbmol 0 05 30 0 95 16 C2H6 C2H6 CH4 CH4 M y M y M m The specific volume of the mixture is 95 CH4 5 C2H6 by volume 3 341 ft lbm 1670 lbmlbmol100 psia psia ft lbmol R520 R 1073 3 3 P M T R m u v The volume flow rate is 7069 ft s 3 10 fts 4 3612 ft 4 2 2 π π D V AV V and the mass flow rate is 2116 lbms ft lbm 3341 ft s 7069 3 3 v V m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1320 1343 The mole numbers temperatures and pressures of two gases forming a mixture are given The final temperature is also given The pressure of the mixture is to be determined using two methods Analysis a Under specified conditions both Ar and N2 will considerably deviate from the ideal gas behavior Treating the mixture as an ideal gas 739 MPa 2280 K 3 MPa 6230 K state Final Initial state 1 1 1 2 2 2 2 2 2 2 1 1 1 1 N T P N T P N R T P N R T P u u V V b Initially 4 kmol N2 190 K 8 MPa 2 kmol Ar 280 K 3 MPa 0 985 0 6173 486 MPa MPa 3 1 854 1510 K K 280 Ar Ar cr 1 Ar cr 1 Z P P P T T T R R Fig A15 or EES Then the volume of the tank is 3 3 Ar 1 529 m 3000 kPa 09852 kmol831 4 kPa m kmol K280 K P R T ZN u V After mixing Ar 496 0 2 96 kPa m kmol K1510 K4860 kPa 8314 m 2 kmol 1529 1 523 1510 K K 230 3 3 crAr Ar cr Ar crAr Ar cr Ar Ar crAr A R u m u R m r R P P R T N P T R T T T V v v Fig A15 or EES N2 43 1 1 235 kPa m kmol K1262 K3390 kPa 8314 m 4 kmol 1529 1 823 1262 K K 230 3 3 crN N cr N crN N cr N N crN N 2 2 2 2 2 2 2 2 2 R u m u R m R P P T R N P T R T T T V v V Fig A15 or EES Thus 485 MPa 1 43 3 39 MPa 241 MPa 0 496 4 86 MPa 2 2 N cr N Ar cr Ar P P P P P P R R and 726 MPa 485 MPa 2 41 MPa N2 Ar P P Pm PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1321 1344 Problem 1343 is reconsidered The effect of the moles of nitrogen supplied to the tank on the final pressure of the mixture is to be studied using the idealgas equation of state and the compressibility chart with Daltons law Analysis The problem is solved using EES and the solution is given below Input Data Ru 8314 kJkmolK universal Gas Constant TAr 280 K PAr 3000 kPa Pressure for only Argon in the tank initially NAr 2 kmol NN2 4 kmol Tmix 230 K TcrAr1510 K Critical Constants are found in Table A1 of the text PcrAr4860 kPa TcrN21262 K PcrN23390 kPa Idealgas Solution PArVTankIG NArRuTAr Apply the ideal gas law the gas in the tank PmixIGVTankIG NmixRuTmix Idealgas mixture pressure NmixNAr NN2 Moles of mixture Real Gas Solution PArVTankRG ZAr1NArRuTAr Real gas volume of tank TRTArTcrAr Initial reduced Temp of Ar PRPArPcrAr Initial reduced Press of Ar ZAr1COMPRESSTR PR Initial compressibility factor for Ar PArmixVTankRG ZArmixNArRuTmix Real gas Ar Pressure in mixture TRArmixTmixTcrAr Reduced Temp of Ar in mixture PRArmixPArmixPcrAr Reduced Press of Ar in mixture ZArmixCOMPRESSTRArmix PRArmix Compressibility factor for Ar in mixture PN2mixVTankRG ZN2mixNN2RuTmix Real gas N2 Pressure in mixture TRN2mixTmixTcrN2 Reduced Temp of N2 in mixture PRN2mixPN2mixPcrN2 Reduced Press of N2 in mixture ZN2mixCOMPRESSTRN2mix PRN2mix Compressibility factor for N2 in mixture PmixPRArmixPcrAr PRN2mixPcrN2 Mixture pressure by Daltons law 23800 NN2 kmol Pmix kPa PmixIG kPa 1 2 3 4 5 6 7 8 9 10 3647 4863 6063 7253 8438 9626 10822 12032 13263 14521 3696 4929 6161 7393 8625 9857 11089 12321 13554 14786 1 2 3 4 5 6 7 8 9 10 3000 5000 7000 9000 11000 13000 15000 NN2 kmol Pmix kPa Solution Method Chart Chart Ideal Gas Ideal Gas PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1322 1345E The mass fractions of gases forming a mixture at a specified pressure and temperature are given The mass of the gas mixture is to be determined using four methods Properties The molar masses of CH4 and C2H6 are 160 and 300 lbmlbmol respectively Table A1E Analysis a We consider 100 lbm of this mixture Then the mole numbers of each component are 0 8333 lbmol 30 lbmlbmol lbm 25 4 6875 lbmol 16 lbmlbmol lbm 75 C2H6 C2H6 C2H6 CH4 CH4 CH4 M m N M m N 75 CH4 25 C2H6 by mass 2000 psia 300F The mole number of the mixture and the mole fractions are 5 5208 lbmol 0 8333 4 6875 Nm 01509 55208 lbmol lbmol 08333 08491 55208 lbmol lbmol 46875 C2H6 C2H6 CH4 CH4 m m N N y N N y Then the apparent molecular weight of the mixture becomes 1811 lbmlbmol 55208 lbmol 100 lbm m m m N m M The apparent gas constant of the mixture is 0 5925 psia ft lbm R lbmlbmol 1811 1073 psia ft lbmol R 3 3 m u M R R The mass of this mixture in a 1 million ft3 tank is 10 lbm 4441 6 psia ft lbm R760 R 05925 psia1 10 ft 2000 3 3 6 RT P m V b To use the Amagats law for this real gas mixture we first need the compressibility factor of each component at the mixture temperature and pressure The compressibility factors are obtained using Fig A15 to be 0 98 2 972 673 psia psia 2000 2 210 3439 R R 760 CH4 crCH4 CH4 crCH4 CH4 Z P P P T T T m R m R 0 77 2 119 708 psia psia 1500 1 382 5498 R R 760 C2H6 C2H6 C2H6 Z P T R R Then 0 9483 0 1509 0 77 0 8491 0 98 C2H6 C2H6 CH4 CH4 Z y Z y y Z Z i i m 10 lbm 4684 6 5925 psia ft lbm R760 R 094830 psia1 10 ft 2000 3 3 6 Z RT P m m V c To use Daltons law with compressibility factors Fig A15 0 98 0 8782 psia ft lbm R3439 R673 psia 06688 075 lbm 1 10 ft 4441 10 210 2 CH4 3 6 3 6 crCH4 crCH4 CH4 CH4 CH4 CH4 Z P T R m T m R R V v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1323 0 92 3 244 psia ft lbm R5498 R708 psia 03574 025 lbm 1 10 ft 4441 10 382 1 CH4 3 6 3 6 crC2H6 C2H6 cr C2H6 C2H6 C2H6 Z P RT m T m R R V v Note that we used m in above calculations the value obtained by ideal gas behavior The solution normally requires iteration until the assumed and calculated mass values match The mass of the component gas is obtained by multiplying the mass of the mixture by its mass fraction Then 025 lbm 10 4441 6 0 9709 0 1509 0 92 0 8491 0 98 C2H6 C2H6 CH4 CH4 Z y Z y y Z Z i i m 10 lbm 4575 6 5925 psia ft lbm R760 R 097090 psia1 10 ft 2000 3 3 6 Z RT P m m V This mass is sufficiently close to the assumed mass value of Therefore there is no need to repeat the calculations at this calculated mass 025 lbm 10 4441 6 d To use Kays rule we need to determine the pseudocritical temperature and pseudocritical pressure of the mixture using the critical point properties of gases 678 3 psia 01509708 psia 0849167 3 psia 375 0 R 015095498 R 084913439 R crC2H6 C2H6 crCh4 Ch4 cr cr crC2H6 C2H6 crCh4 CH4 cr cr P y P y y P P T y T y y T T i i m i i m Then 0 97 2 949 6783 psia psia 2000 2 027 3750 R R 760 cr cr m m m R m m R Z P P P T T T Fig A15 10 lbm 4579 6 925 psia ft lbm R760 R 09705 psia1 10 ft 2000 3 3 6 Z RT P m m V PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1324 1346 The volumetric analysis of a mixture of gases is given The volumetric and mass flow rates are to be determined using three methods Properties The molar masses of O2 N2 CO2 and CH4 are 320 280 440 and 160 kgkmol respectively Table A1 Analysis a We consider 100 kmol of this mixture Noting that volume fractions are equal to the mole fractions mass of each component are 320 kg 20 kmol16 kgkmol 440 kg 10 kmol44 kgkmol 112 0 kg 40 kmol28 kgkmol 96 0 kg 30 kmol32 kgkmol CH4 CH4 CH4 CO2 CO2 CO2 N2 N2 N2 O2 O2 O2 M N m M N m M N m M N m 30 O2 40 N2 10 CO2 20 CH4 by volume The total mass is 2840 kg 320 440 1120 960 CH4 CO2 N2 O2 m m m m mm The apparent molecular weight of the mixture is Mixture 8 MPa 15C 2840 kgkmol 100 kmol 2840 kg m m m N m M The apparent gas constant of the mixture is 02927 kJkg K 2840 kgkmol 8314 kJkmol K m u M R R The specific volume of the mixture is 0 01054 m kg 8000 kPa 02927 kPa m kg K288 K 3 3 P RT v The volume flow rate is 00009425 m s 3 3 ms 4 002 m 4 2 2 π π D V AV V and the mass flow rate is 008942 kgs m kg 001054 m s 00009425 3 3 v V m b To use the Amagats law for this real gas mixture we first need the mole fractions and the Z of each component at the mixture temperature and pressure The compressibility factors are obtained using Fig A15 to be 0 95 1 575 508 MPa MPa 8 1 860 1548 K K 288 O2 crO2 O2 crO2 O2 Z P P P T T T m R m R 0 99 2 360 339 MPa MPa 8 2 282 1262 K K 288 N2 N2 N2 Z P T R R 0 199 1 083 739 MPa MPa 8 0 947 3042 K K 288 CO2 CO2 CO2 Z P T R R 0 85 1 724 464 MPa MPa 8 1 507 1911 K K 288 CH4 CH4 CH4 Z P T R R and 0 8709 0 20 0 85 0 10 0 199 0 40 0 99 0 30 0 95 CH4 CH4 CO2 CO2 O2 O2 O2 O2 Z y Z y Z y Z y y Z Z i i m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1325 Then 0 009178 m kg 8000 kPa 0870902927 kPa m kg K288 K 3 3 P Z m RT v 00009425 m s 3 V 010269 kgs m kg 0009178 m s 00009425 3 3 v V m c To use Kays rule we need to determine the pseudocritical temperature and pseudocritical pressure of the mixture using the critical point properties of mixture gases 4 547 MPa 02046 4 MPa 01073 9 MPa 04033 9 MPa 030508 MPa 165 6 K 0201911 K 0103042 K 0401262 K 030154 8 K crCH4 CH4 crCO2 CO2 crN2 N2 crO2 O2 cr cr crCH4 CH4 crCO2 CO2 crN2 N2 crO2 O2 cr cr P y P y P y P y y P P T y T y T y T y y T T i i m i i m and 0 92 1 759 4547 MPa MPa 8 1 739 1656 K K 288 cr cr m m m R m m R Z P P P T T T Fig A15 Then 0 009694 m kg 8000 kPa 09202927 kPa m kg K288 K 3 3 P Z m RT v 00009425 m s 3 V 0009723 kgs m kg 009694 m s 00009425 3 3 v V m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1326 Properties of Gas Mixtures 1347C Yes Yes extensive property 1348C No intensive property 1349C The answers are the same for entropy 1350C Yes Yes conservation of energy 1351C We have to use the partial pressure 1352C No this is an approximate approach It assumes a component behaves as if it existed alone at the mixture temperature and pressure ie it disregards the influence of dissimilar molecules on each other PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1327 1353 The volume fractions of components of a gas mixture are given This mixture is heated while flowing through a tube at constant pressure The heat transfer to the mixture per unit mass of the mixture is to be determined Assumptions All gases will be modeled as ideal gases with constant specific heats Properties The molar masses of O2 N2 CO2 and CH4 are 320 280 440 and 160 kgkmol respectively Table A1 The constantpressure specific heats of these gases at room temperature are 0918 1039 0846 and 22537 kJkgK respectively Table A2a Analysis We consider 100 kmol of this mixture Noting that volume fractions are equal to the mole fractions mass of each component are 320 kg 20 kmol16 kgkmol 440 kg 10 kmol44 kgkmol 112 0 kg 40 kmol28 kgkmol 96 0 kg 30 kmol32 kgkmol CH4 CH4 CH4 CO2 CO2 CO2 N2 N2 N2 O2 O2 O2 M N m M N m M N m M N m The total mass is 150 kPa 200C 150 kPa 20C 30 O2 40 N2 10 CO2 20 CH4 by volume qin 2840 kg 320 440 1120 960 CH4 CO2 N2 O2 m m m m mm Then the mass fractions are 0 1127 2840 kg 320 kg mf 0 1549 2840 kg 440 kg mf 0 3944 2840 kg 1120 kg mf 0 3380 2840 kg 960 kg mf CH4 CH4 CO2 CO2 N2 N2 O2 O2 m m m m m m m m m m m m The constantpressure specific heat of the mixture is determined from 11051 kJkg K 2 2537 0 1127 0 1549 0846 1 039 0 3944 3380 0918 0 mf mf mf mf CH4 CH4 CO2 CO2 N2 N2 O2 O2 p p p p p c c c c c An energy balance on the tube gives 199 kJkg 20 K 1 1051 kJkg K200 1 2 in T T c q p PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1328 1354E A mixture of helium and nitrogen is heated at constant pressure in a closed system The work produced is to be determined Assumptions 1 Helium and nitrogen are ideal gases 2 The process is reversible Properties The mole numbers of helium and nitrogen are 40 and 280 lbmlbmol respectively Table A1E Analysis One lbm of this mixture consists of 035 lbm of nitrogen and 065 lbm of helium or 035 lbm280 lbmlbmol 00125 lbmol of nitrogen and 065 lbm40 lbmlbmol 01625 lbmol of helium The total mole is 00125016250175 lbmol The constituent mole fraction are then 0 9286 0175 lbmol 1625 lbmol 0 0 07143 0175 lbmol 0125 lbmol 0 total He He total N2 N2 N N y N N y 35 N2 65 He by mass 100 psia 100F Q The effective molecular weight of this mixture is 5 714 lbmlbmol 0 92864 0 0714328 He He N2 N2 M y M y M The work done is determined from 1390 Btulbm 100R 500 5 714 lbmlbmol 1 9858 Btulbmol R 1 2 1 2 1 1 2 2 2 1 T M T R T R T P P Pd w u v v v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1329 1355 The volume fractions of components of a gas mixture are given This mixture is expanded isentropically to a specified pressure The work produced per unit mass of the mixture is to be determined Assumptions All gases will be modeled as ideal gases with constant specific heats Properties The molar masses of H2 He and N2 are 20 40 and 280 kgkmol respectively Table A1 The constant pressure specific heats of these gases at room temperature are 14307 51926 and 1039 kJkgK respectively Table A 2a Analysis We consider 100 kmol of this mixture Noting that volume fractions are equal to the mole fractions mass of each component are 840 kg 30 kmol28 kgkmol 16 0 kg 40 kmol4 kgkmol 6 0 kg 30 kmol2 kgkmol N2 N2 N2 He He He H2 H2 H2 M N m M N m M N m 30 H2 40 He 30 N2 by volume 5 MPa 600C The total mass is 1060 kg 840 160 60 N2 He H2 m m m mm Then the mass fractions are 07925 1060 kg 840 kg mf 01509 1060 kg 160 kg mf 005660 1060 kg 60 kg mf N2 N2 He He H2 H2 m m m m m m m m m The apparent molecular weight of the mixture is 1060 kgkmol 100 kmol 1060 kg m m m N m M The constantpressure specific heat of the mixture is determined from 2 417 kJkg K 1 039 0 7925 5 1926 0 1509 05660 14307 0 mf mf mf N2 N2 He He H2 H2 p p p p c c c c The apparent gas constant of the mixture is 07843 kJkg K 1060 kgkmol 8314 kJkmol K m u M R R Then the constantvolume specific heat is 1 633 kJkg K 0 7843 2 417 R c c p v The specific heat ratio is 1 480 1 633 2 417 vc c k p The temperature at the end of the expansion is 307 K 5000 kPa 200 kPa 873 K 048148 1 1 2 1 2 k k P P T T An energy balance on the adiabatic expansion process gives 1368 kJkg 2 417 kJkg K873 307 K 2 1 out T T c w p PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1330 1356 The mass fractions of components of a gas mixture are given This mixture is enclosed in a rigid wellinsulated vessel and a paddle wheel in the vessel is turned until specified amount of work have been done on the mixture The mixtures final pressure and temperature are to be determined Assumptions All gases will be modeled as ideal gases with constant specific heats Properties The molar masses of N2 He CH4 and C2H6 are 280 40 160 and 300 kgkmol respectively Table A1 The constantpressure specific heats of these gases at room temperature are 1039 51926 22537 and 17662 kJkgK respectively Table A2a Analysis We consider 100 kg of this mixture The mole numbers of each component are 0 6667 kmol 30 kgkmol kg 20 3 75 kmol 16 kgkmol kg 60 1 25 kmol 4 kgkmol kg 5 0 5357 kmol 28 kgkmol kg 15 C2H6 C2H6 C2H6 CH4 CH4 CH4 He He He N2 N2 N2 M m N M m N M m N M m N 15 N2 5 He 60 CH4 20 C2H6 by mass 10 m3 200 kPa 20C Wsh The mole number of the mixture is 6 2024 kmol 0 6667 3 75 1 25 0 5357 C2H6 CH4 He N2 N N N N N m The apparent molecular weight of the mixture is 1612 kgkmol 62024 kmol 100 kg m m m N m M The constantpressure specific heat of the mixture is determined from 2121 kJkg K 1 7662 0 20 2 2537 0 60 5 1926 0 05 1 039 15 0 mf mf mf mf C2H6 C2H6 CH4 CH4 He He N2 N2 p p p p p c c c c c The apparent gas constant of the mixture is 05158 kJkg K 1612 kgkmol 8134 kJkmol K m u M R R Then the constantvolume specific heat is 1605 kJkg K 0 5158 2 121 R c c p v The mass in the container is 1323 kg kPa m kg K293 K 05158 kPa10 m 200 3 3 1 1 RT P m m m V An energy balance on the system gives 2977 K 1323 kg 1 605 kJkg K 100 kJ 293 K shin 1 2 1 2 shin v v c m W T T T T m c W m m Since the volume remains constant and this is an ideal gas 2032 kPa 293 K 200 kPa 297 7 K 1 2 1 2 T P T P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1331 1357 Propane and air mixture is compressed isentropically in an internal combustion engine The work input is to be determined Assumptions Under specified conditions propane and air can be treated as ideal gases and the mixture as an ideal gas mixture Properties The molar masses of C3H8 and air are 440 and 2897 kgkmol respectively TableA1 Analysis Given the airfuel ratio the mass fractions are determined to be 0 05882 17 1 AF 1 1 mf 0 9412 17 16 AF 1 AF mf 3H8 C air Propane Air 95 kPa 30ºC The molar mass of the mixture is determined to be 56 kgkmol 29 440 kgkmol 0 05882 2897 kgkmol 0 9412 1 mf mf 1 8 3 8 3 H C H C air air M M M m The mole fractions are 0 03944 440 kgkmol 0 05882 2956 kgkmol mf 0 9606 2897 kgkmol 0 9412 2956 kgkmol mf 8 3 8 3 8 3 C H C H H C air air air M M y M M y m m The final pressure is expressed from ideal gas relation to be 2 2 1 2 1 2 2 977 27315 K 95 kPa 59 30 T T T P r T P 1 since the final temperature is not known Using Daltons law to find partial pressures the entropies at the initial state are determined from EES to be 6 7697 kJkgK 3 75 kPa 0 03944 95 C 30 5 7417 kJkgK 9126 kPa 0 9606 95 C 30 H 1 C 1 air 3 8 s P T s P T The final state entropies cannot be determined at this point since the final pressure and temperature are not known However for an isentropic process the entropy change is zero and the final temperature and the final pressure may be determined from 0 8 3 3 8 C H C H air air total s mf s mf s and using Eq 1 The solution may be obtained using EES to be T2 6549 K P2 1951 kPa The initial and final internal energies are from EES 1607 kJkg 477 1 kJkg 654 9 K 2404 kJkg 216 5 kJkg 30 C H 2 C air2 2 H 1 C air 1 1 8 3 3 8 u u T u u T Noting that the heat transfer is zero an energy balance on the system gives m m u w u w q in in in where C H 1 2 C H C H air1 air2 air 8 3 8 3 3 8 u u mf u u mf um Substituting 2922 kJkg 2404 0 05882 1607 216 5 0 9412477 1 in um w PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1332 1358 The moles temperatures and pressures of two gases forming a mixture are given The mixture temperature and pressure are to be determined Assumptions 1 Under specified conditions both CO2 and H2 can be treated as ideal gases and the mixture as an ideal gas mixture 2 The tank is insulated and thus there is no heat transfer 3 There are no other forms of work involved Properties The molar masses and specific heats of CO2 and H2 are 440 kgkmol 20 kgkmol 0657 kJkgC and 10183 kJkgC respectively Tables A1 and A2b Analysis a We take both gases as our system No heat work or mass crosses the system boundary therefore this is a closed system with Q 0 and W 0 Then the energy balance for this closed system reduces to H2 75 kmol 400 kPa 40C CO2 25 kmol 200 kPa 27C 2 2 2 2 H 1 CO 1 H CO system out in 0 0 T T mc T T mc U U U E E E m m v v Using cv values at room temperature and noting that m NM the final temperature of the mixture is determined to be K 3088 0 40 C C 2 kg 10183 kJkg 75 27 C C 44 kg 0657 kJkg 25 358C m m m T T T b The volume of each tank is determined from 3 3 H 1 1 H 3 3 CO 1 1 CO 48 79 m 400 kPa kmol831 4 kPa m kmol K313 K 75 31 18 m 200 kPa kmol831 4 kPa m kmol K300 K 25 2 2 2 2 P T NR P T NR u u V V Thus 100 kmol kmol 57 kmol 52 7997 m 4879 m 18 m 31 2 2 2 2 H CO 3 3 3 H CO N N Nm m V V V and 321 kPa 3 3 m 7997 kmol831 4 kPa m kmol K3088 K 100 m m u m m V N R T P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1334 1360 The volume fractions of components of a gas mixture during the expansion process of the ideal Otto cycle are given The thermal efficiency of this cycle is to be determined Assumptions All gases will be modeled as ideal gases with constant specific heats Properties The molar masses of N2 O2 H2O and CO2 are 280 320 180 and 440 kgkmol respectively Table A1 The constantpressure specific heats of these gases at room temperature are 1039 0918 18723 and 0846 kJkgK respectively The air properties at room temperature are cp 1005 kJkgK cv 0718 kJkgK k 14 Table A2a Analysis We consider 100 kmol of this mixture Noting that volume fractions are equal to the mole fractions mass of each component are 1100 kg 25 kmol44 kgkmol 630 kg 35 kmol18 kgkmol 32 0 kg 10 kmol32 kgkmol 840 kg 30 kmol28 kgkmol CO2 CO2 CO2 H2O H2O H2O O2 O2 O2 N2 N2 N2 M N m M N m M N m M N m 30 N2 10 O2 35 H2O 25 CO2 by volume The total mass is kg 2890 630 1100 320 840 CO2 H2O O2 N2 m m m m mm P 4 1 3 2 Then the mass fractions are 0 3806 2890 kg 1100 kg mf 0 2180 2890 kg 630 kg mf 0 1107 2890 kg 320 kg mf 0 2907 2890 kg 840 kg mf CO2 CO2 H2O H2O O2 O2 N2 N2 m m m m m m m m m m m m v The constantpressure specific heat of the mixture is determined from kJkg K 1134 0 846 0 3806 1 8723 0 2180 0 918 0 1107 1 039 2907 0 mf mf mf mf CO2 CO2 H2O H2O O2 O2 N2 N2 p p p p p c c c c c The apparent molecular weight of the mixture is 2890 kgkmol 100 kmol 2890 kg m m m N m M The apparent gas constant of the mixture is 0 2877 kJkg K 2890 kgkmol 8314 kJkmol K m u M R R Then the constantvolume specific heat is 0 846 kJkg K 0 2877 1 134 R c c p v The specific heat ratio is 1 340 0 846 1 134 vc c k p The average of the air properties at room temperature and combustion gas properties are PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1335 1 37 41 1 34 50 0 782 kJkg K 0 718 0 846 50 1 070 kJkg K 1 005 1 134 50 avg avg avg k c c p v These average properties will be used for heat addition and rejection processes For compression the air properties at room temperature and during expansion the mixture properties will be used During the compression process 662 K 288 K8 04 1 1 2 T rk T During the heat addition process 556 kJkg 662 K 0 782 kJkg K1373 2 3 avg in T T c q v During the expansion process 636 1 K 8 1373 K 1 1 037 1 3 4 k r T T During the heat rejection process 272 2 kJkg 288 K 0 782 kJkg K636 1 1 4 avg out T T c q v The thermal efficiency of the cycle is then 0511 556 kJkg 272 2 kJkg 1 1 in out th q q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1336 1361 The thermal efficiency of the cycle in the previous problem is to be compared to that predicted by air standard analysis Assumptions Airstandard assumptions are applicable Properties The air properties at room temperature are cp 1005 kJkgK cv 0718 kJkgK k 14 Table A2a Analysis In the previous problem the thermal efficiency of the cycle was determined to be 0511 511 The thermal efficiency with air standard model is determined from P 4 1 3 2 0565 η 40 1 th 8 1 1 1 1 k r which is greater than that calculated with gas mixture analysis in the previous problem v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1337 1362E The volume fractions of components of a gas mixture passing through the turbine of a simple ideal Brayton cycle are given The thermal efficiency of this cycle is to be determined Assumptions All gases will be modeled as ideal gases with constant specific heats Properties The molar masses of N2 O2 H2O and CO2 are 280 320 180 and 440 lbmlbmol respectively Table A1E The constantpressure specific heats of these gases at room temperature are 0248 0219 0445 and 0203 BtulbmR respectively The air properties at room temperature are cp 0240 BtulbmR cv 0171 BtulbmR k 14 Table A2Ea Analysis We consider 100 lbmol of this mixture Noting that volume fractions are equal to the mole fractions mass of each component are 1760 lbm 40 lbmol44 lbmlbmol 630 lbm 35 lbmol18 lbmlbmol 16 0 lbm 5 lbmol32 lbmlbmol 560 lbm 20 lbmol28 lbmlbmol CO2 CO2 CO2 H2O H2O H2O O2 O2 O2 N2 N2 N2 M N m M N m M N m M N m 20 N2 5 O2 35 H2O 40 CO2 by volume The total mass is 3110 lbm 1760 630 160 560 CO2 H2O O2 N2 m m m m mm 10 psia Then the mass fractions are 0 5659 3110 lbm 1760 lbm mf 0 2026 3110 lbm 630 lbm mf 0 05145 3110 lbm 160 lbm mf 0 1801 3110 lbm 560 lbm mf CO2 CO2 H2O H2O O2 O2 N2 N2 m m m m m m m m m m m m 3 4 1 2 qin qout 500 R 1860 R T s The constantpressure specific heat of the mixture is determined from 2610 Btulbm R 0 0 203 0 5659 0 445 0 2026 0 219 0 05145 0 248 1801 0 mf mf mf mf CO2 CO2 H2O H2O O2 O2 N2 N2 p p p p p c c c c c The apparent molecular weight of the mixture is 3110 lbmlbmol 100 lbmol 3110 lbm m m m N m M The apparent gas constant of the mixture is 0 06385 Btulbm R 3110 lbmlbmol 19858 Btulbmol R m u M R R Then the constantvolume specific heat is 0 1971 Btulbm R 0 06385 0 2610 R c c p v The specific heat ratio is 1 324 0 1971 0 2610 vc c k p The average of the air properties at room temperature and combustion gas properties are PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1338 1 362 41 1 324 50 0 1841 Btulbm R 0 171 0 1971 50 0 2505 Btulbm R 0 240 0 2610 50 avg avg avg k c c p v These average properties will be used for heat addition and rejection processes For compression the air properties at room temperature and during expansion the mixture properties will be used During the compression process 834 3 R 500 R6 0414 1 1 2 1 2 k k P P T T During the heat addition process 256 9 Btulbm 834 3 R 0 2505 Btulbm R1860 2 3 avg in T T c q p During the expansion process 1155 3 R 6 1860 R 1 03621362 1 3 4 3 4 k k P P T T During the heat rejection process 164 2 Btulbm 500 R 0 2505 Btulbm R1155 3 1 4 avg out T T c q p The thermal efficiency of the cycle is then 361 0361 256 9 Btulbm 164 2 Btulbm 1 1 in out th q q η PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1339 1363E The thermal efficiency of the cycle in the previous problem is to be compared to that predicted by air standard analysis Assumptions Airstandard assumptions are applicable Properties The air properties at room temperature are cp 0240 BtulbmR cv 0171 BtulbmR k 14 Table A2Ea Analysis In the previous problem the thermal efficiency of the cycle was determined to be 0361 361 The thermal efficiency with air standard model is determined from 3 4 1 2 qin qout 500 R 1860 R T 401 0401 6 1 1 1 1 41 40 1 th k pr k η which is greater than that calculated with gas mixture analysis in the previous problem s PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1340 1364E The mass fractions of a natural gas mixture at a specified pressure and temperature trapped in a geological location are given This natural gas is pumped to the surface The work required is to be determined using Kays rule and the enthalpydeparture method Properties The molar masses of CH4 and C2H6 are 160 and 300 lbmlbmol respectively The critical properties are 3439 R 673 psia for CH4 and 5498 R and 708 psia for C2H6 Table A1E The constantpressure specific heats of these gases at room temperature are 0532 and 0427 BtulbmR respectively Table A2Ea Analysis We consider 100 lbm of this mixture Then the mole numbers of each component are 0 8333 lbmol 30 lbmlbmol lbm 25 4 6875 lbmol 16 lbmlbmol lbm 75 C2H6 C2H6 C2H6 CH4 CH4 CH4 M m N M m N 75 CH4 25 C2H6 by mass 2000 psia 300F The mole number of the mixture and the mole fractions are 5 5208 lbmol 0 8333 4 6875 Nm 01509 55208 lbmol lbmol 08333 08491 55208 lbmol lbmol 46875 C2H6 C2H6 CH4 CH4 m m N N y N N y Then the apparent molecular weight of the mixture becomes 1811 lbmlbmol 55208 lbmol 100 lbm m m m N m M The apparent gas constant of the mixture is 0 1097 Btulbm R 1811 lbmlbmol 19858 Btulbmol R m u M R R The constantpressure specific heat of the mixture is determined from 0 506 Btulbm R 0 427 0 25 0 532 0 75 mf mf C2H6 C2H6 CH4 CH4 p p p c c c To use Kays rule we need to determine the pseudocritical temperature and pseudocritical pressure of the mixture using the critical point properties of gases 678 3 psia 01509708 psia 0849167 3 psia 375 0 R 015095498 R 084913439 R crC2H6 C2H6 crCh4 Ch4 cr cr crC2H6 C2H6 crCh4 CH4 cr cr P y P y y P P T y T y y T T i i m i i m The compressibility factor of the gas mixture in the reservoir and the mass of this gas are 0 963 2 949 6783 psia psia 2000 2 027 3750 R R 760 cr cr m m m R m m R Z P P P T T T Fig A15 4612 10 lbm 5925 psia ft lbm R760 R 09630 2000 psia1 10 ft 6 3 3 6 Z RT P m m V The enthalpy departure factors in the reservoir and the surface are from EES or Fig A29 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1341 0 703 2 949 6783 psia psia 2000 2 027 3750 R R 760 1 cr 1 cr 1 h m m R m m R Z P P P T T T 0 0112 0 0295 6783 psia psia 20 1 76 3750 R R 660 2 cr 2 cr 2 h m m R m m R Z P P P T T T The enthalpy change for the ideal gas mixture is 50 6 Btulbm 660R 0 506 Btulbm R760 2 1 ideal 2 1 T T c h h p The enthalpy change with departure factors is 12 Btulbm 22 0 0112 0 1096375 0 703 6 50 2 1 cr ideal 2 1 2 1 h h m Z Z RT h h h h The work input is then 10 Btu 102 8 4 612 10 lbm2212 Btulbm 6 2 1 in h m h W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1343 1366E Problem 1365E is reconsidered The problem is first to be solved and then for all other conditions being the same the problem is to be resolved to determine the composition of the nitrogen and carbon dioxide that is required to have an exit velocity of 2200 fts at the nozzle exit Analysis The problem is solved using EES and the solution is given below Given mfN2065 mfCO21mfN2 P160 psia T11400 R Vel10 fts P212 psia etaN088 Vel22200 fts Properties cpN20248 BtulbmR cvN20177 BtulbmR cpCO20203 BtulbmR cvCO20158 BtulbmR MMN228 lbmlbmol MMCO244 lbmlbmol Analysis cpmmfN2cpN2mfCO2cpCO2 cvmmfN2cvN2mfCO2cvCO2 kmcpmcvm T2sT1P2P1km1km etaNT1T2T1T2s 0cpmT2T1Vel22Vel122Convertft2s2 Btulbm NN2mfN2MMN2 NCO2mfCO2MMCO2 NtotalNN2NCO2 yN2NN2Ntotal yCO2NCO2Ntotal SOLUTION of the stated problem cpCO20203 BtulbmR cpm02323 BtulbmR cpN20248 BtulbmR cvCO20158 BtulbmR cvm01704 BtulbmR cvN20177 BtulbmR etaN088 km1363 mfCO2035 mfN2065 MMCO244 lbmlbmol MMN228 lbmlbmol NCO20007955 NN2002321 Ntotal003117 P160 psia P212 psia T11400 R T29703 R T2s9117 R Vel10 fts Vel22236 fts yCO202552 yN207448 SOLUTION of the problem with exit velocity of 2200 fts cpCO20203 BtulbmR cpm02285 BtulbmR cpN20248 BtulbmR cvCO20158 BtulbmR cvm01688 BtulbmR cvN20177 BtulbmR etaN088 km1354 mfCO20434 mfN20566 MMCO244 lbmlbmol MMN228 lbmlbmol NCO20009863 NN2002022 Ntotal003008 P160 psia P212 psia T11400 R T29769 R T2s9193 R Vel10 fts Vel22200 fts yCO203279 yN206721 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1346 1370 Problem 1369 is reconsidered The effect of the mass fraction of methane in the mixture on the mixture temperature and the rate of exergy destruction is to be investigated Analysis The problem is solved using EES and the solution is given below Given 1 C2H6 2 CH4 mdottotal9 kgs mfCH403333 mfC2H61mfCH4 mdot1mfC2H6mdottotal mdot2mfCH4mdottotal T115273 K T260273 K P300 kPa T025273 K Properties cp117662 kJkgK cp222537 kJkgK R102765 kJkgK R205182 kJkgK MM130 kgkmol MM216 kgkmol Analysis 0mdot1cp1T3T1mdot2cp2T3T2 Ndot1mdot1MM1 Ndot2mdot2MM2 NdottotalNdot1Ndot2 y1Ndot1Ndottotal y2Ndot2Ndottotal DELTAs1cp1lnT3T1R1lny1 DELTAs2cp2lnT3T2R2lny2 Sdotgenmdot1DELTAs1mdot2DELTAs2 XdotdestT0Sdotgen 0 02 04 06 08 1 280 290 300 310 320 330 340 mfCH4 T3 K mfF2 T3 K Xdest kW 0 01 02 03 04 05 06 07 08 09 1 288 2936 2989 3039 3087 3132 3176 3217 3256 3294 333 0 3764 5554 6558 7014 7025 6636 5854 4646 2902 009793 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1347 0 02 04 06 08 1 0 100 200 300 400 500 600 700 800 mfCH4 Xdest kW PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1348 1371E In an airliquefaction plant it is proposed that the pressure and temperature of air be adiabatically reduced It is to be determined whether this process is possible and the work produced is to be determined using Kays rule and the departure charts Assumptions Air is a gas mixture with 21 O2 and 79 N2 by mole Properties The molar masses of O2 and N2 are 320 and 280 lbmlbmol respectively The critical properties are 2786 R 736 psia for O2 and 2271 R and 492 psia for N2 Table A1E Analysis To use Kays rule we need to determine the pseudocritical temperature and pseudocritical pressure of the mixture using the critical point properties of gases 543 2 psia 079492 psia 021736 psia 237 9 R 0792271 R 021278 6 R crN2 N2 crO2 O2 cr cr crN2 N2 crO2 O2 cr cr P y P y y P P T y T y y T T i i m i i m 21 O2 79 N2 by mole 1500 psia 40F The enthalpy and entropy departure factors at the initial and final states are from EES 0 339 725 0 3 471 4322 psia psia 1500 2 102 2379 R R 500 1 1 cr 1 1 cr 1 1 s h m m R m m R Z Z P P P T T T 00906 0 0179 0 0 0347 4322 psia psia 15 1 513 2379 R R 360 2 2 cr 2 2 cr 2 2 s h m m R m m R Z Z P P P T T T The enthalpy and entropy changes of the air under the ideal gas assumption is Properties are from Table A17E 33 5 Btulbm 11948 8597 1 ideal 2 h h 0 2370 Btulbm R 1500 0 06855ln 15 0 58233 0 50369 ln 1 2 o 1 o 2 1 ideal 2 P P R s s s s With departure factors the enthalpy change ie the work output and the entropy change are 220 Btulbm 0 0179 0 06855237 9 0 725 5 33 2 1 cr 2 ideal 1 2 1 out h h Z Z RT h h h h w 02596 Btulbm R 0 339 0 06855 0 00906 2370 0 1 2 1 ideal 2 1 2 s s Z R Z s s s s The entropy change in this case is equal to the entropy generation during the process since the process is adiabatic The positive value of entropy generation shows that this process is possible PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1349 1372 Heat is transferred to a gas mixture contained in a piston cylinder device The initial state and the final temperature are given The heat transfer is to be determined for the ideal gas and nonideal gas cases Properties The molar masses of H2 and N2 are 20 and 280 kgkmol Table A1 Analysis From the energy balance relation 6 kg H2 21 kg N2 5 MPa 160 K 2 2 2 2 2 2 1 N 2 N 1 H 2 H N H in out in out in h h N h h N H H H Q U W Q E E E b since Wb and U combine into H for quasiequilibrium constant pressure processes Q N m M N m M H H H N N N 2 2 2 2 2 2 6 kg 2 kg kmol 3 kmol 21 kg 28 kg kmol 075 kmol a Assuming ideal gas behavior the inlet and exit enthalpies of H2 and N2 are determined from the ideal gas tables to be H h h h h N h h h h 2 1 2 2 1 2 160 K 200 K 160 K 200 K 45354 kJ kmol 56692 kJ kmol 4648 kJ kmol 5810 kJ kmol Thus 4273 kJ 4 648 5 810 0 75 4 535 4 5 669 2 3 Qideal b Using Amagats law and the generalized enthalpy departure chart the enthalpy change of each gas is determined to be H2 0 0 6 006 33 3 200 3 846 1 30 5 4 805 33 3 160 2 1 2 2 2 2 2 2 2 1 2 2 1 H cr 2 H crH H H H cr 1 H h h m R m R R m R Z Z T T T P P P P T T T Fig A29 Thus H2 can be treated as an ideal gas during this process N2 70 31 1 58 126 2 200 1 47 3 39 5 1 27 126 2 160 2 1 2 2 2 2 2 2 2 1 2 2 1 N cr 2 N crN N N N cr 1 N h h m R m R R m R Z Z T T T P P P P T T T Fig A29 Therefore 17915kJkmol 4648kJkmol 5810 07 8314kPa m kmol K1262K13 1133 8 kJkmol 4 535 4 669 2 5 3 1 ideal 2 1 N 2 1 H ideal 2 1 H 2 2 1 2 2 2 h h Z Z R T h h h h h h h h u cr Substituting 4745 kJ 075 kmol 17915 kJkmol 3 kmol 11338 kJkmol in Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1351 Therefore 1566 kJK 303 K kJ 4745 kJK 737 487 kJK 04 075 kmol831 4 kPa m kmol K08 21 kJK 18 0 surr surr 3 N ideal N N H ideal H 2 2 1 2 2 2 2 T Q S S Z Z R N S S S s s u and kJ 3006 kJK 992 K 992 kJK 303 303 K 4745 kJ 737 kJK 821 kJK 1 gen 0 destroyed gen T S X S PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1354 1375 Problem 1374 is reconsidered The results obtained by assuming ideal behavior real gas behavior with Amagats law and real gas behavior with EES data are to be compared Analysis The problem is solved using EES and the solution is given below Given yN2079 yO2021 T290 K P12000 kPa P28000 kPa mdot175 kgs Properties Ru8314 kPam3kmolK Mairmolarmassair TcrN21262 K TcrO21548 K PcrN23390 kPa PcrO25080 kPa Analysis Ideal gas NdotmdotMair DELTAhideal0 isothermal process DELTAsidealRulnP2P1 isothermal process QdotinidealNdotTDELTAsideal WdotinidealQdotinideal Amagads law TR1N2TTcrN2 PR1N2P1PcrN2 Zh1N2ENTHDEPTR1N2 PR1N2 the function that returns enthalpy departure factor Zs1N2ENTRDEPTR1N2 PR1N2 the function that returns entropy departure factor TR2N2TTcrN2 PR2N2P2PcrN2 Zh2N2ENTHDEPTR2N2 PR2N2 the function that returns enthalpy departure factor Zs2N2ENTRDEPTR2N2 PR2N2 the function that returns entropy departure factor TR1O2TTcrO2 PR1O2P1PcrO2 Zh1O2ENTHDEPTR1O2 PR1O2 the function that returns enthalpy departure factor Zs1O2ENTRDEPTR1O2 PR1O2 the function that returns entropy departure factor TR2O2TTcrO2 PR2O2P2PcrO2 Zh2O2ENTHDEPTR2O2 PR2O2 the function that returns enthalpy departure factor Zs2O2ENTRDEPTR2O2 PR2O2 the function that returns entropy departure factor DELTAhDELTAhidealyN2RuTcrN2Zh2N2Zh1N2yO2RuTcrO2Zh2O2Zh1O2 DELTAsDELTAsidealyN2RuZs2N2Zs1N2yO2RuZs2O2Zs1O2 QdotinAmagad NdotTDELTAs WdotinAmagadQdotinAmagad NdotDELTAh EES hEES1 yN2enthalpyNitrogenTT PP1 yO2enthalpyOxygenTTPP1 hEES2 yN2enthalpyNitrogenTT PP2 yO2enthalpyOxygenTTPP2 sEES1 yN2entropyNitrogenTT PP1 yO2entropyOxygenTTPP1 sEES2 yN2entropyNitrogenTT PP2 yO2entropyOxygenTTPP2 DELTAhEEShEES2hEES1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1355 DELTAsEESsEES2sEES1 QdotinEESNdotTDELTAsEES WdotinEESQdotinEESNdotDELTAhEES SOLUTION DELTAh3461 kJkmol DELTAhEES3843 kJkmol DELTAhideal0 kJkmol DELTAs1274 kJkmolK DELTAsEES1272 kJkmolK DELTAsideal1153 kJkmolK hEES16473 kJkmol hEES26089 kJkmol Mair2897 kgkmol mdot175 kgs Ndot006041 kmols P12000 kPa P28000 kPa PcrN23390 kPa PcrO25080 kPa PR1N2059 PR1O203937 PR2N2236 PR2O21575 QdotinAmagad2232 kW QdotinEES2229 kW Qdotinideal2019 kW Ru8314 kPam3kmolK sEES11253 kJkmolK sEES21125 kJkmolK T290 K TcrN21262 K TcrO21548 K TR1N22298 TR1O21873 TR2N22298 TR2O21873 WdotinAmagad2023 kW WdotinEES1997 kW Wdotinideal2019 kW yN2079 yO2021 Zh1N201154 Zh1O201296 Zh2N204136 Zh2O204956 Zs1N2005136 Zs1O2005967 Zs2N201903 Zs2O202313 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1356 1376 Two mass streams of two different ideal gases are mixed in a steadyflow chamber while receiving energy by heat transfer from the surroundings Expressions for the final temperature and the exit volume flow rate are to be obtained and two special cases are to be evaluated Assumptions Kinetic and potential energy changes are negligible Analysis a Mass and Energy Balances for the mixing process 1 2 3 1 1 2 2 3 3 1 1 1 2 2 2 3 3 1 2 1 2 3 3 1 1 2 2 3 1 2 3 3 3 in P P P in P P m P P P P in m P m P m m m m m h m h Q m h h C T m C T m C T Q m C T m m C C C m m m C m C Q T T T m C m C m C P m 3 Steadyflow ch r ambe 1 Qin 2 Surroundings b The expression for the exit volume flow rate is obtained as follows 3 3 3 3 3 3 3 1 1 2 2 3 3 3 1 2 3 3 3 3 1 3 2 3 3 1 1 1 2 2 2 3 1 3 2 3 3 3 1 2 P P in P m P m P m P P in P m P m R T V m v m P m C m C m R Q V T T P m C m C m C C R C R P m R Q m RT m R T V C R P C R P PC P P P 1 3 2 3 3 3 1 2 1 2 3 3 3 1 1 1 3 3 2 1 1 2 2 3 1 2 3 3 3 3 P P in P m P m P m u u u P P u in P m P m P m C R C R R Q V V V C R C R PC R R R R 2 3 M M M R M R M R M R M C M C M R Q V V V C M C M P M C The mixture molar mass M3 is found as follows 3 fi i i i i i fi fi i i m M m M y M y m m M m c For adiabatic mixing in is zero and the mixture volume flow rate becomes Q 1 1 2 2 3 1 3 3 P P P m P m C M C M V V C M C M 2 V d When adiabatically mixing the same two ideal gases the mixture volume flow rate becomes 3 1 2 3 1 2 3 1 2 P P P M M M C C C V V V PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1357 Special Topic Chemical Potential and the Separation Work of Mixtures 1377C No a process that separates a mixture into its components without requiring any work exergy input is impossible since such a process would violate the 2nd law of thermodynamics 1378C Yes the volume of the mixture can be more or less than the sum of the initial volumes of the mixing liquids because of the attractive or repulsive forces acting between dissimilar molecules 1379C The person who claims that the temperature of the mixture can be higher than the temperatures of the components is right since the total enthalpy of the mixture of two components at the same pressure and temperature in general is not equal to the sum of the total enthalpies of the individual components before mixing the difference being the enthalpy or heat of mixing which is the heat released or absorbed as two or more components are mixed isothermally 1380C Mixtures or solutions in which the effects of molecules of different components on each other are negligible are called ideal solutions or ideal mixtures The idealgas mixture is just one category of ideal solutions For ideal solutions the enthalpy change and the volume change due to mixing are zero but the entropy change is not The chemical potential of a component of an ideal mixture is independent of the identity of the other constituents of the mixture The chemical potential of a component in an ideal mixture is equal to the Gibbs function of the pure component PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1358 1381 Brackish water is used to produce fresh water The minimum power input and the minimum height the brackish water must be raised by a pump for reverse osmosis are to be determined Assumptions 1 The brackish water is an ideal solution since it is dilute 2 The total dissolved solids in water can be treated as table salt NaCl 3 The environment temperature is also 12C Properties The molar masses of water and salt are Mw 180 kgkmol and Ms 5844 kgkmol The gas constant of pure water is Rw 04615 kJkgK Table A1 The density of fresh water is 1000 kgm3 Analysis First we determine the mole fraction of pure water in brackish water using Eqs 134 and 135 Noting that mfs 000078 and mfw 1 mfs 099922 01 kgkmol 18 18 0 099922 5844 000078 1 mf mf 1 mf 1 m w w s s i i M M M M 0 99976 180 kgkmol 0 99922 1801 kgkmol mf mf w m w w i m i i M M y M M y The minimum work input required to produce 1 kg of freshwater from brackish water is 0 03159 kJkg fresh water 0 4615 kJkg K28515 K ln1099976 0 ln1 min in w w y R T w Therefore 003159 kJ of work is needed to produce 1 kg of fresh water is mixed with seawater reversibly Therefore the required power input to produce fresh water at the specified rate is 885 kW 1kJs 1kW 1000 kgm 0 280 m s00315 9 kJkg 3 3 min in min in w W V ρ The minimum height to which the brackish water must be pumped is 322 m 1kJ 1000 Nm 1N kgms 1 ms 981 0 03159 kJkg 2 2 minin min g w z PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1359 1382 A river is discharging into the ocean at a specified rate The amount of power that can be generated is to be determined Assumptions 1 The seawater is an ideal solution since it is dilute 2 The total dissolved solids in water can be treated as table salt NaCl 3 The environment temperature is also 15C Properties The molar masses of water and salt are Mw 180 kgkmol and Ms 5844 kgkmol The gas constant of pure water is Rw 04615 kJkgK Table A1 The density of river water is 1000 kgm3 Analysis First we determine the mole fraction of pure water in ocean water using Eqs 134 and 135 Noting that mfs 0025 and mfw 1 mfs 0975 32 kgkmol 18 18 0 0975 5844 0025 1 mf mf 1 mf 1 m w w s s i i M M M M 0 9922 180 kgkmol 0 9751832 kgkmol mf mf w m w w i m i i M M y M M y The maximum work output associated with mixing 1 kg of seawater or the minimum work input required to produce 1 kg of freshwater from seawater is 1 046 kJkg fresh water 0 4615 kJkg K28815 Kln109922 0 ln1 max out w w y R T w Therefore 1046 kJ of work can be produced as 1 kg of fresh water is mixed with seawater reversibly Therefore the power that can be generated as a river with a flow rate of 400000 m3s mixes reversibly with seawater is 10 kW 157 6 1kJs 1kW 10 m s1046 kJkg 1000 kgm 51 3 5 3 max out max out w W V ρ Discussion This is more power than produced by all nuclear power plants 112 of them in the US which shows the tremendous amount of power potential wasted as the rivers discharge into the seas PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1360 1383 Problem 1382 is reconsidered The effect of the salinity of the ocean on the maximum power generated is to be investigated Analysis The problem is solved using EES and the solution is given below Given Vdot150000 m3s salinity25 T1527315 K Properties Mw18 kgkmol molarmassH2O Ms5844 kgkmol molar mass of salt Rw04615 kJkgK gas constant of water rho1000 kgm3 Analysis massw100salinity mfssalinity100 mfwmassw100 Mm1mfsMsmfwMw ywmfwMmMw wmaxoutRwTln1yw WdotmaxoutrhoVdotwmaxout Salinity Wmaxout kW 0 05 1 15 2 25 3 35 4 45 5 0 3085E07 6196E07 9334E07 1249E08 1569E08 1891E08 2216E08 2544E08 2874E08 3208E08 0 1 2 3 4 5 000x100 500x107 100x108 150x108 200x108 250x108 300x108 350x108 salinity Wmaxout kW PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1361 1384E Brackish water is used to produce fresh water The mole fractions the minimum work inputs required to separate 1 lbm of brackish water and to obtain 1 lbm of fresh water are to be determined Assumptions 1 The brackish water is an ideal solution since it is dilute 2 The total dissolved solids in water can be treated as table salt NaCl 3 The environment temperature is equal to the water temperature Properties The molar masses of water and salt are Mw 180 lbmlbmol and Ms 5844 lbmlbmol The gas constant of pure water is Rw 01102 BtulbmR Table A1E Analysis a First we determine the mole fraction of pure water in brackish water using Eqs 134 and 135 Noting that mfs 00012 and mfw 1 mfs 09988 015 lbmlbmol 18 18 0 09988 5844 00012 1 mf mf 1 mf 1 m w w s s i i M M M M 099963 180 lbmlbmol 0 9988 18015 lbmlbmol mf mf w m w w i m i i M M y M M y 000037 0 99963 1 1 w s y y b The minimum work input required to separate 1 lbmol of brackish water is brackish water 000037ln 0 00037 0 1102 BtulbmolR525 R099963ln 0 99963 ln ln 0 in min 0191 Btulbm s s w w w y y y y R T w c The minimum work input required to produce 1 lbm of freshwater from brackish water is 00214 Btulbm fresh water 0 1102 Btulbm R525 Rln1099963 0 ln1 min in w w y R T w Discussion Note that it takes about 9 times work to separate 1 lbm of brackish water into pure water and salt compared to producing 1 lbm of fresh water from a large body of brackish water PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1362 1385 A desalination plant produces fresh water from seawater The second law efficiency of the plant is to be determined Assumptions 1 The seawater is an ideal solution since it is dilute 2 The total dissolved solids in water can be treated as table salt NaCl 3 The environment temperature is equal to the seawater temperature Properties The molar masses of water and salt are Mw 180 kgkmol and Ms 5844 kgkmol The gas constant of pure water is Rw 04615 kJkgK Table A1 The density of river water is 1000 kgm3 Analysis First we determine the mole fraction of pure water in seawater using Eqs 134 and 135 Noting that mfs 0032 and mfw 1 mfs 0968 41 kgkmol 18 18 0 0968 5844 0032 1 mf mf 1 mf 1 m w w s s i i M M M M 0 9900 180 kgkmol 0 968 1841 kgkmol mf mf w m w w i m i i M M y M M y The maximum work output associated with mixing 1 kg of seawater or the minimum work input required to produce 1 kg of freshwater from seawater is 1 313 kJkg fresh water 0 4615 kJkg K28315 Kln10990 0 ln1 max out w w y R T w The power that can be generated as 14 m3s fresh water mixes reversibly with seawater is 184 kW 1kJs 1kW m s1313 kJkg 41 1000 kgm 3 3 max out max out Vw W ρ Then the second law efficiency of the plant becomes 216 0 216 85 MW MW 183 in minin II W W η 1386 The power consumption and the second law efficiency of a desalination plant are given The power that can be produced if the fresh water produced is mixed with the seawater reversibly is to be determined Assumptions 1 This is a steadyflow process 2 The kinetic and potential energy changes are negligible Analysis From the definition of the second law efficiency 2875 kW rev rev actual rev II 11500 kW 025 W W W W η which is the maximum power that can be generated PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1363 1387E It is to be determined if it is it possible for an adiabatic liquidvapor separator to separate wet steam at 100 psia and 90 percent quality so that the pressure of the outlet streams is greater than 100 psia Analysis Because the separator divides the inlet stream into the liquid and vapor portions 1 1 3 1 1 2 10 1 90 m x m m m xm m 2 Vapor 3 Liquid 1 Mixture According to the water property tables at 100 psia Table A5E 1 4903 Btulbm R 1 12888 90 0 47427 1 fg f xs s s When the increase in entropy principle is adapted to this system it becomes 4903 Btulbm R 1 10 90 1 1 3 2 1 1 1 3 2 1 1 1 3 3 2 2 s s s m s x m s m s x m s m s m s To test this hypothesis lets assume the outlet pressures are 110 psia Then 48341 Btulbm R 0 5954 Btulbm R 1 3 2 f g s s s s The lefthand side of the above equation is 1 4842 Btulbm R 0 48341 10 1 5954 90 10 90 3 2 s s which is less than the minimum possible specific entropy Hence the outlet pressure cannot be 110 psia Inspection of the water table in light of above equation proves that the pressure at the separator outlet cannot be greater than that at the inlet PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1364 Review Problems PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Zi 1388 Using Daltons law it is to be shown that for a realgas mixture Z y m i i k 1 Analysis Using the compressibility factor the pressure of a component of a realgas mixture and of the pressure of the gas mixture can be expressed as m m u m m m m m u i i i Z N R T P Z N R T P V V and Daltons law can be expressed as m m i m P T P V Substituting m m u i i m m u m m Z N R T Z N R T V V Simplifying i i m m Z N Z N Dividing by Nm i i m y Z Z where Zi is determined at the mixture temperature and volume preparation If you are a student using this Manual you are using it without permission 1365 1389 The volume fractions of components of a gas mixture are given The mole fractions the mass fractions the partial pressures the mixture molar mass apparent gas constant and constantpressure specific heat are to be determined and compared to the values in Table A2a Properties The molar masses of N2 O2 and Ar are 280 320 and 400 kgkmol respectively Table A1 The constant pressure specific heats of these gases at 300 K are 1039 0918 and 05203 kJkgK respectively Table A2a Analysis The volume fractions are equal to the mole fractions 78 N2 21 O2 1 Ar by volume 001 021 078 Ar O2 N2 y y y The volume fractions are equal to the pressure fractions The partial pressures are then kPa 1 kPa 21 kPa 78 0 01100 kPa 0 21100 kPa 0 78100 kPa total Ar Ar total O2 O2 total N2 N2 P y P P y P P y P We consider 100 kmol of this mixture Noting that volume fractions are equal to the mole fractions mass of each component are 40 kg kmol40 kgkmol 1 672 kg 21 kmol32 kgkmol 2184 kg 78 kmol28kgkmol Ar Ar Ar O2 O2 O2 N2 N2 N2 M N m M N m M N m The total mass is 2896 kg 40 672 2184 Ar O2 N2 m m m mm Then the mass fractions are 00138 02320 07541 2896 kg 40 kg mf 2896 kg 672 kg mf 2896 kg 2184 kg mf Ar Ar O2 O2 N2 N2 m m m m m m m m m The apparent molecular weight of the mixture is 2896 kgkmol 100 kmol kg 2896 m m m N m M The constantpressure specific heat of the mixture is determined from 1004 kJkg K 0 5203 0 0138 0 918 0 2320 1 039 7541 0 mf mf mf Ar Ar O2 O2 N2 N2 p p p p c c c c The apparent gas constant of the mixture is 02871 kJkg K 2896 kgkmol kJkmol K 8314 m u M R R This mixture closely correspond to the air and the mixture properies determined mixture molar mass mixture gas constant and mixture specific heat are practically the same as those listed for air in Tables A1 and A2a PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1366 1390 The mole numbers of combustion gases are given The partial pressure of water vapor and the condensation temperature of water vapor are to be determined Properties The molar masses of CO2 H2O O2 and N2 are 440 180 320 and 280 kgkmol respectively Table A1 Analysis The total mole of the mixture and the mole fraction of water vapor are 123 5 kmol 94 12 5 9 8 total N 0 07287 123 5 9 total H2O H2O N N y Noting that molar fraction is equal to pressure fraction the partial pressure of water vapor is 729 kPa 0 07287100 kPa total H2O H2O P y P The temperature at which the condensation starts is the saturation temperature of water at this pressure This is called the dewpoint temperature Then Table A5 397C sat729 kPa cond T T Water vapor in the combustion gases will start to condense when the temperature of the combustion gases drop to 397C PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1367 1391 The masses of gases forming a mixture at a specified pressure and temperature are given The mass of the gas mixture is to be determined using four methods Properties The molar masses of O2 CO2 and He are 320 440 and 40 kgkmol respectively Table A1 Analysis a The given total mass of the mixture is kg 61 50 1 10 He CO2 O2 m m m mm The mole numbers of each component are 01 kg O2 1 kg CO2 05 kg He 0 125 kmol 4 kgkmol kg 05 0 02273 kmol 44 kgkmol kg 1 0 003125 kmol 32 kgkmol kg 01 He He He CO2 CO2 CO2 O2 O2 O2 M m N M m N M m N The mole number of the mixture is 0 1509 kmol 0 125 0 02273 0 003125 He CO2 O2 N N N N m Then the apparent molecular weight of the mixture becomes 1061 kgkmol 01509 kmol 16 kg m m m N m M The mass of this mixture in a 03 m3 tank is 2287 kg kPa m kmol K293 K 8314 kgkmol17500 kPa03 m 1061 3 3 R T P M m u m V b To use the Amagats law for this real gas mixture we first need the mole fractions and the Z of each component at the mixture temperature and pressure 08284 01509 kmol kmol 0125 0 1506 01509 kmol kmol 002273 002071 01509 kmol kmol 0003125 He He CO2 CO2 O2 O2 m m m N N y N N y N N y 0 93 3 445 508 MPa MPa 175 1 893 1548 K K 293 O2 crO2 O2 crO2 O2 Z P P P T T T m R m R Fig A15 0 33 2 368 739 MPa MPa 175 0 963 3042 K K 293 CO2 crCO2 CO2 crCO2 CO2 Z P P P T T T m R m R Fig A15 1 04 76 1 023 MPa MPa 175 55 3 53 K K 293 He crHe He crHe He Z P P P T T T m R m R from EES PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1368 Then 0 9305 0 8284 1 04 0 1506 0 33 0 02071 0 93 He He CO2 CO2 O2 O2 Z y Z y Z y y Z Z i i m 2457 kg 314 kPa m kmol K293 K 093058 kgkmol17500 kPa03 m 1061 3 3 Z R T P M m u m m V c To use Daltons law with compressibility factors 01 26 5 kPa m kg K1548 K5080 kPa 02598 03 m 2287 0116 kg 893 1 O2 3 3 crO2 crO2 O2 O2 O2 O2 Z P T R m T m R R V v 0 86 2 70 kPa K7390 kPa m kg K3042 01889 01 16 kg 03 m 2287 963 0 CO2 3 3 crCO2 crCO2 CO2 CO2 CO2 CO2 Z P T R m T m R R V v 01 0 88 kPa m kg K53 K230 kPa 20769 03 m 2287 0516 kg 3 55 He 3 3 crHe crHe He He He He Z P T R m T m R R V v Note that we used m 2287 kg in above calculations the value obtained by ideal gas behavior The solution normally requires iteration until the assumed and calculated mass values match The mass of the component gas is obtained by multiplying the mass of the mixture by its mass fraction Then 0 9786 0 8284 01 0 1506 0 86 0 02071 01 He He CO2 CO2 O2 O2 Z y Z y Z y y Z Z i i m 2337 kg 314 kPa m kmol K293 K 097868 kgkmol17500 kPa03 m 1061 3 3 Z R T P M m u m m V This mass is sufficiently close to the mass value 2287 kg Therefore there is no need to repeat the calculations at this calculated mass d To use Kays rule we need to determine the pseudocritical temperature and pseudocritical pressure of the mixture using the critical point properties of O2 CO2 and He 1 409 MPa 08284023 MPa 01506739 MPa 002071508 MPa 5341 K 0828453 K 015063042 K 0020711548 K crHe He crCO2 CO2 crO2 O2 cr cr crHe He crCO2 CO2 crO2 O2 cr cr P y P y P y y P P T y T y T y y T T i i m i i m Then 1 194 1242 1409 MPa MPa 175 5 486 5341 K K 293 cr cr m m m R m m R Z P P P T T T from EES 1915 kg 314 kPa m kmol K293 K 11948 kgkmol17500 kPa03 m 1061 3 3 Z R T P M m u m m V PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1369 1392 A mixture of carbon dioxide and nitrogen flows through a converging nozzle The required make up of the mixture on a mass basis is to be determined Assumptions Under specified conditions CO2 and N2 can be treated as ideal gases and the mixture as an ideal gas mixture Properties The molar masses of CO2 and N2 are 440 and 280 kgkmol respectively Table A1 The specific heat ratios of CO2 and N2 at 500 K are kCO2 1229 and kN2 1391 Table A2 Analysis The molar mass of the mixture is determined from 2 2 2 2 N N CO CO M y M y M m CO2 N2 500 K 360 ms The molar fractions are related to each other by 1 2 2 N CO y y The gas constant of the mixture is given by m u m M R R The specific heat ratio of the mixture is expressed as 2 2 2 2 N N CO CO mf mf k k k The mass fractions are m m M M y M M y 2 2 2 2 2 2 N N N CO CO CO mf mf The exit velocity equals the speed of sound at 500 K 1kJkg 1000 m s 2 2 exit kR T V m Substituting the given values and known properties and solving the above equations simultaneously using EES we find 0162 0838 2 2 N CO mf mf PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1370 1393E A mixture of nitrogen and oxygen is expanded isothermally The work produced is to be determined Assumptions 1 Nitrogen and oxygen are ideal gases 2 The process is reversible Properties The mole numbers of nitrogen and oxygen are 280 and 320 lbmlbmol respectively Table A1E Analysis The mole fractions are 0 6667 03 kmol kmol 20 0 3333 03 lbmol lbmol 10 total O2 O2 total N2 N2 N N y N N y 01 lbmol N2 02 lbmol O2 300 psia 5 ft3 The gas constant for this mixture is then 3499 psia ft lbm R 0 1Btu 5 404 psia ft 006475 Btulbm R 006475 Btulbm R 0 6667 32lbmlbmol 0 3333 28 9858 Btulbmol R 1 3 3 O2 O2 N2 N2 M y M y R R u The mass of this mixture of gases is lbm 29 32 20 28 10 O2 O2 N2 N2 M N M N m The temperature of the mixture is 466 0 R lbm0349 9 psia ft lbm R 92 psia5 ft 300 3 3 1 1 1 mR P T V Noting that Pv RT for an ideal gas the work done for this process is then 1924 Btu 3 3 1 2 2 1 2 1 out ft 5 lbm 0 06475 Btulbm R466 R ln 10 ft 29 ln v v v v v mRT d mRT m Pd W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1371 1394 A mixture of nitrogen and carbon dioxide is compressed at constant temperature in a closed system The work required is to be determined Assumptions 1 Nitrogen and carbon dioxide are ideal gases 2 The process is reversible Properties The mole numbers of nitrogen and carbon dioxide are 280 and 440 kgkmol respectively Table A1 Analysis The effective molecular weight of this mixture is 4 kgkmol 30 0 1544 0 8528 CO2 CO2 N2 N2 M y M y M 85 N2 15 CO2 by mole 100 kPa 27C Q The work done is determined from 1320 kJkg 100 kPa 300 Kln 500 kPa 30 4 kgkmol 314 kJkmol K 8 ln ln ln 1 2 1 2 1 2 2 1 2 1 P P M RT R P P RT RT d RT Pd w u v v v v V 1395 The specific heat ratio and an apparent molecular weight of a mixture of ideal gases are given The work required to compress this mixture isentropically in a closed system is to be determined Analysis For an isentropic process of an ideal gas with constant specific heats the work is expressed as 1 1 1 1 1 2 1 1 1 1 1 2 1 1 2 1 1 1 2 1 out k k k k k k k k P k P d P Pd w v v v v v v v v v v Gas mixture k135 M3 ol 2 kgkm 100 kPa 20C since k for an isentropic process Also k P P v v 1 1 2 1 1 2 1 1 1 P P RT P k v v v Substituting we obtain 1776 kJkg 1 100 kPa 1000 kPa 1 35 32 kgkmol1 8 314 kJkmol K293 K 1 1 1 35 1 35 1 1 1 2 1 out k k u P P k M R T w The negative sign shows that the work is done on the system PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1372 1396 A mixture of gases is placed in a springloaded pistoncylinder device The device is now heated until the pressure rises to a specified value The total work and heat transfer for this process are to be determined Properties The molar masses of Ne O2 and N2 are 2018 320 280 kgkmol respectively and the gas constants are 04119 02598 and 02968 kJkgK respectively Table A1 The constantvolume specific heats are 06179 0658 and 0743 kJkgK respectively Table A2a Analysis The total pressure is 200 kPa and the partial pressures are 25 Ne 50 O2 25 N2 by pressure 01 m3 10C 200 kPa 50 kPa 0 25200 kPa 100 kPa 0 50200 kPa 50 kPa 0 25200 kPa N2 N2 O2 O2 Ne Ne m m m P y P P y P P y P The mass of each constituent for a volume of 01 m3 and a temperature of 10C are Q 0 2384 kg 0 05953 0 1360 04289 0 005953 kg kPa m kg K283 K 02968 m kPa01 50 01360 kg kPa m kg K283 K 02598 m kPa01 100 004289 kg kPa m kg K283 K 04119 m kPa01 50 total 3 3 N2 N2 N2 3 3 O2 O2 O2 3 3 Ne Ne Ne m T R P m T R P m T R P m m m m V V V P kPa 1 2 The mass fractions are 500 0 2497 02384 kg 005953 kg mf 0 5705 02384 kg 01360 kg mf 0 1799 02384 kg 004289 kg mf N2 N2 O2 O2 Ne Ne m m m m m m m m m 200 01 V m3 The constantvolume specific heat of the mixture is determined from 0 672 kJkg K 0 743 0 2497 0 658 0 5705 0 6179 1799 0 mf mf mf N2 N2 O2 O2 Ne Ne v v v v c c c c The moles are 008502 kmol 0 0 002126 kmol 28 kgkmol kg 005953 0 00425 kmol 32 kgkmol kg 01360 0 002126 kmol 2018 kgkmol kg 004289 N2 O2 Ne m N2 N2 N2 O2 O2 O2 Ne Ne Ne N N N N M m N M m N M m N Then the apparent molecular weight of the mixture becomes 2804 kgkmol 0008502 kmol 02384 kg m m m N m M The apparent gas constant of the mixture is 0 2964 kJkg K 2805 kgkmol 8314 kJkmol K m u M R R The mass contained in the system is 02384 kg kPa m kg K283 K 02964 kPa01 m 200 3 3 1 1 1 RT P m V Noting that the pressure changes linearly with volume the final volume is determined by linear interpolation to be PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1373 3 2 2 0 4375 m 01 10 01 200 1000 200 500 V V The final temperature is 3096 K kg02964 kPa m kg K 02384 kPa0437 5 m 500 3 3 2 2 2 mR P T V The work done during this process is 118 kJ 3 1 2 2 1 out m 10 0 4375 2 200 kPa 500 2 V V P P W An energy balance on the system gives 569 kJ 283 K 0 2384 kg0672 kJkg K3096 118 1 2 out in T T mc W Q v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1374 1397 A springloaded pistoncylinder device is filled with a mixture of nitrogen and carbon dioxide whose mass fractions are given The gas is heated until the volume has doubled The total work and heat transfer for this process are to be determined Properties The molar masses of N2 and CO2 are 280 and 440 kgkmol respectively Table A1 The constantvolume specific heats of these gases at room temperature are 0743 and 0657 kJkgK respectively Table A2a Analysis We consider 100 kg of this mixture The mole numbers of each component are 1 023 kmol 44 kgkmol kg 45 1 964 kmol 28 kgkmol kg 55 CO2 CO2 CO2 N2 N2 N2 M m N M m N 55 N2 45 CO2 by mass 01 m3 45C 200 kPa The mole number of the mixture is 2 987 kmol 1 023 1 964 CO2 N2 N N N m Q The apparent molecular weight of the mixture is 3348 kgkmol 2987 kmol 100 kg m m m N m M The constantvolume specific heat of the mixture is determined from 0 7043 kJkg K 0 657 0 45 0 743 0 55 mf mf CO2 CO2 N2 N2 v v v c c c The apparent gas constant of the mixture is P kPa 02483 kJkg K 3348 kgkmol 8134 kJkmol K m u M R R 1 2 Noting that the pressure changes linearly with volume the initial volume is determined by linear interpolation using the data of the previous problem to be 200 3 1 1 m 10 01 10 01 200 1000 200 200 V V V m3 The final volume is 3 3 1 2 m 20 m 10 2 2 V V The final pressure is similarly determined by linear interpolation using the data of the previous problem to be 288 9 kPa 01 10 01 0 2 200 1000 200 2 2 P P The mass contained in the system is 0 2533 kg kPa m kg K318 K 02483 m kPa01 200 3 3 1 1 1 RT P m V The final temperature is 918 7 K kg02483 kPa m kg K 02533 kPa02 m 2889 3 3 2 2 2 mR P T V The work done during this process is 244 kJ 3 1 2 2 1 out m 10 20 2 288 9 kPa 200 2 V V P P W An energy balance on the system gives 132 kJ 318 K 0 2533 kg07043 kJkg K918 7 24 4 1 2 out in T T mc W Q v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1375 1398 A springloaded pistoncylinder device is filled with a mixture of nitrogen and carbon dioxide whose mass fractions are given The gas is heated until the pressure has tripled The total work and heat transfer for this process are to be determined Properties The molar masses of N2 and CO2 are 280 and 440 kgkmol respectively Table A1 The constantvolume specific heats of these gases at room temperature are 0743 and 0657 kJkgK respectively Table A2a Analysis We consider 100 kg of this mixture The mole numbers of each component are 1 023 kmol 44 kgkmol kg 45 1 964 kmol 28 kgkmol kg 55 CO2 CO2 CO2 N2 N2 N2 M m N M m N 55 N2 45 CO2 by mass 01 m3 45C 200 kPa The mole number of the mixture is Q 2 987 kmol 1 023 1 964 CO2 N2 N N N m The apparent molecular weight of the mixture is 3348 kgkmol 2987 kmol 100 kg m m m N m M The constantvolume specific heat of the mixture is determined from 0 7043 kJkg K 0 657 0 45 0 743 0 55 mf mf CO2 CO2 N2 N2 v v v c c c The apparent gas constant of the mixture is 02483 kJkg K 3348 kgkmol 8134 kJkmol K m u M R R P kPa 1 2 Noting that the pressure changes linearly with volume the initial volume is determined by linear interpolation using the data of the previous problem to be 3 1 1 m 10 01 10 01 200 1000 200 200 V V 200 The final pressure is V m3 600 kPa 3200 kPa 3 1 2 P P The final volumee is similarly determined by linear interpolation using the data of the previous problem to be 3 2 2 0 55 m 01 10 01 200 1000 200 600 V V The mass contained in the system is 0 2533 kg kPa m kg K318 K 02483 kPa01 m 200 3 3 1 1 1 RT P m V The final temperature is 5247 K kg02483 kPa m kg K 02533 kPa055 m 600 3 3 2 2 2 mR P T V The work done during this process is 180 kJ 3 1 2 2 1 out m 10 0 55 2 600 kPa 200 2 V V P P W An energy balance on the system gives 1059 kJ 318 K 0 2533 kg07043 kJkg K5247 180 1 2 out in T T mc W Q v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1376 1399 The masses pressures and temperatures of the constituents of a gas mixture in a tank are given Heat is transferred to the tank The final pressure of the mixture and the heat transfer are to be determined Assumptions He is an ideal gas and O2 is a nonideal gas Properties The molar masses of He and O2 are 40 and 320 kgkmol Table A1 Analysis a The number of moles of each gas is 125 kmol 025 kmol kmol 1 025 kmol 32 kgkmol kg 8 1 kmol 40 kgkmol kg 4 2 2 2 2 O He O O O He He He N N N M m N M m N m 4 kg He 8 kg O2 170 K 7 MPa Q Then the partial volume of each gas and the volume of the tank are He 3 3 1 1 He He 0202 m 7000 kPa 1 kmol831 4 kPa m kmol K170 K m u P R T N V O2 0 53 1 10 154 8 170 1 38 5 08 7 1 O cr 1 O cr 1 2 1 2 1 Z T T T P P P R m R Fig A15 3 3 3 O He tank 3 3 1 1 O O 0229 m 0027 m m 0202 0027 m 7000 kPa 5 kmol831 4 kPa m kg K170 K 05302 2 2 2 V V V V m u P R T ZN The partial pressure of each gas and the total final pressure is He 7987 kPa m 0229 kmol831 4 kPa m kmol K220 K 1 3 3 tank 2 He He2 V R T N P u O2 39 0 3 616 kPa m kmol K1548 K5080 kPa 8314 m 025 kmol 0229 1 42 154 8 220 3 3 crO O cr O crO O cr O O O cr 2 2 2 2 2 2 2 2 2 2 R u m u R R P P T R N P T R T T T V v v Fig A15 997 MPa 1981 MPa MPa 7987 1981 MPa 1981 kPa 5080 kPa 039 2 2 2 O He 2 m cr O O P P P P P P R PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1377 b We take both gases as the system No work or mass crosses the system boundary therefore this is a closed system with no work interactions Then the energy balance for this closed system reduces to E E E Q U U U in out system in He O2 He 6231 kJ 4 kg 31156 kJkg K 220 170 K 1 He T T mc U m v O2 21 1 963 5 08 9 97 42 1 22 1 38 10 1 2 2 2 1 1 1 h R R h R R Z P T Z P T Fig A29 2742 kJkmol 4949kJkmol 6404 8314 kJkmol K1548 K22 12 1 ideal 2 cr 1 2 2 1 h h Z Z R T h h h h u Also 828 kPa 6172 kPa kPa 7000 6172 kPa m 0229 kmol831 4 kPa m kg K170 K 1 He1 1 1 O 3 3 tank 1 He 1 He 2 P P P R T N P m u V Thus 421 5 kJ 1981 8280229kPa m kmol2742 kJkmol 025 3 tank O 1 2 O 1 2 O 1 1 2 2 1 2 O O 2 2 2 2 2 V V V P P h h N P P h h N U Substituting 1045 kJ 421 5 kJ kJ 6231 in Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1378 13100 A mixture of carbon dioxide and methane expands through a turbine The power produced by the mixture is to be determined using ideal gas approximation and Kays rule Assumptions The expansion process is reversible and adiabatic isentropic Properties The molar masses of CO2 and CH4 are 440 and 160 kgkmol and respectively The critical properties are 3042 K 7390 kPa for CO2 and 1911 K and 4640 kPa for CH4 Table A1 EES may use slightly different values Analysis The molar mass of the mixture is determined to be 37 0 kgkmol 02516 07544 4 4 2 2 CH CH CO CO M y M y M m The gas constant is 75 CO2 25 CH4 1300 K 1000 kPa 180 Ls 0 2246 kJkgK 370 kgkmol 8 314 kJkmolK m u M R R The mass fractions are 0 1083 370 kgkmol 0 25 16 kgkmol mf 0 8917 370 kgkmol 0 75 44 kgkmol mf 4 4 4 2 2 2 CH CH CH CO CO CO m m M M y M M y 100 kPa Ideal gas solution Using Daltons law to find partial pressures the entropies at the initial state are determined from EES to be 1622 kJkgK 250 kPa 0 25 1000 K 1600 6 068 kJkgK 750 kPa 0 75 1000 K 1300 1 CH 1 CO 4 2 s P T s P T The final state entropies cannot be determined at this point since the final temperature is not known However for an isentropic process the entropy change is zero and the final temperature may be determined from 0 CH 1 2 CH CH CO 1 2 CO CO CH CH CO CO total 4 4 4 2 2 2 4 4 2 2 s s mf s s mf s mf s mf s The solution is obtained using EES to be T2 9471 K The initial and final enthalpies and the changes in enthalpy are from EES 2503 kJkg 8248 kJkg 947 1 K 831 kJkg 7803 kJkg 1300 K 2 CH CO 2 2 1 CH CO 1 1 4 2 4 2 h h T h h T Noting that the heat transfer is zero an energy balance on the system gives m m m h W m h W Q out out in where 577 7 kJkg 831 0 1083 2503 7803 0 8917 8248 mf mf CH 1 CH 2 CH CO 1 CO 2 CO 4 4 4 2 2 2 h h h h hm The mass flow rate is 0 6165 kgs 02246 kJkgK1300 K 1000 kPa0180 m s 3 1 1 1 RT P m V Substituting 356 kW 0 6165 577 7 kJkg out m hm W PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1379 Kays rule solution The critical temperature and pressure of the mixture is 6683 kPa 025464 0 kPa 0 kPa 075739 276 K 0251911 K 2 K 075304 4 4 2 2 4 4 2 2 crCH CH crCO CO cr crCH CH crCO CO cr P y P y P T y T y T State 1 properties from EES 0 001197 0064 0 003 1 0 1496 6683 kPa kPa 1000 4 715 276 K K 1300 1 1 1 cr 1 1 cr 1 1 s h R R Z Z Z P P P T T T 0 399 kJkg 0 0064 0 2246 kJkgK276 K cr 1 1 RT Z h h 7047 kJkg 0 399 0 1083 831 0 8917 7803 1 CH 1 CH CO 1 CO 1 4 4 2 2 h h mf h mf h 0 0002688 kJkgK 0 001197 0 2246 kJkgK 1 1 Z R s s 7 1679 kJkgK 0 0002688 0 10831622 0 8917 6 068 mf mf 1 CH 1 CH CO 1 CO 1 4 4 2 2 s s s s The final state entropies cannot be determined at this point since the final temperature is not known However for an isentropic process the entropy change is zero and the final temperature may be determined from mf mf 0 mf mf CH 1 2 CH CH CO 1 2 CO CO CH CH CO CO total 4 4 4 2 2 2 4 4 2 2 s s s s s s s The solution is obtained using EES to be T2 947 K The initial and final enthalpies and the changes in enthalpy are 0 0004057 from EES 0004869 0 0 015 6683 kPa kPa 100 3 434 276 K K 947 2 2 cr 2 2 cr 2 2 s h R R Z Z P P P T T T 0 03015 kJkg 0 0004869 0 2246 kJkgK276 K 2 2 Zh RTcr h 7625 kJkg 0 03015 0 1083 2503 0 8917 8248 mf mf 2 CH 2 CH CO 2 CO 2 4 4 2 2 h h h h Noting that the heat transfer is zero an energy balance on the system gives 1 2 out out in h m h W m h W Q m where the mass flow rate is 0 6149 kgs 100302246 kJkgK1300 K 1000 kPa0180 m s 3 1 1 1 1 Z RT P m V Substituting 356 kW 7047 kJkg 0 6149 kgs 7625 Wout PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1381 13102E The mass percentages of a gas mixture are given This mixture is expanded in an adiabatic steadyflow turbine of specified isentropic efficiency The second law efficiency and the exergy destruction during this expansion process are to be determined Assumptions All gases will be modeled as ideal gases with constant specific heats Properties The molar masses of N2 He CH4 and C2H6 are 280 40 160 and 300 lbmlbmol respectively Table A1E The constantpressure specific heats of these gases at room temperature are 0248 125 0532 and 0427 BtulbmR respectively Table A2Ea Analysis For 1 lbm of mixture the mole numbers of each component are 400 psia 500F N2 He CH4 C2H6 mixture 0 006667 lbmol 30 lbmlbmol lbm 020 0 0375 lbmol 16 lbmlbmol lbm 06 0 0125 lbmol 4 lbmlbmol lbm 005 0 005357 lbmol 28 lbmlbmol lbm 015 C2H6 C2H6 N2 Ch4 CH4 N2 He He N2 N2 N2 N2 M m N M m N M m N M m N 20 psia The mole number of the mixture is 0 06202 lbmol 0 006667 0 0375 0 0125 0 005357 He CO2 O2 N N N N m The apparent molecular weight of the mixture is 1612 lbmlbmol 0065202 lbmol 1lbm m m m N m M The apparent gas constant of the mixture is 01232 Btulbm R 1612 lbmlbmol 19858 lbmlbmol R m u M R R The constantpressure specific heat of the mixture is determined from 0 5043 Btulbm R 0 427 0 20 0 532 0 60 1 25 0 05 0248 15 0 mf mf mf mf C2H6 C2H6 CH4 CH4 He He N2 N2 p p p p p c c c c c Then the constantvolume specific heat is 0 3811 Btulbm R 0 1232 0 5043 R c c p v The specific heat ratio is 1 323 0 3811 0 5043 vc c k p The temperature at the end of the expansion for the isentropic process is 462 0 R 400 psia 20 psia 960 R 03231323 1 1 2 1 2 k k s P P T T Using the definition of turbine isentropic efficiency the actual outlet temperature is 536 7 R 462 0 0 85960 960 R 2 1 turb 1 2 T s T T T η The entropy change of the gas mixture is 0 07583 Btulbm R 400 0 1232ln 20 960 0 5043ln 536 7 ln ln 1 2 1 2 1 2 P P R T T c s s p PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1382 The actual work produced is 213 5 Btulbm R 536 7 0 5043 Btulbm R960 2 1 2 1 out T T c h h w p The reversible work output is 254 2 Btulbm 0 07583 Btulbm R 537 R 213 5 Btulbm 2 1 0 2 1 revout s s T h h w The secondlaw efficiency and the exergy destruction are then 840 0840 254 2 5 213 out rev out II w w η 407 Btulbm 213 5 254 2 out revout dest w w x PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1383 13103 A program is to be written to determine the mole fractions of the components of a mixture of three gases with known molar masses when the mass fractions are given and to determine the mass fractions of the components when the mole fractions are given Also the program is to be run for a sample case Analysis The problem is solved using EES and the solution is given below Procedure FractionsTypeABCABCmfAmfBmfCyAyByC If Type mass fraction OR mole fraction then Call ERRORType must be set equal to mass fraction or mole fraction GOTO 10 endif Sum ABC If ABSSum 1 0 then goto 20 MMA molarmassA MMB molarmassB MMC molarmassC If Type mass fraction then mfA A mfB B mfC C sumMmix mfAMMA mfBMMB mfCMMC yA mfAMMAsumMmix yB mfBMMBsumMmix yC mfCMMCsumMmix GOTO 10 endif if Type mole fraction then yA A yB B yC C MMmix yAMMA yBMMB yCMMC mfA yAMMAMMmix mfB yBMMBMMmix mfC yCMMCMMmix GOTO 10 Endif Call ERRORType must be either mass fraction or mole fraction GOTO 10 20 Call ERRORThe sum of the mass or mole fractions must be 1 10 END Either the mole fraction yi or the mass fraction mfi may be given by setting the parameter Typemole fraction when the mole fractions are given or Typemass fraction is given Input Data in the Diagram Window Typemole fraction A N2 B O2 C Argon A 071 When Typemole fraction A B C are the mole fractions B 028 When Typemass fraction A B C are the mass fractions C 001 Call FractionsTypeABCABCmfAmfBmfCyAyByC SOLUTION A071 AN2 B028 BO2 C001 CArgon mfA0680 mfB0306 mfC0014 Typemole fraction yA0710 yB0280 yC0010 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1384 13104 A program is to be written to determine the entropy change of a mixture of 3 ideal gases when the mole fractions and other properties of the constituent gases are given Also the program is to be run for a sample case Analysis The problem is solved using EES and the solution is given below T1300 K T2600 K P1100 kPa P2500 kPa A N2 B O2 C Argon yA 071 yB 028 yC 001 MMA molarmassA MMB molarmassB MMC molarmassC MMmix yAMMA yBMMB yCMMC mfA yAMMAMMmix mfB yBMMBMMmix mfC yCMMCMMmix DELTAsmixmfAentropyATT2PyBP2 entropyATT1PyAP1mfBentropyBTT2PyBP2 entropyBTT1PyBP1mfCentropyCTT2PyCP2entropyCTT1PyCP1 SOLUTION AN2 BO2 CArgon DELTAsmix1241 kJkgK mfA068 mfB03063 mfC001366 MMA2801 kgkmol MMB32 kgkmol MMC3995 kgkmol MMmix2925 kJkmol P1100 kPa P2500 kPa T1300 K T2600 K yA071 yB028 yC001 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1385 Fundamentals of Engineering FE Exam Problems 13105 An ideal gas mixture whose apparent molar mass is 20 kgkmol consists of nitrogen N2 and three other gases If the mole fraction of nitrogen is 055 its mass fraction is a 015 b 023 c 039 d 055 e 077 Answer e 077 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Mmix20 kgkmol MN228 kgkmol yN2055 mfN2MN2MmixyN2 Some Wrong Solutions with Common Mistakes W1mf yN2 Taking mass fraction to be equal to mole fraction W2mf yN2MmixMN2 Using the molar mass ratio backwords W3mf 1mfN2 Taking the complement of the mass fraction 13106 An ideal gas mixture consists of 2 kmol of N2 and 6 kmol of CO2 The mass fraction of CO2 in the mixture is a 0175 b 0250 c 0500 d 0750 e 0825 Answer e 0825 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values N12 kmol N26 kmol NmixN1N2 MM128 kgkmol MM244 kgkmol mmixN1MM1N2MM2 mf2N2MM2mmix Some Wrong Solutions with Common Mistakes W1mf N2Nmix Using mole fraction W2mf 1mf2 The wrong mass fraction PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1386 13107 An ideal gas mixture consists of 2 kmol of N2 and 4 kmol of CO2 The apparent gas constant of the mixture is a 0215 kJkgK b 0225 kJkgK c 0243 kJkgK d 0875 kJkgK e 124 kJkgK Answer a 0215 kJkgK Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Ru8314 kJkmolK N12 kmol N24 kmol MM128 kgkmol MM244 kgkmol R1RuMM1 R2RuMM2 NmixN1N2 y1N1Nmix y2N2Nmix MMmixy1MM1y2MM2 RmixRuMMmix Some Wrong Solutions with Common Mistakes W1Rmix R1R22 Taking the arithmetic average of gas constants W2Rmix y1R1y2R2 Using wrong relation for Rmixture 13108 A rigid tank is divided into two compartments by a partition One compartment contains 3 kmol of N2 at 400 kPa pressure and the other compartment contains 7 kmol of CO2 at 200 kPa Now the partition is removed and the two gases form a homogeneous mixture at 250 kPa The partial pressure of N2 in the mixture is a 75 kPa b 90 kPa c 125 kPa d 175 kPa e 250 kPa Answer a 75 kPa Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P1 400 kPa P2 200 kPa Pmix250 kPa N13 kmol N27 kmol MM128 kgkmol MM244 kgkmol NmixN1N2 y1N1Nmix y2N2Nmix PN2y1Pmix Some Wrong Solutions with Common Mistakes W1P1 Pmix2 Assuming equal partial pressures W2P1 mf1Pmix mf1N1MM1N1MM1N2MM2 Using mass fractions W3P1 PmixN1P1N1P1N2P2 Using some kind of weighed averaging PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1387 13109 An 80L rigid tank contains an ideal gas mixture of 5 g of N2 and 5 g of CO2 at a specified pressure and temperature If N2 were separated from the mixture and stored at mixture temperature and pressure its volume would be a 32 L b 36 L c 40 L d 49 L e 80 L Answer d 49 L Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Vmix80 L m15 g m25 g MM128 kgkmol MM244 kgkmol N1m1MM1 N2m2MM2 NmixN1N2 y1N1Nmix V1y1Vmix L Some Wrong Solutions with Common Mistakes W1V1Vmixm1m1m2 Using mass fractions W2V1 Vmix Assuming the volume to be the mixture volume 13110 An ideal gas mixture consists of 3 kg of Ar and 6 kg of CO2 gases The mixture is now heated at constant volume from 250 K to 350 K The amount of heat transfer is a 374 kJ b 436 kJ c 488 kJ d 525 kJ e 664 kJ Answer c 488 kJ Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values T1250 K T2350 K Cv103122 Cp105203 kJkgK Cv20657 Cp20846 kJkgK m13 kg m26 kg MM13995 kgkmol MM244 kgkmol Applying Energy balance gives QDeltaUDeltaUArDeltaUCO2 Qm1Cv1m2Cv2T2T1 Some Wrong Solutions with Common Mistakes W1Q m1m2Cv1Cv22T2T1 Using arithmetic average of properties W2Q m1Cp1m2Cp2T2T1Using Cp instead of Cv W3Q m1Cv1m2Cv2T2 Using T2 instead of T2T1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1388 13111 An ideal gas mixture consists of 60 helium and 40 argon gases by mass The mixture is now expanded isentropically in a turbine from 400C and 12 MPa to a pressure of 200 kPa The mixture temperature at turbine exit is a 56C b 195C c 130C d 112C e 400C Answer a 56C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values T1400273K P11200 kPa P2200 kPa mfHe06 mfAr04 k11667 k21667 The specific heat ratio k of the mixture is also 1667 since k1667 for all componet gases kmix1667 T2T1P2P1kmix1kmix273 Some Wrong Solutions with Common Mistakes W1T2 T1273P2P1kmix1kmix Using C for T1 instead of K W2T2 T1P2P1kair1kair273 kair14 Using k value for air W3T2 T1P2P1 Assuming T to be proportional to P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1389 13112 One compartment of an insulated rigid tank contains 2 kmol of CO2 at 20C and 150 kPa while the other compartment contains 5 kmol of H2 gas at 35C and 300 kPa Now the partition between the two gases is removed and the two gases form a homogeneous ideal gas mixture The temperature of the mixture is a 25C b 29C c 22C d 32C e 34C Answer b 29C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values NH25 kmol T1H235 C P1H2300 kPa NCO22 kmol T1CO220 C P1CO2150 kPa CvH210183 CpH214307 kJkgK CvCO20657 CpCO20846 kJkgK MMH22 kgkmol MMCO244 kgkmol mH2NH2MMH2 mCO2NCO2MMCO2 Applying Energy balance gives 0DeltaUDeltaUH2DeltaUCO2 0mH2CvH2T2T1H2mCO2CvCO2T2T1CO2 Some Wrong Solutions with Common Mistakes 0mH2CpH2W1T2T1H2mCO2CpCO2W1T2T1CO2 Using Cp instead of Cv 0NH2CvH2W2T2T1H2NCO2CvCO2W2T2T1CO2 Using N instead of mass W3T2 T1H2T1CO22 Assuming averate temperature PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1390 13113 A pistoncylinder device contains an ideal gas mixture of 3 kmol of He gas and 7 kmol of Ar gas at 50C and 400 kPa Now the gas expands at constant pressure until its volume doubles The amount of heat transfer to the gas mixture is a 62 MJ b 42 MJ c 27 MJ d 10 MJ e 67 MJ Answer e 67 MJ Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values NHe3 kmol NAr7 kmol T150273 C P1400 kPa P2P1 T22T1 since PVTconst for ideal gases and it is given that Pconstant T22T1 K MMHe4 kgkmol MMAr3995 kgkmol mHeNHeMMHe mArNArMMAr CpAr05203 CvAr 3122 kJkgC CpHe51926 CvHe 31156 kJkgK For a Pconst process QDeltaH since DeltaUWb is DeltaH QmArCpArT2T1mHeCpHeT2T1 Some Wrong Solutions with Common Mistakes W1Q mArCvArT2T1mHeCvHeT2T1 Using Cv instead of Cp W2QNArCpArT2T1NHeCpHeT2T1 Using N instead of mass W3QmArCpArT22T1mHeCpHeT22T1 T222T1273273 Using C for T1 W4QmArmHe05CpArCpHeT2T1 Using arithmetic averate of Cp PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 141 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 14 GASVAPOR MIXTURES AND AIR CONDITIONING PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 142 Dry and Atmospheric Air Specific and Relative Humidity 141C Dry air does not contain any water vapor but atmospheric air does 142C Yes by cooling the air at constant pressure 143C Yes 144C Specific humidity will decrease but relative humidity will increase 145C Yes the water vapor in the air can be treated as an ideal gas because of its very low partial pressure 146C The same This is because water vapor behaves as an ideal gas at low pressures and the enthalpy of an ideal gas depends on temperature only 147C Specific humidity is the amount of water vapor present in a unit mass of dry air Relative humidity is the ratio of the actual amount of vapor in the air at a given temperature to the maximum amount of vapor air can hold at that temperature 148C The specific humidity will remain constant but the relative humidity will decrease as the temperature rises in a wellsealed room 149C The specific humidity will remain constant but the relative humidity will decrease as the temperature drops in a wellsealed room 1410C A tank that contains moist air at 3 atm is located in moist air that is at 1 atm The driving force for moisture transfer is the vapor pressure difference and thus it is possible for the water vapor to flow into the tank from surroundings if the vapor pressure in the surroundings is greater than the vapor pressure in the tank PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 143 1411C Insulations on chilled water lines are always wrapped with vapor barrier jackets to eliminate the possibility of vapor entering the insulation This is because moisture that migrates through the insulation to the cold surface will condense and remain there indefinitely with no possibility of vaporizing and moving back to the outside 1412C When the temperature total pressure and the relative humidity are given the vapor pressure can be determined from the psychrometric chart or the relation sat P Pv φ where Psat is the saturation or boiling pressure of water at the specified temperature and φ is the relative humidity 1413 A tank contains dry air and water vapor at specified conditions The specific humidity the relative humidity and the volume of the tank are to be determined Assumptions The air and the water vapor are ideal gases Analysis a The specific humidity can be determined form its definition 00143 kg H Okg dry air 2 21 kg kg 30 a v m m ω 21 kg dry air 03 kg H2O vapor 30C 100 kPa b The saturation pressure of water at 30C is 4 2469 kPa sat 30 C P Pg Then the relative humidity can be determined from 529 0 0143 4 2469 kPa 0 622 0 0143100 kPa 0 622 Pg P ω ω φ c The volume of the tank can be determined from the ideal gas relation for the dry air 187 m3 97755 kPa 21 kg0287 kJkg K303 K 97755 kPa 2 245 100 0 529 4 2469 kPa 2245 kPa a a a v a g v P R T m P P P P P V φ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 144 1414 A tank contains dry air and water vapor at specified conditions The specific humidity the relative humidity and the volume of the tank are to be determined Assumptions The air and the water vapor are ideal gases Analysis a The specific humidity can be determined form its definition 00143 kg H Okg dry air 2 21 kg kg 30 a v m m ω 21 kg dry air 03 kg H2O vapor 20C 100 kPa b The saturation pressure of water at 24C is 2 339 kPa sat 20 C P Pg Then the relative humidity can be determined from 960 0 960 0 0143 2 339 kPa 0 622 0 0143100 kPa 0 622 Pg P ω ω φ c The volume of the tank can be determined from the ideal gas relation for the dry air 181 m3 97755 kPa 21 kg0287 kJkg K293 K 97755 kPa 2 245 100 0 960 2 339 kPa 2245 kPa a a a v a g v P R T m P P P P P V φ 1415 A room contains air at specified conditions and relative humidity The partial pressure of air the specific humidity and the enthalpy per unit mass of dry air are to be determined Assumptions The air and the water vapor are ideal gases Analysis a The partial pressure of dry air can be determined from AIR 20C 98 kPa 85 RH 9601 kPa 1 988 98 0 85 2 3392 kPa 1988 kPa 20 C sat v a g v P P P P P P φ φ b The specific humidity of air is determined from 00129 kg H Okg dry air 2 1 988 kPa 98 0 622 1 988 kPa 622 0 v v P P P ω c The enthalpy of air per unit mass of dry air is determined from 5278 kJkg dry air C20 C 0012925374 kJkg 1 005 kJkg g p v a h c T h h h ω ω PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 145 1416 A room contains air at specified conditions and relative humidity The partial pressure of air the specific humidity and the enthalpy per unit mass of dry air are to be determined Assumptions The air and the water vapor are ideal gases Analysis a The partial pressure of dry air can be determined from AIR 20C 85 kPa 85 RH 8301 kPa 1 988 85 0 85 2 3392 kPa 1988 kPa 20 C sat v a g v P P P P P P φ φ b The specific humidity of air is determined from 00149 kg H Okg dry air 2 1 988 kPa 85 0 622 1 988 kPa 622 0 v v P P P ω c The enthalpy of air per unit mass of dry air is determined from 5790 kJkg dry air C20 C 0014925374 kJkg 1 005 kJkg g p v a h c T h h h ω ω 1417E A room contains air at specified conditions and relative humidity The partial pressure of air the specific humidity and the enthalpy per unit mass of dry air are to be determined Assumptions The air and the water vapor are ideal gases Analysis a The partial pressure of dry air can be determined from AIR 85F 135 psia 60 RH 1314 psia 0 358 5 13 0 60 0 5966 psia 0358 psia 85 F sat v a g v P P P P P P φ φ b The specific humidity of air is determined from 00169 lbm H Olbm dry air 2 0 358 psia 135 0 622 0 358 psia 622 0 v v P P P ω c The enthalpy of air per unit mass of dry air is determined from 3901 Btulbm dry air F85 F 0016910983 Btulbm 0 24 Btulbm g p v a h c T h h h ω ω PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 146 1418 The masses of dry air and the water vapor contained in a room at specified conditions and relative humidity are to be determined Assumptions The air and the water vapor are ideal gases Analysis The partial pressure of water vapor and dry air are determined to be 9659 kPa 1 41 98 0 50 2 811 kPa 141 kPa sat 23 C v a g v P P P P P P φ φ ROOM 240 m3 23C 98 kPa 50 RH The masses are determined to be kg 247 kg 2729 0 4615 kPa m kg K296 K 41 kPa240 m 1 0 287 kPa m kg K296 K 9659 kPa240 m 3 3 3 3 R T P m R T P m v v v a a a V V 1419E Humid air is expanded in an isentropic nozzle The amount of water vapor that has condensed during the process is to be determined Assumptions The air and the water vapor are ideal gases Properties The specific heat ratio of air at room temperature is k 14 Table A2a The saturation properties of water are to be obtained from water tables Analysis Since the mole fraction of the water vapor in this mixture is very small 606 1 R 75 psia 960 R 15 psia 0414 1 1 2 1 2 k k P P T T 15 psia 75 psia 500F ω10018 