Transport Processes and Unit Operations - Christie J Geankoplis - Livro Engenharia Química

Transport Processes and Unit Operations CHRISTIE J GEANKOPLIS University of Minnesota Transport Processes and Unit Operations THIRD EDITION PrenticeHall International Inc ISBN 013045253X This edition may be sold only in those countries to which it is consigned by PrenticeHall International It is not to be reexported and it is not for sale in the USA Mexico or Canada 1993 1983 1978 by PTR PrenticeHall Inc A Simon Schuster Company Englewood Cliffs New Jersey 07632 All rights reserved No part of this book may be reproduced in any form or by any means without permission in writing from the publisher Printed in the United States of America 10 9 ISBN 013045253X PrenticeHall International UK Limited London PrenticeHall of Australia Pty Limited Sydney PrenticeHall Canada Inc Toronto PrenticeHall Hispanoamericana SA Mexico PrenticeHall of India Private Limited New Delhi PrenticeHall of Japan Inc Tokyo Simon Schuster Asia Pte Ltd Singapore Editora PrenticeHall do Brasil Ltda Rio de Janeiro PrenticeHall Inc Englewood Cliffs New Jersey Dedicated to the memory of my beloved mother Helen for her love and encouragement Contents Preface xi PART 1 TRANSPORT PROCESSES MOMENTUM HEAT AND MASS Chapter 1 Introduction to Engineering Principles and Units 1 11 Classification of Unit Operations and Transport Processes 1 12 SI System of Basic Units Used in This Text and Other Systems 3 13 Methods of Expressing Temperatures and Compositions 5 14 Gas Laws and Vapor Pressure 7 15 Conservation of Mass and Material Balances 9 16 Energy and Heat Units 14 17 Conservation of Energy and Heat Balances 19 18 Graphical Numerical and Mathematical Methods 23 Chapter 2 Principles of Momentum Transfer and Overall Balances 31 21 Introduction 31 22 Fluid Statics 32 23 General Molecular Transport Equation for Momentum Heat and Mass Transfer 39 24 Viscosity of Fluids 43 25 Types of Fluid Flow and Reynolds Number 47 26 Overall Mass Balance and Continuity Equation 50 27 Overall Energy Balance 56 28 Overall Momentum Balance 69 29 Shell Momentum Balance and Velocity Profile in Laminar Flow 78 210 Design Equations for Laminar and Turbulent Flow in Pipes 83 211 Compressible Flow of Gases 101 Chapter 3 Principles of Momentum Transfer and Applications 114 31 Flow Past Immersed Objects and Packed and Fluidized Beds 114 32 Measurement of Flow of Fluids 127 33 Pumps and GasMoving Equipment 133 34 Agitation and Mixing of Fluids and Power Requirements 140 35 NonNewtonian Fluids 153 36 Differential Equations of Continuity 164 37 Differential Equations of Momentum Transfer or Motion 170 38 Use of Differential Equations of Continuity and Motion 175 39 Other Methods for Solution of Differential Equations of Motion 184 310 BoundaryLayer Flow and Turbulence 190 311 Dimensional Analysis in Momentum Transfer 202 vii Chapter 4 Principles of SteadyState Heat Transfer 214 41 Introduction and Mechanisms of Heat Transfer 214 42 Conduction Heat Transfer 220 43 Conduction Through Solids in Series 223 44 SteadyState Conduction and Shape Factors 233 45 Forced Convection Heat Transfer Inside Pipes 236 46 Heat Transfer Outside Various Geometries in Forced Convection 247 47 Natural Convection Heat Transfer 253 48 Boiling and Condensation 259 49 Heat Exchangers 267 410 Introduction to Radiation Heat Transfer 276 411 Advanced Radiation HeatTransfer Principles 281 412 Heat Transfer of NonNewtonian Fluids 297 413 Special HeatTransfer Coefficients 300 414 Dimensional Analysis in Heat Transfer 308 415 Numerical Methods for SteadyState Conduction in Two Dimensions 310 Chapter 5 Principles of UnsteadyState Heat Transfer 330 51 Derivation of Basic Equation 330 52 Simplified Case for Systems with Negligible Internal Resistance 332 53 UnsteadyState Heat Conduction in Various Geometries 334 54 Numerical FiniteDifference Methods for UnsteadyState Conduction 350 55 Chilling and Freezing of Food and Biological Materials 360 56 Differential Equation of Energy Change 365 57 BoundaryLayer Flow and Turbulence in Heat Transfer 370 Chapter 6 Principles of Mass Transfer 381 61 Introduction to Mass Transfer and Diffusion 381 62 Molecular Diffusion in Gases 385 63 Molecular Diffusion in Liquids 397 64 Molecular Diffusion in Biological Solutions and Gels 403 65 Molecular Diffusion in Solids 408 66 Numerical Methods for SteadyState Molecular Diffusion in Two Dimensions 413 Chapter 7 Principles of UnsteadyState and Convective Mass Transfer 426 71 UnsteadyState Diffusion 426 72 Convective MassTransfer Coefficients 432 73 MassTransfer Coefficients for Various Geometries 437 74 Mass Transfer to Suspensions of Small Particles 450 75 Molecular Diffusion Plus Convection and Chemical Reaction 453 76 Diffusion of Gases in Porous Solids and Capillaries 462 77 Numerical Methods for UnsteadyState Molecular Diffusion 468 78 Dimensional Analysis in Mass Transfer 474 79 BoundaryLayer Flow and Turbulence in Mass Transfer 475 viii Contents PART 2 UNIT OPERATIONS Chapter 8 Evaporation 489 81 Introduction 489 82 Types of Evaporation Equipment and Operation Methods 491 83 Overall HeatTransfer Coefficients in Evaporators 495 84 Calculation Methods for SingleEffect Evaporators 496 85 Calculation Methods for MultipleEffect Evaporators 502 86 Condensers for Evaporators 511 87 Evaporation of Biological Materials 513 88 Evaporation Using Vapor Recompression 514 Chapter 9 Drying of Process Materials 520 91 Introduction and Methods of Drying 520 92 Equipment for Drying 521 93 Vapor Pressure of Water and Humidity 525 94 Equilibrium Moisture Content of Materials 533 95 Rate of Drying Curves 536 96 Calculation Methods for ConstantRate Drying Period 540 97 Calculation Methods for FallingRate Drying Period 545 98 Combined Convection Radiation and Conduction Heat Transfer in ConstantRate Period 548 99 Drying in FallingRate Period by Diffusion and Capillary Flow 551 910 Equations for Various Types of Dryers 556 911 Freeze Drying of Biological Materials 566 912 UnsteadyState Thermal Processing and Sterilization of Biological Materials 569 Chapter 10 Stage and Continuous GasLiquid Separation Processes 584 101 Types of Separation Processes and Methods 584 102 Equilibrium Relations Between Phases 586 103 Single and Multiple Equilibrium Contact Stages 587 104 Mass Transfer Between Phases 594 105 Continuous Humidification Processes 602 106 Absorption in Plate and Packed Towers 610 107 Absorption of Concentrated Mixtures in Packed Towers 627 108 Estimation of Mass Transfer Coefficients for Packed Towers 632 Chapter 11 VaporLiquid Separation Processes 640 111 VaporLiquid Equilibrium Relations 640 112 SingleStage Equilibrium Contact for VaporLiquid System 642 113 Simple Distillation Methods 644 114 Distillation with Reflux and McCabeThiele Method 649 115 Distillation and Absorption Tray Efficiencies 666 116 Fractional Distillation Using EnthalpyConcentration Method 669 117 Distillation of Multicomponent Mixtures 679 Contents ix Chapter 12 LiquidLiquid and FluidSolid Separation Processes 697 121 Introduction to Adsorption Processes 697 122 Batch Adsorption 700 123 Design of FixedBed Adsorption Columns 701 124 IonExchange Processes 708 125 SingleStage LiquidLiquid Extraction Processes 709 126 Equipment for LiquidLiquid Extraction 715 127 Continuous Multistage Countercurrent Extraction 716 128 Introduction and Equipment for LiquidSolid Leaching 723 129 Equilibrium Relations and SingleStage Leaching 729 1210 Countercurrent Multistage Leaching 733 1211 Introduction and Equipment for Crystallization 737 1212 Crystallization Theory 743 Chapter 13 Membrane Separation Process 754 131 Introduction and Types of Membrane Separation Processes 754 132 Liquid Permeation Membrane Processes or Dialysis 755 133 Gas Permeation Membrane Processes 759 134 CompleteMixing Model for Gas Separation by Membranes 764 135 CompleteMixing Model for Multicomponent Mixtures 769 136 CrossFlow Model for Gas Separation by Membranes 772 137 CountercurrentFlow Model for Gas Separation by Membranes 778 138 Effects of Processing Variables on Gas Separation by Membranes 780 139 ReverseOsmosis Membrane Processes 782 1310 Applications Equipment and Models for Reverse Osmosis 788 1311 Ultrafiltration Membrane Processes 791 Chapter 14 MechanicalPhysical Separation Processes 800 141 Introduction and Classification of MechanicalPhysical Separation Processes 800 142 Filtration in SolidLiquid Separation 801 143 Settling and Sedimentation in ParticleFluid Separation 815 144 Centrifugal Separation Processes 828 145 Mechanical Size Reduction 840 Appendix Appendix A1 Fundamental Constants and Conversion Factors 850 Appendix A2 Physical Properties of Water 854 Appendix A3 Physical Properties of Inorganic and Organic Compounds 864 Appendix A4 Physical Properties of Foods and Biological Materials 889 Appendix A5 Properties of Pipes Tubes and Screens 892 Notation 895 Index 905 x Contents Preface In this third edition the main objectives and the format of the first and second editions remain the same The sections on momentum transfer have been greatly expanded especially in the sections covering differential equations of momentum transfer This now allows full coverage of the transport processes of momentum heat and mass transfer Also a section on adsorption and an expanded chapter on membrane processes have been added to the unit operations sections The field of chemical engineering involved with physical and physicalchemical changes of inorganic and organic materials and to some extent biological materials is overlapping more and more with the other process engineering fields of ceramic engineering process metallurgy agricultural food engineering wastewater treatment civil engineering and bioengineering The principles of momentum heat and mass transport and the unit operations are used in these processing fields The principles of momentum transfer and heat transfer have been taught to all engineers The study of mass transfer has been limited primarily to chemical engineers However engineers in other fields have become more interested in mass transfer in gases liquids and solids Since chemical and other engineering students must study so many topics today a more unified introduction to the transport processes of momentum heat and mass transfer and to the applications of unit operations is provided In this text the principles of the transport processes are covered first and then the unit operations To accomplish this the text is divided into two main parts PART 1 Transport Processes Momentum Heat and Mass This part dealing with fundamental principles includes the following chapters 1 Introduction to Engineering Principles and Units 2 Principles of Momentum Transfer and Overall Balances 3 Principles of Momentum Transfer and Applications 4 Principles of SteadyState Heat Transfer 5 Principles of UnsteadyState Heat Transfer 6 xi Principles of Mass Transfer and 7 Principles of UnsteadyState and Convective Mass Transfer PART 2 Unit Operations This part on applications covers the following unit operations 8 Evaporation 9 Drying of Process Materials 10 Stage and Continuous GasLiquid Separation Processes humidification absorption 11 VaporLiquid Separation Processes distillation 12 LiquidLiquid and FluidSolid Separation Processes adsorption ion exchange extraction leaching crystallization 13 Membrane Separation Processes dialysis gas separation reverse osmosis ultrafiltration 14 MechanicalPhysical Separation Processes filtration settling centrifugal separation mechanical size reduction In Chapter 1 elementary principles of mathematical and graphical methods laws of chemistry and physics material balances and heat balances are reviewed Many especially chemical engineers may be familiar with most of these principles and may omit all or parts of this chapter A few topics involved primarily with the processing of biological materials may be omitted at the discretion of the reader or instructor Sections 55 64 87 911 and 912 Over 230 example or sample problems and over 500 homework problems on all topics are included in the text Some of the homework problems are concerned with biological systems for those readers who are especially interested in that area This text may be used for a course of study using any of the following five suggested plans In all plans Chapter 1 may or may not be included 1 Study of transport processes of momentum heat and mass and unit operations In this plan most of the complete text covering the principles of the transport processes in Part 1 and the unit operations in Part 2 are covered This plan could be applicable primarily to chemical engineering and also to other process engineering fields in a one and onehalf year course of study at the junior andor senior level 2 Study of transport processes of momentum heat and mass and selected unit operations Only the elementary sections of Part 1 the principles chapters2 3 4 5 6 and 7 are covered plus selected unit operations topics in Part 2 applicable to a particular field in a twosemester or threequarter course Those in wastewater treatment engineering food process engineering and process metallurgy could follow this plan 3 Study of transport processes of momentum heat and mass The purpose of this plan in a twoquarter or twosemester course is to obtain a basic understanding of the transport processes of momentum heat and mass transfer This involves studying sections of the principles chapters2 3 4 5 6 and 7 in Part 1and omitting Part 2 the applied chapters on unit operations 4 Study of unit operations If the reader has had courses in the transport processes of momentum heat and mass Chapters 2 through 7 can be omitted and only the unit operations chapters in Part 2 studied in a onesemester or twoquarter course This plan could be used by chemical and certain other engineers 5 Study of mass transfer For those such as chemical or mechanical engineers who have had momentum and heat transfer or those who desire only a background in mass transfer in a onequarter or onesemester course Chapters 6 7 and 10 would be covered Chapters 9 11 12 and 13 might be covered optionally depending on the needs of the reader xii Preface The SI Système International dUnités system of units has been adopted by the scientific community Because of this the SI system of units has been adopted in this text for use in the equations example problems and homework problems However the most important equations derived in the text are also given in a dual set of units SI and English when different Many example and homework problems are also given using English units Christie J Geankoplis PART 1 Transport Processes Momentum Heat and Mass CHAPTER 1 Introduction to Engineering Principles and Units 11 CLASSIFICATION OF UNIT OPERATIONS AND TRANSPORT PROCESSES 11A Introduction In the chemical and other physical processing industries and the food and biological processing industries many similarities exist in the manner in which the entering feed materials are modified or processed into final materials of chemical and biological products We can take these seemingly different chemical physical or biological processes and break them down into a series of separate and distinct steps called unit operations These unit operations are common to all types of diverse process industries For example the unit operation distillation is used to purify or separate alcohol in the beverage industry and hydrocarbons in the petroleum industry Drying of grain and other foods is similar to drying of lumber filtered precipitates and rayon yarn The unit operation absorption occurs in absorption of oxygen from air in a fermentation process or in a sewage treatment plant and in absorption of hydrogen gas in a process for liquid hydrogenation of oil Evaporation of salt solutions in the chemical industry is similar to evaporation of sugar solutions in the food industry Settling and sedimentation of suspended solids in the sewage and the mining industries are similar Flow of liquid hydrocarbons in the petroleum refinery and flow of milk in a dairy plant are carried out in a similar fashion The unit operations deal mainly with the transfer and change of energy and the transfer and change of materials primarily by physical means but also by physicalchemical means The important unit operations which can be combined in various sequences in a process and which are covered in Part 2 of this text are described next 11B Classification of Unit Operations 1 Fluid flow This concerns the principles that determine the flow or transportation of any fluid from one point to another 2 Heat transfer This unit operation deals with the principles that govern accumulation and transfer of heat and energy from one place to another 1 3 Evaporation This is a special case of heat transfer which deals with the evaporation of a volatile solvent such as water from a nonvolatile solute such as salt or any other material in solution 4 Drying In this operation volatile liquids usually water are removed from solid materials 5 Distillation This is an operation whereby components of a liquid mixture are separated by boiling because of their differences in vapor pressure 6 Absorption In this process a component is removed from a gas stream by treatment with a liquid 7 Membrane separation This process involves the separation of a solute from a fluid by diffusion of this solute from a liquid or gas through a semipermeable membrane barrier to another fluid 8 Liquidliquid extraction In this case a solute in a liquid solution is removed by contacting with another liquid solvent which is relatively immiscible with the solution 9 Adsorption In this process a component of a gas or a liquid stream is removed and adsorbed by a solid adsorbent 10 Liquidsolid leaching This involves treating a finely divided solid with a liquid that dissolves out and removes a solute contained in the solid 11 Crystallization This concerns the removal of a solute such as a salt from a solution by precipitating the solute from the solution 12 Mechanicalphysical separations These involve separation of solids liquids or gases by mechanical means such as filtration settling and size reduction which are often classified as separate unit operations Many of these unit operations have certain fundamental and basic principles or mechanisms in common For example the mechanism of diffusion or mass transfer occurs in drying membrane separation absorption distillation and crystallization Heat transfer occurs in drying distillation evaporation and so on Hence the following classification of a more fundamental nature is often made into transfer or transport processes 11C Fundamental Transport Processes 1 Momentum transfer This is concerned with the transfer of momentum which occurs in moving media such as in the unit operations of fluid flow sedimentation and mixing 2 Heat transfer In this fundamental process we are concerned with the transfer of heat from one place to another it occurs in the unit operations heat transfer drying evaporation distillation and others 3 Mass transfer Here mass is being transferred from one phase to another distinct phase the basic mechanism is the same whether the phases are gas solid or liquid This includes distillation absorption liquidliquid extraction membrane separation adsorption and leaching 11D Arrangement in Parts 1 and 2 This text is arranged in two parts Part 1 Transport Processes Momentum Heat and Mass These fundamental principles are covered extensively in Chapters 1 to 7 to provide the basis for study of unit operations 2 Chap 1 Introduction to Engineering Principles and Units Part 2 Unit Operations The various unit operations and their applications to process areas are studied in Part 2 of this text There are a number of elementary engineering principles mathematical techniques and laws of physics and chemistry that are basic to a study of the principles of momentum heat and mass transfer and the unit operations These are reviewed for the reader in this first chapter Some readers especially chemical engineers agricultural engineers civil engineers and chemists may be familiar with many of these principles and techniques and may wish to omit all or parts of this chapter Homework problems at the end of each chapter are arranged in different sections each corresponding to the number of a given section in the chapter 12 SI SYSTEM OF BASIC UNITS USED IN THIS TEXT AND OTHER SYSTEMS There are three main systems of basic units employed at present in engineering and science The first and most important of these is the SI Système International dUnités system which has as its three basic units the meter m the kilogram kg and the second s The others are the English foot ftpound lbsecond s or fps system and the centimeter cmgram gsecond s or cgs system At present the SI system has been adopted officially for use exclusively in engineering and science but the older English and cgs systems will still be used for some time Much of the physical and chemical data and empirical equations are given in these latter two systems Hence the engineer should not only be proficient in the SI system but must also be able to use the other two systems to a limited extent 12A SI System of Units The basic quantities used in the SI system are as follows the unit of length is the meter m the unit of time is the second s the unit of mass is the kilogram kg the unit of temperature is the kelvin K and the unit of an element is the kilogram mole kg mol The other standard units are derived from these basic quantities The basic unit of force is the newton N defined as 1 newton N 1 kgms² The basic unit of work energy or heat is the newtonmeter or joule J 1 joule J 1 newtonm Nm 1 kgm²s² Power is measured in jouless or watts W 1 joules Js 1 watt W The unit of pressure is the newtonm² or pascal Pa 1 newtonm² Nm² 1 pascal Pa Pressure in atmospheres atm is not a standard SI unit but is being used during the transition period The standard acceleration of gravity is defined as 1 g 980665 ms² A few of the standard prefixes for multiples of the basic units are as follows giga G 10⁹ mega M 10⁶ kilo k 10³ centi c 10² milli m 10³ micro μ 10⁶ and nano n 10⁹ The prefix c is not a preferred prefix Sec 12 SI System of Basic Units Used in This Text and Other Systems 3 Temperatures are defined in kelvin K as the preferred unit in the SI system However in practice wide use is made of the degree Celsius C scale which is defined by tC TK 27315 Note that 1C 1 K and that in the case of temperature difference ΔtC ΔT K The standard preferred unit of time is the second s but time can be in nondecimal units of minutes min hours h or days d 12B CGS System of Units The cgs system is related to the SI system as follows 1 g mass g 1 10³ kg mass kg 1 cm 1 10² m 1 dyne dyn 1 gcms² 1 10⁵ newton N 1 erg 1 dyncm 1 10⁷ joule J The standard acceleration of gravity is g 980665 cms² 12C English fps System of Units The English system is related to the SI system as follows 1 lb mass lbₘ 045359 kg 1 ft 030480 m 1 lb force lbᵣ 44482 newtons N 1 ftlbᵣ 135582 newtonm Nm 135582 joules J 1 psia 689476 10³ newtonm² Nm² 18F 1 K 1C centigrade or Celsius g 32174 fts² The proportionality factor for Newtons law is gₐ 32174 ftlbₘlbᵣs² The factor gₐ in SI units and cgs units is 10 and is omitted In Appendix A1 convenient conversion factors for all three systems are tabulated Further discussions and use of these relationships are given in various sections of the text This text uses the SI system as the primary set of units in the equations sample problems and homework problems However the important equations derived in the text are given in a dual set of units SI and English when these equations differ Some 4 Chap 1 Introduction to Engineering Principles and Units example problems and homework problems are also given using English units In some cases intermediate steps andor answers in example problems are also stated in English units 12D Dimensionally Homogeneous Equations and Consistent Units A dimensionally homogeneous equation is one in which all the terms have the same units These units can be the base units or derived ones for example kgs²m or Pa Such an equation can be used with any system of units provided that the same base or derived units are used throughout the equation No conversion factors are needed when consistent units are used The reader should be careful in using any equation and always check it for dimensional homogeneity To do this a system of units SI English etc is first selected Then units are substituted for each term in the equation and like units in each term canceled out 13 METHODS OF EXPRESSING TEMPERATURES AND COMPOSITIONS 13A Temperature There are two temperature scales in common use in the chemical and biological industries These are degrees Fahrenheit abbreviated F and Celsius C It is often necessary to convert from one scale to the other Both use the freezing point and boiling point of water at 1 atmosphere pressure as base points Often temperatures are expressed as absolute degrees K SI standard or degrees Rankine R instead of C or F Table 131 shows the equivalences of the four temperature scales The difference between the boiling point of water and melting point of ice at 1 atm is 100C or 180F Thus a 18F change is equal to a 1C change Usually the value of 27315C is rounded to 2732C and 4597F to 460F The following equations can be used to convert from one scale to another F 32 18C 131 C 118 F 32 132 R F 460 133 K C 27315 134 TABLE 131 Temperature Scales and Equivalents Centigrade Fahrenheit Kelvin Rankine Celsius Boiling water 100C 212F 37315 K 6717R 100C Melting ice 0C 32F 27315 K 4917R 0C Absolute zero 27315C 4597F 0 K 0R 27315C Sec 13 Methods of Expressing Temperatures and Compositions 5 In order that amounts of various gases may be compared standard conditions of temperature and pressure abbreviated STP or SC are arbitrarily defined as 101325 kPa 10 atm abs and 27315 K 0C Under these conditions the volumes are as follows volume of 10 kg mol SC 22414 m3 volume of 10 g mol SC 22414 L liter 22 414 cm3 volume of 10 lb mol SC 35905 ft3 EXAMPLE 141 GasLaw Constant Calculate the value of the gaslaw constant R when the pressure is in psia moles in lb mol volume in ft3 and temperature in R Repeat for SI units Solution At standard conditions p 147 psia V 359 ft3 and T 460 32 492R 27315 K Substituting into Eq 141 for n 10 lb mol and solving for R R pV nT 147 psia359 ft3 10 lb mol492R 1073 ft3 psia lb mol R R pV nT 101325 105 Pa22414 m3 10 kg mol27315 K 8314 m3 Pa kg mol K A useful relation can be obtained from Eq 141 for n moles of gas at conditions p1 V1 T1 and also at conditions p2 V2 T2 Substituting into Eq 141 p1 V1 nRT1 p2 V2 nRT2 Combining gives p1 V1 p2 V2 T1 T2 142 14C Ideal Gas Mixtures Daltons law for mixtures of ideal gases states that the total pressure of a gas mixture is equal to the sum of the individual partial pressures P pA pB pC 143 where P is total pressure and pA pB pC are the partial pressures of the components A B C in the mixture Since the number of moles of a component is