AIR We will assume that the air leaves the nozzle at a relative humidity of 100 will be verified later The vapor pressure and specific humidity at the outlet are then 3 381 psia 3381 psia 01 sat 1461 F 2 2 2 2 P P P g v φ φ 0181 lbm H Olbm dry air 3 381 psia 15 0 622 3 381 psia 0 622 2 2 2 2 v v P P P ω This is greater than the inlet specific humidity 0018 lbmlbm dry air and thus there will be no condensation of water vapor 0 lbm H Olbm dry air 2 2 1 ω ω ω PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 147 1420 Humid air is compressed in an isentropic compressor The relative humidity of the air at the compressor outlet is to be determined Assumptions The air and the water vapor are ideal gases Properties The specific heat ratio of air at room temperature is k 14 Table A2a The saturation properties of water are to be obtained from water tables Analysis At the inlet 800 kPa 0 90 2 3392 kPa 2105 kPa sat 20 C 1 1 1 1 P P P g v φ φ Humid air 00134 kg H Okg dry air 2 105 kPa 100 0 622 2 105 kPa 622 0 2 1 1 1 2 v v P P P ω ω Since the mole fraction of the water vapor in this mixture is very small 531 K 100 kPa 293 K 800 kPa 0414 1 1 2 1 2 k k P P T T 100 kPa 20C 90 RH The saturation pressure at this temperature is 4542 kPa from EES sat 258 C 2 P Pg The vapor pressure at the exit is 1687 kPa 0622 00134 0 0134800 0 622 2 2 2 2 ω ω P Pv The relative humidity at the exit is then 00037 037 4542 87 16 2 2 2 g v P P φ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 148 DewPoint Adiabatic Saturation and Wetbulb Temperatures 1421C Dewpoint temperature is the temperature at which condensation begins when air is cooled at constant pressure 1422C Andys The temperature of his glasses may be below the dewpoint temperature of the room causing condensation on the surface of the glasses 1423C The outer surface temperature of the glass may drop below the dewpoint temperature of the surrounding air causing the moisture in the vicinity of the glass to condense After a while the condensate may start dripping down because of gravity 1424C When the temperature falls below the dewpoint temperature dew forms on the outer surfaces of the car If the temperature is below 0C the dew will freeze At very low temperatures the moisture in the air will freeze directly on the car windows 1425C When the air is saturated 100 relative humidity 1426C These two are approximately equal at atmospheric temperatures and pressure 1427 A house contains air at a specified temperature and relative humidity It is to be determined whether any moisture will condense on the inner surfaces of the windows when the temperature of the window drops to a specified value Assumptions The air and the water vapor are ideal gases 10C 25C φ 65 Analysis The vapor pressure Pv is uniform throughout the house and its value can be determined from 2 06 kPa 0 65 3 1698 kPa 25 C g v P P φ The dewpoint temperature of the air in the house is 180C sat 206 kPa sat dp T T T vP That is the moisture in the house air will start condensing when the temperature drops below 180C Since the windows are at a lower temperature than the dewpoint temperature some moisture will condense on the window surfaces PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 149 1428 A person wearing glasses enters a warm room at a specified temperature and relative humidity from the cold outdoors It is to be determined whether the glasses will get fogged Assumptions The air and the water vapor are ideal gases Analysis The vapor pressure Pv of the air in the house is uniform throughout and its value can be determined from 12C 25C φ 55 1 743 kPa 0 55 3 1698 kPa 25 C g v P P φ The dewpoint temperature of the air in the house is 153C Table A5 or EES sat 1 743 kPa sat dp T T T vP That is the moisture in the house air will start condensing when the air temperature drops below 153C Since the glasses are at a lower temperature than the dewpoint temperature some moisture will condense on the glasses and thus they will get fogged 1429 A person wearing glasses enters a warm room at a specified temperature and relative humidity from the cold outdoors It is to be determined whether the glasses will get fogged Assumptions The air and the water vapor are ideal gases Analysis The vapor pressure Pv of the air in the house is uniform throughout and its value can be determined from 12C 25C φ 30 0 9509 kPa 0 30 3 1698 kPa 25 C g v P P φ The dewpoint temperature of the air in the house is 62C Table A5 or EES sat 0 9509 kPa sat dp T T T vP That is the moisture in the house air will start condensing when the air temperature drops below 62C Since the glasses are at a higher temperature than the dewpoint temperature moisture will not condense on the glasses and thus they will not get fogged 1430E A woman drinks a cool canned soda in a room at a specified temperature and relative humidity It is to be determined whether the can will sweat Assumptions The air and the water vapor are ideal gases Analysis The vapor pressure Pv of the air in the house is uniform throughout and its value can be determined from Cola 40F 80F 50 RH 0 254 psia 0 50 0 50745 psia 80 F g v P P φ The dewpoint temperature of the air in the house is 597F from EES sat 0 254 psia sat dp T T T vP That is the moisture in the house air will start condensing when the air temperature drops below 597C Since the canned drink is at a lower temperature than the dewpoint temperature some moisture will condense on the can and thus it will sweat PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1410 1431 The dry and wetbulb temperatures of atmospheric air at a specified pressure are given The specific humidity the relative humidity and the enthalpy of air are to be determined Assumptions The air and the water vapor are ideal gases Analysis a We obtain the properties of water vapor from EES The specific humidity ω1 is determined from 2 1 2 2 1 2 1 f g fg p h h h T T c ω ω 95 kPa 25C Twb 17C where T2 is the wetbulb temperature and ω2 is determined from 001295 kg H Okg dry air 95 1938 kPa 0 622 1 938 kPa 622 0 2 2 2 2 2 g g P P P ω Thus 000963 kg H Okg dry air 2 7136 kJkg 25465 25 C 00129524606 kJkg C17 005 kJkg 1 ω1 b The relative humidity φ1 is determined from 0 457 or 457 0 00963 3 1698 kPa 0 622 0 0096395 kPa 0 622 1 1 1 1 1 Pg P ω ω φ c The enthalpy of air per unit mass of dry air is determined from 4965 kJkg dry air C25 C 00096325465 kJkg 005 kJkg 1 1 1 1 1 1 1 1 g p v a h c T h h h ω ω PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1411 1432 The dry and wetbulb temperatures of air in room at a specified pressure are given The specific humidity the relative humidity and the dewpoint temperature are to be determined Assumptions The air and the water vapor are ideal gases Analysis a We obtain the properties of water vapor from EES The specific humidity ω1 is determined from 2 1 2 2 1 2 1 f g fg p h h h T T c ω ω 100 kPa 26C Twb 21C where T2 is the wetbulb temperature and ω2 is determined from 001587 kg H Okg dry air 2 488 kPa 100 0 622 2 488 kPa 622 0 2 2 2 2 2 g g P P P ω Thus 001377 kg H Okg dry air 2 8810 kJkg 25483 26 C 00158724512 kJkg C21 005 kJkg 1 ω1 b The relative humidity φ1 is determined from 0 644 or 644 0 01377 3 3638 kPa 0 622 0 01377100 kPa 0 622 1 1 1 1 1 Pg P ω ω φ c The vapor pressure at the inlet conditions is 2 166 kPa 0 644 3 3638 kPa 1 sat 26 C 1 1 1 P P P g v φ φ Thus the dewpoint temperature of the air is 188C sat 2 166 kPa sat dp T T T vP PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1412 1433 EES Problem 1432 is reconsidered The required properties are to be determined using EES at 100 and 300 kPa pressures Analysis The problem is solved using EES and the solution is given below Tdb26 C Twb21 C P1100 kPa P2300 kPa h1enthalpyAirH2OTTdbPP1BTwb v1volumeAirH2OTTdbPP1BTwb Tdp1dewpointAirH2OTTdbPP1BTwb w1humratAirH2OTTdbPP1BTwb Rh1relhumAirH2OTTdbPP1BTwb h2enthalpyAirH2OTTdbPP2BTwb v2volumeAirH2OTTdbPP2BTwb Tdp2dewpointAirH2OTTdbPP2BTwb w2humratAirH2OTTdbPP2BTwb Rh2relhumAirH2OTTdbPP2BTwb SOLUTION h16125 kJkg h23416 kJkg P1100 kPa P2300 kPa Rh106437 Rh204475 Tdb26 C Tdp11876 Tdp21307 Twb21 C v108777 m3kg v202877 m3kg w1001376 kgkg w20003136 kgkg Alternative Solution The following EES routine can also be used to solve this problem The above EES routine uses builtin psychrometric functions whereas the one blow uses analytical expressions together with steam properties Given Tdb26 C Twb21 C P100 kPa Properties Fluidsteamiapws Pg1pressureFluid TTdb x1 Pg2pressureFluid TTwb x1 hg1enthalpyFluid TTdb x1 hg2enthalpyFluid TTwb x1 hf2enthalpyFluid TTwb x0 hfg2hg2hf2 cp1005 kJkgC for air PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1413 Analysis a w20622Pg2PPg2 kg H2Okg dry air w1cpTwbTdbw2hfg2hg1hf2 b phi1w1P0622w1Pg1 c Pv1phi1Pg1 TdptemperatureFluid PPv1 x1 SOLUTION cp1005 kJkgC Fluidsteamiapws hf2881 kJkg hfg224512 kJkg hg125483 kJkg hg225393 kJkg P100 kPa phi106439 Pg133638 kPa Pg22488 kPa Pv12166 kPa Tdb26 C Tdp188 C Twb21 C w1001377 w2001587 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1414 1434E The dry and wetbulb temperatures of air in room at a specified pressure are given The specific humidity the relative humidity and the dewpoint temperature are to be determined Assumptions The air and the water vapor are ideal gases Analysis a The specific humidity ω1 is determined from 147 psia 80F Twb 65F 2 1 2 2 1 2 1 f g fg p h h h T T c ω ω where T2 is the wetbulb temperature and ω2 is determined from 001321 lbm H Olbm dry air 0 30578 psia 147 0 622 0 30578 psia 622 0 2 2 2 2 2 g g P P P ω Thus 000974 lbm H Olbm dry air 2 3308 Btulbm 10961 80 F 00132110565 Btulbm F65 0 24 Btulbm ω1 b The relative humidity φ1 is determined from 0 447 or 447 0 00974 0 50745 psia 0 622 0 0097414 7 psia 0 622 1 1 1 1 1 Pg P ω ω φ c The vapor pressure at the inlet conditions is 0 2268 psia 0 447 0 50745 psia 1 sat 70 F 1 1 1 P P P g v φ φ Thus the dewpoint temperature of the air is from EES 566F sat 0 2268 psia sat dp T T T vP PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1415 1435 Atmospheric air flows steadily into an adiabatic saturation device and leaves as a saturated vapor The relative humidity and specific humidity of air are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible m m m a a a 1 2 Analysis The exit state of the air is completely specified and the total pressure is 98 kPa The properties of the moist air at the exit state may be determined from EES to be 02079 kg H Okg dry air 0 11 kJkg dry air 78 2 2 2 ω h 35C 98 kPa AIR Water 25C Humidifier 25C 98 kPa 100 The enthalpy of makeup water is 10483 kJkg Table A 4 25 C 2 f w h h An energy balance on the control volume gives 7811 kJkg 10483 kJkg 0 02079 1 1 2 1 2 1 ω ω ω h h h h w Pressure and temperature are known for inlet air Other properties may be determined from this equation using EES A hand solution would require a trialerror approach The results are 04511 001654 kg H Okg dry air 2 1 1 1 66 kJkg dry air 77 φ ω h PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1416 Psychrometric Chart 1436C They are very nearly parallel to each other 1437C The saturation states located on the saturation curve 1438C By drawing a horizontal line until it intersects with the saturation curve The corresponding temperature is the dewpoint temperature 1439C No they cannot The enthalpy of moist air depends on ω which depends on the total pressure 1440 The pressure temperature and relative humidity of air in a room are specified Using the psychrometric chart the specific humidity the enthalpy the wetbulb temperature the dewpoint temperature and the specific volume of the air are to be determined Analysis From the psychrometric chart Fig A31 we read a 0 0087 kg H O kg dry air 2 ω b 45 4 kJkg dry air h c C 16 3 wb T d C 12 0 dp T e 0 851 m kg dry air 3 v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1417 1441 Problem 1440 is reconsidered The required properties are to be determined using EES Also the properties are to be obtained at an altitude of 2000 m Analysis The problem is solved using EES and the solution is given below Tdb23 C Rh050 P1101325 kPa Z 2000 m P21013251002256Zconvertmkm5256 Relation giving P as a function of altitude h1enthalpyAirH2OTTdbPP1RRh v1volumeAirH2OTTdbPP1RRh Tdp1dewpointAirH2OTTdbPP1RRh w1humratAirH2OTTdbPP1RRh Twb1wetbulbAirH2OTTdbPP1RRh h2enthalpyAirH2OTTdbPP2RRh v2volumeAirH2OTTdbPP2RRh Tdp2dewpointAirH2OTTdbPP2RRh w2humratAirH2OTTdbPP2RRh Twb2wetbulbAirH2OTTdbPP2RRh SOLUTION h1454 kJkg h25162 kJkg P11013 kPa P27949 kPa Rh05 Tdb23 C Tdp11203 C Tdp21203 C Twb11625 C Twb21567 C v108507 m3kg v21089 m3kg w10008747 kgkg w2001119 kgkg Z2000 m Discussion The atmospheric pressure for a given elevation can also be obtained from Table A16 of the book PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1418 1442 The pressure and the dry and wetbulb temperatures of air in a room are specified Using the psychrometric chart the specific humidity the enthalpy the relative humidity the dewpoint temperature and the specific volume of the air are to be determined Analysis From the psychrometric chart Fig A31 we read a 0 0092 kg H O kg dry air 2 ω b h 47 6 kJ kg dry air c φ 49 6 d C 12 8 dp T e 0 855 m kg dry air 3 v 1443 Problem 1442 is reconsidered The required properties are to be determined using EES Also the properties are to be obtained at an altitude of 3000 m Analysis The problem is solved using EES and the solution is given below Tdb24 C Twb17 C P1101325 kPa Z 3000 m P21013251002256Zconvertmkm5256 Relation giving P as function of altitude h1enthalpyAirH2OTTdbPP1BTwb v1volumeAirH2OTTdbPP1BTwb Tdp1dewpointAirH2OTTdbPP1BTwb w1humratAirH2OTTdbPP1BTwb Rh1relhumAirH2OTTdbPP1BTwb h2enthalpyAirH2OTTdbPP2BTwb v2volumeAirH2OTTdbPP2BTwb Tdp2dewpointAirH2OTTdbPP2BTwb w2humratAirH2OTTdbPP2BTwb Rh2relhumAirH2OTTdbPP2BTwb SOLUTION h14761 kJkg h26168 kJkg P11013 kPa P27011 kPa Rh104956 Rh205438 Tdb24 C Tdp11281 C Tdp21424 C Twb17 C v108542 m3kg v21245 m3kg w10009219 kgkg w2001475 kgkg Z3000 m Discussion The atmospheric pressure for a given elevation can also be obtained from Table A16 of the book PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1419 1444E The pressure temperature and relative humidity of air are specified Using the psychrometric chart the wetbulb temperature specific humidity the enthalpy the dewpoint temperature and the water vapor pressure are to be determined Analysis From the psychrometric chart in Fig A31E or using EES psychrometric functions we obtain a C 81 6 wb T Air 1 atm 90F 70 RH b 0 0214 lbm H O lbm dry air 2 ω c 45 2 Btulbm dry air h d F 78 9 dp T e 0 70 0 69904 psia 0489 psia sat 90 F P P P g v φ φ 1445 The pressure temperature and wetbulb temperature of air are specified Using the psychrometric chart the relative humidity specific humidity the enthalpy the dewpoint temperature and the water vapor pressure are to be determined Analysis From the psychrometric chart in Fig A31 or using EES psychrometric functions we obtain a 24 2 0 242 φ Air 1 atm 32C Twb18C b 0 0072 kg H O kg dry air 2 ω c 50 6 kJkg dry air h d C 19 dp T e 0 242 4 760 kPa 115 kPa sat 32 C P P P g v φ φ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1420 1446 The pressure temperature and wetbulb temperature of air are specified The adiabatic saturation temperature is to be determined Analysis For an adiabatic saturation process we obtained Eq 1414 in the text 2 1 2 2 1 2 1 f g fg p h h h T T c ω ω Water 1 atm 32C Twb18C AIR Humidifier 100 This requires a trialerror solution for the adiabatic saturation temperature T2 The inlet state properties are 0 0072 kg H O kg dry air 2 ω1 2 kJkg 2559 32 C 1 g g h h As a first estimate let us take T2 18C the inlet wetbulb temperature Also at the exit the relative humidity is 100 φ2 1 and the pressure is 1 atm Other properties at the exit state are 0 0130 kg H O kg dry air 2 ω2 3 kJkg Table A 4 2458 54 kJkg Table A 4 75 18 C 2 18 C 2 fg fg f f h h h h Substituting 0 00720 kg H O kg dry air 7554 2559 2 0 01302458 3 32 1 00518 2 2 1 2 2 1 2 1 f g fg p h h h T T c ω ω which is equal to the inlet specific humidity Therefore the adiabatic saturation temperature is T2 18C Discussion This result is not surprising since the wetbulb and adiabatic saturation temperatures are approximately equal to each other for airwater mixtures at atmospheric pressure 1447E The pressure temperature and wetbulb temperature of air are specified Using the psychrometric chart the relative humidity specific humidity the enthalpy the dewpoint temperature and the water vapor pressure are to be determined Analysis From the psychrometric chart in Fig A31E or using EES psychrometric functions we obtain a 61 5 0 615 φ Air 1 atm 90F Tdp75F b 0 0188 lbm H O lbm dry air 2 ω c 42 3 Btulbm dry air h d F 78 9 wb T e 0 615 0 69904 psia 0430 psia sat 28 C P P P g v φ φ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1421 1448E The pressure temperature and wetbulb temperature of air are specified The adiabatic saturation temperature is to be determined Analysis For an adiabatic saturation process we obtained Eq 1414 in the text 2 1 2 2 1 2 1 f g fg p h h h T T c ω ω 1 atm 90F Tdp75F AIR Water Humidifier 100 This requires a trialerror solution for the adiabatic saturation temperature T2 The inlet state properties are 0 0188 lbm H O lbm dry air 2 ω1 Fig A31E 4 Btulbm 1100 90 F 1 g g h h Table A4E As a first estimate let us take T2 789F 79F the inlet wetbulb temperature Also at the exit the relative humidity is 100 φ2 1 and the pressure is 1 atm Other properties at the exit state are 0 0215 lbm H O lbm dry air 2 ω2 6 Btulbm Table A 4E 1048 07 Btulbm Table A 4E 47 79 F 2 79 F 2 fg fg f f h h h h Substituting 0 0189 lbm H O lbm dry air 4707 1100 4 0 02151048 6 90 0 24079 2 2 1 2 2 1 2 1 f g fg p h h h T T c ω ω which is sufficiently close to the inlet specific humidity 00188 Therefore the adiabatic saturation temperature is T2 79F Discussion This result is not surprising since the wetbulb and adiabatic saturation temperatures are approximately equal to each other for airwater mixtures at atmospheric pressure PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1422 Human Comfort and AirConditioning 1449C It humidifies dehumidifies cleans and even deodorizes the air 1450C a Perspires more b cuts the blood circulation near the skin and c sweats excessively 1451C It is the direct heat exchange between the body and the surrounding surfaces It can make a person feel chilly in winter and hot in summer 1452C It affects by removing the warm moist air that builds up around the body and replacing it with fresh air 1453C The spectators Because they have a lower level of activity and thus a lower level of heat generation within their bodies 1454C Because they have a large skin area to volume ratio That is they have a smaller volume to generate heat but a larger area to lose it from 1455C It affects a bodys ability to perspire and thus the amount of heat a body can dissipate through evaporation 1456C Humidification is to add moisture into an environment dehumidification is to remove it 1457C The metabolism refers to the burning of foods such as carbohydrates fat and protein in order to perform the necessary bodily functions The metabolic rate for an average man ranges from 108 W while reading writing typing or listening to a lecture in a classroom in a seated position to 1250 W at age 20 730 at age 70 during strenuous exercise The corresponding rates for women are about 30 percent lower Maximum metabolic rates of trained athletes can exceed 2000 W We are interested in metabolic rate of the occupants of a building when we deal with heating and air conditioning because the metabolic rate represents the rate at which a body generates heat and dissipates it to the room This body heat contributes to the heating in winter but it adds to the cooling load of the building in summer 1458C The metabolic rate is proportional to the size of the body and the metabolic rate of women in general is lower than that of men because of their smaller size Clothing serves as insulation and the thicker the clothing the lower the environmental temperature that feels comfortable PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1423 1459C Sensible heat is the energy associated with a temperature change The sensible heat loss from a human body increases as a the skin temperature increases b the environment temperature decreases and c the air motion and thus the convection heat transfer coefficient increases 1460C Latent heat is the energy released as water vapor condenses on cold surfaces or the energy absorbed from a warm surface as liquid water evaporates The latent heat loss from a human body increases as a the skin wetness increases and b the relative humidity of the environment decreases The rate of evaporation from the body is related to the rate of latent heat loss by where h Q m hfg latent vapor fg is the latent heat of vaporization of water at the skin temperature 1461 A department store expects to have a specified number of people at peak times in summer The contribution of people to the sensible latent and total cooling load of the store is to be determined Assumptions There is a mix of men women and children in the classroom Properties The average rate of heat generation from people doing light work is 115 W and 70 of is in sensible form see Sec 146 Analysis The contribution of people to the sensible latent and total cooling load of the store are W 8450 W 19720 W 28180 03 115 W 245 of people No 07 115 W 245 of people No 115 W 245 of people No person latent latent people person sensible sensible people person total total people Q Q Q Q Q Q 1462E There are a specified number of people in a movie theater in winter It is to be determined if the theater needs to be heated or cooled Assumptions There is a mix of men women and children in the classroom Properties The average rate of heat generation from people in a movie theater is 105 W and 70 W of it is in sensible form and 35 W in latent form Analysis Noting that only the sensible heat from a person contributes to the heating load of a building the contribution of people to the heating of the building is 119420 Btuh 35000 W 70 W 500 No of people person sensible people sensible Q Q since 1 W 3412 Btuh The building needs to be heated since the heat gain from people is less than the rate of heat loss of 130000 Btuh from the building PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1424 1463 The infiltration rate of a building is estimated to be 09 ACH The sensible latent and total infiltration heat loads of the building at sea level are to be determined Assumptions 1 Steady operating conditions exist 2 The air infiltrates at the outdoor conditions and exfiltrates at the indoor conditions 3 Excess moisture condenses at room temperature of 24C 4 The effect of water vapor on air density is negligible Properties The gas constant and the specific heat of air are R 0287 kPam3kgK and cp 1005 kJkgC Table A2 The heat of vaporization of water at 24C is 1 kJkg 2444 24 C fg fg h h Table A4 The properties of the ambient and room air are determined from the psychrometric chart Fig A31 to be 001458 kgkg dryair 35 38º C ambient ambient ambient w T φ 001024 kgkg dryair 55 24º C room room room w T φ Analysis Noting that the infiltration of ambient air will cause the air in the cold storage room to be changed 09 times every hour the air will enter the room at a mass flow rate of 3 3 0 0 ambient 1 135 kgm 0 287 kPam kgK38 273 K 101325 kPa RT P ρ 0 2214 kgs 797 0 kgh 3 m 09 h 1 135 kgm 20 13 ACH 1 3 3 room ambient air V ρ m Then the sensible latent and total infiltration heat loads of the room are determined to be kW 546 kW 235 kW 311 2 35 11 3 0 0102424441 kJkg 0 2214 kgs001458 24 C 0 2214 kgs100 5 kJkg C38 infiltration latent infiltration sensible on total infiltrati room ambient air on latent infiltrati room ambient air on sensible infiltrati Q Q Q h w w m Q T T c m Q fg p Discussion The specific volume of the dry air at the ambient conditions could also be determined from the psychrometric chart at ambient conditions PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1425 1464 The infiltration rate of a building is estimated to be 18 ACH The sensible latent and total infiltration heat loads of the building at sea level are to be determined Assumptions 1 Steady operating conditions exist 2 The air infiltrates at the outdoor conditions and exfiltrates at the indoor conditions 3 Excess moisture condenses at room temperature of 24C 4 The effect of water vapor on air density is negligible Properties The gas constant and the specific heat of air are R 0287 kPam3kgK and cp 1005 kJkgC Table A2 The heat of vaporization of water at 24C is 1 kJkg 2444 24 C fg fg h h Table A4 The properties of the ambient and room air are determined from the psychrometric chart Fig A31 to be 001458 kgkg dryair 35 38º C ambient ambient ambient w T φ 001024 kgkg dryair 55 24º C room room room w T φ Analysis Noting that the infiltration of ambient air will cause the air in the cold storage room to be changed 18 times every hour the air will enter the room at a mass flow rate of 3 3 0 0 ambient 1 135 kgm 0 287 kPam kgK38 273 K 101325 kPa RT P ρ 0 4427 kgs 1594 kgh 3m 18h 1 135 kgm 20 13 ACH 1 3 3 room ambient air V ρ m Then the sensible latent and total infiltration heat loads of the room are determined to be kW 1092 kW 469 kW 623 4 69 23 6 0 0102424441 kJkg 0 4427 kgs001458 24 C 0 4427 kgs100 5 kJkg C38 infiltration latent infiltration sensible on total infiltrati room ambient air on latent infiltrati room ambient air on sensible infiltrati Q Q Q h w w m Q T T c m Q fg p Discussion The specific volume of the dry air at the ambient conditions could also be determined from the psychrometric chart at ambient conditions 1465 An average person produces 025 kg of moisture while taking a shower The contribution of showers of a family of four to the latent heat load of the airconditioner per day is to be determined Assumptions All the water vapor from the shower is condensed by the airconditioning system Properties The latent heat of vaporization of water is given to be 2450 kJkg Analysis The amount of moisture produced per day is mvapor Moisture produced per personNo of persons kg person4 persons day 1 kg day 0 25 Then the latent heat load due to showers becomes Q m hfg latent vapor 1 kg day2450 kJ kg 2450 kJ day PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1426 1466 There are 100 chickens in a breeding room The rate of total heat generation and the rate of moisture production in the room are to be determined Assumptions All the moisture from the chickens is condensed by the airconditioning system Properties The latent heat of vaporization of water is given to be 2430 kJkg The average metabolic rate of chicken during normal activity is 102 W 378 W sensible and 642 W latent Analysis The total rate of heat generation of the chickens in the breeding room is Q q gen total gen totalNo of chickens W chicken100 chickens 10 2 1020 W The latent heat generated by the chicken and the rate of moisture production are 0642 kW 6 42 Wchicken100 chickens 642 W gen latent No of chickens gen latent q Q m Q hfg moisture gen latent kJ s 2430 kJ kg kg s 0 642 0 000264 0264 g s PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1427 Simple Heating and Cooling 1467C Relative humidity decreases during a simple heating process and increases during a simple cooling process Specific humidity on the other hand remains constant in both cases 1468C Because a horizontal line on the psychrometric chart represents a ω constant process and the moisture content ω of air remains constant during these processes 1469 Humid air at a specified state is cooled at constant pressure to the dewpoint temperature The cooling required for this process is to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 2 1 a a a m m m Analysis The amount of moisture in the air remains constant ω 1 ω 2 as it flows through the cooling section since the process involves no humidification or dehumidification The inlet and exit states of the air are completely specified and the total pressure is 1 atm The properties of the air at the inlet state are determined from the psychrometric chart Figure A31 to be C 4 21 0161 kg H Okg dry air 0 3 kJkg dry air 71 dp1 2 2 1 1 T h ω ω 30C 60 RH 1 atm 100 RH AIR 2 1 The exit state enthalpy is 62 4 kJkg dry air 1 C 21 4 atm 1 2 2 dp1 2 h T T P φ From the energy balance on air in the cooling section 89 kJkg dry air 62 4 71 3 2 1 out h h q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1428 1470E Humid air at a specified state is heated at constant pressure to a specified temperature The relative humidity at the exit and the amount of heat required are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 2 1 a a a m m m Analysis The amount of moisture in the air remains constant ω 1 ω 2 as it flows through the heating section since the process involves no humidification or dehumidification The inlet state of the air is completely specified and the total pressure is 40 psia The properties of the air at the inlet and exit states are determined to be 1471 Btulbm dry air F50 F 000250310831 Btulbm 0 240 Btulbm lbm H Olbm dry air 0002503 0 16031 psia 40 0 622 0 16031 psia 622 0 1 Btulbm 1083 0 16031 psia 0 90 0 17812 psia 1 1 1 1 2 1 1 1 1 50 F 1 sat 50 F 1 1 1 1 g p v v g g g v h c T h P P P h h P P P ω ω φ φ 50F 90 RH 40 psia AIR 120F 2 1 3159 Btulbm dry air F120 F 000250311132 Btulbm 0 240 Btulbm 2 Btulbm 1113 0 0946 1 6951 psia 16031 psia 0 6951 psia 1 16031 psia 0 2 2 2 2 1 2 120 F 2 2 2 2 sat 120 F 2 1 2 g p g g g v g v v h c T h h h P P P P P P ω ω ω φ 946 From the energy balance on air in the heating section 169 Btulbm dry air 3159 1471 1 2 in h h q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1429 1471 Air enters a cooling section at a specified pressure temperature velocity and relative humidity The exit temperature the exit relative humidity of the air and the exit velocity are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible m m m a a a 1 2 Analysis a The amount of moisture in the air remains constant ω 1 ω 2 as it flows through the cooling section since the process involves no humidification or dehumidification The inlet state of the air is completely specified and the total pressure is 1 atm The properties of the air at the inlet state are determined from the psychrometric chart Figure A31 or EES to be 8953 m kg dry air 0 01594 kg H Okg dry air 0 14 kJkg dry air 76 3 1 2 2 1 1 v ω ω h 750 kJmin 35C 45 18 ms 1 atm AIR 1 The mass flow rate of dry air through the cooling section is 2 421 kgs 1 03 4 m 18 ms 0 8953 m kg 1 1 2 2 3 1 1 1 π V A ma v From the energy balance on air in the cooling section 35 kJkg dry air 67 7614 kJkg 750 60 kJs 1421 kgs 2 2 1 2 out h h h h m Q a b The exit state of the air is fixed now since we know both h2 and ω2 From the psychrometric chart at this state we read 0 8706 m kg dry air 3 2 2 2 v 731 C 265 φ T c The exit velocity is determined from the conservation of mass of dry air 175 ms 0 8953 18 ms 0 8706 1 1 2 2 2 2 1 1 2 2 1 1 2 1 V V V A V A m m a a v v v v v V v V PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1430 1472 Air enters a cooling section at a specified pressure temperature velocity and relative humidity The exit temperature the exit relative humidity of the air and the exit velocity are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible m m m a a a 1 2 Analysis a The amount of moisture in the air remains constant ω 1 ω 2 as it flows through the cooling section since the process involves no humidification or dehumidification The inlet state of the air is completely specified and the total pressure is 1 atm The properties of the air at the inlet state are determined from the psychrometric chart Figure A31 or EES to be 8953 m kg dry air 0 01594 kg H Okg dry air 0 14 kJkg dry air 76 3 1 2 2 1 1 v ω ω h 1100 kJmin 35C 45 18 ms 1 atm AIR 1 The mass flow rate of dry air through the cooling section is 2 421 kgs 1 03 4 m 18 ms 0 8953 m kg 1 1 2 2 3 1 1 1 π V A ma v From the energy balance on air in the cooling section 24 kJkg dry air 63 7614 kJkg 1100 60 kJs 1421 kgs 2 2 1 2 out h h h h m Q a b The exit state of the air is fixed now since we know both h2 and ω2 From the psychrometric chart at this state we read 0 8591 m kg dry air 3 2 2 2 v 926 C 226 φ T c The exit velocity is determined from the conservation of mass of dry air 173 ms 0 8953 18 ms 0 8591 1 1 2 2 2 2 1 1 2 2 1 1 2 1 V V V A V A m m a a v v v v v V v V PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1431 1473 Saturated humid air at a specified state is heated to a specified temperature The relative humidity at the exit and the rate of heat transfer are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 2 1 a a a m m m Analysis The amount of moisture in the air remains constant ω 1 ω 2 as it flows through the heating section since the process involves no humidification or dehumidification The inlet state of the air is completely specified and the total pressure is 200 kPa The properties of the air at the inlet and exit states are determined to be 2860 kJkg dry air C15 C 000535025283 kJkg 005 kJkg 1 0005350 kg H Okg dry air 1 7057 kPa 200 0 622 1 7057 kPa 622 0 4168 m kg dry air 0 19829 kPa 0 287 kPa m kg K288 K 19829 kPa 1 7057 200 3 kJkg 2528 1 7057 kPa 1 7057 kPa 01 1 1 1 1 2 1 1 1 1 3 3 1 1 1 1 1 1 15 C 1 sat 15 C 1 1 1 1 g p v v a a v a g g g v h c T h P P P P T R P P P h h P P P ω ω φ φ v Heating coils 15C 100 RH 20 ms 200 kPa AIR 30C 2 1 4382 kJkg dry air C30 C 000535025556 kJkg 005 kJkg 1 6 kJkg 2555 0 402 4 2469 kPa 7057 kPa 1 2469 kPa 4 7057 kPa 1 2 2 2 2 1 2 30 C 2 2 2 2 sat 30 C 2 1 2 g p g g g v g v v h c T h h h P P P P P P ω ω ω φ 402 Then 0 06029 kgs 4168 m kg dry air 0 02513 m s 0 0 02513 m s 4 004 m 20 ms 4 3 3 1 1 3 2 2 1 1 1 1 v V V a m D V V A π π From the energy balance on air in the heating section 0918 kW 2860kJkg 0 06029 kgs4382 1 2 in h h m Q a PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1432 1474 Saturated humid air at a specified state is heated to a specified temperature The rate at which the exergy of the humid air is increased is to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 2 1 a a a m m m Analysis The amount of moisture in the air remains constant ω 1 ω 2 as it flows through the heating section since the process involves no humidification or dehumidification The inlet state of the air is completely specified and the total pressure is 200 kPa The properties of the air at the inlet and exit states are determined to be 2860 kJkg dry air C15 C 000535025283 kJkg 005 kJkg 1 0005350 kg H Okg dry air 1 7057 kPa 200 0 622 1 7057 kPa 622 0 0 4168 m kg dry air 19829 kPa 0 287 kPa m kg K288 K 19829 kPa 1 7057 200 7803 kJkg K 8 3 kJkg 2528 1 7057 kPa 1 7057 kPa 01 1 1 1 1 2 1 1 1 1 3 3 1 1 1 1 1 1 15 C 1 15 C 1 sat 15 C 1 1 1 1 g p v v a a v a g g g g g v h c T h P P P P T R P P P s s h h P P P ω ω φ φ v 15C 100 RH 200 kPa 30C AIR 2 1 4382 kJkg dry air C30 C 000535025556 kJkg 005 kJkg 1 4520 kJkg K 8 6 kJkg 2555 19829 kPa 1 7057 200 0 402 4 2469 kPa 7057 kPa 1 2469 kPa 4 7057 kPa 1 2 2 2 2 1 2 30 C 2 30 C 2 2 2 2 2 2 2 sat 30 C 2 1 2 g p g g g g v a g v g v v h c T h s s h h P P P P P P P P P ω ω ω φ 402 The entropy change of the dry air is 0 05103 kJkg K 19829 0 287 ln 19829 288 1 005 ln 303 ln ln 1 2 1 2 1 dry air 2 a a p P P R T T c s s The entropy change of the airwater mixture is 0 04927 kJkg K 8 7803 0005350 8 4520 0 05103 1 water vapor 2 1 dry air 2 1 2 s s s s s s ω The mass flow rate of the dry air is 0 06029 kgs 4168 m kg dry air 0 02513 m s 0 0 02513 m s 4 004 m 20 ms 4 3 3 1 1 3 2 2 1 1 1 1 v V V a m D V V A π π The exergy increase of the humid air during this process is then 0062 kW Φ 2860kJkg 288 K004927 kJkg K 0 06029 kgs 4382 1 2 0 1 2 1 2 s s T h h m m a a ψ ψ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1433 Heating with Humidification 1475C To achieve a higher level of comfort Very dry air can cause dry skin respiratory difficulties and increased static electricity 1476 Air is first heated and then humidified by water vapor The amount of steam added to the air and the amount of heat transfer to the air are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible m m m a a a 1 2 Properties The inlet and the exit states of the air are completely specified and the total pressure is 1 atm The properties of the air at various states are determined from the psychrometric chart Figure A31 to be PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2 h h h 1 2 3 311 0 0064 36 2 581 0 0129 kJ kg dry air kg H O kg dry air kJ kg dry air kJ kg dry air kg H O kg dry air 1 2 3 2 ω ω ω T1 15C φ 1 60 1 atm T2 20C Heating coils 3 AIR T3 25C φ 3 65 2 1 Analysis a The amount of moisture in the air remains constant it flows through the heating section ω 1 ω 2 but increases in the humidifying section ω 3 ω 2 The amount of steam added to the air in the heating section is ω ω ω 3 2 0 0129 0 0064 00065 kg H O kg dry air 2 b The heat transfer to the air in the heating section per unit mass of air is q h h in 2 1 36 2 311 51 kJ kg dry air preparation If you are a student using this Manual you are using it without permission 1434 1477E Air is first heated and then humidified by water vapor The amount of steam added to the air and the amount of heat transfer to the air are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible m m m a a a 1 2 Properties The inlet and the exit states of the air are completely specified and the total pressure is 1 atm The properties of the air at various states are determined from the psychrometric chart Figure A31E or EES to be 008298 lbm H Olbm dry air 0 10 Btulbm dry air 27 002586 lbm H Olbm dry air 0 44 Btulbm dry air 18 002586 lbm H Olbm dry air 0 40 Btulbm dry air 12 2 3 3 2 1 2 2 2 1 1 ω ω ω ω h h h Analysis a The amount of moisture in the air remains constant it flows through the heating section ω1 ω2 but increases in the humidifying section ω 3 ω 2 The amount of steam added to the air in the heating section is T1 40F φ 1 50 147 psia T2 65F Heating coils 3 AIR T3 75F φ 3 45 2 1 0 002586 0 008298 2 3 00057 lbm H Olbm dry air 2 ω ω ω b The heat transfer to the air in the heating section per unit mass of air is 60 Btulbm dry air 1240 1844 1 2 in h h q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1439 1483 Air is cooled and dehumidified at constant pressure The syatem hardware and the psychrometric diagram are to be sketched and the inlet volume flow rate is to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 2 1 a a a m m m Properties a The schematic of the cooling and dehumidifaction process and the the process on the psychrometric chart are given below The psychrometric chart is obtained from the Property Plot feature of EES 0 5 10 15 20 25 30 35 40 0000 0005 0010 0015 0020 0025 0030 0035 0040 0045 0050 T C Humidity Ratio Pressure 1013 kPa 02 04 06 08 AirH2O 1 2 b The inlet and the exit states of the air are completely specified and the total pressure is 1013 kPa 1 atm The properties of the air at the inlet and exit states are determined from the psychrometric chart Figure A31 or EES to be 916 m kg dry air 0 0222 kg H Okg dry air 0 5 kJkg dry air 96 C 7 26 3 1 2 1 1 dp v ω h T T2 167F φ 2 100 1013 kPa Cooling coils 1 2 Condensate Condensate removal 167C T139C φ 150 and 0119 kg H Okg dry air 0 9 kJkg dry air 46 01 C 16 7 10 26 7 10 2 2 2 2 2 2 ω φ h T T Also 7010 kJkg C 16 7 f w h h Table A4 Analysis Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section Water Mass Balance w a a w e w i m m m m m 2 2 1 1 ω ω 2 1 ω ω a w m m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1444 1487 Problem 1486 is reconsidered A general solution of the problem in which the input variables may be supplied and parametric studies performed is to be developed and the process is to be shown in the psychrometric chart for each set of input variables Analysis The problem is solved using EES and the solution is given below Input Data from the Diagram Window D04 P1 10132 kPa T1 32 C RH1 70100 relative humidity Vel1 12060 ms DELTATcw 6 C P2 10132 kPa T2 20 C RH2 100100 Dry air flow rate mdota is constant Voldot1 pi D24Vel1 v1VOLUMEAirH2OTT1PP1RRH1 mdota Voldot1v1 Exit vleocity Voldot2 pi D24Vel2 v2VOLUMEAirH2OTT2PP2RRH2 mdota Voldot2v2 Mass flow rate of the condensed water mdotv1mdotv2mdotw w1HUMRATAirH2OTT1PP1RRH1 mdotv1 mdotaw1 w2HUMRATAirH2OTT2PP2RRH2 mdotv2 mdotaw2 SSSF conservation of energy for the air mdota h1 1w1Vel122Convertm2s2 kJkg Qdot mdotah2 1w2Vel222Convertm2s2 kJkg mdotwhliq2 h1ENTHALPYAirH2OTT1PP1ww1 h2ENTHALPYAirH2OTT2PP2ww2 hliq2ENTHALPYWaterTT2PP2 SSSF conservation of energy for the cooling water Qdot mdotcwCpcwDELTATcw Note QnetwaterQnetair Cpcw SpecHeatwaterT10PP2kJkgK PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1445 RH1 mcw kgs Qout kW Vel2 ms 05 055 06 065 07 075 08 085 09 01475 019 02325 0275 03175 036 04025 0445 04875 3706 4774 5842 691 7978 9046 1011 1118 1225 1921 1916 1911 1907 1902 1897 1893 1888 1884 10 5 0 5 10 15 20 25 30 35 40 0000 0005 0010 0015 0020 0025 0030 0035 0040 0045 0050 T C Humidity Ratio Pressure 1010 kPa 02 04 06 08 0 C 10 C 20 C 30 C 1 2 05 055 06 065 07 075 08 085 09 3 5 7 9 11 13 RH1 Qout kW PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1446 1488 Air is cooled by passing it over a cooling coil The rate of heat transfer the mass flow rate of water and the exit velocity of airstream are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible Analysis a The dew point temperature of the incoming air stream at 35C is C 8 25 3 332 kPa 70 4 76 kPa sat 3 332 kPa sat dp sat 32 C 1 1 1 1 T T T P P P vP g v φ φ Water T T 6C PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Cooling coils 32C 70 120 mmin AIR 20C Saturated 1 Since air is cooled to 20C which is below its dew point temperature some of the moisture in the air will condense The amount of moisture in the air decreases due to dehumidification ω ω 2 1 The inlet and the exit states of the air are completely specified and the total pressure is 95 kPa Then the properties of the air at both states are determined to be 2 01 kJkg dry air 90 C32 C 00226125592 kJkg 005 kJkg 1 002261 kg H Okg dry air 3 332 kPa 95 0 622 3 332 kPa 622 0 0 9549 m kg dry air 9167 kPa 0 287 kPa m kg K305 K 9167 kPa 3 332 95 1 1 1 1 2 1 1 1 1 3 3 1 1 1 1 1 1 g p v v a a v a h c T h P P P P T R P P P ω ω v and 94 kJkg dry air 59 C20 C 0015725374 kJkg 005 kJkg 1 00157 kg H Okg dry air 2 339 kPa 95 0 622 2 339 kPa 622 0 0 9075 m kg dry air 2 339 kPa 95 0 287 kPa m kg K293 K 2 339 kPa 00 1 2 2 2 2 2 2 2 2 2 3 3 2 2 2 sat 20 C 2 2 2 g p v v a a g v h c T h P P P P T R P P P ω ω φ v Also Table A4 83915 kJkg 20 C f w h h Then 1579 kgmin 9549 m kg dry air 0 08 m min 15 1508 m min 4 04 m 120 mmin 4 3 3 1 1 1 3 2 2 1 1 1 1 v V V a m D V V A π π Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section excluding the water preparation If you are a student using this Manual you are using it without permission 1448 1489 Air flows through an air conditioner unit The inlet and exit states are specified The rate of heat transfer and the mass flow rate of condensate water are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible m m m a a a 1 2 Analysis The inlet state of the air is completely specified and the total pressure is 98 kPa The properties of the air at the inlet state may be determined from Fig A31 or using EES psychrometric functions to be we used EES PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 6721 0 01866 kg H Okg dry air 0 88 kJkg dry air 77 1 2 1 1 φ ω h The partial pressure of water vapor at the exit state is 0 9682 kPa Table A 4 sat 65 C 2 P Pv Tdb2 25C Tdp2 65C Cooling coils 1 2 Condensate Condensate removal Tdb1 30C Twb1 25C P 98 kPa 25C The saturation pressure at the exit state is 3 17 kPa Table A 4 sat 25 C 2 P Pg Then the relative humidity at the exit state becomes 0 3054 3 17 9682 0 2 2 2 g v P P φ Now the exit state is also fixed The properties are obtained from EES to be 8820 m kg 0 006206 kg H Okg dry air 0 97 kJkg dry air 40 3 2 2 2 2 v ω h The mass flow rate of dry air is 1133 8 kgmin m kg 08820 m min 1000 3 3 2 2 v V a m The mass flow rate of condensate water is 8472 kgh 1412 kgmin 1133 8 kgmin0018660006206 2 1 ω a ω w m m The enthalpy of condensate water is 10483 kJkg Table A 4 25 C 2 f w h h An energy balance on the control volume gives 6729 kW 377 kJmin 40 1412 kgmin10483 kJkg 1133 8 kgmin4097 kJkg 1133 8 kgmin7788 kJkg out out 2 2 out 1 Q Q m h m h Q m h w w a a preparation If you are a student using this Manual you are using it without permission 1449 1490 Atmospheric air enters the evaporator of an automobile air conditioner at a specified pressure temperature and relative humidity The dew point and wet bulb temperatures at the inlet to the evaporator section the required heat transfer rate from the atmospheric air to the evaporator fluid and the rate of condensation of water vapor in the evaporator section are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible m m m a a a 1 2 Analysis The inlet and exit states of the air are completely specified and the total pressure is 1 atm The properties of the air at the inlet and exit states may be determined from the psychrometric chart Fig A31 or using EES psychrometric functions to be we used EES 1 atm Cooling coils 1 2 Condensate Condensate removal 00686 kg H Okg dry air 0 35 kJkg dry air 27 8655 m kg dry air 0 01115 kg H Okg dry air 0 60 kJkg dry air 55 2 2 2 3 1 2 1 1 wb1 dp1 ω ω h h T T v C 195 C 157 T1 27C φ 1 50 T2 10C φ 2 90 10C The mass flow rate of dry air is 1155 kgmin m 08655 2 m change5 changesmin ACH 3 3 1 car 1 1 v V v V a m The mass flow rates of vapor at the inlet and exit are 01288 kgmin 0 011151155 kgmin 1 1 a v m m ω 007926 kgmin 0 006861155 kgmin 2 2 a v m m ω An energy balance on the control volume gives 2 2 out 1 w w a a m h m h Q m h where the the enthalpy of condensate water is 4202 kJkg Table A 4 10 C 2 f w h h and the rate of condensation of water vapor is 00495 kgmin 0 07926 0 1288 2 1 v v w m m m Substituting 541 kW 4 kJmin 324 0 0495 kgmin4202 kJkg 1155 kgmin2735 kJkg 1155 kgmin5560 kJkg out out 2 2 out 1 Q Q m h m h Q m h w w a a PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1450 0 5 10 15 20 25 30 35 40 0000 0005 0010 0015 0020 0025 0030 0035 0040 0045 0050 T C Humidity Ratio Pressure 1013 kPa 10C 15C 20C 25C 30C 35C 02 04 06 08 AirH2O 1 2 Discussion We we could not show the process line between the states 1 and 2 because we do not know the process path PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1451 1491 Atmospheric air flows into an air conditioner that uses chilled water as the cooling fluid The mass flow rate of the condensate water and the volume flow rate of chilled water supplied to the air conditioner are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible m m m a a a 1 2 Analysis We may assume that the exit relative humidity is 100 percent since the exit temperature 18C is below the dew point temperature of the inlet air 25C The properties of the air at the exit state may be determined from the psychrometric chart Fig A31 or using EES psychrometric functions to be we used EES 01311 kg H Okg dry air 0 34 kJkg dry air 51 2 2 2 ω h Cooling coils 1 2 Condensate Condensate removal The partial pressure of water vapor at the inlet state is Table A4 T1 28C Tdp1 25C 2000 m3h T2 18C 100 RH 100 kPa 3 17 kPa sat 25 C 1 P Pv The saturation pressure at the inlet state is 3 783 kPa Table A 4 sat 28 C 1 P Pg 18C Then the relative humidity at the inlet state becomes 0 8379 3 783 17 3 1 1 1 g v P P φ Now the inlet state is also fixed The properties are obtained from EES to be 8927 m kg 0 02036 kg H Okg dry air 0 14 kJkg dry air 80 3 1 2 1 1 v ω h The mass flow rate of dry air is 3734 kgmin 08927 m kg 2000 60 m h 3 3 1 1 v V a m The mass flow rate of condensate water is 1624 kgh 0 2707 kgmin 3734 kgmin002036 001311 2 1 ω a ω w m m The enthalpy of condensate water is 7554 kJkg Table A 4 18 C 2 f w h h An energy balance on the control volume gives 1759 kW kJmin 1055 0 2707 kgmin7554 kJkg 3734 kgmin5134 kJkg 3734 kgmin8014 kJkg out out 2 2 out 1 Q Q m h m h Q m h w w a a Noting that the rate of heat lost from the air is received by the cooling water the mass flow rate of the cooling water is determined from 2524 kgmin 4 18 kJkg C10 C 1055 kJmin in in cw p cw cw cw p T c Q m T c m Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1452 where we used the specific heat of water value at room temperature Assuming a density of 1000 kgm3 for water the volume flow rate is determined to be 00252 m min 3 1000 kgm3 24 kgmin 25 cw cw cw m ρ V 0 5 10 15 20 25 30 35 40 0000 0005 0010 0015 0020 0025 0030 0035 0040 0045 0050 T C Humidity Ratio Pressure 1000 kPa 10C 15C 20C 25C 30C 35C 02 04 06 08 08 0825 085 m3kg 0875 0925 09 AirH2O 1 2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1453 1492 An automobile air conditioner using refrigerant 134a as the cooling fluid is considered The inlet and exit states of moist air in the evaporator are specified The volume flow rate of the air entering the evaporator of the air conditioner is to be determined Assumptions 1 All processes are steady flow and the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible m m m a a a 1 2 Analysis We assume that the total pressure of moist air is 100 kPa Then the inlet and exit states of the moist air for the evaporator are completely specified The properties may be determined from the psychrometric chart Fig A31 or using EES psychrometric functions to be we used EES PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 006064 kg H Okg dry air 0 31 kJkg dry air 23 8724 m kg dry air 0 01206 kg H Okg dry air 0 88 kJkg dry air 55 2 2 2 3 1 2 1 1 ω ω h h v The mass flow rate of dry air is given by 08724 m kg 3 1 1 1 V v V ma The mass flow rate of condensate water is expressed as 1 1 2 1 0 006873 0 8724 001206 0006064 V V ω a ω w m m T2 8C φ 2 90 AIR Cooling coils 1 2 Condensate R134a 100 kPa Condensate removal T1 25C φ 1 60 8C The enthalpy of condensate water is 3363 kJkg Table A 4 C 8 2 f w h h An energy balance on the control volume gives 3363 0 006873 08724 2311 5588 08724 1 1 out 1 2 2 out 1 V V V Q m h m h Q m h w w a a 1 The properties of the R134a at the inlet of the compressor and the enthalpy at the exit for the isentropic process are R 134a tables 28787 kJkg kPa 1600 9377 kJkgK 0 46 kJkg 244 1 kPa 200 2 1 2 2 1 1 1 1 s R R R R R R R R h s s P s h x P The enthalpies of R134a at the condenser exit and the throttle exit are 93 kJkg 135 93 kJkg 135 3 4 1600 kPa 3 R R f R h h h h The mass flow rate of the refrigerant can be determined from the expression for the compressor power 7 049 kgmin 0 1175 kgs 085 24446 kJkg 28787 kW 6 1 2 R R C R s R R C m m h h m W η The rate of heat absorbed by the R134a in the evaporator is 765 0 kJmin 13593 kJkg 7 049 kgmin24446 4 1 in R R R R h h m Q The rate of heat lost from the air in the evaporator is absorbed by the refrigerant134a That is Then the volume flow rate of the air at the inlet of the evaporator can be determined from Eq 1 to be out in Q QR 2062 m min 3 1 1 1 1 3363 0 006873 08724 2311 765 0 08724 5588 V V V V preparation If you are a student using this Manual you are using it without permission 1455 The mass flow rate of refrigerant134a is 395 0 kg 25277 10732kJkg 450 kJ 57 4 1 h h Q m L R The amount of heat rejected from the condenser is 66350 kg 395 0 kg27529 10732 kJkg 3 2 h h m Q R H Next we calculate the exergy destruction in the components of the refrigeration cycle 1519 kJ 0 39189 kJkg K 395 kg305 K 0 4045 1609 kJ 305 K 66350 kJ 0 92927 kJkg K 305 K 395 kg 0 39189 0 since the process is isentropic 3 4 0 34 destroyed 2 3 0 23 destroyed 1 2 0 12 destroyed s s m T X T Q s s m T X s s m T X R H H R R The entropies of water vapor in the air stream are 5782 kJkg K 8 4114 kJkg K 8 24 C 2 32 C 1 g g g g s s s s The entropy change of water vapor in the air stream is 165 2 kJK 8 4114 0 0292 8 5782 1105 0 0112 1 1 2 2 vapor g g a s s m S ω ω The entropy of water leaving the cooling section is 814 kJK kJkg K 1989 kg 0 4091 28 C f w w m s S The partial pressures of water vapor and dry air for air streams are 9953 kPa 1 792 325 101 1 792 kPa 0 60 2 986 kPa 9680 kPa 4 522 325 101 4 522 kPa 0 95 4 760 kPa 2 2 2 sat 24 C 2 2 2 2 1 1 1 sat 32 C 1 1 1 1 v a g v v a g v P P P P P P P P P P P P φ φ φ φ The entropy change of dry air is 3834 kJkg dry air 9680 0 287 ln 9953 305 1 005 ln 297 1105 ln ln 1 2 1 2 1 2 a a p a a a P P R T T c m s s m S The entropy change of R134a in the evaporator is 207 3 kJK 0 4045 395 kg 0 92927 4 1 41 s s m S R R An entropy balance on the evaporator gives 1190 kJK 8 14 3834 165 2 207 3 vapor 41 genevaporator w a R S S S S S Then the exergy destruction in the evaporator is 3630 kJ 305 K1190 kJK gen evaporator 0 dest T S X Finally the total exergy destruction is 6758 kJ 3630 1609 1519 0 dest evaporator dest throttle dest condenser dest compressor dest total X X X X X The greatest exergy destruction occurs in the evaporator Note that heat is absorbed from humid air and rejected to the ambient air at 32C 305 K which is also taken as the dead state temperature PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1456 Evaporative Cooling 1494C Evaporative cooling is the cooling achieved when water evaporates in dry air It will not work on humid climates 1495C During evaporation from a water body to air the latent heat of vaporization will be equal to convection heat transfer from the air when conduction from the lower parts of the water body to the surface is negligible and temperature of the surrounding surfaces is at about the temperature of the water surface so that the radiation heat transfer is negligible 1496C In steady operation the mass transfer process does not have to involve heat transfer However a mass transfer process that involves phase change evaporation sublimation condensation melting etc must involve heat transfer For example the evaporation of water from a lake into air mass transfer requires the transfer of latent heat of water at a specified temperature to the liquid water at the surface heat transfer 1497 Desert dwellers often wrap their heads with a watersoaked porous cloth The temperature of this cloth on a desert with specified temperature and relative humidity is to be determined Analysis Since the cloth behaves as the wick on a wet bulb thermometer the temperature of the cloth will become the wet bulb temperature According to the pshchrometric chart this temperature is 1 atm 45C 15 AIR Water Humidifier 100 233C wb1 2 T T This process can be represented by an evaporative cooling process as shown in the figure PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1457 1498 Air is cooled by an evaporative cooler The exit temperature of the air and the required rate of water supply are to be determined Analysis a From the psychrometric chart Fig A31 at 36C and 20 relative humidity we read 887 m kg dry air 0 0074 kg H Okg dry air 0 C 5 19 3 1 2 1 wb1 v ω T 1 atm 36C 20 AIR Water mω Humidifier 90 Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream the evaporative cooling process follows a line of constant wetbulb temperature That is C 19 5 wb1 wb2 T T At this wetbulb temperature and 90 relative humidity we read T2 0 0137 205 C ω 2 2 kg H O kg dry air Thus air will be cooled to 205C in this evaporative cooler b The mass flow rate of dry air is 4 51 kgmin 887 m kg dry air 0 m min 4 3 3 1 1 v V a m Then the required rate of water supply to the evaporative cooler is determined from 0028 kgmin 4 51 kgmin00137 00074 1 2 1 2 supply ω a ω w w m m m m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1458 1499E Air is cooled by an evaporative cooler The exit temperature of the air and the required rate of water supply are to be determined Analysis a From the psychrometric chart Fig A31E or EES at 100F and 30 relative humidity we read 4 ft lbm dry air 14 0123 lbm H Olbm dry air 0 F 3 74 3 1 2 1 wb1 v ω T 1 atm 100F 30 AIR Water m ω Humidifier 90 Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream the evaporative cooling process follows a line of constant wetbulb temperature That is F 74 3 wb1 wb2 T T At this wetbulb temperature and 90 relative humidity we read 0 0178 lbm H Olbm dry air 2 2 2 ω 766 F T Thus air will be cooled to 766F in this evaporative cooler b The mass flow rate of dry air is 13 9 lbmmin 4 ft lbm dry air 14 ft min 200 3 3 1 1 v V a m Then the required rate of water supply to the evaporative cooler is determined from 13 9 lbmmin00178 00123 0076 lbmmin 1 2 1 2 supply ω a ω w w m m m m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1459 14100 Air is cooled by an evaporative cooler The final relative humidity and the amount of water added are to be determined Analysis a From the psychrometric chart Fig A31 at 32C and 30 relative humidity we read 877 m kg dry air 0 0089 kg H Okg dry air 0 C 4 19 3 1 2 1 wb1 v ω T 32C 30 2 m3min AIR Water Humidifier 22C Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream the evaporative cooling process follows a line of constant wetbulb temperature That is C 19 4 wb1 wb2 T T At this wetbulb temperature and 22C temperature we read 0 0130 kg H Okg dry air 2 2 2 ω φ 79 b The mass flow rate of dry air is 5 70 kgmin 0 877 m kg dry air 5 m min 3 3 1 1 v V a m Then the required rate of water supply to the evaporative cooler is determined from 5 70 kgmin0013000089 00234 kgmin 1 2 1 2 supply ω a ω w w m m m m 14101 Air enters an evaporative cooler at a specified state and relative humidity The lowest temperature that air can attain is to be determined Analysis From the psychrometric chart Fig A31 at 29C and 40 relative humidity we read 1 atm 29C 40 AIR Water Humidifier 100 C 19 3 wb1 T Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream the evaporative cooling process follows a line of constant wetbulb temperature which is the lowest temperature that can be obtained in an evaporative cooler That is 193C wb1 min T T PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1460 14102 Air is first heated in a heating section and then passed through an evaporative cooler The exit relative humidity and the amount of water added are to be determined Analysis a From the psychrometric chart Fig A31 or EES at 20C and 50 relative humidity we read 0 00726 kg H Okg dry air 2 ω1 15C 60 AIR 1 atm Heating coils 1 2 3 25C Water Humidifier 20C 50 30C 35C The specific humidity ω remains constant during the heating process Therefore ω2 ω1 000726 kg H2O kg dry air At this ω value and 35C we read Twb2 191C Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream the evaporative cooling process follows a line of constant wetbulb temperature That is Twb3 Twb2 191C At this Twb value and 25C we read 0 0114 kg H Okg dry air 2 3 3 ω φ 575 b The amount of water added to the air per unit mass of air is 000413 kg H Okg dry air 2 0 00726 0 0114 2 3 23 ω ω ω PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1461 Adiabatic Mixing of Airstreams 14103C This will occur when the straight line connecting the states of the two streams on the psychrometric chart crosses the saturation line 14104C Yes PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1462 14105 Two airstreams are mixed steadily The specific humidity the relative humidity the drybulb temperature and the volume flow rate of the mixture are to be determined Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 4 The mixing section is adiabatic Properties Properties of each inlet stream are determined from the psychrometric chart Fig A31 to be 882 m kg dry air 0 0119 kg H Okg dry air 0 7 kJkg dry air 62 3 1 2 1 1 v ω h PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course and 819 m kg dry air 0 0079 kg H Okg dry air 0 9 kJkg dry air 31 3 2 2 2 2 v ω h Analysis The mass flow rate of dry air in each stream is 30 5 kgmin 819 m kg dry air 0 m min 25 22 7 kgmin 882 m kg dry air 0 m min 20 3 3 2 2 2 3 3 1 1 1 v V v V a a m m P 1 atm AIR ω3 φ3 T3 32C 40 20 m3min 25 m3min 12C 90 3 2 1 From the conservation of mass m m m a a a 3 1 2 22 7 305 532 kg min kg min The specific humidity and the enthalpy of the mixture can be determined from Eqs 1424 which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams m m h h h h h h a a 1 2 2 3 3 1 2 3 3 1 3 3 3 3 22 7 305 0 0079 0 0119 319 62 7 ω ω ω ω ω ω which yields ω 3 3 kJ kg dry air 00096 kg H O kg dry air 2 h 450 These two properties fix the state of the mixture Other properties of the mixture are determined from the psychrometric chart 0 845 m kg dry air 3 3 3 3 v 634 C 206 φ T Finally the volume flow rate of the mixture is determined from 450 m min 3 53 2 kgmin0845 m kg 3 3 3 3 v V m a preparation If you are a student using this Manual you are using it without permission 1463 14106 Two airstreams are mixed steadily The specific humidity the relative humidity the drybulb temperature and the volume flow rate of the mixture are to be determined Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 4 The mixing section is adiabatic Analysis The properties of each inlet stream are determined to be 6645 kJkg dry air C32 C 0013425592 kJkg 005 kJkg 1 kg H Okg dry air 00134 1 90 kPa 90 0 622 1 90 kPa 622 0 994 m kg dry air 0 8810 kPa 0 287 kPa m kg K305 K 8810 kPa 1 90 90 1 90 kPa 0 40 4 760 kPa 1 1 1 1 2 1 1 1 1 3 3 1 1 1 1 1 1 sat 32 C 1 1 1 1 g p v v a a v a g v h c T h P P P P T R P P P P P P ω ω φ φ v PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course P 90 kPa AIR 1 32C 40 20 m3min ω3 φ3 T3 3 25 m3min 12C 90 2 and 3434 kJkg dry air C12 C 00088325229 kJkg 005 kJkg 1 000883 kg H Okg dry air 1 26 kPa 90 0 622 1 26 kPa 622 0 0 922 m kg dry air 8874 kPa 0 287 kPa m kg K285 K 8874 kPa 1 26 90 1 26 kPa 0 90 1 403 kPa 2 2 2 2 2 2 2 2 2 3 3 2 2 2 2 2 2 sat12 C 2 2 2 2 g p v v a a v a g v h c T h P P P P T R P P P P P P ω ω φ φ v Then the mass flow rate of dry air in each stream is 2711 kgmin 922 m kg dry air 0 m min 25 2012 kgmin 994 m kg dry air 0 m min 20 3 3 2 2 2 3 3 1 1 1 v V v V a a m m From the conservation of mass 4723 kgmin 2711 kgmin 2012 2 1 3 a a a m m m The specific humidity and the enthalpy of the mixture can be determined from Eqs 1424 which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams 6645 3434 0 0134 0 00883 2711 12 20 3 3 3 3 1 3 3 2 1 3 3 2 2 1 h h h h h h m m a a ω ω ω ω ω ω which yields 4802 kJkg dry air 3 3 h 00108 kg H Okg dry air 2 ω These two properties fix the state of the mixture Other properties are determined from 205C 3 3 3 3 3 3 3 3 3 3 kJkg 1 82 001082500 9 C 1005 kJkg kJkg 4802 1 82 2501 3 T T T T c T h c T h p g p ω ω preparation If you are a student using this Manual you are using it without permission 1464 0 639 or 639 241 kPa 54 kPa 1 1 54 kPa 90 0 622 0108 0 622 0 3 sat 3 3 3 3 3 3 3 3 3 3 3 T v g v v v v v v P P P P P P P P P P φ ω Finally 450 m min 3 4723 kgmin0952 m kg 0 952 m kg dry air 8846 kPa 0 287 kPa m kg K2935 K 8846 kPa 1 54 90 3 3 3 3 3 3 3 3 3 3 3 3 v V v a a a v a m P T R P P P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1465 14107 A stream of warm air is mixed with a stream of saturated cool air The temperature the specific humidity and the relative humidity of the mixture are to be determined Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 4 The mixing section is adiabatic Properties The properties of each inlet stream are determined from the psychrometric chart Fig A31 or EES to be 0246 kg H Okg dry air 0 4 kJkg dry air 99 2 1 1 ω h PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course and 00873 kg H Okg dry air 0 1 kJkg dry air 34 2 2 2 ω h Analysis The specific humidity and the enthalpy of the mixture can be determined from Eqs 1424 which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams P 1 atm AIR ω3 φ3 T3 36C 8 kgs Twb1 30C 10 kgs 12C 100 3 2 1 99 4 34 1 0 0246 0 00873 10 8 3 3 3 3 1 3 3 2 1 3 3 2 2 1 h h h h h h m m a a ω ω ω ω ω ω which yields b 00158 kg H Okg dry air 2 ω3 1 kJkg dry air 3 63 h These two properties fix the state of the mixture Other properties of the mixture are determined from the psychrometric chart a 3 228C T c φ3 901 preparation If you are a student using this Manual you are using it without permission 1466 14108 Problem 14107 is reconsidered The effect of the mass flow rate of saturated cool air stream on the mixture temperature specific humidity and relative humidity is to be investigated Analysis The problem is solved using EES and the solution is given below P101325 kPa Tdb1 36 C Twb1 30 C mdot1 8 kgs Tdb2 12 C Rh2 10 mdot2 10 kgs P1P P2P1 P3P1 Energy balance for the steadyflow mixing process We neglect the PE of the flow Since we dont know the cross sectional area of the flow streams we also neglect theKE of the flow Edotin Edotout DELTAEdotsys DELTAEdotsys 0 kW Edotin mdot1h1mdot2h2 Edotout mdot3h3 Conservation of mass of dry air during mixing mdot1mdot2 mdot3 Conservation of mass of water vapor during mixing mdot1w1mdot2w2 mdot3w3 mdot1Vdot1v1convert1min1s mdot2Vdot2v2convert1min1s h1ENTHALPYAirH2OTTdb1PP1BTwb1 Rh1RELHUMAirH2OTTdb1PP1BTwb1 v1VOLUMEAirH2OTTdb1PP1RRh1 w1HUMRATAirH2OTTdb1PP1RRh1 h2ENTHALPYAirH2OTTdb2PP2RRh2 v2VOLUMEAirH2OTTdb2PP2RRh2 w2HUMRATAirH2OTTdb2PP2RRh2 Tdb3TEMPERATUREAirH2Ohh3PP3ww3 Rh3RELHUMAirH2OTTdb3PP3ww3 v3VOLUMEAirH2OTTdb3PP3ww3 Twb2WETBULBAirH2OTTdb2PP2RRH2 Twb3WETBULBAirH2OTTdb3PP3RRH3 mdot3Vdot3v3convert1min1s m2 kgas Tdb3 C Rh3 w3 kgwkga 0 2 4 6 8 10 12 14 16 36 3131 2815 2588 2417 2284 2177 2089 2015 06484 07376 07997 08442 08768 09013 092 09346 09461 002461 002143 001931 00178 001667 001579 001508 00145 001402 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1467 0 2 4 6 8 10 12 14 16 20 22 24 26 28 30 32 34 36 m2 kgas Tdb3 C 0 2 4 6 8 10 12 14 16 065 07 075 08 085 09 095 m2 kgas Rh3 0 2 4 6 8 10 12 14 16 0014 0016 0018 002 0022 0024 m2 kgas w3 kgwkga PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1468 14109E Two airstreams are mixed steadily The mass flow ratio of the two streams for a specified mixture relative humidity and the relative humidity of the mixture are to be determined Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 4 The mixing section is adiabatic Properties Properties of each inlet stream are determined from the psychrometric chart Fig A31E or from EES to be 0076 lbm H Olbm dry air 0 3 Btulbm dry air 20 2 1 1 ω h PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course and 0246 lbm H Olbm dry air 0 7 Btulbm dry air 48 2 2 2 ω h Analysis An application of Eq 1424 which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams gives P 1 atm AIR 50F 100 90F 80 3 70F ω3 T3 2 1 20 3 48 7 0 0076 0246 0 3 3 3 3 2 1 1 3 3 2 1 3 3 2 2 1 h h m m h h h h m m a a a a ω ω ω ω ω ω This equation cannot be solved directly An iterative solution is needed A mixture relative humidity φ3 is selected At this relative humidity and the given temperature 70F specific humidity and enthalpy are read from the psychrometric chart These values are substituted into the above equation If the equation is not satisfied a new value of φ3 is selected This procedure is repeated until the equation is satisfied Alternatively EES software can be used We used the following EES program to get these results 107 100 2 1 3 2 3 3 0 Btulbm dry air 34 lbm H Olbm dry air 00158 a a m m h ω φ Given P14696 psia T150 F phi110 T290 F phi2080 T370 F Analysis FluidAirH2O 1st stream properties h1enthalpyFluid TT1 PP Rphi1 w1humratFluid TT1 PP Rphi1 2nd stream properties h2enthalpyFluid TT2 PP Rphi2 w2humratFluid TT2 PP Rphi2 w2w3w3w1h2h3h3h1 Ratiow2w3w3w1 mixture properties phi3relhumFluid hh3 PP TT3 h3enthalpyFluid Rphi3 PP TT3 preparation If you are a student using this Manual you are using it without permission 1469 14110 Two airstreams are mixed steadily The temperature and the relative humidity of the mixture are to be determined Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 4 The mixing section is adiabatic Properties Properties of each inlet stream are determined from the psychrometric chart Fig A31 or from EES to be 914 m kg dry air 0 0187 kg H Okg dry air 0 5 kJkg dry air 88 3 1 2 1 1 v ω h P 1 atm AIR ω3 φ3 T3 40C 40 3 Ls 1 Ls 15C 80 3 2 1 and 828 m kg dry air 0 0085 kg H Okg dry air 0 7 kJkg dry air 36 3 2 2 2 2 v ω h Analysis The mass flow rate of dry air in each stream is 0 001208 kgs 828 m kg dry air 0 001 m s 0 0 003282 kgs 914 m kg dry air 0 003 m s 0 3 3 2 2 2 3 3 1 1 1 v V v V a a m m From the conservation of mass 0 00449 kgs 0 001208 kgs 0 003282 2 1 3 a a a m m m The specific humidity and the enthalpy of the mixture can be determined from Eqs 1424 which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams 88 5 36 7 0 0187 0 0085 0 001208 003282 0 3 3 3 3 1 3 3 2 1 3 3 2 2 1 h h h h h h m m a a ω ω ω ω ω ω which yields 6 kJkg dry air 74 kg H Okg dry air 00160 3 2 3 h ω These two properties fix the state of the mixture Other properties of the mixture are determined from the psychrometric chart 493 C 334 0 493 3 3 T φ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1470 14111 Two airstreams are mixed steadily The rate of exergy destruction is to be determined Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 4 The mixing section is adiabatic Properties Properties of each inlet stream are determined from the psychrometric chart Fig A31 or from EES to be 914 m kg dry air 0 0187 kg H Okg dry air 0 5 kJkg dry air 88 3 1 2 1 1 v ω h P 1 atm AIR ω3 φ3 T3 40C 40 3 Ls 1 Ls 15C 80 3 2 1 and 828 m kg dry air 0 0085 kg H Okg dry air 0 7 kJkg dry air 36 3 2 2 2 2 v ω h The entropies of water vapor in the air streams are 7803 kJkg K 8 2556 kJkg K 8 15 C 2 40 C 1 g g g g s s s s Analysis The mass flow rate of dry air in each stream is 0 001208 kgs 828 m kg dry air 0 001 m s 0 0 003282 kgs 914 m kg dry air 0 003 m s 0 3 3 2 2 2 3 3 1 1 1 v V v V a a m m From the conservation of mass 0 00449 kgs 0 001208 kgs 0 003282 2 1 3 a a a m m m The specific humidity and the enthalpy of the mixture can be determined from Eqs 1424 which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams 88 5 36 7 0 0187 0 0085 0 001208 003282 0 3 3 3 3 1 3 3 2 1 3 3 2 2 1 h h h h h h m m a a ω ω ω ω ω ω which yields 6 kJkg dry air 74 kg H Okg dry air 00160 3 2 3 h ω These two properties fix the state of the mixture Other properties of the mixture are determined from the psychrometric chart 0493 C 334 3 3 φ T The entropy of water vapor in the mixture is 8 3833 kJkg K C 33 4 3 g g s s An entropy balance on the mixing chamber for the water gives PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1471 kWK 426 10 5 8 7803 0 0085 0 001208 8 2556 0 0187 0 003282 8 3833 0 0160 00449 0 6 2 2 2 1 1 1 3 3 3 s m s m s m S a a a w ω ω ω The partial pressures of water vapor and dry air for all three air streams are 9879 kPa 2 539 325 101 2 539 kPa 0 493 5 150 kPa 9996 kPa 1 365 325 101 1 365 kPa 0 80 1 7057 kPa 9837 kPa 2 954 325 101 2 954 kPa 0 40 7 3851 kPa 3 3 3 C sat 33 4 3 3 3 3 2 2 2 sat 15 C 2 2 2 2 1 1 1 sat 40 C 1 1 1 1 v a g v v a g v v a g v P P P P P P P P P P P P P P P P P P φ φ φ φ φ φ An entropy balance on the mixing chamber for the dry air gives kWK 964 10 4 0 001208 0 06562 0 02264 0 003282 9996 0 287 ln 9879 288 1 005 ln 306 4 0 001208 9837 0 287 ln 9879 313 1 005 ln 306 4 003282 0 ln ln ln ln 6 2 3 2 3 2 1 3 1 3 1 2 3 2 1 3 1 a a p a a a p a a a a P P R T T c m P P R T T c m s s m s s m S The rate of entropy generation is kWK 1039 10 5 426 10 4 964 10 6 6 6 gen w a S S S Finally the rate of exergy destruction is 00031 kW kWK 298 K1039 10 6 gen 0 dest T S X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1472 Wet Cooling Towers 14112C The working principle of a natural draft cooling tower is based on buoyancy The air in the tower has a high moisture content and thus is lighter than the outside air This light moist air rises under the influence of buoyancy inducing flow through the tower 14113C A spray pond cools the warm water by spraying it into the open atmosphere They require 25 to 50 times the area of a wet cooling tower for the same cooling load PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1476 14117 A naturaldraft cooling tower is used to remove waste heat from the cooling water flowing through the condenser of a steam power plant The mass flow rate of the cooling water the volume flow rate of air into the cooling tower and the mass flow rate of the required makeup water are to be determined Assumptions 1 All processes are steadyflow and the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible m m m a a a 1 2 Analysis The inlet and exit states of the moist air for the tower are completely specified The properties may be determined from the psychrometric chart Fig A31 or using EES psychrometric functions to be we used EES 04112 kg H Okg dry air 0 83 kJkg dry air 142 8536 m kg dry air 0 01085 kg H Okg dry air 0 74 kJkg dry air 50 2 2 2 3 1 2 1 1 ω ω h h v AIR 1 2 Makeup water T2 37C φ 2 100 T1 23C Twb1 18C The enthalpies of cooling water at the inlet and exit of the condenser are Table A4 01 kJkg 109 53 kJkg 167 26 C 4 40 C 3 f w f w h h h h The steam properties for the condenser are Steam tables 19181 kJkg 0 kPa 10 2524 3 kJkg 962 kJkgK 7 kPa 10 50471 kJkg 0 kPa 200 3 1 3 2 2 2 1 1 1 s s s s s s s s s h x P h s P h x P The mass flow rate of dry air is given by 08536 m kg 3 1 1 1 V v V ma The mass flow rates of vapor at the inlet and exit of the cooling tower are 1 1 2 2 1 1 1 1 0 04817 0 04112 0 8536 0 01271 0 01085 0 8536 V V V V a v a v m m m m ω ω Mass and energy balances on the cooling tower give 4 2 3 1 cw v cw v m m m m 4 4 2 3 3 1 w cw a w cw a h m m h h m m h The mass flow rate of the makeup water is determined from 4 3 1 2 makeup cw cw v v m m m m m An energy balance on the condenser gives 3 3 3 4 makeup 4 4 2 1 0 82 0 18 w cw s s w w cw s s s s h m m h h m h m m h m h Solving all the above equations simultaneously with known and determined values using EES we obtain kgs 2819 m min 47700 kgs 1413 3 makeup 1 3 m mcw V PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1479 0 9207 lbm waterlbm dry air 0 00391 0 0149 0 9317 1 2 3 4 ω ω m m The entropies of water streams are 09328 Btulbm R 0 12065 Btulbm R 0 80 F 4 95 F 3 f f s s s s The entropy change of water stream is 0 02653 BtuR lbm dry air 0 12065 0 9317 0 09328 0 9207 3 3 4 4 water m s m s s The entropies of water vapor in the air stream are 0352 Btulbm R 2 0788 Btulbm R 2 80 F 2 65 F 1 g g g g s s s s The entropy change of water vapor in the air stream is 0 02220 BtuR lbm dry air 2 0788 0 00391 2 0352 0 0149 1 1 2 2 vapor g g s s s ω ω The partial pressures of water vapor and dry air for air streams are 0 3441 1435 psia 696 14 0 3441 psia 0 80 0 43016 psia 1460 psia 0 0917 696 14 0 0917 psia 0 30 0 30578 psia 2 2 2 sat 75 F 2 2 2 2 1 1 1 1 sat 65 F 1 1 1 v a g v v a g v P P P P P P P P P P P P φ φ φ φ The entropy change of dry air is 0 005712 Btulbm dry air 1460 0 06855ln 1435 525 0 240ln 535 ln ln 1 2 1 2 1 2 a a p a P P R T T c s s s The entropy generation in the cooling tower is the total entropy change 0 001382 BtuR lbm dry air 0 005712 0 02220 0 02653 vapor water gen sa s s s Finally the exergy destruction per unit mass of dry air is 0726 Btulbm dry air 525 R000138 2 BtuR lbm dry air 0 gen dest T s x PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1480 Review Problems 14120 Air is compressed by a compressor and then cooled to the ambient temperature at high pressure It is to be determined if there will be any condensation in the compressed air lines Assumptions The air and the water vapor are ideal gases Properties The saturation pressure of water at 20C is 23392 kPa Table A4 Analysis The vapor pressure of air before compression is 1 17 kPa 05023392 kPa sat 25 C 1 1 1 P P P g v φ φ The pressure ratio during the compression process is 800 kPa92 kPa 870 That is the pressure of air and any of its components increases by 870 times Then the vapor pressure of air after compression becomes 1 17 kPa870 102 kPa Pressure ratio 1 2 v v P P The dewpoint temperature of the air at this vapor pressure is C 46 1 sat 10 2 kPa sat dp 2 T T T vP which is greater than 20C Therefore part of the moisture in the compressed air will condense when air is cooled to 20C 14121E The mole fraction of the water vapor at the surface of a lake and the mole fraction of water in the lake are to be determined and compared Assumptions 1 Both the air and water vapor are ideal gases 2 Air is weakly soluble in water and thus Henrys law is applicable Properties The saturation pressure of water at 60F is 02564 psia Table A4E Henrys constant for air dissolved in water at 290 K 60ºF is given in Table 162 to be H 62000 bar Since we do not have the value at 50ºF we may use this value Analysis The air at the water surface will be saturated Therefore the partial pressure of water vapor in the air at the lake surface will simply be the saturation pressure of water at 60F psia 0 1781 sat 50 F vapor P P Air 145 psi 50F Lake Assuming both the air and vapor to be ideal gases the mole fraction of water vapor in the air at the surface of the lake is determined to be 00123 or 123 percent 14 5 psia psia 01781 vapor vapor P P y The partial pressure of dry air just above the lake surface is 0 1781 1432 psia 14 5 vapor dry air P P P Then the mole fraction of air in the water becomes 5 dryairgasside dryairliquidside 1 593 10 62000 bar1 atm10132 5 bar 1432 psia 1 atm 14696 psia H P y which is very small as expected Therefore the mole fraction of water in the lake near the surface is 10 5 dryairliquidside waterliquid side 1 593 10 1 1 y y Discussion The concentration of air in water just below the airwater interface is 159 moles per 100000 moles The amount of air dissolved in water will decrease with increasing depth PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1481 14122E A room is cooled adequately by a 7500 Btuh airconditioning unit If the room is to be cooled by an evaporative cooler the amount of water that needs to be supplied to the cooler is to be determined Assumptions 1 The evaporative cooler removes heat at the same rate as the air conditioning unit 2 Water evaporates at an average temperature of 70F Properties The enthalpy of vaporization of water at 70F is 10537 Btulbm Table A4E Analysis Noting that 1 lbm of water removes 10537 Btu of heat as it evaporates the amount of water that needs to evaporate to remove heat at a rate of 7500 Btuh is determined from to be Q m hfg water 712 lbmh 10537 Btulbm 7500 Btuh water hfg Q m 14123E The required size of an evaporative cooler in cfm ft3min for an 8ft high house is determined by multiplying the floor area of the house by 4 An equivalent rule is to be obtained in SI units Analysis Noting that 1 ft 03048 m and thus 1 ft2 00929 m2 and 1 ft3 00283 m3 and noting that a flow rate of 4 ft3min is required per ft2 of floor area the required flow rate in SI units per m2 of floor area is determined to 1 22 m min m 1 0 0283 m min 4 0929 m 0 4 ft min ft 1 3 2 3 2 3 2 Therefore a flow rate of 122 m3min is required per m2 of floor area 14124 A cooling tower with a cooling capacity of 220 kW is claimed to evaporate 9500 kg of water per day It is to be determined if this is a reasonable claim Assumptions 1 Water evaporates at an average temperature of 30C 2 The coefficient of performance of the air conditioning unit is COP 3 Properties The enthalpy of vaporization of water at 30C is 24298 kJkg Table A4 Analysis Using the definition of COP the electric power consumed by the air conditioning unit when running is 7333 kW 3 220 kW COP cooling in Q W Then the rate of heat rejected at the cooling tower becomes 293 3 kW 220 7333 cooling rejected Win Q Q Noting that 1 kg of water removes 24298 kJ of heat as it evaporates the amount of water that needs to evaporate to remove heat at a rate of 2933 kW is determined from to be Q m fg rejected water h 0 1207 kgs 4346 kgh 10430 kgday 24298 kJkg 3 kJs 293 rejected water fg h Q m In practice the airconditioner will run intermittently rather than continuously at the rated power and thus the water use will be less Therefore the claim amount of 9500 kg per day is reasonable PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1482 14125E It is estimated that 190000 barrels of oil would be saved per day if the thermostat setting in residences in summer were raised by 6F 33C The amount of money that would be saved per year is to be determined Assumptions The average cooling season is given to be 120 days and the cost of oil to be 20barrel Analysis The amount of money that would be saved per year is determined directly from 190000 barrelday120 daysyear70barrel 1596000000 Therefore the proposed measure will save more than one and half billion dollars a year 14126 Shading the condenser can reduce the airconditioning costs by up to 10 percent The amount of money shading can save a homeowner per year during its lifetime is to be determined Assumptions It is given that the annual airconditioning cost is 500 a year and the life of the airconditioning system is 20 years Analysis The amount of money that would be saved per year is determined directly from 500 year20 years010 1000 Therefore the proposed measure will save about 1000 during the lifetime of the system 14127 Air at a specified state is heated to to a specified temperature The relative humidity after the heating is to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 2 1 a a a m m m Analysis The properties of the air at the ambient state are determined from the psychrometric chart Figure A31 to be 0 0015 kg H Okg dry air 2 2 1 ω ω 0C 40 1 atm 18C AIR 1 As the outside air infiltrates into the dacha it does not gain or lose any water Therefore the humidity ratio inside the dacha is the same as that outside 0 0015 kg H Okg dry air 2 1 2 ω ω 2 Entering the psychrometry chart at this humidity ratio and the temperature inside the dacha gives 118 0 118 φ2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1487 14132E Air is cooled and dehumidified at constant pressure by a simple ideal vaporcompression refrigeration system The systems COP is to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 2 1 a a a m m m Analysis The inlet and the exit states of the air are completely specified and the total pressure is 1 atm The properties of the air at various states are determined from the psychrometric chart Figure A31 to be 44 ft lbm dry air 14 0263 lbm H Olbm dry air 0 6 Btulbm dry air 50 3 1 2 1 1 v ω h 1 atm Cooling coils 1 2 Condensate Condensate removal T1 90F φ 1 85 T2 75F φ 2 50 and 0093 lbm H Olbm dry air 0 2 Btulbm dry air 28 2 2 2 ω h For the desired dehumidification the air at the exit should be saturated with a specific humidity of 00093 lbm waterlbm dry air That is 825F 0093 lbm H Olbm dry air 0 01 2 2 2 ω φ The temperature of the air at this state is the minimum air temperature required during this process 552 F 2 min T From the problem statement the properties of R134a at various states are Tables A11E through A13E or from EES 52 Btulbm throttling 48 4852 Btulbm sat liquid psia 160 11901 kJkg psia 160 F 109 5 19 5 90 0 22156 Btulbm R 10949 Btulbm vapor sat 55 psia F 45 2 10 2 55 3 4 160 psia 3 3 2 1 2 sat 1095 F 2 sat 55 psia 1 55 psia 1 sat 452 F 1 1 h h h h P h s s P P T s s h h P P T f g g QH QL 452F 1 2 3 4 1095F Win 4s s T The COP of this system is then 640 11901 10949 4852 10949 COP 1 2 4 1 in h h h h w qL PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1488 14133 A tank contains saturated air at a specified state The mass of the dry air the specific humidity and the enthalpy of the air are to be determined Assumptions The air and the water vapor are ideal gases Analysis a The air is saturated thus the partial pressure of water vapor is equal to the saturation pressure at the given temperature 8766 kPa 2 339 90 2 339 kPa sat 20 C v a g v P P P P P P 18 m3 20C 90 kPa Treating air as an ideal gas 188 kg 0 287 kPa m kg K293 K 8766 kPa18 m 3 3 R T P m a a a V b The specific humidity of air is determined from 00166 kg H Okg dry air 2 2 339 kPa 90 0 622 2 339 kPa 622 0 v v P P P ω c The enthalpy of air per unit mass of dry air is determined from 622 kJkg dry air C20 C 0016625374 kJkg 005 kJkg 1 g p v a h c T h h h ω ω PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1489 14134 Problem 14133 is reconsidered The properties of the air at the initial state are to be determined and the effects of heating the air at constant volume until the pressure is 110 kPa is to be studied Analysis The problem is solved using EES and the solution is given below Input Data Tdb1 20 C P190 kPa Rh110 P2110 kPa Vol 18 m3 w1HUMRATAirH2OTTdb1PP1RRh1 v1VOLUMEAirH2OTTdb1PP1RRh1 maVolv1 h1ENTHALPYAirH2OTTdb1PP1ww1 Energy Balance for the constant volume tank Ein Eout DELTAEtank DELTAEtankmau2 u1 Ein Qin Eout 0 kJ u1INTENERGYAirH2OTTdb1PP1ww1 u2INTENERGYAirH2OTTdb2PP2ww2 The ideal gas mixture assumption applied to the constant volume process yields P1Tdb1273P2Tdb2273 The mass of the water vapor and dry air are constant thus w2w1 Rh2RELHUMAirH2OTTdb2PP2ww2 h2ENTHALPYAirH2OTTdb2PP2ww2 v2VOLUMEAirH2OTTdb2PP2RRh2 PROPERTIES AT THE INITIAL STATE h16225 kJkga ma1875 kga v109599 m3kga w1001659 kgwkga 90 92 94 96 98 100 102 104 106 108 110 0 10 20 30 40 50 60 70 80 90 100 P2 kPa Qin kJ P2 kPa Qin kJ 90 92 94 96 98 100 102 104 106 108 110 0 9071 1814 2722 3629 4537 5444 6352 7261 8169 9078 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1490 14135E Air at a specified state and relative humidity flows through a circular duct The dewpoint temperature the volume flow rate of air and the mass flow rate of dry air are to be determined Assumptions The air and the water vapor are ideal gases Analysis a The vapor pressure of air is AIR 15 psia 50 fs 60F 50 0 128 psia 050 0 2564 psia sat 60 F P P P g v φ φ Thus the dewpoint temperature of the air is 413F from EES sat 0 128 psia sat dp T T T vP b The volume flow rate is determined from 1745 ft s 3 4 8 12 ft 50 fts 4 2 2 π V πD VA V c To determine the mass flow rate of dry air we first need to calculate its specific volume 1295 ft lbm dry air 14872 psia 0 3704 psia ft lbm R520 R 14872 psia 0 128 15 3 3 1 1 1 a a v a P T R P P P v Thus 135 lbms 95 ft lbm dry air 12 s 45 ft 17 3 3 1 1 1 v V a m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1491 14136 Air enters a cooling section at a specified pressure temperature and relative humidity The temperature of the air at the exit and the rate of heat transfer are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible m m m a a a 1 2 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis a The amount of moisture in the air also remains constant ω ω 1 2 as it flows through the cooling section since the process involves no humidification or dehumidification The total pressure is 97 kPa The properties of the air at the inlet state are 6344 kJkg dry air 005 kJkg C35 C 0011025646 kJkg 1 00110 kg H Okg dry air 1 69 kPa 97 0 622 1 69 kPa 622 0 927 m kg dry air 0 9531 kPa 0 287 kPa m kg K308 K 9531 kPa 1 69 97 1 69 kPa 30 5 629 kPa 1 1 1 1 2 2 1 1 1 1 3 3 1 1 1 1 1 1 1 sat 35 C 1 1 1 g p v v a a v a g v h c T h P P P P T R P P P P P P ω ω ω φ φ v 35C 30 6 m3min 97 kPa Cooling coils AIR 1 2 The air at the final state is saturated and the vapor pressure during this process remains constant Therefore the exit temperature of the air must be the dewpoint temperature 148C sat 1 69 kPa sat dp T T T vP b The enthalpy of the air at the exit is 4278 kJkg dry air C148 C 0011025281 kJkg 005 kJkg 1 2 2 2 2 g p h c T h ω Also 6 47 kgmin 927 m kg dry air 0 m s 6 3 3 1 1 v V a m Then the rate of heat transfer from the air in the cooling section becomes 134 kJmin 4278kJkg 6 47 kgmin6344 2 1 out h h m Q a preparation If you are a student using this Manual you are using it without permission 1492 14137 The outdoor air is first heated and then humidified by hot steam in an airconditioning system The rate of heat supply in the heating section and the mass flow rate of the steam required in the humidifying section are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible m m m a a a 1 2 Properties The amount of moisture in the air also remains constants it flows through the heating section ω ω 1 2 but increases in the humidifying section ω ω 3 2 The inlet and the exit states of the air are completely specified and the total pressure is 1 atm The properties of the air at various states are determined from the psychrometric chart Fig A31 to be 807 m kg dry air 0 0030 kg H Okg dry air 0 7 kJkg dry air 17 3 1 2 2 1 1 v ω ω h 10C 40 22 m3min 1 atm 22C Heating coils 3 AIR 25C 55 2 1 h2 1 29 8 0 0030 kJ kg dry air kg H O kg dry air 2 2 ω ω h3 52 9 0 0109 kJ kg dry air kg H O kg dry air 3 2 ω Analysis a The mass flow rate of dry air is 27 3 kgmin 0 807 m kg 22 m min 3 3 1 1 v V a m Then the rate of heat transfer to the air in the heating section becomes Q m h h a in kg min298 kJ kg 2 1 273 177 3303 kJ min b The conservation of mass equation for water in the humidifying section can be expressed as m m m m m a w a w a 2 2 3 3 2 ω ω ω ω or 3 Thus mw 273 0 0030 kg min00109 0216 kg min PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1495 14140 Air is heated and dehumidified in an airconditioning system consisting of a heating section and an evaporative cooler The temperature and relative humidity of the air when it leaves the heating section the rate of heat transfer in the heating section and the rate of water added to the air in the evaporative cooler are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible m m m a a a 1 2 Analysis a Assuming the wetbulb temperature of the air remains constant during the evaporative cooling process the properties of air at various states are determined from the psychrometric chart Fig A31 or EES to be 824 m kg 0 00581 kg H Okg dry air 0 8 kJkg dry air 29 55 C 15 3 1 2 1 1 1 1 v ω φ h T 15C 55 30 m3min 1 atm T2 Heating coils 3 AIR 2 1 Water 8 kJ kg dry air 47 3 2 2 2 wb3 wb2 1 2 h h T T T 192 325 C φ ω ω 25C 45 C 1 17 00888 kg H Okg dry air 0 8 kJkg dry air 47 45 C 25 3 2 3 3 3 3 wb T h T ω φ b The mass flow rate of dry air is 36 4 kgmin 824 m kg dry air 0 m min 30 3 3 1 1 v V a m Then the rate of heat transfer to air in the heating section becomes 655 kJmin 29 8 kJkg 36 4 kgmin478 1 2 in h h m Q a c The rate of water added to the air in evaporative cooler is 000581 0112 kgmin 36 4 kgmin000888 2 3 2 3 added ω a ω w w w m m m m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1496 14141 Problem 14140 is reconsidered The effect of total pressure in the range 94 to 100 kPa on the results required in the problem is to be studied Analysis The problem is solved using EES and the solution is given below P101325 kPa Tdb1 15 C Rh1 055 Voldot1 30 m3min Tdb3 25 C Rh3 045 P1P P2P1 P3P1 Energy balance for the steadyflow heating process 1 to 2 We neglect the PE of the flow Since we dont know the cross sectional area of the flow streams we also neglect theKE of the flow Edotin Edotout DELTAEdotsys DELTAEdotsys 0 kJmin Edotin mdotah1Qdotin Edotout mdotah2 Conservation of mass of dry air during mixing mdota constant mdota Voldot1v1 Conservation of mass of water vapor during the heating process mdotaw1 mdotaw2 Conservation of mass of water vapor during the evaporative cooler process mdotaw2mdotw mdotaw3 During the evaporative cooler process Twb2 Twb3 Twb3 WETBULBAirH2OTTdb3PP3RRh3 h1ENTHALPYAirH2OTTdb1PP1RRh1 v1VOLUMEAirH2OTTdb1PP1RRh1 w1HUMRATAirH2OTTdb1PP1RRh1 h2ENTHALPYAirH2OTTdb2PP2BTwb2 h2h3 Tdb2TEMPERATUREAirH2Ohh2PP2ww2 w2HUMRATAirH2OTTdb2PP2RRh2 h3ENTHALPYAirH2OTTdb3PP3RRh3 w3HUMRATAirH2OTTdb3PP3RRh3 P kPa mw kgmin Qin kJmin Rh2 Tdb2 C 94 95 96 97 98 99 100 01118 01118 01118 01118 01118 01117 01117 6286 6322 6358 6394 643 6466 6503 01833 01843 01852 0186 01869 01878 01886 3329 332 3311 3303 3294 3286 3278 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1497 94 95 96 97 98 99 100 327 328 329 33 331 332 333 0183 0184 0185 0186 0187 0188 0189 P kPa Tdb2 C Rh2 94 95 96 97 98 99 100 625 630 635 640 645 650 655 P kPa Qin kJmin 94 95 96 97 98 99 100 01116 01117 01118 01119 P kPa mw kgmin PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1498 14142 Air is heated and dehumidified in an airconditioning system consisting of a heating section and an evaporative cooler The temperature and relative humidity of the air when it leaves the heating section the rate of heat transfer in the heating section and the rate at which water is added to the air in the evaporative cooler are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible m m m a a a 1 2 Analysis a Assuming the wetbulb temperature of the air remains constant during the evaporative cooling process the properties of air at various states are determined to be PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 3059 kJkg dry air C15 C 000613825283 kJkg 005 kJkg 1 0006138 kg H Okg dry air 0 8695 kPa 96 0 622 0 8695 kPa 622 0 8695 m kg dry air 0 9506 kPa 0 287 kPa m kg K288 K 9506 kPa 0 938 96 0 938 kPa 0 55 1 7057 kPa 1 1 1 1 2 1 1 1 1 3 3 1 1 1 1 1 1 1 sat 15 C 1 1 1 g p v v a a v a g v h c T h P P P P T R P P P P P P ω ω φ φ v 15C 55 30 m3min 96 kPa T2 Heating coils 3 AIR 2 1 Water 25C 45 and 01 kJkg dry air 49 C25 C 000938125465 kJkg 005 kJkg 1 0009381 kg H Okg dry air 1 426 kPa 96 0 622 1 426 kPa 622 0 9457 kPa 1 426 96 1 426 kPa 0 45 3 17 kPa 3 3 3 3 2 3 3 3 3 3 3 3 sat 25 C 3 3 3 3 g p v v v a g v h c T h P P P P P P P P P ω ω φ φ Also 006138 kg H Okg dry air 0 01 kJkg 49 2 1 2 3 2 ω ω h h Thus 000613825009 182 C 1 005 kJkg 1 82 2500 9 2 2 2 2 2 2 2 2 2 T T T c T h c T h p g p ω ω Solving for T2 5 072 kPa sat331 C 2 2 P P T g C 331 Thus 0185 or 185 0622 00061385072 0 00613896 0 622 2 2 2 2 2 Pg P ω ω φ b The mass flow rate of dry air is 34 5 kgmin 8695 m kg dry air 0 m min 30 3 3 1 1 v V a m Then the rate of heat transfer to air in the heating section becomes 636 kJmin 3059kJkg 34 5 kgmin4901 1 2 in h h m Q a c The rate of water addition to the air in evaporative cooler is 0006138 0112 kgmin 34 5 kgmin0009381 2 3 2 3 added ω a ω w w w m m m m preparation If you are a student using this Manual you are using it without permission 1499 14143 Air is cooled dehumidified and heated at constant pressure The system hardware and the psychrometric diagram are to be sketched and the heat transfer is to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 2 1 a a a m m m Analysis a The schematic of the cooling and dehumidifaction process and the the process on the psychrometric chart are given below The psychrometric chart is obtained from the Property Plot feature of EES 1 atm Cooling coils 1 2 Condensate Condensate removal T130C φ 160 T220C Twb212C 20C 0 5 10 15 20 25 30 35 40 0000 0005 0010 0015 0020 0025 0030 0035 0040 0045 0050 T C Humidity Ratio Pressure 1013 kPa 10C 15C 20C 25C 30C 35C 02 04 06 08 AirH2O 1 2 b The inlet and the exit states of the air are completely specified and the total pressure is 1 atm The properties of the air at the inlet and exit states are determined from the psychrometric chart Figure A31 or EES to be 881 m kg dry air 0 0160 kg H Okg dry air 0 2 kJkg dry air 71 3 1 2 1 1 v ω h and 838 m kg dry air 0 00544 kg H Okg dry air 0 9 kJkg dry air 33 376 0 3 2 2 2 2 2 v ω φ h Also 9 kJkg 83 20 C f w h h Table A4 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 14101 14144 Air is cooled dehumidified and heated at constant pressure The system hardware and the psychrometric diagram are to be sketched and the heat transfer is to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 2 1 a a a m m m Analysis a The schematic of the cooling and dehumidifaction process and the the process on the psychrometric chart are given below The psychrometric chart is obtained from the Property Plot feature of EES 1 atm Cooling coils Heating coils 1 2 Condensate Condensate removal T130C φ 160 Tdp 4C φ 2 50 4C 0 5 10 15 20 25 30 35 40 0000 0005 0010 0015 0020 0025 0030 0035 0040 0045 0050 T C Humidity Ratio Pressure 1013 kPa 10C 15C 20C 25C 30C 35C 02 04 06 08 AirH2O 1 2 b The inlet and the exit states of the air are completely specified and the total pressure is 1 atm The properties of the air at the inlet and exit states are determined from the psychrometric chart Figure A31 or EES to be 881 m kg dry air 0 0160 kg H Okg dry air 0 2 kJkg dry air 71 3 1 2 1 1 dp v ω h T C 214 and 00504 kg H Okg dry air 0 1 kJkg dry air 27 50 0 C 3 14 2 2 2 2 2 ω φ h T PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 14104 0 9824 0 008462 0 7933 0 03065 1 4 3 3 4 3 3 3 4 1 2 3 3 3 4 makeup 3 3 3 waste h h m h m h m h m m m h h m m m h Q a ω ω 1337 kgs 3 3 0 9824 12574 kJkg 17590 70000 kJs m m and 0 79331337 kgs 1061 kgs 0 7933 3 m ma b Then the volume flow rate of air into the cooling tower becomes 902 m s 3 kgs08504 m kg 1061 3 1 1 v V m a c The mass flow rate of the required makeup water is determined from 235 kgs 0008462 1061 kgs003065 1 2 makeup ω ma ω m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 14105 14146 Problem 14145 is reconsidered The effect of air inlet wetbulb temperature on the required air volume flow rate and the makeup water flow rate is to be investigated Analysis The problem is solved using EES and the solution is given below Input Data Patm 101325 kPa Tdb1 23 C Twb1 16 C Tdb2 32 C RH2 100100 relative humidity at state 2 saturated condition Qdotwaste 70 MWConvertMW kW Tcw3 42 C Cooling water temperature at state 3 Tcw4 30 C Cooling water temperature at state 4 Dry air mass flow rates RH1 is the relative humidity at state 1 on a decimal basis v1VOLUMEAirH2OTTdb1PPatmRRH1 Twb1 WETBULBAirH2OTTdb1PPatmRRH1 mdota1 Voldot1v1 Conservaton of mass for the dry air ma in the SSSF mixing device mdotain mdotaout DELTAmdotacv mdotain mdota1 mdotaout mdota2 DELTAmdotacv 0 Steady flow requirement Conservation of mass for the water vapor mv and cooling water for the SSSF process mdotwin mdotwout DELTAmdotwcv mdotwin mdotv1 mdotcw3 mdotwout mdotv2mdotcw4 DELTAmdotwcv 0 Steady flow requirement w1HUMRATAirH2OTTdb1PPatmRRH1 mdotv1 mdota1w1 w2HUMRATAirH2OTTdb2PPatmRRH2 mdotv2 mdota2w2 Conservation of energy for the SSSF cooling tower process The process is adiabatic and has no work done ngelect ke and pe Edotintower Edotouttower DELTAEdottowercv Edotintower mdota1 h1 mdotcw3hw3 Edotouttower mdota2h2 mdotcw4hw4 DELTAEdottowercv 0 Steady flow requirement h1ENTHALPYAirH2OTTdb1PPatmww1 h2ENTHALPYAirH2OTTdb2PPatmww2 hw3ENTHALPYsteamTTcw3x0 hw4ENTHALPYsteamTTcw4x0 Energy balance on the external heater determines the cooling water flow rate Edotinheater Edotoutheater DELTAEdotheatercv Edotinheater Qdotwaste mdotcw4hw4 Edotoutheater mdotcw3 hw3 DELTAEdotheatercv 0 Steady flow requirement Conservation of mass on the external heater gives the makeup water flow rate Note The makeup water flow rate equals the amount of water vaporized in the cooling tower PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 14106 mdotcwin mdotcwout DELTAmdotcwcv mdotcwin mdotcw4 mdotmakeup mdotcwout mdotcw3 DELTAmdotcwcv 0 Steady flow requirement Twb1 C Vol1 m3s mmakeup kgws mcw3 kgws ma1 kgas 14 15 16 17 18 19 20 21 22 23 8282 8627 9017 946 9967 1055 1124 1206 1304 1424 2384 2369 2353 2334 2312 2287 2258 2223 2182 213 1336 1336 1337 1337 1338 1338 1339 1340 1341 1342 9774 1016 1060 1110 1168 1234 1312 1404 1515 1651 14 15 16 17 18 19 20 21 22 23 800 900 1000 1100 1200 1300 1400 1500 21 215 22 225 23 235 24 Twb1 C Vol1 m 3s mmakeup kgws PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 14107 14147 An uninsulated tank contains moist air at a specified state Water is sprayed into the tank until the relative humidity in the tank reaches a certain value The amount of water supplied to the tank the final pressure in the tank and the heat transfer during the process are to be determined Assumptions 1 Dry air and water vapor are ideal gases 2 The kinetic and potential energy changes are negligible Analysis The initial state of the moist air is completely specified The properties of the air at the inlet state may be determined from the psychrometric chart Figure A31 or using EES psychrometric functions to be we used EES 6863 m kg dry air 0 0 005433 kg H Okg dry air 16 kJkg dry air 49 3 1 2 1 1 v ω h The initial mass in the tank is 0 7285 kg m 06863 m 50 3 3 1 1 v V ma The partial pressure of dry air in the tank is 128 8 kPa m 05 273 K 07285 kg0287 kJkgK35 3 2 2 V m R T P a a a Then the pressure of moist air in the tank is determined from 0 622 128 8 kPa 1 0 622 1 2 2 2 2 ω ω Pa P We cannot fix the final state explicitly by a handsolution However using EES which has builtin functions for moist air properties the final state properties are determined to be 9797 kJkg dry air 2 2 h P 13387 kPa 6867 m kg dry air 0 02446 kg H Okg dry air 0 3 2 2 2 v ω The partial pressures at the initial and final states are 5 07 kPa 87 12881 133 12887 kPa 1 126 130 1 126 kPa 20 5 6291 kPa 0 2 2 2 1 1 1 sat35 C 1 1 a v v a v P P P P P P P P φ The specific volume of water at 35ºC is 25205 m kg 3 g 35 C 2 1 v v v w w The internal energies per unit mass of dry air in the tank are 3944 kJkg 25205 1 126 0 005433 0 6863 12887 4916 1 1 1 1 1 1 1 w v a w P P h u v v 6 396 kJkg 25205 5 07 0 02446 0 6867 12881 9797 2 2 2 2 2 2 2 w v a w P P h u v v The enthalpy of water entering the tank from the supply line is 20934 kJkg f 50 C 1 h hw The internal energy of water vapor at the final state is 7 kJkg 2422 g 35 C 2 u uw The amount of water supplied to the tank is 001386 kg 0 7285 kg0024460005433 1 2 ω a ω w m m An energy balance on the system gives 641 kJ in in 2 1 2 1 in tank in 001386 kg24227 kJkg 0 7285 kg 63963944kJkg 0 01386 kg20934 kJkg Q Q m u u u m m h Q E E w w a w w PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 14108 Fundamentals of Engineering FE Exam Problems 14148 A room is filled with saturated moist air at 25C and a total pressure of 100 kPa If the mass of dry air in the room is 100 kg the mass of water vapor is a 052 kg b 197 kg c 296 kg d 204 kg e 317 kg Answer d 204 kg Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values T125 C P100 kPa mair100 kg RH1 PgPRESSURESteamIAPWSTT1x0 RHPvPg PairPPv w0622PvPPv wmvmair Some Wrong Solutions with Common Mistakes W1vmassmairw1 w10622PvP Using P instead of PPv in w relation W2vmassmair Taking mvapor mair W3vmassPvPmair Using wrong relation 14149 A room contains 65 kg of dry air and 06 kg of water vapor at 25C and 90 kPa total pressure The relative humidity of air in the room is a 35 b 415 c 552 d 609 e 730 Answer b 415 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values T125 C P90 kPa mair65 kg mv06 kg w0622PvPPv wmvmair PgPRESSURESteamIAPWSTT1x0 RHPvPg Some Wrong Solutions with Common Mistakes W1RHmvmairmv Using wrong relation W2RHPgP Using wrong relation PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 14109 14150 A 40m3 room contains air at 30C and a total pressure of 90 kPa with a relative humidity of 75 percent The mass of dry air in the room is a 247 kg b 299 kg c 399 kg d 414 kg e 523 kg Answer c 399 kg Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values V40 m3 T130 C P90 kPa RH075 PgPRESSURESteamIAPWSTT1x0 RHPvPg PairPPv Rair0287 kJkgK mairPairVRairT1273 Some Wrong Solutions with Common Mistakes W1massPairVRairT1 Using C instead of K W2massPVRairT1273 Using P instead of Pair W3massmairRH Using wrong relation 14151 A room contains air at 30C and a total pressure of 960 kPa with a relative humidity of 75 percent The partial pressure of dry air is a 820 kPa b 858 kPa c 928 kPa d 906 kPa e 720 kPa Answer c 928 kPa Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values T130 C P96 kPa RH075 PgPRESSURESteamIAPWSTT1x0 RHPvPg PairPPv Some Wrong Solutions with Common Mistakes W1PairPv Using Pv as Pair W2PairPPg Using wrong relation W3PairRHP Using wrong relation PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 14110 14152 The air in a house is at 25C and 65 percent relative humidity Now the air is cooled at constant pressure The temperature at which the moisture in the air will start condensing is a 74C b 163C c 180C d 113C e 202C Answer c 180C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values T125 C RH1065 PgPRESSURESteamIAPWSTT1x0 RH1PvPg TdpTEMPERATURESteamIAPWSx0PPv Some Wrong Solutions with Common Mistakes W1TdpT1RH1 Using wrong relation W2TdpT1273RH1273 Using wrong relation W3TdpWETBULBAirH2OTT1PP1RRH1 P1100 Using wetbulb temperature 14153 On the psychrometric chart a cooling and dehumidification process appears as a line that is a horizontal to the left b vertical downward c diagonal upwards to the right NE direction d diagonal upwards to the left NW direction e diagonal downwards to the left SW direction Answer e diagonal downwards to the left SW direction 14154 On the psychrometric chart a heating and humidification process appears as a line that is a horizontal to the right b vertical upward c diagonal upwards to the right NE direction d diagonal upwards to the left NW direction e diagonal downwards to the right SE direction Answer c diagonal upwards to the right NE direction PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 14111 14155 An air stream at a specified temperature and relative humidity undergoes evaporative cooling by spraying water into it at about the same temperature The lowest temperature the air stream can be cooled to is a the dry bulb temperature at the given state b the wet bulb temperature at the given state c the dew point temperature at the given state d the saturation temperature corresponding to the humidity ratio at the given state e the triple point temperature of water Answer a the dry bulb temperature at the given state 14156 Air is cooled and dehumidified as it flows over the coils of a refrigeration system at 85 kPa from 35C and a humidity ratio of 0023 kgkg dry air to 15C and a humidity ratio of 0015 kgkg dry air If the mass flow rate of dry air is 04 kgs the rate of heat removal from the air is a 4 kJs b 8 kJs c 12 kJs d 16 kJs e 20 kJs Answer d 16 kJs Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P85 kPa T135 C w10023 T215 C w20015 mair04 kgs mwatermairw1w2 h1ENTHALPYAirH2OTT1PPww1 h2ENTHALPYAirH2OTT2PPww2 hwENTHALPYSteamIAPWSTT2x0 Qmairh1h2mwaterhw Some Wrong Solutions with Common Mistakes W1Qmairh1h2 Ignoring condensed water W2QmairCpairT1T2mwaterhw Cpair 1005 Using dry air enthalpies W3Qmairh1h2mwaterhw Using wrong sign PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 151 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 15 CHEMICAL REACTIONS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 152 Fuels and Combustion 151C Nitrogen in general does not react with other chemical species during a combustion process but its presence affects the outcome of the process because nitrogen absorbs a large proportion of the heat released during the chemical process 152C Moisture in general does not react chemically with any of the species present in the combustion chamber but it absorbs some of the energy released during combustion and it raises the dew point temperature of the combustion gases 153C The number of atoms are preserved during a chemical reaction but the total mole numbers are not 154C Airfuel ratio is the ratio of the mass of air to the mass of fuel during a combustion process Fuelair ratio is the inverse of the airfuel ratio 155C No Because the molar mass of the fuel and the molar mass of the air in general are different 156C The dewpoint temperature of the product gases is the temperature at which the water vapor in the product gases starts to condense as the gases are cooled at constant pressure It is the saturation temperature corresponding to the vapor pressure of the product gases 157 Sulfur is burned with oxygen to form sulfur dioxide The minimum mass of oxygen required and the mass of sulfur dioxide in the products are to be determined when 1 kg of sulfur is burned Properties The molar masses of sulfur and oxygen are 3206 kgkmol and 3200 kgkmol respectively Table A1 Analysis The chemical reaction is given by PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2 2 SO O S 2 2 SO O S Hence 1kmol of oxygen is required to burn 1 kmol of sulfur which produces 1 kmol of sulfur dioxide whose molecular weight is 6406 kgkmol 3200 3206 O2 S SO2 M M M Then 0998 kg O kg S 2 1 kmol3206 kgkmol kmol32 kgkmol 1 S S O2 O2 S O2 M N M N m m nd a 1998 kg SO kg S 2 1 kmol3206 kgkmol kmol6406 kgkmol 1 S S SO2 SO2 S SO2 N M M N m m preparation If you are a student using this Manual you are using it without permission 153 158E Methane is burned with diatomic oxygen The mass of water vapor in the products is to be determined when 1 lbm of methane is burned Properties The molar masses of CH4 O2 CO2 and H2O are 16 32 44 and 18 lbmlbmol respectively Table A1E 2H O CO 2O CH 2 2 2 4 Analysis The chemical reaction is given by 2H O CO 2O CH 2 2 2 4 Hence for each lbmol of methane burned 2 lbmol of water vapor are formed Then 4 2 225 lbm H Olbm CH 1 lbmol16 lbmlbmol 2 lbmol18 lbmlbmol CH4 CH4 H2O H2O CH4 H2O M N M N m m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 154 Theoretical and Actual Combustion Processes 159C It represent the amount of air that contains the exact amount of oxygen needed for complete combustion 1510C No The theoretical combustion is also complete but the products of theoretical combustion does not contain any uncombined oxygen 1511C Case b 1512C The causes of incomplete combustion are insufficient time insufficient oxygen insufficient mixing and dissociation 1513C CO Because oxygen is more strongly attracted to hydrogen than it is to carbon and hydrogen is usually burned to completion even when there is a deficiency of oxygen PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 155 1514 Propane is burned with theoretical amount of air The mass fraction of carbon dioxide and the mole and mass fractions of the water vapor in the products are to be determined Properties The molar masses of C3H8 O2 N2 CO2 and H2O are 44 32 28 44 and 18 kgkmol respectively Table A1 Analysis a The reaction in terms of undetermined coefficients is 2 2 2 2 2 8 3 N H O CO 3 76N O C H p z y x Balancing the carbon in this reaction gives C3H8 Air 100 theoretical Combustion chamber CO2 H2O N2 y 3 and the hydrogen balance gives 4 8 2 z z The oxygen balance produces 5 4 2 3 2 2 2 z y x z y x PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course balance of the nitrogen in this reaction gives x balanced form the reaction is 88 N The mass fraction of carbon dioxide is determined from A 2 3 76 2 p p x 18 8 3 76 5 76 3 In 2 2 2 2 8 3 1 4H O 3CO 18 8 N 5O C H 2 0181 4 kg 730 132 kg 18 8 kmol28 kgkmol 4 kmol18 kgkmol 3 kmol44 kgkmol 3 kmol44 kgkmol mf N2 N2 H2O H2O CO2 CO2 CO2 CO2 products CO2 CO2 M N M N M N M N m m The mo ctions of water vapor are b le and mass fra 0155 25 8 kmol 4 kmol 18 8 kmol 4 kmol 3 kmol kmol 4 N2 H2O CO2 H2O products H2O H2O N N N N N N y 00986 730 4 kg 72 kg 18 8 kmol28 kgkmol 4 kmol18 kgkmol 3 kmol44 kgkmol 4 kmol18 kgkmol mf N2 N2 H2O H2O CO2 CO2 H2O H2O products H2O H2O M N M N M N M N m m preparation If you are a student using this Manual you are using it without permission 156 1515 Methane is burned with air The mass flow rates at the two inlets are to be determined Properties The molar masses of CH4 O2 N2 CO2 and H2O are 16 32 28 44 and 18 kgkmol respectively Table A1 Analysis The stoichiometric combustion equation of CH4 is 2 th 2 2 2 2 th 4 N 376 2H O CO 376N O CH a a PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 752N 2H O CO 376N The masses of the reactants are O2 O2 O2 CH4 CH4 CH4 Products CH4 Air O2 balance 2 1 1 th th a a Substituting 2 4 2 O CH 2 2 2 2 211 kg 3 76 kmol28 kgkmol 2 64 kg 2 kmol32 kgkmol 16 kg kmol16 kgkmol 1 N2 N2 N2 M M N m M N m he total m ss is 2 N m T a O CH4 total 291 kg 211 64 16 N2 N Then the mass fractions are m m m 0 7251 kg 291 total N2 m 211 kg mf 0 2199 291 kg 64 kg mf 0 05498 291 kg 16 kg mf N2 total O2 O2 total CH4 CH4 m m m m m or a mixt e flow of 05 kgs the mass flow rates of the reactants are 2 ber of CO in the products per mole of fuel burned are to be determined 2 2 eal Properties The molar masses of C H2 and O2 are 12 kgkmol 2 kgkmol and 32 kgkmol respectively Table A1 tal mole of the products are 45 9 km Then the mole fractions are F ur kgs 04725 kgs 002749 0 02749 50 0 0549805 kgs mf CH4 air CH4 CH4 m m m m m 1516 nButane is burned with stoichiometric amount of oxygen The mole fractions of CO water in the products and the mole num 2 Assumptions 1 Combustion is complete 2 The combustion products contain CO and H O 3 Combustion gases are id gases Analysis The combustion equation in this case is 5H O 4CO 56 O C H 2 2 2 10 O2 Products C4H10 Combustion chamber 4 The to ol 05556 04444 9 kmol kmol 5 9 kmol kmol 4 total H2O CO2 total CO2 CO2 N N y N N y 10 4 2 NCO2 4 kmol CO kmol C H Also preparation If you are a student using this Manual you are using it without permission 157 1517 Propane is burned with stoichiometric amount of air The mass fraction of each product the mass of water and air per unit mass of fuel burned are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O O2 and N2 only Properties The molar masses of C H2 O2 and air are 12 kgkmol 2 kgkmol 32 kgkmol and 29 kgkmol respectively Table A1 Analysis The reaction equation for 100 theoretical air is Products C3H8 Air 100 theoretical Combustion chamber 2 2 2 2 2 th 8 3 N H O CO 376N O C H E D B a where ath is the stoichiometric coefficient for air The coefficient ath and other coefficients are to be determined from the mass balances Carbon balance B 3 Hydrogen balance 4 8 2 D D PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Oxygen balance E ubstitutin the balanced reaction equation is 18 8 N 4 H O The mass of each product and the total mass are 2O N2 N2 N2 H2O H2O H2O CO2 CO2 CO2 m M N m M N m M N m hen the m ss fractions are 5 4 3 50 2 2 2 th th a D B a Nitrogen balance 3 76 th E a 18 8 3 76 5 S g 2 2 2 8 3 3CO 376N 5 O C H 2 2 730 4 kg 526 4 72 132 526 4 kg 18 8 kmol28 kgkmol 72 kg 4 kmol18 kgkmol 132 kg 3 kmol44 kgkmol N2 H CO2 total m m m T a 07207 00986 01807 7304 kg 5264 kg mf 72 kg mf 7304 kg mf total N2 N2 H2O H2O total CO2 m m m m The mass of water per unit mass of fuel burned is 132 kg CO2 m 7304 kg total m 8 3 2 1636 kg H Okg C H 44 kg 1 18 kg 4 C3H8 H2O m m The mass of air required per unit mass of fuel burned is 1569 kg airkg C3H8 44 kg 1 29 kg 4 76 5 C3H8 air m m preparation If you are a student using this Manual you are using it without permission 158 1518 nOctane is burned with stoichiometric amount of air The mass fraction of each product the mass of water in the products and the mass fraction of each reactant are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O O2 and N2 only Properties The molar masses of C H2 O2 and air are 12 kgkmol 2 kgkmol 32 kgkmol and 29 kgkmol respectively Table A1 Analysis The reaction equation for 100 theoretical air is Products C4H10 Air 100 theoretical Combustion chamber 2 2 2 2 2 th 18 8 N H O CO 376N O C H E D B a where ath is the stoichiometric coefficient for air The coefficient ath and other coefficients are to be determined from the mass balances Carbon balance B 8 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course E E a Substituting the balanced reaction equation is 47 N 9 H O 8 CO 376N 12 5 O C H ass of each product and the total mass are m m m m M N m M N m hen the mass fractions are Hydrogen balance D 9 18 2 D Oxygen balance 12 5 9 8 50 2 2 2 th th a D B a Nitrogen balance 47 3 76 12 5 3 76 th 2 2 2 2 2 18 8 The m 1830 kg 1316 162 352 1316 kg 47 kmol28 kgkmol 162 kg 9 kmol18 kgkmol 352 kg 8 kmol44 kgkmol H2O N2 CO2 total N2 N2 N2 H2O H2O H2O CO2 CO2 CO2 M N m T 00885 162 kg mf H2O H2O m 07191 1830 kg mf total N2 N2 m 01923 kg 1316 kg 1830 1830 kg 352 kg mf total total CO2 CO2 m m m m ass of water per unit mass of fuel burned is The m 18 8 2 1421 kg H Okg C H 1 114 kg 18 kg 9 C8H18 H2O m m ass of each reactant and the total mass are The m 1839 5 kg 1725 5 114 1725 5 kg 4 76 kmol29 kgkmol 12 5 114 kg kmol114 kgkmol 1 air C8H18 total air air air C8H18 C8H18 C8H18 m m m M N m M N m Then the mass fractions of reactants are 09380 00620 18395 kg 17255 kg mf 18395 kg 114 kg mf total air air total C8H18 C8H18 m m m m preparation If you are a student using this Manual you are using it without permission 159 1519 Acetylene is burned with 10 percent excess oxygen The mass fractions of each of the products and the mass of oxygen used per unit mass of fuel burned are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O and O2 3 Combustion gases are ideal gases Properties The molar masses of C H2 and O2 are 12 kgkmol 2 kgkmol and 32 kgkmol respectively Table A1 Analysis The stoichiometric combustion equation is O2 Products C2H2 Combustion chamber H O 2CO 52 O C H 2 2 2 2 2 The combustion equation with 10 excess oxygen is 2 2 2 2 2 2 025O H O 2CO 2 75O C H The mass of each product and the total mass are 114 kg 8 18 88 8 kg 0 25 kmol32 kgkmol 18 kg kmol18 kgkmol 1 88 kg 2 kmol44 kgkmol O2 H2O CO2 total O2 O2 O2 H2O H2O H2O CO2 CO2 CO2 m m m m M N m M N m M N m Then the mass fractions are 00702 01579 07719 114 kg mf total CO2 CO2 m 114 kg 8 kg mf 114 kg 18 kg mf kg 88 total O2 O2 total H2O H2O m m m m m The mass of oxygen per unit mass of fuel burned is determined from 2 2 2 3385 kg O kg C H 1 26 kg 32 kg 275 C2H2 O2 m m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1510 1520 nButane is burned with 100 percent excess air The mole fractions of each of the products the mass of carbon dioxide in the products per unit mass of the fuel and the airfuel ratio are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O O2 and N2 only Properties The molar masses of C H2 O2 and air are 12 kgkmol 2 kgkmol 32 kgkmol and 29 kgkmol respectively Table A1 Analysis The combustion equation in this case can be written as 2 th 2 th 2 2 2 2 th 10 4 N 3 76 02 O 01 5H O 4CO 376N O 02 C H a a a where ath is the stoichiometric coefficient for air We have automatically accounted for the 100 excess air by using the factor 20ath instead of ath for air The stoichiometric amount of oxygen athO2 will be used to oxidize the fuel and the remaining excess amount 10athO2 will appear in the products as free oxygen The coefficient ath is determined from the O2 balance O2 balance 56 01 52 4 02 th th th a a a PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 4888N 56 O 5H O 4CO 6N The mole fractions of the products are N Substituting 2 10 4 73 13 O C H 2 2 2 2 2 6438 kmol 4888 56 5 4 m 07592 01010 00777 00621 4 kmol CO2 CO2 N y 6438 kmol kmol 4888 kmol 6438 6438 kmol kmol 5 kmol 6438 N2 N2 O2 H2O H2O m m m m N N y N N N y N The mass of carbon dioxide in the products per unit mass of fuel burned is Products C4H10 Air 100 excess 65 kmol NO2 y 10 4 2 3034 kg CO kg C H 1 58 kg 44 kg 4 C4H10 CO2 m m The airfuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel 3094 kg airkg fuel 1 kmol58 kgkmol 476 kmol29 kgkmol 13 AF fuel air m m preparation If you are a student using this Manual you are using it without permission 1511 1521 nOctane is burned with 50 percent excess air The mole fractions of each of the products the mass of water in the products per unit mass of the fuel and the mass fraction of each reactant are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O O2 and N2 only Properties The molar masses of C H2 O2 and air are 12 kgkmol 2 kgkmol 32 kgkmol and 29 kgkmol respectively Table A1 Analysis The combustion equation in this case can be written as 2 th 2 th 2 2 2 2 th 18 8 N 3 76 51 O 50 9H O 8CO 376N O 51 C H a a a where ath is the stoichiometric coefficient for air We have automatically accounted for the 50 excess air by using the factor 15ath instead of ath for air The stoichiometric amount of oxygen athO2 will be used to oxidize the fuel and the remaining excess amount 05athO2 will appear in the products as free oxygen The coefficient ath is determined from the O2 balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1 th th th a a a m m m m m M N m M N m M N m hen the mass fractions are O2 balance 5 12 5 50 54 8 Substituting 2 2 2 2 2 2 18 8 70 5 N 6 25O 9H O 8CO 3 76N 1875 O C H The mass of each product and the total mass are 2688 kg 1974 200 162 352 1974 kg 70 5 kmol28 kgkmol 200 kg 6 25 kmol32 kgkmol 162 kg 9 kmol18 kgkmol 352 kg 8 kmol44 kgkmol N2 O2 H2O CO2 total N2 N2 N2 O2 O2 O2 H2O H2O H2O CO2 CO2 CO2 M N m Products C8H18 Air 50 excess T 07344 00744 00603 01310 kg 2688 1974 kg mf N2 total m 2688 kg f 162 kg mf 2688 kg 352 kg mf total N2 O2 O2 H2O H2O total CO2 CO2 m m m m m he mass of water per unit mass of fuel burned is kg 200 2688 kg total m m m T 18 8 2 kg H Okg C H 1421 114 kg 1 C8H18 m 18 kg 9 mH2O he mass of each reactant and the total mass are T 2702 kg 2588 114 2588 kg 4 76 kmol29 kgkmol 1775 114 kg kmol114 kgkmol 1 air C8H18 total air air air C8H18 C8H18 C8H18 m m m M N m M N m Then the mass fractions of reactants are 09578 00422 2702 kg 2588 kg mf 2702 kg 114 kg mf total air air total C8H18 C8H18 m m m m preparation If you are a student using this Manual you are using it without permission 1512 1522 Ethyl alcohol is burned with 70 excess air The mole fractions of the products and the reactants the mass of water and oxygen in products per unit mass of fuel are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 CO H2O O2 and N2 only Properties The molar masses of C H2 O2 N2 and air are 12 kgkmol 2 kgkmol 32 kgkmol 28 kgkmol and 29 kgkmol respectively Table A1 Analysis The reaction with stoichiometric air is 2 2 2 2 2 5 2 3 76 N 3H O 2 CO 376N O C H OH th th a a PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 th th a a 3 76 N 3 71 O 3 H O CO x oefficient x is determined from O2 balance 0 x x 1918 N O 12 3 H O 2 CO N he total moles of the products is The mole fractions of the products are CO2 H2O O2 N2 C2H5OH Air 70 excess Combustion chamber where 5 3 51 2 Substituting 2 2 2 2 2 5 2 3 76 N 3 3H O 2 CO 376N 3 O C H OH The reaction with 70 excess air can be written as 2 2 2 2 2 2 5 2 2 376N 3 O 71 C H OH The c 5 12 51 2 3 71 Then 2 5 2 376 O 15 C H OH 2 2 2 2 2 T 26 1918 12 3 2 Nm kmol 28 07298 00799 kmol 2628 O2 O2 Nm y 01142 00761 2628 kmol kmol 1918 kmol 21 kmol 2628 N2 N2 CO2 m m m N N y N N The total m les of the reactants is The mole fractions of the reactants are 2 kmol NCO2 y 2628 kmol 3 kmol H2O H2O N N y o 2528 kmol 4 76 15 1 Nm 09603 00396 2528 kmol 476 kmol 51 2528 kmol kmol 1 air air C2H5OH C2H5OH m m N N y N N y The mass of water and oxygen in the products per unit mass of fuel burned is 1174 kg H Okg C H OH 5 2 2 46 kg 1 18 kg 3 C2H5OH H2O m m kg O kg C H OH 1461 5 2 2 46 kg 1 32 kg 21 C2H5OH O2 m m preparation If you are a student using this Manual you are using it without permission 1513 1523 Ethyl alcohol is burned with 70 excess air The airfuel ratio is to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 CO H2O O2 and N2 only Properties The molar masses of C H2 O2 N2 and air are 12 kgkmol 2 kgkmol 32 kgkmol 28 kgkmol and 29 kgkmol respectively Table A1 Analysis The reaction with stoichiometric air is 2 2 2 2 2 5 2 3 76 N 3H O 2 CO 376N O C H OH th th a a where PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 th th a a 3 76 N 3 71 O 3 H O CO x oefficient x is determined from O2 balance x x hen 1918 N O 12 3 H O 2 CO 376N he airfuel mass ratio is CO2 H2O O2 N2 C2H5OH Air 70 excess Combustion chamber 3 51 2 5 Substituting 2 2 2 2 2 5 2 3 76 N 3 3H O 2 CO 376N 3 O C H OH The reaction with 70 excess air can be written as 2 2 2 2 2 2 5 2 2 376N 3 O 71 C H OH The c 0 12 51 2 3 71 5 T 5 2 O 15 C H OH 2 2 2 2 2 2 T 1530 kg airkg fuel 46 kg 7040 kg 46 kg 1 29 kg 4 76 51 AF fuel air m m 1524 Gasoline is burned steadily with air in a jet engine The AF ratio is given The percentage of exc determined ess air used is to be d 29 kgkmol respectively Table A1 Analysis The theoretical combustion equation in this case can be written as he stoichiometric coefficient for air It is determined from he airfu ratio for the theoretical reaction is determined by taking the ratio of the mass of the air to the mass of the fuel Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O and N2 only Properties The molar masses of C H2 and air are 12 kgkmol 2 kgkmol an N 376 9H O 8CO 376N O C H a a 2 th 2 2 2 2 th 18 8 where ath is t Air Products Gasoline C8H18 Jet engine O2 balance a a 8 45 125 th th T el for 14 kg air kg fuel 15 2 kgkmol 9 kmol 8 kmol 12 kgkmol 476 kmol 29 kgkmol 125 AF fuel airth th m m Then the percent theoretical air used can be determined from 119 1514 kg airkg fuel 18 kg airkg fuel AF AF Percent theoretica l air th act preparation If you are a student using this Manual you are using it without permission 1514 1525E Ethylene is burned with 175 percent theoretical air during a combustion process The AF ratio and the dewpoint temperature of the products are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O O2 and N2 only 3 Combustion gases are ideal gases Properties The molar masses of C H2 and air are 12 lbmlbmol 2 lbmlbmol and 29 lbmlbmol respectively Table A 1E Analysis a The combustion equation in this case can be written as 2 th 2 th 2 2 2 2 th 4 2 N 175 376 O 0 75 2H O 2CO 376N O 1 75 C H a a a where ath is the stoichiometric coefficient for air It is determined from Products C2H4 175 theoretical air O2 balance 3 0 75 2 1 1 75 th th th a a a Substituting 2 2 2 2 2 2 4 2 1974N 225O 2H O 2CO 376N 525 O C H PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course The airfuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel 259 lbm airlbm fuel 2 lbmlbmol 2 lbmol 2 lbmol 12 lbmlbmol fuel m b The dewpoint temperature of a gasvapor 525 476 lbmol 29 lbmlbmol AF mair mixture is the saturation temperature of the water vapor in the product gases corresponding to its partial pressure That is 1 116 psia 145 psia 2599 lbmol 2 lbmol prod prod P P N N v v Thus 1054F sat 1 116 psia dp T T preparation If you are a student using this Manual you are using it without permission 1515 1526 Propylene is burned with 50 percent excess air during a combustion process The AF ratio and the temperature at which the water vapor in the products will start condensing are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O O2 and N2 only 3 Combustion gases are ideal gases Properties The molar masses of C H2 and air are 12 kgkmol 2 kgkmol and 29 kgkmol respectively Table A1 Analysis a The combustion equation in this case can be written as 2 th 2 th 2 2 2 2 th 6 3 N 15 376 O 05 3H O 3CO 376N O 15 C H a a a where ath is the stoichiometric coefficient for air It is determined from PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Products C3H6 50 excess air O2 balance 15 3 15 05 45 a a a th th th Substituting C H O N CO H O O N 3 6 2 2 2 2 2 2 6 75 376 3 3 2 25 2538 The airfuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel 222 kg airkg fuel 675 476 kmol 29 kgkmol AF mair mixture is the saturation temperature of the water vapor in the product gases orrespond g to its partial pressure That is 2 kgkmol 3 kmol 3 kmol 12 kgkmol fuel m b The dewpoint temperature of a gasvapor c in 9367 kPa 105 kPa k 3363 mol 3 kmol prod prod P N N v v hus P T 445C sat9367 kPa dp T T preparation If you are a student using this Manual you are using it without permission 1516 1527 Butane C4H10 is burned with 200 percent theoretical air The kmol of water that needs to be sprayed into the combustion chamber per kmol of fuel is to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O O2 and N2 only Properties The molar masses of C H2 O2 and air are 12 kgkmol 2 kgkmol 32 kgkmol and 29 kgkmol respectively Table A1 Analysis The reaction equation for 200 theoretical air without the additional water is 2 2 2 2 2 2 th 10 4 N O H O CO 376N O 2 C H F E D B a where ath is the stoichiometric coefficient for air We have automatically accounted for the 100 excess air by using the factor 2a instead of a th Carbon balance B 4 th for air The coefficient ath and other coefficients are to be determined from the mass balances PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Hydrogen balance 5 10 2 D D Oxygen balance E D B a 2 2 2 2 th bove equations we find the coefficients E 65 F 4888 and ath 65 and write the balanced reaction n as ith the a itional water sprayed into the combustion chamber the balanced reaction equation is 4888 N O he partial pressure of water in the satu product mixture at the dew point is he vapor mole fraction is Products C4H10 Air 200 theoretical E a th Nitrogen balance F a 3 76 2 th Solving the a equatio 2 2 2 2 2 2 10 4 4888 N O 56 5 H O 4 CO 376N 13 O C H W dd 2 2 2 2 2 10 4 56 H O 5 4 CO H O 376N 13 O C H v v N N 2 2 T rated sat60 C prod P Pv 95 kPa 19 T 0 1995 kPa 100 1995 kPa prod Pv prod yv P The amount of water that needs to be sprayed into the combustion chamber can be determined from 9796 kmol v v v v N N N N N y 4888 56 5 4 5 0 1995 product total water preparation If you are a student using this Manual you are using it without permission 1517 1528 A fuel mixture of 60 by mass methane CH4 and 40 by mass ethanol C2H6O is burned completely with theoretical air The required flow rate of air is to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O and N2 only Properties The molar masses of C H2 O2 and air are 12 kgkmol 2 kgkmol 32 kgkmol and 29 kgkmol respectively Table A1 Analysis For 100 kg of fuel mixture the mole numbers are Products 60 CH4 40 C2H6O Air 100 theoretical 0 8696 kmol 46 kgkmol 40 kg mf 3 75 kmol 16 kgkmol CH4 CH4 M 60 kg mf C2H6O C2H6O C2H6O CH4 M N N ole fraction of methane and ethanol in the fuel mixture are The m 1882 0 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 8696 kmol 3 75 C2H6O CH4 N N The combustion kmol 08696 0 8118 kmol 375 C2H6O CH4 N y N x equation in this case can be written as N H O CO 376N O F D B a where ath c coefficient for air The coefficient ath and other coefficients are to be determined from the mass bal 2 0 8696 kmol 3 75 C2H6O CH4 N N 2 2 2 2 2 th 6 2 4 CH C H O y x is the stoichiometri ances Carbon balance B y x Hydrogen balance D y x 2 6 4 D B y a 2 2 th Oxygen balance F a 3 76 th Nitrogen balance Substituting x and y values into the equations and solving we find the coefficients as 0 1882 8118 0 th D B Then we write the balanced reaction equation as 2 188 1 188 8 228 2 188 F a y x 2 2 2 2 2 6 2 4 8 228 N 2 188 H O 1 188 CO 376N 2 188 O 0 1882 C H O 0 8118 CH The airfuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel 94 kg airkg fuel 13 16kgkmol 6 1 0 1882 kmol2 12 1 kgkmol 4 0 8118 kmol12 4 76 kmol29 kgkmol 2 188 AF fuel air m m Then the required flow rate of air becomes 1394 kgs 139410 kgs AF fuel air m m preparation If you are a student using this Manual you are using it without permission 1518 1529 The volumetric fractions of the constituents of a certain natural gas are given The AF ratio is to be determined if this gas is burned with the stoichiometric amount of dry air Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O and N2 only Properties The molar masses of C H2 N2 O2 and air are 12 kgkmol 2 kgkmol 28 kgkmol 32 kgkmol and 29 kgkmol respectively Table A1 Analysis Considering 1 kmol of fuel the combustion equation can be written as 2 2 2 2 2 th 2 2 2 2 4 N H O CO 376N O 006CO 003O 018N 008H 065CH z y x a The unknown coefficients in the above equation are determined from mass balances PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course C th 2 th th 2 z z a a y x a y y x x 5 0 08 018 0 03 0 06 131 376 0 71 138 5106 4 2 2 2 2 2 2 2 2 2 CH H N O CO O N CO H O N he airfuel ratio for the this reaction is determined by tak ratio of the mass of the air to the mass of the fuel and 5 106 3 76 0 18 N 1 31 2 0 06 0 03 O 1 38 2 2 0 08 4 0 65 H 0 71 0 06 0 65 Dry air Products Natural gas Combustion chamber Thus 0 6 T ing the 192 kg 44 kg 0 06 0 03 32 0 18 28 0 08 2 65 16 0 1808 kg 476 kmol 29 kgkmol 131 fuel air m m 942 kg airkg fuel 192 kg 1808 kg AF fuel airth th m m preparation If you are a student using this Manual you are using it without permission 1519 1530 The composition of a certain natural gas is given The gas is burned with stoichiometric amount of moist air The AF ratio is to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O and N2 only Properties The molar masses of C H2 N2 O2 and air are 12 kgkmol 2 kgkmol 28 kgkmol 32 kgkmol and 29 kgkmol respectively Table A1 Analysis The fuel is burned completely with the stoichiometric amount of air and thus the products will contain only H2O CO2 and N2 but no free O2 The moisture in the air does not react with anything it simply shows up as additional H2O in the products Therefore we can simply balance the combustion equation using dry air and then add the moisture to both sides of the equation Considering 1 kmol of fuel the combustion equation can be written as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2 y z 5 0 08 018 0 03 0 06 131 376 CH H N O CO O N ext we determine the amount of moisture that accompanies 476ath 476131 624 kmol of dry air The partial of the moisture in the air is 065CH 008H 018N 003O 006CO O 376N CO H O N 4 2 2 2 2 th 2 2 2 2 a x The unknown coefficients in the above equation are determined from mass balances 5 106 3 76 0 18 N 1 31 2 0 06 0 03 O 1 38 2 0 08 2 0 65 4 H 0 71 0 06 0 65 C th 2 th th 2 z z a a y x a y y x x Moist air Products Natural gas Combustion chamber Thus 0 71 138 5106 4 2 2 2 2 2 2 2 2 2 CO H O N 0 6 N pressure 2694 kPa 0 85 3 1698 kPa sat25 C air in P Pv φ Assuming ideal gas behavior the number of moles of the moisture in the air Nv in is determined to be 017 kmol 6 24 101325 kPa 2694 kPa air in total total in v N P N in v v v N N P The balanced combustion equation is obtained by substituting the coefficients determined earlier and adding 017 kmol of 5 0 08 018 0 03 0 06 131 376 017 0 71 155 5106 4 2 2 2 2 2 2 2 2 2 2 CH H N O CO O N H O CO H O N The airfuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of the fuel H2O to both sides of the equation 0 6 192 kg 44 kg 0 06 0 03 32 0 18 28 0 08 2 65 16 0 1839 kg 017 kmol 18 kgkmol 476 kmol 29 kgkmol 131 fuel air m m and 958 kg airkg fuel 192kg 1839kg AF fuel airth th m m preparation If you are a student using this Manual you are using it without permission 1520 1531 The composition of a gaseous fuel is given It is burned with 130 percent theoretical air The AF ratio and the fraction of water vapor that would condense if the product gases were cooled are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O O2 and N2 only Properties The molar masses of C H2 N2 and air are 12 kgkmol 2 kgkmol 28 kgkmol and 29 kgkmol respectively Table A1 Analysis a The fuel is burned completely with excess air and thus the products will contain H2O CO2 N2 and some free O2 Considering 1 kmol of fuel the combustion equation can be written as 2 2 th 2 2 2 2 th 2 2 4 N O 03 H O CO 376N O 13 020N 035H 045CH z a y x a The unknown coefficients in the above equation are determined from mass balances 5 332 31 3 76 0 20 N 1 05 30 2 31 O 21 2 2 0 35 4 0 45 H 0 45 0 45 C th 2 th th th 2 z z a a a y x a y y x x Air 30 excess Products Gaseous fuel Combustion chamber Thus 2 2 2 2 2 2 2 2 4 5332N 0315O 12H O 045CO 376N 1365O 020N 035H 045CH PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course The airfuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of the fuel nd 135 kg 28 kg 20 0 35 2 45 16 0 1884 kg 476 kmol 29 kgkmol 1365 fuel air m m a 1396 kg airkg fuel 135 kg 1884 kg AF fuel air m m b For each kmol of fuel burned 045 12 0315 5332 7297 kmol of products are formed including 12 kmol of H2O Assuming that the dewpoint temperature of the products is above 25C some of the water vapor will condense as the products are cooled to 25C If N kmol of H O condenses there will be 12 N kmol of water vapor left in the products The mole number of the products in th w 2 w e gas phase will also decrease to 7297 Nw as a result Treating the product gases ncluding the remaining water vapor as ideal gases Nw is determined by equating the ole fraction of the water vapor to s pressure fraction i m it 1 003 kmol 31698 kPa 21 w w v v N N P N 101325 kPa 7 297 prod prodgas Nw P N since Pv Psat 25C 31698 kPa Thus the fraction of water vapor that condenses is 100312 0836 or 84 preparation If you are a student using this Manual you are using it without permission 1521 1532 Problem 1531 is reconsidered The effects of varying the percentages of CH4 H2 and N2 making up the fuel nalysis The problem is solved using EES and the solution is given below b H2O c376 ay4 axb2 Theoair100 N2 ay4 axb2 ure ewPointMolesH2OvapMolesH2OliqResult vx0 wPprod olesH2Ovap There is condensation am window a r 130 C CH4 16b2c28 kgairkgfuel y4 axb2 Theoair100 esH2OvapMolesH2OliqResult axCO2 ay2b H2O c376 Ath Theoair100 N2 Ath Theoair100 1 O2 and the product gas temperature are to be studied A Lets modify this problem to include the fuels butane ethane methane and propane in pull down menu Adiabatic Combustion of fuel CnHm entering at Tfuel with Stoichiometric Air at Tair Reaction aCxHybH2cN2 ay4 axb2 Theoair100 O2 376 N2 axCO2 ay2 Theoair100 1 O2 Tprod is the product gas temperat Theoair is the theoretical air Procedure H20CondPprodTprodMolesH2OMotherTD Pv MolesH2OMotherMolesH2OPprod TDewPoint temperaturesteamPP IF TDewPoint Tprod then MolesH2Ovap MolesH2O MolesH2Oliq0 ResultNo condensation occurred ELSE PvnewpressuresteamTTprodx0 MolesH2OvapPvnewPprodMother1Pvne MolesH2Oliq MolesH2O M Result ENDIF END Input data from the diagr Pprod 101325 kP Theoai a045 b035 c020 Tprod 25 Fuel x1 y4 Composition of Product gases Ath ay4 a xb2 AFratio 476AthTheoair100molarmassAira MolesO2ay4 a xb2 Theoair100 1 MolesN2c376a MolesCO2ax MolesH2Oay2b MotherMolesO2MolesN2MolesCO2 Call H20CondPprodTprodMolesH2OMotherTDewPointMol Fraccond MolesH2OliqMolesH2OConvert Reaction aCxHybH2cN2 Ath Theoair100 O2 376 N2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1522 AFratio kgair kgfuel Fraccond MolesH2Oliq MolesH2Ovap Tprod C 1427 1427 1427 1427 1427 1427 1427 1427 1427 1427 1427 1427 1427 9567 9316 8942 8392 7594 6444 4792 2406 0 0 0 0 0 1196 1165 1118 1049 09492 08055 0599 03008 0 0 0 0 0 005409 008549 01323 0201 03008 04445 0651 09492 125 125 125 125 125 5 1167 1833 25 3167 3833 45 5167 5833 65 7167 7833 85 0 10 20 30 40 50 60 70 80 90 0 02 04 06 08 1 12 14 Tprod C MolesH2O Liquid Liquid Vapor Vapor Dew Point PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1523 1533 Carbon is burned with dry air The volumetric analysis of the products is given The AF ratio and the percentage of theoretical air used are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 CO O2 and N2 only Properties The molar masses of C H2 and air are 12 kgkmol 2 kgkmol and 29 kgkmol respectively Table A1 Analysis Considering 100 kmol of dry products the combustion equation can be written as 2 2 2 2 2 7883N 1069O 042CO 1006CO 376N a O xC The unknown coefficients x and a are determined from mass balances Dry air Products Carbon Combustion chamber 2096 2096 0 21 1069 1006 CheckO 1048 0 42 1006 C 20965 7883 3 76 N 2 2 a x x a a Thus PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10 48 20 96 376 10 06 0 42 10 69 7883 2 2 2 2 2 C O N CO CO O N The combustion equation for 1 kmol of fuel is obtained by dividing the above equation by 1048 C O N CO CO O N 2 0 376 0 96 0 04 102 7 52 2 2 2 2 2 a The airfuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel 230 kg airkg fuel 1 kmol 12 kgkm 20 476 kmol 29 kgkmol AF fuel air m m ol To find the percent theoretical air used we need to ow the theoretical amount of air which is determined from the theoretical combustion equation of the fuel b kn C O N CO N 1 376 376 2 2 2 2 Then 200 10 476 kmol 20 476 kmol Percent theoretica l air th air act air th air airact N N m m preparation If you are a student using this Manual you are using it without permission 1524 1534 Methane is burned with dry air The volumetric analysis of the products is given The AF ratio and the percentage of theoretical air used are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 CO H2O O2 and N2 only Properties The molar masses of C H2 and air are 12 kgkmol 2 kgkmol and 29 kgkmol respectively Table A1 Analysis Considering 100 kmol of dry products the combustion equation can be written as x a b CH O N CO CO O N H O 4 2 2 2 2 2 2 376 520 0 33 1124 8323 The unknown coefficients x a and b are determined from mass balances Dry air Products CH4 Combustion chamber 1106 2 4 H 5 53 0 33 5 20 C 2214 8323 3 76 N PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2214 2214 2 0 165 1124 5 20 Check O 2 b a 2 b b x x x a a hus T 553 2214 376 52 4 2 2 CH O N 0 0 33 1124 8323 1106 2 2 2 2 CO CO O N H O he combustion equation for 1 kmol of fuel is obtained by dividing the above equation b 53 T y 5 CH 40 O 376N 094CO 006CO 203O 1505N 2H O 4 2 2 2 2 2 2 a The airfuel ratio is determined from its definition 345 kg airkg fuel 2 kgkmol 2 kmol 1 kmol 12 kgkmol 40 476 kmol 29 kgkmol AF fuel air m m b th To find the percent theoretical air used we need to know the theoretical amount of air which is determined from the eoretical combustion equation of the fuel CH O 376N CO 2H O 376 N O 1 1 20 4 th 2 2 2 2 th 2 th th a a a a 2 Then 200 20 476 kmol 40 476 kmol Percent theoretica l air th air act air th air airact N N m m preparation If you are a student using this Manual you are using it without permission 1525 1535 nOctane is burned with 100 excess air The combustion is incomplete The mole fractions of products and the dew point temperature of the water vapor in the products are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 CO H2O O2 and N2 only Properties The molar masses of C H2 O2 N2 and air are 12 kgkmol 2 kgkmol 32 kgkmol 28 kgkmol and 29 kgkmol respectively Table A1 Air 100 excess Products C8H18 Combustion chamber P 1 atm Analysis The combustion reaction for stoichiometric air is 2 2 2 2 2 18 8 125 376N 9H O 8CO 376N 12 5 O C H The combustion equation with 100 excess air and incomplete combustion is 2 2 2 2 2 2 18 8 376 N 2 12 5 O 9H O 0 15 8 CO 0 85 8 CO 376N 2 12 5 O C H x The coefficient for CO is determined from a mass balance O2 balance 13 1 9 50 0 15 8 50 0 85 8 25 x x PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Substituting 94 N 9 H O 13 1 O CO 21 CO 86 376N 25 O C H The mole fractions of the products are 2 2 2 2 2 2 18 8 124 1 kmol 94 9 13 1 21 86 prod N 07575 01056 00725 00097 00548 kmol 1241 prod N The dewpoint temperature of a gasvapor mix kmol 94 1241 kmol kmol 131 1241 kmol kmol 9 1241 kmol kmol 12 1241 kmol kmol 68 N2 N2 prod O2 O2 prod H2O H2O prod CO CO prod CO2 CO2 N y N N y N N y N N y N N y ture is the saturation temperature of the water vapor in the product gases orrespond g to its partial pressure That is c in 7 348 kPa 1241 kmol 101325 kPa 9 kmol prod prod P N N P v v hus Table A5 or EES T 399C sat7348 kPa dp T T preparation If you are a student using this Manual you are using it without permission 1526 1536 Methyl alcohol is burned with 100 excess air The combustion is incomplete The balanced chemical reaction is to be written and the airfuel ratio is to be determined Assumptions 1 Combustion is incomplete 2 The combustion products contain CO2 CO H2O O2 and N2 only Properties The molar masses of C H2 O2 N2 and air are 12 kgkmol 2 kgkmol 32 kgkmol 28 kgkmol and 29 kgkmol respectively Table A1 Analysis The balanced reaction equation for stoichiometric air is 2 th 2 2 2 2 th 3 3 76 N 2 H O CO 376N O CH OH a a CO2 CO H2O O2 N2 CH3OH Air 100 excess Combustion chamber The stoicihiometric coefficient ath is determined from an O2 balance 51 1 1 50 th th a a Substituting 2 2 2 2 2 3 3 76 N 51 2 H O CO 376N O 51 CH OH PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course The reaction with 100 excess air and incomplete combustion can be written as 3 76 N 51 2 O 2 H O 040 CO x nt for 2 is determined from a mass balance x x ubstituting 1128 N 17 O 2 H O 04 CO 06 CO 376N he airfu mass ratio is 2 2 2 2 2 2 3 060 CO 376N O 51 2 CH OH The coefficie O O2 balance 0 71 1 20 60 51 2 5 S 2 3 3 O CH OH 2 2 2 2 2 T el 1294 kg airkg fuel 32 kg 4141 kg 1 32 kg 29 kg 4 76 3 AF fuel air m m preparation If you are a student using this Manual you are using it without permission 1527 1537 Ethyl alcohol is burned with stoichiometric amount of air The combustion is incomplete The apparent molecular weight of the products is to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 CO H2O OH and N2 only Properties The molar masses of C H2 OH N2 and air are 12 kgkmol 2 kgkmol 17 kgkmol 28 kgkmol and 29 kgkmol respectively Table A1 Analysis The reaction with stoichiometric air is 2 2 2 2 2 5 2 3 76 N 3H O 2 CO 376N O C H OH th th a a PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 th th a a 76 N 3 CO2 CO O2 H2O OH N2 C2H5OH Air 100 theoretical Combustion chamber where 3 51 2 5 Substituting 2 2 2 2 2 5 2 3 3H O 2 CO 376N 3 O C H OH The balanced reaction equation with incomplete combustion is 2 2 2 2 2 2 2 010 CO 2090 CO 376N 3 O C 5 3 76 N 3 01 OH 3 0 95H O H OH bO 18013152b b 0025 can be written as 1128 N 0025O OH 30 2 85 H O 020 CO 180 CO he total m les of the products is O2 balance 053 which 2 5 2 376N 3 O C H OH 2 2 2 2 2 T o 0 025 1128 30 2 85 20 81 Nm 1664 kmol The apparent molecular weight of the product gas is 2783 kgkmol 1664 kmol 1128 28 kg 0 025 32 17 30 2 85 18 0 20 28 44 18 m m m N m M preparation If you are a student using this Manual you are using it without permission 1528 1538 Coal whose mass percentages are specified is burned with stoichiometric amount of air The mass fractions of the products and the airfuel ratio are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O SO2 and N2 3 Combustion gases are ideal gases Properties The molar masses of C H2 O2 S and air are 12 2 32 32 and 29 kgkmol respectively Table A1 Analysis We consider 100 kg of coal for simplicity Noting that the mass percentages in this case correspond to the masses of the constituents the mole numbers of the constituent of the coal are determined to be PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 01625 kmol 32 kgkmol kg 052 0 06536 kmol 28 kgkmol N2 m M kg 183 32 kgkmol O2 m M 0 1488 kmol kg 476 kg 7961 S S S N2 O2 O2 C M N N m N m Ash consists of the noncombustible matter in coal Therefore the ber is equal to the mass conte t that leaves Disregarding this nonreacting component for simplicity the combustion equation may be written as Performing mass balances for the constituents gives ce 6 634 th 2 th th 2 2 a w a z y x a z y x on equation without the ash becomes The mass fractions of the products are 7961 C 466 H2 476 O2 183 N2 052 S 862 ash by mass 2 33 kmol 2 kgkmol kg 466 6 634 kmol kgkmol 12 H2 H2 H2 C C M m N M N N2 Air theoretical Coal Combustion chamber CO2 H2O SO2 O2 N2 mass of ash content that enters the combustion cham n 2 2 2 2 2 2 th 2 2 2 N SO H O CO 3 76N O 0 01625S 0 06536N 0 1488O 2 33H 6634C w z y x a 2889 7 667 3 76 0 06536 3 76 0 06536 balance N 7 667 0 1488 0 01625 50 2 33 6 634 50 balance 0 1488 O 0 01625 balance S 2 33 balan H balance C Substituting the balanced combusti 2 2 2 2 2 2 2 2 2 2889N 0 01625SO 2 33H O 6 634CO 3 76N 7 667O 0 01625S 0 06536N 0 1488O 2 33H 6634C 1144 kg 28 22889 64 0 01625 2 33 18 44 6 634 total m 07072 000091 00367 02552 1144 kg 28 kg 2889 mf 1144 kg 64 kg 001625 mf 1144 kg 233 18 kg mf 1144 kg 44 kg 6634 mf total N2 N2 total SO2 SO2 total H2O H2O total CO2 CO2 m m m m m m m m The airfuel mass ratio is then 1158 kg airkg fuel 9138 kg 1058 kg 0 01625 32 kg 0 06536 28 0 1488 32 2 33 2 6634 12 4 76 29 kg 7667 AF fuel air m m preparation If you are a student using this Manual you are using it without permission 1529 1539 Coal whose mass percentages are specified is burned with 40 excess air The airfuel ratio and the apparent molecular weight of the product gas are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 CO H2O SO2 and N2 3 Combustion gases are ideal gases Properties The molar masses of C H2 O2 S and air are 12 2 32 32 and 29 kgkmol respectively Table A1 Analysis We consider 100 kg of coal for simplicity Noting that the mass percentages in this case correspond to the masses of the constituents the mole numbers of the constituent of the coal are determined to be 6740 C 531 H2 1511 O2 144 N2 236 S 838 ash by mass 0 07375 kmol 32 kgkmol kg 236 0 05143 kmol 28 kgkmol kg 144 0 4722 kmol 32 kgkmol kg 1511 2 655 kmol 2 kgkmol kg 531 5 617 kmol 12 kgkmol kg 6740 S S S N2 N2 N2 O2 O2 O2 H2 H2 H2 C C C M m N M m N M m N M m N M m N Air 40 excess Coal Combustion chamber CO2 H2O SO2 O2 N2 The mole number of the mixture and the mole fractions are 8 869 kmol 0 07375 0 05143 0 4722 2 655 5 617 Nm 000832 8869 kmol kmol 007375 000580 8869 kmol kmol 005143 005323 8869 kmol kmol 04722 02994 8869 kmol kmol 2655 0 6333 8869 kmol kmol 5617 S S N2 N2 O2 O2 H2 H2 C C m m m m m N N y N N y N N y N N y N N y Ash consists of the noncombustible matter in coal Therefore the mass of ash content that enters the combustion chamber is equal to the mass content that leaves Disregarding this nonreacting component for simplicity the combustion equation may be written as 2 2 2 2 2 2 2 th 2 2 2 O N SO H O CO 3 76N O 41 0 00832S 0 00580N 0 05323O 02994H 06333C m k z y x a According to the species balances 0 2952 0 7381 40 40 3 891 0 7381 3 76 41 0 00580 3 76 41 0 00580 balance N 0 7381 0 05323 0 00832 0 2994 50 6333 0 50 05323 0 O balance 0 00832 balance S 0 2994 balance H 0 6333 balance C th th 2 th th 2 2 a m a k a z y x a z y x PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1530 Substituting PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 3 76N 1 033O 0 00832S 0 00580N 0 05323O 02994H 06333C The total mass of the products is 2 2 2 2 2 2 2 2 2 2 0 2952O 3 891N 0 00832SO 0 2994H O 0 6333CO 0 2994 18 44 06333 total 152 2 kg 32 0 2952 28 3 891 64 0 00832 The total m le number of the products is m o 5 127 kmol 0 2952 3 891 0 00832 0 2994 06333 Nm he apparent molecular weight of the product gas is T 2968 kgkmol 5127 km m m m N M ol 1522 kg m ass ratio is then The airfuel m 1380 kg airkg fuel 1033 kg kg 1426 0 00832 32 kg 0 00580 28 0 05323 32 0 2994 2 06333 12 4 76 29 kg 1033 AF fuel air m m preparation If you are a student using this Manual you are using it without permission 1531 Enthalpy of Formation and Enthalpy of Combustion 1540C For combustion processes the enthalpy of reaction is referred to as the enthalpy of combustion which represents the amount of heat released during a steadyflow combustion process 1541C Enthalpy of formation is the enthalpy of a substance due to its chemical composition The enthalpy of formation is related to elements or compounds whereas the enthalpy of combustion is related to a particular fuel 1542C The heating value is called the higher heating value when the H2O in the products is in the liquid form and it is called the lower heating value when the H2O in the products is in the vapor form The heating value of a fuel is equal to the absolute value of the enthalpy of combustion of that fuel 1543C If the combustion of a fuel results in a single compound the enthalpy of formation of that compound is identical to the enthalpy of combustion of that fuel 1544C Yes 1545C No The enthalpy of formation of N2 is simply assigned a value of zero at the standard reference state for convenience 1546C 1 kmol of H2 This is evident from the observation that when chemical bonds of H2 are destroyed to form H2O a large amount of energy is released PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1532 1547 The enthalpy of combustion of methane at a 25C and 1 atm is to be determined using the data from Table A26 and to be compared to the value listed in Table A27 Assumptions The water in the products is in the liquid phase Analysis The stoichiometric equation for this reaction is 2 2 2 2 2 4 752N 2H O CO 376N 2 O CH l Both the reactants and the products are at the standard reference state of 25C and 1 atm Also N2 and O2 are stable elements and thus their enthalpy of formation is zero Then the enthalpy of combustion of CH4 becomes 4 2 2 CH H O CO o o o o o f f f f R R f P P R P C Nh Nh Nh N h N h H H h Using h f values from Table A26 o 74850 kJkmol kmol 1 285830 kJkmol 2 kmol 393520 kJkmol kmol 1 890330 kJ per kmol CH4 C h The listed value in Table A27 is 890868 kJkmol which is almost identical to the calculated value Since the water in the products is assumed to be in the liquid phase this hc value corresponds to the higher heating value of CH4 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1533 1548 Problem 1547 is reconsidered The effect of temperature on the enthalpy of combustion is to be studied Analysis The problem is solved using EES and the solution is given below Fuel Methane CH4 Tcomb 25 C Tfuel Tcomb 273 K Tair1 Tcomb 273 K Tprod Tcomb 273 K hbarcombTableA27 890360 kJkmol For theoretical dry air the complete combustion equation is CH4 AthO2376 N21 CO22 H2O Ath 376 N2 Ath21221 theoretical O balance Apply First Law SSSF hfuelEESenthalpyCH4T298 kJkmol hfuelTableA2674850 kJkmol hbarfgH2OenthalpySteamiapwsT298x1enthalpySteamiapwsT298x0 kJkmol HRhfuelEES AthenthalpyO2TTair1Ath376 enthalpyN2TTair1 kJkmol HP1enthalpyCO2TTprod2enthalpyH2OTTprodhbarfgH2OAth376 enthalpyN2TTprod kJkmol hbarCombEESHPHR kJkmol PercentErrorABShbarCombEEShbarcombTableA27ABShbarcombTableA27Convert hCombEES kJkmol TComb C 890335 25 887336 8889 884186 1528 880908 2167 877508 2806 873985 3444 870339 4083 866568 4722 862675 5361 858661 600 0 100 200 300 400 500 600 895000 890000 885000 880000 875000 870000 865000 860000 855000 Tcomb C hCombEES kJkmol PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1534 1549 Ethane is burned with stoichiometric amount of air The heat transfer is to be determined if both the reactants and products are at 25C Assumptions The water in the products is in the vapor phase Products 25C C2H6 25C Air 25C Q Combustion chamber Analysis The stoichiometric equation for this reaction is 2 2 2 2 2 6 2 1316N 3H O 2CO 376N O 53 C H Since both the reactants and the products are at the standard reference state of 25C and 1 atm the heat transfer for this process is equal to enthalpy of combustion Note that N2 and O2 are stable elements and thus their enthalpy of formation is zero Then C2H6 H2O CO2 o o o o o f f f f R R f P P R P C Nh Nh Nh N h N h H H h Q o Using f h values from Table A26 1427820 kJkmol C2H6 1 kmol 84680 kJkmol 3 kmol 241820 kJkmol 2 kmol 393520 kJkmol hC Q 1550 Ethane is burned with stoichiometric amount of air at 1 atm and 25C The minimum pressure of the products which is to be determined ase nalysis The stoichiometric equation for this reaction is t the minimum pressure the product mixture will be saturated with water vapor and he mole fraction of water in the products is will assure that the water in the products will be in vapor form Assumptions The water in the products is in the vapor ph A 2 2 2 2 2 6 2 1316N 3H O 2CO 376N O 53 C H A 3 1698 kPa sat25 C P Pv T 0 1652 3 1316 kmol 2 3 kmol H2O N prod yv N The minimum pressure of the products is then 192 kPa 01652 31698 kPa min v v y P P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1535 1551 The higher and lower heating values of liquid propane are to be determined and compared to the listed values Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O and N2 3 Combustion gases are ideal gases Properties The molar masses of C O2 H2 and air are 12 32 2 and 29 kgkmol respectively Table A1 Analysis The combustion reaction with stoichiometric air is 2 2 2 2 2 8 3 18 8 N 4H O 3CO 376N 5 O C H l Air theoretical Products C3H8 Combustion chamber Both the reactants and the products are taken to be at the standard reference state of 25C and 1 atm for the calculation of heating values The heat transfer for this process is equal to enthalpy of combustion Note that N2 and O2 are stable elements and thus their enthalpy of formation is zero Then C3H8 H2O CO2 o o o o o f f f f R R f P P R P C Nh Nh Nh N h N h H H h q The h f o of liquid propane is obtained by adding hfg of propane at 25C to h f o of gas propane 103850 44097 335 118620 kJkmol For the HHV the water in the products is taken to be liquid Then 205260 kJkmol propane 2 1 kmol 118620 kJkmol 4 kmol 285830 kJkmol 3 kmol 393520 kJkmol hC The HHV of the liquid propane is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 50010 kJkg C3H8 8 3 8 3 44097 kgkmol C H HHV m C M 2205260 kJkmol C H h The listed value from Table A27 is 50330 kJkg For the LHV the water in the products is taken to be vapor Then l 1 kmol 118620 kJkmol 4 kmol 241820 kJkmol 3 kmol 393520 kJkmol h HV of the propane is then 2 029220 kJkmo C propane The L 46020 kJkg C3H8 2029220 kJkmol C H h 8 3 8 3 44097 kgkmol C H LHV m C M The listed value from Table A27 is 46340 kJkg The calculated and listed values are practically identical preparation If you are a student using this Manual you are using it without permission 1536 1552 The higher and lower heating values of gaseous octane are to be determined and compared to the listed values Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O and N2 3 Combustion gases are ideal gases Properties The molar masses of C O2 H2 and air are 12 32 2 and 29 kgkmol respectively Table A1 Analysis The combustion reaction with stoichiometric air is 2 2 2 2 2 18 8 47N 9H O 8CO 376N 12 5 O C H Air theoretical Products C8H18 Combustion chamber Both the reactants and the products are taken to be at the standard reference state of 25C and 1 atm for the calculation of heating values The heat transfer for this process is equal to enthalpy of combustion Note that N2 and O2 are stable elements and thus their enthalpy of formation is zero Then C8H18 H2O PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course CO2 o o o o o f f f f R R f P P R P C Nh Nh Nh N h N h H H h q For the HHV the water in the products is taken to be liquid Then 512180 kJkmol octane 5 1 kmol 208450 kJkmol 9 kmol 285830 kJkmol 8 kmol 393520 kJkmol hC The HHV of the gaseous octane is 48250 kJkg C8H18 18 8 18 8 114231 kgkmol C H 5512180 kJkmol C H HHV m C M The listed value for liquid octane from Table A27 is 47890 kJkg Adding the enthalpy of vaporization of octan h e to this value 4789036348253 the higher heating value of gaseous octane becomes 48253 kJkg octane This value is practically identical to the calculated value For the LHV the water in the products is taken to be vapor hen 1 kmol 208450 kJkmol 9 kmol 241820 kJkmol T 116090 kJkmol octan 5 8 kmol 393520 kJkmol e HV of the gaseous octane is then C h The L 44790 kJkg C8H18 18 8 114231 kgkmol C 5116090 kJkmol C H HHV m C M h 18 8H The listed value for liquid octane from Table A27 is 44430 kJkg Adding the enthalpy of vaporization of octane to this value 4443036344793 the lower heating value of gaseous octane becomes 44793 kJkg octane This value is practically identical to the calculated value preparation If you are a student using this Manual you are using it without permission 1537 1553 The higher and lower heating values of coal from Illinois are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 CO H2O SO2 and N2 3 Combustion gases are ideal gases Properties The molar masses of C H2 O2 S and air are 12 2 32 32 and 29 kgkmol respectively Table A1 Analysis We consider 100 kg of coal for simplicity Noting that the mass percentages in this case correspond to the masses of the constituents the mole numbers of the constituent of the coal are determined to be 0 07375 kmol 32 kgkmol kg 236 0 05143 kmol 28 kgkmol kg 144 0 4722 kmol 32 kgkmol kg 1511 2 655 kmol 2 kgkmol kg 531 5 617 kmol 12 kgkmol kg 6740 S S S N2 N2 N2 O2 O2 O2 H2 H2 H2 C C C M m N M m N M m N M m N M m N 6740 C 531 H2 1511 O2 144 N2 236 S 838 ash by mass Air theoretical Products Coal Combustion chamber The mole number of the mixture and the mole fractions are 8 869 kmol 0 07375 0 05143 0 4722 2 655 5 617 Nm 000832 8869 kmol kmol 007375 000580 8869 kmol kmol 005143 005323 8869 kmol kmol 04722 02994 8869 kmol kmol 2655 0 6333 8869 kmol kmol 5617 S S N2 N2 O2 O2 H2 H2 C C m m m m m N N y N N y N N y N N y N N y Ash consists of the noncombustible matter in coal Therefore the mass of ash content that enters the combustion chamber is equal to the mass content that leaves Disregarding this nonreacting component for simplicity the combustion equation may be written as 2 2 2 2 2 2 th 2 2 2 N SO H O CO 3 76N O 0 00832S 0 00580N 0 05323O 02994H 06333C k z y x a According to the species balances 2 781 0 7381 3 76 0 00580 3 76 0 00580 balance N 0 7381 0 05323 0 00832 0 2994 50 6333 0 50 05323 0 balance O 0 00832 balance S 0 2994 H balance 0 6333 balance C th 2 th th 2 2 a k a z y x a z y x PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1538 Substituting PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 3 76N 0 7381O 0 00832S 0 00580N 0 05323O 02994H 06333C ard reference state of 25C and 1 atm for the calculation of heating values The heat transfer for this process is equal to enthalpy of combustion Note that C S H2 N2 and O2 are stable lements and thus their enthalpy of formation is zero Then 2 2 2 2 2 781N 0 00832SO 0 2994H O 0 6333CO Both the reactants and the products are taken to be at the stand 2 2 2 2 2 e SO2 H2O o o f f Nh Nh CO2 o o o f f R R f P P R P C Nh N h N h H H h q For the HH the water in the products is taken to be liquid Then ol V kJkmol coal 337270 0 00832 kmol 297100 kJkmol 0 2994 kmol 285830 kJkmol 0 6333 kmol 393520 kJkm hC The apparent molecular weight of the coal is 1033 kgkmol coal 1000 kmol kg 1033 000832 kmol 000580 005323 02994 06333 32 kg 0 00832 28 0 00580 32 0 05323 2 0 2994 12 m The HHV of the coal is then 06333 m m N m M 32650 kJkg coal 1033 kgkmol coal 337270 kJkmol coal HHV m C M h For the LHV the water in the products is taken to be vapor Then kJkmol coal 324090 0 00832 kmol 297100 kJkmol 0 2994 kmol 241820 kJkmol 0 6333 kmol 393520 kJkmol C h The LHV of the coal is then 31370 kJkg coal 1033 kgkmol coal 324090 kJkmol coal LHV m C M h preparation If you are a student using this Manual you are using it without permission 1539 First Law Analysis of Reacting Systems 1554C In this case U Wb H and the conservation of energy relation reduces to the form of the steadyflow energy relation 1555C The heat transfer will be the same for all cases The excess oxygen and nitrogen enters and leaves the combustion chamber at the same state and thus has no effect on the energy balance 1556C For case b which contains the maximum amount of nonreacting gases This is because part of the chemical energy released in the combustion chamber is absorbed and transported out by the nonreacting gases PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1540 1557 Propane is burned with an airfuel ratio of 25 The heat transfer per kilogram of fuel burned when the temperature of the products is such that liquid water just begins to form in the products is to be determined Assumptions 1 Steady operating conditions exist 2 Air and combustion gases are ideal gases 3 Kinetic and potential energies are negligible 4 Combustion is complete 5 The reactants are at 25C and 1 atm 6 The fuel is in vapor phase Properties The molar masses of propane and air are 44 kgkmol and 29 kgkmol respectively Table A1 Analysis The mass of air per kmol of fuel is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1100 kg airkmol fuel 25 kg airkg fuel1 44 kgkmol fuel AF fuel air m m The mole number of air per kmol of fuel is then 3793 kmol airkmol fuel 29 kg airkmol air air air M N 1100 kg airkmol fuel air m The combustion equation can be written as 3 76N 3793 4 76 O 4H O 3CO x he coefficient for O2 is obtained from O2 balance 347 x x Products Tdp C3H8 25C Air 25C Q Combustion chamber P 1 atm 2 2 8 3 376N 3793 4 76 O C H 2 2 2 2 T 379 2 968 2 3 6 S g ubstitutin 2 2 2 2 2 2 8 3 2996N 2 968O 4H O 3CO 376N 7 968 O C H The mole fraction of water in the products is 0 1002 3993 kmol 2996 kmol 2 968 4 3 prod N 4 kmol 4 kmol H2O yv N he partia ressure of water vapor at 1 atm total pressure is T l p 1015 kPa 0 1002101325 kPa y P P v v When this mixture is at the dewpoint temperature the wate e same as the saturation pressure Then sat 1015 dp r vapor pressure is th T T 320 K K 3191 C 46 1 kPa We obtain properties at 320 K instead of 3191 K to avoid iterations in the ideal gas tables The heat transfer for this combustion process is determined from the energy balance system out in E E E applied on the combustion chamber with W 0 It reduces to R f f P h h N h N o o o Assuming the air and the com stion products eal gases w h hT From the tables P h h o Qout R h bu to be id e have Substance fo h kJkmol h298K kJkmol h320K kJkmol C H8 103850 3 O2 0 8682 9325 N2 0 8669 9306 H2O g 241820 9904 10639 CO2 393520 9364 10186 Substitutin g 8 3 out 2 017590 kJ kmol C H 0 103850 1 8669 9306 96 0 29 8682 9325 2 968 0 9904 10639 241820 4 9364 10186 393520 3 Q 8 3 out 2 017590 kJ kmol C H Q or Then the heat transfer per kg of fuel is 45850 kJkg C3H8 44 kgkmol 017590 kJkmol fuel 2 fuel out out M Q Q preparation If you are a student using this Manual you are using it without permission 1541 1558 nOctane is burned with 100 percent excess air The heat transfer per kilogram of fuel burned for a product temperature of 257C is to be determined Assumptions 1 Steady operating conditions exist 2 Air and combustion gases are ideal gases 3 Kinetic and potential energies are negligible 4 Combustion is complete 5 The fuel is in vapor phase Properties The molar masses of propane and air are 44 kgkmol and 29 kgkmol respectively Table A1 Analysis The combustion reaction for stoichiometric air is 2 2 2 2 2 18 8 125 376N 9H O 8CO 376N 12 5 O C H The combustion equation with 100 excess air is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 94 12 5 O 9H O 8CO 376N 25 O C H 2 2 2 2 2 2 18 8 N The heat transfer for this combustion process is determined from the energy balance system out in applied on the combustion chamber with W 0 It reduces to E E E R f R P f P h h h N h h h N Qout o o o o ssuming the air and the combustion products to be ideal gases we have h hT From the tables A Substance fo h kJkmol h298K kJkmol h530K kJkmol 18 g 8450 C8H 20 O2 0 8682 15708 N2 0 8669 15469 H2O g CO2 393520 9364 19029 ubstitutin 241820 9904 17889 S g 18 8 out 4 239880 kJ kmol C H 0 0 208450 1 8669 0 15469 94 8682 0 15708 12 5 9904 241820 17889 9 64 393520 19029 8 Q 93 18 8 out 4 239880 kJ kmol C H Q or Then the heat transfer per kg of fuel is 37200 kJkg C8H18 114 kgkmol 239880 kJkmol fuel 4 fuel out out M Q Q Products 257C C8H18 25C 100 excess air 25C Qout Combustion chamber P 1 atm preparation If you are a student using this Manual you are using it without permission 1542 1559 Propane is burned with 50 percent excess air during a steadyflow combustion process The rate of heat transfer in the combustion chamber is to be determined Assumptions 1 Steady operating conditions exist 2 Air and combustion gases are ideal gases 3 Kinetic and potential energies are negligible 4 Combustion is complete Properties The molar masses of propane and air are 44 kgkmol and 29 kgkmol respectively Table A1 Analysis The combustion equation can be written as 2 th 2 th 2 2 2 2 th 8 3 3 76N 51 O 50 4H O 3CO 376N O 51 C H a a a The stoichiometric coefficient is obtained from O2 balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1 th th a a a 28 2 N 52 O 4H O 3CO 376N O 57 C H Products 1500C C3H8 25C Air 500C Q Combustion chamber P 1 atm 5 5 50 2 3 th Substituting 2 2 2 2 2 2 8 3 The specif volume of the air entering the system is ic 2 190 m kg 101 3 kPa 0 287 kJkm air air K773 K ol 3 P T R v nd the mass flow rate of this air is a 0 4566 kg 2 190 m kg 3 air air air v m s 1m s 3 V bustion process is The airfuel ratio for this com 2353 kg airkg fuel 2 kgkmol 4 kmol 3 kmol 12 kgkmol 476 kmol 29 kgkmol 75 AF fuel air m m The mass flow rate of fuel is 0 01941 kgs 2353 0 4566 kgs mair m AF fuel The heat transfer for this combustion process is determined from the energy balance out in E E E system applied on the ombustion chamber with W 0 It reduces to c R f R f P h h h N h h N Q o o o o Assumin the com oducts gases hT tables stance P bustion pr h out g the air and to be ideal we have h From the Sub hf o kmol kJ h298 K ol kJkm K h 773 kJkmol K h1 773 l kJkmo C H 3 8 103850 O2 0 8682 23614 0 8669 22866 56689 g 241820 9904 71177 CO2 393520 9364 87195 59364 N2 H2O Substituting 8 3 298 298 out 596 881 kJ kmol C H 8669 22866 0 28 2 8682 23614 0 57 103850 1 8669 56689 0 28 2 8682 59364 0 52 9904 71177 241820 4 9364 87195 393520 3 h h Q or Then the rate of heat transfer for a mass flow rate of 001941 kgs for the propane becomes 8 3 out 596 881 kJ kmol C H Q 2633 kW 596881 kJkmol 44 kgkmol kgs 001941 out out out M Q m NQ Q preparation If you are a student using this Manual you are using it without permission 1543 1560 Methane is burned completely during a steadyflow combustion process The heat transfer from the combustion chamber is to be determined for two cases Assumptions 1 Steady operating conditions exist 2 Air and combustion gases are ideal gases 3 Kinetic and potential energies are negligible 4 Combustion is complete Analysis The fuel is burned completely with the stoichiometric amount of air and thus the products will contain only H2O CO2 and N2 but no free O2 Considering 1 kmol of fuel the theoretical combustion equation can be written as 2 th 2 2 2 2 th 4 N 376 2H O CO 376N O CH a a PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course a Products 25C CH4 25C Air 100 theoretical Q Combustion chamber P 1 atm where ath is determined from the O2 balance th 2 1 1 Substituting 2 2 2 2 2 4 564N 2H O CO 376N 2 O CH The heat transfer for this combustion process is determined from the energy balance E E in out system E applied on the combustion chamber with W 0 It reduces to o o o o o o f R R f P P R P since both the rea the tables f R f P N h N h h h h N h h h N Q out ctants and the products are at 25C and both the air and the combustion gases can be treated as ideal gases stance From Sub h f o kJkmol CH4 850 74 O2 0 N2 0 H2O l CO2 393520 Thus or 285830 out 890330 kJ kmol CH4 Q If combustion is achiev 4 out 890 330 kJ kmol CH 0 0 74850 1 0 285830 2 393520 1 Q ed with 100 excess air the answer would still be the same since it would enter and leave at 25C and absorb no energy preparation If you are a student using this Manual you are using it without permission 1544 1561E Diesel fuel is burned with 20 percent excess air during a steadyflow combustion process The required mass flow rate of the diesel fuel for a specified heat transfer rate is to be determined Products 800 R C12H26 77F Air 20 excess air 77F Combustion chamber P 1 atm 180 0 Btus Q Assumptions 1 Steady operating conditions exist 2 Air and combustion gases are ideal gases 3 Kinetic and potential energies are negligible 4 Combustion is complete Analysis The fuel is burned completely with the excess air and thus the products will contain only CO2 H2O N2 and some free O2 Considering 1 kmol of C12H26 the combustion equation can be written as 2 th 2 th 2 2 2 2 th 26 12 N 12 376 O 02 13H O 12CO 376N O 12 H C a a a determined from the energy balance where ath is the stoichiometric coefficient and is determined from the O2 balance 12 12 65 02 185 th th th a a a Substituting 2 2 2 2 2 2 26 12 8347N 37O 13H O 12CO 376N 222 O C H The heat transfer for this combustion process is E E E in out system applied on the combustion chamber with W 0 It reduces to o o o o o o o f R R P f P R f R P f P N h h h h N h h h N h h h N Q out since all of the reactants a F re at 77F Assuming the air and the combustion products to be ideal gases we have h hT rom the tables Substance o f h Btulbmol h537 R Btulbmol h800 R Btulbmol H26 5190 C12 12 O2 0 37251 56020 N2 0 37295 55644 H2O g CO2 169300 40275 65529 hus 104040 42580 63969 T 125190 1 3729 5 5564 4 8347 0 3725 1 5602 0 0 73 26 12 out H 3040716 Btu lbmol C 0 0 4258 6396 9 104040 13 4027 5 6552 9 169300 12 Q Then the required mass flow rate of fuel for a heat transfer rate of 1800 Btus becomes 26 12 out H 3040716 Btu lbmol C Q or 01006 lbms 3040716 Btulbmol 170 lbmlbmol 1800 Btus Q M Q NM m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1545 1562 A certain coal is burned steadily with 40 excess air The heat transfer for a given product temperature is to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 CO H2O SO2 and N2 3 Combustion gases are ideal gases Properties The molar masses of C H2 N2 O2 S and air are 12 2 28 32 32 and 29 kgkmol respectively Table A1 Analysis We consider 100 kg of coal for simplicity Noting that the mass percentages in this case correspond to the masses of the constituents the mole numbers of the constituent of the coal are determined to be 0 0247 kmol 32 kgkmol kg 079 0 0257 kmol 28 kgkmol kg 072 1 285 kmol 32 kgkmol kg 4111 3 465 kmol 2 kgkmol kg 693 3 271 kmol 12 kgkmol kg 3925 S S S N2 N2 N2 O2 O2 O2 H2 H2 H2 C C C M m N M m N M m N M m N M m N 3925 C 693 H2 4111 O2 072 N2 079 S 1120 ash by mass Air 40 excess Products 127C Coal Combustion chamber The mole number of the mixture and the mole fractions are 8 071 kmol 0 0247 0 0257 1 285 3 465 3 271 Nm 000306 8071 kmol kmol 00247 000319 8071 kmol kmol 00257 01592 8071 kmol kmol 1285 04293 8071 kmol kmol 3465 0 4052 8071 kmol kmol 3271 S S N2 N2 O2 O2 H2 H2 C C m m m m m N N y N N y N N y N N y N N y Ash consists of the noncombustible matter in coal Therefore the mass of ash content that enters the combustion chamber is equal to the mass content that leaves Disregarding this nonreacting component for simplicity the combustion equation may be written as 2 th 2 2 th 2 2 2 2 th 2 2 2 3 76N 41 0 00306SO O 40 0 4293H O 0 4052CO 3 76N O 41 0 00306S 0 00319N 0 1592O 04293H 04052C a a a ccording to the O2 mass balance 92 th th th a a a Substituting 2 441N 0 00306SO 0 1855O 0 4293H O CO 76N A 0 15 0 4637 0 00306 40 0 4293 50 0 4052 41 2 2 2 2 2 2 2 2 2 2 0 4052 3 0 6492O 0 00306S 0 00319N 0 1592O 04293H 04052C The heat transfer for this combustion process is determined from the energy balance out in E E E system applied on the ombustion chamber with W 0 It reduces to c R f R P f P h h h N h h h N Q o o o o out PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1546 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Assumin he com oducts l gases hT From the tables g the air and t bustion pr to be idea we have h Substance fo h kJkmol h298 K ol kJkm h400 K kJkmol O2 0 8682 11711 N2 0 8669 640 H2O g 241820 9904 13356 SO2 297100 The enthalpy change of sulfur dioxide between the standard temperature and the product temperature using constant specific heat assumption is 11 CO2 393520 9364 13372 4253 kJkmol 25K 41 7 kJkmol K127 SO2 T c h p Substituting into the energy balance relation 253244 kJ kmol C8H18 0 4253 297100 0 00306 8669 2 441 0 11640 1855 0 11711 8682 0 9904 241820 13356 0 4293 9364 372 r out 393520 13 0 4052 Q 253244 kJ kmol fuel out Q o Then the heat transfer per kg of fuel is 23020 kJkg coal 1100 kgkmol 244 kJkmol fuel 253 0 00306 32 kgkmol 28 0 00319 0 1592 32 0 4293 2 04052 12 244 kJkmol fuel 253 fuel out out M Q Q preparation If you are a student using this Manual you are using it without permission 1547 1563 Octane gas is burned with 30 percent excess air during a steadyflow combustion process The heat transfer per Air and combustion gases are ideal gases 3 Kinetic and potential n add the moisture to both sides of the equation Considering 1 kmol of C8H18 the combustion equation can be written as unit mass of octane is to be determined Assumptions 1 Steady operating conditions exist 2 energies are negligible 4 Combustion is complete Properties The molar mass of C8H18 is 114 kgkmol Table A1 Analysis The fuel is burned completely with the excess air and thus the products will contain only CO2 H2O N2 and some free O2 The moisture in the air does not react with anything it simply shows up as additional H2O in the products Therefore for simplicity we will balance the combustion equation using dry air and the 2 th 2 th 2 2 2 2 th 18 8 08 9H O 8CO 376N O 18 g C H a a N 18 376 O a where ath is the stoichiometric coefficient for air It is determined from s Therefore 225 476 1071 kmol of dry air will be used per km air is 125 08 45 8 balance18 O th th th 2 a a a PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Thu 2 2 2 2 2 2 18 8 846N 10O 9H O 8CO 376N 225 O g C H ol of the Products 1000 K C8H18 25C Air 80 excess air 25 q Combustion chamber P 1 atm fuel The partial pressure of the water vapor present in the incoming 1268 kPa 0 40 31698 kPa sat25 C air in P Pv φ Ass th uming ideal gas behavior the number of moles of the moisture at accompanies 1071 kmol of incoming dry air is determined to be C 1 36 kmol 101325 kPa 107 1 1268 kPa in in total in in v v v v N N N P P N total des of the equation 84 6 N 10O 1036H O 8CO O he heat t sfer for this combustion process is determined from the energy balance The balanced combustion equation is obtained by adding 136 kmol of H2O to both si 2 2 18 8 136H 376N 225 O g C H 2 2 2 2 2 E E E T ran in out system applied on the combustion chamber with W 0 It reduces to o o o o o o o f R R P f P R f r and the c R f P N h h h h N h h h N h h h N Q since all of t re at 2 ing the ai ombustion be ideal gases we have h hT From the table P 5C Assum out he reactants a products to s Substance