proportional to its partial pressure the mole fraction of a component is xA pA P pA pA pB pC 144 The volume fraction is equal to the mole fraction Gas mixtures are almost always represented in terms of mole fractions and not weight fractions For engineering purposes Daltons law is sufficiently accurate to use for actual mixtures at total pressures of a few atmospheres or less EXAMPLE 142 Composition of a Gas Mixture A gas mixture contains the following components and partial pressures CO2 75 mm Hg CO 50 mm Hg N2 595 mm Hg O2 26 mm Hg Calculate the total pressure and the composition in mole fraction Solution Substituting into Eq 143 P pA pB pC pD 75 50 595 26 746 mm Hg The mole fraction of CO2 is obtained by using Eq 144 x4CO2 pA P 75 746 0101 In like manner the mole fractions of CO N2 and O2 are calculated as 0067 0797 and 0035 respectively 14D Vapor Pressure and Boiling Point of Liquids When a liquid is placed in a sealed container molecules of liquid will evaporate into the space above the liquid and fill it completely After a time equilibrium is reached This vapor will exert a pressure just like a gas and we call this pressure the vapor pressure of the liquid The value of the vapor pressure is independent of the amount of liquid in the container as long as some is present If an inert gas such as air is also present in the vapor space it will have very little effect on the vapor pressure In general the effect of total pressure on vapor pressure can be considered as negligible for pressures of a few atmospheres or less The vapor pressure of a liquid increases markedly with temperature For example from Appendix A2 for water the vapor pressure at 50C is 12333 kPa 9251 mm Hg At 100C the vapor pressure has increased greatly to 101325 kPa 760 mm Hg The boiling point of a liquid is defined as the temperature at which the vapor pressure of a liquid equals the total pressure Hence if the atmospheric total pressure is 760 mm Hg water will boil at 100C On top of a high mountain where the total pressure is considerably less water will boil at temperatures below 100C A plot of vapor pressure pA of a liquid versus temperature does not yield a straight line but a curve However for moderate temperature ranges a plot of log pA versus 1T is a reasonably straight line as follows log pA m1T b 145 where m is the slope b is a constant for the liquid A and T is the temperature in K 15 CONSERVATION OF MASS AND MATERIAL BALANCES 15A Conservation of Mass One of the basic laws of physical science is the law of conservation of mass This law stated simply says that mass cannot be created or destroyed excluding of course nuclear or atomic reactions Hence the total mass or weight of all materials entering any process must equal the total mass of all materials leaving plus the mass of any materials accumulating or left in the process input output accumulation 151 Sec 15 Conservation of Mass and Material Balances 9 In the majority of cases there will be no accumulation of materials in a process and then the input will simply equal the output Stated in other words what goes in must come out We call this type of process a steadystate process input output steady state 152 15B Simple Material Balances In this section we do simple material weight or mass balances in various processes at steady state with no chemical reaction occurring We can use units of kg lbm lb mol g kg mol etc in our balances The reader is cautioned to be consistent and not to mix several units in a balance When chemical reactions occur in the balances as discussed in Section 15D one should use kg mol units since chemical equations relate moles reacting In Section 26 overall mass balances will be covered in more detail and in Section 36 differential mass balances To solve a materialbalance problem it is advisable to proceed by a series of definite steps as listed below 1 Sketch a simple diagram of the process This can be a simple box diagram showing each stream entering by an arrow pointing in and each stream leaving by an arrow pointing out Include on each arrow the compositions amounts temperatures and so on of that stream All pertinent data should be on this diagram 2 Write the chemical equations involved if any 3 Select a basis for calculation In most cases the problem is concerned with a specific amount of one of the streams in the process which is selected as the basis 4 Make a material balance The arrows into the process will be input items and the arrows going out output items The balance can be a total material balance in Eq 152 or a balance on each component present if no chemical reaction occurs Typical processes that do not undergo chemical reactions are drying evaporation dilution of solutions distillation extraction and so on These can be solved by setting up material balances containing unknowns and solving these equations for the unknowns EXAMPLE 151 Concentration of Orange Juice In the concentration of orange juice a fresh extracted and strained juice containing 708 wt solids is fed to a vacuum evaporator In the evaporator water is removed and the solids content increased to 58 wt solids For 1000 kgh entering calculate the amounts of the outlet streams of concentrated juice and water Solution Following the four steps outlined we make a process flow diagram step 1 in Fig 151 Note that the letter W represents the unknown W kgh water 1000 kgh juice 708 solids evaporator C kgh concentrated juice 58 solids FIGURE 151 Process flow diagram for Example 151 10 Chap 1 Introduction to Engineering Principles and Units amount of water and C the amount of concentrated juice No chemical reactions are given step 2 Basis 1000 kgh entering juice step 3 To make the material balances step 4 a total material balance will be made using Eq152 1000 W C 153 This gives one equation and two unknowns Hence a component balance on solids will be made 1000708100 W0 C58100 154 To solve these two equations we solve Eq 154 first for C since W drops out We get C 1221 kgh concentrated juice Substituting the value of C into Eq 153 1000 W 1221 and we obtain W 8779 kgh water As a check on our calculations we can write a balance on the water component 1000100 708100 8779 1221100 58100 155 Solving 9292 8779 513 9292 In Example 151 only one unit or separate process was involved Often a number of processes in series are involved Then we have a choice of making a separate balance over each separate process andor a balance around the complete overall process 15C Material Balances and Recycle Processes that have a recycle or feedback of part of the product into the entering feed are sometimes encountered For example in a sewage treatment plant part of the activated sludge from a sedimentation tank is recycled back to the aeration tank where the liquid is treated In some fooddrying operations the humidity of the entering air is controlled by recirculating part of the hot wet air that leaves the dryer In chemical reactions the material that did not react in the reactor can be separated from the final product and fed back to the reactor EXAMPLE 152 Crystallization of KNO₃ and Recycle In a process producing KNO₃ salt 1000 kgh of a feed solution containing 20 wt KNO₃ is fed to an evaporator which evaporates some water at 422 K to produce a 50 wt KNO₃ solution This is then fed to a crystallizer at 311 K where crystals containing 96 wt KNO₃ are removed The saturated solution containing 375 wt KNO₃ is recycled to the evaporator Calculate the amount of recycle stream R in kgh and the product stream of crystals P in kgh Solution Figure 152 gives the process flow diagram As a basis we shall use 1000 kgh of fresh feed No chemical reactions are occurring We can Sec 15 Conservation of Mass and Material Balances 11 water W kgh evaporator crystallizer feed 1000 kgh S kgh 422 K 311 K 20 KNO₃ 50 KNO₃ recycle R kgh crystals P kgh 375 KNO₃ 4 H₂O FIGURE 152 Process flow diagram for Example 152 make an overall balance on the entire process for KNO₃ and solve for P directly 1000020 W0 P096 156 P 2083 kg crystalsh To calculate the recycle stream we can make a balance around the evaporator or the crystallizer Using a balance on the crystallizer since it now includes only two unknowns S and R we get for a total balance S R 2083 157 For a KNO₃ balance on the crystallizer S050 R0375 2083096 158 Substituting S from Eq 157 into Eq 158 and solving R 7666 kg recycleh and S 9749 kgh 15D Material Balances and Chemical Reaction In many cases the materials entering a process undergo chemical reactions in the process so that the materials leaving are different from those entering In these cases it is usually convenient to make a molar and not a weight balance on an individual component such as kg mol H₂ or kg atom H kg mol CO₃ ion kg mol CaCO₃ kg atom Na kg mol N₂ and so on For example in the combustion of CH₄ with air balances can be made on kg mol of H₂ C O₂ or N₂ EXAMPLE 153 Combustion of Fuel Gas A fuel gas containing 31 mol H₂ 272 CO 56 CO₂ 05 O₂ and 636 N₂ is burned with 20 excess air ie the air over and above that necessary for complete combustion to CO₂ and H₂O The combustion of CO is only 98 complete For 100 kg mol of fuel gas calculate the moles of each component in the exit flue gas Solution First the process flow diagram is drawn Fig 153 On the 12 Chap 1 Introduction to Engineering Principles and Units A kg mol air burner F kg mol flue gas 100 kg mol fuel gas H₂O 31 H₂ CO 272 CO CO₂ 56 CO₂ O₂ 05 O₂ N₂ 636 N₂ 1000 FIGURE 153 Process flow diagram for Example 153 diagram the components in the flue gas are shown Let A be moles of air and F be moles of flue gas Next the chemical reactions are given CO ½O₂ CO₂ 159 H₂ ½O₂ H₂O 1510 An accounting of the total moles of O₂ in the fuel gas is as follows mol O₂ in fuel gas ½272CO 56CO₂ 05O₂ 197 mol O₂ For all the H₂ to be completely burned to H₂O we need from Eq 1510 ½ mol O₂ for 1 mol H₂ or 31½ 155 total mol O₂ For completely burning the CO from Eq 159 we need 272½ 136 mol O₂ Hence the amount of O₂ we must add is theoretically as follows mol O₂ theoretically needed 155 136 05 in fuel gas 1465 mol O₂ For a 20 excess we add 121465 or 1758 mol O₂ Since air contains 79 mol N₂ the amount of N₂ added is 79211758 or 661 mol N₂ To calculate the moles in the final flue gas all the H₂ gives H₂O or 31 mol H₂O For CO 20 does not react Hence 002272 or 054 mol CO will be unburned A total carbon balance is as follows inlet moles C 272 56 328 mol C In the outlet flue gas 054 mol will be as CO and the remainder of 328 054 or 3226 mol as CO₂ For calculating the outlet mol O₂ we make an overall O₂ balance O₂ in 197 in fuel gas 1758 in air 3728 mol O₂ O₂ out 312 in H₂O 0542 in CO 3226 in CO₂ free O₂ Equating inlet O₂ to outlet the free remaining O₂ 32 mol O₂ For the N₂ balance the outlet 636 in fuel gas 661 in air or 12970 mol N₂ The outlet flue gas contains 310 mol H₂O 054 mol CO 3226 mol CO₂ 320 mol O₂ and 1297 mol N₂ In chemical reactions with several reactants the limiting reactant component is defined as that compound which is present in an amount less than the amount necessary for it to react stoichiometrically with the other reactants Then the percent completion of a reaction is the amount of this limiting reactant actually converted divided by the amount originally present times 100 16 ENERGY AND HEAT UNITS 16A Joule Calorie and Btu In a manner similar to that used in making material balances on chemical and biological processes we can also make energy balances on a process Often a large portion of the energy entering or leaving a system is in the form of heat Before such energy or heat balances are made we must understand the various types of energy and heat units In the SI system energy is given in joules J or kilojoules kJ Energy is also expressed in btu British thermal unit or cal calorie The g calorie abbreviated cal is defined as the amount of heat needed to heat 10 g water 10C from 145C to 155C Also 1 kcal kilocalorie 1000 cal The btu is defined as the amount of heat needed to raise 10 lb water 1F Hence from Appendix A1 1 btu 25216 cal 105506 kJ 161 16B Heat Capacity The heat capacity of a substance is defined as the amount of heat necessary to increase the temperature by 1 degree It can be expressed for 1 g 1 lb 1 g mol 1 kg mol or 1 lb mol of the substance For example a heat capacity is expressed in SI units as Jkg molK in other units as calgC calg molC kcalkg molC btulbmF or btulb molF It can be shown that the actual numerical value of a heat capacity is the same in mass units or in molar units That is 10 calgC 10 btulbmF 162 10 calg molC 10 btulb molF 163 For example to prove this suppose that a substance has a heat capacity of 08 btulbmF The conversion is made using 18F for 1C or 1 K 25216 cal for 1 btu and 4536 g for 1 lbm as follows heat capacity calgC 08 btu lbmF25216 cal btu1 4536 glbm18 F C 08 cal gC The heat capacities of gases sometimes called specific heat at constant pressure cp are functions of temperature and for engineering purposes can be assumed to be independent of pressure up to several atmospheres In most process engineering calculations one is usually interested in the amount of heat needed to heat a gas from one temperature t1 to another at t2 Since the cp varies with temperature an integration must be performed or a suitable mean cpm used These mean values for gases have been obtained for T1 of 298 K or 25C 77F and various T2 values and are tabulated in Table 161 at 101325 kPa pressure or less as cpm in kJkg molK at various values of T2 in K or C EXAMPLE 161 Heating of N2 Gas The gas N2 at 1 atm pressure