o fh kJkmol h298 K kJkmol h1000 K kJkmol C8H18 g 208450 O2 0 8682 31389 N2 0 8669 30129 g 241820 9904 35882 H2O CO2 393520 9364 42769 Substituting 18 8 out 2537130 kJ kmol C H 0 0 241820 1 36 208450 1 8669 30129 0 84 6 8682 31389 10 0 9904 35882 241820 1036 9364 42769 393520 8 Q Thus 2537130 kJ of heat is transferred from the combustion chamber for each kmol 114 kg of C8H18 Then the heat transfer per kg of C8H18 becomes 22260 kJkg C8H18 114kg 2537130 kJ out M Q q preparation If you are a student using this Manual you are using it without permission 1548 1564 Problem 1563 is reconsidered The effect of the amount of excess air on the heat transfer for the combustion nalysis The problem is solved using EES and the solution is given below Excess air100 2376 N2Nw H2O8 CO291ExNw H2O 1Ex Ath 376 N2 Ex Tprod1ExAth376 es what happens to the results as you vary the percent excess air relative humidity and roduct temperature PercentEX kJ process is to be investigated A Fuel Octane C8H18 Tfuel 25273 K PercentEX 80 Ex PercentEX100 Pair1 1013 kPa Tair1 25273 K RH1 40100 Tprod 1000 K Mair 2897 kgkmol Mwater 18 kgkmol MC8H18812181 kgkmol For theoretical dry air the complete combustion equation is C8H18 AthO2376 N28 CO29 H2O Ath 376 N2 Ath28291 theoretical O balance now to find the amount of water vapor associated with the dry air w1HUMRATAirH2OTTair1PPair1RRH1 Humidity ratio kgvkga Nww1Ath476MairMwater Moles of water in the atmoshperic air kmolkmolfuel The balanced combustion equation with Ex excess moist air is C8H18 1EXAthO Ath O2 Apply First Law SSSF Hfuel 208450 kJkmol from Table A26 HRHfuel 1ExAthenthalpyO2TTair11ExAth376 enthalpyN2TTair11ExNwenthalpyH2OTTair1 HP8enthalpyCO2TTprod91ExNwenthalpyH2OT enthalpyN2TTprodExAthenthalpyO2TTprod QnetHPHRkJkmolMC8H18 kgkmol kJkgC8H18 Qout Qnet kJkgC8H18 This solution used the humidity ratio form psychrometric data to determine the moles of water vapor in atomspheric air One should calculate the moles of water contained in the atmospheric air by the method shown in Chapter 14 which uses the relative humidity to find the partial pressure of the water vapor and thus the mol of water vapor Explore p 0 40 80 120 160 200 8000 12000 16000 20000 24000 28000 32000 Q out kgC8H18 0 20 40 60 80 100 120 140 160 180 200 8393 31444 29139 26834 24529 22224 19919 17614 15309 13003 10698 Qout kJkgC8H18 PercentEX PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1549 1565 Propane gas is burned with 100 excess air The combustion is incomplete The balanced chemical reaction is to be written and the dewpoint temperature of the products and the heat transfer from the combustion chamber are to be determined Assumptions 1 Combustion is incomplete 2 The combustion products contain CO2 CO H2O O2 and N2 only Properties The molar masses of C H2 O2 N2 and air are 12 kgkmol 2 kgkmol 32 kgkmol 28 kgkmol and 29 kgkmol respectively Table A1 Analysis a The balanced reaction equation for stoichiometric air is 2 th 2 2 2 2 th 8 3 3 76 N 4 H O 3CO 376N O C H a a The stoicihiometric coefficient ath is determined from an O2 balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course a 3 76 N 5 2 O 4 H O 3CO 010 5 x for 2 is determined from a mass balance ubstitutin 37 6 N 5 15 O 4 H O 03 CO CO b The partial pressure of water vapor is CO2 CO H2O O2 N2 C3H8 Air 100 excess Combustion chamber 5 2 3 th Substituting 2 2 2 2 2 8 3 3 76 N 5 4 H O 3CO 376N 5 O C H The reaction with 100 excess air and incomplete combustion can be written as 2 2 2 2 2 2 8 3 3CO 090 376N O 2 C H The coefficient O O2 balance 5 15 2 3 0 05 3 90 10 x x S g 2 2 8 3 27 376N 10 O C H 2 2 2 2 8 040 kPa 4975 kmol 100 kPa 37 6 5 15 4 30 72 total total N v 4 kmol 100 kPa 4 H2O P N P The dew p re of the product gases is the saturation temperature of water at this pressure determined from the energy balance applied on oint temperatu 415C Tsat 804 kPa Tdp Table A5 c The heat transfer for this combustion process is system out in E E E the combustion chamber with W 0 It reduces to R f R P f P h h h N h h h N Q o o o o out Both the reactants and products are at 25 ir and the combustion products to be ideal gases we have h hT A oC Assuming the a lso since the temperature of products 25 oC is lower than the dewpont temperature some water vapor will condense Noting from Table A5 that 3 17 kPa sat 25 C P the molar amount of water that remain as vapor is determined from kmol 51 37 6 100 kPa 5 15 30 72 3 17 kPa total H2O P N Pv H2Ovapor vapor H2O vapor H2O total N N N N s or Then the heat transfer for a 100 kmol fuel becomes kmol 52 51 4 H2Oliquid N Thu Then using the values given in the table Q 8 3 out 069120 kJ kmol C H 2 1 103850 52 285830 241820 51 30 110530 72 393520 8 3 out 2 069120 kJ kmol C H Q 10 kJ 2069 8 100 kmol fuel2069120 kJkmol fuel out out out N Q m NQ Q preparation If you are a student using this Manual you are using it without permission 1550 1566 A mixture of propane and methane is burned with theoretical air The balanced chemical reaction is to be written and the amount of water vapor condensed and the the required air flow rate for a given heat transfer rate are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 CO H2O O2 and N2 only Properties The molar masses of C H2 O2 N2 and air are 12 kgkmol 2 kgkmol 32 kgkmol 28 kgkmol and 29 kgkmol respectively Table A1 Analysis a The balanced reaction equation for stoichiometric air is 2 th 2 2 2 2 th 4 8 3 3 76 N H O 82 18 CO 376N O CH 60 04 C H a a The stoicihiometric coefficient ath is determined from an O2 balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course a CO2 H2O N2 C3H8 CH4 Air 100 theoretical Combustion chamber 23 41 81 th Substituting 2 2 2 2 2 4 8 3 12032 N H O 82 18 CO 376N O 23 CH 60 04 C H b The partial pressure of water vapor is 1684 kPa 16632 kmol 100 kPa 12032 100 kPa 82 81 total total H2O P N Pv kmol 82 82 N The dew point temperature of the pro e saturation temperature of water at this pressure s The heat transfer for this combustion process is determined from the energy balance applied on duct gases is th 562 C sat 1684 kPa Tdp T Table A5 Since the temperature of the product gases are at 398 K 125C there will be no conden ation of water vapor system out in E E E c the combustion chamber with W 0 It reduces to h h h N h h h N o o o o R f R P f P out ducts are at 125 oC and the ent Q The pro halpy of products can be expressed as T c h h p o re hen using the values given in the table whe 100 K T 100 C 25 125 T 60 74850 40 103850 100 2927 120320 3428 100 82 241820 4116 100 393520 81 out Q 1 246760 kJ kmol fuel r Q 00 kJh the molar flow rate of fuel is o 1 246760 kJ kmol fuel out For a heat transfer rate of 970 0 07780 kmol fuelh 1 246760 kJkmol fuel 000 kJh 97 out Nfuel Qou t Q he molar mass of the fuel mixture is he mass flow rate of fuel is The airfuel ratio is T 16 60 44 40 fuel M 2 kgkmol 27 T 2 116 kgh 0 07780 kmolh27 2 kgkmol fuel fuel fuel M N m 1624 kg airkg fuel 06 16 kg 44 04 29 kg 4 76 32 AF fuel air m m The mass flow rate of air is then 344 kgh 2 116 kgh1624 fuelAF air m m preparation If you are a student using this Manual you are using it without permission 1551 1567 A mixture of ethanol and octane is burned with 10 excess air The combustion is incomplete The balanced chemical reaction is to be written and the dewpoint temperature of the products the heat transfer for the process and the relative humidity of atmospheric air for specified conditions are to be determined Assumptions 1 Combustion is incomplete 2 The combustion products contain CO2 CO H2O O2 and N2 only CO2 CO H2O O2 N2 C8H18 C2H6O Air 10 excess Combustion chamber Properties The molar masses of C H2 O2 N2 and air are 12 kgkmol 2 kgkmol 32 kgkmol 28 kgkmol and 29 kgkmol respectively Table A1 Analysis a The balanced reaction equation for stoichiometric air is 2 th 2 2 2 2 th 18 8 6 2 3 76 N H O 48 74 CO 376N O C H 90 01 C H O a a The stoicihiometric coefficient ath is determined from an O2 balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 th th a a 1155 2 48 47 2 1 Substituting 2 2 2 2 2 18 8 6 2 3 76 N 1155 H O 48 74 CO 376N 1155 O C H 90 01 C H O The reaction with 10 excess air and incomplete combustion can be written as 2 2 2 2 2 2 18 8 6 2 74 CO 09 3 76 N 1155 11 O H O 48 74 CO 01 376N 1155 O 11 C H 90 C H O 01 x nt fo 2 is determined from a mass balance 2 The coefficie r O 1 525 48 50 47 10 50 47 90 1155 11 10 50 x x O balance Substituting 2 2 2 2 2 2 18 8 6 2 4777 N 1 525 O H O 48 074 CO 666 CO 376N 12705 O C H 90 C H O 01 b The partial pressure of water vapor is 12 9 kPa 6510 kmol 100 kPa 4777 1 525 48 0 74 6 66 total N The dew point temperature of the product gases is th kmol 48 100 kPa 48 P e saturation temperature of water at this pressure Table A5 to total H2O P N v 505C Tsat 129 kPa Tdp c The heat transfer for this combustion process is determined from the energy balance system out in E E E applied on the combustion chamber with W 0 It reduces R f R P f P h h h N h h h N Q o o o o out Both ea hT the r ctants and products are at 25 oC Assuming the air and the combustion products to be ideal gases we have h Then using the values given in the table Q or The molar mass of the fuel is 522790 kJ kmol fuel 4 90 208450 10 235310 48 241820 0 74 110530 6 66 393520 out 4 522790 kJ kmol fuel out Q 107 2 kgkmol 114 90 46 10 M preparation If you are a student using this Manual you are using it without permission 1552 Then the heat transfer for a 25 kg of fuel becomes 105480 kJ 1072 kgkmol out out out M d For 957 kmol of water vapor in the products the air must carry 957 84 1 4 522790 kJmol kg 52 m Q NQ Q 17 kmol of water vapor in the tmospheric air The partial pressure of this water vapor in the stmospheric air is a 1 8979 kPa 6165 kmol 100 kPa 1 17 4 76 12705 total total N v The saturat 1 17 kmol 100 kPa 1 17 H2O P N P ion pressure of water at 25C is 317 kPa Table A4 The relative humidity of water vapor in the atmospheric ir is then a 599 0 599 3 17 kPa 8979 kPa 1 Ptotal Pv φ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1553 1568 A mixture of methane and oxygen contained in a tank is burned at constant volume The final pressure in the tank and the heat transfer during this process are to be determined Assumptions 1 Air and combustion gases are ideal gases 2 Combustion is complete Properties The molar masses of CH4 and O2 are 16 kgkmol and 32 kgkmol respectively Table A1 Analysis a The combustion is assumed to be complete and thus all the carbon in the methane burns to CO2 and all of the hydrogen to H2O The number of moles of CH4 and O2 in the tank are PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1875 mol kmol 1875 10 32 kgkmol O2 2 M Then the combustion equation c kg 06 75 mol kmol kg 012 3 O O 3 CH 2 m N m an be written as nal pressure 1200 K Q O2 CH4 25C 200 kPa 75 10 16 kgkmol CH CH 4 4 4 M N 2 2 2 2 4 375O 15H O 75CO 1875O 75CH At 1200 K water exists in the gas phase Assuming both the reactants and the products to be ideal gases the fi in the tank is determined to be 805 kPa 1200 K mol 2625 R R P u P P Substituting 298 K 2625 mol 200 kPa P P P R P R u R R P T T N N P P N R T P N R T P V V te b The h this constant volume c process is determined from the energy balance in out system applie mbustio with W uces to which is relatively low Therefore the ideal gas assumption utilized earlier is appropria eat transfer for E ombustion n chamber E E d on the co 0 It red P h v o R f R P f P h h P h h N v o o out Since bot tants and sum e ideal gas e internal energy and enthalpies depend on temperat ly and the N h Q o h the reac products are as ed to b es all th ure on Pv terms in this eq can be repla uT It yields uation ced by R R u f R P u f P R T h N R T h h h N Q o o 29 8 K 120 0 K out he reactants ar Substance since t e at the standard reference temperature of 25C From the tables hf o kJkmol h298 K kJkmol h1200 K kJkmol CH4 74850 O2 0 8682 38447 H2O g 241820 9904 44380 CO2 393520 9364 53848 Thus 5252 kJ 5251791 J 298 8 314 1875 298 8 314 74850 57 8 314 1200 8682 38447 3 75 0 8 314 1200 9904 44380 241820 15 8 314 1200 9364 53848 393520 57 out Q Thus of heat is transferred from the combustion chamber as 120 g of CH4 burned in this combustion chamber out 5252 kJ Q preparation If you are a student using this Manual you are using it without permission 1554 1569 Problem 1568 is reconsidered The effect of the final temperature on the final pressure and the heat transfer nalysis The problem is solved using EES and the solution is given below K perature kmol mbustion equation is 2 supplied per mole of fuel H4 cess O2 is 21 CO2 2 H2O Ex Ath O2 ber and assume ideal gas ists in the gas phase reac RuTreac lpyH2O TTprod Tprod RuTprod e co prod NCH41 2 ExAth for the combustion process is to be investigated A Input Data Treac 25273 reactant mixture temperature essure Preac 200 kPa reactant mixture pr product mixture tem Tprod 1200 K mO20600 kg initial mass of O2 MwO2 32 kgkmol mCH4 0120 kg initial mass of CH4 MwCH411241 kg Ru 8314 kJkmolK universal gas constant For theoretical oxygen the complete co CH4 Ath O21 CO22 H2O 2Ath1221theoretical O balance s of O now to find the actual mole NO2 mO2MwO2NC NCH4 mCH4MwCH4 The balanced complete combustion equation with Ex ex CH4 1EX Ath O NO2 1ExAth Apply First Law to the closed system combustion cham behavior At 1200 K water ex Ein Eout DELTAEsys Ein 0 Eout Qout kJkmolCH4 No work is done because volume is constant DELTAEsys Uprod Ureac neglect KE and PE and note U H PV Nh Ru T Ureac 1enthalpyCH4 TTreac RuTreac 1EXAthenthalpyO2TT Uprod 1enthalpyCO2 TTprod RuTprod 2entha RuTprodEXAthenthalpyO2T The total heat transfer out in kJ is QouttotQoutkJkmolCH4MwCH4 kgkmolCH4 mCH4kg kJ The final pressure in the tank is the pressure of the product gases Assuming ideal gas behavior for the gases in th ac nstant volume tank the ideal gas law gives PreacV Nreac Ru Tre PprodV Nprod Ru Tprod Nreac NCH41 NO2 N Tprod K Qouttot kJ Pprod kPa 500 5872 3356 700 5712 4698 900 5537 604 1100 5349 7383 1300 5151 8725 1500 4943 1007 500 700 900 1100 1300 1500 4900 5100 5300 5500 5700 5900 300 400 500 600 700 800 900 1000 1100 Tprod K Qouttot kJ Pprod kPa PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1555 1570E Methane is burned with stoichiometric amount of air in a rigid container The heat rejected from the container is to be determined Assumptions 1 Air and combustion gases are ideal gases 2 Combustion is complete Properties The molar masses of CH4 and air are 16 lbmlbmol and 29 lbmlbmol respectively Table A1E Analysis The combustion equation for 1 lbmol of fuel is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 It reduces to 1060F Qout CH4 Theoretical air 77F 144 psia 2 2 2 2 2 4 7 52N 2H O CO 3 76 2O CH N The heat transfer for this constant volume combustion process is determined from the energy balance E E in out system applied on the combustion chamber with E W R f R P f P P h h h N P h h h N Q v v o o o o out Since both the reactants and products are assumed to be ideal gases all the internal energy and enthalpies depend on mperature only and the v P terms in this equation can be replaced by RuT yields te It R u f R P u f P R T h N R T h h h N Q o o 537 R 1520 R out since the reactants are at the standard reference temperature of 77F From the tables tance Subs hf o Btulbmol h537R Btulbmol h1520R Btulbmol 4 210 CH 32 O2 0 37251 111796 N2 0 37295 108004 H2O g CO2 169300 40275 148249 Thus 104040 42580 127388 4 out 284800 Btulbmol CH 537 1 9858 7 52 537 1 9858 2 537 1 9858 32210 1 1 9858 1520 3729 5 10800 4 7 52 0 1 9858 1520 4258 0 12738 8 104040 2 1 9858 1520 4027 5 14824 9 169300 1 Q Thus Qout 284800 Btulbmol CH4 preparation If you are a student using this Manual you are using it without permission 1556 1571 A mixture of benzene gas and 30 percent excess air contained in a constantvolume tank is ignited The heat transfer from the combustion chamber is to be determined Assumptions 1 Both the reactants and products are ideal gases 2 Combustion is complete Analysis The theoretical combustion equation of C6H6 with stoichiometric amount of air is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 6 15 75 stem 1000 K Q C6H6 Air 25C 1 atm 2 th 2 2 2 2 th 6 6 N 376 3H O 6CO 376N O C H g a a where ath is the stoichiometric coefficient and is determined from the O2 balance th a Then the actual combustion equation with 30 excess air becomes 2 2 2 2 2 2 6 6 3666N 249O 3H O 048CO 552CO 376N 975 O C H g The heat transfer for this constant volume combustion process is determined from the energy balance E E in out sy E It reduces to applied on the combustion chamber with W 0 R P Since both the reactants and the products behave as ideal gases f R f P P h h h N P h h h N Q v v o o o o out all the internal energy and enthalpies depend on temperature Pv terms in this equation can be replaced by R only and the uT It yields R P u f R u f P R T h N R T h h h N Q o o 298 K 1000 K out reactants are at the standard reference temperature of 25 C From the tables since the Substance hf o kJkmol h298 K kJkmol h1000 K kJkmol H6 g 82930 C6 O2 0 8682 31389 N2 0 8669 30129 H2O g 241820 9904 35882 CO CO 110530 393520 9 8669 364 30355 42769 2 Thus or 2200433 kJ 8 314 298 4 76 9 75 8 314 298 1 82930 8 314 1000 8669 30129 3666 0 8 314 1000 8682 31389 2 49 0 8 314 1000 9904 35882 241820 3 8 314 1000 8669 30355 110530 0 48 8 314 1000 9364 42769 393520 5 52 out Q Qout 2200433 kJ preparation If you are a student using this Manual you are using it without permission 1557 1572E A mixture of benzene gas and 60 percent excess air contained in a constantvolume tank is ignited The heat transfer from the combustion chamber is to be determined Assumptions 1 Both the reactants and products are ideal gases 2 Combustion is complete Analysis The theoretical combustion equation of C6H6 with stoichiometric amount of air is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 6 15 75 stem 2100 R Q C6H6 Air 77F 1 atm 2 th 2 2 2 2 th 6 6 N 376 3H O 6CO 376N O C H g a a where ath is the stoichiometric coefficient and is determined from the O2 balance th a Then the actual combustion equation with 60 excess air becomes 2 2 2 2 2 2 6 6 4512N 474O 3H O 048CO 552CO 376N 12 O g C H The heat transfer for this constant volume combustion process is determined from the energy balance E E in out sy E It reduces to applied on the combustion chamber with W 0 R P Since both the reactants and the products behave as ideal gases f R f P P h h h N P h h h N Q v v o o o o out all the internal energy and enthalpies depend on temperature only and the Pv terms in this equation can be replaced by R T u It yields R P u f R u f P R T h N R T h h h N Q o o 537R 1800R out reactants are at the standard reference temperature of 77F From the tables since the Substance hf o Btulbmol h537 R Btulbmol h2100R Btulbmol C6H6 g 35680 O2 0 37251 16011 N2 0 3 7295 15334 H2O g 104040 42580 18467 CO CO 4 2 Thus or 7540 169300 4 37251 0275 15463 22353 757400 Btu 537 1986 12 476 537 1986 1 35680 2100 1986 37295 15334 4512 0 2100 1986 37251 16011 474 0 2100 1986 42580 18467 104040 3 2100 1986 37251 15463 47540 048 2100 1986 40275 22353 169300 552 out Q Qout 757400 Btu preparation If you are a student using this Manual you are using it without permission 1558 1573 A high efficiency gas furnace burns gaseous propane C3H8 with 140 percent theoretical air The volume flow rate of water condensed from the product gases is to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O O2 and N2 only Properties The molar masses of C H2 O2 and air are 12 kgkmol 2 kgkmol 32 kgkmol and 29 kgkmol respectively Table A1 Analysis The reaction equation for 40 excess air 140 theoretical air is 2 2 2 2 2 2 th 3 8 N O H O CO 376N O 41 C H F E D B a where ath is the stoichiometric coefficient for air We have automatically accounted for the 40 excess air by using the factor 14ath instead of ath for air The coefficient ath and other coefficients are to be determined from the mass balances Carbon balance B 3 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Products C3H8 25C Air 40 excess Q Combustion chamber Hydrogen balance D 4 8 2 D Oxygen balance E D B a 2 2 41 2 th 4 E a th 0 Nitrogen balance F a 3 76 41 th Solving the above equations we find the coefficients E 2 F 2632 nd a 5 and write the balanced reaction equation as a th he partial pressure of water in the saturated product mixture at the dew point is he vapor mole fraction is 2 2 2 2 2 2 8 3 2632 N 2 O 4 H O 3CO 376N 7 O C H T sat40 C prod P Pv kPa 3851 7 T 0 07385 kPa 100 7 3851 kPa prod Pv prod yv P The kmoles of water condensed is determined from 1 503 kmol 2632 2 4 3 4 0 07385 w w w product total Nwater N N N N y adyflow energy balance is expressed as where v The ste P R H N Q H N fuel fuel fuel 32969 kJh 096 650 kJh 31 furnace out fuel η Q Q 103847 kJkmol 26 70 103847 kJkmol 320 2632 7 N225 C O225 C fuel25 C h h h H o f R 577 10 kJkmol 2 1 503 285830 kJkmol 26320 20 4241820 kJkmol 393520 kJkmol 3 2632 2 4 3 6 H2Oliq w N225 C O225 C H2O25 C C CO225 P h fo N h h h h H tion The molar and mass flow rates of the liquid water are The volume flow rate of liquid water is Substituting into the energy balance equa 0 01333 kmolh 2 577 10 kJkmol 32969 kJh 103847 kJkmol fuel 6 fuel fuel fuel fuel fuel N N N H N Q H N P R 002003 kmolh 503 kmolkmol fuel001333 kmol fuelh 1 fuel w w N N N 03608 kgh kgkmol 0 02003 kmolh18 w w w N M m 87 Lday 00003619 m h 0 001003 m kg03608 kgh 3 3 w 25 C w m f v V preparation If you are a student using this Manual you are using it without permission 1559 1574 Wheat straw that is being considered as an alternative fuel is tested in a bomb calorimeter The heating value of this straw is to be determined and compared to the higher heating value of propane Assumptions 1 Combustion is complete Analysis The heat released by the combustion is 180 kJ 100 kJK18 K T mc Q v The heating value is then 18000 kJkg 0010 kg 180 kJ HV m Q From Table A27 the higher heating value of propane is HHV 50330 kJkg PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1560 Adiabatic Flame Temperature 1575C For the case of stoichiometric amount of pure oxygen since we have the same amount of chemical energy released but a smaller amount of mass to absorb it 1576C Under the conditions of complete combustion with stoichiometric amount of air PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1561 1577 Hydrogen is burned with 50 percent excess air during a steadyflow combustion process The exit temperature etic and potential er adiabatic conditions steadyflow conditions the energy balance applied on the combustion chamber reduces to of product gases is to be determined Assumptions 1 Steady operating conditions exist 2 Air and combustion gases are ideal gases 3 Kin energies are negligible 4 There are no work interactions 5 The combustion chamber is adiabatic Analysis Adiabatic flame temperature is the temperature at which the products leave the combustion chamber und Q 0 with no work interactions W 0 Under system out in E E E e combustion equation of H2 with 50 excess air is 282N 025O H O 376N rom the tables bstance e combustion equation of H2 with 50 excess air is 2 2 2 2 282N 025O H O 376N rom the tables bstance R f R P f P h h h N h h h N o o o o R f R P f P h h h N h h h N o o o o Products TP H2 27C Air 50 excess air 27C Combustion chamber Th Th 2 2 075 O H 2 2 075 O H 2 2 2 2 FF Su Su hf o kJkmol h300K kJkmol h 298 K ol kJkm H2 0 8522 8468 O2 0 8736 8682 N2 0 8723 8669 H2O g 241820 9966 9904 hus T 0 8669 8723 2 82 0 8682 8736 0 75 0 8468 8522 1 8669 2 82 0 8682 0 25 0 9904 241820 1 2 2 2 N O H O h h h It yields 278590 kJ 2 82 0 25 2 2 2 N O H O h h h The adiabatic flame temperature is obtained from a trial and error solution A first guess is obtained by dividing the right hand side of the equation by the total number of moles which yields 2785901 025 282 68450 kJkmol This e corresponds to about 2100 K for N2 Noting that the majority of the moles are N2 TP will be close to 2100 K but somewhat under it because of the higher specific heat of H2O enthalpy valu 330 kJ Higher than 278590 kJ 282 2 82 64810 0 25 67881 1 82593 2 82 0 25 2 2 2 N O O PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course At 2000 K H h h h At 1960 K 880 kJ Lower than 275 2 82 63381 0 25 66374 1 80555 2 826 0 25 2 2 2 N O H O h h h 278590 kJ By interpolation TP 1977 K Discussion The adiabatic flame temperature cam be obtained by using EES without a trial and error approach We found the temperature to be 1978 K by EES The results are practically identical preparation If you are a student using this Manual you are using it without permission 1562 1578 Problem 1577 is reconsidered This problem is to be modified to include the fuels butane ethane methane and propane as well as H2 to include the effects of inlet air and fuel temperatures and the percent theoretical air supplied Analysis The problem is solved using EES and the solution is given below Adiabatic Combustion of fuel CnHm entering at Tfuel with Stoichiometric Air at Tair Reaction CxHy y4 x Theoair100 O2 376 N2 xCO2 y2 H2O 376 y4 x Theoair100 N2 y4 x Theoair100 1 O2 Tprod is the adiabatic combustion temperature assuming no dissociation Theoair is the theoretical air The initial guess value of Tprod 450K Procedure FuelFuelxyName This procedure takes the fuel name and returns the moles of C and moles of H If fuelC2H6 then x2y6 Nameethane else If fuelC3H8 then x3 y8 Namepropane else If fuelC4H10 then x4 y10 Namebutane else if fuelCH4 then x1 y4 Namemethane else if fuelH2 then x0 y2 Namehydrogen endif endif endif endif endif end Input data from the diagram window Tfuel 300 K Tair 300 K Theoair 150 FuelH2 Call FuelfuelxyName HRenthalpyFuelTTfuel y4 x Theoair100 enthalpyO2TTair376y4 x Theoair100 enthalpyN2TTair HPHR Adiabatic HPxenthalpyCO2TTprody2enthalpyH2OTTprod376y4 x Theoair100enthalpyN2TTprody4 x Theoair100 1enthalpyO2TTprod MolesO2y4 x Theoair100 1 MolesN2376y4 x Theoair100 MolesCO2x MolesH2Oy2 T1Tprod xa1Theoair array variable are plotted in Plot Window 1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1563 Theoair Tprod K 100 150 200 250 300 350 400 450 500 2528 1978 1648 1428 1271 1153 1060 9863 9255 100 150 200 250 300 350 400 450 500 750 1150 1550 1950 2350 2750 Theoair Tprod K Calculated point Calculated point PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1564 1579 Acetylene is burned with stoichiometric amount of oxygen The adiabatic flame temperature is to be determined Assumptions 1 Steady operating conditions exist 2 Air and combustion gases are ideal gases 3 Kinetic and potential energies are negligible 4 There are no work interactions 5 The combustion chamber is adiabatic Analysis Under steadyflow conditions the energy balance system out in E E E applied on the combustion chamber with Q W 0 reduces to o o o o o o o f R R P f P R f R P f P N h h h h N h h h N h h h N since all the reactants are at the standard reference temperature of 25C Then for the stoichiometric oxygen 2 2 2 From the tables hus C H 2 2 1H O 2 CO O 52 Products TP C2H2 25C 100 theoretical O2 25C Combustion chamber PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course T 0 0 226730 1 9904 241820 1 9364 520 393 2 H2O CO2 h h It yields 1 284220 kJ 1 2 H2O CO2 h h The adiabatic flame temperature is obtained from a trial and error solution A first guess is obtained by dividing the right hand side of the equation by the total number of moles which yields 228422021 428074 kJkmol The ideal gas tables do not list enthalpy values this high Therefore we cannot use the tables to estimate the adiabatic flame temperature Table A2b the highest available value of specific heat is cp 1234 kJkgK or CO2 at 1000 K The specific heat of water vapor is c 18723 kJkgK Table A2a Using these specific heat values In f p 0 0 226730 1 241820 1 393520 2 T c T c p p here The specific heats on a molar base are w 25 C af T T 337 kJkmol K 8723 kJkg K18 kgkmol 1 543 kJkmol K 234 kJkg K44 kgkmol 1 H2O CO 2 c M c M c p p c p p Substituting 8824 K kJkmol K 33 7 54 3 2 1 255590 kJkmol 1 255590 33 7 54 3 2 226730 33 7 241820 1 54 3 393520 2 T T T T T Then the adiabatic flame temperature is estimated as e Substanc o f h kJkmol h298K kJkmol 2 g 226730 C2H O2 2 g 393520 9364 0 8682 N2 0 8669 H O 241820 9904 CO2 8849C 25 8824 25 af T T preparation If you are a student using this Manual you are using it without permission 1565 1580 Propane is burned with stoichiometric and 50 percent excess air The adiabatic flame temperature is to be determined for both cases Assumptions 1 Steady operating conditions exist 2 Air and combustion gases are ideal gases 3 Kinetic and potential energies are negligible 4 There are no work interactions 5 The combustion chamber is adiabatic Analysis Under steadyflow conditions the energy balance system out in E E E applied on the combustion chamber with Q W 0 reduces to o o o o o o o f R R P f P R f R P f P N h h h h N h h h N h h h N PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Products TP C3H8 25C Air 100 theoretical air Combustion chamber since all the reactants are at the standard reference temperature of 25C Then for the stoicihiometric air here ath is the stoichio etric coefficient and is determined from the O2 balance Thus 3 76 N 5 O C H Fr s Thus 2 th 2 2 2 2 th 8 3 3 76 N 4 H O 3CO 3 76 N O C H a a w m 5 2 3 th a 2 2 2 2 2 8 3 18 8 N 4 H O CO 3 om the table e Substanc o f h kJkmol h298K ol H8 g 3850 kJkm C3 10 O2 0 8682 N2 0 8669 H2O g 241820 9904 393520 9364 CO2 0 0 103850 1 8669 18 8 0 9904 241820 4 9364 393520 3 N2 H2O CO2 h h h It yields 2 274680 kJ 18 8 4 3 N2 H2O CO2 h h h The adiabatic flame temperature is obtained from a trial and error solution A first guess is obtained by dividing the right e equation by the total number of moles which yields 22746803 4 188 88166 kJkmol This e corresponds to about 2650 K for N2 Noting that the majority of the moles are N2 TP will be close to 2650 K but somewhat under it because of the higher specific heat of H2O At 2500 K hand side of th enthalpy valu 389380 kJ Higher than 2274680 kJ 2 82981 18 8 4 108868 3 131290 18 8 4 3 CO2 h N2 H2O h h At 2450 K 334990 kJ Higher than 2274680 kJ 2 81149 18 8 4 106183 3 128219 18 8 4 3 N2 H2O CO2 h h h At 2400 K 280704 kJ Higher than 2274680 kJ 2 79320 18 8 4 103508 3 125152 18 8 4 3 N2 H2O CO2 h h h preparation If you are a student using this Manual you are using it without permission PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 1566 At 2350 K 226580 kJ Lower than 2274680 kJ 2 77496 18 8 4 100846 3 122091 18 8 4 3 N2 H2O hCO2 Products TP C3H8 25C Air 50 excess air 25C Combustion chamber h h By inter TP 2394 K 2121C When propane is burned with 50 excess air the reaction equation may be written as 3 76N a here ath is the stoichiometric coefficient and is determined from the O2 balance 28 2 N 52 O 4H O 3CO 3 76N 57 O C H he values in the table polation of the two results 2 th 2 2 2 2 th 8 3 51 O 50 4H O 3CO 3 76N O 51 C H a a 2 th w 5 05 2 3 15 th th th a a a Thus 2 2 2 2 2 2 8 3 Using t 8682 52 0 9904 241820 4 9364 393520 3 h h h 0 0 103850 1 8669 28 2 0 N2 O2 H2O CO2 h It yields 2 377870 kJ 28 2 52 4 3 N2 O2 H2O CO2 h h h h The adiabati hand side of th c flame temperature is obtained from a trial and error solution A first guess is obtained by dividing the right e equation by the total number of moles which yields 23778703425282 63073 kJkmol This enthalpy v e corresponds to about 1960 K for N2 Noting that the majority of the moles are N2 TP will be close to 1960 K under it because of the higher specific heat of H2O At 1800 K alu but somewhat 333160 kJ Lower than 2377870 kJ 2 57651 28 2 60371 52 72513 4 88806 3 28 2 52 4 3 N2 O2 H2O CO2 h h h h At 1840 K 392190 kJ Higher than 2377870 kJ 2 59075 28 2 61866 52 74506 4 91196 3 28 2 52 4 3 N2 O2 H2O CO2 h h h h By interpolation TP 1830 K 1557C 1567 1581 Octane is burned with 40 percent excess air adiabatically during a steadyflow combustion process The exit temperature of product gases is to be determined Assumptions 1 Steady operating conditions exist 2 Air and combustion gases are ideal gases 3 Kinetic and potential energies are negligible 4 There are no work interactions 5 The combustion chamber is adiabatic Products TP C8H18 25C Air 30 excess air 307C Combustion chamber Analysis Under steadyflow conditions the energy balance system out in E E E applied on the combustion chamber with Q W 0 reduces to R f R P f P h h h N h h h N o o o o since all the reactants are at the standard reference temperature of 25C Then 2 th 2 th 2 2 2 2 th 18 8 14 O 04 9H O 8CO 376N O 14 C H a a l 376 a N where ath is the stoichiometric coefficient and is determined from the O2 balance 658N 5O 9H O 8CO 376N 17 5 O From the tables hus 125 04 45 8 14 th th th a a a Thus 18 8 g C H 2 2 2 2 2 2 T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 8682 5 0 9904 241820 9 9364 393520 8 O2 H2O CO2 h h h 8669 65 8 16962 8682 17290 0 17 5 249950 1 8669 0 65 8 N2 h It yields 6 548788 kJ 65 8 5 9 8 N2 O2 H2O CO2 h h h h The adiabatic flame temperature is obtained from a trial and error solution A first guess is obtained by dividing the right e equation by the total number of moles which yields 65487888 9 5 658 74588 kJkmol This enthalpy value corresponds to about 2250 K for N2 Noting that the majority of the moles are N2 TP will be close to 2250 K but som what under it because of the higher specific heat of H2O hand side of th e At 2100 K 6 504706 kJ Lower than 6548788 kJ 68417 65 8 5 71668 9 87735 8 106864 65 8 5 9 8 N2 O2 H2O CO2 h h h h At 2150 K 6 680890 kJ Higher than 6548788 kJ 70226 65 8 5 73573 9 90330 8 109898 65 8 5 9 8 O2 H2O CO2 h h h N2 h By interpolation TP 2113 K 1840C Substance o f h kJkmol h298K kJkmol h580K kJkmol 18 l 9950 C8H 24 O2 0 8682 17290 N2 0 8669 16962 g CO2 393520 9364 H2O 241820 9904 preparation If you are a student using this Manual you are using it without permission 1568 1582 A certain coal is burned with 100 percent excess air adiabatically during a steadyflow combustion process The temperature of product gases is to be determined for complete combustion and incomplete combustion cases Assumptions 1 Steady operating conditions exist 2 Air and combustion gases are ideal gases 3 Kinetic and potential energies are negligible 4 There are no work interactions 5 The combustion chamber is adiabatic Properties The molar masses of C H2 N2 O2 S and air are 12 2 28 32 32 and 29 kgkmol respectively Table A1 Analysis We consider 100 kg of coal for simplicity Noting that the mass percentages in this case correspond to the masses of the constituents the mole numbers of the constituent of the coal are determined to be PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 0278 kmol 32 kgkmol kg 089 0 0225 kmol kgkmol 28 kg 063 0 1375 kmol kgkmol 32 N2 O2 O2 O2 m N M N kg 440 0 945 kmol 2 kgkmol kg 189 7 03 kmol kg 8436 S S S N2 N2 H2 H2 H2 C M m N M m M m N m d the mole fractions are N 12 kgkmol C C M 8436 C 189 H2 440 O2 063 N2 89 S 783 ash by mass 0 The mole number of the mixture an 8 163 kmol 0 0278 0 0225 0 1375 0 945 7 03 Nm 0003407 8163 kmol kmol 00278 000276 8163 kmol kmol 00225 001684 8163 kmol kmol 01375 01158 8163 kmol kmol 0945 0 8611 8163 kmol kmol 703 S S N2 N2 O2 O2 H2 H2 C C m m m m m N N y N N y N N y N N y N N y CO2 CO H2O SO2 O2 N2 Tprod Coal 25C 100 excess air 25C Combustion chamber Ash consists of the noncombustible matter in coal Therefore the mass of ash content that enters the combustion chamber is equal to the mass content that leaves Disregarding this nonreacting component for simplicity the combustion equation may be written as According to the species balances 2 th 2 2 2 2 2 2 th 2 2 2 O N SO H O 0 03CO 0 97CO 3 76N O 2 0 00341S 0 00276N 0 01684O 01158H 08611C a k z y x a 6 72 0 8927 3 76 2 0 00276 3 76 2 balance 0 00276 N 0 8927 0 01684 0 00341 50 0 1158 0 015 0 8611 0 97 0 8611 50 0 015 0 97 2 01684 0 balance O 0 00341 balance S 0 1158 balance H 0 8611 balance C th 2 th th th 2 2 k k a a a z y x x a z y x preparation If you are a student using this Manual you are using it without permission 1569 Substituting 2 2 2 2 2 2 2 2 2 2 0 8927O 6 72N 0 00341SO 0 1158H O 0 0258CO 0 8353CO 3 76N 1 785O 0 003407S 0 00276N 0 01684O 01158H 08611C Under steadyflow conditions the energy balance system out in E E E applied on the combustion chamber with Q W 0 reduces to o o o o o o o f R R P f P R f R P f N h h h h N h h h N h h h NP From the tables tance Subs o f h kJkmol h298K kJkmol O2 0 8682 N2 0 8669 H2O g CO2 110530 8669 hus 241820 9904 CO 110530 8669 T 0 8669 6 72 0 8682 0 8927 0 N2 O2 h h It yields 9904 241 0 1158 8669 110530 0 0258 9364 393520 0 8353 H2O CO CO2 h h h 820 434760 kJ 6 72 0 8927 0 1158 0 0258 0 8353 h h h N2 O2 H2O CO CO2 h h 76008353002580115800034167208927 50595 kJkmol This enthalpy value corresponds to about 1600 K for N2 Noting that the majority of the moles are N2 TP will be close to 1600 K but somewhat under it because of the higher specific heat of H2O The product temperature is obtained from a trial and error solution A first guess is obtained by dividing the righthand side of the equation by the total number of moles which yields 434 At 1500 K 427647 kJ Lower than 434760 kJ 6 7247073 0 892749292 0 115857999 0 025847517 0 835371078 6 842 0 9095 0 1158 0 0258 0 8353 N2 O2 H2O CO CO2 h h h h h At 1520 K 094 kJ Lower than 434760 kJ PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 434 6 7247771 0 892750024 58942 8 842 N2 h 22 K 1249C We repeat the calculations for the complete combustion now The combustion equation in this case may be written as According to the species balances 0 115 0 025848222 0 835372246 H2O CO CO2 6 0 9095 0 1158 0 0258 0 8353 O2 h h h h By extrapolation TP 15 2 th 2 2 2 2 2 2 th 2 2 2 O N SO H O CO 3 76N O 2 0 00341S 0 00276N 0 01684O 01158H 08611C a k z y x a preparation If you are a student using this Manual you are using it without permission 1570 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course balance O 0 00341 0 8611 C th 2 th th th 2 k k a a a z y x a z x Substituti 2 2 2 2 0 9056O 6 81N 0 00341SO H O 0 8611CO 3 76N 1 819O 0 003407S N 0 0 01684O 01158H 08611C Under ste dition y balan E E balance S 0 1158 H2 balance y 6 81 0 9056 3 76 2 0 00276 3 76 2 balance 0 00276 N 0 9056 0 01684 0 00341 50 0 1158 0 8611 50 2 01684 0 balance ng 2 0 1158 2 00276 adyflow con s the energ ce in E 2 2 2 2 system out applied on the combustion chamber with Q W 0 reduces to o o o o o o f R R P f P f R P N h h h h N h h h N h h N o R P hf he tables Substance From t o f h kJkmol h298K kJkmol O 0 8682 2 N2 0 8669 H2O g 241820 9904 CO 110530 8669 CO2 0 8682 Thus 0 8669 6 81 0 8682 9056 0 0 9904 241820 0 1158 9364 393520 8611 0 N2 O2 H2O CO2 h h h h 442971 kJ 6 81 0 9056 0 1158 0 8611 N2 O2 H2O CO2 h h h h It yields The produc perature is obtained from a trial and error solution A first guess is obtained by dividing the righthand side of the equa on by the total number of moles which yields 8611 56 50940 kJkmol This enthalpy value corresponds to about 1600 K for N2 Noting that the majority of the moles are N2 TP will be close to 1600 K but somewhat under it because of the higher specific heat of H2O At 1500 K t tem ti 4429710 01158000341681090 128 kJ Lower than 442971 kJ 433 6 8147073 0 905649292 0 115857999 0 861171078 6 81 0 9056 0 1158 0 8611 N2 O2 H2O CO2 h h h h At 1520 K 658 kJ Lower than 442971 kJ 439 6 8147771 0 905650024 0 115858942 0 861172246 6 81 0 9056 0 1158 0 8611 N2 O2 H2O CO2 h h h h By extrapolation TP 1530 K 1257C preparation If you are a student using this Manual you are using it without permission 1571 1583 A mixture of hydrogen and the stoichiometric amount of air contained in a constantvolume tank is ignited The final temperature in the tank is to be determined Assumptions 1 The tank is adiabatic 2 Both the reactants and products are ideal gases 3 There are no work interactions 4 Combustion is complete Analysis The combustion equation of H2 with stoichiometric amount of air is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course s W 0 TP H2 AIR 25C 1 atm 2 2 2 2 2 188N H O 376N 05 O H The final temperature in the tank is determined from the energy balance relation E E E in out system for reacting closed systems under adiabatic conditions Q 0 with no work interaction R f R P f P P h h h N P h h h N v v o o o o Since both the reactants and the products behave as ideal gases all the internal energy and enthalpies depend on temperature only and the v P terms in this equation can be replaced by RuT It yields R u f R P u T f P h R T N R T h h h N P o o 298 K since the reactants are at the standard reference temperature of 25C From the tables Substance hf o kJkmol h298 K kJkmol H2 0 8468 O2 0 8682 N2 0 8669 H2O g 241820 9904 Thus 298 8 314 1 88 0 298 4 8 0 50 298 8 314 1 0 8 314 8669 1 88 0 8 314 9904 241820 1 2 2 N H O P P T h T h 31 It yields h h TP H2O N2 kJ 188 2394 259 648 The temperature of the product gases is obtained from a trial and error solution 163 kJ Higher than 259648 kJ 260 2394 3050 1 88 103260 1 139051 2394 1 88 2 2 N H O TP h h At 3050 K 089 kJ Lower than 259648 kJ 255 2394 3000 1 88 101407 1 136264 2394 1 88 2 2 N H O TP h h At 3000 K By interpolation TP 3045 K preparation If you are a student using this Manual you are using it without permission 1572 1584 Methane is burned with 300 percent excess air adiabatically in a constant volume container The final pressure and temperature of product gases are to be determined Assumptions 1 Air and combustion gases are ideal gases 3 Kinetic and potential energies are negligible 4 There are no work interactions 5 The combustion chamber is adiabatic Analysis The combustion equation is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course O 3 3 76 N 4 2 H O CO 376N O 4 CH a a a a a a 3008 N 2 H O e process the energy balance 2 th 2 th 2 2 2 2 th 4 where ath is the stoichiometric coefficient and is determined from the O2 balance 2 3 1 1 4 th th th Substituting 2 2 2 2 4 6 O CO 376N 8 O CH For this constantvolum 2 2 system out in E E E applied on the combustion chamber with Q W 0 duces to re R f R P f P P h h h N P h h h N v v o o o o Since bot reactants an assumed to be ideal gases all the internal ene h the d products are rgy and enthalpies depend on and the v P terms in this equation can be replaced by RuT It yields temperature only R u R P f P R h h N R T h h h N o o o From the hus f h o u T tables Substance o f h kJkmol h29 K 8 ol 4 g 850 kJkm CH 74 O2 0 8682 N2 0 8669 H2O g 241820 9904 CO2 393520 9364 Tp Air CH4 25C 100 kPa T 8 314 8669 3008 0 8 314 8682 0 6 8 314 9904 241820 2 8 314 9364 393520 1 N2 O2 H2O CO2 p p p p T h T h T h T h 298 8 314 3008 298 8 314 8 0 298 8 314 74850 1 It yields 1 04 514 kJ 1 219188 171674 324 9 3008 6 2 T h h h h 7 N2 O2 H2O CO2 p e adiabatic flame temperature is obtained m a trial and error solution A first guess may be obtained by assuming all Th fro the products are nitrogen and using nitrogen enthalpy in the above equation That is 1 047514 kJ 324 9 3908 T h N2 p An investigation of Table A18 shows that this equation is satisfied at a temperature close to 1200 K but it will be somewhat use of the higher specific heat of H2O 985858 324 9 1100 300833426 634899 240071 48258 under it beca At 1100 K Lower than 1047514 kJ t 1200 K 1089662 324 9 1200 0836777 30 638447 244380 53848 A Higher than 1047514 kJ terpolation TP 1159 K he volume of reactants when 1 kmol of fuel is burned is By in T 3 air fuel air fuel 955 8 m 1013 kPa 3808 kmol 8314 kJkmol K298 K 1 P R T N N u V V V The final pressure is then 394 kPa 3 prod m 9558 3908 kmol 8314 kJkmol K1159 K V R T N P u preparation If you are a student using this Manual you are using it without permission 1574 1589 Ethylene gas is burned steadily with 20 percent excess air The temperature of products the entropy generation and the exergy destruction or irreversibility are to be determined Assumptions 1 Combustion is complete 2 Steady operating conditions exist 3 Air and the combustion gases are ideal gases 4 Changes in kinetic and potential energies are negligible Analysis a The fuel is burned completely with the excess air and thus the products will contain only CO2 H2O N2 and some free O2 Considering 1 kmol of C2H4 the combustion equation can be written as 2 th 2 th 2 2 2 2 th 4 2 N 12 376 O 02 2H O 2CO 376N O 12 C H g a a a be determined from the steadyflow energy equation which reduces where ath is the stoichiometric coefficient and is determined from the O2 balance 12 2 1 02 3 th th th a a a Products TP C2H4 25C Air 20 excess air 25C Combustion chamber Thus 2 2 2 2 2 2 4 2 1354N 06O 2H O 2CO 376N 36 O C H g Under steadyflow conditions the exit temperature of the product gases can to C2H4 f f R R P f P o o o o Nh N h h h h N nce all the reactants are at the standard reference state and for O2 and N2 From the tables Substance si hf o kJkmol h298 K mol H4 g 280 kJk C2 52 O2 0 8682 N2 0 8669 H2O g CO2 393520 9364 ubstitutin 241820 9904 S g 1 52280 8669 1354 0 8682 0 60 9904 241820 2 9364 520 93 2 2 2 2 N O H O CO h h h h or 3 2 1 484083 kJ 1354 60 2 2 2 CO h 2 2 2 N O H O h h h T 22696 K By trial and error P b The entropy generation during this adiabatic process is determined from S S S N s N s P R P P R R gen The C in ir an 2H4 is at 25C and 1 atm and thus its absolute entropy is 21983 kJkmolK Table A26 The entropy values listed the ideal gas tables are for 1 atm pressure Both the a d the product gases are at a total pressure of 1 atm but the entropies are to be calculated at the partial pressure of the components which is equal to Pi yi Ptotal where yi is the mole fraction of component i Also m i u i i i i i i y P R T P N s N s T P S ln 0 o PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1575 The entropy calculations can be presented in tabular form as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Ni yi T1atm si o m i u R ln y P N s i i C2H4 1 100 21983 21983 O2 36 021 20514 1298 78487 N2 1354 079 19161 196 262094 SR 362564 kJK CO2 2 01103 316881 18329 67042 H2O 2 01103 271134 18329 57893 O2 06 00331 273467 28336 18108 N2 1354 07464 256541 2432 350649 SP 493692 kJK Thus 131128 kJkmol K 362564 493692 gen R P S S S and c 4 2 4 2 gen 0 destroyed per kmol C H 298 K 131128 kJkmol K C H 390760 kJ T S X preparation If you are a student using this Manual you are using it without permission 1576 1590 Liquid octane is burned steadily with 50 percent excess air The heat transfer rate from the combustion chamber the entropy generation rate and the reversible work and exergy destruction rate are to be determined Assumptions 1 Combustion is complete 2 Steady operating conditions exist 3 Air and the combustion gases are ideal gases 4 Changes in kinetic and potential energies are negligible Analysis a The fuel is burned completely with the excess air and thus the products will contain only CO2 H2O N2 and some free O2 Considering 1 kmol C8H18 the combustion equation can be written as 2 th 2 th 2 2 2 2 th 8 18 N 15 376 O 05 9H O 8CO 376N O 15 C H a a a l where ath is the stoichiometric coefficient and is determined from the O2 balance 15 8 45 05 125 th th th a a a Thus 2 2 2 2 2 2 8 18 705N 625O 9H O 8CO 376N 1875 O C H l Under steadyflow conditions the energy balance E E Ein out system applied on the combustion chamber with W 0 reduces to o o o o o o f R R f P P R f R P f P N h N h h h h N h h h N Q out si F nce all of the reactants are at 25C Assuming the air and the combustion products to be ideal gases we have h hT rom the tables Products 25C C8H18 l 25C Air 50 excess air 25C Combustion chamber T0 298 K Q Substance o f h kJkmol H18 l 950 C8 249 O2 0 N2 0 H2O l 285830 2 393520 Substituting CO 5 470680 kJkmol of C8H18 0 0 249950 1 0 0 out 285830 9 393520 8 Q or The C8H18 is burned at a rate of 025 kgmin or 18 8 out 5 470680 kJkmol of C H Q Thus 11997 kJmin ol kmolmin 2193 10 18 1 kgkmol 8 12 025 kgmin 3 M m N f combustion of liquid C8H18 which could easily be de etermined from Table A27 to be kmolmin 5470680 kJkm 10 2193 3 out out NQ Q The heat transfer for this process is also equivalent to the enthalpy o hC d 5470740 kJkmol C8H18 b The entropy generation during this process is determined from PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course surr out gen surr out gen T Q N s N s S T Q S S S R R P P R P preparation If you are a student using this Manual you are using it without permission 1577 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course he C8H18 is at 25C and 1 atm and thus its absolute en py is sC8H18 T tro 36079 kJkmolK Table A26 The entropy the air and the product gases are at a total pressure of 1 atm the components which is equal to Pi yi Ptotal where yi is the ole fraction of component i Also values listed in the ideal gas tables are for 1 atm pressure Both but the entropies are to be calculated at the partial pressure of m m i u i i i i i i P R T N s N s T S ln y 0 o The e calcula s can be p ed in tab as P P ntropy tion resent ular form Ni yi T1atm si o m i Ruln y P Nisi C8H18 1 100 36079 36079 O2 1875 50 6 69 R 180 K 021 20514 1298 408975 N2 70 079 19161 19 13646 S 9723 kJ CO2 8 00944 21380 1962 18673 H2O l 9 6992 O2 625 00737 20504 2168 14176 2 7050 08319 19161 153 136163 SP 17531 kJK 6293 N Thus nd 3903 kJmin K 17798 kJk 5470523 kJ 17531 18097 surr gen Q S S S R P a kmolmin 17798 kJkmol K 10 2193 mol K K 298 3 gen gen surr NS S T c The exergy destruction rate associated with this process is determined from 1939 kW 11632 kJmin 298 K 3903 kJmin K gen 0 destroyed T S X preparation If you are a student using this Manual you are using it without permission 1578 1591E Benzene gas is burned steadily with 90 percent theoretical air The heat transfer rate from the combustion chamber and the exergy destruction are to be determined Assumptions 1 Steady operating conditions exist 2 Air and the combustion gases are ideal gases 3 Changes in kinetic and potential energies are negligible Analysis a The fuel is burned with insufficient amount of air and thus the products will contain some CO as well as CO2 H2O and N2 The theoretical combustion equation of C6H6 is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 6 15 75 Products 1900 R C6H6 77F Air 90 theoretical 77F Combustion chamber Q 2 th 2 2 2 2 th 6 6 N 376 3H O 6CO 376N O C H a a where ath is the stoichiometric coefficient and is determined from the O2 balance th a Then the actual combustion equation can be written as 2 2 2 2 2 6 6 2538N 3H O CO 6 CO 376N 090 75 O C H x x The value of x is determined from an O2 balance Thus 2 2 2 2 2 6 6 2538N 3H O 51 CO 45CO 376N 76 5 O H C 45 15 2 6 75 x x x 090 Under steadyflow conditions the energy balance E E E in out system applied on the combustion chamber with W 0 reduces to o o o o o o o f R R P f P R f R P f P N h h h h N h h h N h h h N Q out since all of the reactants are at 77F Assuming the air and the combustion products to be ideal gases we have h hT From the bstance tables Su o f h ulbmol Bt h537 R Btulbmol 1 00R h 9 Btulbmol C6H6 g 35680 O2 0 37251 14322 N2 0 37295 13742 H2O g 104040 42580 16428 CO 47540 37251 13850 2 169300 40275 19698 hus CO T 804630 Btulbmol of C6H6 0 0 1 35680 3729 5 1 742 2538 0 4258 16428 104040 3 b The entropy generation during this process is determined from 3 3725 1 13850 47540 51 4027 5 19698 169300 54 out Q surr surr out out gen T N s N s T Q S S S R R P P R P Q preparation If you are a student using this Manual you are using it without permission 1579 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course The C H6 is at 77F and 1 atm and thus its absolu entropy is sC H 6 6 6 te 6434 B lbmo R Table A26E The entropy values listed in the id gas tables are for 1 atm p re Both th product at a total pressure of 1 atm but th pies are be calculated at the partial pressure of the c ponents which to Pi yi Ptotal where yi is the mole fraction of com nt i Al tu l eal to ressu e air and the om gases are is equal e entro pone so yi Pm u i i i i i R N s N s S ln The entropy calculations can be presented in tabular form as T iP T P 0 o Ni yi T1atm s o i m i u R ln y P Nisi C6H6 1 100 6434 6434 O2 N2 2538 079 4577 047 SR 158959 BtuR 675 021 4900 310 35168 117357 CO2 45 01309 64999 4038 31067 CO 15 00436 56509 6222 9410 56097 4843 18282 2 54896 0603 140856 SP 199615 BtuR H2O g 3 00873 N 2538 07382 Thus 1904 9 Btu R 537 804630 158959 199615 surr out gen T Q S S S R P Then the exergy destroyed is determined from 6 6 gen 0 destroyed per lbmol C H 537 R 19049 Btulbmol R 1022950 BtuR T S X preparation If you are a student using this Manual you are using it without permission 1580 1592 Liquid propane is burned steadily with 150 percent excess air The mass flow rate of air the heat transfer rate from the combustion chamber and the rate of entropy generation are to be determined Assumptions 1 Combustion is complete 2 Steady operating conditions exist 3 Air and the combustion gases are ideal gases 4 Changes in kinetic and potential energies are negligible Properties The molar masses of C3H8 and air are 44 kgkmol and 29 kgkmol respectively Table A1 Analysis a The fuel is burned completely with the excess air and thus the products will contain only CO2 H2O N2 and some free O2 Considering 1 kmol of C3H8 the combustion equation can be written as 2 th 2 th 2 2 2 2 th 3 8 N 25 376 O 15 4H O 3CO 376N O 25 C H a a a l where a is the stoichiometric coefficient and is determined from the O balance th 2 47N 75O 4H O The airfuel ratio for this combustion process is 25 3 2 15 5 th th th a a a Substituting 2 2 3 8 3CO 376N 125 O C H l 2 2 2 2 392 kg airkg fuel 2 kgkmol 4 kmol 3 kmol 12 kgkmol 125 476 kmol 29 kgkmol AF fuel air m m Thus kg airmin 392 kg airkg fuel 04 AF fuel air 157 kg fuelmin m m b Under steadyflow conditions the energy balance E E E in out system applied on the combustion chamber with W 0 reduces to R f R P f P h h h N h h h N Q o o o o out Assuming the air and the combustion products to be ideal gases we have h hT From the tables The h f o of liquid propane is btained by adding the hfg at 25C to o h f o of gaseous propane e Substanc hf o kJkmol h285 K l kJkmo h298 K ol kJkm h1200 K l H8 l 8910 kJkmo C3 11 O2 0 82965 65 H2O g 241820 9904 44380 CO2 393520 9364 53848 hus 8682 38447 N2 0 828 8669 36777 T 298 298 out 118910 1 8669 36777 47 0 8682 38447 0 57 9904 44380 241820 4 9364 53848 393520 3 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 190464 kJkmol of C3H8 8669 8286 5 47 0 8682 8296 5 0 12 5 h h Q s transferred from the combustion chamber for each kmol 44 kg of propane This corresponds to 904644 3287 kJ of heat transfer per kg of propane Then the rate of heat transfer for a mass flow rate of 04 kgmin Thus 190464 kJ of heat i 1 4 4 for the propane becomes preparation If you are a student using this Manual you are using it without permission 1581 1732 kJmin 04 kgmin 43287 kJkg out out mq Q PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course c The entropy generation during this process is determined from S S S Q T Q P R gen out N s N s T P P R R surr out surr he C3H8 is at 25C and 1 atm and thus its absolute entropy for the gas phase is 26991 C3H8 s k T JkmolK Table A26 Then the entropy of C3H8l is obtained from 219 4 kJkmol K 15 298 15060 26991 g g 8 3 8 3 8 3 C H C H C H T h s s s s fg fg l The entropy values listed in the ideal gas tables are for 1 atm pressure Both the air and the product gases are at a total partial pressure of the components which is equal to Pi yi total where yi is the mole fraction of component i Then pressure of 1 atm but the entropies are to be calculated at the P m i u i i i i i i y P R P N s N s T P S ln 0 o The e r calculat n be presented in tabul as T nt opy ions ca ar form Ni yi T1atm si o m i Ruln y P N s i i C3H8 1 21940 21940 O2 125 SR 8 kJK 021 20370 1298 270850 N2 47 079 19018 196 903058 119584 CO2 3 00488 279307 25112 91326 H2O g 4 00650 240333 22720 O2 75 01220 249906 17494 200550 2 47 07642 234115 2236 1110850 SP 1507947 kJK 105221 N Thus 8 3 surr out gen 3760 1 kJ K per kmol C H 298 190464 1195848 1507947 T Q S S S R P Then the rate of entropy generation becomes 34 2 kJmin K kJkmol K 44 kmolmin 37601 04 gen gen N S S preparation If you are a student using this Manual you are using it without permission 1582 1593 Problem 1592 is reconsidered The effect of the surroundings temperature on the rate of exergy destruction is nalysis The problem is solved using EES and the solution is given below inConvertkgmin kgs olC3H8 2O Ath 376 N2 O2 4 H2O 1Ex Ath 376 N2 Ex Ath O2 r1MwC3H8 kgairkgfuel combustion process per kilomole of fuel t DELTAEcv s the enthalpy of the liquid fuel TTfuel hfgfuel 1ExAthenthalpyO2TTair1ExAth376 d1ExAth376 2TTprod cess and heat rejection to the surroundings to be studied A Fuel Propane C3 Tfuel 25 27315 H8liq K Pfuel 1013 kPa mdotfuel 04 kgm Ex 15 Excess air Pair 1013 kPa Tair 1227315 K Tprod 1200 K Pprod 1013 kPa Mwair 2897 lbmlbmolair MwC3H831281 kgkm TsurrC 25 C Tsurr TsurrC27315 K For theoretical dry air the complete combustion equation is C3H8 AthO2376 N23 CO24 H 2Ath3241theoretical O balance The balanced combustion equation with Ex100 excess moist air is C3H8 1EXAthO2376 N23 C The airfuel ratio on a mass basis is AF 1ExAth476Mwai The air mass flow rate is mdotair mdotfuel AF t Law SSSF to the Apply Firs Ein Eou Ein HR Since EES gives the enthalpy of gasesous components we adjust the EES calculated enthalpy to get the liquid enthalpy Subtracting the enthalpy of vaporization from the gaseous enthalpy give hfuelliq hfuelgas hfgfuel hfgfuel 15060 kJkmol from Table A27 HR 1enthalpyC3H8 enthalpyN2TTair Eout HP Qout HP3enthalpyCO2TTprod4enthalpyH2OTTpro enthalpyN2TTprodExAthenthalpyO DELTAEcv 0 Steadyflow requirement The heat transfer rate from the combustion chamber is QdotoutQoutkJkmolfuelMwC3H8 kgkmolfuelmdotfuelkgs kW Entopy Generation due to the combustion pro Entopy of the reactants per kilomole of fuel PO2reac 1476Pair Daltons law of partial pressures for O2 in air sO2reacentropyO2TTairPPO2reac PN2reac 376476Pair Daltons law of partial pressures for N2 in air N2reac sN2reacentropyN2TTairPP PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1583 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course TTfuelPPfuel sfgfuel Adjust the EES gaseous value by sfg fgfuel hfgfuelTfuel reac 1ExAth376sN2reac e of fuel l kmol of products p oducts H2Oprod 4NprodPprod Patrial pressure H2O in products ExAthsO2prod t rejected to the surroundings per kilomole fuel e entropy of the surroundings is dotgen SP SR SsurrkJkmolfuelMwC3H8 kgkmolfuelmdotfuelkgs kWK dotdest TsurrSd Ts C sC3H8reacentropyC3H8 For phase change sfg is given by s SR 1sC3H8reac 1ExAthsO2 Entopy of the products per kiloml By Daltons law the partial pressures of the product gases is the product of the mole fraction and Pprod Nprod 3 4 1ExAth376 ExAth tota PO2prod ExAthNprodPprod Patrial pressure O2 in products sO2prodentropyO2TTprodPPO2prod PN2prod 1ExAth376N rodPprod Patrial pressure N2 in pr sN2prodentropyN2TTprodPPN2prod PCO2prod 3NprodPprod Patrial pressure CO2 in products sCO2prodentropyCO2 TTprodPPCO2prod P sH2OprodentropyH2O TTprodPPH2Oprod SP 3sCO2prod 4sH2Oprod 1ExAth376sN2prod Since Qout is the hea th Ssurr QoutTsurr Rate of entropy generation S X otgenkW urrC Xdest kW 0 1578 4 1597 8 1616 12 1635 16 1654 20 1673 24 1692 28 1711 32 173 36 1749 38 1758 0 5 10 15 20 25 30 35 40 1575 1615 1655 1695 1735 1775 TsurrC C Xdest kW preparation If you are a student using this Manual you are using it without permission 1584 1594 Liquid octane is burned steadily with 70 percent excess air The entropy generation and exergy destruction per unit mass of the fuel are to be determined Assumptions 1 Combustion is complete 2 Steady operating conditions exist 3 Air and the combustion gases are ideal gases 4 Changes in kinetic and potential energies are negligible Properties The molar masses of C8H18 and air are 114 kgkmol and 29 kgkmol respectively Table A1 Analysis The fuel is burned completely with the excess air and thus the products will contain only CO2 H2O N2 and some free O2 Considering 1 kmol C8H18 the combustion equation can be written as 2 th 2 th 2 2 2 2 th 18 8 N 17 376 O 70 9H O 8CO 376N O 71 C H a a a l where ath is the stoichiometric coefficient and is determined from the O2 balance Products 1500 K C8H18 l 25C Air 70 excess air 600 K Combustion chamber 125 70 45 8 17 th th th a a a Thus 2 2 2 2 2 2 18 8 79 9 N 8 75O 9H O 8CO 376N 2125 O C H l b Under steadyflow conditions the energy balance system out in E E E applied on the combustion chamber with W 0 duces to re PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course R f R P f P h h h N h h h N Q o o o o out Assuming the air and the combustion products to be ideal gases we have h hT From the tables Substance o h f kJkmol h298K kJkmol K h 600 mol kJk 1 00K h 5 kJkmol 18 l 9950 C H 8 24 O2 0 8682 17929 49292 N2 H 0 8669 17563 g CO2 393520 9364 71078 hus 47073 2O 241820 9904 57999 T out 8669 17563 0 79 9 8682 17929 2125 0 249950 1 8669 47073 0 79 9 8682 49292 8 75 0 9904 999 57 241820 9 9364 71078 393520 8 Q 631335 kJkmol of C8H18 1 The entropy generation during this process is determined from S S S Q Q T N s N s T P R P P R R n out surr out surr The entropy values listed in the ideal gas tables are for 1 atm pressure Both the air and the product gases are at a total pressure of Pm 600 kPa 600101325592 atm but the entropies are to be calculated at the partial pressure of the components which is equal to Pi yi Ptotal where yi is the mole fraction of component i Then ge m i u i i i i i i y P R T P N s N s T P S ln 0 o preparation If you are a student using this Manual you are using it without permission 1585 The entropy calculations can be presented in tabular form as Ni yi T1atm si o m i u R ln y P Nisi C8H18 1 46673 1479 45194 O2 2125 021 22635 181 477148 N2 799 079 21207 1283 1591928 SR 2114270 kJK CO2 8 00757 29211 6673 239026 H2O g 9 00852 25045 5690 230526 O2 875 00828 25797 5928 230911 N2 799 07563 24177 1246 1832187 SP 2532650 kJK Thus 18 8 surr out gen R P 9658 1 kJK per kmol C H 298 1 631335 2114270 2532650 T Q S S S The exergy destruction is 18 8 gen 0 dest 2 878114 kJK per kmol C 2989658 1 kJK H T S X ation and exergy destruction per unit mass of the fuel are The entropy gener 8472 kJK kg C8H18 114 kgkmol 1 kJK kmol 9658 fuel gen gen M S S 25250 kJkg C8H18 114 kgkmol 878114 kJK kmol 2 fuel dest dest M X X PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1586 1595 Methyl alcohol is burned steadily with 200 percent excess air in an automobile engine The maximum amount of work that can be produced by this engine is to be determined Assumptions 1 Combustion is complete 2 Steady operating conditions exist 3 Air and the combustion gases are ideal gases 4 Changes in kinetic and potential energies are negligible Analysis The fuel is burned completely with the excess air and thus the products will contain only CO2 H2O N2 and some free O2 Considering 1 kmol CH3OH the combustion equation can be written as 2 th 2 th 2 2 2 2 th 3 3 76N 3 O 2 2H O CO 376N O 3 CH OH a a a PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2 2 16 3O 2H O yflow conditions the energy balance where ath is the stoichiometric coefficient and is determined from the O2 balance 15 2 1 1 3 05 th th th a a a Thus 2 2 2 2 3 N CO 376N O 54 CH OH Under stead Products 77C CH3OH 25C 200 excess air 25C Combustion Chamber 1 atm Qout 92 system out in E E E applied on the combustion chamber with W 0 reduces to R f R P f P h h h N h h h N Q o o o o out Assuming the air and the combustion products to be ideal gases we have h hT From the tables e Substanc o f h kJkmol h298K kJkmol h350K kJkmol 3OH 0670 CH 20 O2 0 8682 10213 N2 0 8669 10180 H2O g CO2 393520 9364 11351 hus 241820 9904 11652 T 663550 kJkmol of fuel 200670 1 8669 10180 1692 0 8682 10213 3 0 9904 11652 241820 2 9364 11351 393520 1 out Q The entropy generation during this process is determined from surr out out gen T Q N s N s T Q S S S R R P P R P surr ntropy values listed in the ideal gas tables are for 1 atm pressure Both the air and the product gases are at a total pressure of 1 atm but the entropies are to be calculated at the partial pressure of the components which is equal to Pi yi Ptotal where yi is the mole fraction of component i Then The e m i u i i i i i i y P R T P N s N s T P S ln 0 o preparation If you are a student using this Manual you are using it without permission 1587 The entropy calculations can be presented in tabular form as Ni yi T1atm si o m i u R ln y P Nisi CH3OH 1 23970 23970 O2 45 021 20504 1298 98109 N2 1692 079 19161 1960 327520 SR 4496 kJK CO2 1 00436 219831 2605 24588 H2O g 2 00873 194125 2027 42879 O2 3 01309 209765 1691 68003 N2 1692 07382 196173 252 336189 SP 4717 kJK Thus 2448 kJK per kmol fuel 298 surr gen T R P 663550 4496 4717 out Q S S S The maximum work is equal to the exergy destruction 729400 kJK per kmol fuel 2982448 kJK gen 0 de max st T S a X W Per unit m ss basis 22794 kJkg fuel 32 kgkmol 729400 kJK kmol Wmax PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 1588 Review Problems 1596 A sample of a certain fluid is burned in a bomb calorimeter The heating value of the fuel is to be determined Properties The specific heat of water is 418 kJkgC Table A3 Analysis We take the water as the system which is a closed system for which the energy balance on the system with W 0 can be written as E E E in out system 1 g WATER 2 kg T 25C Reaction chamber Fuel Qin U or 90 kJ per gram of fuel 20 C 25 C kg 418 kJkg 2 in mc T Q Therefore heat transfer per kg of the fuel would be 20900 kJkg fuel Disregarding the slight energy stored in the gases of the combustion chamber this value corresponds to the heating value of the fuel 1589 1597E Hydrogen is burned with 100 percent excess air The AF ratio and the volume flow rate of air are to be determined Assumptions 1 Combustion is complete 2 Air and the combustion gases are ideal gases Properties The molar masses of H2 and air are 2 kgkmol and 29 kgkmol respectively Table A1 Analysis a The combustion is complete and thus products will contain only H2O O2 and N2 The moisture in the air does not react with anything it simply shows up as additional H2O in the products Therefore for simplicity we will balance the combustion equation using dry air and then add the moisture to both sides of the equation The combustion equation in this case can be written as 2 th 2 th 2 2 2 th 2 N 2 376 O H O 376N O 2 H a a a 5 he partial pressure of the water vapor present in the incoming r is Products H2 Air 90F Q Combustion chamber P 145 psia where ath is the stoichiometric coefficient for air It is determined from O2 balance 2 05 0 th th th a a a Substituting 2 2 2 2 2 2 376N 05O H O 376N O H Therefore 476 lbmol of dry air will be used per kmol of the fuel T ai 0419 psia 0 60 069904 psi sat90 F air in P Pv φ T he number of m oles of the moisture that accompanies 476 lbmol of incoming dry air Nv in is determined to be PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 142 lbmol 4 76 0419 psia total total in in v v N N N P P N 145 psia in in v v he balanced combustion equation is obtained by substituting the coefficients termined earlier and adding 0142 lbmol of 376N 05O 1142H O 0142H O 376N O H T de H2O to both sides of the equation 2 2 2 2 2 2 2 The airfuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel 703 lbm airlbmfuel 2 lbmlbmol 1 lbmol 0142 lbmol 18 lbmlbmol 476 lbmol 29 lbmlbmol AF fuel air m m b The mass flow rate of H2 is given to be 10 lbmh Since we need 703 lbm air per lbm of H2 the required mass flow rate f air is o 1758 lbmh 25 lbmh 70 3 AF fuel air m m The mole fractions of water vapor and the dry air in the incoming air are 0 971 0 029 1 0 029 and 0 142 4 76 0 142 dryair total H O H O 2 2 y N N y Thus 24928 ft h 3 lbmh 1418 ft lbm 1758 1418 ft lbm 145 psia psia ft lbm R 550 R 1073287 28 7 lbmlbmol 0 971 29 029 18 0 3 3 3 dryair O H2 v V v m P RT yM yM M preparation If you are a student using this Manual you are using it without permission 1590 1598 A gaseous fuel with a known composition is burned with dry air and the volumetric analysis of products gases is determined The AF ratio the percent theoretical air used and the volume flow rate of air are to be determined Assumptions 1 Combustion is complete 2 Air and the combustion gases are ideal gases Properties The molar masses of C H2 N2 O2 and air are 12 2 28 32 and 29 kgkmol respectively Table A1 Analysis Considering 100 kmol of dry products the combustion equation can be written as H O 8164N 1491O 009CO PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 336CO 2 The unknown coefficients 376N O 010O 025N 2 2 2 2 2 2 2 4 b a x x a and b are determined from mass balances 2 a a x b b x x x alance is not completely satisfied Yet we solve the problem with an a value of 2136 being aware f this situ ion Then 065CH 2136 8164 3 76 0 25 N 6 90 2 4 0 65 H 5 31 0 09 3 36 0 65 C 2123 2 1491 0 045 3 36 0 10 Check O 2 a b a x The N2 balance and O2 balance gives two different a values There must be a small error in the volumetric analysis of the products and the mass b o at 376N 2136 O 010O 025N 531 065CH 2 2 2 2 4 69H O 8164N 1491O 009CO 336CO 2 2 2 2 The combustion equation for 1 kmol of fuel is obtained by dividing the above equation by 531 13H O 1537N 281O 0017CO 0633CO 376N 402 O 010O 025N 065CH 2 2 2 4 2 2 2 2 2 a The airfuel ratio is determined from its definition 269 kg airkg fuel 32 0 10 28 0 25 0 65 16 476 kmol 29 kgkmol 402 AF fuel air m m b To find the percent theoretical air used we need to know the theoretical amount of air which is determined from the theoretical ombustion equation of the fuel c 37 025 13H O 065CO 376N O 010O 025N 065CH 2 2 2 2 th 2 2 4 a 6 th th 2 2 th a a a hen 12 065 065 010 O N 335 3 35 12 476 kmol 402 476 kmol Percent theoretica l air th air act air th air airact N N m m T c The specific volume mass flow rate and the volume flow rate of air at the inlet conditions are 805 m min 3 kgmin 0855 m kg 9415 min 94 15 m 269 kg airkg fuel 35 kg fuelmin AF 0855 m kg 100 kPa kPa m kg K 298 K 0287 3 air air 3 fuel air 3 3 v V v m m m P RT 336 CO2 009 CO 1491 O2 8164 N2 65 CH4 25 N2 10 O2 Air Combustion chamber preparation If you are a student using this Manual you are using it without permission 1591 1599E Propane is burned with stoichiometric amount of air The fraction of the water in the products that is vapor is to be determined Assumptions 1 Combustion is complete 2 Steady operating conditions exist 3 Air and the combustion gases are ideal gases Analysis The fuel is burned completely with the air and thus the products will contain only CO2 H2O and N2 Considering 1 kmol C3H8 the combustion equation can be written as 2 2 2 2 2 8 3 18 8 N 4H O 3CO 376N 5 O C H C3H8 Theoretical air Combustion chamber 1 atm CO2 H2O N2 120F The mole fraction of water in the products is 0 1550 18 8 kmol 4 3 kmol 4 prod H2O N N y The saturation pressure for the water vapor is psia 1 6951 sat120 F v P P When the combustion gases are saturated the mole fraction of the water vapor will be 0 1153 696 kPa 14 1 6951 kPa P P y v hus the action of water vapor in the combustion products is g T fr 0744 0 1550 0 1153 vapor y y f g PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1592 15100 Coal whose mass percentages are specified is burned with 20 excess air The dewpoint temperature of the products is to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 CO H2O SO2 and N2 3 Combustion gases are ideal gases Properties The molar masses of C H2 O2 S and air are 12 2 32 32 and 29 kgkmol respectively Table A1 Analysis We consider 100 kg of coal for simplicity Noting that the mass percentages in this case correspond to the masses of the constituents the mole numbers of the constituent of the coal are determined to be 0 04406 kmol 32 kgkmol kg 141 0 03893 kmol 28 kgkmol kg 109 0 7909 kmol 32 kgkmol kg 2531 2 895 kmol 2 kgkmol kg 579 5 117 kmol 12 kgkmol kg 6140 S S S N2 N2 N2 O2 O2 O2 H2 H2 H2 C C C M m N M m N M m N M m N M m N 6140 C 579 H2 2531 O2 109 N2 141 S 500 ash by mass Air 20 excess Products Coal Combustion chamber The mole number of the mixture and the mole fractions are 8 886 kmol 0 04406 0 03893 0 7909 2 895 5 117 Nm 000496 8886 kmol kmol 004406 000438 8886 kmol kmol 003893 00890 8886 kmol kmol 07909 03258 8886 kmol kmol 2895 0 5758 8886 kmol kmol 5117 S S N2 N2 O2 O2 H2 H2 C C m m m m m N N y N N y N N y N N y N N y Ash consists of the noncombustible matter in coal Therefore the mass of ash content that enters the combustion chamber is equal to the mass content that leaves Disregarding this nonreacting component for simplicity the combustion equation may be written as th A PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2 2 th 2 2 2 2 2 th 2 2 3 76N 1 25 O 0 25 0 00496SO 0 3258H O 0 5758CO 3 76N O 1 25 0 00496S 0 00438N 0 0890O 03258H 05758C a a a ccording to the oxygen balance th th th a a a 3 077N 0 1637O 0 00496SO 0 3258H O 0 5758CO 3 76N 0 8184O 0 00496S 00438N The dewpoint temperature of a gasvapor mixture is the saturation temperature of the water vapor in the product gases hat is 2 O balance 0 6547 0 25 0 00496 0 3258 50 0 5758 1 25 0 0890 2 Substituting 2 2 0 0 0890O 03258H 05758C 2 2 2 2 2 2 2 2 corresponding to its partial pressure T 7 96 kPa 3 077 kmol 1013 kPa 01637 000496 03258 05758 03258 kmol prod prod P N N P v v Thus Table A5 413C sat 7 96 kPa dp T T preparation If you are a student using this Manual you are using it without permission 1593 15101 Methane is burned steadily with 50 percent excess air The dewpoint temperature of the water vapor in the products is to be determined Assumptions 1 Combustion is complete 2 Steady operating conditions exist 3 Air and the combustion gases are ideal gases Properties The molar masses of CH4 and air are 16 kgkmol and 29 kgkmol respectively Table A1 Analysis The fuel is burned completely with the excess air and thus the products will contain only CO2 H2O N2 and some free O2 Considering 1 kmol CH4 the combustion equation can be written as 2 th 2 th 2 2 2 2 th 4 N 15 376 O 50 2H O CO 376N O 51 CH a a a where ath is the stoichiometric coefficient and is determined from the O2 balance Products Tdp CH4 Air 50 excess Combustion chamber 2 50 1 1 15 th th th a a a Thus 2 2 2 2 2 2 4 1128N O 2H O CO 376N 3 O CH The dewpoint temperature of a gasvapor mixture is the saturation temperature of the water vapor in the product gases corresponding to its partial pressure That is 1326 kPa 2 1 1128 kmol 101325 kPa 1 2 kmol prod prod P N N Pv v Thus 514C sat1326 kPa dp T T from EES PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1594 15102 A mixture of 40 by volume methane CH4 and 60 by volume propane C3H8 is burned completely with theoretical air The amount of water formed during combustion process that will be condensed is to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O and N2 only Products 100ºC 40 CH4 60 C3H8 Air 100 theoretical Properties The molar masses of C H2 O2 and air are 12 kgkmol 2 kgkmol 32 kgkmol and 29 kgkmol respectively Table A1 Analysis The combustion equation in this case can be written as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 D B a 4 2 2 2 2 2 th 8 3 4 N H O CO 376N O C H 60 CH F where ath is the stoichiometric coefficient for air The coefficient ath and other coefficients are to be determined from the mass balances Carbon balance 22 60 3 40 B Hydrogen balance Oxygen balance F F hen we write the balanced reaction equation as 1429 N H O 23 CO 22 76N he vapor mole fraction in the products is 23 2 60 8 40 4 2 D D D 83 23 2 22 2 2 2 th th th a a D B a Nitrogen balance 3 76 3 76 th F a 1429 83 T 2 8 3 4 3 O 83 C H 60 CH 40 2 2 2 2 T 0 1625 23 y 1429 23 22 v The partial pressure of water in the products is 1625 kPa 0 1625100 kPa prod vprod y P P v The partial pressure of the water vapor remaining in oducts at the The dew point temperature of the products is 5564 C T T 0 20 40 60 80 100 120 140 160 180 200 0 100 200 300 400 s kJkmolK T C 1625 kPa 6997 kPa Steam 1 2 3 sat 1625 kPa dp the products at the product temperature is sat 39 C P P kPa 07 v The kmol of water vapor in the pr product temperature is 241 kmol 1 1429 22 kPa 07 prod product total v v v v v N N N P N N P The kmol of water condensed is 196 kmol waterkmo l fuel 1 241 23 Nw preparation If you are a student using this Manual you are using it without permission 1595 15103 A gaseous fuel mixture of 60 propane C3H8 and 40 butane C4H10 on a volume basis is burned with an airfuel ratio of 25 The moles of nitrogen in the air supplied to the combustion process the moles of water formed in the combustion process and the moles of oxygen in the product gases are to be determined Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O and N2 only Properties The molar masses of C H2 O2 and air are 12 kgkmol 2 kgkmol 32 kgkmol and 29 kgkmol respectively Table A1 Analysis The theoretical combustion equation in this case can be written as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 D B a 2 2 2 2 2 th 10 4 8 3 N H O CO 376N O C H 40 6 C H F where ath is the stoichiometric coefficient for air The coefficient ath and other coefficients are to be determined from the mass balances Carbon balance 43 40 4 60 3 B Hydrogen balance F F hen we write the balanced theoretical reaction equation as 2106 N H O 44 he airfu ratio for the theoretical reaction is determined from 44 2 40 10 60 8 D D 65 44 43 2 2 2 2 th th th a a D B a Oxygen balance Nitrogen balance 3 76 3 76 th F a 2106 65 T 2 2 10 4 8 3 CO 43 376N O 65 C H 40 C H 60 2 2 2 T el 1559 kg airkg fuel 58 kg 40 44 60 4 76 kmol29 kgkmol 65 AF air m fuel th m The percent theoretical air is 160 4 100 1559 25 AF AF PercentTH actual th air The moles of nitrogen supplied is per kmol fuel 100 65 3 76 160 4 3 76 100 th N2 PercentTH N air 338 kmol a he moles of water formed in the combustion process is The moles of oxygen in the product gases is T per kmol fuel N H2O D 44 kmol per kmol fuel 1 65 100 160 4 1 100 PercentTH N th air O2 338 kmol a Products 60 C3H8 40 C4H10 Air preparation If you are a student using this Manual you are using it without permission 1596 15104 Ethane is completely burned with air Various parameters are to be determined for the given reaction Assumptions The water in the products is in the vapor phase Analysis a The reaction equation is given as 2 2 2 2 2 2 6 2 18N 1 288O 3H O 2CO 376N 4 788 O C H The partial pressure of water vapor is 1235 kPa 24288 kmol 100 kPa 18 100 kPa 1 288 3 2 total total P N Pv 3 kmol 3 H2O N The dew point temperature of the pro e saturation temperature of water at this pressure Table A5 duct gases is th 496C Tsat1235 kPa Tdp b The partial p ressure of oxygen is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 5 303 kPa 288 kmol 100 kPa 24 1 288 kmo NO2 l total total O2 P N P The specific volume of oxygen is then 183 m kg 3 0 2598 kJkmol K373 K T R O2 v 2 5 3 76N oth the reactants and the products are taken to be at the standard reference state of 25C and atm for the calculation of of combustion Note that N2 and O2 are stable elements and thus their enthalpy of formation is zero Then Products C2H6 Air Combustion Chamber 100 kPa 5 303 kPa O2 P c The combustion reaction with stoichiometric air is 2 2 2 2 6 2 3 3H O 2CO 376N O 53 C H O 2 B 1 heating values The heat transfer for this process is equal to enthalpy Nhfo C2H6 H2O CO2 o o o o f f f R R f P P R P C Nh Nh N h N h H H h q or the LHV the water in the products is taken to b vapor Then ol F e 427820 kJkmol ethane 1 1 kmol 84680 kJkmol 3km 241820 kJkmol 2 kmol 393520 kJkmol hC stion d The average molar mass of the product gas is The LHV per unit kmol of the fuel is the negative of the enthalpy of combu 1427820 kJkmol C2H6 hC LHV 2829 kgkmol 288 kmol 24 687 2 kmol 288 kmol 24 18 kmol28 kgkmol 1 288 kmol32 kgkmol 3 kmol18 kgkmol kmol total N2 N2 O2 O2 H2O H2O CO2 CO2 N M N M N M N M N M e The average molar constant pressure specific heat of the product gas is 2 kmol44 kg 3091 kJkmol K 24288 kmol 750 8 kJK 24288 kmol 1 288 3014 18 2927 kJK 3 3428 2 4116 total N2 N2 O2 O2 H2O H2O CO2 CO2 N c N c N c N c N c p p p p p f The airfuel mass ratio is 2203 kg airkg fuel 30 kg 6609 kg 1 30 kg 29 kg 4 76 4788 AF fuel air m m g For a molar fuel flow rate is 01 kmolmin the mass flow rate of water in the product gases is 54 kgmin 1kmol 18 kgkmol kmolmin 3 kmol 10 H2O C2H6 H2O C2H6 H2O M N N N m preparation If you are a student using this Manual you are using it without permission 1597 15105 CO gas is burned with air during a steadyflow combustion process The rate of heat transfer from the combustion chamber is to be determined Assumptions 1 Steady operating conditions exist 2 Air and combustion gases are ideal gases 3 Kinetic and potential energies are negligible 4 There are no work interactions 5 Combustion is complete Properties The molar masses of CO and air are 28 kgkmol and 29 kgkmol respectively Table A1 Analysis We first need to calculate the amount of air used per kmol of CO before we can write the combustion equation PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0478 kgmin m kg 0836 m min 04 0836 m kg kPa kPa m kg K 310 K 02968 3 3 CO CO CO 3 3 v V m RT 110 CO v P Products 900 K CO 37C Air Then the molar airfuel ratio becomes 303 kmol airkmol fuel 0478 kgmin 28 kgkmol 15 kgmin 29 kgkmol AF fuel fuel air air fuel air M m M m N N 25C Q Combustion chamber Thus the number of moles of O2 used per mole of C 37 Then the combustion equation in this case can be written as O is 303476 06 240N 0137O CO 376N 0637 O CO flow conditions the energy balance 2 2 2 2 2 E E E Under steady system out in applied on the combustion chamber with W 0 o reduces t N Qout R f P h N h h h o o o o Assuming the air and the combustion products to be ideal gases we hT From the tables tance f R h h P have h Subs hf o kJkmol h298 K ol kJkm h310 K mol kJk h900 K l CO 110530 8669 9014 27066 kJkmo O2 0 8682 27928 N2 0 8669 26890 CO2 393520 9364 37405 Thus 208927 kJkmol of CO 0 0 8669 9014 110530 1 8669 26890 0 42 8682 27928 0 137 0 9364 37405 393520 1 out Q Then the rate of heat transfer for a mass flow rate of 0956 kgmin for CO becomes 3567 kJmin 208927 kJkmol 28 kgkmol 0478 kgmin out out out N Q m NQ Q preparation If you are a student using this Manual you are using it without permission 1598 15106 Ethanol gas is burned with 10 excess air The combustion is incomplete The theoretical kmols of oxygen in the reactants the balanced chemical reaction and the rate of heat transfer are to be determined Assumptions 1 Combustion is incomplete 2 The combustion products contain CO2 CO H2O O2 and N2 only Properties The molar masses of C H2 O2 N2 and air are 12 kgkmol 2 kgkmol 32 kgkmol 28 kgkmol and 29 kgkmol respectively Table A1 Analysis a The balanced reaction equation for stoichiometric air is 2 th 2 2 2 2 th 6 2 3 76 N 3 H O 2 CO 376N O C H O a a The stoicihiometric coefficient ath is determined from an O2 balance PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 th th a a 3 76 N 3 11 O 3 H O 01 CO x for 2 is determined from a mass balance ubstitutin N determined from the energy balance applied on e combustion chamber with W 0 It reduces to CO2 CO H2O O2 N2 C2H6O Air 10 excess Combustion chamber 3 51 2 5 Substituting 2 2 2 2 2 6 2 1128 N 3 H O 2 CO 376N 3 O C H O Therefore 3 kmol of oxygen is required to burn 1 kmol of ethanol b The reaction with 10 excess air and incomplete combustion can be written as 2 2 2 6 2 209 CO 376N 3 O 11 C H O 2 2 2 The coefficient O 40 3 50 2 10 50 2 90 3 11 50 x x O2 balance S g 2 2 2 2 2 6 2 12408 O 40 3 H O 02 CO 18 CO 376N O 33 C H O b The heat transfer for this combustion process is 2 system out in E E E th f R P f P h h h N h h h N Q o o o o out R Both the reactants and products are at 25 oC Assuming the air and the combustion products to be ideal gases we have h T Then using the values given in the table or a 35 kgh of fuel burned the rate of heat transfer is h 220590 kJ kmol fuel 1 1 235310 3 241820 20 110530 393520 81 out Q or 1 220590 kJ kmol fuel out Q F 2580 kW 92870 kJh 1 220590 kJkmol 46 kgkmol kgh 53 out out out M Q m NQ Q preparation If you are a student using this Manual you are using it without permission 1599 15107 Propane gas is burned with air during a steadyflow combustion process The adiabatic flame temperature is to be determined for different cases Assumptions 1 Steady operating conditions exist 2 Air and combustion gases are ideal gases 3 Kinetic and potential energies are negligible 4 There are no work interactions 5 The combustion chamber is adiabatic PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course mbustion chamber reduces to Analysis Adiabatic flame temperature is the temperature at which the products leave the combustion chamber under adiabatic conditions Q 0 with no work interactions W 0 Under steadyflow conditions the energy balance system out in E E E applied on the co C3H8 o o o o o o o f P T f P R f R P f P Nh h h h N h h h N h h h N since all the reactants are at the standard reference temperature of 25C and o 0 for O and hf 2 N2 a The theoretical combustion equation of C3H8 with stoichiometric amount of air is 188N 4H O 3CO 376N 5 O rom the tables Substance C3H8 g 2 2 2 2 2 Products TP C3H8 25C Air 25C Combustion chamber F hf o kJkmol h298 K mol H8 g 3850 kJk C3 10 O2 0 8682 N2 0 8669 H2O g CO2 393520 9364 Thus 241820 9904 CO 110530 8669 103850 1 8669 0 18 8 9904 241820 4 9364 393520 3 2 2 2 N H O CO h h h It yields 2 274675 kJ 18 8 4 3 2 2 2 N H O CO h h h The adiabatic flame temperature is obtained from a trial and error solution A first guess is obtained by dividing the right hand side of the equation by the total number of moles which yields 2274675 3 4 188 88165 kJkmol This enthalpy value corresponds to about 2650 K for N Noting that the majority of the m 2 oles are N2 TP will be close to 2650 r it because of the higher specific heats of CO2 and H2O K but somewhat unde 2 280704 kJ Higher than 2274675 kJ 79320 18 8 4 103508 3 125152 18 8 4 3 2 2 2 N H O CO h h h At 2400 K At 2350 K 226582 kJ Lower than 2274675 kJ 2 2 2 2 N H O CO 77496 18 8 4 100846 3 122091 h y interpo tion TP 2394 K for complete combustion with 200 theoretical air is Substituting known numerical values 18 8 4 3 h h B la b The balanced combustion equation 2 2 2 2 2 2 8 3 376N 5O 4H O 3CO 376N 10 O C H g preparation If you are a student using this Manual you are using it without permission 15100 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 9904 241820 4 9364 393520 2 2 H O CO h h 103850 1 8669 0 37 6 8682 0 5 2 2 N O h h which yields 3 2 481060 kJ 37 6 5 4 3 N O H O CO h h h h 2 2 2 2 c flame temperature is obtained from a trial and error solution A first guess is obtained by dividing the right hand side of the equation by the total number of moles which yields 2481060 3 4 5 376 50021 kJkmol This ue corresponds to about 1580 K for N2 Noting that the majority of the moles are N2 TP will be close to 1580 what under it because of the higher specific heats of CO2 and H2O The adiabati enthalpy val K but some 2 536055 kJ Higher than 2481060 kJ 37 5 50756 4 59888 3 73417 37 6 5 4 3 2 2 2 2 N O H O CO h h h h At 1540 K 48470 6 0 K At 150 2 461630 kJ Lower than 2481060 kJ 47073 37 6 5 49292 4 57999 3 71078 37 6 5 4 3 2 2 N h h h h By interpolation TP 1510 K ced combustion equation for incomplete combustion with 95 theoretical air is g 2 2 O H O CO c The balan 2 2 2 2 2 3 8 1786N 4H O 05CO 52 CO 376N 475 O C H Substituting known numerical values 10385 1 8669 1786 0 9904 241820 4 8669 110530 50 9364 393520 52 2 2 2 N O H CO CO h h h h 0 which yields 2124684 kJ 1786 4 50 52 2 2 2 N H O CO CO h h h h The adiabatic flame temperature is obtained from a trial and error solution A first guess is obtained by divid the right hand side of the equation by the total number of moles which yields 2124684 25 4 05 1786 85 66 kJkmol This enthalpy value corresponds to about 2550 K for N2 Noting that the majority of the moles are N2 TP will be close to hat und the higher specific heats of CO2 and H2O At 2350 K ing 4 2550 K but somew er it because of 2131779 kJ Higher than 2124684 kJ 1786 77496 4 100846 78178 50 122091 52 1786 4 50 52 2 2 2 N H O CO CO h h h h At 2300 K 2 080129 kJ Lower than 2124684 kJ 1786 75676 4 98199 76345 50 119035 52 1786 4 50 52 2 2 2 N H O CO CO h h h h By interpolation TP 2343 K preparation If you are a student using this Manual you are using it without permission 15101 15108 The highest possible temperatures that can be obtained when liquid gasoline is burned steadily with air and with pure oxygen are to be determined Assumptions 1 Steady operating conditions exist 2 Air and combustion gases are ideal gases 3 Kinetic and potential energies are negligible 4 There are no work interactions 5 The combustion chamber is adiabatic Analysis The highest possible temperature that can be achieved during a combustion process is the temperature which occurs when a fuel is burned completely with stoichiometric amount of air in an adiabatic combustion chamber It is determined from C8H18 f P T f P R f R P f P o o o o o o o Nh h h h N h h h N h h h N at the standard reference temperature of 25C and for O2 and N2 The theoretical combustion n of C8H18 air is 47N 9H O 8CO 376N 125 O From the tables Substance since all the reactants are equatio 2 2 2 2 2 C8H18 hf o kJkmol h298 K kJkmol H18 l 9950 C8 24 O2 0 8682 N2 0 8669 H2O g CO2 393520 9364 Thus 241820 9904 Products TP max C8H18 25C Air 25C Combustion chamber 249950 1 8669 47 0 9904 241820 9 9364 393520 2 2 2 N H O CO h h h 8 It yields 5 646081 kJ 47 9 8 2 2 2 N H O CO h h h The adiabatic flame temperature is obtained from a trial and error solution A first guess is obtained by dividing the right hand side of the equation by the total number of moles which yields 56460818 9 47 88220 kJkmol This enthalpy value corresponds to about 2650 K for N2 Noting that the majority of the moles are N2 TP will be close to 2650 K but somewhat un er it because of the higher specific heat of H2O d 5 660828 kJ Higher than 5646081 kJ 47 79320 9 103508 8 125152 47 9 8 2 2 2 N H O CO h h h At 2400 K 5 526654 kJ Lower than 5646081 kJ 47 77496 9 100846 8 122091 47 2 2 2 N H O CO h 9 8 h h At 2350 K B la y interpo tion TP 2395 K burned with stoichiometric amount of pure O2 the combustion equation would b If the fuel is e C 9H O 8CO 125O H 2 2 2 18 8 Thus 249950 1 9904 241820 9 9364 393520 8 H O CO 2 2 h h It yields 5 238638 kJ 9 8 H O CO 2 2 h h The adiabati f c flame temperature is obtained from a trial and error solution A irst guess is obtained by dividing the right hand side of the equation by the total number of moles which yields 52386388 9 308155 kJkmol This enthalpy er than e highest enthalpy value listed for H2O and CO2 Thus an estimate of the adiabatic flame temperature value is high th can be obtained by extrapolation At 3200 K 2 724673 kJ 147457 2 2 At 3250 K 9 8 174695 9 8 H O CO h h 2 775024 kJ 9 150272 8 177822 9 8 H O CO 2 2 h h By extrapolation we get TP 3597 K However the solution of this problem using EES gives 5645 K The large difference between these two values is due to extrapolation PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15102 15109 Methyl alcohol vapor is burned with the stoichiometric amount of air in a combustion chamber The maximum pressure that can occur in the combustion chamber if the combustion takes place at constant volume and the maximum volume of the combustion chamber if the combustion occurs at constant pressure are to be determined Assumptions 1 Combustion is complete 2 Air and the combustion gases are ideal gases 4 Changes in kinetic and potential energies are negligible Analysis a The combustion equation of CH3OHg with stoichiometric amount of air is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course a a E CH3OHg AIR 25C 101 kPa 2 th 2 2 2 2 th 3 N 376 2H O CO 376N O CH OH a a where ath is the stoichiometric coefficient and is determined from the O2 balance 1 2 2 2 15 th th Thus 2 2 2 2 2 3 564N 2H O CO 376N 15 O CH OH The final temperature in the tank is determined from the energy balance relation E E system out in for reacting closed stems under adiabatic conditions Q 0 with no work interactions W 0 sy R f R P f P Assuming both the reactants and the products to behave as ideal gases all the interna P h h h N P h h h N v v o o o o 0 l energy and enthalpies depend on mperature only and the Pv te terms in this equation can be replaced by RuT It yields R u f R P u K T f P P 298 R T h N R T h h h N o o since the reactants are at the standard reference temperature of C From the tables Substance 25 hf o kJkmol h298 K mol 3OH 0670 kJk CH 20 O2 0 8682 N2 0 8669 H2O g CO2 393520 9364 hus 241820 9904 T 8 314 298 5 64 0 8 314 298 0 51 8 314 298 200670 1 8 314 8669 64 0 5 8 314 9904 241820 2 8 314 9364 393520 1 2 2 2 N H O CO P P P T h T h T h It yields h h h TP CO2 H2O N2 kJ 2 564 71833 734 388 ture of the product gases is obtained from a trial and error solution 0 K The tempera At 285 744733 kJ Higher than 734388 kJ 71833 2850 5 64 95859 2 127952 1 152908 71833 5 64 2 2 2 2 N H O CO TP h h h At 2800 K 729311 kJ Lower than 734388 kJ 71833 2800 5 64 94014 2 125198 1 149808 71833 5 64 2 N H O CO TP h h h 2 2 2 preparation If you are a student using this Manual you are using it without permission 15103 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course termined to be By interpolation TP 2816 K Since both the reactants and the products behave as ideal gases the final maximum pressure that can occur in the combustion chamber is de 1013 kPa 864 kmol 2816 K 101 kPa 2 2 1 1 1 N T P P N R T P u V the same in the case of constant pressure Further the boundary work this case an be combined with the u terms so that the first law relation can be expressed in terms of enthalpies just like 814 kmol 298 K 1 1 1 2 2 2 2 N T N R T P u V b The combustion equation of CH3OHg remains in c the steadyflow process R f R P f P h h h N h h h N Q o o o o Since both the reactants and the products behave as ideal gases we have h hT Also noting that Q 0 for an adiabatic combustion process the 1st law relation reduces to R f R P T f P h N h h h N P o o 298 K since the reactants are at the standard reference temperature of 25C Then using data from the mini table above we get 5 64 0 0 51 200670 1 8669 5 64 0 9904 241820 2 9364 393520 2 2 2 N H O CO h h h 1 It yields 5 kJ 754 5 64 2 N CO h h h 5 5 H O 0 K 2 2 2 The temperature of the product gases is obtained from a trial and error solution 760860 kJ Higher than 754555 kJ 5 64 77496 2 100846 1 122091 5 64 2 2 2 2 N H O CO h h h At 235 742246 kJ Lower than 754555 kJ 5 64 75676 2 98199 1 119035 5 64 2 2 2 2 N H O CO h h h At 2300 K By interpolation TP 2333 K Treating both the reactants and the products as ideal gases the final maximum volume that the combustion chamber can have is determined to be 125 L 15 L 814 kmol 298 K 864 kmol 2333 K 1 1 1 2 2 2 2 2 1 1 2 1 V V V V N T N T N R T N R T P P u u preparation If you are a student using this Manual you are using it without permission 15104 15110 Problem 15109 is reconsidered The effect of the initial volume of the combustion chamber on the maximum pressure of the chamber for constant volume combustion or the maximum volume of the chamber for constant nalysis The problem is solved using EES and the solution is given below kJkmolK ath O2376N2 CO2 2 H2O 376ath N2 of reactants and products in kmol 376 le A26 in kJkmol uT2a 2hO2RuT1NN2hN2RRuT1 T1P1 Final pressure ts are in kJkmol NN2hN2R 2NPNRT2bT1V1 Final pressure pressure combustion is to be investigated A Given V115 L T125273 K P1101 kPa T025273 K Properties Ru8314 Analysis The stoichiometric combustion equation is CH3OH O balance 12ath22 Mol numbers NCH3OH1 NO2ath NN2ath NCO21 NH2O2 Enthalpy of formation data from Tab hfCH3OH200670 Enthalpies of reactants in kJkmol hO2enthalpyO2 TT1 hN2RenthalpyN2 TT1 Enthalpies of products in kJkmol hN2PaenthalpyN2 TT2a hCO2aenthalpyCO2 TT2a hH2OaenthalpyH2O TT2a HPaNCO2hCO2aRuT2aNH2OhH2OaRuT2aNN2hN2PaR HRaNCH3OHhfCH3OHRuT1NO HPaHRa P2NPNRT2a NR1476ath NP12376ath b Now ideal gas enthalpies of produc hN2PbenthalpyN2 TT2b hCO2benthalpyCO2 TT2b hH2ObenthalpyH2O TT2b HPbNCO2hCO2bNH2OhH2ObNN2hN2Pb HRbNCH3OHhfCH3OHNO2hO2 HPbHRb V PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15105 V1 L V2 L P2 kPa 02 04 06 08 1 12 14 16 18 2 1663 3325 4988 6651 8313 9976 1164 133 1496 1663 1013 1013 1013 1013 1013 1013 1013 1013 1013 1013 02 04 06 08 1 12 14 16 18 2 0 2 4 6 8 10 12 14 16 18 800 850 900 950 1000 1050 1100 1150 1200 V1 L V2 L P2 kPa PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15106 15111 Methane is burned with the stoichiometric amount of air in a combustion chamber The maximum pressure that can occur in the combustion chamber if the combustion takes place at constant volume and the maximum volume of the combustion chamber if the combustion occurs at constant pressure are to be determined Assumptions 1 Combustion is complete 2 Air and the combustion gases are ideal gases 4 Changes in kinetic and potential energies are negligible Analysis a The combustion equation of CH4g with stoichiometric amount of air is CH4 g AIR 25C101 kPa 2 th 2 2 2 2 th 4 N 376 2H O CO 376N O CH a a where ath is the stoichiometric coefficient and is determined from the O2 balance a a th th 1 1 2 Thus 2 2 2 2 2 4 752N 2H O CO 376N 2 O CH The final temperature in the tank is determined from the energy balance relation E E E in out system for reacting closed systems un er adiabatic conditions Q 0 with no work interactions W 0 d R f R P f P P h h h N P h h h N v v o o o o 0 h and the products behave as ideal gases all the inte Since bot the reactants o rnal energy and enthalpies depend on temperature nly and the Pv terms in this equation can be replaced by RuT It yields u f R u T f P R T h N R T h h h N o o 298K R P P since the reactants are ndard reference temperature of C From the tables tance at the sta 25 Subs ho f kJkmol h298 K ol 4 850 kJkm CH 74 O2 0 8682 N2 0 8669 H2O g 241820 9904 CO2 393520 9364 Thus 8 314 298 7 52 0 8 314 298 2 0 8 314 298 74850 1 8 314 8669 52 0 7 8 314 9904 241820 2 8 314 9364 393520 1 2 2 2 N H O CO P P P T h T h T h It yields 870609 kJ 87463 7 52 2 2 2 2 N H O hCO TP h h perature of the product gases is obtained from a trial and error solution At 2850 K The tem 880402 kJ Higher than 870609 kJ 87463 2850 7 52 95859 2 127952 1 152908 87463 7 52 2 2 CO h h 2 2 N H O TP h At 2800 K 87463 2800 7 52 94014 2 125198 1 149808 87463 7 52 2 2 2 2 N H O CO TP h h h 862293 kJ Lower than 870609 kJ PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15107 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course TP 2823 K Treating both the reactants and the products as ideal gases the final maximum pressure that can occur in the combustion By interpolation chamber is determined to be 957 kPa 1052 kmol 2823 K 101 kPa 2 2 2 1 1 2 1 N T P P N R N R T P P u V V 1052 kmol 298 K 1 1 1 2 2 N T T u The co bustion equation of CH g remains the same in the case of constant pressure Further the boundary work in the b m 4 this case can be combined with the u terms so that the first law relation can be expressed in terms of enthalpies just like steadyflow process R f R P f P h h h N h h h N Q o o o o diabatic c mbustion process the energy balance relation reduces to Again since both the reactants and the products behave as ideal gases we have h hT Also noting that Q 0 for an a o R f R P T f P h N h h h N P o o 298 K since the reactants are at the standard reference temperature of 25C Then using data from the mini table above we get 7 52 0 2 0 74850 1 8669 N2 7 52 0 9904 241820 2 9364 393520 1 2 2 H O CO h h h It yields 896673 kJ 7 52 2 2 2 2 N H O CO h h h The temperature of the product gases is ob tained from a trial and error solution 7 52 77496 2 100846 1 122091 7 52 2 CO2 h h h At 2350 K 553 kJ Higher than 896673 kJ 906 2 2 N H O t 2300 K 884517 kJ Lower than 896673 kJ 7 52 75676 2 98199 1 119035 7 52 2 2 2 2 N H O CO h h h A By interpolation TP 2328 K Treating both the reactants and the products as ideal gases the final maximum volume that the combustion chamber can have is determined to be 117 L 15 L 1052 kmol 298 K 1052 kmol 2328 K 1 1 1 2 2 2 2 2 1 1 2 1 V V V V N T N T N R T N R T P P u u preparation If you are a student using this Manual you are using it without permission 15108 15112 nOctane is burned with 100 percent excess air The combustion is incomplete The maximum work that can be produced is to be determined Assumptions 1 Combustion is incomplete 2 Steady operating conditions exist 3 Air and the combustion gases are ideal gases 4 Changes in kinetic and potential energies are negligible Analysis The combustion equation with 100 excess air and 10 CO is 2 2 2 2 2 2 18 8 3 76 N 2 12 5 O 9 H O 0 10 CO 8 0 90 CO 376N 2 12 5 O C H x The coefficient for O2 is determined from its mass balance as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course x x 12 9 54 40 27 25 Substituting 2 2 2 2 2 2 18 8 94 N 12 9 O 9 H O CO 80 CO 27 376N 25 O C H The reactants and products are at 25C and 1 atm which is the standard reference state and also the state of the surroundings Therefore the reversible work in this case is simply the difference between the Gibbs function of formation of the reactants and that of the products 9 228590 80 137150 27 394360 6 530 1 5 22952 kJ per kmol of fuel 1 rev o o f P P f R R N g N g W nce the of stable elements at 25C and 1 atm is zero Per unit mass basis Products 1 atm 25C C8H18 1 atm 25C Air 100 excess 1 atm 0 si o f g 44060 kJkg fuel 114 kgkmol 5022952 kJkmol Wrev 15113E Methane is burned with stoichiometric air The maxim m work that can be produced is to be determined Assumptions 1 Combustion is incomplete 2 Steady operating conditions exist 3 Air and the combustion gases are ideal The reactants and products are at 77F and 1 atm which is the standa sta reversible work in this case is simply the difference between the Gibbs nction o ation of the reactants and that of the products 25C Combustion chamber Products 1 atm 77F CH4 atm 77F Air 100 theoretical 1 atm 77F Combustion chamber u gases 4 Changes in kinetic and potential energies are negligible Analysis The combustion equation is 7 52N 2H O CO 376N 2 O CH 2 2 2 2 2 4 rd reference te and also the state of the surroundings Therefore the 1 fu f form 344520 Btu per lbmol of fuel 2 98350 1 169680 21860 1 rev o o f P P f R R N g N g W since the of stable elements at 77F and 1 atm is zero Per unit mass basis o f g 21530 Btulbm fuel 16 lbmlbmol 344520 Btulbmol Wrev preparation If you are a student using this Manual you are using it without permission 15109 15114E Methane is burned with 100 excess air The maximum work that can be produced is to be determined and compared to when methane is burned with stoichiometric air Assumptions 1 Combustion is incomplete 2 Steady operating conditions exist 3 Air and the combustion gases are ideal gases 4 Changes in kinetic and potential energies are negligible Analysis The combustion equation with 100 excess air is 2 2 2 2 2 2 4 1504N 2O 2H O CO 376N 4 O CH Products 1 atm 77F CH4 1 atm 77F Air 100 excess 1 atm 77F Combustion chamber The reactants and products are at 77F and 1 atm which is the standard reference state and also the state of the surroundings Therefore the reversible work in this case is simply the difference between the Gibbs function of formation of the reactants and that of the products PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2 98350 1 169680 21860 1 Btu per lbmol of fuel 344520 o o f P P f R R N g N g Wrev since the o g f of stable elements at 77F and 1 atm is zero Per unit mass basis 21530 Btulbm fuel 16 lbmlbmol 34452 Btulbmol Wrev The excess air only adds oxygen and nitrogen to the reactants and products The excess air then does not change the maximum work preparation If you are a student using this Manual you are using it without permission 15110 15115 Methane is burned steadily with 50 percent excess air in a steam boiler The amount of steam generated per unit of fuel mass burned the change in the exergy of the combustion streams the change in the exergy of the steam stream and the lost work potential are to be determined Assumptions 1 Combustion is complete 2 Steady operating conditions exist 3 Air and the combustion gases are ideal gases 4 Changes in kinetic and potential energies are negligible Properties The molar masses of CH4 and air are 16 kgkmol and 29 kgkmol respectively Table A1 Analysis a The fuel is burned completely with the excess air and thus the products will contain only CO2 H2O N2 and some free O2 Considering 1 kmol CH4 the combustion equation can be written as 2 2 2 2 2 2 4 1128N O 2H O CO 376N 3 O CH PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course E Products 227C CH4 25C Air 50 excess air 25C Combustion chamber Under steadyflow conditions the energy balance sy out in E E stem 0 reduces to applied on the combustion chamber with W R f R P f P h h h N h h h N Q o o o o out Assuming the air and the combustion products to be ideal gases we have h hT From the tables Substance o f h kJkmol h298K kJkmol h500K kJkmol CH4 74850 O2 0 8682 14770 N2 0 8669 14581 H O g 2 CO 241820 9904 16828 2 393520 9364 17678 Thus 707373 kJkmol of fuel 74850 1 1128 0 14581 8669 8682 1 0 14770 9904 241820 16828 2 9364 393520 17678 1 out Q The heat loss per unit mass of the fuel is 44211 kJkg fuel 16 kgkmol of fuel out Q 373 kJkmol of fuel 707 The amount of steam generated per unit mass of fuel burned is determined from an energy balance to be Enthalpies of steam are from tables A4 and A6 1872 kg steamkg fuel 85226 kJkg steam 3214 5 out s f s h m 44211 kJkg fuel Q m The en y generation during this process is determined from b trop S S S Q T N s N s Q T P R P P R R gen out surr out su rr he entropy values listed in the ideal gas tables are for 1 atm pressure Both the air and the product gases are at a total pressure of 1 atm but the entropies are to be calculated at the partial pressure of the components which is equal to Pi yi Ptotal where yi is the mole fraction of component i Then T m i u i i i i i i y P R T P N s N s T P S ln 0 o preparation If you are a student using this Manual you are using it without permission 15111 The entropy calculations can be presented in tabular form as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Ni yi T1atm si o m i u R ln y P Nisi CH4 1 18616 0 18616 O2 3 021 20504 1298 65406 N2 1128 079 19161 1960 218347 SR 302369 kJK CO2 1 00654 234814 2267 25748 H2O g 2 01309 206413 1691 44665 O2 1 00654 220589 2267 24326 N2 1128 07382 206630 2524 235926 SP 330665 kJK Thus 2657 kJK per kmol fuel 298 707373 302369 330665 surr out gen T Q S S S R P The exergy change of the combustion streams is equal to the exergy destruction since there is no actual work output That is 791786 kJkmol fuel 298 K2657 kJK gen 0 dest gases T S X X Per unit mass basis 49490 kJkg fuel 16 kgkmol 791786 kJkmol fuel X gases Note that the exergy change is negative since the exergy of combustion gases decreases c The exergy change of the steam stream is 1039 kJkg steam 2 3305 298 6 7714 85226 3214 5 0 steam s T h X d The lost work potential is the negative of the net exergy change of both streams 30040 kJkg fuel 49490 kJkg fuel 1872 kg steamkg fuel1039 kJkg steam gases steam dest X X m m X f s preparation If you are a student using this Manual you are using it without permission 15112 15116 A coal from Utah is burned steadily with 50 percent excess air in a steam boiler The amount of steam generated per unit of fuel mass burned the change in the exergy of the combustion streams the change in the exergy of the steam stream and the lost work potential are to be determined Assumptions 1 Combustion is complete 2 Steady operating conditions exist 3 Air and the combustion gases are ideal gases 4 Changes in kinetic and potential energies are negligible 5 The effect of sulfur on the energy and entropy balances is negligible Properties The molar masses of C H2 N2 O2 S and air are 12 2 28 32 32 and 29 kgkmol respectively Table A1 Analysis a We consider 100 kg of coal for simplicity Noting that the mass percentages in this case correspond to the masses of the constituents the mole numbers of the constituent of the coal are determined to be PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 04406 kmol 32 kgkmol kg 141 0 03893 kmol kgkmol 28 kg 109 0 7909 kmol kgkmol 32 N2 O2 O2 O2 m N M N kg 2531 2 895 kmol 2 kgkmol kg 579 5 117 kmol kg 6140 S S S N2 N2 H2 H2 H2 C M m N M m M m N m d the mole fractions are N 12 kgkmol C C M 6140 C 579 H2 2531 O2 109 N2 41 S 500 ash by mass 1 The mole number of the mixture an 8 886 kmol 0 04406 0 03893 0 7909 2 895 5 117 Nm 000496 8886 kmol S m N kmol 004406 000438 kmol 8886 00890 8886 kmol kmol 07909 03258 kmol 2895 0 5758 8886 kmol kmol 5117 S O2 H2 C C m m N N N N y N y N N y sh consists of the noncombustible matter in coal Therefore the mass of ash content that enters the combustion chamber is equal to the mass content that leaves Disregar ng this nonreacting c mponent for simplicity the combustion equation may be written as 2 2 3 76N 51 O 50 00 8H O 0 5758CO 3 76N O 0 0049 0 00438 O 05758C a a According oxygen bala e 0 0 3258 0 5758 0 0890 O th th a a 3 76N 0 9821O 0 00496S 0 00438N 0 0890O 03258H 05758C The apparent molecular weight of the coal is Products 227C Coal 25C Air 50 Combustion chamber 8886 kmol H2 m N 003893 kmol N2 N2 m N y O2 excess air 25C y A di o 2 th 2 th 2 2 2 2 2 th 51 S 6 a 2 0 0890 0496SO 325 0 N 03258H to the nc 00496 50 51 a balance th 2 0 6547 50 Substituting 2 2 2 2 2 2 2 2 2 2 3 693N 0 3274O 0 00496SO 0 3258H O 0 5758CO preparation If you are a student using this Manual you are using it without permission 15113 kgkmol coal 1069 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course kmol 10 kg 1069 000496 kmol 000438 00890 03258 758 flow conditions the energy balance 05 0 00496 32 kg 0 00438 28 0 0890 32 0 3258 2 05758 12 m m m N m M system out in E E E Under steady applied on the combustion chamber with W 0 o reduces t R f R P f P h h h N h h h N Q o o o o out Assuming the air and the combustion products to be ideal gases we have h hT From the tables Substance o f h kJkmol h298K kJkmol h500K kJkmol O2 0 8682 14770 N2 0 8669 14581 H2O g 241820 9904 16828 CO2 393520 9364 17678 Thus 274505 kJkmol of fuel 0 3 693 0 14581 8669 8682 0 3274 0 14770 9904 241820 16828 0 3258 9364 393520 17678 0 5758 out Q The heat loss per unit mass of the fuel is 25679 kJkg fuel 1069 kgkmol of fuel 274505 kJkmol of fuel out Q The amount of steam generated per unit mass of fuel burned is determined from an energy balance to be Enthalpies of steam are from tables A4 and A6 1087 kg steamkg fuel 85226 kJkg steam 3214 5 25679 kJkg fuel out s f s h Q m m b The entropy generation during this process is determined from S S S Q T N s N s Q T P R P P R R gen out surr out surr The entropy values listed in the ideal gas tables are for 1 atm pressure Both the air and the product gases are at a total pressure of 1 atm but the entropies are to be calculated at the partial pressure of the components which is equal to Pi yi Ptotal where yi is the mole fraction of component i Then m i u i i i i i i y P R T P N s N s T P S ln 0 o preparation If you are a student using this Manual you are using it without permission 15114 The entropy calculations can be presented in tabular form as Ni yi T1atm si o m i u R ln y P Nisi C 05758 05758 574 4589 595 H2 03258 03258 13068 9324 4561 O2 00890 00890 20504 2011 2004 N2 000438 000438 19161 4515 104 O2 09821 021 20504 1298 21412 N2 3693 079 19161 1960 71485 SR 100161 kJK CO2 05758 01170 234814 1784 14548 H2O g 03258 00662 206413 2257 7460 O2 03274 00665 220589 2254 7960 N2 3693 07503 206630 2388 77190 SP 107158 kJK Thus 991 1 kJK per kmol fuel 298 surr gen T R P 274505 107158 100161 out Q S S S The exergy change of the combustion streams is equal to the exergy destruction since ere is no actual work output That is th 295348 kJkmol fuel 298 K991 1 kJK gen 0 dest gases T S X er unit m X P ass basis PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 27630 kJkg fuel 295348 kJK 1069 kgkmol gases X Note that t e he xergy change is negative since the exergy of combustion gases decreases c The exergy change of the steam stream is 1039 kJkg steam 2 3305 298 6 7714 85226 3214 5 0 X steam s T h The lost work potential is the negative of the net exergy change of both streams d 16340 kJkg fuel 27630 kJkg fuel 1087 kg steamkg fuel1039 kJkg steam gases steam dest X X m m X f s preparation If you are a student using this Manual you are using it without permission 15115 15117 An expression for the HHV of a gaseous alkane CnH2n2 in terms of n is to be developed Assumptions 1 Combustion is complete 2 The combustion products contain CO2 H2O and N2 3 Combustion gases are ideal gases Analysis The complete reaction balance for 1 kmol of fuel is 2 2 2 2 2 2 2 3 76N 2 1 3 1 H O CO 376N O 2 1 3 C H n n n n n n Both the reactants and the products are taken to be at the standard reference state of 25C and 1 atm for the calculation of heating values The heat transfer for this process is equal to enthalpy of combustion Note that N2 and O2 are stable elements and thus their enthalpy of formation is zero Then fuel H2O CO2 o o o o o f f f f R R f P P R P C Nh Nh Nh N h N h H H h q For the HHV the water in the products is taken to be liquid Then fuel 1 285830 393520 o f C h n n h Air theoretical Products Fuel Combustion chamber The HHV of the fuel is fuel fuel 1 285830 393520 HHV h n n h f C fuel M M o or the LHV the water in the products is taken to be vapor Then F fuel fuel 1 241820 393520 LHV M h n n f o PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15116 15118 It is to be shown that the work output of the Carnot engine will be maximum when T T T p af 0 It is also to be shown that the maximum work output of the Carnot engine in this case becomes 2 af 0 af 1 T T CT w her is the negative of the heat supplied to the heat engine That is hen the work output of the Carnot heat engine ca press Analysis The combustion gases will leave the combustion chamber and enter the heat exchanger at the adiabatic flame temperature Taf since the chamber is adiabatic and the fuel is burned completely The combustion gases experience no change in their chemical composition as they flow through the heat exchanger Therefore we can treat the combustion gases as a gas stream with a constant specific heat cp Noting that the heat exchanger involves no work interactions the balance equation for this singlestream steadyflow device can be written as energy Taf mC T h m h Q p i e e Q Surroundings T0 Q W Adiabatic combustion chamber Fuel Air Heat Exchanger TP const TP T0 w p H T mC T Q Q af T n be ex ed as p p p H T T mC T T Q W af 1 1 1 T T 0 0 Taking the partial derivative of with respect to Tp while holding Taf and T0 onstant gives W c 0 1 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 Tp 2 0 af 0 p p p T T T mC T T T mC W olving fo we obtain S r Tp T T T p 0 af which the temperature at which the work output of the Carnot engine will be a mum The maximum work output is determined by substituting the relation bove into Eq 1 maxi a p af 0 0 0 af af 0 af 1 1 T T T T T mC T T T T C T p W m It simplifies to 2 af 0 af 1 T T mCT W or 2 af 0 af 1 T T CT w which is the desired relation preparation If you are a student using this Manual you are using it without permission 15118 15120 The combustion of a hydrocarbon fuel CnHm with excess air and incomplete combustion is considered The coefficients of the reactants and products are to be written in terms of other parameters Analysis The balanced reaction equation for stoichiometric air is 2 th 2 2 2 2 th N 3 76 2 H O CO 376N O C H A m n A m n PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course m n A Thus G 2 is determined from a mass balance O2 balance CO2 CO H2O O2 N2 CnHm Excess air Combustion chamber The stoichiometric coefficient Ath is determined from an O2 balance 4 th The reaction with excess air and incomplete combustion is 2 th 2 2 2 2 2 th N 3 76 1 O 2 H O CO CO 376N O 1 C H B A G m bn an B A m n The given reaction is 2 2 2 2 2 2 th N O H O CO CO 376N O 1 C H J G F E D B A m n th 3 76 1 2 B A J m F bn E an D The coefficient for O th th th th th th 2 2 2 1 2 4 2 4 4 2 4 1 4 2 1 BA bn BA bn nb bn BA a n bn an BA n G G m bn an BA m n G m bn an m n B G m bn an A B preparation If you are a student using this Manual you are using it without permission 15119 15121 The combustion of an alcohol fuel CnHmOx with excess air and incomplete combustion is considered The coefficients of the reactants and products are to be written in terms of other parameters Analysis The balanced reaction equation for stoichiometric air is 2 th 2 2 2 2 th N 3 76 2 H O CO 376N O C H O A m n A x m n PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course th th x m n A m n A x Thus m 2 F bn E an D nt G 2 is determined from a mass balance O2 balance CO2 CO H2O O2 N2 CnHmOx Excess air Combustion chamber The stoichiometric coefficient Ath is determined from an O2 balance 2 4 4 2 The reaction with excess air and incomplete combustion is 2 th 2 2 2 2 2 th N 3 76 1 O 2 H O CO CO 376N O 1 C H O B A G m bn an B A x m n The given reaction is 2 2 2 2 2 2 th N O H O CO CO 376N O 1 C H O J G F E D B A x m n th 3 76 1 B A J The coefficie for O th th th th th th 2 2 2 1 2 4 2 2 4 2 4 2 2 4 2 4 2 4 2 2 4 1 2 4 2 1 2 BA bn BA bn nb BA bn a n bn BA an n G G m bn an BA x m n x G m bn an x m B n x m n x G m bn an x m n B x G m bn an B A x preparation If you are a student using this Manual you are using it without permission 15120 15122 The combustion of a mixture of an alcohol fuel CnHmOx and a hydrocarbon fuel CwHz with excess air and incomplete combustion is considered The coefficients of the reactants and products are to be written in terms of other parameters Analysis The balanced reaction equation for stoichiometric air is 2 th 2 2 1 2 2 1 2 2 th 2 1 N 3 76 H O 50 CO 376N O C H C H O A y z y m y w y n A y y z w x m n The stoichiometric coefficient Ath is determined from an O2 balance 2 4 4 2 1 2 1 2 1 th 2 1 2 1 th 1 y x y z y m y w y n A y z y m y w y n A y x The reaction with excess air and incomplete combustion is 2 th 2 2 2 1 2 1 2 2 1 2 2 th 2 1 N 3 76 1 O H O 50 CO CO 376N O 1 C H C H O B A G y z y m y w b y n y w y n a B A y y z w x m n The given reaction is 2 2 2 2 2 2 th 2 1 N O H O CO CO 376N O 1 C H C H O J G F E D B A y y z w x m n Thus 50 y z y m F y w b y n E y w a y n D nt G 2 is determined from a mass balance alance CO2 CO H2O O2 N2 CnHmOx CwHz Excess air Combustion chamber th 3 76 1 B A J 2 1 2 1 2 1 The coefficie for O O2 b th 2 1 th 2 1 2 1 th 2 1 2 1 th 2 1 2 1 2 1 2 1 2 1 2 1 th 1 2 1 2 1 1 2 1 2 1 2 1 th 1 50 50 50 1 50 0 25 50 50 0 25 50 0 25 50 1 50 BA y w y n b BA y w b y n y w y n b BA y w b y n a y w n y BA y w b y n y w a y n y w y n G G y z y m y w b y n y w a y n BA y x y z y m y w y n x y G y z y m y w b y n y w a y n B A x y PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15121 15123 The effect of the amount of air on the adiabatic flame temperature of liquid octane C8H18 is to be investigated Analysis The problem is solved using EES and the solution is given below Adiabatic Combustion of fuel CnHm entering at Tfuel with Stoichiometric Air at Tair Reaction CxHyOz y4 xz2 Theoair100 O2 376 N2 xCO2 y2 H2O 376 y4 xz2 Theoair100 N2 y4 xz2 Theoair100 1 O2 For theoretical oxygen the complete combustion equation for CH3OH is CH3OH Ath O21 CO22 H2O 1 2Ath1221theoretical O balance Adiabatic Incomplete Combustion of fuel CnHm entering at Tfuel with Stoichiometric Air at Tair Reaction CxHyOz y4 xz2 Theoair100 O2 376 N2 xwCO2 wCO y2 H2O 376 y4 xz2 Theoair100 N2 y4 xz2 Theoair100 1 w2O2 Tprod is the adiabatic combustion temperature assuming no dissociation Theoair is the theoretical air The initial guess value of Tprod 450K Procedure FuelFuelTfuelxyzhfuelName This procedure takes the fuel name and returns the moles of C and moles of H If fuelC2H2g then x2y2 z0 NameAcetylene hfuel 226730 else If fuelC3H8l then x3 y8 z0 NamePropaneliq hfuel 10385015060 else If fuelC8H18l then x8 y18 z0 NameOctaneliq hfuel 249950 else if fuelCH4g then x1 y4 z0 NameMethane hfuel enthalpyCH4TTfuel else if fuelCH3OHg then x1 y4 z1 NameMethyl alcohol hfuel 200670 endif endif endif endif endif end Procedure MolesxyzThairAthwMolO2SolMeth ErrTh 2x y2 z x2Ath100 IF Thair 1 then SolMeth 100 the solution assumes complete combustion MolCO 0 MolCO2 x w0 MolO2 AthThair 1 GOTO 10 ELSE w 2x y2 z 2AthThair PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15122 IF w x then Call ERRORThe moles of CO2 are negative the percent theoretical air must be xxxF3 ErrTh Else SolMeth 100 the solution assumes incomplete combustion with no O2 in products MolO2 0 endif endif 10 END Input data from the diagram window Tair 298 K Theoair 200 FuelCH4g Tfuel 298 K Call FuelFuelTfuelxyzhfuelName Ath x y4 z2 Thair Theoair100 Call MolesxyzThairAthwMolO2SolMeth HRhfuel xy4z2 Theoair100 enthalpyO2TTair376xy4z2 Theoair100 enthalpyN2TTair HPHR Adiabatic HPxwenthalpyCO2TTprodwenthalpyCOTTprody2enthalpyH2OTTprod376xy4 z2 Theoair100enthalpyN2TTprodMolO2enthalpyO2TTprod MolesO2MolO2 MolesN2376xy4z2 Theoair100 MolesCO2xw MolesCOw MolesH2Oy2 Theoair Tprod K 75 2077 90 2287 100 2396 120 2122 150 1827 200 1506 300 1153 500 8401 800 6484 0 100 200 300 400 500 600 700 800 500 900 1300 1700 2100 2500 Theoair Tprod K Adiabatic Flame Temp for C8 H18 liquid PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15123 15124 A general program is to be written to determine the adiabatic flame temperature during the comple combustion of a hydrocarbo te n fuel CnHm at 25C in a steadyflow combustion chamber when the percent of excess air and its nalysis The problem is solved using EES and the solution is given below y4 xz2 Theoair100 1 O2 stion equation for CH3OH is y2 H2O 376 y4 xz2 Theoair100 N2 y4 xz2 ture assuming no dissociation el name and returns the moles of C and moles of H cetylene hfuel 226730 hfuel 10385015060 taneliq hfuel 249950 hfuel enthalpyCH4TTfuel l alcohol hfuel 200670 f endif endif endif olO2SolMeth x2Ath100 100 the solution assumes complete combustion temperature are specified A Adiabatic Combustion of fuel CnHm entering at Tfuel with Stoichiometric Air at Tair Reaction CxHyOz y4 xz2 Theoair100 O2 376 N2 xCO2 y2 H2O 376 y4 xz2 Theoair100 N2 For theoretical oxygen the complete combu CH3OH Ath O21 CO22 H2O 1 2Ath1221theoretical O balance Adiabatic Incomplete Combustion of fuel CnHm entering at Tfuel with Stoichiometric Air at Tair Reaction CxHyOz y4 xz2 Theoair100 O2 376 N2 xwCO2 wCO Theoair100 1 w2O2 Tprod is the adiabatic combustion tempera Theoair is the theoretical air The initial guess value of Tprod 450K Procedure FuelFuelTfuelxyzhfuelName This procedure takes the fu If fuelC2H2g then 2y2 z0 x Namea else If fuelC3H8l then 3 y8 z0 x Namepropaneliq else If fuelC8H18l then 8 y18 z0 x Nameoc else if fuelCH4g then 1 y4 z0 x Namemethane else if fuelCH3OHg then 1 y4 z1 x Namemethy endif endi end Procedure MolesxyzThairAthwM ErrTh 2x y2 z IF Thair 1 then SolMeth MolCO 0 MolCO2 x PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15124 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course AthThair 1 x then ErrTh 100 the solution assumes incomplete combustion with no O2 in products 2 0 endif 0 diagram window uelCH4g yzhfuelName thwMolO2SolMeth 2 Theoair100 enthalpyO2TTair376xy4z2 Theoair100 CO2TTprodwenthalpyCOTTprody2enthalpyH2OTTprod376xy4 olO2enthalpyO2TTprod 6xy4z2 Theoair100 olesCOw olesH2Oy2 mple calculatio kg 3000 eliq 100 the solution ass combustion 1200 8 K l298 K d2112 K w0 3 y8 z0 w0 MolO2 GOTO 10 ELSE w 2x y2 z 2AthThair IF w Call ERRORThe moles of CO2 are negative the percent theoretical air must be xxxF3 Else SolMeth MolO endif 1 END Input data from the Tair 298 K Theoair 120 F Tfuel 298 K Call FuelFuelTfuelx Ath x y4 z2 Thair Theoair100 Call MolesxyzThairA HRhfuel xy4z enthalpyN2TTair HPHR Adiabatic HPxwenthalpy z2 Theoair100enthalpyN2TTprodM MolesO2MolO2 MolesN237 MolesCO2xw M M SOLUTION for the sa n Ath5 fuelC3H8l HP119035 kJkg HR119035 kJ hfuel118910 MolesCO0000 MolesCO2 MolesH2O4 MolesN222560 MolesO21000 MolO21 Namepropan SolMeth umes complete Theoair120 Thair Tair29 Tfue Tpro x preparation If you are a student using this Manual you are using it without permission 15125 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15125 The minimum percent of excess air that needs to be used for the fuels CH4g C2H2g CH3OHg C3H8g and C8H18l if the adiabatic flame temperature is not to exceed 1500 K is to be determined Tprod is the adiabatic combustion temperature assuming no dissociation Theoair is the theoretical air The initial guess value of Tprod 450K Procedure FuelFuelTfuelxyzhfuelName This procedure takes the fuel name and returns the moles of C and moles of H Namepropane hfuel 249950 Namemethane x1 y4 z1 endif endif endif endif endif Tair 298 K ExcessairTheoair 100 Analysis The problem is solved using EES and the solution is given below Adiabatic Combustion of fuel CnHm entering at Tfuel with Stoichiometric Air at Tair Reaction CxHyOz y4 xz2 Theoair100 O2 376 N2 xCO2 y2 H2O 376 y4 xz2 Theoair100 N2 y4 xz2 Theoair100 1 O2 For theoretical oxygen the complete combustion equation for CH3OH is CH3OH Ath O21 CO22 H2O 1 2Ath1221theoretical O balance If fuelC2H2g then x2y2 z0 Nameacetylene hfuel 226730 else If fuelC3H8g then x3 y8 z0 hfuel enthalpyC3H8TTfuel else If fuelC8H18l then x8 y18 z0 Nameoctane else if fuelCH4g then x1 y4 z0 hfuel enthalpyCH4TTfuel else if fuelCH3OHg then Namemethyl alcohol hfuel 200670 end Input data from the diagram window FuelCH4g Tfuel 298 K Call FuelFuelTfuelxyzhfuelName Ath y4 xz2 Thair Theoair100 HRhfuel y4 xz2 Theoair100 enthalpyO2TTair376y4 xz2 Theoair100 enthalpyN2TTair HPHR Adiabatic preparation If you are a student using this Manual you are using it without permission 15126 HPxenthalpyCO2TTprody2enthalpyH2OTTprod376y4 xz2 Theoair100enthalpyN2TTprody4 xz2 Theoair100 1enthalpyO2TTprod MolesO2y4 xz2 Theoair100 1 MolesN2376y4 xz2 Theoair100 MolesCO2x MolesH2Oy2 T1Tprod xa1Theoair SOLUTION for a sample calculation Ath25 Excessair156251 fuelC2H2g HP226596 kJkg HR226596 kJkg hfuel226730 MolesCO22 MolesH2O1 MolesN22409 MolesO23906 Nameacetylene Theoair2563 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Tair298 K Tfuel298 K Thair2563 T11500 K Tprod1500 K x2 xa12563 y2 z0 preparation If you are a student using this Manual you are using it without permission 15127 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15126 The minimum percentages of excess air that need to be used for the fuels CH4g C2H2g CH3OHg C3H8g and C8H18l AFOR adiabatic flame temperatures of 1200 K 1750 K and 2000 K are to be determined nalysis The problem is solved using EES and the solution is given below Adiabatic Combustion of fuel CnHm entering at Tfuel with Stoichiometric Air at Tair y4 xz2 Theoair100 1 O2 ustion equation for CH3OH is ture assuming no dissociation uel name and returns the moles of C and moles of H cetylene hfuel 226730 lse hfuel enthalpyC3H8TTfuel octane hfuel 249950 hfuel enthalpyCH4TTfuel l alcohol hfuel 200670 f endif endif endif end the diagram window yzhfuelName Theoair100 enthalpyO2TTair376y4 xz2 Theoair100 r HPHR Adiabatic A Reaction CxHyOz y4 xz2 Theoair100 O2 376 N2 xCO2 y2 H2O 376 y4 xz2 Theoair100 N2 For theoretical oxygen the complete comb CH3OH Ath O21 CO22 H2O 1 2Ath1221theoretical O balance Tprod is the adiabatic combustion tempera Theoair is the theoretical air The initial guess value of Tprod 450K Procedure FuelFuelTfuelxyzhfuelName This procedure takes the f If fuelC2H2g then 2y2 z0 x Namea e If fuelC3H8g then 3 y8 z0 x Namepropane else If fuelC8H18l then 8 y18 z0 x Name else if fuelCH4g then 1 y4 z0 x Namemethane else if fuelCH3OHg then 1 y4 z1 x Namemethy ndif endi e Input data from Tair 298 K FuelCH4g Tfuel 298 K ExcessairTheoair 100 Call FuelFuelTfuelx Ath y4 xz2 Thair Theoair100 HRhfuel y4 xz2 enthalpyN2TTai preparation If you are a student using this Manual you are using it without permission 15128 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course pyH2OTTprod376y4 xz2 xz2 Theoair100 1enthalpyO2TTprod xz2 Theoair100 1 Theoair100 olesCO2x olesH2Oy2 SOLUTION for a sample calculation air1314 Thair1314 T12000 K Tair298 K Tfuel298 K Tprod2000 K x3 xa11314 y8 z0 HPxenthalpyCO2TTprody2enthal Theoair100enthalpyN2TTprody4 MolesO2y4 MolesN2376y4 xz2 M M T1Tprod xa1Theoair Ath5 Excessair31395 fuelC3H8g HP103995 kJkg HR103995 kJkg hfuel103858 MolesCO23 MolesH2O4 MolesN2247 MolesO21570 Namepropane Theo preparation If you are a student using this Manual you are using it without permission 15129 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15127 The adiabatic flame temperature of CH4g is to be determined when both the fuel and the air enter the combustion chamber at 25C for the cases of 0 20 40 60 80 100 200 500 and 1000 percent excess air Adiabatic Combustion of fuel CnHm entering at Tfuel with Stoichiometric Air at Tair 1 2Ath1221theoretical O balance Theoair is the theoretical air If fuelC2H2g then If fuelC3H8g then SolMeth 100 the solution assumes complete combustion Analysis The problem is solved using EES and the solution is given below Reaction CxHyOz y4 xz2 Theoair100 O2 376 N2 xCO2 y2 H2O 376 y4 xz2 Theoair100 N2 y4 xz2 Theoair100 1 O2 For theoretical oxygen the complete combustion equation for CH3OH is CH3OH Ath O21 CO22 H2O Adiabatic Incomplete Combustion of fuel CnHm entering at Tfuel with Stoichiometric Air at Tair Reaction CxHyOz y4 xz2 Theoair100 O2 376 N2 xwCO2 wCO y2 H2O 376 y4 xz2 Theoair100 N2 y4 xz2 Theoair100 1 w2O2 Tprod is the adiabatic combustion temperature assuming no dissociation The initial guess value of Tprod 450K Procedure FuelFuelTfuelxyzhfuelName This procedure takes the fuel name and returns the moles of C and moles of H x2y2 z0 Nameacetylene hfuel 226730 else x3 y8 z0 Namepropane hfuel enthalpyC3H8TTfuel else If fuelC8H18l then x8 y18 z0 Nameoctane hfuel 249950 else if fuelCH4g then x1 y4 z0 Namemethane hfuel enthalpyCH4TTfuel else if fuelCH3OHg then x1 y4 z1 Namemethyl alcohol hfuel 200670 endif endif endif endif endif end Procedure MolesxyzThairAthwMolO2SolMeth ErrTh 2x y2 z x2Ath100 IF Thair 1 then MolCO 0 MolCO2 x w0 MolO2 AthThair 1 preparation If you are a student using this Manual you are using it without permission 15130 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course ELSE Else 10 Tair 298 K Call FuelFuelTfuelxyzhfuelName HRhfuel xy4z2 Theoair100 enthalpyO2TTair376xy4z2 Theoair100 enthalpyN2TTair Theoair Tprod GOTO 10 w 2x y2 z 2AthThair IF w x then Call ERRORThe moles of CO2 are negative the percent theoretical air must be xxxF3 ErrTh SolMeth 100 the solution assumes incomplete combustion with no O2 in products MolO2 0 endif endif END Input data from the diagram window Theoair 200 FuelCH4g Tfuel 298 K Ath x y4 z2 Thair Theoair100 Call MolesxyzThairAthwMolO2SolMeth HPHR Adiabatic HPxwenthalpyCO2TTprodwenthalpyCOTTprody2enthalpyH2OTTprod376xy4 z2 Theoair100enthalpyN2TTprodMolO2enthalpyO2TTprod MolesO2MolO2 MolesN2376xy4z2 Theoair100 MolesCO2xw MolesCOw MolesH2Oy2 100 200 300 400 500 600 700 800 900 1000 1100 0 500 1000 1500 2000 2500 3000 Theoair Tprod K Product temperature vs excess air for CH4 K 100 2329 120 2071 140 1872 160 1715 180 1587 200 1480 300 1137 600 7495 1100 553 preparation If you are a student using this Manual you are using it without permission 15131 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15128 The fuel among CH4g C2H2g C2H6g C3H8g and C8H18l that gives the highest temperature when burned completely in an adiabatic constantvolume chamber with the theoretical amount of air is to be determined nalysis The problem is solved using EES and the solution is given below el TairTreac in a constant volume closed system y4z2 Theoair100 1 O2 stion equation for CH3OH is ic Air at Tfuel TairTreac in a constant CO2 wCO y2 H2O 376 xy4z2 Theoair100 N2 xy4z2 Theoair100 1 temperature assuming no dissociation el name and returns the moles of C and moles of H Nameacetylene hfuel 226730Table A26 Namepropane hfuel enthalpyC3H8TTfuel 8 y18 z0 Nameoctane hfuel 249950Table A26 Namemethane hfuel enthalpyCH4TTfuel lse Namemethyl alcohol hfuel 200670Table A26 endif endif endif endif zThairAthwMolO2SolMeth n assumes complete combustion thThair 1 A Adiabatic Combustion of fuel CnHm with Stoichiometric Air at Tfu Reaction CxHyOz xy4z2 Theoair100 O2 376 N2 xCO2 y2 H2O 376 xy4z2 Theoair100 N2 x For theoretical oxygen the complete combu CH3OH Ath O21 CO22 H2O 1 2Ath1221theoretical O balance Adiabatic Incomplete Combustion of fuel CnHm with Stoichiometr volume closed system Reaction CxHyOz xy4z2 Theoair100 O2 376 N2 xw w2O2 Tprod is the adiabatic combustion Theoair is the theoretical air The initial guess value of Tprod 450K Procedure FuelFuelTfuelxyzhfuelName This procedure takes the fu If fuelC2H2g then 2y2 z0 x else If fuelC3H8g then 3 y8 z0 x else If fuelC8H18l then x else if fuelCH4g then 1 y4 z0 x e if fuelCH3OHg then 1 y4 z1 x endif end Procedure Molesxy ErrTh 2x y2 z x2Ath100 IF Thair 1 then SolMeth 100 the solutio w0 MolO2 A preparation If you are a student using this Manual you are using it without permission 15132 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 10 w 2x y2 z 2AthThair RRORThe moles of CO2 are negative the percent theoretical air must be xxxF3 ErrTh 100 the solution assumes incomplete combustion with no O2 in products f endif ND diagram window uelCH4g elxyzhfuelName Ath x y4 z2 enthalpyO2TTairRuTair376xy4z2 Theoair100enthalpyN2TTprodRuTprodMolO2enthalpyO2TTprodRuTprod constant volume conservation of energy MolesN2376xy4z2 Theoair100 w olesH2Oy2 MolesCO0000 MolesCO21000 MolesH2O2 MolO20 Theoair100 Thair1000 Tair298 K 1 y4 z0 1 MolO20 2 y2 z0 GOTO ELSE IF w x then Call E Else SolMeth MolO2 0 endi 10 E Input data from the Theoair 200 F Treac 298 K Tair Treac Tfuel Treac Ru 8314 kJkmolK Call FuelFuelTfu Thair Theoair100 Call MolesxyzThairAthwMolO2SolMeth URhfuelRuTfuel xy4z2 Theoair100 Theoair100 enthalpyN2TTairRuTair UPxwenthalpyCO2TTprodRuTprodwenthalpyCOTTprod RuTprody2enthalpyH2OTTprodRuTprod376xy4z2 UR UP Adiabatic MolesO2MolO2 MolesCO2x MolesCOw M SOLUTION for CH4 Ath2 fuelCH4g hfuel74875 MolesN27520 MolesO20000 Namemethane Ru8314 kJkmolK SolMeth 100 the solution assumes complete combustion Tfuel298 K Tprod2824 K Treac298 K UP100981 UR100981 w0 x SOLUTION for C2H2 Ath25 fuelC2H2g hfuel226730 MolesCO0000 MolesCO22000 MolesH2O MolesN29400 MolesO20000 Nameacetylene Ru8314 kJkmolK SolMeth 100 the solution assumes complete combustion Theoair100 Thair1000 Tair298 K Tfuel298 K Tprod3535 K Treac298 K UP194717 UR194717 w0 x preparation If you are a student using this Manual you are using it without permission 15133 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course SOLUTION for CH3OH MolesN25640 MolesO20000 MolO20 Namemethyl alcohol Ru8314 kJkmolK Theoair100 Thair1000 Tair298 K Tfuel298 K Tprod2817 K Treac298 K x1 y4 z1 SOLUTION for C3H8 Ath15 fuelCH3OHg hfuel200670 MolesCO0000 MolesCO21000 MolesH2O2 SolMeth 100 the solution assumes complete combustion UP220869 UR220869 w0 Ath5 fuelC3H8g hfuel103858 MolesCO0000 MolesCO23000 MolesH2O4 MolesN218800 MolesO20000 MolO20 Namepropane Ru8314 kJkmolK SolMeth 100 the solution assumes complete combustion Theoair100 Thair1000 Tair298 K Tfuel298 K Tprod2909 K Treac298 K UP165406 UR165406 w0 x3 y8 z0 SOLUTION for C8H18 Ath125 fuelC8H18l hfuel249950 MolesCO0000 MolesCO28000 MolesH2O9 MolesN247000 MolesO20000 MolO20 Nameoctane Ru8314 kJkmolK SolMeth 100 the solution assumes complete combustion Theoair100 Thair1000 Tair298 K Tfuel298 K Tprod2911 K Treac298 K UP400104 UR400104 w0 x8 y18 z0 preparation If you are a student using this Manual you are using it without permission 15134 Fundamentals of Engineering FE Exam Problems 15129 A fuel is burned with 70 percent theoretical air This is equivalent to a 30 excess air b 70 excess air c 30 deficiency of air d 70 deficiency of air e stoichiometric amount of air PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Answer c 30 deficiency ofair 15130 Propane C3H8 is burned with 150 percent theoretical air The airfuel mass ratio for this combustion process is a 53 b 105 c 157 d 234 e 393 athnCnH4 mairnO232nN228 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values airth07 airthairaccess1 airth1airdeficiency Answer d 234 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values nC3 nH8 mfuelnH1nC12 coeff15 coeff1 for theoretical combustion 15 for 50 excess air nO2coeffath nN2376nO2 AFmairmfuel preparation If you are a student using this Manual you are using it without permission 15135 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15131 One kmol of methane CH4 is burned with an unknown amount of air during a combustion process If the combustion is complete and there are 1 kmol of free O2 in the products the airfuel mass ratio is a 346 b 257 c 172 d 143 e 119 nN2376nO2 mairnO232nN228 W3AFAFcoeff Ignoring excess air c the air is dry Answer b 257 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values nC1 nH4 mfuelnH1nC12 athnCnH4 coeff1ath1 O2 balance Coeff1 for theoretical combustion 15 for 50 excess air nO2coeffath AFmairmfuel Some Wrong Solutions with Common Mistakes W1AF1AF Taking the inverse of AF W2AFnO2nN2 Finding airfuel mole ratio 15132 A fuel is burned steadily in a combustion chamber The combustion temperature will be the highest except when a the fuel is preheated b the fuel is burned with a deficiency of air d the combustion chamber is well insulated e the combustion is complete Answer b the fuel is burned with a deficiency of air preparation If you are a student using this Manual you are using it without permission 15136 15133 An equimolar mixture of carbon dioxide and water vapor at 1 atm and 60C enter a dehumidifying section where the entire water vapor is condensed and removed from the mixture and the carbon dioxide leaves at 1 atm and 60C The entropy change of carbon dioxide in the dehumidifying section is a 28 kJkgK b 013 kJkgK c 0 d 013 kJkgK e 28 kJkgK PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Answer b 013 kJkgK T2T1 P21 atm W2DsCpCO2lnT2T1RCO2lnP1CO2P2CO2 Using pressure fractions backwards 15134 Methane CH4 is burned completely with 80 excess air during a steadyflow combustion process If both the reactants and the products are maintained at 25C and 1 atm and the water in the products exists in the liquid form the heat transfer from the combustion chamber per unit mass of methane is HHVCH4 5553 MJkg Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values CpCO20846 RCO201889 T160273 K P1 1 atm y1CO205 P1CO2y1CO2P1 y2CO21 P2CO2y2CO2P2 DsCO2CpCO2lnT2T1RCO2lnP2CO2P1CO2 Some Wrong Solutions with Common Mistakes W1Ds0 Assuming no entropy change a 890 MJkg b 802 MJkg c 75 MJkg d 56 MJkg e 50 MJkg Answer d 56 MJkg Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values T 25 C P1 atm EXCESS08 Heat transfer in this case is the HHV at room temperature LHVCH4 5005 MJkg Some Wrong Solutions with Common Mistakes W1QLHVCH4 Assuming lower heating value W2QEXCESShHVCH4 Assuming Q to be proportional to excess air preparation If you are a student using this Manual you are using it without permission 15137 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15135 The higher heating value of a hydrocarbon fuel CnHm with m 8 is given to be 1560 MJkmol of fuel Then its lower heating value is Answer a 1384 MJkmol 15136 Acetylene gas C2H2 is burned completely during a steadyflow combustion process The fuel and the air enter the combustion chamber at 25C and the products leave at 1500 K If the enthalpy of the products relative to the standard reference state is 404 MJkmol of fuel the heat transfer from the combustion chamber is Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Hprod404 MJkmol fuel W2Qout HreactHprod Adding enthalpies instead of subtracting them a 1384 MJkmol b 1208 MJkmol c 1402 MJkmol d 1540 MJkmol e 1550 MJkmol Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values HHV1560 MJkmol fuel hfg24423 MJkg Enthalpy of vaporization of water at 25C nH8 nwaternH2 mwaternwater18 LHVHHVhfgmwater Some Wrong Solutions with Common Mistakes W1LHVHHV hfgnwater Using mole numbers instead of mass W2LHV HHV hfgmwater2 Taking mole numbers of H2O to be m instead of m2 W3LHV HHV hfgnwater2 Taking mole numbers of H2O to be m instead of m2 and using mole numbers a 177 MJkmol b 227 MJkmol c 404 MJkmol d 631 MJkmol e 751 MJkmol Answer d 631 MJkmol hffuel2267301000 MJkmol fuel Hreacthffuel QoutHreactHprod Some Wrong Solutions with Common Mistakes W1Qout Hprod Taking Qout to be Hprod preparation If you are a student using this Manual you are using it without permission 15138 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 15137 Benzene gas C6H6 is burned with 95 percent theoretical air during a steadyflow combustion process The mole fraction of the CO in the products is Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values 2nCO2nCOnH2O2nO2 Oxygen balance nprodnCO2nCOnH2OnN2 Total mole numbers of product gases Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values XdestToSgen a 83 b 47 c 21 d 19 e 143 Answer c 21 nC6 nH6 athnCnH4 coeff095 coeff1 for theoretical combustion 15 for 50 excess air Assuming all the H burns to H2O the combustion equation is C6H6coeffathO2376N2 nCO2 CO2nCOCOnH2O H2OnN2 N2 nO2coeffath nN2376nO2 nH2OnH2 nCO2nCOnC yCOnCOnprod mole fraction of CO in product gases Some Wrong Solutions with Common Mistakes W1yCOnCOn1prod n1prodnCO2nCOnH2O Not including N2 in nprod W2yCOnCO2nCOnprod Using both CO and CO2 in calculations 15138 A fuel is burned during a steadyflow combustion process Heat is lost to the surroundings at 300 K at a rate of 1120 kW The entropy of the reactants entering per unit time is 17 kWK and that of the products is 15 kWK The total rate of exergy destruction during this combustion process is a 520 kW b 600 kW c 1120 kW d 340 kW e 739 kW Answer a 520 kW To300 K Qout1120 kW Sreact17 kWK Sprod 15 kWK SreactSprodQoutToSgen0 Entropy balance for steady state operation SinSoutSgen0 Some Wrong Solutions with Common Mistakes W1XdestSgen Taking Sgen as exergy destruction W2XdestToSgen1 SreactSprodSgen10 Ignoring QoutTo preparation If you are a student using this Manual you are using it without permission 15139 15139 15144 Design and Essay Problems PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis The mass flow rate of the liquid ethanolwater solution is given to be 10 kgs Considering that the mass fraction of ethanol in the solution is 02 Noting that the molar masses Methanol 46 and Mwater 18 kgkmol and that mole numbers N mM the mole flow rates become 15139 A certain industrial process generates a liquid solution of ethanol and water as the waste product The solution is to be burned using methane A combustion process is to be developed to accomplish this incineration process with minimum amount of methane 8 kgs 10 kgs 80 2 kgs 10 kgs 20 water ethanol m m 044444 kmols 18 kgkmol kgs 8 004348 kmols 46 kgkmol kgs 2 water water water ethanol ethanol ethanol M m N M m N Note that 10222 kmol H Okmol C H OH 0 04348 0 44444 5 2 2 ethanol water N N That is 10222 moles of liquid water is present in the solution for each mole of ethanol Assuming complete combustion the combustion equation of C2H5OH l with stoichiometric amount of air is where ath is the stoichiometric coefficient and is determined from the O2 balance s 1128N 3H O 2CO 376N 3 O C H OH l ten as 2 th 2 2 2 2 th 5 2 N 376 3H O 2CO 376N O C H OH a a l Thu th th 3 3 4 2 1 a a 2 2 2 2 2 5 2 Noting that 10222 kmol of liquid water accompanies each kmol of ethanol the actual combustion equation can be writ 10222H O l 1128N 3H O g 2CO 10222H O 376N 3 O C H OH 2 2 2 2 2 2 2 5 2 l l The heat transfer for this combustion process is determined from the steadyflow energy balance equation with W 0 R f R P f P h h h N h h h N Q o o o o Assuming the air and the combustion products to be ideal gases we have h hT We assume all the reactants to enter the combustion chamber at the standard reference temperature of 25C Furthermore we assume the products to leave the combustion chamber at 1400 K which is a little over the required temperature of 1100C From the tables preparation If you are a student using this Manual you are using it without permission 161 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 16 CHEMICAL AND PHASE EQUILIBRIUM PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 162 Kp and Equilibrium Composition of Ideal Gases 161C No the wooden table is NOT in chemical equilibrium with the air With proper catalyst it will reach with the oxygen in the air and burn 162C They are ν ν ν ν ν ν ν total and N P N N N N K e K P P P P K B A D C u B A D C B A D C p R T T G p v B A D v C p where B A D C ν ν ν ν ν The first relation is useful in partial pressure calculations the second in determining the Kp from gibbs functions and the last one in equilibrium composition calculations 163C a No because Kp depends on temperature only b In general the total mixture pressure affects the mixture composition The equilibrium constant for the reaction can be expressed as N O 2NO 2 2 total O N NO O2 N2 NO 2 O 2 2 2 NO ν ν ν ν ν ν N P N N N K N p The value of the exponent in this case is 211 0 Therefore changing the total mixture pressure will have no effect on the number of moles of N2 O2 and NO 164C a The equilibrium constant for the reaction 2 2 2 1 CO O CO can be expressed as total O CO CO O2 CO 2 CO 2 O 2 CO 2 CO 2 ν ν ν ν ν ν N P N N N K p Judging from the values in Table A28 the Kp value for this reaction decreases as temperature increases That is the indicated reaction will be less complete at higher temperatures Therefore the number of moles of CO2 will decrease and the number moles of CO and O2 will increase as the temperature increases b The value of the exponent in this case is 110505 which is negative Thus as the pressure increases the term in the brackets will decrease The value of Kp depends on temperature only and therefore it will not change with pressure Then to keep the equation balanced the number of moles of the products CO2 must increase and the number of moles of the reactants CO O2 must decrease PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 163 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course N 165C a The equilibrium constant for the reaction N 2 2 can be expressed as total N N N2 N 2 N 2 N ν ν ν ν N P N N K p Judging from the values in Table A28 the Kp value for this reaction increases as the temperature increases That is the indicated reaction will be more complete at higher temperatures Therefore the number of moles of N will increase and the number moles of N2 will decrease as the temperature increases b The value of the exponent in this case is 21 1 which is positive Thus as the pressure increases the term in the brackets also increases The value of Kp depends on temperature only and therefore it will not change with pressure Then to keep the equation balanced the number of moles of the products N must decrease and the number of moles of the reactants N2 must increase 166C The equilibrium constant for the reaction 2 2 2 1 CO O CO can be expressed as total O CO CO O2 CO 2 CO 2 O 2 CO 2 CO 2 ν ν ν ν ν ν N P N N N K p Adding more N2 an inert gas at constant temperature and pressure will increase Ntotal but will have no direct effect on other terms Then to keep the equation balanced the number of moles of the products CO2 must increase and the number of moles of the reactants CO O2 must decrease 167C The values of the equilibrium constants for each dissociation reaction at 3000 K are from Table A28 3 685 greater than 22359 ln 2H H 22359 ln 2N N 2 2 p p K K Thus H2 is more likely to dissociate than N2 168C a This reaction is the reverse of the known CO reaction The equilibrium constant is then 1 KP b This reaction is the reverse of the known CO reaction at a different pressure Since pressure has no effect on the equilibrium constant 1 KP c This reaction is the same as the known CO reaction multiplied by 2 The quilibirium constant is then 2 P K d This is the same as reaction c occurring at a different pressure Since pressure has no effect on the equilibrium constant 2 P K preparation If you are a student using this Manual you are using it without permission 164 169C a This reaction is the reverse of the known H2O reaction The equilibrium constant is then 1 KP b This reaction is the reverse of the known H2O reaction at a different pressure Since pressure has no effect on the equilibrium constant 1 KP c This reaction is the same as the known H2O reaction multiplied by 3 The quilibirium constant is then 3 P K d This is the same as reaction c occurring at a different pressure Since pressure has no effect on the equilibrium constant 3 P K 1610 The partial pressures of the constituents of an ideal gas mixture is given The Gibbs function of the nitrogen in this mixture at the given mixture pressure and temperature is to be determined Analysis The partial pressure of nitrogen is N2 CO2 NO PN2 110 kPa 293 K 1 086 atm 110 101325 110 kPa N2 P The Gibbs function of nitrogen at 293 K and 1086 atm is 200 kJkmol 8 314 kJkmolK293 Kln1086 atm 0 ln 293 K 1 atm 293 K 1086 atm PN2 R T g g u 1611 The mole fractions of the constituents of an ideal gas mixture is given The Gibbs function of the N2 in this mixture at the given mixture pressure and temperature is to be determined Analysis From Tables A18 and A26 at 1 atm pressure 61278 kJkmol 298 191502 8669 212066 600 17563 0 600 K 1 atm o o T Ts h T g g f 30 N2 30 O2 40 H2O 5 atm 600 K The partial pressure of N2 is atm 51 0 30 5 atm N2 CO P y P The Gibbs function of N2 at 600 K and 15 atm is 59260 kJkmol 8 314 kJkmol600 Kln15 atm 278 kJkmol 61 ln 600 K 1 atm 600 K 15 atm PCO R T g g u PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 165 1612 The temperature at which 02 percent of diatomic oxygen dissociates into monatomic oxygen at two pressures is to be determined Assumptions 1 The equilibrium composition consists of N2 and N 2 The constituents of the mixture are ideal gases Analysis a The stoichiometric and actual reactions can be written as Stoichiometric 2 1 and 2N thus N N N2 2 ν ν Actual 43 42 1 4243 1 prod react 2 2 0 004N 0 998N N N2 2N 02 1 kPa The equilibrium constant Kp can be determined from 7 2 1 2 total N2 N 1 579 10 0 004 0 998 1101325 0 998 0 004 N2 N N2 N ν ν ν ν N P N N K p and 1566 ln K p From Table A28 the temperature corresponding to this lnKp value is T 3628 K b At 10 kPa 6 2 1 2 total N2 N 1 579 10 0 004 0 998 10 101325 0 998 0 004 N2 N N2 N ν ν ν ν N P N N K p 1336 ln K p From Table A28 the temperature corresponding to this lnKp value is T 3909 K PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 166 1613 The equilibrium constant of the reaction 2 2 1 2 2 O H H O is to be determined using Gibbs function Analysis a The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K e K G T R T p G T R T p u u ln or H2O H2 ½O2 500 K where H2O H2O O2 O2 H2 H2 T g T g T g T G ν ν ν At 500 K 219067 kJkmol 206413 500 9904 16828 241820 1 220589 500 8682 14770 0 50 500 145628 8468 14350 0 1 H2O 298 500 H2O O2 298 500 O2 H2 298 500 H2 H2O H2O O2 O2 H2 H2 H2O H2O O2 O2 H2 H2 Ts h h h Ts h h h Ts h h h Ts h Ts h Ts h T g T g T g T G f f f ν ν ν ν ν ν ν ν ν Substituting 5270 219067 kJkmol8314 kJkmol K500 K ln K p or 5270 23 Table A 28 ln p p K K 10 130 At 2000 K 135556 kJkmol 264571 2000 9904 82593 241820 1 268655 2000 8682 67881 0 50 2000 188297 8468 61400 0 1 H2O 298 2000 H2O O2 298 2000 O2 H2 298 2000 H2 H2O H2O O2 O2 H2 H2 H2O H2O O2 O2 H2 H2 Ts h h h Ts h h h Ts h h h Ts h Ts h Ts h T g T g T g T G f f f ν ν ν ν ν ν ν ν ν Substituting 815 135556 kJkmol8314 kJkmol K2000 K ln K p or 8 15 4 Table A 28 ln p p K K 10 288 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 167 1614 The reaction C O2 CO2 is considered The mole fraction of the carbon dioxide produced when this reaction occurs at a1 atm and 3800 K are to be determined Assumptions 1 The equilibrium composition consists of CO2 C and O2 2 The constituents of the mixture are ideal gases Analysis The stoichiometric and actual reactions in this case are C O2 CO2 3800 K 1 atm Stoichiometric 1 1 and 1 CO thus O C CO2 O2 C 2 2 ν ν ν Actual 3 2 1 4243 1 products 2 react 2 2 CO O C O C z y x C balance x z z x 1 1 O balance x x z y z y 1 1 1 2 2 2 Total number of moles x z y x N 1 total The equilibrium constant relation can be expressed as total O2 C CO2 O2 C CO2 O2 C CO2 ν ν ν ν ν ν N P N N N K p From the problem statement at 3800 K 0 461 ln K p Then 0 6307 0 461 exp K p Substituting 1 1 1 1 1 1 6307 0 x x x x Solving for x x 07831 Then y x 07831 z 1 x 02169 Therefore the equilibrium composition of the mixture at 3800 K and 1 atm is 2 2 0 2169 CO 07831 C 07831 O The mole fraction of carbon dioxide is 01216 0 7831 1 2169 0 total CO2 CO2 N N y PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 168 1615 The reaction C O2 CO2 is considered The mole fraction of the carbon dioxide produced when this reaction occurs at a1 atm and 3800 K and 700 kPa and 3800 K are to be determined Assumptions 1 The equilibrium composition consists of CO2 C and O2 2 The constituents of the mixture are ideal gases Analysis We first solve the problem for 1 atm pressure The stoichiometric and actual reactions in this case are C O2 CO2 3800 K 1 atm Stoichiometric 1 1 and 1 CO thus O C CO2 O2 C 2 2 ν ν ν Actual 3 2 1 4243 1 products 2 react 2 2 CO O C O C z y x C balance x z z x 1 1 O balance x x z y z y 1 1 1 2 2 2 Total number of moles x z y x N 1 total The equilibrium constant relation can be expressed as total O2 C CO2 O2 C CO2 O2 C CO2 ν ν ν ν ν ν N P N N N K p From the problem statement at 3800 K 0 461 ln K p Then 0 6307 0 461 exp K p Substituting 1 1 1 1 1 1 6307 0 x x x x Solving for x x 07831 Then y x 07831 z 1 x 02169 Therefore the equilibrium composition of the mixture at 3800 K and 1 atm is 2 2 0 2169 CO 07831 C 07831 O The mole fraction of carbon dioxide is 01216 0 7831 1 2169 0 total CO2 CO2 N N y We repeat the calculations at 700 kPa pressure The pressure in this case is 700 kPa101325 kPaatm 6908 atm Then total O2 C CO2 O2 C CO2 O2 C CO2 ν ν ν ν ν ν N P N N N K p C O2 CO2 3800 K 700 kPa 1 1 1 1 6 908 1 6307 0 x x x x x 04320 y x 04320 z 1 x 05680 Therefore the equilibrium composition of the mixture at 3800 K and 700 kPa is 2 2 0 5680 CO 04320 C 04320 O The mole fraction of carbon dioxide is 03966 0 4320 1 5680 0 total CO2 CO2 N N y PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 169 1616 The reaction C O2 CO2 is considered The mole fraction of the carbon dioxide produced when this reaction occurs at a1 atm and 3800 K and 700 kPa and 3800 K are to be determined Assumptions 1 The equilibrium composition consists of CO2 C and O2 2 The constituents of the mixture are ideal gases Analysis We first solve the problem for the reaction C O2 CO2 The stoichiometric and actual reactions in this case are C O2 CO2 3800 K 1 atm Stoichiometric 1 1 and 1 CO thus O C CO2 O2 C 2 2 ν ν ν Actual 3 2 1 4243 1 products 2 react 2 2 CO O C O C z y x C balance x z z x 1 1 O balance x x z y z y 1 1 1 2 2 2 Total number of moles x z y x N 1 total The equilibrium constant relation can be expressed as total O2 C CO2 O2 C CO2 O2 C CO2 ν ν ν ν ν ν N P N N N K p From the problem statement at 3800 K 0 461 ln K p Then 0 6307 0 461 exp K p Substituting 1 1 1 1 1 1 6307 0 x x x x Solving for x x 07831 Then y x 07831 z 1 x 02169 Therefore the equilibrium composition of the mixture at 3800 K and 1 atm is 2 2 0 2169 CO 07831 C 07831 O The mole fraction of carbon dioxide is 01216 0 7831 1 2169 0 total CO2 CO2 N N y PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1610 If the reaction is C O2 376 N2 CO2 376 N2 The stoichiometric and actual reactions in this case are Stoichiometric 3 76 1 and 3 76 1 1 3 76 N thus CO 3 76 N O C N2 CO2 N2 O2 C 2 2 2 2 ν ν ν ν ν Actual 4 4 4 3 14 2 4243 1 products 2 2 react 2 2 2 3 76N CO O C 6 N 73 O C z y x CO2376N2 CO2376N2 3800 K 1 atm C balance x z z x 1 1 O balance x x z y z y 1 1 1 2 2 2 Total number of moles x z y x N 4 76 3 76 total The equilibrium constant relation can be expressed as total N2 O2 C N2 CO2 N2 O2 C N2 CO2 N2 O2 C N2 CO2 ν ν ν ν ν ν ν ν ν ν N P N N N N N K p or O2 C CO2 O2 C CO2 total O2 C CO2 ν ν ν ν ν ν N P N N N K p From the problem statement at 3800 K 1249 ln K p Then 265670 exp1249 p K Substituting 1 4 76 1 1 670 265 x x x x Solving for x x 0004226 Then y x 0004226 z 1 x 09958 Therefore the equilibrium composition of the mixture at 3800 K and 1 atm is 2 2 2 2 3 76 N 0 9958 CO 3 76 N 0004226 C 0004226 O The mole fraction of carbon dioxide is 02090 0 004226 4 76 9958 0 total CO2 CO2 N N y PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1611 1617 A gaseous mixture consisting of methane and carbon dioxide is heated The equilibrium composition by mole fraction of the resulting mixture is to be determined Assumptions 1 The equilibrium composition consists of CH4 C H2 and CO2 2 The constituents of the mixture are ideal gases Analysis The stoichiometric and actual reactions in this case are Stoichiometric 2 1 and 1 2H thus C CH H2 C CH4 2 4 ν ν ν Actual 43 42 1 4243 1 3 2 1 inert 2 products 2 react 4 2 4 07CO C H CH 70 CO 03CH z y x CH4 CO2 1200 K 1 atm C balance x y y x 30 30 H balance x z x 2 60 z 2 4 21 Total number of moles x z y x N 2 61 1 total The equilibrium constant relation can be expressed as H2 C CH4 H2 C CH4 total H2 C CH4 ν ν ν ν ν ν N P N N N K p From the problem statement at 1200 K 4 147 ln K p Then 63244 exp 4 147 p K For the reverse reaction that we consider 0 01581 1 63244 p K Substituting 1 2 1 2 2 61 1 2 60 30 01581 0 x x x x Solving for x x 00006637 Then y 03 x 02993 z 06 2x 05987 Therefore the equilibrium composition of the mixture at 1200 K and 1 atm is 2 2 4 CO 70 0 5987 H 02993 C 00006637 CH The mole fractions are 04379 03745 01872 0000415 1 599 70 1 599 5987 0 1 599 2993 0 1 599 0 0006637 0 0006637 2 61 0006637 0 total CO2 CO2 total H2 H2 total C C total CH4 CH4 N N y N N y N N y N N y PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1612 1618 The dissociation reaction CO2 CO O is considered The composition of the products at given pressure and temperature is to be determined Assumptions 1 The equilibrium composition consists of CO2 CO and O 2 The constituents of the mixture are ideal gases Analysis For the stoichiometric reaction O CO CO 2 2 1 2 from Table A28 at 2500 K CO2 2500 K 1 atm 3 331 ln K p For the oxygen dissociation reaction 05O2 O from Table A28 at 2500 K 4 255 8 509 2 ln p K For the desired stoichiometric reaction 1 1 and 1 O thus CO CO O CO CO2 2 ν ν ν 7 586 4 255 3 331 ln p K and 0 0005075 7 586 exp p K Actual 4243 1 3 2 1 products react 2 2 CO O CO CO z y x C balance x y y x 1 1 O balance x z y x 1 z 2 2 Total number of moles x z y x N 2 total The equilibrium constant relation can be expressed as CO2 O CO CO2 O CO total CO2 O CO ν ν ν ν ν ν N P N N N K p Substituting 1 1 1 2 1 1 1 0005075 0 x x x x Solving for x x 09775 Then y 1 x 00225 z 1 x 00225 Therefore the equilibrium composition of the mixture at 2500 K and 1 atm is 00225 O 09775 CO 00225 CO 2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1613 1619 The dissociation reaction CO2 CO O is considered The composition of the products at given pressure and temperature is to be determined when nitrogen is added to carbon dioxide Assumptions 1 The equilibrium composition consists of CO2 CO O and N2 2 The constituents of the mixture are ideal gases Analysis For the stoichiometric reaction O CO CO 2 2 1 2 from Table A28 at 2500 K CO2 3N2 2500 K 1 atm 3 331 ln K p For the oxygen dissociation reaction 05O2 O from Table A28 at 2500 K 4 255 8 509 2 ln p K For the desired stoichiometric reaction 1 1 and 1 O thus CO CO O CO CO2 2 ν ν ν 7 586 4 255 3 331 ln p K and 0 0005075 7 586 exp p K Actual inert 2 products react 2 2 2 3N CO O CO 3N CO 4243 1 3 2 1 z y x C balance x y y x 1 1 O balance x z y x 1 z 2 2 Total number of moles x z y x N 5 3 total The equilibrium constant relation can be expressed as CO2 O CO CO2 O CO total CO2 O CO ν ν ν ν ν ν N P N N N K p Substituting 1 1 1 5 1 1 1 0005075 0 x x x x Solving for x x 09557 Then y 1 x 00443 z 1 x 00443 Therefore the equilibrium composition of the mixture at 2500 K and 1 atm is 2 2 3N 00443 O 09557 CO 00443 CO PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1614 1620 The reaction N2 O2 2NO is considered The equilibrium mole fraction of NO 1600 K and 1 atm is to be determined Assumptions 1 The equilibrium composition consists of N2 O2 and NO 2 The constituents of the mixture are ideal gases Analysis The stoichiometric and actual reactions in this case are Stoichiometric 2 1 and 1 2NO thus O N NO O2 N2 2 2 ν ν ν N2 O2 1600 K 1 atm Actual products react 2 2 2 2 NO O N O N z y x 4243 1 N balance x z z x 2 2 2 2 O balance x y z y 2 2 Total number of moles 2 total z y x N The equilibrium constant relation can be expressed as total O2 N2 NO O2 N2 NO O2 N2 NO ν ν ν ν ν ν N P N N N K p From Table A28 at 1600 K 5 294 Since the stoichiometric reaction being considered is double this reaction ln p K 2 522 10 5 5 294 2 exp p K Substituting 1 1 2 2 2 5 2 1 2 2 522 10 2 x x Solving for x x 09975 Then y x 09975 z 2 2x 0005009 Therefore the equilibrium composition of the mixture at 1000 K and 1 atm is 0 005009 NO 0 9975 O 09975 N 2 2 The mole fraction of NO is then 0002505 2 005009 0 total NO NO N N y PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1615 1621E The equilibrium constant of the reaction H2 12O2 H2O is listed in Table A28 at different temperatures The data are to be verified at two temperatures using Gibbs function data Analysis a The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K e K G T R T p G T R T p u u ln or H2 ½O2 H2O 537 R where 2 2 2 2 2 2 O O H H H O H O T g T g T g T G ν ν ν At 537 R G T 1 98 350 1 0 05 0 98 350 Btu lbmol Substituting ln Kp 98 350 Btu lbmol 1986 Btu lbmol R537 R 9222 or K K p p 112 1040 Table A 28 ln 92 21 b At 4320 R 451 Btulbmol 48 65831 4320 3725 1 35746 0 50 46554 4320 3640 3 32647 2 0 1 65504 4320 4258 44533 104040 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 O 298 4320 O H 298 4320 H H O 537 4320 O H O O H H H O O H O O H H H O O H Ts h h h Ts h h h Ts h h h Ts h Ts h Ts h T g T g T g T G f f f ν ν ν ν ν ν ν ν ν Substituting 48451 Btulbmol1986 BtulbmolR4320 R 5647 ln p K or 5 619 Table A 28 ln p p K K 283 Discussion Solving this problem using EES with the builtin ideal gas properties give Kp 1041040 for part a and Kp 278 for part b PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1616 1622 The equilibrium constant of the reaction CO 12O2 CO2 at 298 K and 2000 K are to be determined and compared with the values listed in Table A28 Analysis a The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K e K G T R T p G T R T p u u ln or CO2 2 O2 1 CO 298 K where O2 O2 CO CO CO2 CO2 T g T g T g T G ν ν ν At 298 K 257210 kJkmol 50 0 137150 1 394360 1 T G where the Gibbs functions are obtained from Table A26 Substituting 10381 8314 kJkmol K298 K 257210 kJkmol ln K p From Table A28 K p 10376 ln b At 2000 K 110409 kJkmol 200026853 59193 50 200025848 1 53826 200030900 302128 1 O2 O2 CO CO CO2 CO2 O2 O2 CO CO CO2 CO2 Ts h Ts h Ts h T g T g T g T G ν ν ν ν ν ν The enthalpies at 2000 K and entropies at 2000 K and 1013 kPa 1 atm are obtained from EES Substituting 664 8314 kJkmol K2000 K 110409 kJkmol ln K p From Table A28 ln K p 6635 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1617 1623 The effect of varying the percent excess air during the steadyflow combustion of hydrogen is to be studied Analysis The combustion equation of hydrogen with stoichiometric amount of air is 2 2 2 2 2 50 3 76 N H O 376N O 50 H For the incomplete combustion with 100 excess air the combustion equation is 2 2 2 2 2 2 2 N O H 0 97 H O 376N 50 O 1 H c b a Ex The coefficients are to be determined from the mass balances Hydrogen balance 0 03 2 2 0 97 2 a a Oxygen balance 2 0 97 2 50 1 b Ex Nitrogen balance 2 3 76 2 50 1 c Ex Solving the above equations we find the coefficients Ex 1 a 003 b 0515 c 376 and write the balanced reaction equation as 2 2 2 2 2 2 2 3 76 N 0 515 O 0 03 H 0 97 H O 376N O H Total moles of products at equilibrium are 5 275 3 76 0 515 0 03 0 97 tot N The assumed equilibrium reaction is 2 2 2 50 O H H O The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K e K G T R T p G T R T p u u ln or where prod H2O H2O prod O2 O2 prod H2 H2 T g T g T g T G ν ν ν and the Gibbs functions are defined as H2O prod prod H2O O2 prod prod O2 H2 prod prod H2 s T h T g s T h T g s T h T g The equilibrium constant is also given by 0 009664 0 97 0 03 0 515 5 275 1 97 0 50 50 1 50 1 50 1 ab N P K tot p and 4 647 ln 0 009664 ln p K The corresponding temperature is obtained solving the above equations using EES to be Tprod 2600 K This is the temperature at which 97 percent of H2 will burn into H2O The copy of EES solution is given next Input Data from parametric table PercentEx 10 Ex PercentEx100 EX Excess air100 Pprod 1013kPa PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1618 Ru8314 kJkmolK The combustion equation of H2 with stoichiometric amount of air is H2 05O2 376N2H2O 05376N2 For the incomplete combustion with 100 excess air the combustion equation is H2 1EX05O2 376N2097 H2O aH2 bO2cN2 Specie balance equations give the values of a b and c H hydrogen 2 0972 a2 O oxygen 1Ex052097 b2 N nitrogen 1Ex05376 2 c2 Ntot 097a b c Total kilomoles of products at equilibrium The assumed equilibrium reaction is H2OH205O2 The following equations provide the specific Gibbs function ghTs for each H2mponent in the product gases as a function of its temperature Tprod at 1 atm pressure 1013 kPa gH2OEnthalpyH2OTTprod Tprod EntropyH2OTTprod P1013 gH2EnthalpyH2TTprod Tprod EntropyH2TTprod P1013 gO2EnthalpyO2TTprod Tprod EntropyO2TTprod P1013 The standardstate Gibbs function is DELTAG 1gH205gO21gH2O The equilibrium constant is given by Eq 1514 KP expDELTAG RuTprod PPprod 1013atm The equilibrium constant is also given by Eq 1515 K P PNtot1051a1b050971 sqrtPNtot a sqrtb KP 097 lnKp lnkP ln Kp PercentEx Tprod K 5414 10 2440 5165 20 2490 5019 30 2520 4918 40 2542 4844 50 2557 4786 60 2570 4739 70 2580 47 80 2589 4667 90 2596 4639 100 2602 10 20 30 40 50 60 70 80 90 100 2425 2465 2505 2545 2585 2625 PercentEx Tprod PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1619 1624 The equilibrium constant of the reaction CH4 2O2 CO2 2H2O at 25C is to be determined Analysis The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K e K G T R T p G T R T p u u ln or CH4 2O2 CO2 2H2O 25C where 2 2 4 4 2 2 2 2 O O CH CH H O H O CO CO T g T g T g T g T G ν ν ν ν At 25C G T 1 394 360 2 228 590 1 50 790 2 0 800 750 kJ kmol Substituting 800750 kJkmol8314 kJkmol K298 K 32304 ln K p or K p 196 10140 1625 The equilibrium constant of the reaction CO2 CO 12O2 is listed in Table A28 at different temperatures It is to be verified using Gibbs function data Analysis a The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using R T T G K e K u p R T T G p u or ln CO2 CO ½O2 298 K where 2 2 2 2 CO CO O O CO CO T g T g T g T G ν ν ν At 298 K 257210 kJkmol 394360 1 50 0 137150 1 T G Substituting 257210 kJkmol8314 kJkmol K298 K 10381 ln K p or 10376 Table A 28 ln p p K K 1046 824 b At 1800 K 240 2 kJkmol 127 9364 1800 302884 88806 1 393520 60371 8682 1800 264701 0 50 58191 8669 1800 254797 110530 1 2 2 2 2 2 2 2 2 2 2 2 2 CO 298 1800 CO O 298 1800 O CO 298 1800 CO CO CO O O CO CO CO CO O O CO CO Ts h h h Ts h h h Ts h h h Ts h Ts h Ts h T g T g T g T G f f f ν ν ν ν ν ν ν ν ν Substituting 8502 127240 2 kJkmol8314 kJkmol K1800 K ln K p or 8 497 Table A 28 ln p p K K 104 203 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1620 1626 Carbon monoxide is burned with 100 percent excess air The temperature at which 93 percent of CO burn to CO2 is to be determined Assumptions 1 The equilibrium composition consists of CO2 CO O2 and N2 2 The constituents of the mixture are ideal gases Analysis Assuming N2 to remain as an inert gas the stoichiometric and actual reactions can be written as Stoichiometric 1 and 1 CO thus O CO 2 1 O CO CO 2 2 2 1 2 2 ν ν ν Actual 43 42 1 444 3 4 44 2 1 4243 1 inert 2 reactants 2 product 2 2 2 3 76N 0 07CO 0535O 0 93CO 3 76N CO 1O The equilibrium constant Kp can be determined from 4180 3 76 0 535 0 07 0 93 1 0 535 07 0 0 93 51 1 50 total O CO CO O2 CO 2 CO 2 O 2 CO 2 CO 2 ν ν ν ν ν ν N P N N N K p CO ½O2 CO2 93 1 atm and 3 733 ln p K From Table A28 the temperature corresponding to this Kp value is T 2424 K PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1621 1627 Problem 1626 is reconsidered The effect of varying the percent excess air during the steadyflow process from 0 to 200 percent on the temperature at which 93 percent of CO burn into CO2 is to be studied Analysis The problem is solved using EES and the solution is given below To solve this problem we need to give EES a guess value for Tprop other than the default value of 1 Set the guess value of Tprod to 1000 K by selecting Variable Information in the Options menu Then press F2 or click the Calculator icon Input Data from the diagram window PercentEx 100 Ex PercentEx100 EX Excess air100 Pprod 1013 kPa Ru8314 kJkmolK f093 The combustion equation of CO with stoichiometric amount of air is CO 05O2 376N2CO2 05376N2 For the incomplete combustion with 100 excess air the combustion equation is CO 1EX05O2 376N2097 CO2 aCO bO2cN2 Specie balance equations give the values of a b and c C Carbon 1 f a O oxygen 1 1Ex052f2 a 1 b2 N nitrogen 1Ex05376 2 c2 Ntot fa b c Total kilomoles of products at equilibrium The assumed equilibrium reaction is CO2CO05O2 The following equations provide the specific Gibbs function ghTs for each component in the product gases as a function of its temperature Tprod at 1 atm pressure 1013 kPa gCO2EnthalpyCO2TTprod Tprod EntropyCO2TTprod P1013 gCOEnthalpyCOTTprod Tprod EntropyCOTTprod P1013 gO2EnthalpyO2TTprod Tprod EntropyO2TTprod P1013 The standardstate Gibbs function is DELTAG 1gCO05gO21gCO2 The equilibrium constant is given by Eq 1514 KP expDELTAG RuTprod PPprod 1013atm The equilibrium constant is also given by Eq 1515 K P PNtot1051a1b050971 sqrtPNtot a sqrtb KP f lnKp lnkP Compare the value of lnKp calculated by EES with the value of lnKp from table A28 in the text PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1622 PercentEx Tprod K 0 20 40 60 80 100 120 140 160 180 200 2247 2342 2377 2398 2411 2421 2429 2435 2440 2444 2447 0 40 80 120 160 200 2250 2300 2350 2400 2450 PercentEx Tprod K PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1623 1628E Carbon monoxide is burned with 100 percent excess air The temperature at which 93 percent of CO burn to CO2 is to be determined Assumptions 1 The equilibrium composition consists of CO2 CO O2 and N2 2 The constituents of the mixture are ideal gases Analysis Assuming N2 to remain as an inert gas the stoichiometric and actual reactions can be written as Stoichiometric 1 and 1 CO thus O CO 2 1 O CO CO 2 2 2 1 2 2 ν ν ν Actual 43 42 1 444 3 4 44 2 1 4243 1 inert 2 reactants 2 product 2 2 2 3 76N 0 07CO 0535O 0 93CO 3 76N CO 1O The equilibrium constant Kp can be determined from 4180 3 76 0 535 0 07 0 93 1 0 535 07 0 0 93 51 1 50 total O CO CO O2 CO 2 CO 2 O 2 CO 2 CO 2 ν ν ν ν ν ν N P N N N K p CO ½O2 CO2 93 1 atm and 3 733 ln p K From Table A28 the temperature corresponding to this Kp value is T 2424 K 4363 R 1629 Hydrogen is burned with 150 percent theoretical air The temperature at which 98 percent of H2 will burn to H2O is to be determined Assumptions 1 The equilibrium composition consists of H2O H2 O2 and N2 2 The constituents of the mixture are ideal gases Analysis Assuming N2 to remain as an inert gas the stoichiometric and actual reactions can be written as Stoichiometric H O H O thus and 2 1 2 2 H O H O 2 2 2 2 1 2 1 1 ν ν ν Actual H 075O N H O H 026O N 2 2 2 product 2 2 reactants inert 376 0 98 0 02 2 82 2 2 1 2 4 3 4 1 2 444 3 444 12 4 3 4 The equilibrium constant Kp can be determined from 19411 2 82 0 26 0 02 0 98 1 0 26 02 0 0 98 51 1 50 total O H H O O2 H2 2O H 2 O 2 2 H 2 2O H 2 ν ν ν ν ν ν N P N N N K p Combustion chamber H2 H2O H2 O2 N2 Air From Table A28 the temperature corresponding to this Kp value is T 2472 K PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1624 1630 Air is heated to a high temperature The equilibrium composition at that temperature is to be determined Assumptions 1 The equilibrium composition consists of N2 O2 and NO 2 The constituents of the mixture are ideal gases Analysis The stoichiometric and actual reactions in this case are Stoichiometric and 1 NO thus O N 2 1 O 2 1 N NO 2 2 1 2 2 1 2 2 ν ν ν AIR 2000 K 2 atm Actual 376 N O NO N O 2 2 prod 2 2 reactants x y z 123 1 42 3 4 N balance 752 x 2y or y 376 05x O balance 2 x 2z or z 1 05x Total number of moles Ntotal x y z x 476 x 476 The equilibrium constant relation can be expressed as total O N NO O2 N2 NO 2 O 2 2 N 2 NO ν ν ν ν ν ν N P N N N K p From Table A28 ln Kp 3931 at 2000 K Thus Kp 001962 Substituting 1 1 50 50 4 76 2 50 1 50 3 76 01962 0 x x x Solving for x x 00376 Then y 37605x 37412 z 105x 09812 Therefore the equilibrium composition of the mixture at 2000 K and 2 atm is 0 0376 0 9812 NO 37412N O 2 2 The equilibrium constant for the reactions O2 2O ln Kp 14622 and N 2 2N ln Kp 41645 are much smaller than that of the specified reaction ln Kp 3931 Therefore it is realistic to assume that no monatomic oxygen or nitrogen will be present in the equilibrium mixture Also the equilibrium composition is in this case is independent of pressure since ν 1 05 05 0 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1625 1631 Hydrogen is heated to a high temperature at a constant pressure The percentage of H2 that will dissociate into H is to be determined Assumptions 1 The equilibrium composition consists of H2 and H 2 The constituents of the mixture are ideal gases Analysis The stoichiometric and actual reactions can be written as Stoichiometric 2 1 and 2H thus H H H 2 2 ν ν H2 4000 K 5 atm Actual prod react 2 2 H H H y x H balance 2 2x y or y 2 2x Total number of moles Ntotal x y x 2 2x 2 x The equilibrium constant relation can be expressed as H2 H 2 H 2 H total H H ν ν ν ν N P N N K p From Table A28 ln Kp 0934 at 4000 K Thus Kp 2545 Substituting 2 1 2 2 5 2 2 545 2 x x x Solving for x x 0664 Thus the percentage of H2 which dissociates to H at 3200 K and 8 atm is 1 0664 0336 or 336 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1626 1632E A mixture of CO O2 and N2 is heated to a high temperature at a constant pressure The equilibrium composition is to be determined Assumptions 1 The equilibrium composition consists of CO2 CO O2 and N2 2 The constituents of the mixture are ideal gases Analysis The stoichiometric and actual reactions in this case are PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Stoichiometric 1 and 1 CO thus O CO 2 1 O CO CO 2 2 2 1 2 2 ν ν ν Actual 2CO 2O N CO CO O N 2 2 2 products 2 reactants 2 inert 6 6 x y z 123 1 42 3 44 2 CO 2 O2 6 N2 4320 R 3 atm C balance 2 2 x y y x x O balance 6 2 2 2 05 x y z z x Total number of moles N x y z total 6 10 05 The equilibrium constant relation can be expressed as total O CO CO O2 CO 2 CO 2 O 2 CO 2 CO 2 ν ν ν ν ν ν N P N N N K p From Table A28 ln 47465 2400 K Thus 4320 3 860 at p p K R T K Substituting 51 1 50 50 10 3 50 2 2 465 47 x x x x Solving for x x 1930 Then y 2 x 0070 z 2 05x 1035 Therefore the equilibrium composition of the mixture at 2400 K and 3 atm is 1930CO 0070CO 1035O 6N 2 2 2 preparation If you are a student using this Manual you are using it without permission 1627 1633 A mixture of N2 O2 and Ar is heated to a high temperature at a constant pressure The equilibrium composition is to be determined Assumptions 1 The equilibrium composition consists of N2 O2 Ar and NO 2 The constituents of the mixture are ideal gases Analysis The stoichiometric and actual reactions in this case are PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Stoichiometric and 1 NO thus O N 2 1 O 2 1 N NO 2 2 1 2 2 1 2 2 ν ν ν Actual 3 01 0 N O Ar NO N O Ar 2 2 prod 2 2 reactants inert x y z 123 1 42 43 123 3 N2 1 O2 01 Ar 2400 K 10 atm 1 N balance 6 2 3 05 x y y x O balance 2 2 1 05 x z z x Total number of moles N x y z total 01 41 The equilibrium constant relation becomes 0 5 0 5 1 total 0 5 0 5 total 2 2 2 2 2 2 ν ν ν ν O ν N ν NO p N P z y x N P N N N K O N NO O N NO From Table A28 ln 0 04885 3 019 at 2400 K Thus p p K K Substituting 1 50 1 50 3 0 04885 50 50 x x x Solving for x x 00823 Then y 3 05x 29589 z 1 05x 09589 Therefore the equilibrium composition of the mixture at 2400 K and 10 atm is 01Ar 09589O 00823NO 29589N 2 2 preparation If you are a student using this Manual you are using it without permission 1628 1634 The mole fraction of sodium that ionizes according to the reaction Na Na e at 2000 K and 15 atm is to be determined Assumptions All components behave as ideal gases Analysis The stoichiometric and actual reactions can be written as