absolute is being heated in a heat exchanger Calculate the amount of heat needed in J to heat 30 g mol N2 in the 14 Chap 1 Introduction to Engineering Principles and Units TABLE 161 Mean Molar Heat Capacities of Gases Between 298 and T K 25 and TC at 101325 kPa or Less SI Units cp kJkg molK TK TC H2 N2 CO Air O2 H2O CO2 CH4 SO2 298 25 2886 2914 2916 2919 2938 3359 3720 358 399 373 100 2899 2919 2924 2929 2966 3385 3873 376 412 473 200 2913 2929 2938 2940 3007 3424 4062 403 429 573 300 2918 2946 2960 2961 3053 3439 4232 431 445 673 400 2923 2968 2988 2994 3101 3521 4380 459 458 773 500 2929 2997 3019 3025 3146 3575 4512 488 470 873 600 2935 3027 3052 3056 3189 3633 4628 514 479 973 700 2944 3056 3084 3087 3226 3691 4732 540 488 1073 800 2956 3085 3116 3118 3262 3753 4827 564 496 1173 900 2963 3116 3149 3148 3297 3814 4915 588 503 1273 1000 2984 3143 3177 3179 3325 3871 4991 610 509 1473 1200 3018 3197 3230 3232 3378 3988 5129 649 519 1673 1400 3051 3240 3273 3276 3419 4090 5234 Mean Molar Heat Capacities of Gases Between 25 and TC at 1 atm Pressure or Less English Units cp btulb molF TC H2 N2 CO Air O2 NO H2O CO2 HCl Cl2 CH4 SO2 C2H4 SO3 C2H6 25 6894 6961 6965 6972 7017 7134 8024 8884 696 812 855 954 1045 1211 1263 100 6924 6972 6983 6996 7083 7144 8084 9251 697 824 898 985 1135 1284 1376 200 6957 6996 7017 7021 7181 7224 8177 9701 698 837 962 1025 1253 1374 1527 300 6970 7036 7070 7073 7293 7252 8215 10108 700 848 1029 1062 1365 1454 1672 400 6982 7089 7136 7152 7406 7301 8409 10462 702 855 1097 1094 1467 1522 1811 500 6995 7159 7210 7225 7515 7389 8539 10776 706 861 1165 1122 1560 1582 1939 600 7011 7229 7289 7299 7616 7470 8678 11053 710 866 1227 1145 1645 1633 2058 700 7032 7298 7365 7374 7706 7549 8816 11303 715 870 1290 1166 1722 1677 2168 800 7060 7369 7443 7447 7792 7630 8963 1153 721 873 1348 1184 1795 1717 2272 900 7076 7443 7521 7520 7874 7708 9109 1174 727 877 1404 1201 1863 1752 2369 1000 7128 7507 7587 7593 7941 7773 9246 1192 733 880 1456 1215 1923 1786 2456 1100 7169 7574 7653 7660 8009 7839 9389 1210 739 882 1504 1228 1981 1817 2540 1200 7209 7635 7714 7719 8068 7898 9524 1225 745 894 1549 1239 2033 1844 2615 1300 7252 7692 7772 7778 8123 7952 966 1239 1400 7288 7738 7818 7824 8166 7994 977 1250 1500 7326 7786 7866 7873 8203 8039 989 1269 1600 7386 7844 7922 7929 8269 8092 995 1275 1700 7421 7879 7958 7965 8305 8124 1013 1270 1800 7467 7924 8001 8010 8349 8164 1024 1294 1900 7505 7957 8033 8043 8383 8192 1034 1301 2000 7548 7994 8069 8081 8423 8225 1043 1310 2100 7588 8028 8101 8115 8460 8255 1052 1317 2200 7624 8054 8127 8144 8491 8277 1061 1324 Source O A Hougen K W Watson and R A Ragatz Chemical Process Principles Part I 2nd ed New York John Wiley Sons Inc 1954 With permission following temperature ranges a 298673 K 25400C b 2981123 K 25850C c 6731123 K 400850C Sec 16 Energy and Heat Units 15 Solution For case a Table 161 gives cpm values at 1 atm pressure or less and can be used up to several atm pressures For N2 at 673 K cpm 2968 kJkg molK or 2968 Jg molK This is the mean heat capacity for the range 298673 K heat required M g mol cpm J g molK T2 T1K 164 Substituting the known values heat required 302968673 298 33 390 J For case b the cpm at 1123 K obtained by linear interpolation between 1073 and 1173 K is 3100 Jg molK heat required 3031001123 298 76 725 J For case c there is no mean heat capacity for the interval 6731123 K However we can use the heat required to heat the gas from 298 to 673 K in case a and subtract it from case b which includes the heat to go from 298 to 673 K plus 673 to 1123 K heat required 6731123 K heat required 2981123 K heat required 298673 165 Substituting the proper values into Eq 165 heat required 76 725 33 390 43 335 J On heating a gas mixture the total heat required is determined by first calculating the heat required for each individual component and then adding the results to obtain the total The heat capacities of solids and liquids are also functions of temperature and independent of pressure Data are given in Appendix A2 Physical Properties of Water A3 Physical Properties of Inorganic and Organic Compounds and A4 Physical Properties of Foods and Biological Materials More data are available in P1 EXAMPLE 162 Heating of Milk Rich cows milk 4536 kgh at 44C is being heated in a heat exchanger to 544C by hot water How much heat is needed Solution From Appendix A4 the average heat capacity of rich cows milk is 385 kJkgK Temperature rise T 544 44C 50 K heat required 4536 kgh385 kJkgK13600 hs50 K 2425 kW The enthalpy H of a substance in Jkg represents the sum of the internal energy plus the pressurevolume term For no reaction and a constantpressure process with a change in temperature the heat change as computed from Eq 164 is the difference in enthalpy H of the substance relative to a given temperature or base point In other units H btulbm or calg 16C Latent Heat and Steam Tables Whenever a substance undergoes a change of phase relatively large amounts of heat changes are involved at a constant temperature For example ice at 0C and 1 atm pressure can absorb 60134 kJkg mol This enthalpy change is called the latent heat of fusion Data for other compounds are available in various handbooks P1 W1 16 Chap 1 Introduction to Engineering Principles and Units When a liquid phase vaporizes to a vapor phase under its vapor pressure at constant temperature an amount of heat called the latent heat of vaporization must be added Tabulations of latent heats of vaporization are given in various handbooks For water at 25C and a pressure of 2375 mm Hg the latent heat is 44 020 kJkg mol and at 25C and 760 mm Hg 44 045 kJkg mol Hence the effect of pressure can be neglected in engineering calculations However there is a large effect of temperature on the latent heat of water Also the effect of pressure on the heat capacity of liquid water is small and can be neglected Since water is a very common chemical the thermodynamic properties of it have been compiled in steam tables and are given in Appendix A2 in SI and in English units EXAMPLE 163 Use of Steam Tables Find the enthalpy change ie how much heat must be added for each of the following cases using SI and English units a Heating 1 kg lbm water from 2111C 70F to 60C 140F at 101325 kPa 1 atm pressure b Heating 1 kg lbm water from 2111C 70F to 1156C 240F and vaporizing at 1722 kPa 2497 psia c Vaporizing 1 kg lbm water at 1156C 240F and 1722 kPa 2497 psia Solution For part a the effect of pressure on the enthalpy of liquid water is negligible From Appendix A2 H at 2111C 8860 kJkg or at 70F 3809 btulbm H at 60C 25113 kJkg or at 140F 10796 btulbm change in H ΔH 25113 8860 16253 kJkg 10796 3809 6987 btulbm In part b the enthalpy at 1156C 240F and 1722 kPa 2497 psia of the saturated vapor is 26999 kJkg or 11607 btulbm change in H ΔH 26999 8860 26113 kJkg 11607 3809 11226 btulbm The latent heat of water at 1156C 240F in part c is 26999 4849 22150 kJkg 11607 20844 95226 btulbm 16D Heat of Reaction When chemical reactions occur heat effects always accompany these reactions This area where energy changes occur is often called thermochemistry For example when HCl is neutralized with NaOH heat is given off and the reaction is exothermic Heat is absorbed in an endothermic reaction This heat of reaction is dependent on the chemical nature of each reacting material and product and on their physical states For purposes of organizing data we define a standard heat of reaction ΔH as the change in enthalpy when 1 kg mol reacts under a pressure of 101325 kPa at a temper ature of 298 K 25C For example for the reaction H2g 12 O2g H2Ol 166 the ΔH is 285840 103 kJkg mol or 68317 kcalg mol The reaction is exothermic and the value is negative since the reaction loses enthalpy In this case the H2 gas reacts with the O2 gas to give liquid water all at 298 K 25C Special names are given to ΔH depending upon the type of reaction When the product is formed from the elements as in Eq 166 we call the ΔH heat of formation of the product water ΔHf For the combustion of CH4 to form CO2 and H2O we call it heat of combustion ΔHc Data are given in Appendix A3 for various values of ΔHc EXAMPLE 164 Combustion of Carbon A total of 100 g mol of carbon graphite is burned in a calorimeter held at 298 K and 1 atm The combustion is incomplete and 90 of the C goes to CO2 and 10 to CO What is the total enthalpy change in kJ and kcal Solution From Appendix A3 the ΔHc for carbon going to CO2 is 393513 103 kJkg mol or 940518 kcalg mol and for carbon going to CO is 110523 103 kJkg mol or 264157 kcalg mol Since 9 mol CO2 and 1 mol CO are formed total ΔH 9393513 1110523 3652 kJ 9940518 1264157 8729 kcal If a table of heats of formation ΔHf of compounds is available the standard heat of the reaction ΔH can be calculated by ΔH ΔHfproducts ΔHfreactants 167 In Appencix A3 a short table of some values of ΔHf is given Other data are also available H1 P1 S1 EXAMPLE 165 Reaction of Methane For the following reaction of 1 kg mol of CH4 at 10132 kPa and 298 K CH4g H2Ol COg 3H2g calculate the standard heat of reaction ΔH at 298 K in kJ Solution From Appendix A3 the following standard heats of formation are obtained at 298 K ΔHf kJkg mol CH4g 74848 103 H2Ol 285840 103 COg 110523 103 H2g 0 Note that the ΔHf of all elements is by definition zero Substituting into Eq 167 ΔH 110523 103 30 74848 103 285840 103 250165 103 kJkg mol endothermic 18 Chap 1 Introduction to Engineering Principles and Units 17 CONSERVATION OF ENERGY AND HEAT BALANCES 17A Conservation of Energy In making material balances we used the law of conservation of mass which states that the mass entering is equal to the mass leaving plus the mass left in the process In a similar manner we can state the law of conservation of energy which says that all energy entering a process is equal to that leaving plus that left in the process In this section elementary heat balances will be made More elaborate energy balances will be considered in Sections 27 and 56 Energy can appear in many forms Some of the common forms are enthalpy electrical energy chemical energy in terms of ΔH reaction kinetic energy potential energy work and heat inflow In many cases in process engineering which often takes place at constant pressure electrical energy kinetic energy potential energy and work either are not present or can be neglected Then only the enthalpy of the materials at constant pressure the standard chemical reaction energy ΔH at 25C and the heat added or removed must be taken into account in the energy balance This is generally called a heat balance 17B Heat Balances In making a heat balance at steady state we use methods similar to those used in making a material balance The energy or heat coming into a process in the inlet materials plus any net energy added to the process is equal to the energy leaving in the materials Expressed mathematically HR ΔH298 q Hp 171 where HR is the sum of enthalpies of all materials entering the reaction process relative to the reference state for the standard heat of reaction at 298 K and 10132 kPa If the inlet temperature is above 298 K this sum will be positive ΔH298 standard heat of the reaction at 298 K and 10132 kPa The reaction contributes heat to the process so the negative of ΔH298 is taken to be positive input heat for an exothermic reaction q net energy or heat added to the system If heat leaves the system this item will be negative Hp sum of enthalpies of all leaving materials referred to the standard reference state at 298 K 25C Note that if the materials coming into a process are below 298 K HR will be negative Care must be taken not to confuse the signs of the items in Eq 171 If no chemical reaction occurs then simple heating cooling or phase change is occurring Use of Eq 171 will be illustrated by several examples For convenience it is common practice to call the terms on the lefthand side of Eq 171 input items and those on the right output items EXAMPLE 171 Heating of Fermentation Medium A liquid fermentation medium at 30C is pumped at a rate of 2000 kgh through a heater where it is heated to 70C under pressure The waste heat water used to heat this medium enters at 95C and leaves at 85C The average heat capacity of the fermentation medium is 406 kJkg K and that for water is 421 kJkg K Appendix A2 The fermentation stream and the wastewater stream are separated by a metal surface through which heat is transferred and do not physically mix with each other Make a complete heat balance on the system Calculate the water flow and the amount of heat added to the fermentation medium assuming no heat losses The process flow is given in Fig 171 Sec 17 Conservation of Energy and Heat Balances 19 q heat added 2000 kgh liquid 2000 kgh liquid 30C 70C W kgh W kgh water 85C 95C FIGURE 171 Process flow diagram for Example 171 Solution It is convenient to use the standard reference state of 298 K 25C as the datum to calculate the various enthalpies From Eq 171 the input items are as follows Input items HR of the enthalpies of the two streams relative to 298 K 25C note that Δt 30 25C 5C 5 K Hliquid 2000 kgh406 kJkg K5 K 4060 104 kJh Hwater W42195 25 2947 102 W kJh W kgh ΔH298 0 since there is no chemical reaction q 0 there are no heat losses or additions Output items HP of the two streams relative to 298 K 25C Hliquid 200040670 25 365 105 kJh Hwater W42185 25 2526 102 W kJh Equating input to output in Eq 171 and solving for W 4060 104 2947 102 W 3654 105 2526 102 W W 7720 kgh water flow The amount of heat added to the fermentation medium is simply