Na Na e 2000 K 15 atm Stoichiometric 1 1 and 1 e thus Na Na e Na Na ν ν ν Actual 4 4 43 142 products react e Na Na Na y y x Na balance 1 1 x y or y x Total number of moles N x y x total 2 2 The equilibrium constant relation becomes 1 1 1 total 2 total Na Na Na e Na Na e Na N P x y N P N N N K e p ν ν ν ν ν ν Substituting x x x 2 51 1 668 0 2 Solving for x x 04449 Thus the fraction of Na which dissociates into Na and e is 1 04449 0555 or 555 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1629 1635 Oxygen is heated from a specified state to another state The amount of heat required is to be determined without and with dissociation cases Assumptions 1 The equilibrium composition consists of O2 and O 2 The constituents of the mixture are ideal gases Analysis a Obtaining oxygen properties from table A19 an energy balance gives 50989 kJkmol 6203 192 57 1 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u u q E E E 43 42 1 4243 1 O2 2200 K 1 atm b The stoichiometric and actual reactions in this case are Stoichiometric 2 1 and 2O thus O O O2 2 ν ν Actual products react 2 2 O O O y x O balance x y y x 2 2 2 2 Total number of moles x y x N 2 total The equilibrium constant relation can be expressed as O2 O O2 O total O2 O ν ν ν ν N P N N K p From Table A28 at 2200 K 11827 Then ln p K 7 305 10 6 11827 exp p K Substituting 2 1 2 6 2 1 2 2 305 10 7 x x x Solving for x x 099865 Then y 2 2x 00027 Therefore the equilibrium composition of the mixture at 2200 K and 1 atm is 0 0027 O 099865 O 2 Hence the oxygen ions are negligible and the result is same as that in part a in 50989 kJkmol q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1630 1636 Air is heated from a specified state to another state The amount of heat required is to be determined without and with dissociation cases Assumptions 1 The equilibrium composition consists of O2 and O and N2 2 The constituents of the mixture are ideal gases Analysis a Obtaining air properties from table A17 an energy balance gives 1660 kJkg 21264 4 1872 1 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in u u q E E E 43 42 1 4243 1 O2 376N2 2200 K 1 atm b The stoichiometric and actual reactions in this case are Stoichiometric 2 1 and 2O thus O O O2 2 ν ν Actual 43 42 1 inert 2 products react 2 2 2 3 76N O O 3 76N O y x O balance x y y x 2 2 2 2 Total number of moles x y x N 5 76 3 76 total The equilibrium constant relation can be expressed as O2 O O2 O total O2 O ν ν ν ν N P N N K p From Table A28 at 2200 K 11827 Then ln p K 7 305 10 6 11827 exp p K Substituting 2 1 2 6 5 76 1 2 2 305 10 7 x x x Solving for x x 099706 Then y 2 2x 000588 Therefore the equilibrium composition of the mixture at 2200 K and 1 atm is 2 2 3 76 N 0 00588 O 099706 O Hence the atomic oxygen is negligible and the result is same as that in part a in 1660 kJkg q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1631 1637 Liquid propane enters a combustion chamber The equilibrium composition of product gases and the rate of heat transfer from the combustion chamber are to be determined Assumptions 1 The equilibrium composition consists of CO2 H2O CO N2 and O2 2 The constituents of the mixture are ideal gases 1200 K C3H8 25C Air 12C Combustion chamber 2 atm CO CO2 H2O O2 N2 Analysis a Considering 1 kmol of C3H8 the stoichiometric combustion equation can be written as 2 th 2 2 2 2 th 8 3 N 4H O 376 3CO 376N O C H a a l where ath is the stoichiometric coefficient and is determined from the O2 balance 25 3 2 15 5 th th th a a a Then the actual combustion equation with 150 excess air and some CO in the products can be written as C H O N CO 9 05 O H O 47N 3 8 2 2 2 2 2 2 CO l 125 376 3 4 x x x After combustion there will be no C3 H8 present in the combustion chamber and H2O will act like an inert gas The equilibrium equation among CO2 CO and O2 can be expressed as 1 and 1 O thus CO CO 2 1 O CO CO 2 2 1 2 2 2 ν ν ν and total CO O CO CO2 O2 CO 2 CO 2 2 O 2 CO ν ν ν ν ν ν N P N N N K p where N x x x total x 3 9 05 4 47 63 05 p p K K From Table A28 ln 1 73 10 8 Substituting 871 at 1200 K Thus 17 1 51 50 8 50 63 2 50 9 3 73 10 1 x x x x Solving for x 03 2 9999999 x Therefore the amount CO in the product gases is negligible and it can be disregarded with no loss in accuracy Then the combustion equation and the equilibrium composition can be expressed as C H O N CO O H O 47N 3 8 2 2 2 2 2 2 l 12 5 376 3 7 5 4 and 3CO 75O 4H O 47N 2 2 2 2 b The heat transfer for this combustion process is determined from the steadyflow energy balance E E E in out system on the combustion chamber with W 0 R f R P f P h h h N h h h N Q o o o o out Assuming the air and the combustion products to be ideal gases we have h hT From the tables The h f o of liquid propane is obtained by adding the hfg at 25C to h f o of gaseous propane PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1632 Substance o f h kJkmol h285 K kJkmol h298 K kJkmol h1200 K kJkmol C3H8 l 118910 O2 0 86965 8682 38447 N2 0 82865 8669 36777 H2O g 241820 9904 44380 CO2 393520 9364 53848 Substituting Q h h out 3 8 kJ kmol of C H 3 393 520 53 848 9364 4 241820 44 380 9904 75 0 38 447 8682 47 0 36 777 8669 1 118 910 125 0 82965 8682 47 0 81865 8669 185 764 298 298 or Qout 3 185 764 kJ kmol of C H8 The mass flow rate of C3H8 can be expressed in terms of the mole numbers as N m M 12 44 0 02727 kg min kg kmol kmol min Thus the rate of heat transfer is 5066 kJmin 0 02727 kmolmin185746 kJkmol out out Q N Q The equilibrium constant for the reaction NO O N 2 2 1 2 2 1 is ln Kp 7569 which is very small This indicates that the amount of NO formed during this process will be very small and can be disregarded PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1633 1638 Problem 1637 is reconsidered It is to be investigated if it is realistic to disregard the presence of NO in the product gases Analysis The problem is solved using EES and the solution is given below To solve this problem the Gibbs function of the product gases is minimized Click on the MinMax icon For this problem at 1200 K the moles of CO are 0000 and moles of NO are 0000 thus we can disregard both the CO and NO However try some product temperatures above 1286 K and observe the sign change on the Qout and the amout of CO and NO present as the product temperature increases The reaction of C3H8liq with excess air can be written C3H8l 1ExAth O2376N2 a C02 b CO c H2O d N2 e O2 f NO The coefficients Ath and EX are the theoretical oxygen and the percent excess air on a decimal basis Coefficients a b c d e and f are found by minimiming the Gibbs Free Energy at a total pressure of the product gases PProd and the product temperature TProd The equilibrium solution can be found by applying the Law of Mass Action or by minimizing the Gibbs function In this problem the Gibbs function is directly minimized using the optimization capabilities built into EES To run this program click on the MinMax icon There are six compounds present in the products subject to four specie balances so there are two degrees of freedom Minimize the Gibbs function of the product gases with respect to two molar quantities such as coefficients b and f The equilibrium mole numbers a b c d e and f will be determined and displayed in the Solution window PercentEx 150 Ex PercentEx100 EX Excess air100 Pprod 2Patm TProd1200 K mdotfuel 05 kgs FuelC3H8 Tair 12273 K Tfuel 25273 K Patm 101325 kPa Ru8314 kJkmolK Theoretical combustion of C3H8 with oxygen C3H8 Ath O2 3 C02 4 H2O 2Ath 32 41 Balance the reaction for 1 kmol of C3H8 C3H8l 1ExAth O2376N2 a C02 b CO c H2O d N2 e O2 f NO bmax 3 fmax 1ExAth3762 eguessExAth 13 a1b1 Carbon balance 18c2 Hydrogen balance 1ExAth2a2b1c1e2f1 Oxygen balance 1ExAth3762d2f1 Nitrogen balance PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1634 Total moles and mole fractions NTotalabcdef yCO2aNTotal yCObNTotal yH2OcNTotal yN2dNTotal yO2eNTotal yNOfNTotal The following equations provide the specific Gibbs function for each component as a function of its molar amount gCO2EnthalpyCO2TTProdTProdEntropyCO2TTProdPPProdyCO2 gCOEnthalpyCOTTProdTProdEntropyCOTTProdPPProdyCO gH2OEnthalpyH2OTTProdTProdEntropyH2OTTProdPPProdyH2O gN2EnthalpyN2TTProdTProdEntropyN2TTProdPPProdyN2 gO2EnthalpyO2TTProdTProdEntropyO2TTProdPPProdyO2 gNOEnthalpyNOTTProdTProdEntropyNOTTProdPPProdyNO The extensive Gibbs function is the sum of the products of the specific Gibbs function and the molar amount of each substance GibbsagCO2bgCOcgH2OdgN2egO2fgNO For the energy balance we adjust the value of the enthalpy of gaseous propane given by EES hfgfuel 15060kJkmol Table A27 hfuel enthalpyFuelTTfuelhfgfuel Energy balance for the combustion process C3H8l 1ExAth O2376N2 a C02 b CO c H2O d N2 e O2 f NO HR QoutHP HRhfuel 1ExAthenthalpyO2TTair376enthalpyN2TTair HPaenthalpyCO2TTprodbenthalpyCOTTprodcenthalpyH2OTTproddenthalpyN2TTpro deenthalpyO2TTprodfenthalpyNOTTprod The heat transfer rate is QdotoutQoutmolarmassFuelmdotfuel kW SOLUTION a3000 kmol Ath5 b0000 kmol bmax3 c4000 kmol d47000 kmol e7500 kmol Ex15 eguess75 f0000 kmol FuelC3H8 fmax94 Gibbs17994897 kJ gCO703496 kJkmol gCO2707231 kJkmol gH2O515974 kJkmol gN2248486 kJkmol gNO342270 kJkmol gO2284065 kJkmol HP330516747 kJkmol HR141784529 kJkmol hfgfuel15060 kJkmol hfuel118918 kJkmol mdotfuel05 kgs NTotal615 kmolkmolfuel PercentEx150 Patm1013 kPa Pprod2027 kPa Qdotout2140 kW Qout188732 kJkmolfuel Ru8314 kJkmolK Tair285 K Tfuel298 K TProd120000 K yCO1626E15 yCO2004878 yH2O006504 yN207642 yNO7857E08 yO20122 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1635 1639 Oxygen is heated during a steadyflow process The rate of heat supply needed during this process is to be determined for two cases Assumptions 1 The equilibrium composition consists of O2 and O 2 All components behave as ideal gases Analysis a Assuming some O2 dissociates into O the dissociation equation can be written as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course O O O 2 2 2 1 x x Q O2 298 K O2 O 3000 K The equilibrium equation among O2 and O can be expressed as 2 1 and 2O thus O O O 2 2 ν ν Assuming ideal gas behavior for all components the equilibrium constant relation can be expressed as O2 O 2 O 2 O total O O ν ν ν ν N P N N K p where N x x total 2 1 2 x From Table A28 ln 0 01282 4 357 at 3000 K Thus p p K K Substituting 2 1 2 2 1 2 2 01282 0 x x x Solving for x gives x 0943 Then the dissociation equation becomes O O 2 2 0 943 0114 O The heat transfer for this combustion process is determined from the steadyflow energy balance E E E in out system on the combustion chamber with W 0 R f R P f P h h h N h h h N Q o o o o in Assuming the O2 and O to be ideal gases we have h hT From the tables Substance hf o kJkmol h298 K kJkmol h3000 K kJkmol O 249190 6852 63425 O2 0 8682 106780 Substituting Qin kJ kmol O2 0943 0 106 780 8682 0114 249190 63 425 6852 0 127 363 The mass flow rate of O2 can be expressed in terms of the mole numbers as 0 01563 kmolmin 32 kgkmol kgmin 50 M m N Thus the rate of heat transfer is 1990 kJmin 0 01563 kmolmin127363 kJkmol in in Q N Q b If no O2 dissociates into O then the process involves no chemical reactions and the heat transfer can be determined from the steadyflow energy balance for nonreacting systems to be 1533 kJmin 0 01563 kmolmin1067808682 kJkmol 1 2 1 2 in h N h h m h Q preparation If you are a student using this Manual you are using it without permission 1636 1640 The equilibrium constant Kp is to be estimated at 3000 K for the reaction CO H2O CO2 H2 Analysis a The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using R T T G K e K u p R T T G p u or ln where H2O H2O CO CO H2 H2 CO2 CO2 T g T g T g T g T G ν ν ν ν At 3000 K 49291 kJkmol 3000286273 9904 136264 1 241820 3000273508 8669 102210 1 110530 3000202778 8468 97211 1 0 3000334084 9364 162226 393520 1 H2O H2O CO CO H2 H2 CO2 CO2 H2O H2O CO CO H2 H2 CO2 CO2 Ts h Ts h Ts h Ts h T g T g T g T g T G ν ν ν ν ν ν ν ν Substituting 01386 p p K K 1 9762 8314 kJkmol K3000 K 49291 kJkmol ln The equilibrium constant may be estimated using the integrated vant Hoff equation 01307 est est 1 est 3000 K 1 2000 K 1 8 314 kJkmolK 26176 kJkmol 0 2209 ln 1 1 ln p p R u R p p K K T T R h K K PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1637 1641 A constant volume tank contains a mixture of H2 and O2 The contents are ignited The final temperature and pressure in the tank are to be determined Analysis The reaction equation with products in equilibrium is 2 2 2 2 2 O H O H O H c b a The coefficients are determined from the mass balances Hydrogen balance b a 2 2 2 Oxygen balance c b 2 2 The assumed equilibrium reaction is 2 2 2 50 O H H O The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K e K G T R T p G T R T p u u ln or where prod H2O H2O prod O2 O2 prod H2 H2 T g T g T g T G ν ν ν and the Gibbs functions are given by H2O prod prod H2O O2 prod prod O2 H2 prod prod H2 s T h T g s T h T g s T h T g The equilibrium constant is also given by 50 2 50 1 50 1 tot 1 50 1 101 3 c b a P b ac N P b a c K p An energy balance on the tank under adiabatic conditions gives P R U U where 4958 kJkmol 8 314 kJkmolK29815 K 0 8 314 kJkmolK29815 K 0 1 1 reac O225 C reac H225 C R T h R T h U u u R prod O2 prod H2O prod H2 prod prod prod R T c h R T b h R T a h U u T u T u T P The relation for the final pressure is 29815 K 101 3 kPa 2 prod 1 reac prod 1 tot 2 T c b a P T T N N P Solving all the equations simultaneously using EES we obtain the final temperature and pressure in the tank to be kPa 1043 K 3857 2 prod P T PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1638 1642 It is to be shown that as long as the extent of the reaction α for the disassociation reaction X2 2X is smaller than one α is given by P P K K 4 α Assumptions The reaction occurs at the reference temperature Analysis The stoichiometric and actual reactions can be written as Stoichiometric 2 1 and 2X thus X X X2 2 ν ν Actual prod react 2 2 2 X 1 X αX α 4243 1 The equilibrium constant Kp is given by 1 1 4 1 1 1 2 2 2 1 2 total X2 X X2 X X2 X α α α α α α ν ν ν ν N P N N K p Solving this expression for α gives P P K K 4 α PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1639 Simultaneous Reactions 1643C It can be expresses as dGTP 0 for each reaction Or as the Kp relation for each reaction must be satisfied 1644C The number of Kp relations needed to determine the equilibrium composition of a reacting mixture is equal to the difference between the number of species present in the equilibrium mixture and the number of elements PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1640 1645 Two chemical reactions are occurring in a mixture The equilibrium composition at a specified temperature is to be determined Assumptions 1 The equilibrium composition consists of H2O OH O2 and H2 2 The constituents of the mixture are ideal gases 2H2 O H2OOH H2O 3400 K 1 atm Analysis The reaction equation during this process can be expressed as H O H O H O OH 2 2 2 2 x y z w Mass balances for hydrogen and oxygen yield H balance 1 2 2 2 x y w O balance 1 2 x z w 2 The mass balances provide us with only two equations with four unknowns and thus we need to have two more equations to be obtained from the Kp relations to determine the equilibrium composition of the mixture They are 2 2 1 2 2 O H H O reaction 1 OH H H O 2 2 1 2 reaction 2 The equilibrium constant for these two reactions at 3400 K are determined from Table A28 to be ln ln K K K K P P P P 1 1 2 2 1891 015092 1576 020680 The Kp relations for these two simultaneous reactions are total O H OH H 2 total O H O H 1 H2O OH 2 H 2O H 2 OH 2 H 2 H2O O2 2 H 2O H 2 2 O 2 2 H 2 and ν ν ν ν ν ν ν ν ν ν ν ν N P N N N K N P N N N K P P where N N N N N x y z total H O H O OH 2 2 2 w Substituting 1 2 1 2 1 0 15092 w z y x x y z 3 1 2 1 2 1 0 20680 w z y x x w y 4 Solving Eqs 1 2 3 and 4 simultaneously for the four unknowns x y z and w yields x 0574 y 0308 z 0095 w 0236 Therefore the equilibrium composition becomes 0574H O 0308H 0095O 0236OH 2 2 2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1641 1646 Two chemical reactions are occurring in a mixture The equilibrium composition at a specified temperature is to be determined Assumptions 1 The equilibrium composition consists of CO2 CO O2 and O 2 The constituents of the mixture are ideal gases Analysis The reaction equation during this process can be expressed as CO2 CO O2 O 2000 K 4 atm 2CO O CO CO O O 2 2 2 2 x y z w Mass balances for carbon and oxygen yield C balance 1 2 x y O balance 2 6 2 2 x y z w The mass balances provide us with only two equations with four unknowns and thus we need to have two more equations to be obtained from the KP relations to determine the equilibrium composition of the mixture They are 2 2 1 2 O CO CO reaction 1 reaction 2 O 2 2 O The equilibrium constant for these two reactions at 2000 K are determined from Table A28 to be 7 2 2 1 1 4 464 10 14622 ln 0 001314 6 635 ln P P P P K K K K The KP relations for these two simultaneous reactions are O2 O 2 O 2 O CO2 O2 CO 2 CO 2 2 O 2 CO total O O 2 total CO O CO 1 ν ν ν ν ν ν ν ν ν ν N P N N K N P N N N K P P where N N N N N x y z total CO O CO O 2 2 w Substituting 1 2 1 2 4 0 001314 w z y x x y z 3 2 1 2 7 4 10 464 4 w z y x z w 4 Solving Eqs 1 2 3 and 4 simultaneously using an equation solver such as EES for the four unknowns x y z and w yields x 1998 y 0002272 z 1001 w 0000579 Thus the equilibrium composition is 0000579O 1001O 0002272CO 1998CO 2 2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1642 1647 Two chemical reactions are occurring at hightemperature air The equilibrium composition at a specified temperature is to be determined Assumptions 1 The equilibrium composition consists of O2 N2 O and NO 2 The constituents of the mixture are ideal gases Analysis The reaction equation during this process can be expressed as AIR Reaction chamber 2 atm Heat O2 N2 O NO 3000 K O 376 N N NO O O 2 2 2 2 x y z w Mass balances for nitrogen and oxygen yield N balance 1 52 2x y 7 O balance 2 w z y 2 2 The mass balances provide us with only two equations with four unknowns and thus we need to have two more equations to be obtained from the Kp relations to determine the equilibrium composition of the mixture They are 1 2 1 2 N O N 2 2 O O reaction 1 reaction 2 O 2 2 The equilibrium constant for these two reactions at 3000 K are determined from Table A28 to be ln ln K K K K P P P P 1 1 2 2 2114 012075 4 357 001282 The KP relations for these two simultaneous reactions are O2 O 2 O 2 O O2 N2 NO 2 O 2 2 N 2 NO total O O 2 total O N NO 1 ν ν ν ν ν ν ν ν ν ν N P N N K N P N N N K P P where w z y x N N N N N O O NO N total 2 2 Substituting 50 50 1 50 50 2 12075 0 w z y x z x y 3 2 1 2 2 01282 0 w z y x z w 4 Solving Eqs 1 2 3 and 4 simultaneously using EES for the four unknowns x y z and w yields x 3656 y 02086 z 08162 w 01591 Thus the equilibrium composition is 01591O 08162O 02086NO 3656N 2 2 The equilibrium constant of the reaction N 2 2 N at 3000 K is lnKP 22359 which is much smaller than the KP values of the reactions considered Therefore it is reasonable to assume that no N will be present in the equilibrium mixture PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1643 1648E Two chemical reactions are occurring in air The equilibrium composition at a specified temperature is to be determined Assumptions 1 The equilibrium composition consists of O2 N2 O and NO 2 The constituents of the mixture are ideal gases Analysis The reaction equation during this process can be expressed as AIR Reaction chamber 1 atm Heat O2 N2 O NO 5400 R O 376 N N NO O O 2 2 2 2 x y z w Mass balances for nitrogen and oxygen yield N balance 752 2 x y 1 O balance 2 2 2 y z w The mass balances provide us with only two equations with four unknowns and thus we need to have two more equations to be obtained from the Kp relations to determine the equilibrium composition of the mixture They are NO O N 2 2 1 2 2 1 reaction 1 reaction 2 O 2 2 O The equilibrium constant for these two reactions at T 5400 R 3000 K are determined from Table A28 to be ln ln K K K K P P P P 1 1 2 2 2114 012075 4 357 001282 The KP relations for these two simultaneous reactions are O2 O 2 O 2 O O2 N2 NO 2 O 2 2 N 2 NO total O O 2 total O N NO 1 ν ν ν ν ν ν ν ν ν ν N P N N K N P N N N K P P where N N N N N x y z total N NO O O 2 2 w Substituting 50 50 1 50 50 1 12075 0 w z y x z x y 3 2 1 2 1 01282 0 w z y x z w 4 Solving Eqs 1 2 3 and 4 simultaneously for the four unknowns x y z and w yields x 3658 y 02048 z 07868 w 02216 Thus the equilibrium composition is 02216O 07868O 02048NO 3658N 2 2 The equilibrium constant of the reaction N 2 2 N at 5400 R is lnKP 22359 which is much smaller than the KP values of the reactions considered Therefore it is reasonable to assume that no N will be present in the equilibrium mixture PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1644 1449E Problem 1648E is reconsidered Using EES or other software the equilibrium solution is to be obtained by minimizing the Gibbs function by using the optimization capabilities built into EES This solution technique is to be compared with that used in the previous problem Analysis The problem is solved using EES and the solution is given below This example illustrates how EES can be used to solve multireaction chemical equilibria problems by directly minimizing the Gibbs function 021 O2079 N2 a O2b O c N2 d NO Two of the four coefficients a b c and d are found by minimiming the Gibbs function at a total pressure of 1 atm and a temperature of 5400 R The other two are found from mass balances The equilibrium solution can be found by applying the Law of Mass Action to two simultaneous equilibrium reactions or by minimizing the Gibbs function In this problem the Gibbs function is directly minimized using the optimization capabilities built into EES To run this program select MinMax from the Calculate menu There are four compounds present in the products subject to two elemental balances so there are two degrees of freedom Minimize Gibbs with respect to two molar quantities such as coefficients b and d The equilibrium mole numbers of each specie will be determined and displayed in the Solution window Minimizing the Gibbs function to find the equilibrium composition requires good initial guesses Data from Data Input Window T5400 R P1 atm AO2021 BN2079 Composition of air AO22a2bd Oxygen balance BN22c2d Nitrogen balance The total moles at equilibrium are Ntotabcd yO2aNtot yObNtot yN2cNtot yNOdNtot The following equations provide the specific Gibbs function for three of the components gO2EnthalpyO2TTTEntropyO2TTPPyO2 gN2EnthalpyN2TTTEntropyN2TTPPyN2 gNOEnthalpyNOTTTEntropyNOTTPPyNO EES does not have a builtin property function for monatomic oxygen so we will use the JANAF procedure found under OptionsFunction InfoExternal Procedures The units for the JANAF procedure are kgmole K and kJ so we must convert h and s to English units TKTConvertRK Convert R to K Call JANAFOTKCphS Units from JANAF are SI SOSConvertkJkgmoleK BtulbmoleR hOhConvertkJkgmole Btulbmole The entropy from JANAF is for one atmosphere so it must be corrected for partial pressure gOhOTSORulnYO Ru19858 The universal gas constant in BtumoleR The extensive Gibbs function is the sum of the products of the specific Gibbs function and the molar amount of each substance GibbsagO2bgOcgN2dgNO PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1645 d lbmol b lbmol Gibbs Btulbmol yO2 yO yNO yN2 T R 0002698 000001424 162121 02086 00000 00027 07886 3000 0004616 000006354 178354 02077 00001 00046 07877 3267 0007239 00002268 194782 02062 00002 00072 07863 3533 001063 0000677 211395 02043 00007 00106 07844 3800 001481 0001748 228188 02015 00017 00148 07819 4067 001972 0004009 245157 01977 00040 00197 07786 4333 002527 0008321 262306 01924 00083 00252 07741 4600 003132 001596 279641 01849 00158 00311 07682 4867 003751 002807 297179 01748 00277 00370 07606 5133 004361 004641 314941 01613 00454 00426 07508 5400 3000 3500 4000 4500 5000 5500 0000 0010 0020 0030 0040 0050 T R Mole fraction of NO and O NO O Discussion The equilibrium composition in the above table are based on the reaction in which the reactants are 021 kmol O2 and 079 kmol N2 If you multiply the equilibrium composition mole numbers above with 476 you will obtain equilibrium composition for the reaction in which the reactants are 1 kmol O2 and 376 kmol N2This is the case in problem 1643E PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1646 1650 Water vapor is heated during a steadyflow process The rate of heat supply for a specified exit temperature is to be determined for two cases Assumptions 1 The equilibrium composition consists of H2O OH O2 and H2 2 The constituents of the mixture are ideal gases Analysis a Assuming some H2O dissociates into H2 O2 and O the dissociation equation can be written as Q H2O 298 K H2O H2 O2 OH 2400 K H O H O H O OH 2 2 2 2 x y z w Mass balances for hydrogen and oxygen yield H balance 1 2 2 2 x y w O balance 1 2 x z w 2 The mass balances provide us with only two equations with four unknowns and thus we need to have two more equations to be obtained from the KP relations to determine the equilibrium composition of the mixture They are 2 2 1 2 2 O H H O reaction 1 OH H H O 2 2 1 2 reaction 2 The equilibrium constant for these two reactions at 2400 K are determined from Table A28 to be 0 002932 5 832 ln 0 003628 5 619 ln 2 2 1 1 P P P P K K K K The KP relations for these three simultaneous reactions are total O H OH H 2 total O H O H 1 H2O OH 2 H 2O H 2 OH 2 H 2 H2O O2 2 H 2O H 2 2 O 2 2 H 2 ν ν ν ν ν ν ν ν ν ν ν ν N P N N N K N P N N N K P P where w z y x N N N N N OH O H H O total 2 2 2 Substituting 1 2 1 2 1 0 003628 w z y x x y z 3 1 2 1 2 1 0 002932 w z y x x w y 4 Solving Eqs 1 2 3 and 4 simultaneously for the four unknowns x y z and w yields x 0960 y 003204 z 001205 w 001588 Thus the balanced equation for the dissociation reaction is 001588OH 001205O 003204H 0960H O H O 2 2 2 2 The heat transfer for this dissociation process is determined from the steadyflow energy balance E E E in out system with W 0 R f R P f P h h h N h h h N Q o o o o in PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1647 Assuming the O2 and O to be ideal gases we have h hT From the tables Substance hf o kJkmol h298 K kJkmol h3000 K kJkmol H2O 241820 9904 103508 H2 0 8468 75383 O2 0 8682 83174 OH 39460 9188 77015 Substituting 103380 kJkmol H O 241820 9188 77015 0 0158839460 8682 83174 0 012050 8468 75383 032040 0 9904 103508 960 241820 0 2 in Q The mass flow rate of H2O can be expressed in terms of the mole numbers as 0 03333 kmolmin 18 kgkmol kgmin 60 M m N Thus 3446 kJmin 0 03333 kmolmin103380 kJkmol in in Q N Q b If no dissociates takes place then the process involves no chemical reactions and the heat transfer can be determined from the steadyflow energy balance for nonreacting systems to be 3120 kJmin 9904 kJkmol 0 03333 kmolmin103508 1 2 1 2 in h N h h m h Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1648 1651 Problem 1650 is reconsidered The effect of the final temperature on the rate of heat supplied for the two cases is to be studied Analysis The problem is solved using EES and the solution is given below Given T1298 K T22400 K P1 atm mdot06 kgmin T0298 K The equilibrium constant for these two reactions at 2400 K are determined from Table A28 Kp1exp5619 Kp2exp5832 Properties MMH2OmolarmassH2O Analysis a Actual reaction H2O NH2O H2O NH2 H2 NO2 O2 NOH OH 22NH2O2NH2NOH H balance 1NH2O2NO2NOH O balance NtotalNH2ONH2NO2NOH Stoichiometric reaction 1 H2O H2 12 O2 Stoichiometric coefficients for reaction 1 nuH2O11 nuH211 nuO2112 Stoichiometric reaction 2 H2O 12 H2 OH Stoichiometric coefficients for reaction 2 nuH2O21 nuH2212 nuOH21 Kp relations are Kp1NH2nuH21NO2nuO21NH2OnuH2O1PNtotalnuH21nuO21nuH2O1 Kp2NH2nuH22NOHnuOH2NH2OnuH2O2PNtotalnuH22nuOH2nuH2O2 Enthalpy of formation data from Table A26 hfOH39460 Enthalpies of products hH2ORenthalpyH2O TT1 hH2OPenthalpyH2O TT2 hH2enthalpyH2 TT2 hO2enthalpyO2 TT2 hOH98763 at T2 from the ideal gas tables in the text Standard state enthalpies hoOH9188 at T0 from the ideal gas tables in the text Heat transfer HPNH2OhH2OPNH2hH2NO2hO2NOHhfOHhOHhoOH HRNH2ORhH2OR NH2OR1 QinaHPHR QdotinamdotMMH2OQina b QinbNH2ORhH2OPhH2OR PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course QdotinbmdotMMH2OQinb preparation If you are a student using this Manual you are using it without permission 1649 Tprod K QinDissoc kJmin QinNoDissoc kJmin 2500 2600 2700 2800 2900 3000 3100 3200 3300 3400 3500 3660 3839 4019 4200 4382 4566 4750 4935 5121 5307 5494 3295 3475 3657 3840 4024 4210 4396 4583 4771 4959 5148 2500 2700 2900 3100 3300 3500 3200 3600 4000 4400 4800 5200 T2 K Qin kJmin dissociation no dissociation PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1650 1652 Ethyl alcohol C2H5OH gas is burned in a steadyflow adiabatic combustion chamber with 40 percent excess air The adiabatic flame temperature of the products is to be determined and the adiabatic flame temperature as a function of the percent excess air is to be plotted Analysis The complete combustion reaction in this case can be written as 2 2 th 2 2 2 2 th 5 2 N O 3 H O 2 CO 376N O 1 C H OH gas f Ex a Ex a where ath is the stoichiometric coefficient for air The oxygen balance gives 2 3 1 2 2 2 1 1 th th Ex a Ex a The reaction equation with products in equilibrium is NO N O H O CO CO 376N O 1 C H OH gas 2 2 2 2 2 2 th 5 2 g f e d b a Ex a The coefficients are determined from the mass balances Carbon balance a b 2 Hydrogen balance 3 2 6 d d Oxygen balance g e d b a Ex a 2 2 2 1 1 th Nitrogen balance g f Ex a 2 2 3 76 1 th Solving the above equations we find the coefficients to be Ex 04 ath 3 a 1995 b 0004712 d 3 e 117 f 1576 g 006428 Then we write the balanced reaction equation as 0 06428 NO 1576 N 1 17 O 3 H O 0 004712 CO 1 995 CO 376N O 24 H OH gas C 2 2 2 2 2 2 5 2 Total moles of products at equilibrium are 2199 1 17 1576 3 0 004712 1 995 tot N The first assumed equilibrium reaction is 2 2 50 O CO CO The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using prod prod 1 1 exp T R T G K u p Where prod CO2 CO2 prod O2 O2 prod CO CO prod 1 T g T g T g T G ν ν ν and the Gibbs functions are defined as CO2 prod prod CO2 O2 prod prod O2 CO prod prod CO s T h T g s T h T g s T h T g The equilibrium constant is also given by 0 0005447 2199 1 1 995 0 004712 1 17 50 50 1 50 1 tot 50 1 N P a be K p The second assumed equilibrium reaction is NO 50 O N 50 2 2 Also for this reaction we have PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1651 N2 prod prod N2 NO prod prod NO s T h T g s T h T g O2 prod prod O2 s T h T g prod O2 O2 prod N2 N2 prod NO NO prod 2 T g T g T g T G ν ν ν prod prod 2 2 exp T R T G K u p 0 01497 1576 17 1 0 06428 2199 1 50 50 0 50 5 50 50 1 tot 2 f e g N P K o p A steady flow energy balance gives P R H H where 235310 kJkmol 15790 24 0 235310 kJkmol 1579 24 N225 C O225 C fuel25 C h h h H o f R prod prod prod prod prod prod NO N2 O2 H2O CO CO2 0 06428 1576 1 17 3 0 004712 1 995 T T T T T T P h h h h h h H Solving the energy balance equation using EES we obtain the adiabatic flame temperature K prod 1901 T The copy of entire EES solution including parametric studies is given next The reactant temperature is Treac 25273 K For adiabatic combustion of 1 kmol of fuel Qout 0 kJ PercentEx 40 Percent excess air Ex PercentEx100 EX Excess air100 Pprod 1013kPa Ru8314 kJkmolK The complete combustion reaction equation for excess air is C2H5OHgas 1ExAth O2 376N22 CO2 3 H2O ExAth O2 f N2 Oxygen Balance for complete combustion 1 1ExAth22231 ExAth2 The reaction equation for excess air and products in equilibrium is C2H5OHgas 1ExAth O2 376N2a CO2 b CO d H2O e O2 f N2 g NO Carbon Balance 2a b Hydrogen Balance 62d Oxygen Balance 1 1ExAth2a2b d e2 g Nitrogen Balance 1ExAth376 2 f2 g Ntot a b d e f g Total kilomoles of products at equilibrium The first assumed equilibrium reaction is CO2CO05O2 The following equations provide the specific Gibbs function ghTs for each component in the product gases as a function of its temperature Tprod at 1 atm pressure 1013 kPa gCO2EnthalpyCO2TTprod Tprod EntropyCO2TTprod P1013 gCOEnthalpyCOTTprod Tprod EntropyCOTTprod P1013 gO2EnthalpyO2TTprod Tprod EntropyO2TTprod P1013 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1652 The standardstate Gibbs function is DELTAG1 1gCO05gO21gCO2 The equilibrium constant is given by Eq 1514 KP1 expDELTAG1 RuTprod PPprod 1013atm The equilibrium constant is also given by Eq 1515 K P1 PNtot1051b1e05a1 sqrtPNtot b sqrte KP1a The econd assumed equilibrium reaction is 05N205O2NO gNOEnthalpyNOTTprod Tprod EntropyNOTTprod P1013 gN2EnthalpyN2TTprod Tprod EntropyN2TTprod P1013 The standardstate Gibbs function is DELTAG2 1gNO05gO205gN2 The equilibrium constant is given by Eq 1514 KP2 expDELTAG2 RuTprod The equilibrium constant is also given by Eq 1515 K P2 PNtot10505g1e05f05 gKP2 sqrtef The steadyflow energy balance is HR QoutHP hbarfC2H5OHgas235310 kJkmol HR1hbarfC2H5OHgas 1ExAthENTHALPYO2TTreac1ExAth376ENTHALPYN2TTreac kJkmol HPaENTHALPYCO2TTprodbENTHALPYCOTTproddENTHALPYH2OTTprodeENTHAL PYO2TTprodfENTHALPYN2TTprodgENTHALPYNOTTprod kJkmol ath a b d e f g PercentEx Tprod K 3 1922 007779 3 03081 1238 00616 10 2184 3 1971 00293 3 05798 135 006965 20 2085 3 1988 001151 3 08713 1463 006899 30 1989 3 1995 0004708 3 117 1576 006426 40 1901 3 1998 0001993 3 1472 1689 005791 50 1820 3 1999 00008688 3 1775 1802 005118 60 1747 3 2 00003884 3 2078 1915 004467 70 1682 3 2 00001774 3 2381 2028 003867 80 1621 3 2 000008262 3 2683 2142 00333 90 1566 3 2 000003914 3 2986 2255 002856 100 1516 10 20 30 40 50 60 70 80 90 100 1500 1600 1700 1800 1900 2000 2100 2200 PercentEx Tprod K PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1653 Variations of Kp with Temperature 1653C It enables us to determine the enthalpy of reaction hR from a knowledge of equilibrium constant KP 1654C At 2000 K since combustion processes are exothermic and exothermic reactions are more complete at lower temperatures 1655 The hR value for the dissociation process O2 2O at a specified temperature is to be determined using enthalpy and KP data Assumptions Both the reactants and products are ideal gases Analysis a The dissociation equation of O2 can be expressed as O 2 2 O The hR of the dissociation process of O2 at 3100 K is the amount of energy absorbed or released as one kmol of O2 dissociates in a steadyflow combustion chamber at a temperature of 3100 K and can be determined from R f R P f P R h h h N h h h N h o o o o Assuming the O2 and O to be ideal gases we have h h T From the tables Substance hf o kJkmol h298 K kJkmol h2900 K kJkmol O 249190 6852 65520 O2 0 8682 110784 Substituting 513614 kJkmol 8682 110784 1 0 6852 65520 2249190 hR b The hR value at 3100 K can be estimated by using KP values at 3000 K and 3200 K the closest two temperatures to 3100 K for which KP data are available from Table A28 2 1 1 2 2 1 1 2 1 1 ln 1 or ln 1 ln T T R h K K T T R h K K u R P P u R P P 512808 kJkmol R R h h 3200 K 1 3000 K 1 8 314 kJkmol K 4 357 3 072 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1654 1656 The hR at a specified temperature is to be determined using the enthalpy and KP data Assumptions Both the reactants and products are ideal gases Analysis a The complete combustion equation of CO can be expressed as 2 2 2 1 CO O CO The hR of the combustion process of CO at 2200 K is the amount of energy released as one kmol of CO is burned in a steadyflow combustion chamber at a temperature of 2200 K and can be determined from R f R P f P R h h h N h h h N h o o o o Assuming the CO O2 and CO2 to be ideal gases we have h hT From the tables Substance hf o kJkmol h298 K kJkmol h2200 K kJkmol CO2 393520 9364 112939 CO 110530 8669 72688 O2 0 8682 75484 Substituting hR 1 393 520 112 939 9364 1 110 530 72 688 8669 05 0 75 484 8682 276835 kJ kmol b The hR value at 2200 K can be estimated by using KP values at 2000 K and 2400 K the closest two temperatures to 2200 K for which KP data are available from Table A28 2 1 1 2 2 1 1 2 1 1 ln 1 or ln 1 ln T T R h K K T T R h K K u R P P u R P P 276856 kJkmol R R h h 2400 K 1 2000 K 1 8 314 kJkmol K 6 635 3 860 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1655 1657E The hR at a specified temperature is to be determined using the enthalpy and KP data Assumptions Both the reactants and products are ideal gases Analysis a The complete combustion equation of CO can be expressed as 2 2 2 1 CO O CO The hR of the combustion process of CO at 3960 R is the amount of energy released as one kmol of H2 is burned in a steady flow combustion chamber at a temperature of 3960 R and can be determined from R f R P f P R h h h N h h h N h o o o o Assuming the CO O2 and CO2 to be ideal gases we have h h T From the tables Substance hf o Btulbmol h537 R Btulbmol h3960 R Btulbmol CO2 169300 40275 48647 CO 47540 37251 312565 O2 0 37251 324405 Substituting hR 1 169 300 48 647 4027 5 1 47 540 31 2565 37251 05 0 32 4405 37251 119030 Btu lbmol b The hR value at 3960 R can be estimated by using KP values at 3600 R and 4320 R the closest two temperatures to 3960 R for which KP data are available from Table A28 2 1 1 2 2 1 1 2 1 1 ln 1 or ln 1 ln T T R h K K T T R h K K u R P P u R P P 119041 Btulbmol R R h h 4320 R 1 3600 R 1 1 986 Btulbmol R 6 635 3 860 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1656 1658 The KP value of the combustion process H2 12O2 H2O is to be determined at a specified temperature using hR data and KP value Assumptions Both the reactants and products are ideal gases Analysis The hR and KP data are related to each other by 2 1 1 2 2 1 1 2 1 1 ln 1 or ln 1 ln T T R h K K T T R h K K u R P P u R P P The hR of the specified reaction at 3000 K is the amount of energy released as one kmol of H2 is burned in a steadyflow combustion chamber at a temperature of 3000 K and can be determined from R f R P f P R h h h N h h h N h o o o o Assuming the H2O H2 and O2 to be ideal gases we have h h T From the tables Substance hf o kJkmol h298 K kJkmol h3000K kJkmol H2O 241820 9904 136264 H2 0 8468 97211 O2 0 8682 106780 Substituting 250 kJkmol 253 8682 106780 0 50 8468 97211 0 1 9904 136264 241820 1 R h The KP value at 3200 K can be estimated from the equation above by using this hR value and the KP value at 2800 K which is ln KP1 3812 or KP1 4524 3200 K 1 2800 K 1 8 314 kJkmol K 253250 kJkmol 4524 ln KP2 2 451 2 452 Table A 28 ln ln 2 2 P P K K or 2 P 116 K PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1657 1659 The hR value for the dissociation process CO2 CO 12O2 at a specified temperature is to be determined using enthalpy and Kp data Assumptions Both the reactants and products are ideal gases Analysis a The dissociation equation of CO2 can be expressed as 2 2 1 2 O CO CO The hR of the dissociation process of CO2 at 2200 K is the amount of energy absorbed or released as one kmol of CO2 dissociates in a steadyflow combustion chamber at a temperature of 2200 K and can be determined from R f R P f P R h h h N h h h N h o o o o Assuming the CO O2 and CO2 to be ideal gases we have h h T From the tables Substance hf o kJkmol h298 K kJkmol h2 00 K 2 kJkmol CO2 393520 9364 112939 CO 110530 8669 72688 O2 0 8682 75484 Substituting hR 1 110 530 72 688 8669 05 0 75 484 8682 1 393 520 112 939 9364 276835 kJ kmol b The hR value at 2200 K can be estimated by using KP values at 2000 K and 2400 K the closest two temperatures to 2200 K for which KP data are available from Table A28 2 1 1 2 2 1 1 2 1 1 ln 1 or ln 1 ln T T R h K K T T R h K K u R P P u R P P 276856 kJkmol R R h h 2400 K 1 2000 K 1 8 314 kJkmol K 6 635 3 860 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1658 1660 The enthalpy of reaction for the equilibrium reaction CH4 2O2 CO2 2H2O at 2000 K is to be estimated using enthalpy data and equilibrium constant Kp data Analysis The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using R T T G K e K u p R T T G p u or ln where O2 O2 CH4 CH4 H2O H2O CO2 CO2 T g T g T g T g T G ν ν ν ν At T1 2000 10 1990 K 797938 kJkmol 2 475399 559718 1 2 695638 917176 1 1 O2 O2 1 CH4 CH4 1 H2O H2O 1 CO2 CO2 1 T g T g T g T g T G ν ν ν ν At T2 2000 10 2010 K 839 kJkmol 797 2 480771 565835 1 2 700929 923358 1 2 O2 O2 2 CH4 CH4 2 H2O H2O 2 CO2 CO2 2 T g T g T g T g T G ν ν ν ν The Gibbs functions are obtained from enthalpy and entropy properties using EES Substituting 20 1 10 8 820 8314 kJkmol K1990 K 797938 kJkmol exp K p 20 2 5 426 10 8314 kJkmol K2010 K 797839 kJkmol exp K p The enthalpy of reaction is determined by using the integrated vant Hoff equation 807752 kJkmol R R u R p p h h T T R h K K 2010 K 1 1990 K 1 8 314 kJkmolK 820 10 8 5 426 10 ln 1 1 ln 20 20 2 1 1 2 The enthalpy of reaction can also be determined from an energy balance to be R P R H H h where 640419 kJkmol 2 169162 302094 2 1 167333 kJkmol 259193 48947 2 1 H2O 2000 K 2000 K CO2 O2 2000 K 2000 K CH4 h h H h h H P R The enthalpies are obtained from EES Substituting 807752 kJkmol 167333 640419 R P R H H h which is identical to the value obtained using Kp data PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1659 Phase Equilibrium 1661C No Because the specific gibbs function of each phase will not be affected by this process ie we will still have gf gg 1662C Yes Because the number of independent variables for a twophase PH2 twocomponent C2 mixture is from the phase rule IV C PH 2 2 2 2 2 Therefore two properties can be changed independently for this mixture In other words we can hold the temperature constant and vary the pressure and still be in the twophase region Notice that if we had a single component C1 two phase system we would have IV1 which means that fixing one independent property automatically fixes all the other properties 1163C Using solubility data of a solid in a specified liquid the mass fraction w of the solid A in the liquid at the interface at a specified temperature can be determined from liquid solid solid mf m m m A where msolid is the maximum amount of solid dissolved in the liquid of mass mliquid at the specified temperature 1164C The molar concentration Ci of the gas species i in the solid at the interface Ci solid side 0 is proportional to the partial pressure of the species i in the gas Pi gas side0 on the gas side of the interface and is determined from kmolm 0 S 0 i gas side isolid side P C 3 where S is the solubility of the gas in that solid at the specified temperature 1165C Using Henrys constant data for a gas dissolved in a liquid the mole fraction of the gas dissolved in the liquid at the interface at a specified temperature can be determined from Henrys law expressed as y P H i liquid side i gas side 0 0 where H is Henrys constant and Pigas side0 is the partial pressure of the gas i at the gas side of the interface This relation is applicable for dilute solutions gases that are weakly soluble in liquids PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1660 1666E The maximum partial pressure of the water evaporated into the air as it emerges from a porous media is to be determined Assumptions The air and waterair solution behave as ideal solutions so that Raoults law may be used Analysis The saturation temperature of water at 70F is 0 36334 psia sat70 F P Since the mole fraction of the air in the liquid water is essentially zero 036334 psia sat70F max 1 P Pv 1667 The number of independent properties needed to fix the state of a mixture of oxygen and nitrogen in the gas phase is to be determined Analysis In this case the number of components is C 2 and the number of phases is PH 1 Then the number of independent variables is determined from the phase rule to be IV C PH 2 2 1 2 3 Therefore three independent properties need to be specified to fix the state They can be temperature the pressure and the mole fraction of one of the gases 1668 It is to be shown that a saturated liquidvapor mixture of refrigerant134a at 20C satisfies the criterion for phase equilibrium Analysis Using the definition of Gibbs function and enthalpy and entropy data from Table A11 0 9842 kJkg 25315 K094564 kJkg K 23841 kJkg 0 9967 kJkg 25315 K010463 kJkg K 2549 kJkg g g g f f f Ts h g Ts h g which are sufficiently close Therefore the criterion for phase equilibrium is satisfied 1669 It is to be shown that a mixture of saturated liquid water and saturated water vapor at 300 kPa satisfies the criterion for phase equilibrium Analysis The saturation temperature at 300 kPa is 4067 K Using the definition of Gibbs function and enthalpy and entropy data from Table A5 118 6 kJkg 406 7 K69917 kJkg K 2724 9 kJkg 118 5 kJkg 406 7 K16717 kJkg K 56143 kJkg g g g f f f Ts h g Ts h g which are practically same Therefore the criterion for phase equilibrium is satisfied PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1661 1670 The values of the Gibbs function for saturated refrigerant134a at 280 kPa are to be calculated Analysis The saturation temperature of R134a at 280 kPa is 125C Table A12 Obtaining other properties from Table A12 the Gibbs function for the liquid phase is 374 kJkg 27315 K 0 19829 kJkg K 1 25 5018 kJkg f f f Ts h g R134a 280 kPa x 07 For the vapor phase 372 kJkg 27315 K 0 93210 kJkg K 1 25 24972 kJkg g g g Ts h g The results agree and demonstrate that phase equilibrium exists 1671E The values of the Gibbs function for saturated steam at 300F as a saturated liquid saturated vapor and a mixture of liquid and vapor are to be calculated Analysis Obtaining properties from Table A4E the Gibbs function for the liquid phase is 6240 Btulbm 75967 R 0 43720 Btulbm R 26973 Btulbm f f f Ts h g Steam 300F For the vapor phase 6236 Btulbm 75967 R 1 6354 Btulbm R 0 Btulbm 1180 g g g Ts h g For the saturated mixture with a quality of 60 6238 Btulbm 75967 R 1 1561 Btulbm R 87 Btulbm 815 1 1561 Btulbm R 060119818 Btulbm R 43720 Btulbm R 0 81587 Btulbm 06091024 Btulbm 73 Btulbm 269 Ts h g xs s s xh h h fg f fg f The results agree and demonstrate that phase equilibrium exists 1672 A liquidvapor mixture of ammonia and water in equilibrium at a specified temperature is considered The pressure of ammonia is to be determined for two compositions of the liquid phase Assumptions The mixture is ideal and thus Raoults law is applicable H2O NH3 10C Analysis According to Raoultss law when the mole fraction of the ammonia liquid is 20 kPa 1231 0 20615 3 kPa satNH3 NH3 NH3 T P y P f When the mole fraction of the ammonia liquid is 80 4922 kPa 0 80615 3 kPa satNH3 NH3 NH3 T P y P f PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1662 1673 Using the liquidvapor equilibrium diagram of an oxygennitrogen mixture the composition of each phase at a specified temperature and pressure is to be determined Analysis From the equilibrium diagram Fig 1621 we read Liquid 2 2 N and 35 O 65 Vapor 2 2 N and 10 O 90 1674 Using the liquidvapor equilibrium diagram of an oxygennitrogen mixture at a specified pressure the temperature is to be determined for a specified composition of the vapor phase Analysis From the equilibrium diagram Fig 1621 we read T 82 K 1675 Using the liquidvapor equilibrium diagram of an oxygennitrogen mixture at a specified pressure the temperature is to be determined for a specified composition of the nitrogen Properties The molar masses of O2 is 32 kgkmol and that of N2 is 28 kgkmol Table A1 Analysis For 100 kg of liquid phase the mole numbers are 3 393 kmol 2 143 25 1 2 143 kmol 28 kgkmol kg 60 1 25 kmol 32 kgkmol kg 40 total N2 N2 N2 O2 O2 O2 f f f f f N M m N M m N The mole fractions in the liquid phase are 0 6316 3393 kmol 143 kmol 2 0 3684 3393 kmol 25 kmol 1 total N2 N2 total O2 O2 f f f f f f N N y N N y From the equilibrium diagram Fig 1621 we read T 805 K PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1663 1676 Using the liquidvapor equilibrium diagram of an oxygennitrogen mixture at a specified pressure the mass of the oxygen in the liquid and gaseous phases is to be determined for a specified composition of the mixture Properties The molar masses of O2 is 32 kgkmol and that of N2 is 28 kgkmol Table A1 Analysis From the equilibrium diagram Fig 1621 at T 84 K the oxygen mole fraction in the vapor phase is 34 and that in the liquid phase is 70 That is and 0 70 O2 y f 0 34 O2 g y The mole numbers are 2 366 kmol 1 429 9375 0 1 429 kmol 28 kgkmol kg 40 0 9375 kmol 32 kgkmol kg 30 total N2 N2 N2 O2 O2 O2 N M m N M m N The total number of moles in this system is 1 2 366 g f N N The total number of moles of oxygen in this system is 2 0 9375 0 34 70 g f N N Solving equations 1 and 2 simultneously we obtain 996 1 3696 0 g f N N Then the mass of oxygen in the liquid and vapor phases is kg 2172 kg 828 0 34 1 996 kmol32 kgkmol 70 0 3696 kmol32 kgkmol O2 O2 O2 O2 O2 O2 N M y m N M y m g g g f f f PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1664 1677 Using the liquidvapor equilibrium diagram of an oxygennitrogen mixture at a specified pressure the total mass of the liquid phase is to be determined Properties The molar masses of O2 is 32 kgkmol and that of N2 is 28 kgkmol Table A1 Analysis From the equilibrium diagram Fig 1621 at T 84 K the oxygen mole fraction in the vapor phase is 34 and that in the liquid phase is 70 That is and 0 70 O2 y f 0 34 O2 g y Also and 0 30 N2 y f 0 66 N2 g y The mole numbers are 2 366 kmol 1 429 9375 0 1 429 kmol 28 kgkmol kg 40 0 9375 kmol 32 kgkmol kg 30 total N2 N2 N2 O2 O2 O2 N M m N M m N The total number of moles in this system is 1 2 366 g f N N The total number of moles of oxygen in this system is 2 0 9375 0 34 70 g f N N Solving equations 1 and 2 simultneously we obtain 996 1 3696 0 g f N N The total mass of liquid in the mixture is then 1138 kg 30 0 3696 kmol28 kgkmol 70 0 3696 kmol32 kgkmol N2 N2 O2 O2 O2 O2 total N M y N M y m m m f f f f f f f PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1665 1678 A rubber wall separates O2 and N2 gases The molar concentrations of O2 and N2 in the wall are to be determined Assumptions The O2 and N2 gases are in phase equilibrium with the rubber wall Properties The molar mass of oxygen and nitrogen are 320 and 280 kgkmol respectively Table A1 The solubility of oxygen and nitrogen in rubber at 298 K are 000312 and 000156 kmolm3bar respectively Table 163 Analysis Noting that 300 kPa 3 bar the molar densities of oxygen and nitrogen in the rubber wall are determined to be Rubber plate CO2 CN2 000936 kmolm3 0 00312 kmolm bar 3 bar S 0 3 O gas side solid side O 2 2 P C O2 25C 300 kPa N2 25C 300 kPa 000468 kmolm3 0 00156 kmolm bar 3 bar S 0 3 N gas side solid side N 2 2 P C That is there will be 000936 kmol of O2 and 000468 kmol of N2 gas in each m3 volume of the rubber wall 1679 A liquidvapor mixture of ammonia and water in equilibrium at a specified temperature is considered The composition of the vapor phase is given The composition of the liquid phase is to be determined Assumptions The mixture is ideal and thus Raoults law is applicable Properties At 2033 5 kPa 12352 kPa and 50 C 3 2 satNH satH O P P Analysis We have 1 and H O 2 yg 99 NH3 yg For an ideal twophase mixture we have 1 3 2 3 3 3 2 2 2 NH H O satNH NH NH H O satH O H O f f f m g f m g y y T P y P y T P y P y H2O NH3 50C Solving for y f H O 2 0 9912352 kPa 1 0 012033 5 kPa 1 H O H O satH O NH satNH H O H O 2 2 2 3 3 2 2 f f g g f y y P y P y y It yields 0376 0624 3 2 NH H O and f f y y PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1666 1680 A mixture of water and ammonia is considered The mole fractions of the ammonia in the liquid and vapor phases are to be determined Assumptions The mixture is ideal and thus Raoults law is applicable Properties At 1003 5 kPa 3 1698 kPa and 25 C satNH3 satH2O P P Tables A4 Analysis According to Raoultss law the partial pressures of ammonia and water in the vapor phase are given by 3 1698 kPa 1003 5 kPa H2O H2O H2O satH2O H2O H2O H2O NH3 NH3 NH3 satNH3 NH3 f f f f g f f f f g N N N P y P N N N P y P H2O NH3 100 kPa 25C The sum of these two partial pressures must equal the total pressure of the vapor mixture In terms of NH3 H2O f f N N x this sum is 100 1 3 1698 1 1003 5 x x x Solving this expression for x gives x 9331 kmol H2Okmol NH3 In the vapor phase the partial pressure of the ammonia vapor is 9713 kPa 9 331 1 1003 5 1 1003 5 NH3 x Pg The mole fraction of ammonia in the vapor phase is then 09713 100 kPa 9713 kPa NH3 NH3 P P y g g According to Raoults law 00968 1003 5 kPa 13 kPa 97 NH3 sat NH3 NH3 P P y g f PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1667 1681 An ammoniawater absorption refrigeration unit is considered The operating pressures in the generator and absorber and the mole fractions of the ammonia in the strong liquid mixture being pumped from the absorber and the weak liquid solution being drained from the generator are to be determined Assumptions The mixture is ideal and thus Raoults law is applicable Properties At 1010 kPa 0 6112 kPa and at 46 C 0 C satH2O satH2O P P Table A4 The saturation pressures of ammonia at the same temperatures are given to be 4306 kPa and 18302 kPa respectively Analysis According to Raoultss law the partial pressures of ammonia and water are given by satH2O NH3 satH2O H2O H2O g satNH3 NH3 gNH3 1 P y P y P P y P f f f Using Daltons partial pressure model for ideal gas mixtures the mole fraction of the ammonia in the vapor mixture is 003294 NH3 NH3 NH3 NH3 satH2O NH3 satNH3 NH3 satNH3 NH3 NH3 0 6112 1 6 430 430 6 96 0 1 f f f f f f f g y y y y P y P y P y y Then 1478 kPa 0 03294 0 6112 1 0 03294430 6 1 satH2O NH3 satNH3 NH3 P y P y P f f Performing the similar calculations for the regenerator 01170 NH3 NH3 NH3 NH3 1010 1 2 1830 1830 2 0 96 f f f f y y y y kPa 2231 0 11701010 1 0 11701830 2 P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1668 1682 An ammoniawater absorption refrigeration unit is considered The operating pressures in the generator and absorber and the mole fractions of the ammonia in the strong liquid mixture being pumped from the absorber and the weak liquid solution being drained from the generator are to be determined Assumptions The mixture is ideal and thus Raoults law is applicable Properties At 7 3851 kPa 0 9353 kPa and at 40 C 6 C satH2O satH2O P P Table A4 or EES The saturation pressures of ammonia at the same temperatures are given to be 5348 kPa and 15567 kPa respectively Analysis According to Raoultss law the partial pressures of ammonia and water are given by satH2O NH3 satH2O H2O H2O g satNH3 NH3 gNH3 1 P y P y P P y P f f f Using Daltons partial pressure model for ideal gas mixtures the mole fraction of the ammonia in the vapor mixture is 004028 NH3 NH3 NH3 NH3 satH2O NH3 satNH3 NH3 satNH3 NH3 NH3 0 9353 1 8 534 534 8 96 0 1 f f f f f f f g y y y y P y P y P y y Then 2244 kPa 0 04028 0 9353 1 0 04028534 8 1 satH2O NH3 satNH3 NH3 P y P y P f f Performing the similar calculations for the regenerator 01022 NH3 NH3 NH3 NH3 7 3851 1 7 1556 1556 7 0 96 f f f f y y y y 1657 kPa 0 1022 7 3851 1 0 10221556 7 P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1669 1683 A liquid mixture of water and R134a is considered The mole fraction of the water and R134a vapor are to be determined Assumptions The mixture is ideal and thus Raoults law is applicable Properties At 57207 kPa 2 3392 kPa and 20 C satR satH2O P P Tables A4 A11 The molar masses of water and R 134a are 18015 and 10203 kgkmol respectively Table A1 Analysis The mole fraction of the water in the liquid mixture is H2O R134a 20C 0 9808 10203 10 18015 90 18015 90 mf mf mf R R H2O H2O H2O H2O total H2O H2O M M M N N y f f f f f According to Raoultss law the partial pressures of R134a and water in the vapor mixture are 1098 kPa 0 980857207 kPa 1 satR R R P y P f g 2294 kPa 0 9808 2 3392 kPa satH2O H2O H2O P y P f g The total pressure of the vapor mixture is then 13274 kPa 2 294 1098 H2O R total g g P P P Based on Daltons partial pressure model for ideal gases the mole fractions in the vapor phase are 01728 13274 kPa 294 kPa 2 total H2O H2O P P y g g 08272 13274 kPa 98 kPa 10 total R R P P y g g PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1670 1684 A glass of water is left in a room The mole fraction of the water vapor in the air and the mole fraction of air in the water are to be determined when the water and the air are in thermal and phase equilibrium Assumptions 1 Both the air and water vapor are ideal gases 2 Air is saturated since the humidity is 100 percent 3 Air is weakly soluble in water and thus Henrys law is applicable Properties The saturation pressure of water at 27C is 3568 kPa Table A4 Henrys constant for air dissolved in water at 27ºC 300 K is given in Table 162 to be H 74000 bar Molar masses of dry air and water are 29 and 18 kgkmol respectively Table A1 Analysis a Noting that air is saturated the partial pressure of water vapor in the air will simply be the saturation pressure of water at 27C 3 568 kPa Table A4 sat 27 C vapor P P Air 27ºC 92 kPa φ 100 Water 27ºC Assuming both the air and vapor to be ideal gases the mole fraction of water vapor in the air is determined to be 00388 92 kPa 3 568 kPa vapor vapor P P y b Noting that the total pressure is 92 kPa the partial pressure of dry air is 8843 kPa 08843 bar 3 568 92 vapor dry air P P P From Henrys law the mole fraction of air in the water is determined to be 10 5 120 74000 bar 0 8843 bar dryairgas side dryairliquidside H P y Discussion The amount of air dissolved in water is very small as expected PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1671 1685 A carbonated drink in a bottle is considered Assuming the gas space above the liquid consists of a saturated mixture of CO2 and water vapor and treating the drink as a water determine the mole fraction of the water vapor in the CO2 gas and the mass of dissolved CO2 in a 300 ml drink are to be determined when the water and the CO2 gas are in thermal and phase equilibrium Assumptions 1 The liquid drink can be treated as water 2 Both the CO2 and the water vapor are ideal gases 3 The CO2 gas and water vapor in the bottle from a saturated mixture 4 The CO2 is weakly soluble in water and thus Henrys law is applicable Properties The saturation pressure of water at 27C is 3568 kPa Table A4 Henrys constant for CO2 dissolved in water at 27ºC 300 K is given in Table 162 to be H 1710 bar Molar masses of CO2 and water are 44 and 18 kgkmol respectively Table A1 Analysis a Noting that the CO2 gas in the bottle is saturated the partial pressure of water vapor in the air will simply be the saturation pressure of water at 27C 3 568 kPa sat 27 C vapor P P more accurate EES value compared to interpolation value from Table A4 Assuming both CO2 and vapor to be ideal gases the mole fraction of water vapor in the CO2 gas becomes 00274 130 kPa 3 568 kPa vapor vapor P P y b Noting that the total pressure is 130 kPa the partial pressure of CO2 is 126 4 kPa 1264 bar 3 568 130 vapor CO2 gas P P P From Henrys law the mole fraction of CO2 in the drink is determined to be y P H CO liquid side CO gas side 2 2 bar 1710bar 1264 739 10 4 Then the mole fraction of water in the drink becomes y y water liquid side CO liquid side 2 1 1 7 39 10 0 9993 4 The mass and mole fractions of a mixture are related to each other by m i i m m i i m i i M M y N M N M m m mf where the apparent molar mass of the drink liquid water CO2 mixture is M y M y M y M m i i liquid water water CO CO 2 2 kg kmol 09993 18 0 739 10 44 18 02 4 Then the mass fraction of dissolved CO2 gas in liquid water becomes 000180 1802 44 7 39 10 0 mf 4 CO CO liquidside liquidside CO 2 2 2 m M M y Therefore the mass of dissolved CO2 in a 300 ml 300 g drink is 054 g 0 00180300 g mf 2 2 CO CO mm m PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1672 Review Problems 1686 The equilibrium constant of the dissociation process O2 2O is given in Table A28 at different temperatures The value at a given temperature is to be verified using Gibbs function data Analysis The KP value of a reaction at a specified temperature can be determined from the Gibbs function data using K e K G T R T p G T R T p u u ln or O2 2O 2000 K where 375 kJkmol 243 2000 268655 67881 8682 1 0 2000 201135 6852 42564 249190 2 2 2 2 2 2 2 O 298 2000 O O 298 2000 O O O O O O O O O Ts h h h Ts h h h Ts h Ts h T g T g T G f f ν ν ν ν ν ν Substituting 14636 243375 kJkmol8314 kJkmol K2000 K ln K p or Table A28 ln K K p 44 10 7 P 14622 1687 A mixture of H2 and Ar is heated is heated until 10 of H2 is dissociated The final temperature of mixture is to be determined Assumptions 1 The constituents of the mixture are ideal gases 2 Ar in the mixture remains an inert gas Analysis The stoichiometric and actual reactions can be written as Stoichiometric 2 1 and 2H thus H H H 2 2 ν ν H H 2 2 Ar 1 atm Actual inert react 2 prod 2 Ar 0 90H 20 H Ar H 43 42 1 The equilibrium constant KP can be determined from 0 02116 1 20 90 1 90 20 2 1 2 total H H H2 H 2 H 2 H ν ν ν ν N P N N K p and 3 855 ln p K From Table A28 the temperature corresponding to this KP value is T 2974 K PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1673 1688 The equilibrium constant for the reaction CH4 2O2 CO2 2H2O at 100 kPa and 2000 K is to be determined Assumptions 1 The constituents of the mixture are ideal gases Analysis This is a simultaneous reaction We can begin with the dissociation of methane and carbon dioxide 2 4 2H C CH e 9 685 K P CH42O2 CO22H2O 3000 K 690 kPa 2 2 CO O C e15869 K P When these two reactions are summed and the common carbon term cancelled the result is 2 2 2 4 2H CO CH O 6 184 9 685 15869 e e K P Next we include the water dissociation reaction Table A28 2H O O 2H 2 2 2 6 172 2 3 086 e e K P which when summed with the previous reaction and the common hydrogen term is cancelled yields 2H O CO 2O CH 2 2 2 4 12356 6 172 6 184 e e K P Then KP 12356 ln PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1674 1689 A mixture of H2O O2 and N2 is heated to a high temperature at a constant pressure The equilibrium composition is to be determined Assumptions 1 The equilibrium composition consists of H2O O2 N2 and H2 2 The constituents of the mixture are ideal gases Analysis The stoichiometric and actual reactions in this case are 1 H2O 2 O2 5 N2 2200 K 5 atm Stoichiometric 1 and 1 O thus H H O 2 1 O H H O 2 2 1 2 2 2 2 2 ν ν ν Actual inert 2 products 2 2 react 2 2 2 2 5N O H H O 5N H O 2O 14243 3 2 1 z y x H balance 2 2 2 1 x y y x O balance x z z x 50 52 2 5 Total number of moles x z y x N 50 58 5 total The equilibrium constant relation can be expressed as 1 50 1 total 50 total O H O H H2O O2 2 H 2O H 2 2 O 2 2 H 2 N P x z y N P N N N K p ν ν ν ν ν ν From Table A28 lnKP 6768 at 2200 K Thus KP 000115 Substituting 50 50 50 58 5 50 51 1 0 00115 x x x x Solving for x x 09981 Then y 1 x 00019 z 25 05x 200095 Therefore the equilibrium composition of the mixture at 2200 K and 5 atm is 2 2 2 2 5N 200095O 09981H O 00019H The equilibrium constant for the reaction H O OH H 2 1 2 2 is lnKP 7148 which is very close to the KP value of the reaction considered Therefore it is not realistic to assume that no OH will be present in equilibrium mixture PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1675 1690 Methane gas is burned with stoichiometric amount of air during a combustion process The equilibrium composition and the exit temperature are to be determined Assumptions 1 The product gases consist of CO2 H2O CO N2 and O2 2 The constituents of the mixture are ideal gases 3 This is an adiabatic and steadyflow combustion process Analysis a The combustion equation of CH4 with stoichiometric amount of O2 can be written as CH O N CO 05 05 O H O 752N 4 2 2 2 2 2 2 376 1 2 CO x x x 2 After combustion there will be no CH4 present in the combustion chamber and H2O will act like an inert gas The equilibrium equation among CO2 CO and O2 can be expressed as 1 and 1 O thus CO CO 2 1 O CO CO 2 2 1 2 2 2 ν ν ν CH4 25C and total CO O CO CO2 O2 CO 2 CO 2 2 O 2 CO ν ν ν ν ν ν N P N N N K p CO CO2 H2O O2 N2 Combustion chamber 1 atm PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Air 25C where N x x x total x 1 15 05 2 752 12 02 05 Substituting 1 51 50 50 1202 1 50 50 1 x x x x K p The value of KP depends on temperature of the products which is yet to be determined A second relation to determine KP and x is obtained from the steadyflow energy balance expressed as f R R P f P R f R P f P N h h h h N h h h N h h h N o o o o o o o 0 0 since the combustion is adiabatic and the reactants enter the combustion chamber at 25C Assuming the air and the combustion products to be ideal gases we have h h T From the tables Substance hf o kJkmol h298 K kJkmol CH4g 74850 N2 0 8669 O2 0 8682 H2Og 241820 9904 CO 110530 8669 CO2 393520 9364 Substituting 0 393 520 9364 1 110 530 8669 2 241820 9904 05 05 0 8682 7 52 0 8669 1 74 850 0 0 298 298 x h x h h x h h h h CO CO H O O N 2 2 2 2 which yields xh x h h x h h x CO CO H O O N 2 2 2 2 1 2 05 05 752 279 344 617 329 Now we have two equations with two unknowns TP and x The solution is obtained by trial and error by assuming a temperature TP calculating the equilibrium composition from the first equation and then checking to see if the second equation is satisfied A first guess is obtained by assuming there is no CO in the products ie x 1 It yields TP 2328 K The adiabatic combustion temperature with incomplete combustion will be less Take K Take K T K x RH T K x RH p p p p 2300 4 49 0870 641 093 2250 4805 0893 612 755 ln ln S S 2 By interpolation T x p 2258 K and 0889 Thus the composition of the equilibrium mixture is 0889CO 0111CO 00555O 2H O 752N 2 2 2 preparation If you are a student using this Manual you are using it without permission 1676 1691 Problem 1690 is reconsidered The effect of excess air on the equilibrium composition and the exit temperature by varying the percent excess air from 0 to 200 percent is to be studied Analysis The problem is solved using EES and the solution is given below Often for nonlinear problems such as this one good gusses are required to start the solution First run the program with zero percent excess air to determine the net heat transfer as a function of Tprod Just press F3 or click on the Solve Table icon From Plot Window 1 where Qnet is plotted vs Tprod determnine the value of Tprod for Qnet0 by holding down the Shift key and move the cross hairs by moving the mouse Qnet is approximately zero at Tprod 2269 K From Plot Window 2 at Tprod 2269 K a b and c are approximately 089 010 and 0056 respectively For EES to calculate a b c and Tprod directly for the adiabatic case remove the in the last line of this window to set Qnet 00 Then from the Options menu select Variable Info and set the Guess Values of a b c and Tprod to the guess values selected from the Plot Windows Then press F2 or click on the Calculator icon Input Data PercentEx 0 Ex PercentEX100 Pprod 1013 kPa Ru8314 kJkmolK Tfuel298 K Tair298 K The combustion equation of CH4 with stoichiometric amount of air is CH4 1Ex2O2 376N2CO2 2H2O1Ex2376N2 For the incomplete combustion process in this problem the combustion equation is CH4 1Ex2O2 376N2aCO2 bCO cO22H2O1Ex2376N2 Specie balance equations O 4a 2b c 22 C 1a b Ntot a b c 21Ex2376 Total kilomoles of products at equilibrium We assume the equilibrium reaction is CO2CO05O2 The following equations provide the specific Gibbs function ghTs for each component as a function of its temperature at 1 atm pressure 1013 kPa gCO2EnthalpyCO2TTprod Tprod EntropyCO2TTprod P1013 gCOEnthalpyCOTTprod Tprod EntropyCOTTprod P1013 gO2EnthalpyO2TTprod Tprod EntropyO2TTprod P1013 The standardstate Gibbs function is DELTAG 1gCO05gO21gCO2 The equilibrium constant is given by Eq 1614 KP expDELTAG RuTprod PPprod 1013atm The equilibrium constant is also given by Eq 1615 K P PNtot1051b1c05a1 sqrtPNtot b sqrtc KP a Conservation of energy for the reaction assuming SSSF neglecting work ke and pe Ein Eout DELTAEcv Ein Qnet HR The enthalpy of the reactant gases is HRenthalpyCH4TTfuel 1Ex2 enthalpyO2TTair1Ex2376 enthalpyN2TTair Eout HP The enthalpy of the product gases is HPa enthalpyCO2TTprod b enthalpyCOTTprod 2enthalpyH2OTTprod 1Ex2376enthalpyN2TTprod c enthalpyO2TTprod DELTAEcv 0 Steadyflow requirement Qnet0 For an adiabatic reaction the net heat added is zero PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1677 PercentEx Tprod K 0 2260 20 2091 40 1940 60 1809 80 1695 100 1597 120 1511 140 1437 160 1370 180 1312 200 1259 0 40 80 120 160 200 1200 1400 1600 1800 2000 2200 2400 Percent Excess Air Tprod K 1200 1400 1600 1800 2000 2200 2400 2600 010 010 030 050 070 090 110 Tprod K Coefficients a b c Coefficients for CO2 CO and O2 vs Tprod a CO2 a CO2 b CO b CO c O2 c O2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1678 1692 The equilibrium partial pressure of the carbon dioxide for the reaction CH4 2O2 CO2 2H2O at 700 kPa and 3000 K is to be determined Assumptions 1 The equilibrium composition consists of CH4 O2 CO2 and H2O 2 The constituents of the mixture are ideal gases Analysis This is a simultaneous reaction We can begin with the dissociation of methane and carbon dioxide 2 4 2H C CH e 9 685 K P CH42O2 CO22H2O 3000 K 700 kPa 2 2 CO O C e15869 K P When these two reactions are summed and the common carbon term cancelled the result is 2 2 2 4 2H CO CH O 6 184 9 685 15869 e e K P Next we include the water dissociation reaction 2H O O 2H 2 2 2 6 172 2 3 086 e e K P which when summed with the previous reaction and the common hydrogen term is cancelled yields 2H O CO 2O CH 2 2 2 4 12356 6 172 6 184 e e K P Then 12356 ln P K Actual reeaction 4 4 4 3 14 2 4 4 4 3 14 2 products 2 2 react 2 4 2 4 H O CO O CH 2O CH m z y x C balance x z z x 1 1 H balance x m m x 2 2 2 4 4 O balance x y m z y 2 2 2 4 Total number of moles 3 total m z y x N The equilibrium constant relation can be expressed as O2 CH4 H2O CO2 O2 CH4 H2O CO2 total O2 CH4 H2O CO2 ν ν ν ν ν ν ν ν N P N N N N K p Substituting 2 1 2 1 2 2 12356 3 101325 700 2 2 2 1 x x x x e Solving for x x 001601 Then y 2x 003202 z 1 x 098399 m 2 2x 196798 Therefore the equilibrium composition of the mixture at 3000 K and 700 kPa is 196798 H O 098399 CO 003202 O CH 001601 2 2 2 4 The mole fraction of carbon dioxide is 03280 3 0 98399 CO2 y and the partial pressure of the carbon dioxide in the product mixture is 230 kPa 0 3280700 kPa CO2 CO2 P y P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1679 1693 Methane is heated from a specified state to another state The amount of heat required is to be determined without and with dissociation cases Properties The molar mass and gas constant of methane are 16043 kgkmol and 05182 kJkgK Table A1 Assumptions 1 The equilibrium composition consists of O2 and O 2 The constituents of the mixture are ideal gases Analysis a An energy balance for the process gives 1 2 1 2 1 2 in potential etc energies in internal kinetic Change system by heat work and mass Net energy transfer out in T T R h h N u N u Q E E E u 43 42 1 4243 1 CH4 1000 K 1 atm Using the empirical coefficients of Table A2c 239 kJkmol 38 298 1000 4 01 10 11 298 1000 3 1 269 10 298 1000 2 0 05024 298 891000 19 4 3 2 4 4 9 3 3 5 2 2 4 1 4 2 3 1 3 2 2 1 2 2 1 2 2 1 1 2 T d T T c T T b T T a T c dT h h p Substituting 324000 kJ 298K 8314 kJkmol K1000 10 kmol 38239 kJkmol Qin b The stoichiometric and actual reactions in this case are Stoichiometric 2 4 2H C CH 2 1 and 1 thus H2 C CH4 ν ν ν Actual 4243 1 3 2 1 products 2 react 4 4 H C CH CH z y x C balance x y y x 1 1 H balance x z z x 2 2 2 4 4 Total number of moles x z y x N 2 3 total The equilibrium constant relation can be expressed as CH4 H2 C CH4 H2 C total CH4 H2 C ν ν ν ν ν ν N P N N N K p From the problem statement at 1000 K 2 328 ln K p Then 0 09749 2 328 e K P Substituting 1 2 1 2 2 3 1 2 2 1 09749 0 x x x x Solving for x x 06414 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1680 Then y 1 x 03586 z 2 2x 07172 Therefore the equilibrium composition of the mixture at 1000 K and 1 atm is 2 4 07172 H 0 3586 C 06414 CH The mole fractions are 04177 1 7172 7172 0 02088 1 7172 3586 0 03735 1 7172 0 6414 0 7172 0 3586 0 6414 6414 0 total H2 H2 total C C total CH4 CH4 N N y N N y N N y The heat transfer can be determined from 245700 kJ 1027 8 298 0 2088 0 7111000 0 417721 7 1000 10 0 373563 3 1000 CH4 1 2 C C 2 H2 H2 2 CH4 CH4 in T Nc T y c T c y T c N y Q v v v v PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1681 1694 Solid carbon is burned with a stoichiometric amount of air The number of moles of CO2 formed per mole of carbon is to be determined Assumptions 1 The equilibrium composition consists of CO2 CO O2 and N2 2 The constituents of the mixture are ideal gases Analysis Inspection of Table A28 reveals that the dissociation equilibrium constants of CO2 O2 and N2 are quite small and therefore may be neglected We learned from another source that the equilibrium constant for CO is also small The combustion is then complete and the reaction is described by Carbon Air 25C 2 2 2 2 3 76N CO 376N O C The number of moles of CO2 in the products is then 1 C CO2 N N 1695 Solid carbon is burned with a stoichiometric amount of air The amount of heat released per kilogram of carbon is to be determined Assumptions 1 The equilibrium composition consists of CO2 CO O2 and N2 2 The constituents of the mixture are ideal gases Analysis Inspection of Table A28 reveals that the dissociation equilibrium constants of CO2 O2 and N2 are quite small and therefore may be neglected We learned from another source that the equilibrium constant for CO is also small The combustion is then complete and the reaction is described by Carbon Air 25C 2 2 2 2 3 76N CO 376N O C The heat transfer for this combustion process is determined from the energy balance system out in E E E applied on the combustion chamber with W 0 It reduces to R f R P f P h h h N h h h N Q o o o o out Assuming the air and the combustion products to be ideal gases we have h hT From the tables Substance fo h kJkmol h298K kJkmol h1240K kJkmol N2 0 8669 38129 CO2 393520 9364 56108 Substituting kJkmol C 236000 8669 38129 3 76 0 9364 56108 393520 1 out Q or 19670 kJkg C 12 kgkmol 236000 kJkmol Qout PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1682 1696 Methane gas is burned with 30 percent excess air The equilibrium composition of the products of combustion and the amount of heat released by this combustion are to be determined Assumptions 1 The equilibrium composition consists of CO2 H2O O2 NO and N2 2 The constituents of the mixture are ideal gases Analysis Inspection of the equilibrium constants of the possible reactions indicate that only the formation of NO need to be considered in addition to other complete combustion products Then the stoichiometric and actual reactions in this case are Stoichiometric 2 1 and 1 2NO thus O N NO O2 N2 2 2 ν ν ν Actual 2 2 2 2 2 2 4 N O NO 2H O CO 3 76N 62 O CH z y x N balance x z z x 50 9 776 2 9 776 2 O balance x y y x 50 60 2 2 2 25 Qout Combustion chamber 1 atm 30 excess air 25C CH4 25C Total number of moles 1338 2 1 total z y x N CO2 H2O NO O2 N2 1600 K The equilibrium constant relation can be expressed as total O2 N2 NO O2 N2 NO O2 N2 NO ν ν ν ν ν ν N P N N N K p From Table A28 at 1600 K 5 294 Since the stoichiometric reaction being considered is double this reaction ln p K 2 522 10 5 5 294 2 exp p K Substituting 2 1 1 2 5 1338 1 50 9 766 50 60 522 10 2 x x x Solving for x x 00121 Then y 06 05x 0594 z 9776 05x 977 Therefore the equilibrium composition of the products mixture at 1600 K and 1 atm is 2 2 2 2 2 2 4 977N 0594O 00121NO 2H O CO 376N 26O CH The heat transfer for this combustion process is determined from the energy balance system out in E E E applied on the combustion chamber with W 0 It reduces to R f R P f P h h h N h h h N Q o o o o out Assuming the air and the combustion products to be ideal gases we have h hT From the tables Substance fo h kJkmol h298K kJkmol h1600K kJkmol CH4 74850 O2 0 8682 52961 N2 0 8669 50571 H2O 241820 9904 62748 CO2 393520 9364 76944 Neglecting the effect of NO in the energy balance and substituting 4 out 500 kJkmol CH 193 74850 9 7750571 8669 0 59452961 8682 9904 62748 2 241820 9364 76944 393520 1 Q or out 193500 kJkmol CH4 Q PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1683 1697E Gaseous octane gas is burned with 40 excess air The equilibrium composition of the products of combustion is to be determined Assumptions 1 The equilibrium composition consists of CO2 H2O O2 NO and N2 2 The constituents of the mixture are ideal gases Analysis The stoichiometric and actual reactions in this case are Stoichiometric 2 1 and 1 2NO thus O N NO O2 N2 2 2 ν ν ν Actual 2 2 2 2 2 2 18 8 N O NO 9H O 8CO 3 76N 12 5 O 41 C H z y x N balance x z z x 50 65 8 2 131 6 O balance x y y x 50 5 2 9 16 35 Combustion chamber 600 psia 40 excess air C8H18 Total number of moles 87 8 9 8 total z y x N CO2 H2O NO O2 N2 3600 R The equilibrium constant relation can be expressed as total O2 N2 NO O2 N2 NO O2 N2 NO ν ν ν ν ν ν N P N N N K p From Table A28 at 2000 K 3600 R 3 931 ln K p Since the stoichiometric reaction being considered is double this reaction 3 851 10 4 3 931 2 exp p K Substituting 2 1 1 2 4 87 8 600 14 7 50 65 8 50 5 851 10 3 x x x Solving for x x 03492 Then y 5 05x 4825 z 658 05x 6563 Therefore the equilibrium composition of the products mixture at 2000 K and 4 MPa is 2 2 2 2 2 2 8 18 6563N 4825O 03492NO 9H O 8CO 376N 175O C H PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1684 1698 Propane gas is burned with 20 excess air The equilibrium composition of the products of combustion on a mass basis and the amount of heat released by this combustion are to be determined Assumptions 1 The equilibrium composition consists of CO2 H2O O2 NO and N2 2 The constituents of the mixture are ideal gases Analysis a The stoichiometric and actual reactions in this case are Stoichiometric 2 1 and 1 2NO thus O N NO O2 N2 2 2 ν ν ν Actual 2 2 2 2 2 2 8 3 N O NO 4H O 3CO 3 76N 5O 21 C H z y x N balance x z z x 50 2256 2 4512 Qout Combustion chamber 1 atm 20 excess air 25C C3H8 25C Products 2000 K O balance x y y x 50 1 2 4 6 12 Total number of moles 3056 4 3 total z y x N The equilibrium constant relation can be expressed as total O2 N2 NO O2 N2 NO O2 N2 NO ν ν ν ν ν ν N P N N N K p From Table A28 at 2000 K 3 931 Since the stoichiometric reaction being considered is double this reaction ln p K 3 851 10 4 3 931 2 exp p K Substituting 2 1 1 2 4 3056 1 50 2256 50 1 851 10 3 x x x Solving for x x 009097 Then y 1 05x 09545 z 2256 05x 2251 Therefore the equilibrium composition of the products mixture at 2000 K and 1 atm is 2 2 2 2 2 2 8 3 2251N 09545O 009097NO 4H O 3CO 376N 6O C H The mass of each product and the total mass of the products is 86755 kg 63028 3054 2 73 72 132 63028 kg 2251 kmol28 kgkmol 3054 kg 0 9545 kmol32 kgkmol 2 73 kg 0 09097 kmol30 kgkmol 72 kg 4 kmol18 kgkmol 132 kg 3 kmol44 kgkmol total N2 N2 N2 O2 O2 O2 NO NO NO H2O H2O H2O CO2 CO2 CO2 m M N m M N m M N m M N m M N m The mass fractions of the products are PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1685 07265 00352 00031 00830 01522 86755 kg 63028 kg mf 86755 kg 3054 kg mf 86755 kg 273 kg mf 86755 kg 72 kg mf 86755 kg 132 kg mf total N2 N2 total O2 O2 total NO NO total H2O H2O total CO2 CO2 m m m m m m m m m m b The heat transfer for this combustion process is determined from the energy balance applied on the combustion chamber with W 0 It reduces to system out in E E E R f R P f P h h h N h h h N Q o o o o out Assuming the air and the combustion products to be ideal gases we have h hT From the tables Substance fo h kJkmol h298K kJkmol h1600K kJkmol C3H8 103850 O2 0 8682 67881 N2 0 8669 64810 H2O 241820 9904 82593 CO2 393520 9364 100804 Neglecting the effect of NO in the energy balance and substituting 8 3 out 675 kJkmol C H 158 103850 8669 225164810 8682 0 954567881 9904 82593 4 241820 9364 100804 393520 3 Q or 3606 kJkg C3H8 44 kgkmol 158675 kJkmol Qout PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1686 1699 Propane gas is burned with stoichiometric air in an adiabatic manner The temperature of the products and the equilibrium composition of the products are to be determined Assumptions 1 The equilibrium composition consists of CO2 H2O O2 NO and N2 2 The constituents of the mixture are ideal gases Analysis a The stoichiometric and actual reactions in this case are Stoichiometric 2 1 and 1 2NO thus O N NO O2 N2 2 2 ν ν ν Actual 2 2 2 2 2 2 8 3 N O NO 4H O 3CO 3 76N 5O 11 C H z y x N balance x z z x 50 2068 2 4136 O balance x y y x 50 50 2 4 6 11 Combustion Chamber 1 atm Air 10 excess air 25C C3H8 25C Products TP Total number of moles 2118 4 3 total z y x N The equilibrium constant relation can be expressed as total O2 N2 NO O2 N2 NO O2 N2 NO ν ν ν ν ν ν N P N N N K p We assume that the products will be at 2000 K Then from Table A28 at 2000 K Since the stoichiometric reaction being considered is double this reaction 3 931 ln p K 3 851 10 4 3 931 2 exp p K Substituting 2 1 1 2 4 2118 1 50 2068 50 50 851 10 3 x x x Solving for x x 00611 Then y 05 05x 04695 z 2068 05x 2065 Therefore the equilibrium composition of the products mixture at 2000 K and 1 atm is 2 2 2 2 2 2 3 8 2065N 04695O 00611NO 4H O 3CO 376N 6O C H b From the tables Substance o f h kJkmol h298K kJkmol C3H8 g 103850 O2 0 8682 N2 0 8669 H2O g 241820 9904 CO2 393520 9364 Thus 0 0 103850 1 8669 2065 0 8682 0 4695 0 9188 0 0611 39460 9904 241820 4 9364 393520 3 N2 O2 OH H2O CO2 h h h h h PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1687 It yields 2 292940 kJ 2065 0 4695 0 0611 4 3 N2 O2 OH H2O CO2 h h h h h The adiabatic flame temperature is obtained from a trial and error solution A first guess is obtained by dividing the right hand side of the equation by the total number of moles which yields 22929403400611046952065 81366 kJkmol This enthalpy value corresponds to about 2450 K for N2 Noting that the majority of the moles are N2 TP will be close to 2450 K but somewhat under it because of the higher specific heat of H2O At 2200 K 2 088620 kJ Lower than 2292940 206564810 469575484 0 0 061169932 492940 3112939 2065 0 4695 0 0611 4 3 N2 O2 OH H2O CO2 h h h h h At 2400 K 2 471200 kJ Higher than 2292940 206579320 469583174 0 0 061177015 4103508 3125152 2065 0 4695 0 0611 4 3 N2 O2 OH H2O CO2 h h h h h By interpolation of the two results TP 2307 K 2034C PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1688 16100 A mixture of H2O and O2 is heated to a high temperature The equilibrium composition is to be determined Assumptions 1 The equilibrium composition consists of H2O OH O2 and H2 2 The constituents of the mixture are ideal gases Analysis The reaction equation during this process can be expressed as H2O OH H2 O2 3600 K 8 atm 2H O 3O H O H O OH 2 2 2 2 2 x y z w Mass balances for hydrogen and oxygen yield H balance 1 w y x 2 2 4 O balance 2 w z x 2 8 The mass balances provide us with only two equations with four unknowns and thus we need to have two more equations to be obtained from the KP relations to determine the equilibrium composition of the mixture They are H O H O 2 2 1 2 2 reaction 1 H O H OH 2 1 2 2 reaction 2 The equilibrium constant for these two reactions at 3600 K are determined from Table A28 to be ln ln K K K K P P P P 1 1 2 2 1392 0 24858 1088 033689 The KP relations for these two simultaneous reactions are total O H OH H 2 total O H O H 1 H2O OH 2 H 2O H 2 OH 2 H 2 H2O O2 2 H 2O H 2 2 O 2 2 H 2 ν ν ν ν ν ν ν ν ν ν ν ν N P N N N K N P N N N K P P where w z y x N N N N N OH O H H O total 2 2 2 Substituting 1 2 1 2 8 0 24858 w z y x x y z 3 1 2 1 2 8 0 33689 w z y x x w y 4 Solving Eqs 1 2 3 and 4 simultaneously for the four unknowns x y z and w yields x 1371 y 01646 z 285 w 0928 Therefore the equilibrium composition becomes 1371H O 0165H 285O 0928OH 2 2 2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1689 16101 A mixture of CO2 and O2 is heated to a high temperature The equilibrium composition is to be determined Assumptions 1 The equilibrium composition consists of CO2 CO O2 and O 2 The constituents of the mixture are ideal gases Analysis The reaction equation during this process can be expressed as CO2 CO O2 O 2600 K 15 atm O O CO CO 3O 3CO 2 2 2 2 w z y x Mass balances for carbon and oxygen yield C balance 1 3 x y O balance 2 w z y x 2 2 12 The mass balances provide us with only two equations with four unknowns and thus we need to have two more equations to be obtained from the KP relations to determine the equilibrium composition of the mixture They are 2 2 1 2 O CO CO reaction 1 reaction 2 O 2 2 O The equilibrium constant for these two reactions at 2600 K are determined from Table A28 to be 0 0005416 7 521 ln 0 06075 2 801 ln 2 2 1 1 P P P P K K K K The KP relations for these two simultaneous reactions are O2 O 2 O 2 O CO2 O2 CO 2 CO 2 2 O 2 CO total O O 2 total CO O CO 1 ν ν ν ν ν ν ν ν ν ν N P N N K N P N N N K P P where w z y x N N N N N O CO O CO total 2 2 Substituting 1 2 1 2 51 0 06075 w z y x x y z 3 2 1 2 51 0005416 0 w z y x z w 4 Solving Eqs 1 2 3 and 4 simultaneously for the four unknowns x y z and w yields x 2803 y 0197 z 3057 w 008233 Thus the equilibrium composition is 00823O 3057O 0197CO 2803CO 2 2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1690 16102 Problem 16101 is reconsidered The effect of pressure on the equilibrium composition by varying pressure from 1 atm to 10 atm is to be studied Analysis The problem is solved using EES and the solution is given below Given T2600 K P15 atm The equilibrium constant for these two reactions at 2600 K are determined from Table A28 Kp1exp2801 Kp2exp7521 Analysis Actual reaction 3 CO2 3 O2 NCO2 CO2 NCO CO NO2 O2 NO O 3NCO2NCO C balance 122NCO2NCO2NO2NO O balance NtotalNCO2NCONO2NO Stoichiometric reaction 1 CO2 CO 12 O2 Stoichiometric coefficients for reaction 1 nuCO211 nuCO11 nuO2112 Stoichiometric reaction 2 O2 2 O Stoichiometric coefficients for reaction 2 nuO221 nuO22 Kp relations are Kp1NCOnuCO1NO2nuO21NCO2nuCO21PNtotalnuCO1nuO21nuCO21 Kp2NOnuO2NO2nuO22PNtotalnuO2nuO22 Patm atm b kmolCO 1 2 3 4 5 6 7 8 9 10 02379 01721 01419 01237 01111 01017 009442 00885 008357 00794 1 2 3 4 5 6 7 8 9 10 006 008 01 012 014 016 018 02 022 024 P atm NCO kmol PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1691 16103 The hR at a specified temperature is to be determined using enthalpy and Kp data Assumptions Both the reactants and products are ideal gases Analysis a The complete combustion equation of H2 can be expressed as H O O H 2 2 2 1 2 The hR of the combustion process of H at 2400 K is the amount of energy released as one kmol of H 2 2 is burned in a steady flow combustion chamber at a temperature of 2400 K and can be determined from R f R P f P R h h h N h h h N h o o o o Assuming the H2O H2 and O2 to be ideal gases we have h h T From the tables Substance hf o kJkmol h298 K kJkmol h2400 K kJkmol H2O 241820 9904 103508 H2 0 8468 75383 O2 0 8682 83174 Substituting hR 1 241820 103 508 9904 1 0 75 383 8468 05 0 83174 8682 252377 kJ kmol b The hR value at 2400 K can be estimated by using KP values at 2200 K and 2600 K the closest two temperatures to 2400 K for which KP data are available from Table A28 2 1 1 2 2 1 1 2 1 1 ln 1 or ln 1 ln T T R h K K T T R h K K u R P P u R P P 252047 kJkmol R R h h 2600 K 1 2200 K 1 8 314 kJkmol K 6 768 4 648 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1692 16104 Problem 16103 is reconsidered The effect of temperature on the enthalpy of reaction using both methods by varying the temperature from 2000 to 3000 K is to be investigated Analysis The problem is solved using EES and the solution is given below Input Data Tprod2400 K DELTATprod 25 K Ru8314 kJkmolK Tprod1 Tprod DELTATprod Tprod2 Tprod DELTATprod The combustion equation is 1 H2 05 O2 1 H2O The enthalpy of reaction HbarR using enthalpy data is hbarREnthalpy HP HR HP 1EnthalpyH2OTTprod HR 1EnthalpyH2TTprod 05EnthalpyO2TTprod The enthalpy of reaction HbarR using enthalpy data is found using the following equilibruim data The following equations provide the specific Gibbs function ghTs for each component as a function of its temperature at 1 atm pressure 1013 kPa gH2O1EnthalpyH2OTTprod1 Tprod1 EntropyH2OTTprod1 P1013 gH21EnthalpyH2TTprod1 Tprod1 EntropyH2TTprod1 P1013 gO21EnthalpyO2TTprod1 Tprod1 EntropyO2TTprod1 P1013 gH2O2EnthalpyH2OTTprod2 Tprod2 EntropyH2OTTprod2 P1013 gH22EnthalpyH2TTprod2 Tprod2 EntropyH2TTprod2 P1013 gO22EnthalpyO2TTprod2 Tprod2 EntropyO2TTprod2 P1013 The standardstate at 1 atm Gibbs functions are DELTAG1 1gH2O105gO211gH21 DELTAG2 1gH2O205gO221gH22 The equilibrium constants are given by Eq 1514 Kp1 expDELTAG1RuTprod1 From EES data KP2 expDELTAG2RuTprod2 From EES data the entahlpy of