the difference of the outlet and inlet liquid enthalpies Houtlet liquid Hinlet liquid 3654 105 4060 104 3248 105 kJh 9025 kW Note in this example that since the heat capacities were assumed constant a simpler balance could have been written as follows heat gained by liquid heat lost by water 200040670 30 W42195 85 Then solving W 7720 kgh This simple balance works well when cp is constant However when the cp varies with temperature and the material is a gas cpm values are only available between 298 K 25C and t K and the simple method cannot be used without obtaining new cpm values over different temperature ranges EXAMPLE 172 Heat and Material Balance in Combustion The waste gas from a process of 1000 g molh of CO at 473 K is burned at 1 atm pressure in a furnace using air at 373 K The combustion is complete and 90 excess air is used The flue gas leaves the furnace at 1273 K Calculate the heat removed in the furnace Solution First the process flow diagram is drawn in Fig 172 and then a material balance is made COg 12 O2g CO2g ΔH298 282989 103 kJkg mol from Appendix A3 mol CO 1000 g molh moles CO2 100 kg molh mol O2 theoretically required 12 100 0500 kg molh mol O2 actually added 050019 0950 kg molh mol N2 added 0950 079021 3570 kg molh air added 0950 3570 4520 kg molh A O2 in outlet flue gas added used 0950 0500 0450 kg molh CO2 in outlet flue gas 100 kg molh N2 in outlet flue gas 3570 kg molh For the heat balance relative to the standard state at 298 K we follow Eq 171 Input items HCO 100cpm473 298 1002938473 298 5142 kJh The cpm of CO of 2938 kJkg mol K between 298 and 473 K is obtained from Table 151 Hair 4520cpm373 298 45202929373 298 9929 kJh q heat added kJh 1000 g molh CO 473 K A g molh air 373 K furnace flue gas 1273 K heat removed q FIGURE 172 Process flow diagram for Example 172 This will give a negative value here indicating that heat was removed ΔH298 282989 103 kJkg mol100 kg molh 282 990 kJh Output items HCO2 100cpm1273 298 10049911273 298 48 660 kJh HO2 0450cpm1273 298 045033251273 298 14 590 kJh HN2 3570cpm1273 298 357031431273 298 109 400 kJh Equating input to output and solving for q 5142 9929 q 282 990 48 660 14 590 109 400 q 125 411 kJh Hence heat is removed 34 837 W Often when chemical reactions occur in the process and the heat capacities vary with temperature the solution in a heat balance can be trial and error if the final temperature is the unknown EXAMPLE 173 Oxidation of Lactose In many biochemical processes lactose is used as a nutrient which is oxidized as follows C12 H22 O11s 12 O2 g 12 CO2g 11H2 Ol The heat of combustion ΔHc in Appendix A3 at 25C is 56488 103 Jg mol Calculate the heat of complete oxidation combustion at 37C which is the temperature of many biochemical reactions The cpm of solid lactose is 120 Jg K and the molecular weight is 3423 g massg mol Solution This can be treated as an ordinary heatbalance problem First the process flow diagram is drawn in Fig 173 Next the datum temperature of 25C is selected and the input and output enthalpies calculated The temperature difference Δt 37 25C 37 25 K ΔH37C 1 g mol lactose s 37C 12 g mol O2 g 37C 1 atm combustion 11 g mol H2 O l 37C 12 g mol CO2 g 37C FIGURE 173 Process flow diagram for Example 173 Input items Hlactose 3423 gcpm Jg K3725 K 34231203725 4929 J HO2 gas 12 g molcpm Jg mol K3725 K 1229383725 4230 J The cpm of O2 was obtained from Table 161 ΔH2s0 56488 x 103 Output items HH2O liquid 111802 gcpm Jg K3725 K 11180241837 25 9943 J The cpm of liquid water was obtained from Appendix A2 HCO2 gas 12 g molcpm Jg mol K3725 K 1237453725 5393 J The cpm of CO2 is obtained from Table 161 ΔH37C Setting input output and solving 4929 4230 56488 x 103 9943 5393 ΔH37C ΔH37C 56426 x 103 Jg mol ΔH310 K 18 GRAPHICAL NUMERICAL AND MATHEMATICAL METHODS 18A Graphical Integration Often the mathematical function fx to be integrated is too complex and we are not able to integrate it analytically Or in some cases the function is one that has been obtained from experimental data and no mathematical equation is available to represent the data so that they can be integrated analytically In these cases we can use graphical integration Integration between the limits x a to x b can be represented graphically as shown in Fig 181 Here a function y fx has been plotted versus x The area under the curve y fx between the limits x a to x b is equal to the integral This area is then equal to the sum of the areas of the rectangles as follows xab fx dx A1 A2 A3 A4 A5 181 Sec 18 Graphical Numerical and Mathematical Methods 23 y y fx A1 A2 A3 A4 A5 x a x b FIGURE 181 Graphical integration of xab fx dx 18B Numerical Integration and Simpsons Rule Often it is desired or necessary to perform a numerical integration by computing the value of a definite integral from a set of numerical values of the integrand fx This of course can be done graphically but if data are available in large quantities numerical methods suitable for the digital computer are desired The integral to be evaluated is as follows xab fx dx 182 where the interval is b a The most generally used numerical method is the parabolic rule often called Simpsons rule This method divides the total interval b a into an even number of subintervals m where m b a h 183 The value of h a constant is the spacing in x used Then approximating fx by a parabola on each subinterval Simpsons rule is xab fx dx h3 f0 4f1 f3 f5 fm1 2f2 f4 f6 fm2 fm 184 where f0 is the value of fx at x a f1 the value of fx at x x1 fm the value of fx at x b The reader should note that m must be an even number and the increments evenly spaced This method is well suited for digital computation 24 Chap 1 Introduction to Engineering Principles and Units PROBLEMS 121 Temperature of a Chemical Process The temperature of a chemical reaction was found to be 3532 K What is the temperature in F C and R Ans 176F 80C 636R 122 Temperature for Smokehouse Processing of Meat In smokehouse processing of sausage meat a final temperature of 155F inside the sausage is often used Calculate this temperature in C K and R 131 Molecular Weight of Air For purposes of most engineering calculations air is assumed to be composed of 21 mol oxygen and 79 mol nitrogen Calculate the average molecular weight Ans 289 g massg mol lb masslb mol or kg masskg mol 132 Oxidation of CO and Mole Units The gas CO is being oxidized by O2 to form CO2 How many kg of CO2 will be formed from 56 kg of CO Also calculate the kg of O2 theoretically needed for this reaction Hint First write the balanced chemical equation to obtain the mol O2 needed for 10 kg mol CO Then calculate the kg mol of CO in 56 kg CO Ans 880 kg CO2 320 kg O2 133 Composition of a Gas Mixture A gaseous mixture contains 20 g of N2 83 g of O2 and 45 g of CO2 Calculate the composition in mole fraction and the average molecular weight of the mixture Ans Average mol wt 342 g massg mol 342 kg masskg mol 134 Composition of a Protein Solution A liquid solution contains 115 wt of a protein 027 wt KCl and the remainder water The average molecular weight of the protein by gel permeation is 525 000 g massg mol Calculate the mole fraction of each component in solution 135 Concentration of NaCl Solution An aqueous solution of NaCl has a concentration of 240 wt NaCl with a density of 1178 gcm3 at 25C Calculate the following a Mole fraction of NaCl and water b Concentration of NaCl as g molliter lbmft3 lbmgal and kgm3 141 Conversion of Pressure Measurements in Freeze Drying In the experimental measurement of freeze drying of beef an absolute pressure of 24 mm Hg was held in the chamber Convert this pressure to atm in of water at 4C μm of Hg and Pa Hint See Appendix A1 for conversion factors Ans 316 x 103 atm 1285 in H2O 2400 μm Hg 320 Pa 142 Compression and Cooling of Nitrogen Gas A volume of 650 ft3 of N2 gas at 90F and 290 psig is compressed to 75 psig and cooled to 65F Calculate the final volume in ft3 and the final density in lbmft3 Hint Be sure to convert all pressures to psia first and then to atm Substitute original conditions into Eq 141 to obtain n lb mol 143 Gas Composition and Volume A gas mixture of 013 g mol NH3 127 g mol N2 and 0025 g mol H2O vapor is contained at a total pressure of 830 mm Hg and 323 K Calculate the following a Mole fraction of each component b Partial pressure of each component in mm Hg c Total volume of mixture in m3 and ft3 144 Evaporation of a HeatSensitive Organic Liquid An organic liquid is being evaporated from a liquid solution containing a few percent nonvolatile dissolved solids Since it is heatsensitive and may discolor at high temperatures it will be evaporated under vacuum If the lowest absolute pressure that can be obtained in the apparatus is 120 mm Hg what will be the temperature of evaporation in K It will be assumed that the small amount of solids does not affect the vapor Chap 1 Problems 25 pressure which is given as follows log PA 22501T 905 where PA is in mm Hg and T in K Ans T 2823 K or 91C 151 Evaporation of Cane Sugar Solutions An evaporator is used to concentrate cane sugar solutions A feed of 10000 kgd of a solution containing 38 wt sugar is evaporated producing a 74 wt solution Calculate the weight of solution produced and amount of water removed Ans 5135 kgd of 74 wt solution 4865 kgd water 152 Processing of Fish Meal Fish are processed into fish meal and used as a supplementary protein food In the processing the oil is first extracted to produce wet fish cake containing 80 wt water and 20 wt bonedry cake This wet cake feed is dried in rotary drum dryers to give a dry fish cake product containing 40 wt water Finally the product is finely ground and packed Calculate the kgh of wet cake feed needed to produce 1000 kgh of dry fish cake product Ans 3000 kgh wet cake feed 153 Drying of Lumber A batch of 100 kg of wet lumber containing 11 wt moisture is dried to a water content of 638 kg water10 kg bonedry lumber What is the weight of dried lumber and the amount of water removed 154 Processing of Paper Pulp A wet paper pulp contains 68 wt water After the pulp was dried it was found that 55 of the original water in the wet pulp was removed Calculate the composition of the dried pulp and its weight for a feed of 1000 kgmin of wet pulp 155 Production of Jam from Crushed Fruit in Two Stages In a process producing jam Cl crushed fruit containing 14 wt soluble solids is mixed in a mixer with sugar 122 kg sugar100 kg crushed fruit and pectin 00025 kg pectin100 kg crushed fruit The resultant mixture is then evaporated in a kettle to produce a jam containing 67 wt soluble solids For a feed of 1000 kg crushed fruit calculate the kg mixture from the mixer kg water evaporated and kg jam produced Ans 22225 kg mixture 189 kg water 20335 kg jam 156 Drying of Cassava Tapioca Root Tapioca flour is used in many countries for bread and similar products The flour is made by drying coarse granules of the cassava root containing 66 wt moisture to 5 moisture and then grinding to produce a flour How many kg of granules must be dried and how much water removed to produce 5000 kgh of flour 157 Processing of Soybeans in Three Stages A feed of 10000 kg of soybeans is processed in a sequence of three stages or steps El The feed contains 35 wt protein 271 wt carbohydrate 94 wt fiber and ash 105 wt moisture and 180 wt oil In the first stage the beans are crushed and pressed to remove oil giving an expressed oil stream and a stream of pressed beans containing 6 oil Assume no loss of other constituents with the oil stream In the second step the pressed beans are extracted with hexane to produce an extracted meal stream containing 05 wt oil and a hexaneoil stream Assume no hexane in the extracted meal Finally in the last step the extracted meal is dried to give a dried meal of 8 wt moisture Calculate a Kg of pressed beans from the first stage b Kg of extracted meal from stage 2 c Kg of final dried meal and the wt protein in the dried meal Ans a 8723 kg b 8241 kg c 7816 kg 448 wt protein 26 Chap1 Problems 158 Recycle in a Dryer A solid material containing 150 wt moisture is dried so that it contains 70 wt water by blowing fresh warm air mixed with recycled air over the solid in the dryer The inlet fresh air has a humidity of 001 kg waterkg dry air the air from the drier that is recycled has a humidity of 01 kg waterkg dry air and the mixed air to the dryer 003 kg waterkg dry air For a feed of 100 kg solidh fed to the dryer calculate the kg dry airh in the fresh air the kg dry airh in the recycle air and the kgh of dried product Ans 956 kgh dry air in fresh air 273 kgh dry air in recycle air and 914 kgh dried product 159 Crystallization and Recycle It is desired to produce 1000 kgh of Na3PO412H2O crystals from a feed solution containing 56 wt Na3PO4 and traces of impurity The original solution is first evaporated in an evaporator to a 35 wt Na3PO4 solution and then cooled to 293 K in a crystallizer where the hydrated crystals and a mother liquor solution are removed One out of every 10 kg of mother liquor is discarded to waste to get rid of the impurities and the remaining mother liquor is recycled to the evaporator The solubility of Na3PO4 at 293 K is 991 wt Calculate the kgh of feed solution and kgh of water evaporated Ans 7771 kgh feed 6739 kgh water 1510 Evaporation and Bypass in Orange Juice Concentration In a process for concentrating 1000 kg of freshly extracted orange juice Cl containing 125 wt solids the juice is strained yielding 800 kg of strained juice and 200 kg of pulpy juice The strained juice is concentrated in a vacuum evaporator to give an evaporated juice of 58 solids The 200 kg of pulpy juice is bypassed around the evaporator and mixed with the evaporated juice in a mixer to improve the flavor This final concentrated juice contains 42 wt solids Calculate the concentration of solids in the strained juice the kg of final concentrated juice and the concentration of solids in the pulpy juice bypassed Hint First make a total balance and then a solids balance on the overall process Next make a balance on the evaporator Finally make a balance on the mixer Ans 342 wt solids in pulpy juice 1511 Manufacture of Acetylene For the making of 6000 ft³ of acetylene CHCH gas at 70F and 750 mm Hg solid calcium carbide CaC2 which contains 97 wt CaC2 and 3 wt solid inerts is used along with water The reaction is CaC2 2H2O CHCH CaOH2 The final lime slurry contains water solid inerts and CaOH2 lime In this slurry the total wt solids of inerts plus CaOH2 is 20 How many lb of water must be added and how many lb of final lime slurry is produced Hint Use a basis of 6000 ft³ and convert to lb mol This gives 1530 lb mol C2H2 1530 lb mol CaOH2 and 1530 lb mol CaC2 added Convert lb mol CaC2 feed to lb and calculate lb inerts added The total lb solids in the slurry is then the sum of the CaOH2 plus inerts In calculating the water added remember that some is consumed in the reaction Ans 5200 lb water added 2359 kg 5815 lb lime slurry 2638 kg 1512 Combustion of Solid Fuel A fuel analyzes 740 wt C and 120 ash inert Air is added to burn the fuel producing a flue gas of 124 CO2 12 CO 57 O2 and 807 N2 Calculate the kg of fuel used for 100 kg mol of outlet flue gas and the kg mol of air used Hint First calculate the mol O2 added in the air using the fact that the N2 in the flue gas equals the N2 added in the air Then make a carbon balance to obtain the total moles of C added 1513 Burning of Coke A furnace burns a coke containing 810 wt C 08 H and the rest inert ash The furnace uses 60 excess air air over and above that needed to burn all C to CO2 and H to H2O Calculate the moles of all components in the flue gas if only 95 of the carbon goes to CO2 and 5 to CO Chap 1 Problems 27 1514 Production of Formaldehyde Formaldehyde CH2O is made by the catalytic oxidation of pure methanol vapor and air in a reactor The moles from this reactor are 631 N2 134 O2 59 H2O 41 CH2O 123 CH3OH and 12 HCOOH The reaction is CH3OH 12 O2 CH2O H2O A side reaction occurring is CH2O 12 O2 HCOOH Calculate the mol methanol feed mol air feed and percent conversion of methanol to formaldehyde Ans 176 mol CH3OH 798 mol air 233 conversion 161 Heating of CO2 Gas A total of 250 g of CO2 gas at 373 K is heated to 623 K at 10132 kPa total pressure Calculate the amount of heat needed in cal btu and kJ Ans 15050 cal 597 btu 6298 kJ 162 Heating a Gas Mixture A mixture of 25 lb mol N2 and 75 lb mol CH4 is being heated from 400F to 800F at 1 atm pressure Calculate the total amount of heat needed in btu 163 Final Temperature in Heating Applesauce A mixture of 454 kg of applesauce at 10C is heated in a heat exchanger by adding 121300 kJ Calculate the outlet temperature of the applesauce Hint In Appendix A4 a heat capacity for applesauce is given at 328C Assume that this is constant and use this as the average cpm Ans 765C 164 Use of Steam Tables Using the steam tables determine the enthalpy change for 1 lb water for each of the following cases a Heating liquid water from 40F to 240F at 30 psia Note that the effect of total pressure on the enthalpy of liquid water can be neglected b Heating liquid water from 40F to 240F and vaporizing at 240F and 2497 psia c Cooling and condensing a saturated vapor at 212F and 1 atm abs to a liquid at 60F d Condensing a saturated vapor at 212F and 1 atm abs Ans a 20042 btulbm b 11527 btulbm c 11224 btulbm d 9703 btulbm 22569 kJkg 165 Heating and Vaporization Using Steam Tables A flow rate of 1000 kgh of water at 211C is heated to 110C when the total pressure is 2442 kPa in the first stage of a process In the second stage at the same pressure the water is heated further until it is all vaporized at its boiling point Calculate the total enthalpy change in the first stage and in both stages 166 Combustion of CH4 and H2 For 100 g mol of a gas mixture of 75 mol CH4 and 25 H2 calculate the total heat of combustion of the mixture at 298 K and 10132 kPa assuming that combustion is complete 167 Heat of Reaction from Heats of Formation For the reaction 4NH3g 50 g 4NOg 6H2Og calculate the heat of reaction ΔH at 298 K and 10132 kPa for 4 g mol of NH3 reacting Ans ΔH heat of reaction 9047 kJ 171 Heat Balance and Cooling of Milk In the processing of rich cows milk 4540 kgh of milk is cooled from 60C to 444C by a refrigerant Calculate the heat removed from the milk Ans Heat removed 2696 kW 28 Chap 1 Problems 172 Heating of Oil by Air A flow of 2200 lbmh of hydrocarbon oil at 100F enters a heat exchanger where it is heated to 150F by hot air The hot air enters at 300F and is to leave at 200F Calculate the total lb mol airh needed The mean heat capacity of the oil is 045 btulbm F Ans 701 lb mol airh 318 kg molh 173 Combustion of Methane in a Furnace A gas stream of 10000 kg molh of CH4 at 10132 kPa and 373 K is burned in a furnace using air at 313 K The combustion is complete and 50 excess air is used The flue gas leaves the furnace at 673 K Calculate the heat removed in the furnace Hint Use a datum of 298 K and liquid water at 298 K The input items will be the following the enthalpy of CH4 at 373 K referred to 298 K the enthalpy of the air at 313 K referred to 298 K delta H subscript c superscript o the heat of combustion of CH4 at 298 K which is referred to liquid water and q the heat added The output items will include the enthalpies of CO2 O2 N2 and H2O gases at 673 K referred to 298 K and the latent heat of H2O vapor at 298 K and 10132 kPa from Appendix A2 It is necessary to include this latent heat since the basis of the calculation and of the delta H subscript c superscript o is liquid water 174 Preheating Air by Steam for Use in a Dryer An air stream at 322C is to be used in a dryer and is first preheated in a steam heater where it is heated to 655C The air flow is 1000 kg molh The steam enters the heater saturated at 1489C is condensed and cooled and leaves as a liquid at 1378C Calculate the amount of steam used in kgh Ans 450 kg steamh 175 Cooling of Cans of Potato Soup After Thermal Processing A total of 1500 cans of potato soup undergo thermal processing in a retort at 240F The cans are then cooled to 100F in the retort before being removed from the retort by cooling water which enters at 75F and leaves at 85F Calculate the lb of cooling water needed Each can of soup contains 10 lb of liquid soup and the empty metal can weigh 016 lb The mean heat capacity of the soup is 094 btulbm F and that of the metal can is 012 btulbm F A metal rack or basket which is used to hold the cans in the retort weighs 350 lb and has a heat capacity of 012 btulbm F Assume that the metal rack is cooled from 240F to 85F the temperature of the outlet water The amount of heat removed from the retort walls in cooling from 240 to 100F is 10000 btu Radiation loss from the retort during cooling is estimated as 5000 btu Ans 21320 lb water 9670 kg 181 Graphical Integration and Numerical Integration Using Simpsons Method The following experimental data of y f x were obtained x f x x f x 0 100 04 53 01 75 05 60 02 605 06 725 03 535 It is desired to determine the integral A integral from x 0 to x06 of f x dx a Do this by a graphical integration b Repeat using Simpsons numerical method Ans a A 3855 b A 3845 Chap 1 Problems 29 43 182 Graphical and Numerical Integration to Obtain Wastewater Flow The rate of flow of wastewater in an open channel has been measured and the following data obtained Time min Flow m3min Time min Flow m3min 0 655 70 800 10 705 80 725 20 780 90 670 30 830 100 640 40 870 110 620 50 890 120 610 60 870 a Determine the total flow in m3 for the first 60 min and also the total for 120 min by graphical integration b Determine the flow for 120 min using Simpsons numerical method Ans a 48460 m3 for 60 min 90390 m3 for 120 m REFERENCES C1 CHARM SE The Fundamentals of Food Engineering 2nd ed Westport Conn Avi Publishing Co Inc 1971 E1 EARLE RL Unit Operations in Food Processing Oxford Pergamon Press Inc 1966 H1 HOUGEN OA WATSON KM and RAGATZ RA Chemical Process Principles Part I 2nd ed New York John Wiley Sons Inc 1954 O1 OKOS MR MS thesis Ohio State University Columbus Ohio 1972 P1 PERRY RH and GREEN D Perrys Chemical Engineers Handbook 6th ed New York McGrawHill Book Company 1984 S1 SOBER HA Handbook of Biochemistry Selected Data for Molecular Biology 2nd ed Boca Raton Fla Chemical Rubber Co Inc 1970 W1 WEAST RC and SELBY SM Handbook of Chemistry and Physics 48th ed Boca Raton Fla Chemical Rubber Co Inc 19671968 30 Chap 1 References 44 CHAPTER 2 Principles of Momentum Transfer and Overall Balances 21 INTRODUCTION The flow and behavior of fluids is important in many of the unit operations in process engineering A fluid may be defined as a substance that does not permanently resist distortion and hence will change its shape In this text gases liquids and vapors are considered to have the characteristics of fluids and to obey many of the same laws In the process industries many of the materials are in fluid form and must be stored handled pumped and processed so it is necessary that we become familiar with the principles that govern the flow of fluids and also with the equipment used Typical fluids encountered include water air CO2 oil slurries and thick syrups If a fluid is inappreciably affected by changes in pressure it is said to be incompressible Most liquids are incompressible Gases are considered to be compressible fluids However if gases are subjected to small percentage changes in pressure and temperature their density changes will be small and they can be considered to be incompressible Like all physical matter a fluid is composed of an extremely large number of molecules per unit volume A theory such as the kinetic theory of gases or statistical mechanics treats the motions of molecules in terms of statistical groups and not in terms of individual molecules In engineering we are mainly concerned with the bulk or macroscopic behavior of a fluid rather than with the individual molecular or microscopic behavior In momentum transfer we treat the fluid as a continuous distribution of matter or as a continuum This treatment as a continuum is valid when the smallest volume of fluid contains a large enough number of molecules so that a statistical average is meaningful and the macroscopic properties of the fluid such as density pressure and so on vary smoothly or continuously from point to point The study of momentum transfer or fluid mechanics as it is often called can be divided into two branches fluid statics or fluids at rest and fluid dynamics or fluids in motion In Section 22 we treat fluid statics in the remaining sections of Chapter 2 and in Chapter 3 fluid dynamics Since in fluid dynamics momentum is being transferred the term momentum transfer or transport is usually used In Section 23 momentum transfer is related to heat and mass transfer 31 22 FLUID STATICS 22A Force Units and Dimensions In a static fluid an important property is the pressure in the fluid Pressure is familiar as a surface force exerted by a fluid against the walls of its container Also pressure exists at any point in a volume of a fluid In order to understand pressure which is defined as force exerted per unit area we must first discuss a basic law of Newtons This equation for calculation of the force exerted by a mass under the influence of gravity is F mg SI units 221 F mggc English units where in SI units F is the force exerted in newtons Nkgms² m the mass in kg and g the standard acceleration of gravity 980665 ms² In English units F is in lbₓ m in lbₘ g is 321740 fts² and gₓ a gravitational conversion factor is 32174 lbₓftlbₘs² The use of the conversion factor gₓ means that ggₓ has a value of 10 lbₓlbₘ and that 1 lbₘ conveniently gives a force equal to 1 lbₓ Often when units of pressure are given the word force is omitted such as in lbin² psi instead of lbₓin² When the mass m is given in g mass F is g force g 980665 cms² and gₓ 980665 g masscmg forces² However the units g force are seldom used Another system of units sometimes used in Eq 221 is that where the gₓ is omitted and the force F mg is given as lbₘfts² which is called poundals Then 1 lbₘ acted on by gravity will give a force of 32174 poundals lbₓfts² Or if 1 g mass is used the force F mg is expressed in terms of dynes gcms² This is the centimetergramsecond cgs systems of units Conversion factors for different units of force and of force per unit area pressure are given in Appendix A1 Note that always in the SI system and usually in the cgs system the term gₓ is not used EXAMPLE 221 Units and Dimensions of Force Calculate the force exerted by 3 lb mass in terms of the following a Lb force English units b Dynes cgs units c Newtons SI units Solution For part a using Eq 221 F force m ggₓ 3 lbₘ32174 fts²132174 lbₘftlbₓs² 3 lb force lbₓ For part b F mg 3 lbₘ45359 glbₘ980665 cms² 1332 10⁶ gcms² 1332 10⁶ dyn 32 Chap 2 Principles of Momentum Transfer and Overall Balances As an alternative method for part b from Appendix A1 1 dyn 22481 10⁶ lbₓ F 3 lbₓ122481 10⁶ lbₓdyn 1332 10⁶ dyn To calculate newtons in part c F mg 3 lbₘ 1 kg22046 lbₘ980665 ms² 1332 kgms² 1332 N As an alternative method using values from