reaction is estimated from the equilibrium constant Kp by using EQ 1518 as lnKP2KP1hbarRKpRu1Tprod1 1Tprod2 PercentError ABShbarRenthalpy hbarRKphbarRenthalpyConvert Percent Error Tprod K hREnthalpy kJkmol hRKp kJkmol 00002739 2000 251723 251722 00002333 2100 251920 251919 0000198 2200 252096 252095 00001673 2300 252254 252254 00001405 2400 252398 252398 00001173 2500 252532 252531 000009706 2600 252657 252657 000007957 2700 252778 252777 000006448 2800 252897 252896 000005154 2900 253017 253017 00000405 3000 253142 253142 2000 2200 2400 2600 2800 3000 253250 252900 252550 252200 251850 251500 Tprod k hR kJkmol Enthalpy Data Enthalpy Data Kp Data Kp Data DELTATprod 25 K PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1693 16105 The KP value of the dissociation process O2 2O at a specified temperature is to be determined using the hR data and KP value at a specified temperature Assumptions Both the reactants and products are ideal gases Analysis The hR and KP data are related to each other by 2 1 1 2 2 1 1 2 1 1 ln or ln 1 1 ln T T R h K K T T R h K K u R P P u R P P The hR of the specified reaction at 2800 K is the amount of energy released as one kmol of O2 dissociates in a steadyflow combustion chamber at a temperature of 2800 K and can be determined from R f R P f P R h h h N h h h N h o o o o Assuming the O2 and O to be ideal gases we have h h T From the tables Substance hf o kJkmol h298 K kJkmol h2800 K kJkmol O 249190 6852 59241 O2 0 8682 98826 Substituting hR 2 249 190 59 241 6852 1 0 98 826 8682 513 014 kJ kmol The KP value at 3000 K can be estimated from the equation above by using thishR value and the KP value at 2600 K which is ln KP1 7521 3000 K 1 2600 K 1 8 314 kJkmol K 513014 kJkmol 7 521 ln KP2 4 357 4 357 Table A 28 ln ln 2 2 P P K K or KP2 00128 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1694 16106 A mixture of CO and O2 contained in a tank is ignited The final pressure in the tank and the amount of heat transfer are to be determined Assumptions 1 The equilibrium composition consists of CO2 and O2 2 Both the reactants and the products are ideal gases Analysis The combustion equation can be written as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2 5 CO2 CO O2 25C 3 atm CO O CO O 2 2 3 2 The heat transfer can be determined from R f R P f P P h h h N P h h h N Q v v o o o o out Both the reactants and the products are assumed to be ideal gases and thus all the internal energy and enthalpies depend on temperature only and the v P terms in this equation can be replaced by RuT It yields R u f R P u f P R T h N R T h h h N Q o o 29 8 K 80 0 K out since reactants are at the standard reference temperature of 25C From the tables Substance hf o kJkmol h298 K kJkmol h800K kJkmol CO 110530 8669 23844 O2 0 8682 24523 CO2 393520 9364 32179 Substituting 298 8 314 110530 1 298 8 314 0 3 800 8 314 8682 24523 0 52 800 8 314 9364 32179 393520 1 out Q 233940 kJkmol CO r he final pressure in the tank is determined from o Qout 233940 kJkmol CO T 705 atm 298 K 3atm 800 K 4 53 1 1 1 2 2 2 2 2 1 1 2 1 N T P N T P N R T N R T P P u u V V CO O CO 1 2 2 2 The equilibrium constant for the reaction at 800 K is ln KP 372 by interpolation which is much greater than 705 Therefore it is not realistic to assume that no CO will be present in equilibrium mixture preparation If you are a student using this Manual you are using it without permission 1695 16107 A 2L bottle is filled with carbonated drink that is fully charged saturated with CO2 gas The volume that the CO2 gas would occupy if it is released and stored in a container at room conditions is to be determined Assumptions 1 The liquid drink can be treated as water 2 Both the CO2 gas and the water vapor are ideal gases 3 The CO2 gas is weakly soluble in water and thus Henrys law is applicable Properties The saturation pressure of water at 17C is 1938 kPa Table A4 Henrys constant for CO2 dissolved in water at 17ºC 290 K is H 1280 bar Table 162 Molar masses of CO2 and water are 4401 and 18015 kgkmol respectively Table A1 The gas constant of CO2 is 01889 kPam3kgK Also 1 bar 100 kPa Analysis In the charging station the CO2 gas and water vapor mixture above the liquid will form a saturated mixture Noting that the saturation pressure of water at 17C is 1938 kPa the partial pressure of the CO2 gas is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2 CO 59806 kPa 59806 bar 1 938 600 sat 17 C vapor gas side P P P P P From Henrys law the mole fraction of CO2 in the liquid drink is determined to be 000467 1280 bar 5 9806 bar CO gas side CO liquidside 2 2 H P y Then the mole fraction of water in the drink becomes y y water liquid side CO liquid side 2 1 1 0 00467 0 99533 The mass and mole fractions of a mixture are related to each other by w m m N M N M y M M i i m m m m i i i i where the rent molar mass of the drink liquid water CO2 mixture is kg kmol 0 99533 18 015 0 00467 44 01 1814 Then the mass fraction of dissolved CO2 in liquid drink becomes appa M y M y M y M m i i liquid water water CO CO 2 2 w y MCO2 0 0 00467 44 Mm O liquidside CO liquidside 2 2 00113 01 1814 Then the volume occupied by this CO2 at the room conditions of 20C and 100 kPa becomes C Therefore the mass of dissolved CO2 in a 2 L 2 kg drink is m w mm CO CO 2 2 kg 00226 kg 00113 2 125 L 00125 m 3 100 kPa 0 0226 kg01889 kPa m kg K293 K 3 P mRT V Discussion Note that the amount of dissolved CO2 in a 2L pressurized drink is large enough to fill 6 such bottles at room temperature and pressure Also we could simplify the calculations by assuming the molar mass of carbonated drink to be the same as that of water and take it to be 18 kgkmol because of the very low mole fraction of CO2 in the drink preparation If you are a student using this Manual you are using it without permission 1696 16108 Ethyl alcohol C2H5OH gas is burned in a steadyflow adiabatic combustion chamber with 90 percent excess air The adiabatic flame temperature of the products is to be determined and the adiabatic flame temperature as a function of nalysis The complete combustion reaction in this case can be written as N O f Ex a here ath is the stoichiometric coefficient for air The oxygen balance gives the percent excess air is to be plotted A 2 2 2 2 th 5 2 3 H O 2 CO 376N O 1 C H OH gas Ex a 2 2 th w 2 3 1 2 2 2 1 1 th th Ex a Ex a The reaction equation with products in equilibrium is N O H O CO CO f e d b a determined from the mass balances 2 2 th 5 2 376N O 1 C H OH gas Ex a 2 2 2 2 The coefficients are 2 a b Carbon balance Hydrogen balance 3 2 d d 6 2 2 2 1 1 th e d b a Ex a Oxygen balance Solving th th d 3 e 27 f 2143 Then we write the balanced reaction equation as are Nitrogen balance f Ex a 3 76 1 th e above equations we find the coefficients to be Ex 09 a 3 a 2 b 000008644 2 2 2 2 2 2 5 2 2143 N O 72 3 H O 0 00008644 CO 2 CO 376N O 75 C H OH gas Total moles of products at equilibrium 2913 2143 72 3 0 00008644 2 tot N The assumed equilibrium reaction is 2 2 50 O CO CO The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using G T R T p u u ln or Kp e K G T R T where prod CO2 CO2 prod O2 O2 prod CO CO T g T g T g T G ν ν ν and the Gibbs functions are defined as O2 prod prod O2 s T h T g PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course CO2 prod prod CO2 CO prod prod CO s T h T g s T h T g e equilibrium constant is also given by Th 0 00001316 2913 1 2 0 00008644 72 50 50 1 50 1 tot 50 N P a be K p A steady flow energy balance gives P R H H preparation If you are a student using this Manual you are using it without permission 1697 where PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 235310 kJkmol 21430 75 0 235310 kJkmol fuel25 C 2143 75 N225 C O225 C h h h H o f R prod prod prod prod CO2 prod 0 00008644 2 T P h H N2 O2 H2O CO 2143 72 3 T T T T h h h h using EES we obtain the adiabatic flame temperature to be parametric studies is given next re isTprod is reac 2527315 K x PercentEx100 EX Excess air100 plete combustion reaction equation for excess air is xAth O2 376N22 CO2 3 H2O ExAth O2 f N2 n Balance for complete combustion 231 ExAth2 n for excess air and products in equilibrium is xAth O2 376N2a CO2 b CO d H2O e O2 f N2 a b e2 2CO05O2 ific Gibbs function ghTs for ach component in the product gases as a function of its temperature Tprod rod EntropyCO2TTprod P1013 Tprod Tprod EntropyCOTTprod P1013 Tprod P1013 he standardstate Gibbs function is CO2 stant is given by Eq 1514 P expDELTAG RuTprod he equilibrium constant is also given by Eq 1515 e05a1 Solving the energy balance equation Tprod 1569 K The copy of entire EES solution including The product temperatu The reactant temperature T For adiabatic combustion of 1 kmol of fuel Qout 0 kJ PercentEx 90 Percent excess air E Pprod 1013kPa Ru8314 kJkmolK The com C2H5OHgas 1E Oxyge 1 1ExAth22 The reaction equatio C2H5OHgas 1E Carbon Balance 2 Hydrogen Balance 62d Oxygen Balance 1 1ExAth2a2b d Nitrogen Balance 1ExAth376 f Ntot a b d e f Total kilomoles of products at equilibrium The assumed equilibrium reaction is CO The following equations provide the spec e at 1 atm pressure 1013 kPa gCO2EnthalpyCO2TTprod Tp gCOEnthalpyCOT gO2EnthalpyO2TTprod Tprod EntropyO2T T DELTAG 1gCO05gO21g The equilibrium con K PPprod 1013atm T K P PNtot1051b1 sqrtPNtot b sqrte KP a The steadyflow energy balance is preparation If you are a student using this Manual you are using it without permission 1698 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course R QoutHP barfC2H5OHgas235310 kJkmol HR1h Hgas 1ExA ENTH O2T reac Ath376E Y T kJkmol HPaEN ALPY db YCO Tp E Y Tprod eENTH YO2 fENTHA Tp PercentEx H h barfC2H5O th ALPY T 1Ex NTHALP N2T reac TH CO2TTpro ENTHALP T rodd NTHALP H2OT ALP TTprod LPYN2T rod kJkmol a ath b d e f Tprod K 10 20 30 40 50 60 70 80 90 100 1921 197 1988 1995 1998 1999 2 2 2 2 3 3 3 3 3 3 3 3 001212 0004983 0002111 00009184 00004093 00001863 000008644 000004081 3 3 3 3 3 3 3 3 03393 06152 09061 1202 1501 18 21 24 27 3 1466 1579 1692 1805 1918 203 2143 2256 1996 1907 1826 1752 1685 1625 1569 1518 3 3 007868 003043 3 3 1241 1354 2191 2093 10 20 30 40 50 60 70 80 90 100 1500 1600 1700 1800 1900 2000 2100 2200 PercentEx Tprod K preparation If you are a student using this Manual you are using it without permission 1699 16109 The percent theoretical air required for the combustion of octane such that the volume fraction of CO in the products is less than 01 and the heat transfer are to be determined Also the percent theoretical air required for 01 CO nalysis The complete combustion reaction equation for excess air is N O 1 9 H O 8 CO 376N O f a P he oxyge lance is in the products as a function of product pressure is to be plotted A 2 2 th th 2 2 2 2 th th C8H18 P a T n ba 2 1 9 1 2 8 2 th th th th a P P a The reaction equation for excess air and products in equilibrium is N O H O f e d to be de rmined from the mass balances 2 2 2 th th 18 8 CO CO 376N O C H b a P a 2 2 2 The coefficients are te 8 a b Carbon balance Hydrogen balance 9 2 d d 18 Oxygen balance 2 2 2 th th e d b a P a ume fraction of CO must be less than 01 That is Nitrogen balance f P a 3 76 th th Vol 0 001 tot CO f e d b a b N b y The assumed equilibrium reaction is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 2 2 The K value of a reaction at a specified temperature can be determined from the Gibbs function data 50 O CO CO p 920121 kJkmol 200030900 128 302 CO2 prod prod CO2 s T h T g 477876 kJkmol 200026853 59193 570781 kJkmol 200025848 53826 O2 prod prod O2 CO prod prod CO s T h T s T h T g The enthalpies at 2000 K and entropies at 2000 K and 1013 kPa are obtained from EES Substituting g 110402 kJkmol 920121 50 477876 570781 1 prod CO2 CO2 prod O2 O2 prod CO CO prod T g T g T g T G ν ν ν 0 001308 8 3142000 110402 exp K p exp prod prod T R T G u The equilibrium constant is also given by 1 50 1 prod 50 1 50 1 tot 50 101 3 f e d b a P a be N P a be K p The steady flow energy balance gives where P R H Q H out preparation If you are a student using this Manual you are using it without permission 16100 56115 59193 169171 53826 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 30212 a 8 b 208459 kJkmol 3 760 0 3 76 1 N2 2000 K O2 2000 K H2O 2000 K 2000 K th th th N2 298 K th th O2 298 K th th f e d f h eh h d a P h P a P a h h H uations simultaneously using EES we obtain th C8H18 298 K a P R 208459 CO CO2 2000 K bh ah H P The enthalpies are obtained from EES Solving all the eq 7 935 5 12 a a th 1 024 P 1024 th 1 024 100 100 PercentTh P 995500 kJkmol C8H18 out th 4811 0 3289 9 0 06544 f e d b s re is tical air Pth theoretical air100 Pth1Ath O2 f N2 91 Pth1Ath2 rium is ts at equilibrium cts is to be less than 01 For ideal gas mixtures volume fractions equal CO2CO05O2 O2EnthalpyO2TTprod Tprod EntropyO2TTprod P1013 he stan Gibb DELTAG CO0 CO2 The equ constant i en by Eq 1514 KP expDELTAG RuT d PPpro 3atm The equ stan given by Eq 1515 K P PNtot105 e05a1 sqrtPNtot b sqrte The stea energy b e is Q The copy of entire EES solution including parametric studies is given next The product temperature i Tprod 2000 K The reactant temperatu Treac 25273 K PercentTH is Percent theore Pth PercentTh100 Pprod 5 atm convertatmkPakPa Ru8314 kJkmolK The complete combustion reaction equation for excess air is C8H18 PthAth O2 376N28 CO2 9 H2O Oxygen Balance for complete combustion PthAth282 The reaction equation for excess air and products in equilib C8H18 PthAth O2 376N2a CO2 b CO d H2O e O2 f N2 Carbon Balance 8a b Hydrogen Balance 182d Oxygen Balance PthAth2a2b d e2 Nitrogen Balance PthAth376 f Ntot a b d e f Total kilomoles of produc The volume faction of CO in the produ mole fractions The mole fraction of CO in the product gases is yCO 0001 yCO bNtot The assumed equilibrium reaction is The following equations provide the specific Gibbs function ghTs for each component in the product gases as a function of its temperature Tprod at 1 atm pressure 1013 kPa gCO2EnthalpyCO2TTprod Tprod EntropyCO2TTprod P1013 gCOEnthalpyCOTTprod Tprod EntropyCOTTprod P1013 g T dardstate 1g s function is 5gO21g ilibrium s giv pro d 101 ilibrium con t is also 1b1 KP a dyflow alanc preparation If you are a student using this Manual you are using it without permission 16101 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course HR Q P HR1E LPYC8H18TTreacPthAthENTHALPYO2TTreacPthAth376ENTHALPYN2T Treac ol HPaE LPYCO2 rodbENTHALPYCOTTproddENTHALPYH2OTTprod eENTH O2TT ENTHALPYN2TTprod kJkmol Pprod kPa PercentTh outH NTHA kJkm NTHA TTp ALPY prodf 100 112 300 1041 500 1024 700 1017 900 1012 1100 101 1300 1008 1500 1006 1700 1005 1900 1005 2100 1004 2300 1003 0 500 1000 1500 2000 2500 100 102 104 106 108 110 112 Pprod kPa PercentTh preparation If you are a student using this Manual you are using it without permission 16102 16110 It is to be shown that when the three phases of a pure substance are in equilibrium the specific Gibbs function of each phase is the same Analysis The total Gibbs function of the three phase mixture of a pure substance can be expressed as g g s s m g m g m g G l l g s g l l 0 ubstituting g l l l earranging l g varied independently Thus each term on the right hand side must be ero to satisfy the equilibrium criteria It yields ombining hese two conditions gives the desired result where the subscripts s l and g indicate solid liquid and gaseous phases Differentiating by holding the temperature and pressure thus the Gibbs functions g constant yields g g s s g dm g dm g dm dG l l ms ml mg From conservation of mass dms dm dm dm dm dm S dG g dm dm g dm g dm s g g R dG g g dm g g dm s g s g l l For equilibrium dG 0 Also dm and dm can be z g g g g s g s l and C t g g g s s l PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 16103 16111 It is to be shown that when the two phases of a twocomponent system are in equilibrium the specific Gibbs function of each phase of each component is the same Analysis The total Gibbs function of the two phase mixture can be expressed as PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course G m g m g m g m g g g g g l l l l 1 1 1 1 2 2 2 2 where the subscripts l and g indicate liquid and gaseous phases Differentiating by holding the temperature and pressure thus the Gibbs functions constant yields dG g dm g dm g dm g dm g g g g l l l l 1 1 1 1 2 2 2 2 2 1 2 2 ried independently Thus each term on the right hand side must be satisfy the equilibrium criteria Then we have 2 2 and hich is the desired result own that the dissolved gases in a liquid can be driven off by heating the liquid nalysis Henrys law is expressed as ml1 ml2 mg1 mg2 From conservation of mass dm dm dm dm g g 1 1 2 l l and Substituting dG g g dm g g dm g g l l l l 1 1 2 For equilibrium dG 0 Also dml1 and dml2 can be va zero to g g g g g g l l 1 1 w 16112 Using Henrys law it is to be sh A y P H i liquidside i gas side 0 0 Henrys constant H increases with temperature and thus the fraction of gas i in the liquid yiliquid side decreases Therefore heating a liquid will drive off the dissolved gases in a liquid preparation If you are a student using this Manual you are using it without permission 16104 Fundamentals of Engineering FE Exam Problems 16113 If the equilibrium constant for the reaction H2 ½O2 H2O is K the equilibrium constant for the reaction 2H2O 2H2 O2 at the same temperature is a 1K b 12K c 2K d K2 e 1K2 Answer e 1K2 16114 If the equilibrium constant for the reaction CO ½O2 CO2 is K the equilibrium constant for the reaction CO2 3N2 CO ½O2 3N2 at the same temperature is a 1K b 1K 3 c 4K d K e 1K2 Answer a 1K 16115 The equilibrium constant for the reaction H2 ½O2 H2O at 1 atm and 1500C is given to be K Of the reactions given below all at 1500C the reaction that has a different equilibrium constant is a H2 ½O2 H2O at 5 atm b 2H2 O2 2H2O at 1 atm c H2 O2 H2O ½O2 at 2 atm d H2 ½O2 3N2 H2O 3N2 at 5 atm e H2 ½O2 3N2 H2O 3N2 at 1 atm Answer b 2H2 O2 2H2O at 1 atm 16116 Of the reactions given below the reaction whose equilibrium composition at a specified temperature is not affected by pressure is a H2 ½O2 H2O b CO ½O2 CO2 c N2 O2 2NO d N2 2N e all of the above Answer c N2 O2 2NO PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 16105 16117 Of the reactions given below the reaction whose number of moles of products increases by the addition of inert gases into the reaction chamber at constant pressure and temperature is a H2 ½O2 H2O b CO ½O2 CO2 c N2 O2 2NO d N2 2N e none of the above Answer d N2 2N 16118 Moist air is heated to a very high temperature If the equilibrium composition consists of H2O O2 N2 OH H2 and NO the number of equilibrium constant relations needed to determine the equilibrium composition of the mixture is a 1 b 2 c 3 d 4 e 5 Answer c 3 16119 Propane C3H8 is burned with air and the combustion products consist of CO2 CO H2O O2 N2 OH H2 and NO The number of equilibrium constant relations needed to determine the equilibrium composition of the mixture is a 1 b 2 c 3 d 4 e 5 Answer d 4 16120 Consider a gas mixture that consists of three components The number of independent variables that need to be specified to fix the state of the mixture is a 1 b 2 c 3 d 4 e 5 Answer d 4 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 171 Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGrawHill 2011 Chapter 17 COMPRESSIBLE FLOW PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 172 Stagnation Properties 171C No there is not significant error because the velocities encountered in airconditioning applications are very low and thus the static and the stagnation temperatures are practically identical Discussion If the air stream were supersonic however the error would indeed be significant 172C Stagnation enthalpy combines the ordinary enthalpy and the kinetic energy of a fluid and offers convenience when analyzing highspeed flows It differs from the ordinary enthalpy by the kinetic energy term Discussion Most of the time we mean specific enthalpy ie enthalpy per unit mass when we use the term enthalpy 173C Dynamic temperature is the temperature rise of a fluid during a stagnation process Discussion When a gas decelerates from high speed to zero speed at a stagnation point the temperature of the gas rises 174C The temperature of the air rises as it approaches the nose because of the stagnation process Discussion In the frame of reference moving with the aircraft the air decelerates from high speed to zero at the nose stagnation point and this causes the air temperature to rise 175 The inlet stagnation temperature and pressure and the exit stagnation pressure of air flowing through a compressor are specified The power input to the compressor is to be determined Assumptions 1 The compressor is isentropic 2 Air is an ideal gas 100 kPa 27C AIR 006 kgs 900 kPa W Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Analysis The exit stagnation temperature of air T02 is determined from 5624 K 100 300 2 K 900 41 1 41 1 01 02 01 02 k k P P T T From the energy balance on the compressor 01 20 in h m h W or 3002K 158 kW 0 06 kgs100 5 kJkg K5624 01 02 in T T mc W p Discussion Note that the stagnation properties can be used conveniently in the energy equation PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 173 176 Air at 320 K is flowing in a duct The temperature that a stationary probe inserted into the duct will read is to be determined for different air velocities Assumptions The stagnation process is isentropic Properties The specific heat of air at room temperature is cp 1005 kJkgK Analysis The air which strikes the probe will be brought to a complete stop and thus it will undergo a stagnation process The thermometer will sense the temperature of this stagnated air which is the stagnation temperature T0 It is determined from cp V T T 2 2 0 The results for each case are calculated below a 3200 K m2 s2 1000 1kJkg 1 005 kJkg K 2 1 ms2 320 K T0 AIR 320 K V b K 3201 2 2 2 0 m s 1000 1kJkg 1 005 kJkg K 2 10 ms 320 K T c 3250 K 2 2 2 0 m s 1000 1kJkg 1 005 kJkg K 2 100 ms 320 K T d 8175 K 2 2 2 0 m s 1000 1kJkg 1 005 kJkg K 2 1000 ms 320 K T Discussion Note that the stagnation temperature is nearly identical to the thermodynamic temperature at low velocities but the difference between the two is significant at high velocities PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 174 177 The states of different substances and their velocities are specified The stagnation temperature and stagnation pressures are to be determined Assumptions 1 The stagnation process is isentropic 2 Helium and nitrogen are ideal gases Analysis a Helium can be treated as an ideal gas with cp 51926 kJkgK and k 1667 Then the stagnation temperature and pressure of helium are determined from 555C 2 2 2 2 0 s m 1000 1kJkg C 5 1926 kJkg 2 240 ms 50 C 2 cp V T T 0261 MPa 1 667 1 1 667 1 0 0 3232 K 0 25 MPa 3287 K k k T P T P b Nitrogen can be treated as an ideal gas with cp 1039 kJkgK and k 1400 Then the stagnation temperature and pressure of nitrogen are determined from 933C 2 2 2 2 0 s m 1000 1kJkg C 1 039 kJkg 2 300 ms 50 C 2 cp V T T 0233 MPa 1 41 41 1 0 0 3232 K 0 15 MPa 3665 K k k T P T P c Steam can be treated as an ideal gas with cp 1865 kJkgK and k 1329 Then the stagnation temperature and pressure of steam are determined from 685 K 4118 C 2 2 2 2 0 s m 1000 1kJkg C 1 865 kJkg 2 480 ms 350 C 2 c p V T T 0147 MPa 1 1 329 1 329 1 0 0 6232 K 685 K MPa 10 k k T P T P Discussion Note that the stagnation properties can be significantly different than thermodynamic properties 178 The state of air and its velocity are specified The stagnation temperature and stagnation pressure of air are to be determined Assumptions 1 The stagnation process is isentropic 2 Air is an ideal gas Properties The properties of air at room temperature are cp 1005 kJkgK and k 14 Analysis The stagnation temperature of air is determined from 348 K 347 9 K m s 1000 1kJkg 1 005 kJkg K 2 470 ms 238 K 2 2 2 2 2 0 c p V T T Other stagnation properties at the specified state are determined by considering an isentropic process between the specified state and the stagnation state 136 kPa 135 9 kPa 238 K 36 kPa 3479 K 1 41 41 1 0 0 k k T P T P Discussion Note that the stagnation properties can be significantly different than thermodynamic properties PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 175 179E Steam flows through a device The stagnation temperature and pressure of steam and its velocity are specified The static pressure and temperature of the steam are to be determined Assumptions 1 The stagnation process is isentropic 2 Steam is an ideal gas Properties Steam can be treated as an ideal gas with cp 04455 BtulbmR and k 1329 Analysis The static temperature and pressure of steam are determined from 6637F 2 2 2 2 0 s ft 25037 1Btulbm F 0 4455 Btulbm 2 900 fts 700 F 2 cp V T T 1055 psia 1 1 329 1 329 1 0 0 1160 R 120 psia 11237 R k k T T P P Discussion Note that the stagnation properties can be significantly different than thermodynamic properties 1710 Air flows through a device The stagnation temperature and pressure of air and its velocity are specified The static pressure and temperature of air are to be determined Assumptions 1 The stagnation process is isentropic 2 Air is an ideal gas Properties The properties of air at an anticipated average temperature of 600 K are cp 1051 kJkgK and k 1376 Analysis The static temperature and pressure of air are determined from 5186 K 2 2 2 2 0 s m 1000 1kJkg 1 051 kJkg K 2 570 ms 673 2 2 c p V T T and 023 MPa 1 1 376 1 376 1 02 2 02 2 6732 K MPa 5186 K 60 k k T T P P Discussion Note that the stagnation properties can be significantly different than thermodynamic properties PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 176 1711 The inlet stagnation temperature and pressure and the exit stagnation pressure of products of combustion flowing through a gas turbine are specified The power output of the turbine is to be determined Assumptions 1 The expansion process is isentropic 2 Products of combustion are ideal gases Properties The properties of products of combustion are cp 1157 kJkgK R 0287 kJkgK and k 133 Analysis The exit stagnation temperature T02 is determined to be 577 9 K 1 1023 2 K 01 1 33 1 1 33 1 01 02 01 02 k k P P T T 1 MPa 750C 100 kPa STEAM Also W 1 1 33 0 287 kJkg K 1 33 1 1 157 kJkg K p v p p kR c kc k c R c k From the energy balance on the turbine 01 20 out h h w or out 01 02 1 157 kJkg K 10232 5779 K 5152 kJkg p w c T T 515 kJkg Discussion Note that the stagnation properties can be used conveniently in the energy equation 177 Speed of Sound and Mach Number 1712C Sound is an infinitesimally small pressure wave It is generated by a small disturbance in a medium It travels by wave propagation Sound waves cannot travel in a vacuum Discussion Electromagnetic waves like light and radio waves can travel in a vacuum but sound cannot 1713C Yes the propagation of sound waves is nearly isentropic Because the amplitude of an ordinary sound wave is very small and it does not cause any significant change in temperature and pressure Discussion No process is truly isentropic but the increase of entropy due to sound propagation is negligibly small 1714C The sonic speed in a medium depends on the properties of the medium and it changes as the properties of the medium change Discussion The most common example is the change in speed of sound due to temperature change 1715C Sound travels faster in warm higher temperature air since kRT c Discussion On the microscopic scale we can imagine the air molecules moving around at higher speed in warmer air leading to higher propagation of disturbances 1716C Sound travels fastest in helium since kRT c and helium has the highest kR value It is about 040 for air 035 for argon and 346 for helium Discussion We are assuming of course that these gases behave as ideal gases a good approximation at room temperature 1717C Air at specified conditions will behave like an ideal gas and the speed of sound in an ideal gas depends on temperature only Therefore the speed of sound is the same in both mediums Discussion If the temperature were different however the speed of sound would be different 1718C In general no because the Mach number also depends on the speed of sound in gas which depends on the temperature of the gas The Mach number remains constant only if the temperature and the velocity are constant Discussion It turns out that the speed of sound is not a strong function of pressure In fact it is not a function of pressure at all for an ideal gas PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 178 1719 The Mach number of an aircraft and the speed of sound in air are to be determined at two specified temperatures Assumptions Air is an ideal gas with constant specific heats at room temperature Properties The gas constant of air is R 0287 kJkgK Its specific heat ratio at room temperature is k 14 Analysis From the definitions of the speed of sound and the Mach number 347 ms 1kJkg 0 287 kJkg K300 K 1000 m s 41 2 2 kRT c and 0692 347 ms 240 ms Ma c V b At 1000 K 634 ms 1kJkg 0 287 kJkg K1000 K 1000 m s 41 2 2 kRT c and 0379 634 ms 240 ms Ma c V Discussion Note that a constant Mach number does not necessarily indicate constant speed The Mach number of a rocket for example will be increasing even when it ascends at constant speed Also the specific heat ratio k changes with temperature and the accuracy of the result at 1000 K can be improved by using the k value at that temperature it would give k 1386 c 619 ms and Ma 0388 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 179 1720 Carbon dioxide flows through a nozzle The inlet temperature and velocity and the exit temperature of CO2 are specified The Mach number is to be determined at the inlet and exit of the nozzle Assumptions 1 CO2 is an ideal gas with constant specific heats at room temperature 2 This is a steadyflow process Properties The gas constant of carbon dioxide is R 01889 kJkgK Its constant pressure specific heat and specific heat ratio at room temperature are cp 08439 kJkgK and k 1288 Analysis a At the inlet 540 3 ms 1kJkg 288 0 1889 kJkg K1200 K 1000 m s 1 2 2 1 1 1 k RT c Thus 1200 K 50 ms PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 400 K Carbon dioxide 00925 540 3 ms 50 ms Ma 1 1 1 c V b At the exit 312 0 ms 1kJkg 288 0 1889 kJkg K400 K 1000 m s 1 2 2 2 2 2 k RT c The nozzle exit velocity is determined from the steadyflow energy balance relation 2 0 2 1 2 2 1 2 V V h h 2 0 2 1 2 2 1 2 V V T c p T 1163 ms s m 1000 1kJkg 2 50 ms 0 8439 kJkg K400 1200 K 0 2 2 2 2 2 2 V V Thus 373 312 ms 1163 ms Ma 2 2 2 c V Discussion The specific heats and their ratio k change with temperature and the accuracy of the results can be improved by accounting for this variation Using EES or another property database At 1200 K cp 1278 kJkgK k 1173 c1 516 ms V1 50 ms Ma1 00969 At 400 K cp 09383 kJkgK k 1252 c2 308 ms V2 1356 ms Ma2 441 Therefore the constant specific heat assumption results in an error of 45 at the inlet and 155 at the exit in the Mach number which are significant preparation If you are a student using this Manual you are using it without permission 1710 1721 Nitrogen flows through a heat exchanger The inlet temperature pressure and velocity and the exit pressure and velocity are specified The Mach number is to be determined at the inlet and exit of the heat exchanger Assumptions 1 N2 is an ideal gas 2 This is a steadyflow process 3 The potential energy change is negligible Properties The gas constant of N2 is R 02968 kJkgK Its constant pressure specific heat and specific heat ratio at room temperature are cp 1040 kJkgK and k 14 Analysis 342 9 ms 1 kJkg s 400 0 2968 kJkg K283 K 1000 m 1 2 2 1 1 1 k RT c Thus Nitrogen 120 kJkg 0292 342 9 ms 100 ms Ma 1 1 1 c V 150 kPa 10C 100 ms 100 kPa 200 ms From the energy balance on the heat exchanger 2 2 1 2 2 1 2 in V V T T c q p 2 2 2 2 2 s m 1000 1 kJkg 2 100 ms 200 ms 10 C 1 040 kJkg C 120 kJkg T It yields T2 111C 384 K 399 ms 1 kJkg s 0 2968 kJkg K384 K 1000 m 41 2 2 2 2 2 k RT c Thus 0501 399 ms 200 ms Ma 2 2 2 c V Discussion The specific heats and their ratio k change with temperature and the accuracy of the results can be improved by accounting for this variation Using EES or another property database At 10C cp 1038 kJkgK k 1400 c1 343 ms V1 100 ms Ma1 0292 At 111C cp 1041 kJkgK k 1399 c2 399 ms V2 200 ms Ma2 0501 Therefore the constant specific heat assumption results in an error of 45 at the inlet and 155 at the exit in the Mach number which are almost identical to the values obtained assuming constant specific heats 1722 The speed of sound in refrigerant134a at a specified state is to be determined Assumptions R134a is an ideal gas with constant specific heats at room temperature Properties The gas constant of R134a is R 008149 kJkgK Its specific heat ratio at room temperature is k 1108 Analysis From the idealgas speed of sound relation 173 ms 1kJkg 273 K 1000 m s 108 0 08149 kJkg K60 1 2 2 kRT c Discusion Note that the speed of sound is independent of pressure for ideal gases PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1711 1723 The Mach number of a passenger plane for specified limiting operating conditions is to be determined Assumptions Air is an ideal gas with constant specific heats at room temperature Properties The gas constant of air is R 0287 kJkgK Its specific heat ratio at room temperature is k 14 Analysis From the speed of sound relation 293 ms 1 kJkg s 273 K 1000 m 0 287 kJkg K60 41 2 2 kRT c Thus the Mach number corresponding to the maximum cruising speed of the plane is 0897 293 ms ms 63 945 Ma max c V Discussion Note that this is a subsonic flight since Ma 1 Also using a k value at 60C would give practically the same result 1724E Steam flows through a device at a specified state and velocity The Mach number of steam is to be determined assuming ideal gas behavior Assumptions Steam is an ideal gas with constant specific heats Properties The gas constant of steam is R 01102 BtulbmR Its specific heat ratio is given to be k 13 Analysis From the idealgas speed of sound relation 2040 fts 1Btulbm s 0 1102 Btulbm R1160 R 25037 ft 31 2 2 kRT c Thus 0441 2040 fts 900 fts Ma c V Discussion Using property data from steam tables and not assuming ideal gas behavior it can be shown that the Mach number in steam at the specified state is 0446 which is sufficiently close to the idealgas value of 0441 Therefore the ideal gas approximation is a reasonable one in this case PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1712 1725E Problem 1724E is reconsidered The variation of Mach number with temperature as the temperature changes between 350 and 700F is to be investigated and the results are to be plotted Analysis The EES Equations window is printed below along with the tabulated and plotted results TTemperature460 R01102 V900 k13 cSQRTkRT25037 MaVc 350 400 450 500 550 600 650 700 044 045 046 047 048 049 05 051 052 053 Temperature F Ma Temperature T F Mach number Ma 350 375 400 425 450 475 500 525 550 575 600 625 650 675 700 0528 0520 0512 0505 0498 0491 0485 0479 0473 0467 0462 0456 0451 0446 0441 Discussion Note that for a specified flow speed the Mach number decreases with increasing temperature as expected 1726 The expression for the speed of sound for an ideal gas is to be obtained using the isentropic process equation and the definition of the speed of sound Analysis The isentropic relation Pvk A where A is a constant can also be expressed as k k A A v P ρ 1 Substituting it into the relation for the speed of sound kRT k P k A kA A P c k k s k s 1 2 ρ ρ ρ ρ ρ ρ ρ since for an ideal gas P ρRT or RT Pρ Therefore kRT c which is the desired relation Discussion Notice that pressure has dropped out the speed of sound in an ideal gas is not a function of pressure PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1713 1727 The inlet state and the exit pressure of air are given for an isentropic expansion process The ratio of the initial to the final speed of sound is to be determined Assumptions Air is an ideal gas with constant specific heats at room temperature Properties The properties of air are R 0287 kJkgK and k 14 The specific heat ratio k varies with temperature but in our case this change is very small and can be disregarded Analysis The final temperature of air is determined from the isentropic relation of ideal gases 228 4 K 15 MPa 333 2 K 04 MPa 41 1 41 1 1 2 1 2 k k P P T T Treating k as a constant the ratio of the initial to the final speed of sound can be expressed as 121 4 228 333 2 Ratio 2 1 2 2 1 1 1 2 T T RT k RT k c c Discussion Note that the speed of sound is proportional to the square root of thermodynamic temperature 1728 The inlet state and the exit pressure of helium are given for an isentropic expansion process The ratio of the initial to the final speed of sound is to be determined Assumptions Helium is an ideal gas with constant specific heats at room temperature Properties The properties of helium are R 20769 kJkgK and k 1667 Analysis The final temperature of helium is determined from the isentropic relation of ideal gases 196 3 K 15 333 2 K 04 1 667 1 1 667 1 1 2 1 2 k k P P T T The ratio of the initial to the final speed of sound can be expressed as 130 3 196 333 2 Ratio 2 1 2 2 1 1 1 2 T T RT k RT k c c Discussion Note that the speed of sound is proportional to the square root of thermodynamic temperature PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1714 1729E The inlet state and the exit pressure of air are given for an isentropic expansion process The ratio of the initial to the final speed of sound is to be determined Assumptions Air is an ideal gas with constant specific heats at room temperature Properties The properties of air are R 006855 BtulbmR and k 14 The specific heat ratio k varies with temperature but in our case this change is very small and can be disregarded Analysis The final temperature of air is determined from the isentropic relation of ideal gases 489 9 R 170 659 7 R 60 41 1 41 1 1 2 1 2 k k P P T T Treating k as a constant the ratio of the initial to the final speed of sound can be expressed as 116 489 9 659 7 Ratio 2 1 2 2 1 1 1 2 T T RT k RT k c c Discussion Note that the speed of sound is proportional to the square root of thermodynamic temperature PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1715 One Dimensional Isentropic Flow 1730C a The velocity increases b c d The temperature pressure and density of the fluid decrease Discussion The velocity increase is opposite to what happens in supersonic flow 1731C a The velocity decreases b c d The temperature pressure and density of the fluid increase Discussion The velocity decrease is opposite to what happens in supersonic flow 1732C a The exit velocity remains constant at sonic speed b the mass flow rate through the nozzle decreases because of the reduced flow area Discussion Without a diverging portion of the nozzle a converging nozzle is limited to sonic velocity at the exit 1733C a The velocity decreases b c d The temperature pressure and density of the fluid increase Discussion The velocity decrease is opposite to what happens in subsonic flow 1734C a The velocity increases b c d The temperature pressure and density of the fluid decrease Discussion The velocity increase is opposite to what happens in subsonic flow 1735C The pressures at the two throats are identical Discussion Since the gas has the same stagnation conditions it also has the same sonic conditions at the throat 1736C No it is not possible Discussion The only way to do it is to have first a converging nozzle and then a diverging nozzle PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1716 1737 The Mach number of scramjet and the air temperature are given The speed of the engine is to be determined Assumptions Air is an ideal gas with constant specific heats at room temperature Properties The gas constant of air is R 0287 kJkgK Its specific heat ratio at room temperature is k 14 Analysis The temperature is 20 27315 25315 K The speed of sound is 2 140287 kJkg K25315 K 1000 m s2 31893 ms 1 kJkg c kRT and 36 kmh Ma 31893 ms7 8037 kmh 1 ms V c 8040 kmh Discussion Note that extremely high speed can be achieved with scramjet engines We cannot justify more than three significant digits in a problem like this 1738E The Mach number of scramjet and the air temperature are given The speed of the engine is to be determined Assumptions Air is an ideal gas with constant specific heats at room temperature Properties The gas constant of air is R 006855 BtulbmR Its specific heat ratio at room temperature is k 14 Analysis The temperature is 0 45967 45967 R The speed of sound is 2 14006855 Btulbm R45967 R 25037 ft s2 105095 fts 1 Btulbm c kRT and 1 mih Ma 105095 fts7 50159 mih 146667 fts V c 5020 mih Discussion Note that extremely high speed can be achieved with scramjet engines We cannot justify more than three significant digits in a problem like this PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1717 1739 The speed of an airplane and the air temperature are give It is to be determined if the speed of this airplane is subsonic or supersonic Assumptions Air is an ideal gas with constant specific heats at room temperature Properties The gas constant of air is R 0287 kJkgK Its specific heat ratio at room temperature is k 14 Analysis The temperature is 50 27315 22315 K The speed of sound is 2 1000 m s2 36 kmh 140287 kJkg K22315 K 107797 kmh 1 kJkg 1 ms c kRT and 920 kmh Ma 085346 107797 kmh V c 0853 The speed of the airplane is subsonic since the Mach number is less than 1 Discussion Subsonic airplanes stay sufficiently far from the Mach number of 1 to avoid the instabilities associated with transonic flights PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1718 1740 The critical temperature pressure and density of air and helium are to be determined at specified conditions Assumptions Air and Helium are ideal gases with constant specific heats at room temperature Properties The properties of air at room temperature are R 0287 kJkgK k 14 and cp 1005 kJkgK The properties of helium at room temperature are R 20769 kJkgK k 1667 and cp 51926 kJkgK Analysis a Before we calculate the critical temperature T pressure P and density ρ we need to determine the stagnation temperature T0 pressure P0 and density ρ0 C 131 1 m s 1000 1 kJkg C 1 005 kJkg 2 250 ms 100 2 100 C 2 2 2 2 0 cp V T 264 7 kPa 3732 K 200 kPa 4043 K 1 41 41 1 0 0 k k T T P P 3 3 0 0 0 kgm 2 281 0 287 kPa m kg K4043 K 264 7 kPa RT P ρ Thus 337 K 14 1 2 404 3 K 1 2 T0 k T 140 kPa 1 41 41 1 0 14 1 2 264 7 kPa 1 2 k k k P P 145 kgm3 1 1 41 3 1 1 0 14 1 2 2 281 kgm 1 2 k k ρ ρ b For helium 487 C m s 1000 1 kJkg C 5 1926 kJkg 2 300 ms 40 2 2 2 2 2 0 cp V T T 214 2 kPa 3132 K 200 kPa 3219 K 1 1 667 1 667 1 0 0 k k T T P P 3 3 0 0 0 0 320 kgm 2 0769 kPa m kg K3219 K 214 2 kPa RT P ρ Thus 241 K 1667 1 2 321 9 K 1 2 T0 k T 1043 kPa 1 1 667 1 667 1 0 1667 1 2 214 2 kPa 1 2 k k k P P 0208 kgm3 1 1 1 667 3 1 1 0 1667 1 2 0 320 kgm 1 2 k k ρ ρ Discussion These are the temperature pressure and density values that will occur at the throat when the flow past the throat is supersonic PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1719 1741 Quiescent carbon dioxide at a given state is accelerated isentropically to a specified Mach number The temperature and pressure of the carbon dioxide after acceleration are to be determined Assumptions Carbon dioxide is an ideal gas with constant specific heats at room temperature Properties The specific heat ratio of the carbon dioxide at room temperature is k 1288 Analysis The inlet temperature and pressure in this case is equivalent to the stagnation temperature and pressure since the inlet velocity of the carbon dioxide is said to be negligible That is T0 Ti 400 K and P0 Pi 1200 kPa Then 0 2 2 2 2 600 K 57043 K 2 1Ma 21288106 T T k 570 K and 1 12881288 1 0 0 1200 kPa 57043 K 95723 K 600 K k k T P P T 957 kPa Discussion Note that both the pressure and temperature drop as the gas is accelerated as part of the internal energy of the gas is converted to kinetic energy 1742 Air enters a convergingdiverging nozzle at specified conditions The lowest pressure that can be obtained at the throat of the nozzle is to be determined Assumptions 1 Air is an ideal gas with constant specific heats at room temperature 2 Flow through the nozzle is steady onedimensional and isentropic Properties The specific heat ratio of air at room temperature is k 14 Analysis The lowest pressure that can be obtained at the throat is the critical pressure P which is determined from 423 kPa 1 41 41 1 0 14 1 2 800 kPa 1 2 k k k P P Discussion This is the pressure that occurs at the throat when the flow past the throat is supersonic PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1720 1743 Helium enters a convergingdiverging nozzle at specified conditions The lowest temperature and pressure that can be obtained at the throat of the nozzle are to be determined Assumptions 1 Helium is an ideal gas with constant specific heats 2 Flow through the nozzle is steady onedimensional and isentropic Properties The properties of helium are k 1667 and cp 51926 kJkgK Analysis The lowest temperature and pressure that can be obtained at the throat are the critical temperature T and critical pressure P First we determine the stagnation temperature T0 and stagnation pressure P0 801K m s 1000 1 kJkg C 5 1926 kJkg 2 100 ms 800 K 2 2 2 2 2 0 cp V T T Helium 0 702 MPa 800 K MPa 801 K 70 1 1 667 1 667 1 0 0 k k T T P P Thus 601 K 1667 1 2 801 K 1 2 T0 k T and 0342 MPa 1 1 667 1 667 1 0 1667 1 2 0 702 MPa 1 2 k k k P P Discussion These are the temperature and pressure that will occur at the throat when the flow past the throat is supersonic PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1721 1744 Air flows through a duct The state of the air and its Mach number are specified The velocity and the stagnation pressure temperature and density of the air are to be determined Assumptions Air is an ideal gas with constant specific heats at room temperature Properties The properties of air at room temperature are R 0287 kPam3kgK and k 14 Analysis The speed of sound in air at the specified conditions is 3872 ms 1 kJkg s 0 287 kJkg K3732 K 1000 m 41 2 2 kRT c Thus AIR 80 387 2 ms 310 ms Ma c V Also 3 3 1 867 kgm 0 287 kPa m kg K3732 K 200 kPa RT P ρ Then the stagnation properties are determined from K 421 2 1 4108 373 2 K 1 2 1 Ma 1 2 2 0 k T T 305 kPa 1 41 41 1 0 0 3732 K 200 kPa 4210 K k k T P T P 252 kgm3 1 1 41 3 1 1 0 0 3732 K 867 kgm 4210 K 1 k T ρ T ρ Discussion Note that both the pressure and temperature drop as the gas is accelerated as part of the internal energy of the gas is converted to kinetic energy PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1722 1745 Problem 1744 is reconsidered The effect of Mach number on the velocity and stagnation properties as the Ma is varied from 01 to 2 are to be investigated and the results are to be plotted Analysis The EES Equations window is printed below along with the tabulated and plotted results P200 0 04 08 12 16 2 0 200 400 600 800 1000 1200 1400 1600 Ma V T0 P0 and 100ρ0 T0 P0 ρ0 V T10027315 R0287 k14 cSQRTkRT1000 MaVc rhoPRT Stagnation properties T0T1k1Ma22 P0PT0Tkk1 rho0rhoT0T1k1 Mach num Ma Velocity V ms Stag Temp T0 K Stag Press P0 kPa Stag Density ρ0 kgm3 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 387 774 1162 1549 1936 2323 2710 3098 3485 3872 4259 4647 5034 5421 5808 6195 6583 6970 7357 7744 3739 3761 3799 3851 3918 4000 4097 4209 4336 4478 4635 4806 4993 5194 5411 5642 5888 6150 6426 6717 2014 2057 2129 2233 2372 2551 2774 3049 3383 3786 4270 4850 5541 6365 7342 8501 9872 11492 13401 15649 1877 1905 1953 2021 2110 2222 2359 2524 2718 2946 3210 3516 3867 4269 4728 5250 5842 6511 7267 8118 Discussion Note that as Mach number increases so does the flow velocity and stagnation temperature pressure and density PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1723 1746 An aircraft is designed to cruise at a given Mach number elevation and the atmospheric temperature The stagnation temperature on the leading edge of the wing is to be determined Assumptions Air is an ideal gas with constant specific heats at room temperature Properties The properties of air are R 0287 kPam3kgK cp 1005 kJkgK and k 14 Analysis The speed of sound in air at the specified conditions is 3080 ms 1 kJkg s 0 287 kJkg K23615 K 1000 m 41 2 2 kRT c Thus 308 0 ms 4312 ms 41 Ma c V Then 329 K 2 2 2 2 0 s m 1000 1kJkg 1 005 kJkg K 2 4312 ms 23615 2 c p V T T Discussion Note that the temperature of a gas increases during a stagnation process as the kinetic energy is converted to enthalpy 1747E Air flows through a duct at a specified state and Mach number The velocity and the stagnation pressure temperature and density of the air are to be determined Assumptions Air is an ideal gas with constant specific heats at room temperature Properties The properties of air are R 006855 BtulbmR 03704 psiaft3lbmR and k 14 Analysis First T 320 45967 77967 K The speed of sound in air at the specified conditions is 2 14006855 Btu1bm R77967 R 25037 ft s2 136872 fts 1 Btu1bm c kRT Thus Ma 07136872 fts 95810 V c 958 fts Also 3 3 25 psia 0086568 1bmft 03704 psia ft lbm R77967 R P ρ RT Then the stagnation properties are determined from 2 2 0 1Ma 14107 1 77967 R 1 85608 R 2 2 k T T 856 R 1 1414 1 0 0 25 psia 85608 R 34678 psia 77967 R k k T P P T 347 psia 1 1 114 1 3 3 0 0 008656 1bmft 85608 R 010936 lbmft 77967 R k T T ρ ρ 3 0109 lbmft Discussion Note that the temperature pressure and density of a gas increases during a stagnation process PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1724 Isentropic Flow Through Nozzles 1748C The fluid would accelerate even further instead of decelerating Discussion This is the opposite of what would happen in subsonic flow 1749C The fluid would accelerate even further as desired Discussion This is the opposite of what would happen in subsonic flow 1750C a The exit velocity reaches the sonic speed b the exit pressure equals the critical pressure and c the mass flow rate reaches the maximum value Discussion In such a case we say that the flow is choked 1751C a No effect on velocity b No effect on pressure c No effect on mass flow rate Discussion In this situation the flow is already choked initially so further lowering of the back pressure does not change anything upstream of the nozzle exit plane 1752C If the back pressure is low enough so that sonic conditions exist at the throats the mass flow rates in the two nozzles would be identical However if the flow is not sonic at the throat the mass flow rate through the nozzle with the diverging section would be greater because it acts like a subsonic diffuser Discussion Once the flow is choked at the throat whatever happens downstream is irrelevant to the flow upstream of the throat 1753C Maximum flow rate through a converging nozzle is achieved when Ma 1 at the exit of a nozzle For all other Ma values the mass flow rate decreases Therefore the mass flow rate would decrease if hypersonic velocities were achieved at the throat of a converging nozzle Discussion Note that this is not possible unless the flow upstream of the converging nozzle is already hypersonic 1754C Ma is the local velocity nondimensionalized with respect to the sonic speed at the throat whereas Ma is the local velocity nondimensionalized with respect to the local sonic speed Discussion The two are identical at the throat when the flow is choked 1755C a The velocity decreases b the pressure increases and c the mass flow rate remains the same Discussion Qualitatively this is the same as what we are used to in previous chapters for incompressible flow PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1725 1756C No if the flow in the throat is subsonic If the velocity at the throat is subsonic the diverging section would act like a diffuser and decelerate the flow Yes if the flow in the throat is already supersonic the diverging section would accelerate the flow to even higher Mach number Discussion In duct flow the latter situation is not possible unless a second convergingdiverging portion of the duct is located upstream and there is sufficient pressure difference to choke the flow in the upstream throat 1757 It is to be explained why the maximum flow rate per unit area for a given ideal gas depends only on P T 0 0 Also for an ideal gas a relation is to be obtained for the constant a in a mmax A 0 0 T P Properties The properties of the ideal gas considered are R 0287 kPam3kgK and k 14 Analysis The maximum flow rate is given by 1 2 1 0 0 max 1 2 k k k k RT P A m or 1 2 1 0 0 max 1 2 k k k R k T P A m For a given gas k and R are fixed and thus the mass flow rate depends on the parameter P T 0 0 Thus can be expressed as max A m 0 0 max T a P A m where 1 2 1 2408 2 2 2 14 2 1 14 1 0287 kJkgK 1000 m s 1 kJkg k k a k R k 00404 ms K Discussion Note that when sonic conditions exist at a throat of known crosssectional area the mass flow rate is fixed by the stagnation conditions 1758 For an ideal gas an expression is to be obtained for the ratio of the speed of sound where Ma 1 to the speed of sound based on the stagnation temperature cc0 Analysis For an ideal gas the speed of sound is expressed as kRT c Thus 1 2 0 0 0 c kRT T c T kRT 12 2 1 k Discussion Note that a speed of sound changes the flow as the temperature changes PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 1726 1759 Air enters a convergingdiverging nozzle at a specified pressure The back pressure that will result in a specified exit Mach number is to be determined Assumptions 1 Air is an ideal gas with constant specific heats 2 Flow through the nozzle is steady onedimensional and isentropic Properties The specific heat ratio of air is k 14 Analysis The stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible It remains constant throughout the nozzle since the flow is isentropic AIR i e P0 Pi 12 MPa Vi 0 Mae 18 From Table A32 at Mae 18 we read Pe P0 01740 Thus P 01740P0 0174012 MPa 0209 MPa Discussion If we solve this problem using the relations for compressible isentropic flow the results would be identical 1727 1760 Air enters a nozzle at specified temperature pressure and velocity The exit pressure exit temperature and exitto inlet area ratio are to be determined for a Mach number of Ma 1 at the exit Assumptions 1 Air is an ideal gas with constant specific heats at room temperature 2 Flow through the nozzle is steady onedimensional and isentropic Properties The properties of air are k 14 and cp 1005 kJkgK Analysis The properties of the fluid at the location where Ma 1 are the critical properties denoted by superscript We first determine the stagnation temperature and pressure which remain constant throughout the nozzle since the flow is isentropic 2 2 0 2 2 150 ms 1 kJkg 420 K 431194 K 2 2 1005 kJkg K 1000 m s i i p V T T c 150 ms i AIR Ma 1 and 1 1414 1 0 0 06 MPa 431194 K 065786 MPa 420 K k k i i T P P T From Table A32 or from Eqs 1718 and 1719 at Ma 1 we read TT0 08333 PP0 05283 Thus T 08333T0 08333431194 K 35931 K 359 K and P 05283P0 05283065786 MPa 034754 MPa 0348 MPa 348 kPa Also 2 140287 kJkg K420 K 1000 m s2 410799 ms 1 kJkg i ic kRT and 150 ms Ma 03651 410799 ms i i i V c From Table A32 at this Mach number we read Ai A 17452 Thus the ratio of the throat area to the nozzle inlet area is 1 057300 17452 i A A 0573 Discussion We can also solve this problem using the relations for compressible isentropic flow The results would be identical PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1728 1761 Air enters a nozzle at specified temperature and pressure with low velocity The exit pressure exit temperature and exittoinlet area ratio are to be determined for a Mach number of Ma 1 at the exit Assumptions 1 Air is an ideal gas 2 Flow through the nozzle is steady onedimensional and isentropic Properties The specific heat ratio of air is k 14 Vi 0 AIR Analysis The properties of the fluid at the location where Ma 1 are the critical properties denoted by superscript The stagnation temperature and pressure in this case are identical to the inlet temperature and pressure since the inlet velocity is negligible They remain constant throughout the nozzle since the flow is isentropic i Ma 1 T0 Ti 350 K and P0 Pi 02 MPa From Table A32 or from Eqs 1718 and 1719 at Ma 1 we read TT0 08333 PP0 05283 Thus T 08333T0 08333350 K 292 K and P 05283P0 0528302 MPa 0106 MPa The Mach number at the nozzle inlet is Ma 0 since Vi 0 From Table A32 at this Mach number we read AiA Thus the ratio of the throat area to the nozzle inlet area is 0 1 iA A Discussion If we solve this problem using the relations for compressible isentropic flow the results would be identical PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1729 1762E Air enters a nozzle at specified temperature pressure and velocity The exit pressure exit temperature and exit toinlet area ratio are to be determined for a Mach number of Ma 1 at the exit Assumptions 1 Air is an ideal gas with constant specific heats at room temperature 2 Flow through the nozzle is steady onedimensional and isentropic Properties The properties of air are k 14 and cp 0240 BtulbmR Table A2Ea Analysis The properties of the fluid at the location where Ma 1 are the critical properties denoted by superscript We first determine the stagnation temperature and pressure which remain constant throughout the nozzle since the flow is isentropic 646 9 R ft s 25037 1Btu1bm 0 240 Btulbm R 2 450 fts 630 R 2 2 2 2 2 0 p i c V T T i PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 450 fts Ma 1 AIR 32 9 psia 630 K 30 psia 6469 K 1 41 41 1 0 0 k k i Ti P T P From Table A32 or from Eqs 1718 and 1719 at Ma 1 we read TT0 08333 PP0 05283 Thus T 08333T0 083336469 R 539 R and P 05283P0 05283329 psia 174 psia Also 1230 fts 1Btu1bm s 0 06855 Btu1bm R630 R 25037 ft 41 2 2 i i kRT c and 0 3657 1230 fts 450 fts Ma i i i c V From Table A32 at this Mach number we read AiA 17426 Thus the ratio of the throat area to the nozzle inlet area is 0574 1 7426 1 iA A Discussion If we solve this problem using the relations for compressible isentropic flow the results would be identical preparation If you are a student using this Manual you are using it without permission 1730 1763 For subsonic flow at the inlet the variation of pressure velocity and Mach number along the length of the nozzle are to be sketched for an ideal gas under specified conditions Assumptions 1 The gas is an ideal gas 2 Flow through the nozzle is steady one dimensional and isentropic 3 The flow is choked at the throat Analysis Using EES and CO2 as the gas we calculate and plot flow area A velocity V and Mach number Ma as the pressure drops from a stagnation value of 1400 kPa to 200 kPa Note that the curve for A is related to the shape of the nozzle with horizontal axis serving as the centerline The EES equation window and the plot are shown below Mai 1 200 400 600 800 1000 1200 1400 0 1 2 P kPa A Ma V500 A m2 Ma V500 ms k1289 Cp0846 kJkgK R01889 kJkgK P01400 kPa T0473 K m3 kgs rho0P0RT0 rhoPRT rhonormrhorho0 Normalized density TT0PP0k1k TnormTT0 Normalized temperature VSQRT2CpT0T1000 VnormV500 AmrhoV500 CSQRTkRT1000 MaVC Discussion We are assuming that the back pressure is sufficiently low that the flow is choked at the throat and the flow downstream of the throat is supersonic without any shock waves Mach number and velocity continue to rise right through the throat into the diverging portion of the nozzle since the flow becomes supersonic PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1731 1764 We repeat the previous problem but for supersonic flow at the inlet The variation of pressure velocity and Mach number along the length of the nozzle are to be sketched for an ideal gas under specified conditions Analysis Using EES and CO2 as the gas we calculate and plot flow area A velocity V and Mach number Ma as the pressure rises from 200 kPa at a very high velocity to the stagnation value of 1400 kPa Note that the curve for A is related to the shape of the nozzle with horizontal axis serving as the centerline 200 400 600 800 1000 1200 1400 0 1 2 P kPa A Ma V500 A m2 Ma V500 ms k1289 Cp0846 kJkgK R01889 kJkgK P01400 kPa T0473 K m3 kgs rho0P0RT0 rhoPRT rhonormrhorho0 Normalized density TT0PP0k1k TnormTT0 Normalized temperature VSQRT2CpT0T1000 VnormV500 AmrhoV500 CSQRTkRT1000 MaVC Discussion Note that this problem is identical to the proceeding one except the flow direction is reversed In fact when plotted like this the plots are identical PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1732 1765 Nitrogen enters a convergingdiverging nozzle at a given pressure The critical velocity pressure temperature and density in the nozzle are to be determined Assumptions 1 Nitrogen is an ideal gas 2 Flow through the nozzle is steady onedimensional and isentropic Properties The properties of nitrogen are k 14 and R 02968 kJkgK Analysis The stagnation pressure in this case are identical to the inlet properties since the inlet velocity is negligible They remain constant throughout the nozzle P0 Pi 700 kPa T0 Ti 400 K i N2 Vi 0 3 3 0 0 0 5 896 kgm 0 2968 kPa m kg K400 K 700 kPa RT P ρ Critical properties are those at a location where the Mach number is Ma 1 From Table A32 at Ma 1 we read TT0 08333 PP0 05283 and ρρ0 06339 Then the critical properties become T 08333T0 08333400 K 333 K P 05283P0 05283700 kPa 370 MPa ρ 06339ρ0 063395896 kgm3 374 kgm3 Also 372 ms 1kJkg 0 2968 kJkg K333 K 1000 m s 41 2 2 kRT c V Discussion We can also solve this problem using the relations for compressible isentropic flow The results would be identical 1766 An ideal gas is flowing through a nozzle The flow area at a location where Ma 24 is specified The flow area where Ma 12 is to be determined Assumptions Flow through the nozzle is steady onedimensional and isentropic Properties The specific heat ratio is given to be k 14 Analysis The flow is assumed to be isentropic and thus the stagnation and critical properties remain constant throughout the nozzle The flow area at a location where Ma2 12 is determined using A A data from Table A32 to be 2 2 1 1 1 1498 cm 2 4031 36 cm 2 4031 2 4031 42 Ma A A A A 154 cm2 1 03041498 cm 1 0304 1 0304 21 Ma 2 2 2 2 A A A A Discussion We can also solve this problem using the relations for compressible isentropic flow The results would be identical PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1733 1767 An ideal gas is flowing through a nozzle The flow area at a location where Ma 24 is specified The flow area where Ma 12 is to be determined Assumptions Flow through the nozzle is steady onedimensional and isentropic Analysis The flow is assumed to be isentropic and thus the stagnation and critical properties remain constant throughout the nozzle The flow area at a location where Ma2 12 is determined using the A A relation 1 2 1 Ma2 2 1 1 1 2 Ma 1 k k k k A A For k 133 and Ma1 24 2 570 24 2 1 33 1 1 1 33 1 2 24 1 0 33 2 33 2 2 1 A A and 2 2 1 1401 cm 2 570 36 cm 2 570 A A For k 133 and Ma2 12 1 0316 12 2 1 33 1 1 1 33 1 2 12 1 0 33 2 33 2 2 2 A A and 1445 cm2 1 03161401 cm 0316 1 2 2 A A Discussion Note that the compressible flow functions in Table A32 are prepared for k 14 and thus they cannot be used to solve this problem PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1734 1768E Air enters a convergingdiverging nozzle at a specified temperature and pressure with low velocity The pressure temperature velocity and mass flow rate are to be calculated in the specified test section Assumptions 1 Air is an ideal gas 2 Flow through the nozzle is steady onedimensional and isentropic Properties The properties of air are k 14 and R 006855 BtulbmR 03704 psiaft3lbmR Analysis The stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible They remain constant throughout the nozzle since the flow is isentropic P0 Pi 150 psia and T0 Ti 100F 560 R i e AIR Vi 0 Then R 311 2 2 0 1412 2 2 560 R 1 Ma 2 2 k T Te psia 191 40 41 1 0 0 560 150 psia 311 k k e T T P P 3 3 0 166 1bmft 0 3704 psiaft 1bm R311 R 19 1 psia e e e RT P ρ The nozzle exit velocity can be determined from Ve Maece where ce is the speed of sound at the exit conditions Ve Maece 2 2 25037 ft s Ma 2 1 4 0 06855 Btu1bm R 311 R 1729 fts 1 Btu1bm e e kRT 1730 fts Finally e e e AV m ρ 0 166 1bmft35 ft2 1729 fts 1435 lbms 1440 lbms Discussion Air must be very dry in this application because the exit temperature of air is extremely low and any moisture in the air will turn to ice particles PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1735 1769 Air enters a converging nozzle at a specified temperature and pressure with low velocity The exit pressure the exit velocity and the mass flow rate versus the back pressure are to be calculated and plotted Assumptions 1 Air is an ideal gas with constant specific heats at room temperature 2 Flow through the nozzle is steady onedimensional and isentropic Properties The properties of air are k 14 R 0287 kJkgK and cp 1005 kJkgK Analysis The stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible They remain constant throughout the nozzle since the flow is isentropic P0 Pi 900 kPa T0 Ti 400 K The critical pressure is determined to be i e AIR Vi 0 475 5 kPa 14 1 2 900 kPa 1 2 40 41 1 0 k k k P P Then the pressure at the exit plane throat will be Pe Pb for Pb 4755 kPa Pe P 4755 kPa for Pb 4755 kPa choked flow Thus the back pressure will not affect the flow when 100 Pb 4755 kPa For a specified exit pressure Pe the temperature the velocity and the mass flow rate can be determined from Temperature 41 40 e 1 0 0 900 P 400 K k k e e P P T T Pe Velocity 1 kJkg 2 1 005 kJkg K400 T 1000 m s 2 2 2 e 0 e p T T c V Pb Density e e e e e kg K T P RT P 0 287 kPa m 3 ρ Ve Mass flow rate m 0 001 2 e e e e e V V A m ρ ρ c The results of the calculations are tabulated as Pb Pb kPa Pe kPa Te K Ve ms ρe kgm3 m kg s 900 900 400 0 7840 0 800 800 3868 1629 7206 1174 700 700 3723 2360 6551 1546 600 600 3562 2967 5869 1741 500 500 3382 3524 5151 1815 4755 4755 3333 3662 4971 1820 400 4755 3333 3662 4971 1820 300 4755 3333 3662 4971 1820 200 4755 3333 3662 4971 1820 100 4755 3333 3662 4971 1820 Pb kPa max m m 900 100 4755 Discussion We see from the plots that once the flow is choked at a back pressure of 4755 kPa the mass flow rate remains constant regardless of how low the back pressure gets PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1736 1770 We are to reconsider the previous problem Using EES or other software we are to solve the problem for the inlet conditions of 08 MPa and 1200 K Analysis Air at 800 kPa 1200 K enters a converging nozzle with a negligible velocity The throat area of the nozzle is 10 cm2 Assuming isentropic flow calculate and plot the exit pressure the exit velocity and the mass flow rate versus the back pressure Pb for 08 Pb 01 MPa Procedure ExitPressPbackPcrit Pexit Condition If PbackPcrit then PexitPback Unchoked Flow Condition Conditionunchoked else PexitPcrit Choked Flow Condition Conditionchoked Endif End GasAir Acm210 Throat area cm2 Pinlet 800kPa Tinlet 1200K Pback 4227 kPa Aexit Acm2Convertcm2m2 CpspecheatGasTTinlet CpCvR kCpCv MMOLARMASSGas Molar mass of Gas R 8314M Gas constant for Gas Since the inlet velocity is negligible the stagnation temperature Tinlet and since the nozzle is isentropic the stagnation pressure Pinlet PoPinlet Stagnation pressure ToTinlet Stagnation temperature Pcrit Po2k1kk1 Critical pressure from Eq 1622 Call ExitPressPbackPcrit Pexit Condition Texit ToPexitPok1k Exit temperature for isentopic flow K Vexit 22CpToTexit1000 Exit velocity ms RhoexitPexitRTexit Exit density kgm3 mdotRhoexitVexitAexit Nozzle mass flow rate kgs If you wish to redo the plots hide the diagram window and remove the from the first 4 variables just under the procedure Next set the desired range of back pressure in the parametric table Finally solve the table F3 The table of results and the corresponding plot are provided below PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1737 EES SOLUTION Acm210 Aexit0001 Conditionchoked Cp1208 Cv09211 GasAir k1312 M2897 mdot09124 Pback4227 Pcrit4349 Pexit4349 Pinlet800 Po800 R0287 Rhoexit1459 Texit1038 Tinlet1200 To1200 Vexit6252 PbackkPa Pexit kPa Vexit ms m kgs Texit K ρexit kgm3 100 4349 6252 09124 1038 1459 200 4349 6252 09124 1038 1459 300 4349 6252 09124 1038 1459 400 4349 6252 09124 1038 1459 4227 4349 6252 09124 1038 1459 500 500 5535 08984 1073 1623 600 600 4377 08164 1121 1865 700 700 3009 06313 1163 2098 800 800 0001523 0000003538 1200 2323 100 200 300 400 500 600 700 800 0 02 04 06 08 1 Pback kPa m kgs PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1738 100 200 300 400 500 600 700 800 400 450 500 550 600 650 700 750 800 Pback kPa Pexit kPa 100 200 300 400 500 600 700 800 0 100 200 300 400 500 600 700 Vexit ms Pback kPa Discussion We see from the plot that once the flow is choked at a back pressure of 4227 kPa the mass flow rate remains constant regardless of how low the back pressure gets PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1739 Shock Waves and Expansion Waves 1771C No because the flow must be supersonic before a shock wave can occur The flow in the converging section of a nozzle is always subsonic Discussion A normal shock if it is to occur would occur in the supersonic diverging section of the nozzle 1772C The Fanno line represents the states that satisfy the conservation of mass and energy equations The Rayleigh line represents the states that satisfy the conservation of mass and momentum equations The intersections points of these lines represent the states that satisfy the conservation of mass energy and momentum equations Discussion Ts diagrams are quite helpful in understanding these kinds of flows 1773C No the second law of thermodynamics requires the flow after the shock to be subsonic Discussion A normal shock wave always goes from supersonic to subsonic in the flow direction 1774C a velocity decreases b static temperature increases c stagnation temperature remains the same d static pressure increases and e stagnation pressure decreases Discussion In addition the Mach number goes from supersonic Ma 1 to subsonic Ma 1 1775C Oblique shocks occur when a gas flowing at supersonic speeds strikes a flat or inclined surface Normal shock waves are perpendicular to flow whereas inclined shock waves as the name implies are typically inclined relative to the flow direction Also normal shocks form a straight line whereas oblique shocks can be straight or curved depending on the surface geometry Discussion In addition while a normal shock must go from supersonic Ma 1 to subsonic Ma 1 the Mach number downstream of an oblique shock can be either supersonic or subsonic 1776C Yes the upstream flow has to be supersonic for an oblique shock to occur No the flow downstream of an oblique shock can be subsonic sonic and even supersonic Discussion The latter is not true for normal shocks For a normal shock the flow must always go from supersonic Ma 1 to subsonic Ma 1 1777C Yes the claim is correct Conversely normal shocks can be thought of as special oblique shocks in which the shock angle is β π2 or 90o Discussion The component of flow in the direction normal to the oblique shock acts exactly like a normal shock We can think of the flow parallel to the oblique shock as going along for the ride it does not affect anything PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1740 1778C When the wedge halfangle δ is greater than the maximum deflection angle θmax the shock becomes curved and detaches from the nose of the wedge forming what is called a detached oblique shock or a bow wave The numerical value of the shock angle at the nose is β 90o Discussion When δ is less than θmax the oblique shock is attached to the nose 1779C When supersonic flow impinges on a blunt body like the rounded nose of an aircraft the wedge halfangle δ at the nose is 90o and an attached oblique shock cannot exist regardless of Mach number Therefore a detached oblique shock must occur in front of all such bluntnosed bodies whether twodimensional axisymmetric or fully threedimensional Discussion Since δ 90o at the nose δ is always greater than θmax regardless of Ma or the shape of the rest of the body 1780C The isentropic relations of ideal gases are not applicable for flows across a normal shock waves and b oblique shock waves but they are applicable for flows across c PrandtlMeyer expansion waves Discussion Flow across any kind of shock wave involves irreversible losses hence it cannot be isentropic PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1741 1781 Air flowing through a nozzle experiences a normal shock Various properties are to be calculated before and after the shock Assumptions 1 Air is an ideal gas with constant specific heats 2 Flow through the nozzle is steady onedimensional and isentropic before the shock occurs Properties The properties of air at room temperature are k 14 R 0287 kJkgK and cp 1005 kJkgK Analysis The stagnation temperature and pressure before the shock are shock wave 477 4 K m s 1000 1kJkg 2 1 005 kJkg K 740 ms 205 2 2 2 2 2 1 1 01 cp V T T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course AIR 1 2 347 0 kPa 205 K 18 kPa 4774 K 1 41 41 1 1 01 1 01 k k T P T P The velocity and the Mach number before the shock are determined from 2870 ms 1kJkg 0 287 kJkg K205 K 1000 m s 41 2 2 1 1 kRT c and 2578 2870 ms 740 ms Ma 1 1 1 c V The fluid properties after the shock denoted by subscript 2 are related to those before the shock through the functions listed in Table A33 For Ma1 2578 we read We obtained the following values using analytical relations in Table A33 2 2158 7 5871 and 9 0349 Ma 1 2 1 2 1 02 2 T T P P P P 05058 Then the stagnation pressure P02 static pressure P2 and static temperature T2 are determined to be P02 90349P1 9034918 kPa 1626 kPa P2 75871P1 7587118 kPa 1366 kPa T2 22158T1 22158205 K 4542 K The air velocity after the shock can be determined from V2 Ma2c2 where c2 is the speed of sound at the exit conditions after the shock ms 2161 1kJkg 0 287 kJkgK4542 K 1000 m s 41 0 5058 Ma Ma 2 2 2 2 2 2 2 kRT c V Discussion This problem could also be solved using the relations for compressible flow and normal shock functions The results would be identical preparation If you are a student using this Manual you are using it without permission 1742 1782 Air flowing through a nozzle experiences a normal shock The entropy change of air across the normal shock wave is to be determined Assumptions 1 Air is an ideal gas with constant specific heats 2 Flow through the nozzle is steady onedimensional and isentropic before the shock occurs Properties The properties of air at room temperature are R 0287 kJkgK and cp 1005 kJkgK Analysis The entropy change across the shock is determined to be 02180 kJkg K 0287 kJkg Kln75871 005 kJkg Kln22158 1 ln ln 1 2 1 2 1 2 P P R T T c s s p Discussion A shock wave is a highly dissipative process and the entropy generation is large during shock waves PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1743 1783 Air flowing through a convergingdiverging nozzle experiences a normal shock at the exit The effect of the shock wave on various properties is to be determined Assumptions 1 Air is an ideal gas with constant specific heats 2 Flow through the nozzle is steady onedimensional and isentropic before the shock occurs 3 The shock wave occurs at the exit plane Properties The properties of air are k 14 and R 0287 kJkgK Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible Then P01 Pi 1 MPa 1 AIR Shock wave 2 T01 Ti 300 K Then i Vi 0 139 4 K 14 124 2 2 300 K 1 Ma 2 2 2 2 1 01 1 k T T and 0 06840 MPa 300 MPa 1394 1 40 41 1 0 1 01 1 k k T T P P The fluid properties after the shock denoted by subscript 2 are related to those before the shock through the functions listed in Table A33 For Ma1 24 we read 02 2 2 2 01 1 1 Ma 05231 05401 65533 and 20403 P P T P P T 0523 Then the stagnation pressure P02 static pressure P2 and static temperature T2 are determined to be P02 05401P01 0540110 MPa 0540 MPa 540 kPa P2 65533P1 65533006840 MPa 0448 MPa 448 kPa T2 20403T1 204031394 K 284 K The air velocity after the shock can be determined from V2 Ma2c2 where c2 is the speed of sound at the exit conditions after the shock V2 Ma2c2 177 ms 1kJkg 0 287 kJkg K284 K 1000 m s 41 0 5231 Ma 2 2 2 2 kRT Discussion We can also solve this problem using the relations for normal shock functions The results would be identical PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 1744 1784 Air enters a convergingdiverging nozzle at a specified state The required back pressure that produces a normal shock at the exit plane is to be determined for the specified nozzle geometry Assumptions 1 Air is an ideal gas 2 Flow through the nozzle is steady onedimensional and isentropic before the shock occurs 3 The shock wave occurs at the exit plane Analysis The inlet stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible Since the flow before the shock to be isentropic shock wave P01 Pi 2 MPa Vi 0 AIR 1 i 2 It is specified that AA 35 From Table A32 Mach number and the pressure ratio which corresponds to this area ratio are the Ma1 280 and P1P01 00368 The pressure ratio across the shock for this Ma1 value is from Table A33 P2P1 898 Thus the back pressure which is equal to the static pressure at the nozzle exit must be Pb P2 898P1 89800368P01 898003682 MPa 0661 MPa Discussion We can also solve this problem using the relations for compressible flow and normal shock functions The results would be identical 1785 Air enters a convergingdiverging nozzle at a specified state The required back pressure that produces a normal shock at the exit plane is to be determined for the specified nozzle geometry Assumptions 1 Air is an ideal gas 2 Flow through the nozzle is steady onedimensional and isentropic before the shock occurs Analysis The inlet stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible Since the flow before the shock to be isentropic P0x Pi 2 MPa shock wave It is specified that AA 2 From Table A32 the Mach number and the pressure ratio which corresponds to this area ratio are the Ma1 220 and P1P01 00935 The pressure ratio across the shock for this M1 value is from Table A33 P2P1 548 Thus the back pressure which is equal to the static pressure at the nozzle exit must be Vi 0 AIR 1 i 2 Pb P2 548P1 54800935P01 548009352 MPa 102 MPa Discussion We can also solve this problem using the relations for compressible flow and normal shock functions The results would be identical 1745 1786 Air flowing through a nozzle experiences a normal shock The effect of the shock wave on various properties is to be determined Analysis is to be repeated for helium under the same conditions Assumptions 1 Air and helium are ideal gases with constant specific heats 2 Flow through the nozzle is steady one dimensional and isentropic before the shock occurs Properties The properties of air are k 14 and R 0287 kJkgK and the properties of helium are k 1667 and R 20769 kJkgK Analysis The air properties upstream the shock are shock wave PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course AIR 1 i 2 Ma1 32 P1 58 kPa and T1 270 K Fluid properties after the shock denoted by subscript 2 are related to those before the shock through the functions in Table A33 For Ma1 32 Ma1 32 2 9220 11780 and 13656 Ma 1 2 1 2 1 02 2 T T P P P 04643 P We obtained these values using analytical relations in Table A33 Then the stagnation pressure P02 static pressure P2 and static temperature T2 are determined to be P02 13656P1 1365658 kPa 7920 kPa P2 11780P1 1178058 kPa 6832 kPa T2 29220T1 29220270 K 7889 K The air velocity after the shock can be determined from V2 Ma2c2 where c2 is the speed of sound at the exit conditions after the shock 2614 ms 1kJkg 0 287 kJkg K7889 K 1000 m s 41 0 4643 Ma Ma 2 2 2 2 2 2 2 kRT c V We now repeat the analysis for helium This time we cannot use the tabulated values in Table A33 since k is not 14 Therefore we have to calculate the desired quantities using the analytical relations 05136 2 1 2 2 2 1 2 1 2 1 2 1 1 1 667 1 667 23 2 1 2 1 667 23 1 1 Ma 2 1 2 Ma Ma k k k 12551 0 5136 1 667 1 23 1 667 1 Ma 1 Ma 1 2 2 2 2 2 1 1 2 k k P P 4 0580 2 1 1 667 0 5136 1 2 1 1 667 23 1 2 1 Ma 1 2 1 Ma 1 2 2 2 2 2 1 1 2 k k T T 1 2 2 2 2 2 1 1 02 1 Ma 2 1 Ma 1 Ma 1 k k k k k P P 15495 0 5136 2 1 1 667 1 0 5136 1 667 1 23 1 667 1 1 667 0 667 2 2 2 Thus P02 15495P1 1549558 kPa 8987 kPa P2 12551P1 1255158 kPa 7280 kPa T2 40580T1 40580270 K 1096 K 1000 ms 1kJkg 1 667 2 0769 kJkg K1096 K 1000 m s 0 5136 Ma Ma 2 2 2 2 2 2 kRTy c V Discussion The velocity and Mach number are higher for helium than for air due to the different values of k and R preparation If you are a student using this Manual you are using it without permission 1746 1787 Air flowing through a nozzle experiences a normal shock The entropy change of air across the normal shock wave is to be determined Assumptions 1 Air and helium are ideal gases with constant specific heats 2 Flow through the nozzle is steady one dimensional and isentropic before the shock occurs Properties The properties of air are R 0287 kJkgK and cp 1005 kJkgK and the properties of helium are R 20769 kJkgK and cp 51926 kJkgK Analysis The entropy change across the shock is determined to be 0370 kJkg K 1 005 kJkg Kln29220 0287 kJkg Kln11780 ln ln 1 2 1 2 1 2 P P R T T c s s p For helium the entropy change across the shock is determined to be 202 kJkg K 5 1926 kJkg Kln40580 20769 kJkg Kln12551 ln ln 1 2 1 2 1 2 P P R T T c s s p Discussion Note that shock wave is a highly dissipative process and the entropy generation is large during shock waves PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1747 1788E Air flowing through a nozzle experiences a normal shock Effect of the shock wave on various properties is to be determined Analysis is to be repeated for helium Assumptions 1 Air and helium are ideal gases with constant specific heats 2 Flow through the nozzle is steady one dimensional and isentropic before the shock occurs Properties The properties of air are k 14 and R 006855 BtulbmR and the properties of helium are k 1667 and R 04961 BtulbmR Analysis The air properties upstream the shock are shock wave Ma1 25 P1 10 psia and T1 4405 R PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course AIR 1 i 2 Fluid properties after the shock denoted by subscript 2 are related to those before the shock through the functions listed in Table A33 For Ma1 25 Ma1 25 2 1375 7 125 and 8 5262 Ma 1 2 1 2 1 02 2 T T P P P 0513 P Then the stagnation pressure P02 static pressure P2 and static temperature T2 are determined to be P02 85262P1 8526210 psia 853 psia P2 7125P1 712510 psia 713 psia T2 21375T1 213754405 R 942 R The air velocity after the shock can be determined from V2 Ma2c2 where c2 is the speed of sound at the exit conditions after the shock 772 fts 1Btu1bm s 0 06855 Btu1bm R9416 R 25037 ft 41 0 513 Ma Ma 2 2 2 2 2 2 2 kRT c V We now repeat the analysis for helium This time we cannot use the tabulated values in Table A33 since k is not 14 Therefore we have to calculate the desired quantities using the analytical relations 0553 2 1 2 2 2 1 2 1 2 1 2 1 1 1 667 1 667 52 2 1 2 1 667 52 1 1 Ma 2 1 2 Ma Ma k k k 7 5632 0 553 1 667 1 52 1 667 1 Ma 1 Ma 1 2 2 2 2 2 1 1 2 k k P P 2 7989 2 1 1 667 0 553 1 2 1 1 667 52 1 2 1 Ma 1 2 1 Ma 1 2 2 2 2 2 1 1 2 k k T T 1 2 2 2 2 2 1 1 02 1 Ma 2 1 Ma 1 Ma 1 k k k k k P P 9 641 0 553 2 1 1 667 1 0 553 1 667 1 52 1 667 1 1 667 0 667 2 2 2 Thus P02 11546P1 1154610 psia 115 psia P2 75632P1 7563210 psia 756 psia T2 27989T1 279894405 R 1233 R 2794 fts 1Btu1bm s 1 667 0 4961 Btu1bmR12329 R 25037 ft 0 553 Ma Ma 2 2 2 2 2 2 2 kRT c V Discussion This problem could also be solved using the relations for compressible flow and normal shock functions The results would be identical preparation If you are a student using this Manual you are using it without permission 1748 1789E We are to reconsider Prob 1788E Using EES or other software we are to study the effects of both air and helium flowing steadily in a nozzle when there is a normal shock at a Mach number in the range 2 Mx 35 In addition to the required information we are to calculate the entropy change of the air and helium across the normal shock and tabulate the results in a parametric table Analysis We use EES to calculate the entropy change of the air and helium across the normal shock The results are given in the Parametric Table for 2 Mx 35 Procedure NormalShockMxkMyPyOPx TyOTxRhoyORhox PoyOPox PoyOPx If Mx 1 Then My 1000PyOPx1000TyOTx1000RhoyORhox1000 PoyOPox1000PoyOPx1000 else Mysqrt Mx22k1 2Mx2kk11 PyOPx1kMx21kMy2 TyOTx 1Mx2k12 1My2k12 RhoyORhoxPyOPxTyOTx PoyOPoxMxMy 1My2k12 1Mx2k12 k12k1 PoyOPx1kMx21My2k12kk11kMy2 Endif End Function ExitPressPbackPcrit If PbackPcrit then ExitPressPback Unchoked Flow Condition If PbackPcrit then ExitPressPcrit Choked Flow Condition End Procedure GetPropGasCpkR Cp and k data are from Text Table A2E MMOLARMASSGas Molar mass of Gas R 1545M Particular gas constant for Gas ftlbflbmR k Ratio of Cp to Cv Cp Specific heat at constant pressure if GasAir then Cp024BtulbmR k14 endif if GasCO2 then Cp0203BtulbmR k1289 endif if GasHelium then Cp125BtulbmR k1667 endif End Variable Definitions M flow Mach Number Pratio PPo for compressible isentropic flow Tratio TTo for compressible isentropic flow Rhoratio RhoRhoo for compressible isentropic flow AratioAA for compressible isentropic flow Fluid properties before the shock are denoted with a subscript x Fluid properties after the shock are denoted with a subscript y My Mach Number down stream of normal shock PyOverPx PyPx Pressue ratio across normal shock TyOverTx TyTx Temperature ratio across normal shock RhoyOverRhoxRhoyRhox Density ratio across normal shock PoyOverPox PoyPox Stagantion pressure ratio across normal shock PoyOverPx PoyPx Stagnation pressure after normal shock ratioed to pressure before shock PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1749 Input Data Px 10 psia Values of Px Tx and Mx are set in the Parametric Table Tx 4405 R Mx 25 GasAir This program has been written for the gases Air CO2 and Helium Call GetPropGasCpkR Call NormalShockMxkMyPyOverPx TyOverTxRhoyOverRhox PoyOverPox PoyOverPx PoyairPxPoyOverPx Stagnation pressure after the shock PyairPxPyOverPx Pressure after the shock TyairTxTyOverTx Temperature after the shock MyairMy Mach number after the shock The velocity after the shock can be found from the product of the Mach number and speed of sound after the shock Cyair sqrtkRftlbflbmRTyairR322 lbmftlbfs2 VyairMyairCyair DELTAsairentropyairTTyair PPyair entropyairTTxPPx Gas2Helium Gas2 can be either Helium or CO2 Call GetPropGas2Cp2k2R2 Call NormalShockMxk2My2PyOverPx2 TyOverTx2RhoyOverRhox2 PoyOverPox2 PoyOverPx2 PoyhePxPoyOverPx2 Stagnation pressure after the shock PyhePxPyOverPx2 Pressure after the shock TyheTxTyOverTx2 Temperature after the shock MyheMy2 Mach number after the shock The velocity after the shock can be found from the product of the Mach number and speed of sound after the shock Cyhe sqrtk2R2ftlbflbmRTyheR322 lbmftlbfs2 VyheMyheCyhe DELTAsheentropyheliumTTyhe PPyhe entropyheliumTTxPPx The parametric table and the corresponding plots are shown below Vyhe fts Vyair fts Tyhe R Tyair R Tx R Pyhe psia Pyair psia Px psia Poyhe psia Poyair psia Myhe Myair Mx she Btulbm R sair Btulbm R 2644 7719 9156 7433 4405 475 45 10 6346 564 0607 05774 2 01345 00228 2707 7671 1066 8376 4405 6079 574 10 7901 7002 05759 05406 225 02011 00351 2795 7719 1233 9416 4405 7563 7125 10 9641 8526 0553 0513 25 02728 004899 3022 8004 1616 1180 4405 110 1033 10 1367 1206 05223 04752 3 04223 008 3292 8454 2066 1460 4405 1506 1413 10 1845 1624 05032 04512 35 05711 01136 20 22 24 26 28 30 32 34 36 044 046 048 050 052 054 056 058 060 062 Mx My Helium Air Mach Number After Shock vs Mx PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1750 20 22 24 26 28 30 32 34 36 002 012 021 031 041 050 060 Mx s BtulbmR Helium Air Entropy Change Across Shock vs Mx 20 22 24 26 28 30 32 34 36 700 980 1260 1540 1820 2100 Mx Ty R Helium Air Temperature After Shock vs Mx 20 22 24 26 28 30 32 34 36 2600 2700 2800 2900 3000 3100 3200 3300 600 650 700 750 800 850 900 Mx Vyhe fts Vyair fts Velocity After shock vs Mx 20 22 24 26 28 30 32 34 36 40 60 80 100 120 140 160 Mx Py psia Helium Air Pressure After Shock vs Mx Discussion In all cases regardless of the fluid or the Mach number entropy increases across a shock wave This is because a shock wave involves