Appendix A1 1 gcms² dyn 10⁵ kgms² newton F 1332 10⁶ dyn10⁵ newtondyn 1332 N 22B Pressure in a Fluid Since Eq 221 gives the force exerted by a mass under the influence of gravity the force exerted by a mass of fluid on a supporting area or forceunit area pressure also follows from this equation In Fig 221 a stationary column of fluid of height h₂ m and constant crosssectional area A m² where A A₀ A₁ A₂ is shown The pressure above the fluid is P₀ Nm² that is this could be the pressure of the atmosphere above the fluid The fluid at any point say h₁ must support all the fluid above it It can be shown that the forces at any given point in a nonmoving or static fluid must be the same in all directions Also for a fluid at rest the forceunit area or pressure is the same at all points with the same elevation For example at h₁ m from the top the pressure is the same at all points shown on the crosssectional area A₁ The use of Eq 221 will be shown in calculating the pressure at different vertical points in Fig 221 The total mass of fluid for h₂ m height and density ρ kgm³ is total kg fluid h₂ mA m²ρ kgm³ h₂ Aρ kg 222 FIGURE 221 Pressure in a static fluid P₀ A₀ P₁ h₁ A₁ h₂ h₃ P₂ A₂ Sec 22 Fluid Statics 33 Substituting into Eq 222 the total force F of the fluid on area A₁ due to the fluid only is F h₂ Aρ kgg ms² h₂ Aρg kgms² N 223 The pressure P is defined as forceunit area P FA h₂ Aρg1A h₂ ρg Nm² or Pa 224 This is the pressure on A₂ due to the mass of the fluid above it However to get the total pressure P₂ on A₂ the pressure P₀ on the top of the fluid must be added P₂ h₂ ρg P₀ Nm² or Pa 225 Equation 225 is the fundamental equation to calculate the pressure in a fluid at any depth To calculate P₁ P₁ h₁ ρg P₀ 226 The pressure difference between points 2 and 1 is P₂ P₁ h₂ ρg P₀ h₁ ρg P₀ h₂ h₁ ρg SI units 227 P₂ P₁ h₂ h₁ ρg gₓ English units Since it is the vertical height of a fluid that determines the pressure in a fluid the shape of the vessel does not affect the pressure For example in Fig 222 the pressure P₁ at the bottom of all three vessels is the same and equal to h₁ ρg P₀ EXAMPLE 222 Pressure in Storage Tank A large storage tank contains oil having a density of 917 kgm³ 0917 gcm³ The tank is 366 m 120 ft tall and is vented open to the atmosphere of 1 atm abs at the top The tank is filled with oil to a depth of 305 m 10 ft and also contains 061 m 20 ft of water in the bottom of the tank Calculate the pressure in Pa and psia 305 m from the top of the tank and at the bottom Also calculate the gage pressure at the tank bottom Solution First a sketch is made of the tank as shown in Fig 223 The pressure P₀ 1 atm abs 14696 psia from Appendix A1 Also P₀ 101325 10⁵ Pa P₀ ρ P₁ P₀ h₁ P₁ P₀ P₁ P₀ h₁ FIGURE 222 Pressure in vessels of various shapes 34 Chap 2 Principles of Momentum Transfer and Overall Balances FIGURE 223 Storage tank in Example 222 From Eq 226 using English and then SI units P1 h1 ρoil ggc P0 10 ft 0917 x 6243 lbmft310 lbflbm1144 in2ft2 14696 lbfin2 1868 psia P1 h1 ρoil g P0 305 m917 kgm398066 ms2 10132 x 105 1287 x 105 Pa To calculate P2 at the bottom of the tank ρwater 100 gcm3 and P2 h2 ρwater ggc P1 20100 x 6243101144 1868 1955 psia h2 ρwater g P1 061100098066 1287 x 105 1347 x 105 Pa The gage pressure at the bottom is equal to the absolute pressure P2 minus 1 atm pressure Pgage 1955 psia 14696 psia 485 psig 22C Head of a Fluid Pressures are given in many different sets of units such as psia dyncm2 and newtonsm2 as given in Appendix A1 However a common method of expressing pressures is in terms of head in m or feet of a particular fluid This height or head in m or feet of the given fluid will exert the same pressure as the pressures it represents Using Eq 224 which relates pressure P and height h of a fluid and solving for h which is the head in m hhead Pρg m SI h P gc ρg ft English 228 Sec 22 Fluid Statics 35 EXAMPLE 223 Conversion of Pressure to Head of a Fluid Given the pressure of 1 standard atm as 101325 kNm2 Appendix A1 do as follows a Convert this pressure to head in m water at 4C b Convert this pressure to head in m Hg at 0C Solution For part a the density of water at 4C in Appendix A2 is 1000 gcm3 From A1 a density of 1000 gcm3 equals 1000 kgm3 Substituting these values into Eq 228 hhead Pρg 101325 x 103 1000980665 1033 m of water at 4C For part b the density of Hg in Appendix A1 is 135955 gcm3 For equal pressures P from different fluids Eq 228 can be rewritten as P ρHg hHg g ρH2O hH2O g 229 Solving for hHg in Eq 229 and substituting known values hHghead ρH2O ρHg hH2O 10001359551033 0760 m Hg 22D Devices to Measure Pressure and Pressure Differences In chemical and other industrial processing plants it is often important to measure and control the pressure in a vessel or process andor the liquid level in a vessel Also since many fluids are flowing in a pipe or conduit it is necessary to measure the rate at which the fluid is flowing Many of these flow meters depend upon devices to measure a pressure or pressure difference Some common devices are considered in the following paragraphs 1 Simple Utube manometer The Utube manometer is shown in Fig 224a The FIGURE 224 Manometers to measure pressure differences a U tube b twofluid U tube Chap 2 Principles of Momentum Transfer and Overall Balances 36 pressure pa Nm2 is exerted on one arm of the U tube and pb on the other arm Both pressures pa and pb could be pressure taps from a fluid meter or pa could be a pressure tap and pb the atmospheric pressure The top of the manometer is filled with liquid B having a density of ρB kgm3 and the bottom with a more dense fluid A having a density of ρA kgm3 Liquid A is immiscible with B To derive the relationship between pa and pb pa is the pressure at point 1 and pb at point 5 The pressure at point 2 is p2 pa Z RρB g Nm2 2210 where R is the reading of the manometer in m The pressure at point 3 must be equal to that at 2 by the principles of hydrostatics p3 p2 2211 The pressure at point 3 also equals the following p3 pb Z ρB g R ρA g 2212 Equating Eq 2210 to 2212 and solving pa Z RρB g pb Z ρB g R ρA g 2213 pa pb RρA ρB g SI 2214 pa pb RρA ρB ggc English The reader should note that the distance Z does not enter into the final result nor do the tube dimensions provided that pa and pb are measured in the same horizontal plane EXAMPLE 224 Pressure Difference in a Manometer A manometer as shown in Fig 224a is being used to measure the head or pressure drop across a flow meter The heavier fluid is mercury with a density of 136 gcm3 and the top fluid is water with a density of 100 gcm3 The reading on the manometer is R 327 cm Calculate the pressure difference in Nm2 using SI units Solution Converting R to m R 327100 0327 m Also converting ρA and ρB to kgm3 and substituting into Eq 2214 pa pb RρA ρBg 0327 m136 101000 kgm398066 ms2 4040 x 104 Nm2 585 psia 2 Twofluid U tube In Fig 224b a twofluid U tube is shown which is a sensitive device to measure small heads or pressure differences Let a m2 be the crosssectional area of each of the large reservoirs and a m2 be the crosssectional area of each of the tubes forming the U Proceeding and making a pressure balance as for the U tube pa pb R R0ρA ρB aA ρB aA ρCg 2215 where R0 is the reading when pa pb R is the actual reading ρA is the density of the heavier fluid and ρB is the density of the lighter fluid Usually aA is made sufficiently Sec 22 Fluid Statics 37 small to be negligible and also R₀ is often adjusted to zero then pa pb RρA ρBg SI pa pb RρA ρB g gc English 2216 If ρA and ρB are close to each other the reading R is magnified EXAMPLE 225 Pressure Measurement in a Vessel The Utube manometer in Fig 225a is used to measure the pressure pA in a vessel containing a liquid with a density ρA Derive the equation relating the pressure pA and the reading on the manometer as shown Solution At point 2 the pressure is p₂ patm h₂ ρB g Nm² 2217 At point 1 the pressure is p₁ pA h₁ ρA g 2218 Equating p₁ p₂ by the principles of hydrostatics and rearranging pA patm h₂ ρB g h₁ ρA g 2219 Another example of a Utube manometer is shown in Fig 225b This device is used in this case to measure the pressure difference between two vessels 3 Bourdon pressure gage Although manometers are used to measure pressures the most common pressuremeasuring device is the mechanical Bourdontube pressure gage A coiled hollow tube in the gage tends to straighten out when subjected to internal pressure and the degree of straightening depends on the pressure difference between the inside and outside pressures The tube is connected to a pointer on a calibrated dial 4 Gravity separator for two immiscible liquids In Fig 226 a continuous gravity separator decanter is shown for the separation of two immiscible liquids A heavy liquid and B light liquid The feed mixture of the two liquids enters at one end of the separator vessel and the liquids flow slowly to the other end and separate into two distinct layers Each liquid flows through a separate overflow line as shown Assuming pA patm h₂ h₁ 2 1 ρA ρB pB ρB ρC a b FIGURE 225 Measurements of pressure in vessels a measurement of pressure in a vessel b measurement of differential pressure 38 Chap 2 Principles of Momentum Transfer and Overall Balances feed vent light liquid B overflow liquid B liquid A heavy liquid A overflow FIGURE 226 Continuous atmospheric gravity separator for immiscible liquids the frictional resistance to the flow of the liquids is essentially negligible the principles of fluid statics can be used to analyze the performance In Fig 226 the depth of the layer of heavy liquid A is hA1 m and that of B is hB The total depth hT hA1 hB and is fixed by position of the overflow line for B The heavy liquid A discharges through an overflow leg hA2 m above the vessel bottom The vessel and the overflow lines are vented to the atmosphere A hydrostatic balance gives hB ρB g hA1 ρA g hA2 ρA g 2220 Substituting hB hT hA1 into Eq 2220 and solving for hA1 hA1 hA2 hT ρB ρA 1 ρB ρA 2221 This shows that the position of the interface or height hA1 depends on the ratio of the densities of the two liquids and on the elevations hA2 and hT of the two overflow lines Usually the height hA2 is movable so that the interface level can be adjusted 23 GENERAL MOLECULAR TRANSPORT EQUATION FOR MOMENTUM HEAT AND MASS TRANSFER 23A General Molecular Transport Equation and General Property Balance 1 Introduction to transport processes In molecular transport processes in general we are concerned with the transfer or movement of a given property or entity by molecular movement through a system or medium which can be a fluid gas or liquid or a solid This property that is being transferred can be mass thermal energy heat or momentum Each molecule of a system has a given quantity of the property mass thermal energy or momentum associated with it When a difference of concentration of the property exists for any of these properties from one region to an adjacent region a net transport of this property occurs In dilute fluids such as gases where the molecules are relatively far apart the rate of transport of the property should be relatively fast since few molecules are present to block the transport or interact In dense fluids such as liquids the molecules are close together and transport or diffusion proceeds more slowly The molecules in solids are even more closepacked than in liquids and molecular migration is even more restricted Sec 23 General Molecular Transport Equation 39 2 General molecular transport equation All three of the molecular transport processes of momentum heat or thermal energy and mass are characterized in the elementary sense by the same general type of transport equation First we start by noting the following rate of a transfer process driving force resistance 231 This states what is quite obviousthat we need a driving force to overcome a resistance in order to transport a property This is similar to Ohms law in electricity where the rate of flow of electricity is proportional to the voltage drop driving force and inversely proportional to the resistance We can formalize Eq 231 by writing an equation as follows for molecular transport or diffusion of a property ψz δ dΓdz 232 where ψz is defined as the flux of the property as amount of property being transferred per unit time per unit crosssectional area perpendicular to the z direction of flow in amount of propertys m² δ is a proportionality constant called diffusivity in m²s Γ is concentration of the property in amount of propertym³ and z is the distance in the direction of flow in m If the process is at steady state then the flux ψz is constant Rearranging Eq 232 and integrating ψz z₁ to z₂ dz δ Γ₁ to Γ₂ dΓ 233 ψz δΓ₁ Γ₂ z₂ z₁ 234 A plot of the concentration Γ versus z is shown in Fig 231a and is a straight line Since the flux is in the direction 1 to 2 of decreasing concentration the slope dΓdz is negative and the negative sign in Eq 232 gives a positive flux in the direction 1 to 2 In Section 23B the specialized equations for momentum heat and mass transfer will be shown to be the same as Eq 234 for the general property transfer unit area Concentration of property Γ ψz flux in ψzz out ψzz Δz z₁ z₂ z z Δz Distance z Δz a b FIGURE 231 Molecular transport of a property a plot of concentration versus distance for steady state b unsteadystate general property balance 40 Chap 2 Principles of Momentum Transfer and Overall Balances EXAMPLE 231 Molecular Transport of a Property at Steady State A property is being transported by diffusion through a fluid at steady