irreversibilities PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1751 1790 For an ideal gas flowing through a normal shock a relation for V2V1 in terms of k Ma1 and Ma2 is to be developed Analysis The conservation of mass relation across the shock is 2 2 1 1 V V ρ ρ and it can be expressed as 1 2 2 1 2 2 1 1 2 1 1 2 T T P P RT P RT P V V ρ ρ From Eqs 1735 and 1738 2 1 Ma 1 2 1 Ma 1 Ma 1 Ma 1 2 2 2 1 2 1 2 2 1 2 k k k k V V Discussion This is an important relation as it enables us to determine the velocity ratio across a normal shock when the Mach numbers before and after the shock are known 1791 The entropy change of air across the shock for upstream Mach numbers between 05 and 15 is to be determined and plotted Assumptions 1 Air is an ideal gas 2 Flow through the nozzle is steady onedimensional and isentropic before the shock occurs Properties The properties of air are k 14 R 0287 kJkgK and cp 1005 kJkgK Analysis The entropy change across the shock is determined to be PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1 2 1 2 1 2 ln ln P P R T T c s s p where 2 1 2 1 2 1 2 1 1 Ma 2 1 2 Ma Ma k k k 2 2 2 1 1 2 Ma 1 Ma 1 k k P P and 2 1 Ma 1 2 1 Ma 1 2 2 2 1 1 2 k k T T The results of the calculations can be tabulated as Ma1 Ma2 T2T1 P2P1 s2 s1 05 26458 01250 04375 1853 s2 sx 0 1 Ma 06 18778 02533 06287 1247 07 15031 04050 07563 0828 08 12731 05800 08519 0501 09 11154 07783 09305 0231 10 10000 10000 10000 00 11 09118 10649 12450 00003 12 08422 11280 15133 00021 13 07860 11909 18050 00061 14 07397 12547 21200 00124 15 07011 13202 24583 00210 Discussion The total entropy change is negative for upstream Mach numbers Ma1 less than unity Therefore normal shocks cannot occur when Ma1 1 preparation If you are a student using this Manual you are using it without permission 1752 1792 Supersonic airflow approaches the nose of a twodimensional wedge and undergoes a straight oblique shock For a specified Mach number the minimum shock angle and the maximum deflection angle are to be determined Assumptions Air is an ideal gas with a constant specific heat ratio of k 14 so that Fig 1741 is applicable Oblique shock θ β Ma1 Ma2 δ Analysis For Ma 5 we read from Fig 1741 Minimum shock or wave angle βmin 12 Maximum deflection or turning angle θmax 415 Discussion Note that the minimum shock angle decreases and the maximum deflection angle increases with increasing Mach number Ma1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1753 1793E Air flowing at a specified supersonic Mach number is forced to undergo a compression turn an oblique shock The Mach number pressure and temperature downstream of the oblique shock are to be determined Assumptions 1 The flow is steady 2 The boundary layer on the wedge is very thin 3 Air is an ideal gas with constant specific heats βweak δ 15o Weak shock Ma1 Properties The specific heat ratio of air is k 14 Analysis On the basis of Assumption 2 we take the deflection angle as equal to the wedge halfangle ie θ δ 15o Then the two values of oblique shock angle β are determined from 2 cos 2 Ma tan 1 2Ma sin tan 2 1 2 2 1 β β β θ k 2 cos 2 41 2 tan 1 22 sin tan15 2 2 2 β β β which is implicit in β Therefore we solve it by an iterative approach or with an equation solver such as EES It gives βweak 4534o and βstrong 7983o Then the upstream normal Mach number Ma1n becomes Weak shock 1 423 2sin 4534 Ma sin Ma 1 1n β PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Strong shock 1 969 2sin 7983 Ma sin Ma 1 1n β Also the downstream normal Mach numbers Ma2n become Weak shock 0 7304 1 41 1 423 41 2 2 1 1 423 41 1 Ma 2 2 1 Ma Ma 2 2 2 n 1 2 1n 2n k k k Ma1 δ 15o Strong shock βstrong Strong shock 0 5828 1 41 1 969 41 2 2 1 1 969 41 1 Ma 2 2 1 Ma Ma 2 2 2 n 1 2 1n 2n k k k The downstream pressure and temperature for each case are determined to be Weak shock 132 psia 1 41 1 41 1 423 41 6 psia 2 1 1 2 Ma 2 2 1n 1 2 k k k P P 609 R 2 2 2 n 1 2 n 1 1 2 1 2 1 1 2 1 2 1 1 423 41 1 1 423 41 2 6 psia 480 R 13 2 psia 1 Ma 1 Ma 2 k k P T P P T P T ρ ρ Strong shock 261 psia 1 41 1 41 1 969 41 6 psia 2 1 1 2 Ma 2 2 1n 1 2 k k k P P 798 R 2 2 2 n 1 2 n 1 1 2 1 2 1 1 2 1 2 1 1 969 41 1 1 969 41 2 6 psia 480 R 26 1 psia 1 Ma 1 Ma 2 k k P T P P T P T ρ ρ The downstream Mach number is determined to be Weak shock 145 15 sin4534 0 7304 sin Ma Ma 2n 2 θ β Strong shock 0644 15 sin7983 0 5828 sin Ma Ma 2n 2 θ β Discussion Note that the change in Mach number pressure temperature across the strong shock are much greater than the changes across the weak shock as expected For both the weak and strong oblique shock cases Ma1n is supersonic and Ma2n is subsonic However Ma2 is supersonic across the weak oblique shock but subsonic across the strong oblique shock preparation If you are a student using this Manual you are using it without permission 1754 1794 Air flowing at a specified supersonic Mach number undergoes an expansion turn over a tilted wedge The Mach number pressure and temperature downstream of the sudden expansion above the wedge are to be determined Assumptions 1 The flow is steady 2 The boundary layer on the wedge is very thin 3 Air is an ideal gas with constant specific heats Ma1 24 10 25 Ma2 δ θ Properties The specific heat ratio of air is k 14 Analysis On the basis of Assumption 2 the deflection angle is determined to be θ δ 25 10 15o Then the upstream and downstream PrandtlMeyer functions are determined to be 1 Ma tan 1 1 Ma 1 1 tan 1 Ma 2 1 2 1 k k k k ν Upstream 3675 1 24 tan 1 1 42 41 1 41 1 tan 41 1 41 Ma 2 1 2 1 1 ν Then the downstream PrandtlMeyer function becomes 5175 3675 15 Ma Ma 1 2 ν θ ν Now Ma2 is found from the PrandtlMeyer relation which is now implicit Downstream 5175 1 Ma tan 1 1 Ma 41 1 41 1 tan 41 1 41 Ma 2 2 1 2 2 1 2 ν It gives Ma2 3105 Then the downstream pressure and temperature are determined from the isentropic flow relations 238 kPa 70 kPa 2 1 41 42 1 2 1 41 3 105 1 2 1 Ma 1 2 1 Ma 1 40 41 2 40 41 2 1 1 2 1 1 2 2 1 0 1 0 2 2 P k k P P P P P P k k k k 2 1 2 1 2 0 2 2 1 1 2 1 2 1 1 0 1 1 Ma 1 2 1 3105 14 1 2 260 K 1 Ma 1 2 1 24 14 1 2 T T k T T T T T k 191 K Note that this is an expansion and Mach number increases while pressure and temperature decrease as expected Discussion There are compressible flow calculators on the Internet that solve these implicit equations that arise in the analysis of compressible flow along with both normal and oblique shock equations eg see wwwaoevtedudevenporaoe3114calchtml PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1755 1795 Air flowing at a specified supersonic Mach number undergoes a compression turn an oblique shock over a tilted wedge The Mach number pressure and temperature downstream of the shock below the wedge are to be determined Assumptions 1 The flow is steady 2 The boundary layer on the wedge is very thin 3 Air is an ideal gas with constant specific heats PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The specific heat ratio of air is k 14 Analysis On the basis of Assumption 2 the deflection angle is determined to be θ δ 25 10 35o Then the two values of oblique shock angle β are determined from 2 cos 2 Ma tan 1 2Ma sin tan 2 1 2 2 1 β β β θ k 2 cos 2 41 34 tan 1 sin 2 43 tan12 2 2 2 β β β Ma1 5 10 25 Ma2 δ θ which is implicit in β Therefore we solve it by an iterative approach or with an equation solver such as EES It gives βweak 4986o and βstrong 7766o Then for the case of strong oblique shock the upstream normal Mach number Ma1n becomes 4 884 5sin 7766 Ma sin Ma 1 1n β Also the downstream normal Mach numbers Ma2n become 0 4169 1 41 4 884 41 2 2 1 4 884 41 1 Ma 2 2 1 Ma Ma 2 2 2 n 1 2 1n 2n k k k The downstream pressure and temperature are determined to be 1940 kPa 1 41 1 41 4 884 41 70 kPa 2 1 1 2 Ma 2 2 1n 1 2 k k k P P 1450 K 2 2 2 n 1 2 n 1 1 2 1 2 1 1 2 1 2 1 4 884 41 1 4 884 41 2 70 kPa 260 K 1940 kPia 1 Ma 1 Ma 2 k k P T P P T P T ρ ρ The downstream Mach number is determined to be 0615 35 sin7766 0 4169 sin Ma Ma 2n 2 θ β Discussion Note that Ma1n is supersonic and Ma2n and Ma2 are subsonic Also note the huge rise in temperature and pressure across the strong oblique shock and the challenges they present for spacecraft during reentering the earths atmosphere preparation If you are a student using this Manual you are using it without permission 1756 1796E Air flowing at a specified supersonic Mach number is forced to turn upward by a ramp and weak oblique shock forms The wave angle Mach number pressure and temperature after the shock are to be determined Assumptions 1 The flow is steady 2 The boundary layer on the wedge is very thin 3 Air is an ideal gas with constant specific heats PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Properties The specific heat ratio of air is k 14 Analysis On the basis of Assumption 2 we take the deflection angle as equal to the ramp ie θ δ 8o Then the two values of oblique shock angle β are determined from 2 cos 2 Ma tan 1 2Ma sin tan 2 1 2 2 1 β β β θ k 2 cos 2 41 2 tan 1 22 sin tan 8 2 2 2 β β β βweak δ 8o Weak shock Ma1 which is implicit in β Therefore we solve it by an iterative approach or with an equation solver such as EES It gives βweak 3721o and βstrong 8505o Then for the case of weak oblique shock the upstream normal Mach number Ma1n becomes 1 209 2sin 3721 Ma sin Ma 1 1n β Also the downstream normal Mach numbers Ma2n become 0 8363 1 41 1 209 41 2 2 1 1 209 41 1 Ma 2 2 1 Ma Ma 2 2 2 n 1 2 1n 2n k k k The downstream pressure and temperature are determined to be 185 psia 1 41 1 41 1 209 41 12 psia 2 1 1 2 Ma 2 2 1n 1 2 k k k P P 556 R 2 2 2 n 1 2 n 1 1 2 1 2 1 1 2 1 2 1 1 209 41 1 1 209 41 2 12 psia 490 R18 5 psia 1 Ma 1 Ma 2 k k P T P P T P T ρ ρ The downstream Mach number is determined to be 171 8 sin3721 0 8363 sin Ma Ma 2n 2 θ β Discussion Note that Ma1n is supersonic and Ma2n is subsonic However Ma2 is supersonic across the weak oblique shock it is subsonic across the strong oblique shock preparation If you are a student using this Manual you are using it without permission 1757 1797 Air flowing at a specified supersonic Mach number undergoes an expansion turn The Mach number pressure and temperature downstream of the sudden expansion along a wall are to be determined Assumptions 1 The flow is steady 2 The boundary layer on the wedge is very thin 3 Air is an ideal gas with constant specific heats Ma1 36 δ 15o Ma2 θ Properties The specific heat ratio of air is k 14 Analysis On the basis of Assumption 2 we take the deflection angle as equal to the wedge halfangle ie θ δ 15o Then the upstream and downstream PrandtlMeyer functions are determined to be 1 Ma tan 1 1 Ma 1 1 tan 1 Ma 2 1 2 1 k k k k ν Upstream 6009 1 36 tan 1 1 36 41 1 41 1 tan 41 1 41 Ma 2 1 2 1 1 ν Then the downstream PrandtlMeyer function becomes 7509 6009 15 Ma Ma 1 2 ν θ ν Ma2 is found from the PrandtlMeyer relation which is now implicit Downstream 7509 1 Ma tan 1 1 Ma 41 1 41 1 tan 41 1 41 Ma 2 2 1 2 2 1 2 ν Solution of this implicit equation gives Ma2 481 Then the downstream pressure and temperature are determined from the isentropic flow relations kPa 831 40 kPa 2 1 41 63 1 2 1 41 4 81 1 2 1 Ma 1 2 1 Ma 1 40 41 2 40 41 2 1 1 2 1 1 2 2 1 0 1 0 2 2 P k k P P P P P P k k k k 179 K 280 K 2 1 41 63 1 2 1 41 4 81 1 2 1 Ma 1 2 1 Ma 1 1 2 1 2 1 1 2 1 1 2 2 1 0 1 0 2 2 T k k T T T T T T Note that this is an expansion and Mach number increases while pressure and temperature decrease as expected Discussion There are compressible flow calculators on the Internet that solve these implicit equations that arise in the analysis of compressible flow along with both normal and oblique shock equations eg see wwwaoevtedudevenporaoe3114calchtml PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1758 1798 Air flowing at a specified supersonic Mach number impinges on a twodimensional wedge The shock angle Mach number and pressure downstream of the weak and strong oblique shock formed by a wedge are to be determined βweak δ 8o Weak shock Ma1 Ma1 δ 8o Strong shock βstrong Assumptions 1 The flow is steady 2 The boundary layer on the wedge is very thin 3 Air is an ideal gas with constant specific heats Properties The specific heat ratio of air is k 14 Analysis On the basis of Assumption 2 we take the deflection angle as equal to the wedge halfangle ie θ δ 8o Then the two values of oblique shock angle β are determined from 2 cos 2 Ma tan 1 2Ma sin tan 2 1 2 2 1 β β β θ k 2 cos 2 41 34 tan 1 sin 2 43 tan 8 2 2 2 β β β which is implicit in β Therefore we solve it by an iterative approach or with an equation solver such as EES It gives βweak 2315o and βstrong 8745o Then the upstream normal Mach number Ma1n becomes Weak shock 1 336 sin 2315 43 Ma sin Ma 1 1n β Strong shock 3 397 sin 8745 43 Ma sin Ma 1 1n β Also the downstream normal Mach numbers Ma2n become Weak shock 0 7681 1 41 1 336 41 2 2 1 1 336 41 1 Ma 2 2 1 Ma Ma 2 2 2 n 1 2 1n 2n k k k Strong shock 0 4553 1 41 3 397 41 2 2 1 3 397 41 1 Ma 2 2 1 Ma Ma 2 2 2 n 1 2 1n 2n k k k The downstream pressure for each case is determined to be Weak shock 1150 kPa 1 41 1 41 1 336 41 60 kPa 2 1 1 2 Ma 2 2 1n 1 2 k k k P P Strong shock 7976 kPa 1 41 1 41 3 397 41 60 kPa 2 1 1 2 Ma 2 2 1n 1 2 k k k P P The downstream Mach number is determined to be Weak shock 294 8 sin2315 0 7681 sin Ma Ma 2n 2 θ β Strong shock 0463 8 sin8745 0 4553 sin Ma Ma 2n 2 θ β Discussion Note that the change in Mach number and pressure across the strong shock are much greater than the changes across the weak shock as expected For both the weak and strong oblique shock cases Ma1n is supersonic and Ma2n is subsonic However Ma2 is supersonic across the weak oblique shock but subsonic across the strong oblique shock PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1759 Duct Flow with Heat Transfer and Negligible Friction Rayleigh Flow 1799C The characteristic aspect of Rayleigh flow is its involvement of heat transfer The main assumptions associated with Rayleigh flow are the flow is steady onedimensional and frictionless through a constantarea duct and the fluid is an ideal gas with constant specific heats Discussion Of course there is no such thing as frictionless flow It is better to say that frictional effects are negligible compared to the heating effects 17100C The points on the Rayleigh line represent the states that satisfy the conservation of mass momentum and energy equations as well as the property relations for a given state Therefore for a given inlet state the fluid cannot exist at any downstream state outside the Rayleigh line on a Ts diagram Discussion The Ts diagram is quite useful since any downstream state must lie on the Rayleigh line 17101C In Rayleigh flow the effect of heat gain is to increase the entropy of the fluid and the effect of heat loss is to decrease the entropy Discussion You should recall from thermodynamics that the entropy of a system can be lowered by removing heat 17102C In Rayleigh flow the stagnation temperature T0 always increases with heat transfer to the fluid but the temperature T decreases with heat transfer in the Mach number range of 0845 Ma 1 for air Therefore the temperature in this case will decrease Discussion This at first seems counterintuitive but if heat were not added the temperature would drop even more if the air were accelerated isentropically from Ma 092 to 095 17103C Heating the fluid increases the flow velocity in subsonic Rayleigh flow but decreases the flow velocity in supersonic Rayleigh flow Discussion These results are not necessarily intuitive but must be true in order to satisfy the conservation laws 17104C The flow is choked and thus the flow at the duct exit remains sonic Discussion There is no mechanism for the flow to become supersonic in this case PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1760 17105 Fuel is burned in a tubular combustion chamber with compressed air For a specified exit Mach number the exit temperature and the rate of fuel consumption are to be determined Assumptions 1 The assumptions associated with Rayleigh flow ie steady onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects are valid 2 Combustion is complete and it is treated as a heat addition process with no change in the chemical composition of flow 3 The increase in mass flow rate due to fuel injection is disregarded Properties We take the properties of air to be k 14 cp 1005 kJkgK and R 0287 kJkgK Analysis The inlet density and mass flow rate of air are PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 3 1 1 1 2 942 kgm 0287 kJkgK45 0 K 380 kPa RT P ρ 3 254 kgs 2 942 kgm 016 m 455 ms 2 3 1 1 1 π ρ A V m c air The stagnation temperature and Mach number at the inlet are 451 5 K m s 1000 1kJkg 1 005 kJkg K 2 55 ms 450 K 2 2 2 2 2 1 1 01 cp V T T Q COMBUSTOR TUBE T2 V2 P1 380 kPa T1 450 K V1 55 ms 425 2 ms 1kJkg 0 287 kJkg K450 K 1000 m s 41 2 2 1 1 kRT c 0 1293 425 2 ms 55 ms Ma 1 1 1 c V The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are Table A34 We used analytical functions Ma1 01293 T1T 009201 T01T 007693 V1V 003923 Ma2 08 T2T 10255 T02T 09639 V2V 08101 The exit temperature stagnation temperature and velocity are determined to be 11146 0 09201 1 0255 1 2 1 2 T T T T T T 5016 K 11146450 K 11146 1 2 T T 12530 0 07693 0 9639 01 02 1 0 02 T T T T T T 5658 K 12530451 5 K 12530 01 02 T T 20650 0 03923 0 8101 1 2 1 2 V V V V V V 1136 ms 2065055 ms 20650 1 2 V V Then the mass flow rate of the fuel is determined to be 5232 kJkg 451 5 K 1005 kJkg K5658 01 02 T T c q p 17024 kW 3254 kgs5232 kJkg air q m Q 04365 kgs 39000 kJkg 17024 kJs fuel HV Q m Discussion Note that both the temperature and velocity increase during this subsonic Rayleigh flow with heating as expected This problem can also be solved using appropriate relations instead of tabulated values which can likewise be coded for convenient computer solutions preparation If you are a student using this Manual you are using it without permission 1761 17106 Air is heated in a duct during subsonic flow until it is choked For specified pressure and velocity at the exit the temperature pressure and velocity at the inlet are to be determined Assumptions The assumptions associated with Rayleigh flow ie steady onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects are valid Properties We take the properties of air to be k 14 cp 1005 kJkgK and R 0287 kJkgK Analysis Noting that sonic conditions exist at the exit the exit temperature is PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 620 ms 620 ms1 2Ma2 2 V c 2 2 kRT c 620 ms 1 kJkg s 1000 m 0 287 kJkg K 41 2 2 2 T It gives T2 9567 K Then the exit stagnation temperature becomes 1148 K m s 1000 1 kJkg 1 005 kJkg K 2 620 ms 956 7 K 2 2 2 2 2 2 2 02 cp V T T P2 270 kPa V2 620 ms Ma2 1 q 52 kJkg P1 T1 Ma1 The inlet stagnation temperature is from the energy equation 01 02 T T c q p 1096 K 1005 kJkg K 52 kJkg 1148 K 02 01 cp q T T The maximum value of stagnation temperature T0 occurs at Ma 1 and its value in this case is T02 since the flow is choked Therefore T0 T02 1148 K Then the stagnation temperature ratio at the inlet and the Mach number corresponding to it are from Table A34 0 9547 1148 K 1096 K 0 01 T T Ma1 07792 0779 The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are Table A34 Ma1 07792 T1T 1022 P1P 1297 V1V 07877 Ma2 1 T2T 1 P2P 1 V2V 1 Then the inlet temperature pressure and velocity are determined to be 1 017 1 1 2 1 2 T T T T T T 9775 K 1 022956 7 K 1 022 2 1 T T 1 319 1 1 2 1 2 P P P P P P 3503 kPa 1 297270 kPa 1 319 2 1 P P 0 7719 1 1 2 1 2 V V V V V V 4884 ms 0 7877620 ms 0 7877 2 1 V V Discussion Note that the temperature and pressure decreases with heating during this subsonic Rayleigh flow while velocity increases This problem can also be solved using appropriate relations instead of tabulated values which can likewise be coded for convenient computer solutions preparation If you are a student using this Manual you are using it without permission 1762 17107E Air flowing with a subsonic velocity in a round duct is accelerated by heating until the flow is choked at the exit The rate of heat transfer and the pressure drop are to be determined Assumptions 1 The assumptions associated with Rayleigh flow ie steady onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects are valid 2 The flow is choked at the duct exit 3 Mass flow rate remains constant Properties We take the properties of air to be k 14 cp 02400 BtulbmR and R 006855 BtulbmR 03704 psiaft3lbmR Q Ma2 1 T2 680 R P1 30 psia T1 800 R m 5 lbms Analysis The inlet density and velocity of air are 3 3 1 1 1 0 1012 lbmft psia ft lbm R800 R 03704 30 psia RT P ρ 565 9 fts 0 1012 lbmft 412 ft 4 lbms 5 2 3 1 1 1 π ρ c air A m V The stagnation temperature and Mach number at the inlet are 826 7 R ft s 25037 1Btulbm 0 2400 Btulbm R 2 565 9 fts 800 R 2 2 2 2 2 1 1 01 cp V T T 1386 fts 1Btulbm s 0 06855 Btulbm R800 R 25037 ft 41 2 2 1 1 kRT c 0 4082 1386 fts 565 9 fts Ma 1 1 1 c V The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are Table A34 Ma1 04082 T1T 06310 P1P 1946 T01T0 05434 Ma2 1 T2T 1 P2P 1 T02T0 1 Then the exit temperature pressure and stagnation temperature are determined to be 0 6310 1 1 2 1 2 T T T T T T 2 1268 R 0 6310 800 R 0 6310 1 T T 1 946 1 1 2 1 2 P P P P P P 2 15 4 psia 1 946 30 psia 2 272 1 P P 0 5434 1 01 02 1 0 02 T T T T T T 0 R 1521 0 5434 826 7 R 0 1743 01 T 2 T Then the rate of heat transfer and the pressure drop become 834 Btus 5lbms 0 2400 Btulbm R1521 826 7 R 01 02 air T T c m Q p 146 psia 15 4 30 2 1 P P P Discussion Note that the entropy of air increases during this heating process as expected PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1763 17108 Air flowing with a subsonic velocity in a duct The variation of entropy with temperature is to be investigated as the exit temperature varies from 600 K to 5000 K in increments of 200 K The results are to be tabulated and plotted Analysis We solve this problem using EES making use of Rayleigh functions The EES Equations window is printed below along with the tabulated and plotted results k14 cp1005 R0287 P1350 T1600 V170 C1sqrtkRT11000 Ma1V1C1 T01T1105k1Ma12 P01P1105k1Ma12kk1 F1105k1Ma12 T01Ts2k1Ma12F11kMa122 P01Ps1k1kMa122F1k1kk1 T1TsMa11k1kMa122 P1Ps1k1kMa12 V1VsMa121k1kMa12 F2105k1Ma22 T02Ts2k1Ma22F21kMa222 P02Ps1k1kMa222F2k1kk1 T2TsMa21k1kMa222 P2Ps1k1kMa22 V2VsMa221k1kMa22 T02T02TsT01TsT01 P02P02PsP01PsP01 T2T2TsT1TsT1 P2P2PsP1PsP1 V2V2VsV1VsV1 DeltascplnT2T1RlnP2P1 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1764 Exit temperature T2 K Exit Mach number Ma2 Exit entropy relative to inlet s2 kJkgK 6001 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000 3200 3400 3600 3800 4000 4200 4400 4600 4800 5000 0143 0166 0188 0208 0227 0245 0263 0281 0299 0316 0333 0351 0369 0387 0406 0426 0446 0467 0490 0515 0541 0571 0606 0000 0292 0519 0705 0863 1001 1123 1232 1331 1423 1507 1586 1660 1729 1795 1858 1918 1975 2031 2085 2138 2190 2242 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 0 045 09 135 18 225 T2 K δs kJkgK Discussion Note that the entropy of air increases during this heating process as expected PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1765 17109E Air flowing with a subsonic velocity in a square duct is accelerated by heating until the flow is choked at the exit The rate of heat transfer and the entropy change are to be determined Assumptions 1 The assumptions associated with Rayleigh flow ie steady onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects are valid 2 The flow is choked at the duct exit 3 Mass flow rate remains constant Properties We take the properties of air to be k 14 cp 02400 BtulbmR and R 006855 BtulbmR 03704 psiaft3lbmR Analysis The inlet density and mass flow rate of air are 3 3 1 1 1 0 3085 lbmft psia ft lbm R700 R 03704 80 psia RT P ρ 2006 lbms 6144 ft 260 fts 0 3085 lbmft 6 2 3 1 1 1 A V m c air ρ The stagnation temperature and Mach number at the inlet are 705 6 R ft s 25037 1Btulbm 0 2400 Btulbm R 2 260 fts 700 R 2 2 2 2 2 1 1 01 c p V T T 1297 fts 1Btulbm s 0 06855 Btulbm R700 R 25037 ft 41 2 2 1 1 kRT c 0 2005 260 fts Ma 1 1 V 1297 fts 1c The Ray ctio ding rs are Table A34 Ma2 1 T2T 1 P2P 1 T02T 1 Then the exit tem pressure and stagnation temperature are be leigh flow fun ns correspon to the inlet and exit Mach numbe Ma1 02005 T1T 02075 P1P 2272 T01T0 01743 0 perature determined to 0 2075 1 2 2 T T 2075 1 1 T T T T 0 700 R 0 2075 1 2 3374 R T T 2 272 1 1 2 1 P P P P P 2 P 35 2 psia 2 272 80 psia 2 272 1 2 P P 0 1743 1 01 02 1 0 02 T T T T T T 4048 R 0 1743 705 6 R 0 1743 01 02 T T Then the rate of heat transfer and entropy change become 90 01 02 air p 160 R 705 6 2006 lbms 0 2400 Btulbm R4048 T T c m Q Btus 0434 Btulbm R 80 psia 0 06855 Btulbm R ln 35 2 psia 700 R 0 2400 Btulbm R ln 3374 R ln ln 1 2 1 2 P P R T T c s p Discussion Note that the entropy of air increases during this heating process as expected Q Ma2 1 P1 80 psia T1 700 R V1 260 fts PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1766 17110 Air enters the combustion chamber of a gas turbine at a subsonic velocity For a specified rate of heat transfer the Mach number at the exit and the loss in stagnation pressure to be determined Assumptions 1 The assumptions associated with Rayleigh flow ie steady onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects are valid 2 The cross sectional area of the combustion chamber is constant 3 The increase in mass flow rate due to fuel injection is disregarded Properties We take the properties of air to be k 14 cp 1005 kJkgK and R 0287 kJkgK Analysis The inlet stagnation temperature and pressure are 554 4 K 20 2 141 550 K 1 2 Ma 1 1 2 2 1 1 01 k T T PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1 1 4 0 4 2 2 01 1 1 1 141 1 Ma 600 kPa 1 0 2 2 2 k k k P P The exit st n temperature is determined from 617 0 kPa agnatio 01 02 air T T c m Q p K 554 4 30 kgs 1 005 kJkg K 200 kJs 02 T It gives At Ma 0 from T T 01736 Table A34 Therefore T02 1218 K 1 01 0 2 we read 3193 5 K 1736 0 4 K 554 01 T 0 1736 0 T Then the stagnation temperature ratio at the exit and the Mach number corresponding to it are Table A34 0 3814 319 T0 Ma2 K 53 1218 K 02 T 03187 0319 Also M 3187 P02P0 1191 the stagnation pressure at the exit and the pressure drop become Ma1 02 P01P0 12346 a2 0 Then 0 9647 1 2346 0 01 01 P P P 595 2 kPa 0 9647617 kPa 0 9647 01 02 1 191 0 02 2 P P P0 P P and 218 kPa 595 2 617 0 02 01 0 P P P Discussion This problem can also be solved using appropriate relations instead of tabulated values which can likewise be coded for convenient computer solutions 200 kJs Q COMBUSTOR TUBE Ma2 P1 600 kPa T1 550 K Ma1 02 preparation If you are a student using this Manual you are using it without permission 1767 17111 Air enters the combustion chamber of a gas turbine at a subsonic velocity For a specified rate of heat transfer the Mach number at the exit and the loss in stagnation pressure to be determined Assumptions 1 The assumptions associated with Rayleigh flow ie steady onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects are valid 2 The cross sectional area of the combustion chamber is constant 3 The increase in mass flow rate due to fuel injection is disregarded Properties We take the properties of air to be k 14 cp 1005 kJkgK and R 0287 kJkgK Analysis The inlet stagnation temperature and pressure are PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 554 4 K 20 2 141 550 K 1 2 Ma 1 1 2 2 1 1 01 k T T 1 1 4 0 4 2 2 1 1410 2 kPa k k k The exit stagnation temperature is determined from T01 01 1 1 1 Ma 600 kPa 1 2 2 P P 300 kJs Q COMBUSTOR TUBE Ma2 P1 600 kPa T1 550 K Ma1 02 617 0 02 air T c m Q p K 554 4 30 kgs 1 005 kJkg K 300 kJs 02 T It gives T02 1549 K At Ma1 02 we read from T01T0 01736 Table A34 Therefore 3193 5 K 554 4 K 01 T T 0 1736 0 1736 0 Then the stagnation temperature ratio at the exit and the Mach number corresponding to it are Table A34 0 4850 15 02 T Ma K 49 K 2 03753 0375 Ma1 02 P01P0 123 Ma2 03753 P02P0 1167 the stagnation pressure at the exit and the pressure drop become 5 3193 0 T Also 46 Then 0 9452 1 2346 0 01 01 P P P 01 02 1 167 0 02 02 P P P 583 3 kPa 0 9452617 kPa 0 9452 P P Discussion This problem can also be solved using appropriate relations instead of tabulated values which can likewise be coded for convenient computer solutions and 337 kPa 583 3 617 0 02 01 0 P P P preparation If you are a student using this Manual you are using it without permission 1768 17112 Fuel is burned in a rectangular duct with compressed air For specified heat transfer the exit temperature and Mach number are to be determined Assumptions The assumptions associated with Rayleigh flow ie steady onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects are valid Properties We take the properties of air to be k 14 cp 1005 kJkgK and R 0287 kJkgK Analysis The stagnation temperature and Mach number at the inlet are PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 347 2 ms 1 kJkg s 0 287 kJkg K300 K 1000 m 41 2 2 1 1 kRT c q 55 kJkg 694 4 ms 2347 2 ms Ma 1 1 1 c V 539 9 K m s 1000 1 kJkg 694 4 ms 2 2 V 1 005 kJkg K 2 300 K 2 2 1 1 01 cp T T 2 The exit stagnation temperature is from the energy equation 01 02 T T c q p 594 6 K 1005 kJkg K 55 kJkg 5399 K 01 02 cp q T T The maximum value of stagnation temperature T0 occurs at Ma 1 and its value can be determined from Table A34 or from the a e read T01T0 07934 Therefore ppropriate relation At Ma1 2 w 680 5 K 7934 0 7934 0 539 9 K 01 0 T T The stag io at ch number corresponding to it are from Table A34 nation temperature rat the exit and the Ma 0 8738 594 6 K 02 T 5 K 680 T0 Ma2 1642 164 Also emperature becomes Ma1 2 T1T 05289 Ma2 1642 T2T 06812 Then the exit t 1 288 0 5289 0 6812 1 2 1 2 T T T T T T 386 K 1 288300 K 1 288 1 2 T T Discussion Note that the temperature increases during this supersonic Rayleigh flow with heating This problem can also be solved using appropriate relations instead of tabulated values which can likewise be coded for convenient computer solutions P1 420 k T Pa 00 K 1 3 M T2 Ma2 a1 2 preparation If you are a student using this Manual you are using it without permission 1769 17113 Compressed air is cooled as it flows in a rectangular duct For specified heat rejection the exit temperature and Mach number are to be determined Assumptions The assumptions associated with Rayleigh flow ie steady onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects are valid Properties We take the properties of air to be k 14 cp 1005 kJkgK and R 0287 kJkgK Analysis The stagnation temperature and Mach number at the inlet are 347 2 ms 1 kJkg s 0 287 kJkg K300 K 1000 m 41 2 2 1 1 kRT c q 55 kJkg T2 Ma2 P1 420 k T Pa 300 K a 2 1 M 1 1 694 4 ms 2347 2 ms Ma 1 1 c V 539 9 K m s 1000 1 kJkg 1 005 kJkg K 2 694 4 ms 300 K 2 2 2 2 2 1 1 01 cp V T T The exit stagnation temperature is from the energy equation 01 02 T T c q p 485 2 K 1005 kJkg K 55 kJkg 5399 K 01 02 p The maximum value of stagnation temperature T0 c q T T e can be determined from Table A34 or from the appropriate relation At Ma1 2 we read T01T0 07934 Therefore occurs at Ma 1 and its valu 680 5 K 0 7934 539 9 K 0 7934 01 0 T T The stagnation temperature ratio a rresponding to it are from Table A34 t the exit and the Mach number co 0 7130 680 5 K 485 2 K 0 02 T T Ma2 2479 248 Also 1 Ma T2T 03838 Then the exit te perature becomes T T 05289 Ma1 2 2 2479 m 0 7257 0 5289 0 3838 1 2 1 2 T T T T T T 218 K 0 7257300 K 0 7257 1 2 T T Discussion Note that the temperature decreases and Mach number increases during this supersonic Rayleigh flow with cooling This problem can also be solved using appropriate relations instead of tabulated values which can likewise be coded for convenient computer solutions PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1770 17114 Argon flowing at subsonic velocity in a constantdiameter duct is accelerated by heating The highest rate of heat transfer without reducing the mass flow rate is to be determined Assumptions 1 The assumptions associated with Rayleigh flow ie steady onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects are valid 2 Mass flow rate remains constant Properties We take the properties of argon to be k 1667 cp 05203 kJkgK and R 02081 kJkgK Analysis Heat transfer stops when the flow is choked and thus Ma2 V2c2 1 The inlet stagnation temperature is 405 3 K 20 2 2 1 1 01 The Rayleigh 16671 400 K 1 1 Ma 1 2 2 k T T flow functions corresponding to the inlet and exit Mach numbers T02T0 1 since Ma2 1 are 0 1900 20 1 667 1 20 1 1 667 2 20 1 667 1 Ma 1 1 Ma 1 Ma 2 2 2 2 2 2 2 1 2 1 2 1 0 T 01 k k k T Therefore PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1900 0 1 0 02 02 T T T 2133 K 0 1900 405 3 K 0 1900 01 02 T T 0 01 1 T T 0 T Then the rate of heat transfer becomes 1080 kW 400 K kgs 0 5203 kJkg K2133 21 01 02 air T T c m Q p Discussion It can also be shown that T2 1600 K which is the highest thermodynamic temperature that can be attained under stated conditions If more heat is transferred the additional temperature rise will cause the mass flow rate to decrease Also in the solution of this problem we cannot use the values of Table A34 since they are based on k 14 Q Ma2 1 P1 320 kPa T1 400 K Ma1 02 preparation If you are a student using this Manual you are using it without permission 1771 17115 Air flowing at a supersonic velocity in a duct is decelerated by heating The highest temperature air can be heated by heat addition and the rate of heat transfer are to be determined Assumptions 1 The assumptions associated with Rayleigh flow ie steady onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects are valid 2 Mass flow rate remains constant Properties We take the properties of air to be k 14 cp 1005 kJkgK and R 0287 kJkgK Analysis Heat transfer will stop when the flow is choked and thus Ma2 V2c2 1 Knowing stagnation properties the static properties are determined to be 364 1 K 81 2 14 1 600 K 1 2 Ma 1 1 1 2 1 2 1 01 1 k T T 1 1 4 0 4 2 2 1 01 1 1 141 1 Ma 210 kPa 1 1 8 2 2 36 55 kPa k k k P P PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 3 1 1 0 3498 kgm 3655 kPa ρ P 1 K 0287 kJkgK3641 RT Then the inlet velocity and the mass flow rate beco e m 382 5 ms 1 kJkg 0 287 kJkg K3641 K 41 1 1 kRT c s 1000 m 2 2 2 The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are Table A34 Ma1 18 T 89 T T0 08363 Ma2 1 T2T 1 T02T0 1 Then the exit temperature and stagn n temp ature are determined to be 688 5 ms 382 5 ms Ma 1 1 1 c V 81 1 891 0 3498 kgm 010 m 4688 5 ms 1 1 1 A V m c air 3 π ρ kgs 1T 060 01 atio er 0 6089 1 2 1 2 T T T 1 T T T 598 K 0 6089 364 1 K 0 6089 1 2 T T 0 8363 0 01 01 T T T 1 0 02 02 T T T 02 01 0 8363 600 K 0 8363 717 4 K T T 717 K Finally the rate of heat transfer is Q m 223 kW 600 K 1 891 kgs1005 kJkg K717 4 01 02 air T c p T Discussion Note that this is the highest temperature that can be attained under stated conditions If more heat is transferred the additional temperature will cause the mass flow rate to decrease Also once the sonic conditions are reached the thermodynamic temperature can be increased further by cooling the fluid and reducing the velocity see the Ts diagram for Rayleigh flow Q Ma2 1 P01 210 kPa T01 600 K Ma1 18 preparation If you are a student using this Manual you are using it without permission 1772 Steam Nozzles 17116C The delay in the condensation of the steam is called supersaturation It occurs in highspeed flows where there isnt sufficient time for the necessary heat transfer and the formation of liquid droplets PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1779 Review Problems 17121 A leak develops in an automobile tire as a result of an accident The initial mass flow rate of air through the leak is to be determined Assumptions 1 Air is an ideal gas with constant specific heats 2 Flow of air through the hole is isentropic Properties The gas constant of air is R 0287 kPam3kgK The specific heat ratio of air at room temperature is k 14 Table A2a Analysis The absolute pressure in the tire is 314 kPa 94 220 atm gage P P P The critical pressure is from Table 172 94 kPa 0 5283314 kPa 166 kPa 0 5283 0 P P Therefore the flow is choked and the velocity at the exit of the hole is the sonic speed Then the flow properties at the exit becomes PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 3 1 1 41 3 1 1 0 3 3 0 0 0 2 327 kgm 1 41 2 3 671 kgm 1 2 3 671 kgm 0 287 kPa m kg K29 8 K kPa 314 k k RT P ρ ρ ρ 248 3 K 1 298 K 41 2 2 1 0 T k T 315 9 ms 248 3 K kJkg 1 s 0 287 kJkg K 1000 m 41 2 2 kRT c V Then the initial mass flow rate through the hole becomes Discussion The mass flow rate will decrease with time as the pressure inside the tire drops 2 327 kgm 0004 m 43159 ms 000924 kgs 0 554 kgmin 2 3 π ρ AV m preparation If you are a student using this Manual you are using it without permission 1780 17122 The thrust developed by the engine of a Boeing 777 is about 380 kN The mass flow rate of air through the nozzle is to be determined Assumptions 1 Air is an ideal gas with constant specific properties 2 Flow of combustion gases through the nozzle is isentropic 3 Choked flow conditions exist at the nozzle exit 4 The velocity of gases at the nozzle inlet is negligible Properties The gas constant of air is R 0287 kPam3kgK Table A1 and it can also be used for combustion gases The specific heat ratio of combustion gases is k 133 Table 172 Analysis The velocity at the nozzle exit is the sonic velocity which is determined to be 318 0 ms 265 K 1kJkg 1 33 0 287 kJk kRT c V K 1000 m s g 2 2 Noting that thrust F is related to velocity by mV F the mass flow rate of combustion gases is determined to be 11948 kg s N 1 1kg 380000 N m F ms ms 3180 2 V iscussion The combustion gases are mostly nitrogen due to the 78 of N2 in air and thus they can be treated as air with good degree of approximation perature of air is to be Properties The specific heat of air at room temperature is cp 1005 kJkgK Table A2a Analysis The air that strikes the probe will be brought to a complete stop and thus it will undergo a stagnation process The thermometer will sense the temperature of this stagnated air which is the stagnation temperature The actual air temperature D a 17123 A stationary temperature probe is inserted into an air duct reads 50C The actual tem determined Assumptions 1 Air is an ideal gas with constant specific heats at room temperature 2 The stagnation process is isentropic is determined from T 125 ms 422C 2 2 1kJkg 125 ms V 2 2 0 1 005 kJkg K 1000 m s 2 50 C 2 cp T T Discussion Temperature rise due to stagnation is very significant in highspeed flows and should always be considered when compressibility effects are not negligible PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1782 17125 An expression for the speed of sound based on van der Waals equation of state is to be derived Using this relation the speed of sound in carbon dioxide is to be determined and compared to that obtained by ideal gas behavior Properties The properties of CO2 are R 01889 kJkgK and k 1279 at T 50C 3232 K Table A2b Analysis Van der Waals equation of state can be expressed as v 2 v b a RT P Differentiating 3 2 v v b T 2 v a RT P Noting that the speed of sound relation becomes Substituting 2 1 v v v d d ρ ρ v v b c 2 2 v v v ak kRT P k r P k c T T 2 2 2 2 2 M 44 kgkmol the constant a and b can be expressed per unit mass as he specif volume of CO2 is determined to be Using the molar mass of CO m kg 9 705 10 3 4 0 1882 kPa m6kg2 a and b T ic 0 3031 m kg 0 1882 kPa m kg 2 0009705 m kg 0 200 kPa 01889 kPa m kg K3232 K 3 2 2 6 3 3 v v v Substituting 2775 ms 2 2 3 6 00 009705 m kg 03031 c 2 1 3 2 3 2 2 2 3 2 3 kPa m kg 1 s m 1000 0 3031 m kg kg 1 279 0 1882 kPam 2 1kJkg s m kg 127901889 kJkg K3232 K 1000 m 03031 If we treat CO2 as an ideal gas the speed of sound becomes 2794 ms 1kJkg s 279 0 1889 kJkg K3232 K 1000 m 1 2 2 kRT c Discussion Note that the ideal gas relation is the simplest equation of state and it is very accurate for most gases encountered in practice At high pressures andor low temperatures however the gases deviate from ideal gas behavior and it becomes necessary to use more complicated equations of state PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1783 17126 The equivalent relation for the speed of sound is to be verified using thermodynamic relations Analysis The two relations are s P c ρ 2 and T P k c ρ 2 From 2 Thus 1 v v v d dr r PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course s s s s r s c T T P T T P P P v v v v v v 2 2 2 rom the cyclic rule 2 F T P s T P s s P T s T P P s s T T P P T s 1 v v v v v v v s T s T T s s T s T T s T s 1 Substituting T P T T T P s P s T T s s T s s P s c v v v v v 2 2 Recall that 2 p c v v T s T c P T s and T Substituting T T k c T c p P P T c 2 2 eplacing by dρ 2 v v v v v dv v 2 R T P k c ρ 2 Discussion Note that the differential thermodynamic property relations are very useful in the derivation of other property relations in differential form preparation If you are a student using this Manual you are using it without permission 1784 17127 For ideal gases undergoing isentropic flows expressions for PP TT and ρρ as functions of k and Ma are to be obtained Analysis Equations 1718 and 1721 are given to be 2 1 Ma 2 2 0 k T T and PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1 0 k T 2 T Multiplying the two T0 T 1 2 2 1 Ma 2 2 0 k k T T Simplifying and inverting 1 Ma2 2 1 k k T T 1 From 1 2 1 1 Ma 2 1 P T k k k k k k P T P 2 P From 1 2 1 1 Ma 2 1 k k k k k k ρ ρ ρ ρ ρ ρ 3 Discussion Note that some very useful relations can be obtained by very simple manipulations preparation If you are a student using this Manual you are using it without permission 1785 17128 It is to be verified that for the steady flow of ideal gases dT0T dAA 1Ma2 dVV The effect of heating and area changes on the velocity of an ideal gas in steady flow for subsonic flow and supersonic flow are to be explained Analysis We start with the relation 2 0 2 T T c V p 1 Differentiating 0 dT dT c dV p V 2 We also have 0 V dV A dA d ρ ρ 3 0 V dV dP and 4 n ρ dP PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Differentiating the ideal gas relatio P ρRT P T d dT ρ ρ 0 5 c T k kRT c p From the speed of sound relation kP ρ 6 1 2 0 V dV A dA T dT P dP 7 Combining Eqs 3 and 5 VdV kP c dP dP 2 ρ Combining Eqs 4 and 6 V dV k V dV c k V V dV c k P dP 2 2 2 2 Ma or 8 bining Eqs 2 and 6 Com V cp dT dT 0 dV V dV k T dT V dV k c V T dT T dT V dV c T V T dT T dT p 2 0 2 2 0 2 0 1 Ma 1 or 9 Combining Eqs 7 8 and 9 0 1 Ma 1 Ma 2 0 2 d k dT dV k V dV A dA V V T V V k k A T 1 1 Ma Ma 2 2 0 dV dA dT or Thus V dV A dA T dT Ma 1 2 0 10 Differentiating the steadyflow energy equation 01 02 01 02 T T c h h q p c dT0 q p δ 11 Eq 11 relates the stagnation temperature change dT0 to the net heat transferred to the fluid Eq 10 relates the velocity changes to area changes dA and the stagnation temperature change dT0 or the heat transferred a When Ma 1 subsonic flow the fluid will accelerate if the duck converges dA 0 or the fluid is heated dT0 0 or δq 0 The fluid will decelerate if the duck converges dA 0 or the fluid is cooled dT0 0 or δq 0 b When Ma 1 supersonic flow the fluid will accelerate if the duck diverges dA 0 or the fluid is cooled dT0 0 or δq 0 The fluid will decelerate if the duck converges dA 0 or the fluid is heated dT0 0 or δq 0 preparation If you are a student using this Manual you are using it without permission 1786 17129 A pitot tube measures the difference between the static and stagnation pressures for a subsonic airplane The speed of the airplane and the flight Mach number are to be determined Assumptions 1 Air is an ideal gas with constant specific heat ratio 2 The stagnation process is isentropic Properties The properties of air are R 0287 kJkgK and k 14 Table A2a Analysis The stagnation pressure of air at the specified conditions is 50 8 kPa 20 30 8 0 P P P Then 40 41 2 1 2 0 2 1 Ma 41 1 30 8 50 8 2 1 Ma 1 k k k P P It yields he speed of sound in air at the specified conditions is Ma 0877 T 310 5 ms 1kJkg 0 287 kJkg K240 K 1000 m s 41 2 2 kRT c iscussion Note that the flow velocity can be measured in a simple and accurate way by simply measuring pressure 272 ms 0 877310 5 ms Ma c V Thus D 17130 The mass flow parameter m RT s the Mach number fo 1 Ma 0 is to be plotted AP 0 0 versu r k 12 14 and 16 in the range of l The mass flow rate parameter m RT P A 0 0 Ana ysis can be expressed as 1 2 1 2 1 M k 0 A P 0 2 Ma k k RT m Thu 2 k s Ma k 12 k 14 k 16 00 0 0 0 01 01089 01176 01257 02 02143 02311 02465 03 03128 03365 03582 04 04015 04306 04571 05 06 0 04782 5411 05111 05763 05407 06077 07 05894 06257 06578 7106 10 06485 06847 07164 Discussion Note that the mass flow rate increases with increasing Mach number and specific heat ratio It levels off at Ma 1 and remains constant choked flow 04 Ma 02 k 16 14 12 08 06 10 0 030 045 015 075 060 08 06230 06595 06916 09 06424 06787 0 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1787 17131 Helium gas is accelerated in a nozzle The pressure and temperature of helium at the location where Ma 1 and the ratio of the flow area at this location to the inlet flow area are to be determined Assumptions 1 Helium is an ideal gas with constant specific heats 2 Flow through the nozzle is steady onedimensional and isentropic Properties The properties of helium are R 20769 kJkgK cp 51926 kJkgK and k 1667 Table A2a Analysis The properties of the fluid at the location where Ma 1 are the critical properties denoted by superscript We first determine the stagnation temperature and pressure which remain constant throughout the nozzle since the flow is isentropic 501 4 K s m 1000 1kJkg 5 1926 kJkg K 2 120 ms 500 K 2 2 2 2 2 0 p i i c V T T i PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 120 ms Ma 1 He and 0 806 MPa 500 K MPa 80 0 Pi Ti P 5014 K 1 1 667 1 667 1 0 k k T The Mach number at the nozzle exit is given to be Ma 1 Therefore the properties at the nozzle exit are the critical roperties determined from p 376 K 16671 2 501 4 K 1 2 T0 k T 0393 MPa 1 667 1 667 1 k k 1 0 1667 1 2 0 806 MPa 1 2 k P P The speed of sound and the Mach number at the nozzle inlet are 1316 ms 1 kJkg s K500 K 1000 m 1 667 2 0769 k i i kRT c Jkg 2 2 0 0912 1316 ms 120 ms Ma i i i c V The ratio of the entrancetothroat area is 20 6 2 1 1 667 0912 0 1 667 1 0 0912 1 2 1 1 Ma 1 2 1 0 667 667 2 2 1 1 2 2 k k i i k A Then the ratio of the throat area to the entrance area becomes 2 1 Ma 2 i k A A Ai 1 6 20 0161 Discussion The compressible flow functions are essential tools when determining the proper shape of the compressible flow duct preparation If you are a student using this Manual you are using it without permission 1788 17132 Helium gas enters a nozzle with negligible velocity and is accelerated in a nozzle The pressure and temperature of helium at the location where Ma 1 and the ratio of the flow area at this location to the inlet flow area are to be determined Assumptions 1 Helium is an ideal gas with constant specific heats 2 Flow through the nozzle is steady onedimensional and isentropic 3 The entrance velocity is negligible Properties The properties of helium are R 20769 kJkgK cp 51926 kJkgK and k 1667 Table A2a Analysis We treat helium as an ideal gas with k 1667 The properties of the fluid at the location where Ma 1 are the critical properties denoted by superscript The stagnation temperature and pressure in this case are identical to the inlet temperature and pressure since the inlet velocity is negligible They remain constant throughout the nozzle since the flow is isentropic T0 Ti 500 K P0 Pi 08 MPa The Mach number at the nozzle exit is given to be Ma 1 Therefore the properties at the nozzle exit are the critical properties determined from i PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Vi 0 Ma 1 He 375 K 16671 2 500 K 1 2 T0 k T 0390 MPa 0 MPa 1667 1 80 1 k P P 1 1 667 1 667 1 2 2 k k The ratio of the nozzle inlet area to the throat area is determined from 1 1 2 2 2 1 Ma i i k A 1 Ma 1 2 1 k k i k A ut the M h number at the nozzle inlet is Ma 0 since Vi 0 Thus the ratio of the throat area to the nozzle inlet area is B ac A 1 0 iA The compressible flow functions are essential tools when determining the proper shape of the compressible flow duct Discussion preparation If you are a student using this Manual you are using it without permission 1789 17133 Air enters a converging nozzle The mass flow rate the exit velocity the exit Mach number and the exit pressurestagnation pressure ratio versus the back pressurestagnation pressure ratio for a specified back pressure range are to be calculated and plotted Assumptions 1 Air is an ideal gas with constant specific heats at room temperature 2 Flow through the nozzle is steady onedimensional and isentropic Properties The properties of air at room temperature are R 0287 kJkgK cp 1005 kJkgK and k 14 Table A2a Analysis The stagnation properties remain constant throughout the nozzle since the flow is isentropic They are determined from 526 3 K m s 1000 1kJkg 1 005 kJkg K 2 230 ms 500 K 2 2 2 2 2 0 p i i c V T T and PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 230 ms Air i e 1077 kPa 500 K 900 kPa 5263 K 0 Pi Ti P 1 41 41 1 0 k k T The critical pressure is determined to be 569 0 kPa 1077 kPa 14 1 1 0 k P P 2 2 1 k k 40 41 Then the ne throat w Pe P 5690 kPa for Pb 5690 kPa choked flow pressure will not affect the flow when 100 Pb 90 kPa For a specified exit pressure Pe the temperature the velocity and the mass flow rate can be determined from Temperature pressure at the exit pla ill be Pe Pb for Pb 5690 kPa Thus the back 56 41 40 e 1 0 0 1077 P 526 3 K k k e e P P T T Velocity 1kJkg 1000 m s 2 1 005 kJkg K5263 2 2 2 0 e e p T T T c V Speed of sound 1kJkg 0 287 kJkg K 1000 m s 41 2 2 e e kRT c ach number e e e V c Ma M Density e e e P P ρ e e T RT 0 287 kPa m kg K 3 ρ ρ he EES solution and the results are given below Given Pi900 kPa Ti500 K Veli230 ms Ae10E4 m2 m 0 001 2 e e e e e V V A m Mass flow rate T preparation If you are a student using this Manual you are using it without permission 1790 Properties Cp1005 k14 R0287 Analysis T0TiVeli22CpConvertm2s2 kJkg P0PiT0Tikk1 PstarP02k1kk1 TeT0PeP0k1k Velesqrt2CpT0TeConvertkJkg m2s2 CesqrtkRTeConvertkJkg m2s2 MeVeleCe rhoePeRTe mdotrhoeVeleAe RatioPePeP0 RatioPbPbP0 PePb for Pb Pstar and PePstar for Pb Pstar Pb kPa RatioPb Pe kPa RatioPe Te K Ve ms Me ρe kgm3 m kg s 900 800 700 600 569 500 400 300 200 100 0836 0743 0650 0557 0528 0464 0371 0279 0186 0093 900 800 700 600 5459 5459 5459 5459 5459 5459 0836 0743 0650 0557 0507 0507 0507 0507 0507 0507 500 4835 4654 4453 4334 4334 4334 4334 4334 4334 230 2935 350 4035 4321 4321 4321 4321 4321 4321 051 067 081 095 104 104 104 104 104 104 6272 5766 5241 4695 4388 4388 4388 4388 4388 4388 1443 1692 1835 1894 1896 1896 1896 1896 1896 1896 0 01 02 03 04 05 06 07 08 09 045 05 055 06 065 07 075 08 085 09 Pb Pe PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1791 100 200 300 400 500 600 700 800 900 200 240 280 320 360 400 440 Pb Vele 100 200 300 400 500 600 700 800 900 14 15 16 17 18 19 2 Pb m kgs PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1792 17134 Steam enters a converging nozzle The exit pressure the exit velocity and the mass flow rate versus the back pressure for a specified back pressure range are to be plotted Assumptions 1 Steam is to be treated as an ideal gas with constant specific heats 2 Flow through the nozzle is steady one dimensional and isentropic 3 The nozzle is adiabatic Properties The ideal gas properties of steam are given to be R 0462 kJkgK cp 1872 kJkgK and k 13 Analysis The stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible Since the flow is isentropic they remain constant throughout the nozzle P0 Pi 6 MPa T0 Ti 700 K The critical pressure is determined from to be i e STEAM Vi 0 3 274 MPa 6 MPa 13 1 1 0 k P P 2 2 30 31 1 k k b e he temperature the velocity and the mass flow rom Temperature Then the pressure at the exit plane throat will be Pe Pb for Pb 3274 MPa P Pe C P V Pe P 3274 MPa for Pb 3274 MPa choked flow Thus the back pressure will not affect the flow when 3 P 3274 MPa For a specified exit pressure P t rate can be determined f e 31 30 1 0 0 6 700 K e k k e e P P P T T Velocity 1 kJkg 1000 m s 2 1 872 kJkg K700 2 2 2 0 e e p T T T c V PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course max m P m 3274 3 6 MPa Density e e e e T P RT P 0 462 kPa m kg K 3 ass flow rate 0 e A The res f the calculations can bulated follows Pa Pa 3 e ρ M 008 m 2 e 0 e e ρ V eV m ρ ults o be ta as Pb M Pe M Te K Ve ms ρe kgm m kg s 60 60 700 0 1855 0 55 55 6861 2281 1735 3166 50 50 6712 3284 1612 4235 45 45 6550 4105 1487 4883 40 40 6375 4837 1358 5255 35 35 6181 5537 1226 5431 3274 3274 6087 5847 1164 5445 30 3274 6087 5847 1164 5445 preparation If you are a student using this Manual you are using it without permission 1793 17135 An expression for the ratio of the stagnation pressure after a shock wave to the static pressure before the shock wave as a function of k and the Mach number upstream of the shock wave is to be found Analysis The relation between P1 and P2 is 1 2 2 1 1 1 Ma 1 k P P k P 2 2 2 1 2 2 Ma Ma 1 Ma 1 k k P ubstituting this into the isentropic relation 2 S 1 2 2 1 02 1 Ma 2 1 k k k P P Then 1 2 2 2 2 2 1 1 02 2 1 Ma 1 Ma 1 Ma 1 k k k k k P P where 1 1 Ma 2 1 2 Ma Ma 2 2 2 1 2 2 k k k Substituting 1 2 1 2 1 2 1 2 1 2 1 1 02 1 1 Ma 2 1 2 1 Ma 1 3 1 Ma 1 Ma 2 Ma 1 k k k k k k k k k k k P P PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1794 17136 Nitrogen entering a convergingdiverging nozzle experiences a normal shock The pressure temperature velocity Mach number and stagnation pressure downstream of the shock are to be determined The results are to be compared to those of air under the same conditions Assumptions 1 Nitrogen is an ideal gas with constant specific heats 2 Flow through the nozzle is steady onedimensional and isentropic 3 The nozzle is adiabatic Properties The properties of nitrogen are R 02968 kJkgK and k 14 Table A2a Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible Assuming the flow before the shock to be isentropic shock wave PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course N2 1 i 2 K 300 kPa 700 01 01 i i T T P P Vi 0 Then Ma1 30 107 1 K 1413 2 2 300 K 1 Ma 2 2 2 2 1 01 1 k T T and 1906 kPa 300 700 kPa 01 1 01 1 T P P 1071 40 41 1 k k T he fluid properties after the shock denoted by subscript 2 are related to those before the shock through the functions sted in T le A33 For Ma 30 we read T li ab 1 2 679 10333 and 0 32834 Ma 1 2 1 2 01 02 2 T P 04752 P Then the stagnation pressure P02 static pressure P2 and static temperature T2 are determined to be 2 679107 1 K 2 679 1 2 1 T T T P P kPa 197 kPa 230 103331906 kPa 333 10 0 32834700 kPa 32834 0 2 01 02 P P P P K 287 The velocity after the shock can be determined from V2 Ma2c2 where c2 is the speed of sound at the exit conditions after the shock 164 ms 1kJkg 0 2968 kJkg K287 K 1000 m s 41 0 4752 Ma Ma 2 2 2 2 2 2 2 kRT c V Discussion For air at specified conditions k 14 same as nitrogen and R 0287 kJkgK Thus the only quantity which will be different in the case of air is the velocity after the normal shock which happens to be 1613 ms preparation If you are a student using this Manual you are using it without permission 1795 17137 The diffuser of an aircraft is considered The static pressure rise across the diffuser and the exit area are to be determined Assumptions 1 Air is an ideal gas with constant specific heats at room temperature 2 Flow through the diffuser is steady onedimensional and isentropic 3 The diffuser is adiabatic Ma2 025 Diffuser 2 AIR Ma1 07 1 Properties Air properties at room temperature are R 0287 kJkgK cp 1005 kJkgK and k 14 Table A2a Analysis The inlet velocity is 218 6 ms 1kJkg 0 287 kJkg K2427 K 1000 m s 41 70 Ma 2 2 1 1 1 1 1 kRT M c V Then the stagnation temperature and pressure at the diffuser inlet become 266 5 K m s 1000 1kJkg 2 1 005 kJkg K 2186 ms 242 7 2 2 2 2 2 1 1 01 cp V T T kPa 07 41 1 kPa 2427 K 1 1 01 T P P 5 2665 K 1 41 41 1 01 k k T or an adiabatic diffuser the energy equation reduces to h01 h02 Noting that h cpT and the specific heats are assumed to e constant we have he isentropic relation between states 1 and 02 gives F b 266 5 K T T T 0 02 01 T 5559 kPa 2427 K 41 1 kPa 2665 K 1 02 1 02 T P P 1 41 41 1 k k T xit velocity can be expressed as The e 2 2 2 2 2 2 2 2 2 5 01 1kJkg 1000 m s 0 287 kJkg K 41 0 25 Ma Ma T T kRT c V 263 2 K m s 1000 1kJkg K Thus 2 1 005 kJkg 2 02 2 c p m s 501 266 5 2 2 2 2 2 2 2 22 T T V T T Then the static exit pressure becomes 5323 kPa 2665 K 5559 kPa 2632 K 1 41 41 1 02 2 02 2 k k T T P P Thus th ta e s tic pressure rise across the diffuser is Also 1213 kPa 41 1 5323 1 2 P P P 3 3 2 2 2 0 7047 kgm 0 287 kPa m kg K2632 K 5323 kPa P RT ρ 81 3 ms 5 01 263 2 6 01 2 2 T V 0524 m2 0 7047 kgm 81 3 ms kgs 30 3 2 2 2 V m A ρ Thus Discussion The pressure rise in actual diffusers will be lower because of the irreversibilities However flow through well designed diffusers is very nearly isentropic PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1796 17138 Helium gas is accelerated in a nozzle isentropically For a specified mass flow rate the throat and exit areas of the nozzle are to be determined Assumptions 1 Helium is an ideal gas with constant specific heats 2 Flow through the nozzle is steady onedimensional and isentropic 3 The nozzle is adiabatic Properties The properties of helium are R 20769 kJkgK cp 51926 kJkgK k 1667 Table A2a Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible T T P P 01 1 01 1 500 10 K MPa PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 01 500 K he critical pressure and temperature are determined fro He The flow is assumed to be isentropic thus the stagnation temperature and pressure remain constant throughout the nozzle 1 2 Vi 0 P P 02 01 10 MPa T T 02 T m 375 0 K 1667 1 2 500 K 1 2 0 k T T 0 487 MPa 1667 1 2 MPa 01 1 2 1 1 667 1 667 1 0 k k k P P 3 3 0 625 kgm 2 0769 kPa m kg K375 K ρ RT 487 kPa P 1139 4 ms 1 kJkg s 1 667 2 0769 kJkg K375 K 1000 m 2 2 kRT c V hus the throat area is T 351 cm2 2 4 3 m 3 51 10 0 25 kgs V m A ρ 0 625 kgm 1139 4 ms t the nozzle exit the pressure is P2 01 MPa Then the other properties at the zzle exit are determined to be A no 1 667 0 667 2 2 1 2 2 2 0 Ma 2 1 667 1 1 01 MPa MPa 01 Ma 2 1 1 k k k P P ields Ma2 2130 which is greater than 1 Therefore the nozzle must be convergingdiverging It y 199 0 K 2 13 1 1 667 2 2 500 K 1 Ma 2 0 2 T T 2 2 2 2 k 3 3 2 2 2 0 242 kgm 2 0769 kPa m kg K199 K RT ρ 100 kPa P 1768 0 ms 1 kJkg s 1 667 2 0769 kJkg K199 K 1000 m 2 13 Ma Ma 2 2 2 2 2 2 2 kRT c V Thus the exit area is 584 cm2 2 4 3 2 2 2 m 5 84 10 0 242 kgm 1768 ms 0 25 kgs V m A ρ Discussion Flow areas in actual nozzles would be somewhat larger to accommodate the irreversibilities preparation If you are a student using this Manual you are using it without permission 1797 17139E Helium gas is accelerated in a nozzle For a specified mass flow rate the throat and exit areas of the nozzle are to be determined for the cases of isentropic and 97 efficient nozzles Assumptions 1 Helium is an ideal gas with constant specific heats 2 Flow through the nozzle is steady onedimensional and isentropic 3 The nozzle is adiabatic Properties The properties of helium are R 04961 BtulbmR 26809 psiaft3lbmR cp 125 BtulbmR and k 1667 Table A2Ea Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible T T P P 01 1 01 1 900 150 R psia PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 01 900 R he critical pressure and temperature are determined from He The flow is assumed to be isentropic thus the stagnation temperature and pressure remain constant throughout the nozzle 1 2 T02 T Vi 0 P P 02 01 150 psia T 674 9 R 1667 1 2 900 R 1 2 0 k T T 73 1 psia 1667 1 2 150 psia 1 2 1 1 667 1 667 1 0 k k k P P 3 3 0 0404 1bmft 2 6809 psia ft lbm R6749 R 73 1 psia RT P ρ 3738 fts 1Btu1bm 1 667 0 4961 Btulbm R6749 R 25037 ft s kRT c V 2 2 and 000132 ft2 1bms 20 3 V m A ρ 0 0404 1bmft 3738 fts t the nozzle exit the pressure is P2 15 psia Then the other properties at the nozzle exit are determined to be A 1 667 0 667 2 2 1 2 2 2 0 Ma 2 1 p Ma 2 1 667 1 1 15 psia 150 psia 1 k k k p elds Ma2 2130 which is greater than 1 Therefore the nozzle must be convergingdiverging It yi 358 1 R 2 13 1 1 667 2 2 900 R 1 Ma 2 0 2 T T 2 2 2 2 k 3 3 2 2 2 0 0156 1bmft R 2 6809 psia ft lbm R3581 RT ρ 15 psia P 5800 fts 1Btu1bm s 1 667 0 4961 Btulbm R3581 R 25037 ft 2 13 Ma Ma 2 2 2 2 2 2 2 kRT c V Thus the exit area is 000221 ft2 0 0156 lbmft 5800 fts lbms 20 3 2 2 2 V m A ρ Discussion Flow areas in actual nozzles would be somewhat larger to accommodate the irreversibilities preparation If you are a student using this Manual you are using it without permission 1798 17140 Using the compressible flow relations the onedimensional compressible flow functions are to be evaluated and tabulated as in Table A32 for an ideal gas with k 1667 Properties The specific heat ratio of the ideal gas is given to be k 1667 Analysis The compressible flow functions listed below are expressed in EES and the results are tabulated 2 1 Ma 2 1 Ma Ma k k 1 1 50 2 Ma 2 1 1 1 2 Ma 1 k k k k A A 1 2 0 Ma 2 1 1 k k k P P 1 1 2 0 Ma 2 1 1 k k ρ ρ 1 a2 1 k T 0 2 T M 1 k P 1667 P0 k1M2 kk1 TT 1k1M 2 DD k1M 1k1 Mc SQRTk AA 2k1 k1 5k M 1 01 2 2 01 22 12k1 rM M2 cr 105 M20 1k1 Ma Ma AA PP 0 ρ ρ 0 TT0 00 01 02 03 04 05 06 07 08 09 10 12 14 16 18 20 22 24 26 28 30 50 16643 17007 17318 18895 19996 980 22893 26222 29990 97920 86 00524 00403 00313 00038 0 05 01705 01457 01251 00351 0 24 03073 02767 02499 01071 0 0 01153 02294 03413 04501 05547 06547 07494 08386 09222 10000 11390 12572 13570 14411 15117 15713 16216 56624 28879 19891 15602 13203 11760 10875 10351 10081 10000 10267 10983 12075 13519 15311 17459 19 10000 09917 09674 09288 08782 08186 07532 06850 06166 05501 04871 03752 02845 02138 01603 01202 00906 006 10000 09950 09803 09566 09250 08869 08437 07970 07482 06987 06495 05554 04704 03964 03334 02806 02368 020 10000 09967 09868 09709 09493 09230 08928 08595 08241 07873 07499 06756 06047 05394 04806 04284 03825 034 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1799 17141 Using the normal shock relations the normal shock functions are to be evaluated and tabulated as in Table A 33 for an ideal gas with k 1667 Properties The specific heat ratio of the ideal gas is given to be k 1667 Analysis The normal shock relations listed below are expressed in EES and the results are tabulated 1 Ma 2 2 1 Ma a 2 2 1 2 k M 1 k k 1 1 Ma 2 Ma 1 Ma 1 2 1 2 2 2 1 1 2 k k k k k P P 1 Ma 2 1 Ma 2 2 2 1 2 k 2 1 k T T 2 1 2 1 2 1 1 2 1 2 1 2 1 Ma 2 1 Ma V V k k T T P P ρ ρ 1 2 1 2 1 2 2 2 1 01 02 2 1 Ma 1 2 1 1 Ma Ma Ma k k k k P P 2 2 1 2 2 2 1 1 02 Ma 1 2 1 Ma Ma 1 1 k k k P P k k x2kk11 121My2k12 k1k1 0yPx1kMx21My2k12kk11kMy2 P01 k1667 MySQRTMx22k12M PyPx1kMx21kMy2 TyTx1Mx2k RyRxPyPxTyTx P0yP0xMxMy1My2k121Mx2k1205 P Ma1 Ma2 P2P1 ρ2ρ1 T2T1 P02 P02P1 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 40 50 04473 022 39985 834 98 0 723 10000 09131 08462 07934 07508 07157 06864 06618 06407 06227 06070 05933 05814 05708 05614 05530 05455 05388 05327 05273 05223 04905 04753 10000 12625 15500 18626 22001 25626 29501 33627 38002 42627 47503 52628 58004 63629 69504 75630 82005 88631 95506 102632 110007 197514 310 10000 11496 12972 14413 15805 17141 18415 19624 20766 21842 22853 23802 24689 25520 26296 27021 27699 28332 28923 29476 29993 33674 35703 10000 10982 11949 12923 13920 14950 16020 17135 18300 19516 20786 22111 23493 24933 26432 27989 29606 31283 33021 34819 36678 58654 86 1 0999 09933 09813 09626 0938 09085 08752 08392 08016 0763 07243 06861 06486 06124 05775 05442 05125 04824 04541 04274 02374 013 20530 23308 26473 29990 33838 38007 42488 47278 52371 57767 63462 69457 75749 82339 89225 96407 103885 111659 119728 128091 136750 239530 371 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 17100 17142 The critical temperature pressure and density of an equimolar mixture of oxygen and nitrogen for specified stagnation properties are to be determined Assumptions Both oxygen and nitrogen are ideal gases with constant specific heats at room temperature Properties The specific heat ratio and molar mass are k 1395 and M 32 kgkmol for oxygen and k 14 and M 28 kgkmol for nitrogen Tables A1 and A2 Analysis The gas constant of the mixture is 30 kgkmol 28 50 32 50 2 2 2 2 N N O O M y M y M m 02771 kJkg K 30 kgkmol 8314 kJkmol K m u m M R R The specific heat ratio is 14 for nitrogen and nearly 14 for oxygen Therefore the specific heat ratio of the mixture is also 14 Then the critical temperature pressure and density of the mixture become 5000 K 1 14 2 600 K 1 2 T0 k T 1585 kPa 1 41 41 1 0 14 1 2 300 kPa 1 2 k k k P P PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 1144 kgm3 158 5 kPa 3 RT P ρ 0 2771 kPa m kg K500 K Discussion If the specific heat ratios k of the two gases were different then we would need to determine the k of the mixture f o k cpmcvm where the specific heats of the mixture are determined from p c r m 2 2 2 2 2 2 2 2 2 2 N N N O O O N N O O mf mf m p m p p p m M M y c M M y c c c 2 2 2 2 2 2 2 2 2 2 N N N O O O N N O O mf mf v v v v v c M M y c M M y c c c m m m where mf is the mass fraction and y is the mole fraction In this case it would give 0 9745 kJkgK 1 039 2830 50 0 918 3230 50 c m p 0 6977 kJkgK 0 743 28 30 50 0 658 32 30 50 vc m and k 0974506977 1397 preparation If you are a student using this Manual you are using it without permission 17101 17143 Using EES or other software the shape of a convergingdiverging nozzle is to be determined for specified flow rate and stagnation conditions The nozzle and the Mach number are to be plotted Assumptions 1 Air is an ideal gas with constant specific heats 2 Flow through the nozzle is steady onedimensional and isentropic 3 The nozzle is adiabatic Properties The specific heat ratio of air at room temperature is 14 Table A2a Analysis The problem is solved using EES and the results are tabulated and plotted below k14 Cp1005 kJkgK R0287 kJkgK P01400 kPa T0200273 K m3 kgs rho0P0RT0 rhoPRT TT0PP0k1k VSQRT2CpT0T1000 0 200 400 600 800 1000 1200 1400 0 05 1 15 2 25 P kPa Ma AmrhoV10000 cm2 CSQRTkRT1000 MaVC Pressure P kPa Flow area A cm2 Mach number Ma 1400 1350 1300 1250 1200 1150 1100 1050 1000 950 900 850 800 750 700 650 600 550 500 450 400 350 300 250 200 150 100 301 217 181 160 147 137 130 125 122 119 117 116 115 115 116 118 120 123 128 133 140 150 164 183 214 270 0 0229 0327 0406 0475 0538 0597 0655 0710 0766 0820 0876 0931 0988 1047 1107 1171 1237 1308 1384 1467 1559 1663 1784 1929 2114 2373 0 200 400 600 800 1000 1200 1400 10 15 20 25 30 35 40 45 50 P kPa Flow area A cm 2 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 17102 17144 Using the compressible flow relations the onedimensional compressible flow functions are to be evaluated Analysis The compressible flow functions listed below are expressed in EES and the results are tabulated and tabulated as in Table A32 for air Properties The specific heat ratio is given to be k 14 for air 2 1 Ma 2 1 Ma Ma k k 1 1 50 2 Ma 2 1 1 1 2 Ma 1 k k k k A A 1 2 0 Ma 2 1 1 k k k P P 1 1 2 0 Ma 2 1 1 k k ρ ρ 1 a2 1 k T 0 M 2 1 T P01k1M22kk1 T011k1M22 DD k1M 1k1 Mc SQRTk k1 AA 2k1 1 5k Air k14 P T 01 22 rM 12 M2 cr 105k M20 1k1M Ma Ma AA PP0 ρρ0 TT0 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 23591 23689 23772 23843 23905 1901094 2510862 3271893 4211314 5359375 00001 00001 00000 00000 00000 00014 00011 00008 00006 00005 00725 00647 00581 00525 00476 10000 13646 16330 18257 19640 20642 21381 21936 22361 22691 22953 23163 23333 23474 10000 11762 16875 26367 42346 67896 107188 165622 250000 368690 531798 751343 1041429 1418415 05283 02724 01278 00585 00272 00131 00066 00035 00019 00011 00006 00004 00002 00002 06339 03950 02300 01317 00762 00452 00277 00174 00113 00076 00052 00036 00026 00019 08333 06897 05556 04444 03571 02899 02381 01980 01667 01418 01220 01058 00926 00816 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 17103 17145 Using the compressible flow relations the onedimensional compressible flow functions are to be evaluated Analysis The compressible flow functions listed below are expressed in EES and the results are tabulated and tabulated as in Table A32 for methane Properties The specific heat ratio is given to be k 13 for methane 2 1 Ma 2 1 Ma Ma k k 1 1 50 2 Ma 2 1 1 1 2 Ma 1 k k k k A A 1 2 0 Ma 2 1 1 k k k P P 1 1 2 0 Ma 2 1 1 k k ρ ρ 1 2 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 2 T 1 Ma 1 k P01k1M22kk1 T011k1M22 DD k1M 1k1 Mc SQRTk k1 AA 2k1 1 5k T Methane k13 P T 01 22 rM 12 M2 cr 105k M20 1k1M Ma Ma AA PP0 ρρ0 TT0 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 26350 26493 26615 26719 26810 6231235 8955077 12656040 17612133 24161184 00000 00000 00000 00000 00000 00004 00003 00002 00001 00001 00943 00845 00760 00688 00625 10000 13909 16956 19261 20986 22282 23263 24016 24602 25064 25434 25733 25978 26181 10000 11895 17732 29545 51598 91098 159441 273870 459565 752197 1200965 1872173 2853372 4258095 05457 02836 01305 00569 00247 00109 00050 00024 00012 00006 00003 00002 00001 00001 06276 03793 02087 01103 00580 00309 00169 00095 00056 00033 00021 00013 00008 00006 08696 07477 06250 05161 04255 03524 02941 02477 02105 01806 01563 01363 01198 01060 preparation If you are a student using this Manual you are using it without permission 17104 17146 Using the normal shock relations the normal shock functions are to be evaluated and tabulated as in Table Analysis The normal shock relations listed below are expressed in EES and the results are tabulated A33 for air Properties The specific heat ratio is given to be k 14 for air 1 Ma 2 2 1 Ma Ma 2 1 2 1 2 k k k 1 1 Ma 2 Ma 1 Ma 1 2 1 2 2 2 1 1 2 k k k k k P P 1 Ma 2 1 Ma 2 2 2 2 1 1 2 k k T T 2 1 2 1 2 1 1 2 1 2 1 2 1 Ma 2 1 Ma V V k k T T P P ρ ρ 1 2 1 2 1 2 2 2 1 01 02 2 1 Ma 1 2 1 1 Ma Ma Ma k k k k P P 2 2 1 2 2 2 1 1 02 Ma 1 2 1 Ma Ma 1 1 k k k P P k k Air 1 k1k1 0yPx1kMx21My2k12kk11kMy2 k14 MySQRTMx22k12Mx2kk1 PyPx1kMx21kMy2 TyTx1Mx2k121My2k12 RyRxPyPxTyTx P0yP0xMxMy1My2k121Mx2k1205 P Ma1 Ma2 P P 2 1 ρ2ρ1 T T 2 1 P02P01 P P 02 1 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 03876 1165000 57143 203875 0003045 1292170 10000 07011 05774 05130 04752 04512 04350 04236 04152 04090 04042 04004 03974 03949 03929 03912 03898 03886 10000 24583 45000 71250 103333 141250 185000 234583 290000 351250 418333 491250 570000 654583 745000 841250 943333 1051250 10000 18621 26667 33333 38571 42609 45714 48119 50000 51489 52683 53651 54444 55102 55652 56117 56512 56850 10000 13202 16875 21375 26790 33151 40469 48751 58000 68218 79406 91564 104694 118795 133867 149911 166927 184915 1 09298 07209 0499 03283 02129 01388 00917 006172 004236 002965 002115 001535 001133 0008488 0006449 0004964 0003866 18929 34133 56404 85261 120610 162420 210681 265387 326535 394124 468152 548620 635526 728871 828655 934876 1047536 1166634 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 17105 17147 Using the normal shock relations the normal shock functions are to be evaluated and tabulated as in Table A33 for methane Properties The specific heat ratio is given to be k 13 for methane Analysis The normal shock relations listed below are expressed in EES and the results are tabulated 1 Ma 2 2 1 Ma a 2 1 2 k M 2 1 k k 1 1 Ma 2 Ma 1 Ma 1 2 1 2 2 2 1 1 2 k k k k k P P 1 Ma 2 1 Ma 2 2 2 1 2 k 2 1 k T T 2 1 2 1 2 1 1 2 1 2 1 2 1 Ma 2 1 Ma V V k k T T P P ρ ρ 1 2 1 2 1 2 2 2 1 01 02 2 1 Ma 1 2 1 1 Ma Ma Ma k k k k P P 2 2 1 2 2 2 1 1 02 Ma 1 2 1 Ma Ma 1 1 k k k P P k k ethane x2kk11 121My2k12 k1k1 0yPx1kMx21My2k12kk11kMy2 P01 M k13 MySQRTMx22k12M PyPx1kMx21kMy2 TyTx1Mx2k RyRxPyPxTyTx P0yP0xMxMy1My2k121Mx2k1205 P Ma1 Ma2 P2P1 ρ2ρ1 T2T1 P02 P02P1 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 03510 1129130 71875 157096 0000740 122239 10000 06942 05629 04929 04511 04241 04058 03927 03832 03760 03704 03660 03625 03596 03573 03553 03536 03522 10000 24130 43913 69348 100435 137174 179565 227609 281304 340652 405652 476304 552609 634565 722174 815435 914348 1018913 10000 19346 28750 37097 44043 49648 54118 57678 60526 62822 64688 66218 67485 68543 69434 70190 70837 71393 10000 12473 15274 18694 22804 27630 33181 39462 46476 54225 62710 71930 81886 92579 104009 116175 129079 142719 1 09261 07006 0461 02822 01677 009933 005939 003613 002243 001422 0009218 0006098 0004114 0002827 0001977 0001404 0001012 18324 32654 53700 80983 114409 153948 199589 251325 309155 373076 443087 519188 601379 689658 784027 884485 991032 110367 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 17106 17148 Air flowing at a supersonic velocity in a duct is accelerated by cooling For a specified exit Mach number the rate of heat transfer is to be determined Assumptions The assumptions associated with Rayleigh flow ie steady onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects are valid Properties We take the properties of air to be k 14 cp 1005 kJkgK and R 0287 kJkgK Table A2a Analysis Knowing stagnation properties the static properties are determined to be 271 7 K 21 2 14 1 350 K 1 2 Ma 1 1 1 2 1 2 1 01 1 k T T 9897 kPa 21 2 14 1 240 kPa 1 2 Ma 1 1 40 41 2 1 2 1 01 1 k k k P P 3 1 1 1 269 kgm 9897 kPa ρ1 P 0287 kJkgK2717 K RT hen the inlet velocity and the mass flow rate become T 330 4 ms 1 kJkg s 0 287 kJkg K2717 K 1000 m 41 2 2 1 1 kRT c 396 5 ms 21 330 4 ms Ma 1 1 1 c V 269 kgm 0 1 3 1 air π ρ A m c 1581 kgs 20 m 4330 4 ms 2 1 1 V The Ray ctio ng to the inlet and exit Mach numbers are Table A34 M T02T0 07934 e is determined to be leigh flow fun ns T0T0 correspondi Ma1 18 T01T0 09787 a2 2 Then the exit stagnation temperatur PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 0 8107 0 9787 0 01 01 T T T 283 7 K 0 8107350 K 0 8107 01 02 0 7934 0 02 02 T T T T T Finally the rate of heat transfer is 1053 kW 350 K kgs1005 kJkg K283 7 1581 01 02 air T T c m Q p Discussion The negative sign confirms that the gas needs to be cooled in order to be accelerated Also it can be shown that the thermodynamic temperature drops to 158 K at the exit which is extremely low Therefore the duct may need to be heavily insulated to maintain indicated flow conditions Q P01 240 kPa T01 350 K Ma1 12 Ma2 2 preparation If you are a student using this Manual you are using it without permission 17107 17149 Air flowing at a subsonic velocity in a duct is accelerated by heating The highest rate of heat transfer without affecting the inlet conditions is to be determined Assumptions 1 The assumptions associated with Rayleigh flow ie steady onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects are valid 2 Inlet conditions and thus the mass flow rate remain constant Properties We take the properties of air to be k 14 cp 1005 kJkgK and R 0287 kJkgK Table A2a PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course Analysis Heat transfer will stop when the flow is choked and thus Ma2 V2c2 1 The inlet density and stagnation temperature are 3 1 1 1 4 259 kgm 0287 kJkgK45 0 K 550 kPa RT P ρ 458 1 K 30 2 2 1 1 01 14 1 450 K 1 1 Ma 1 2 2 k T T Then the inlet velocity and the mass flow rate become Q Ma2 1 P1 550 kPa T1 450 K Ma1 03 425 2 ms 1kJkg 0 287 kJkg K450 K 1000 m s 41 2 2 1 1 kRT c The Ray ding to the inlet and exit Mach numbers are T nce Ma2 1 127 6 ms 30 425 2 ms Ma 1 1 1 c V 3 477 kgs 4 259 kgm 008 008 m 127 6 ms 2 3 1 1 1 air A V m c ρ leigh flow functions correspon 02T0 1 si 0 3469 30 41 1 30 1 41 2 30 1 41 Ma 1 1 Ma 2 1 Ma 2 2 2 2 2 2 2 1 1 2 1 0 01 k k k T T Ther efore 0 3469 1 0 01 1 0 0 02 02 T T T T T T 1320 7 K 0 3469 458 1 K 0 3469 01 02 T T Then the rate of heat transfer becomes K 1 458 3 477 kgs 1 005 kJkg K1320 7 T T c m Q 3014 kW 01 02 air p Discussion It can also be shown that T2 1101 K which is the highest thermodynamic temperature that can be attained under stated conditions If more heat is transferred the additional temperature rise will cause the mass flow rate to decrease We can also solve this problem using the Rayleigh function values listed in Table A34 preparation If you are a student using this Manual you are using it without permission 17108 17150 Helium flowing at a subsonic velocity in a duct is accelerated by heating The highest rate of heat transfer without affecting the inlet conditions is to be determined Assumptions 1 The assumptions associated with Rayleigh flow ie steady onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects are valid 2 Inlet conditions and thus the mass flow rate remain constant Properties We take the properties of helium to be k 1667 cp 5193 kJkgK and R 2077 kJkgK Table A2a Q Ma2 1 P1 550 kPa T1 450 K Ma1 03 Analysis Heat transfer will stop when the flow is choked and thus Ma2 V2c2 1 The inlet density and stagnation temperature are 3 1 1 1 0 5885 kgm 2077 kJkgK45 0 K 550 kPa RT P ρ 463 5 K 30 2 450 K 1 2 Ma 1 1 1 01 T T 1667 1 1 2 2 k hen the i et velocity and the mass flow rate become T nl 1248 ms 1kJkg 667 2 077 kJkg K450 K 1000 m s 1 2 2 1 1 kRT c The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are T0 1 since Ma2 1 374 5 ms 30 1248 ms Ma c V 1 1 1 1 410 kgs 0 5885 kgm 008 008 m 374 5 ms 2 3 1 1 1 air A V m c ρ T02 0 3739 30 1 667 1 30 1 1 667 1 0 33 2 667 1 Ma 1 2 2 1 0 k T 1 Ma 2 2 2 2 2 2 1 2 1 k T Therefore 1 Ma 01 k 0 3739 1 0 01 0 02 1 0 02 T T T T T T 1239 8 K 0 3739 463 5 K 0 3739 01 02 T T Then the rate of heat transfer becomes Discussion It can also be shown that T2 930 K which is the highest thermodynamic temperature that can be attained under stated conditions If more heat is transferred the additional temperature rise will cause the mass flow rate to decrease Also in the solution of this problem we cannot use the values of Table A34 since they are based on k 14 5685 kW 463 5 K 1 410 kgs 5 193 kJkg K1239 8 01 02 air T T c m Q p PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 17109 17151 Air flowing at a subsonic velocity in a duct is accelerated by heating For a specified exit Mach number the heat transfer for a specified exit Mach number as well as the maximum heat transfer are to be determined Assumptions 1 The assumptions associated with Rayleigh flow ie steady onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects are valid 2 Inlet conditions and thus the mass flow rate remain constant Properties We take the properties of air to be k 14 cp 1005 kJkgK and R 0287 kJkgK Table A2a Analysis The inlet Mach number and stagnation temperature are 400 9 ms 1 kJkg s 0 287 kJkg K400 K 1000 m 41 2 2 1 1 kRT c 0 2494 400 9 ms 100 ms Ma 1 1 1 c V PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 405 0 K 0 2494 2 14 1 400 K 1 2 Ma 1 1 2 2 1 1 01 k T T The Ray e inlet and exit Mach numbers are Table A34 Then the exit stagnation temperature and the heat transfer are determ to be q Ma2 08 P1 35 kPa T1 400 K V1 100 ms leigh flow functions corresponding to th Ma1 02494 T01T 02559 Ma2 08 T02T 09639 ined 3 7667 0 2559 01 01 T T T 01 2 0 0 9639 02 2 T T T0 1526 K 3 7667405 0 K 3 7667 T T 405 K g K1526 100 01 02 T T c q p Maximum heat transfer will occur when the flow is choked and thus Ma2 1 and thus T02T 1 Then 1126 kJkg kJk 5 0 2559 01 01 T T T 01 2 0 1 02 02 T T T 1583 K 0 2559 405 0 K 0 2559 T T Discussion This is the maximum heat that can be transferred to the gas without affecting the mass flow rate If more heat is transferred the additional temperature rise will cause the mass flow rate to decrease 1184 kJkg 405 K 1005 kJkg K1583 01 02 max T T c q p preparation If you are a student using this Manual you are using it without permission 17110 17152 Air flowing at sonic conditions in a duct is accelerated by cooling For a specified exit Mach number the amount of heat transfer per unit mass is to be determined Assumptions The assumptions associated with Rayleigh flow ie steady onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects are valid Properties We take the properties of air to be k 14 cp 1005 kJkgK and R 0287 kJkgK Table A2a Analysis Noting that Ma1 1 the inlet stagnation temperature is 600 K 1 2 500 1 14 2 Ma 1 2 1 1 01 T T K 1 1 2 k The Ray io ding to the inlet and Ma1 1 T01T0 1 Ma2 16 T02T0 08842 Then the exit stagnation temperature and heat transfer are determined to be leigh flow funct ns T0T0 correspon exit Mach numbers are Table A34 0 8842 1 0 01 01 T T T 530 5 K 0 8842600 K 0 8842 01 02 0 8842 0 02 02 T T T T T 698 kJkg 600 K 1005 kJkg K530 5 01 02 T T c q p Discussion The negative sign confirms that the gas needs to be cooled in order to be accelerated Also it can be shown that the thermodynamic temperature drops to 351 K at the exit q P01 420 kPa T01 500 K Ma1 1 Ma2 16 PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 17111 17153 Saturated steam enters a convergingdiverging nozzle with a low velocity The throat area exit velocity mass flow rate and exit Mach number are to be determined for isentropic and 90 percent efficient nozzle cases Assumptions 1 Flow through the nozzle is steady and onedimensional 2 The nozzle is adiabatic Analysis a The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible Thus h10 h1 At the inlet 1 Steam 2 t Vi 0 a ηN 1 b η 00 92 4 87816 090 19171 26035 kJkg 1 75 MPa 1 N 1 4 0033 5987 2 3844 090 1 75 MPa 1 1 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course kJkg K fg f x h h h At the exit P2 12 MPa and s2 s2s s1 59874 kJkgK Thus 4 2537 4 kJkg 0 8759 1985 4 33 798 0 8759 4 3058 2 2159 9874 5 3 2 2 2 2 2 2 fg f fg f x h h h x x x s s s hen the e t velocity is determined from the steadyflow energy balance to be fg f x s s s 0 1431 0 001138 0 16326 0 8759 0 001138 2 2 fg f x v v v m kg T xi 2 0 2 2 2 1 2 2 1 2 2 2 2 2 V1 1 V V h h V h h Solving for V2 3637 ms 22603 5 2 h h V 1kJkg kg 1000 m s 25374kJ 2 2 2 1 2 The mass flow rate is determined from m 363 7 ms 635 kgs 25 10 14314 m kg 0 1 1 2 4 3 2 2 2 A V m v The velocity of sound at the exit of the nozzle is determined from 1 2 2 1 1 P r P c v The specific volume of steam at s2 59874 kJkgK and at pressures just below and just above the specified pressure 11 nd 13 MPa are determined to be 01547 nd 01333 m3kg Substituting s s a a 438 9 ms kPa m 1 1000 m s 1 1 1100 kPa 1300 3 2 c 0 1547 kgm 1333 0 2 2 3 ber becomes Then the exit Mach num 0829 438 9 ms 363 7 ms Ma 2 2 2 c V he steam saturated and thus the critical pressure which occurs at the throat is taken to be T is 1 008 MPa 1 75 0 576 0 576 01 P P P t hen at the throat T 5 9874 kJkg K 1 008 MPa and 1 s s P t t hus T preparation If you are a student using this Manual you are using it without permission 17114 Fundamentals of Engineering FE Exam Problems 17154 An aircraft is cruising in still air at 5C at a velocity of 400 ms The air temperature at the nose of the aircraft where stagnation occurs is a 5C b 25C c 55C d 80C e 85C Answer e 85C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values k14 Cp1005 kJkgK T15 C Vel1 400 ms T1stagT1Vel122Cp1000 Some Wrong Solutions with Common Mistakes W1TstagT1 Assuming temperature rise W2TstagVel122Cp1000 Using just the dynamic temperature W3TstagT1Vel12Cp1000 Not using the factor 2 17155 Air is flowing in a wind tunnel at 25C 80 kPa and 250 ms The stagnation pressure at a probe inserted into the flow stream is a 87 kPa b 93 kPa c 113 kPa d 119 kPa e 125 kPa Answer c 113 kPa Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values k14 Cp1005 kJkgK T125 K P180 kPa Vel1 250 ms T1stagT1273Vel122Cp1000 C T1stagT1273P1stagP1k1k Some Wrong Solutions with Common Mistakes T11stagT1W1P1stagP1k1k T11stagT1Vel122Cp1000 Using deg C for temperatures T12stagT1273W2P1stagP1k1k T12stagT1273Vel12Cp1000 Not using the factor 2 T13stagT1273W3P1stagP1k1 T13stagT1273Vel122Cp1000 Using wrong isentropic relation PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 17115 17156 An aircraft is reported to be cruising in still air at 20C and 40 kPa at a Mach number of 086 The velocity of the aircraft is a 91 ms b 220 ms c 186 ms d 280 ms e 378 ms Answer d 280 ms Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values k14 Cp1005 kJkgK R0287 kJkgK T120273 K P140 kPa Mach086 VS1SQRTkRT11000 MachVel1VS1 Some Wrong Solutions with Common Mistakes W1velMachVS2 VS2SQRTkRT1 Not using the factor 1000 W2velVS1Mach Using Mach number relation backwards W3velMachVS3 VS3kRT1 Using wrong relation 17157 Air is flowing in a wind tunnel at 12C and 66 kPa at a velocity of 230 ms The Mach number of the flow is a 054 b 087 c 33 d 036 e 068 Answer e 068 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values k14 Cp1005 kJkgK R0287 kJkgK T112273 K P166 kPa Vel1230 ms VS1SQRTkRT11000 MachVel1VS1 Some Wrong Solutions with Common Mistakes W1MachVel1VS2 VS2SQRTkRT12731000 Using C for temperature W2MachVS1Vel1 Using Mach number relation backwards W3MachVel1VS3 VS3kRT1 Using wrong relation PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 17116 17158 Consider a converging nozzle with a low velocity at the inlet and sonic velocity at the exit plane Now the nozzle exit diameter is reduced by half while the nozzle inlet temperature and pressure are maintained the same The nozzle exit velocity will a remain the same b double c quadruple d go down by half e go down to onefourth Answer a remain the same 17159 Air is approaching a convergingdiverging nozzle with a low velocity at 12C and 200 kPa and it leaves the nozzle at a supersonic velocity The velocity of air at the throat of the nozzle is a 338 ms b 309 ms c 280 ms d 256 ms e 95 ms Answer b 309 ms Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values k14 Cp1005 kJkgK R0287 kJkgK Properties at the inlet T112273 K P1200 kPa Vel10 ms ToT1 since velocity is zero PoP1 Throat properties Tthroat2Tok1 PthroatPo2k1kk1 The velocity at the throat is the velocity of sound VthroatSQRTkRTthroat1000 Some Wrong Solutions with Common Mistakes W1VthroatSQRTkRT11000 Using T1 for temperature W2VthroatSQRTkRT2throat1000 T2throat2To273k1 Using C for temperature W3VthroatkRTthroat Using wrong relation PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 17117 17160 Argon gas is approaching a convergingdiverging nozzle with a low velocity at 20C and 120 kPa and it leaves the nozzle at a supersonic velocity If the crosssectional area of the throat is 0015 m2 the mass flow rate of argon through the nozzle is a 041 kgs b 34 kgs c 53 kgs d 17 kgs e 22 kgs Answer c 53 kgs Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values k1667 Cp05203 kJkgK R02081 kJkgK A0015 m2 Properties at the inlet T120273 K P1120 kPa Vel10 ms ToT1 since velocity is zero PoP1 Throat properties Tthroat2Tok1 PthroatPo2k1kk1 rhothroatPthroatRTthroat The velocity at the throat is the velocity of sound VthroatSQRTkRTthroat1000 mrhothroatAVthroat Some Wrong Solutions with Common Mistakes W1massrhothroatAV1throat V1throatSQRTkRT1throat1000 T1throat2To273k1 Using C for temp W2massrho2throatAVthroat rho2throatP1RT1 Using density at inlet PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course 17118 17161 Carbon dioxide enters a convergingdiverging nozzle at 60 ms 310C and 300 kPa and it leaves the nozzle at a supersonic velocity The velocity of carbon dioxide at the throat of the nozzle is a 125 ms b 225 ms c 312 ms d 353 ms e 377 ms Answer d 353 ms Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values k1289 Cp0846 kJkgK R01889 kJkgK Properties at the inlet T1310273 K P1300 kPa Vel160 ms ToT1Vel122Cp1000 ToT1PoP1k1k Throat properties Tthroat2Tok1 PthroatPo2k1kk1 The velocity at the throat is the velocity of sound VthroatSQRTkRTthroat1000 Some Wrong Solutions with Common Mistakes W1VthroatSQRTkRT11000 Using T1 for temperature W2VthroatSQRTkRT2throat1000 T2throat2Tthroat273k1 Using C for temperature W3VthroatkRTthroat Using wrong relation 17162 Consider gas flow through a convergingdiverging nozzle Of the five statements below select the one that is incorrect a The fluid velocity at the throat can never exceed the speed of sound b If the fluid velocity at the throat is below the speed of sound the diversion section will act like a diffuser c If the fluid enters the diverging section with a Mach number greater than one the flow at the nozzle exit will be supersonic d There will be no flow through the nozzle if the back pressure equals the stagnation pressure e The fluid velocity decreases the entropy increases and stagnation enthalpy remains constant during flow through a normal shock Answer c If the fluid enters the diverging section with a Mach number greater than one the flow at the nozzle exit will be supersonic PROPRIETARY MATERIAL preparation If you are a student using this Manual you are using it without permission 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course