state At a given point 1 the concentration is 137 x 102 amount of property m3 and 072 x 102 at point 2 at a distance z2 040 m The diffusivity δ 0013 m2s and the crosssectional area is constant a Calculate the flux b Derive the equation for Γ as a function of distance c Calculate Γ at the midpoint of the path Solution For part a substituting into Eq 234 ψz δΓ1 Γ2 z2 z1 0013137 x 102 072 x 102 040 0 2113 x 104 amount of propertys m2 For part b integrating Eq 232 between Γ1 and Γ and z1 and z and rearranging ψz z1z dz δ Γ1Γ dΓ 235 Γ Γ1 ψz δz1 z 236 For part c using the midpoint z 020 m and substituting into Eq 236 Γ 137 x 102 2113 x 104 00130 02 1045 x 102 amount of property m3 3 General property balance for unsteady state In calculating the rates of transport in a system using the molecular transport equation 232 it is necessary to account for the amount of this property being transported in the entire system This is done by writing a general property balance or conservation equation for the property momentum thermal energy or mass at unsteady state We start by writing an equation for the z direction only which accounts for all the property entering by molecular transport leaving being generated and accumulating in a system shown in Fig 231b which is an element of volume Δz1 m3 fixed in space rate of property in rate of generation of property rate of property out rate of accumulation of property 237 The rate of input is ψzz11 amount of propertys and the rate of output is ψzzΔz1 where the crosssectional area is 10 m2 The rate of generation of the property is RΔz 1 where R is rate of generation of propertys m3 The accumulation term is rate of accumulation of property Γt Δz 1 238 Sec 23 General Molecular Transport Equation 41 Substituting the various terms into Eq 237 ψzz1 RΔz 1 ψzzΔz1 Γt Δz 1 239 Dividing by Δz and letting Δz go to zero Γt ψzz R 2310 Substituting Eq 232 for ψz into 2310 and assuming that δ is constant Γt δ 2Γz2 R 2311 For the case where no generation is present Γt δ 2Γz2 2312 This final equation relates the concentration of the property Γ to position z and time t Equations 2311 and 2312 are general equations for the conservation of momentum thermal energy or mass and will be used in many sections of this text The equations consider here only molecular transport occurring and the equations do not consider other transport mechanisms such as convection and so on which will be considered when the specific conservation equations are derived in later sections of this text for momentum energy or mass 23B Introduction to Molecular Transport The kinetic theory of gases gives us a good physical interpretation of the motion of individual molecules in fluids Because of their kinetic energy the molecules are in rapid random movement often colliding with each other Molecular transport or molecular diffusion of a property such as momentum heat or mass occurs in a fluid because of these random movements of individual molecules Each individual molecule containing the property being transferred moves randomly in all directions and there are fluxes in all directions Hence if there is a concentration gradient of the property there will be a net flux of the property from high to low concentration This occurs because equal numbers of molecules diffuse in each direction between the highconcentration and lowconcentration regions 1 Momentum transport and Newtons law When a fluid is flowing in the x direction parallel to a solid surface a velocity gradient exists where the velocity vx in the x direction decreases as we approach the surface in the z direction The fluid has xdirected momentum and its concentration is vx ρ momentumm3 where the momentum has units of kg ms Hence the units of vx ρ are kg msm3 By random diffusion of molecules there is an exchange of molecules in the z direction an equal number moving in each direction z and z directions between the fastermoving layer of molecules and the slower adjacent layer Hence the xdirected momentum has been transferred in the z direction from the faster to the slowermoving layer The equation for this transport of 42 Chap 2 Principles of Momentum Transfer and Overall Balances momentum is similar to Eq 232 and is Newtons law of viscosity written as follows for constant density ρ τzx v dvx ρdz 2313 where τzx is flux of xdirected momentum in the z direction kg mss m2 v is μρ the momentum diffusivity in m2s z is the direction of transport or diffusion in m ρ is the density in kgm3 and μ is the viscosity in kgm s 2 Heat transport and Fouriers law Fouriers law for molecular transport of heat or heat conduction in a fluid or solid can be written as follows for constant density ρ and heat capacity cp qz A α dρcp T dz 2314 where qzA is the heat flux in Js m2 α is the thermal diffusivity in m2s and ρcp T is the concentration of heat or thermal energy in Jm3 When there is a temperature gradient in a fluid equal numbers of molecules diffuse in each direction between the hot and the colder region In this way energy is transferred in the z direction 3 Mass transport and Ficks law Ficks law for molecular transport of mass in a fluid or solid for constant total concentration in the fluid is JAz DAB dcA dz 2315 where JAz is the flux of A in kg mol As m2 DAB is the molecular diffusivity of the molecule A in B in m2s and cA is the concentration of A in kg mol Am3 In a manner similar to momentum and heat transport when there is a concentration gradient in a fluid equal numbers of molecules diffuse in each direction between the high and the lowconcentration region and a net flux of mass occurs Hence Eqs 2313 2314 and 2315 for momentum heat and mass transfer are all similar to each other and to the general molecular transport equation 232 All equations have a flux on the lefthand side of each equation a diffusivity in m2s and the derivative of the concentration with respect to distance All three of the molecular transport equations are mathematically identical Thus we state we have an analogy or similarity among them It should be emphasized however that even though there is a mathematical analogy the actual physical mechanisms occurring can be totally different For example in mass transfer two components are often being transported by relative motion through one another In heat transport in a solid the molecules are relatively stationary and the transport is done mainly by the electrons Transport of momentum can occur by several types of mechanisms More detailed considerations of each of the transport processes of momentum energy and mass are presented in this and succeeding chapters 24 VISCOSITY OF FLUIDS 24A Newtons Law and Viscosity When a fluid is flowing through a closed channel such as a pipe or between two flat plates either of two types of flow may occur depending on the velocity of this fluid At Sec 24 Viscosity of Fluids 43 low velocities the fluid tends to flow without lateral mixing and adjacent layers slide past one another like playing cards There are no cross currents perpendicular to the direction of flow nor eddies or swirls of fluid This regime or type of flow is called laminar flow At higher velocities eddies form which leads to lateral mixing This is called turbulent flow The discussion in this section is limited to laminar flow A fluid can be distinguished from a solid in this discussion of viscosity by its behavior when subjected to a stress force per unit area or applied force An elastic solid deforms by an amount proportional to the applied stress However a fluid when subjected to a similar applied stress will continue to deform ie to flow at a velocity that increases with increasing stress A fluid exhibits resistance to this stress Viscosity is that property of a fluid which gives rise to forces that resist the relative movement of adjacent layers in the fluid These viscous forces arise from forces existing between the molecules in the fluid and are of similar character as the shear forces in solids The ideas above can be clarified by a more quantitative discussion of viscosity In Fig 241 a fluid is contained between two infinite very long and very wide parallel plates Suppose that the bottom plate is moving parallel to the top plate and at a constant velocity Δvz ms faster relative to the top plate because of a steady force F newtons being applied This force is called the viscous drag and it arises from the viscous forces in the fluid The plates are Δy m apart Each layer of liquid moves in the z direction The layer immediately adjacent to the bottom plate is carried along at the velocity of this plate The layer just above is at a slightly slower velocity each layer moving at a slower velocity as we go up in the y direction This velocity profile is linear with y direction as shown in Fig 241 An analogy to a fluid is a deck of playing cards where if the bottom card is moved all the other cards above will slide to some extent It has been found experimentally for many fluids that the force F in newtons is directly proportional to the velocity Δvz in ms to the area A in m² of the plate used and inversely proportional to the distance Δy in m Or as given by Newtons law of viscosity when the flow is laminar FA μ ΔvzΔy 241 where μ is a proportionality constant called the viscosity of the fluid in Pas or kgms If we let Δy approach zero then using the definition of the derivative τyz μ dvzdy SI units 242 where τyz FA and is the shear stress or force per unit area in newtonsm² Nm² In the cgs system F is in dynes μ in gcms vz in cms and y in cm We can also write Eq 222 as τyz gc μ dvzdy English units 243 where τyz is in units of lbfft² The units of viscosity in the cgs system are gcms called poise or centipoise cp In the SI system viscosity is given in Pas Nsm² or kgms 1 cp 1 x 103 kgms 1 x 103 Pas 1 x 103 Nsm² SI 1 cp 001 poise 001 gcms 1 cp 67197 x 104 lbmfts Other conversion factors for viscosity are given in Appendix A1 Sometimes the viscosity is given as μρ kinematic viscosity in m²s or cm²s where ρ is the density of the fluid EXAMPLE 241 Calculation of Shear Stress in a Liquid Referring to Fig 241 the distance between plates is Δy 05 cm Δvz 10 cms and the fluid is ethyl alcohol at 273 K having a viscosity of 177 cp 00177 gcms a Calculate the shear stress τyz and the velocity gradient or shear rate dvzdy using cgs units b Repeat using lb force s and ft units English units c Repeat using SI units Solution We can substitute directly into Eq 241 or we can integrate Eq 242 Using the latter method rearranging Eq 242 calling the bottom plate point 1 and integrating from y205 to y10 dy μ from vz0 to vz10 dvz 244 τyz μ v1 v2y2 y1 245 Substituting the known values τyz μ v1 v2y2 y1 00177 gcms 10 0 cms 05 0 cm 0354 gcms²cm² 0354 dyncm² 246 To calculate the shear rate dvzdy since the velocity change is linear with y shear rate dvzdy ΔvzΔy 10 0 cms 05 0 cm 200 s¹ 247 For part b using lb force units and the viscosity conversion factor from Appendix A1 μ 177 cp 67197 x 104 lbmftscp 177 67197 x 104 lbmfts Integrating Eq 243 τyz μ lbmfts v1 v2 fts gc lbmftlbfs² y2 y1 ft 248 Substituting known values into Eq 248 and converting Δvz to fts and Δy to ft τyz 739 x 104 lbfft² Also dvzdy 20 s¹ For part c Δy 05100 0005 m Δvz 10100 01 ms and μ 177 x 103 kgms 177 x 103 Pas Substituting into Eq 245 τyz 177 x 1030100005 00354 Nm² The shear rate will be the same at 200 s¹ 24B Momentum Transfer in a Fluid The shear stress τyz in Eqs 241243 can also be interpreted as a flux of zdirected momentum in the y direction which is the rate of flow of momentum per unit area The units of momentum are mass times velocity in kgms The shear stress can be written τyz kgmsm²s momentumm²s 249 This gives an amount of momentum transferred per second per unit area This can be shown by considering the interaction between two adjacent layers of a fluid in Fig 241 which have different velocities and hence different momentum in the z direction The random motions of the molecules in the fastermoving layer send some of the molecules into the slowermoving layer where they collide with the slowermoving molecules and tend to speed them up or increase their momentum in the z direction Also in the same fashion molecules in the slower layer tend to retard those in the faster layer This exchange of molecules between layers produces a transfer or flux of zdirected momentum from highvelocity to lowvelocity layers The negative sign in Eq 242 indicates that momentum is transferred down the gradient from high to lowvelocity regions This is similar to the transfer of heat from high to lowtemperature regions 24C Viscosities of Newtonian Fluids Fluids that follow Newtons law of viscosity Eqs 241243 are called Newtonian fluids For a Newtonian fluid there is a linear relation between the shear stress τyz and the velocity gradient dvzdy rate of shear This means that the viscosity μ is a constant and independent of the rate of shear For nonNewtonian fluids the relation between τyz and dvzdy is not linear ie the viscosity μ does not remain constant but is a function of shear rate Certain liquids do not obey this simple Newtons law These are primarily pastes slurries high polymers and emulsions The science of the flow and deformation of fluids is often called rheology A discussion of nonNewtonian fluids will not be given here but will be included in Section 35 The viscosity of gases which are Newtonian fluids increases with temperature and is approximately independent of pressure up to a pressure of about 1000 kPa At higher pressures the viscosity of gases increases with increase in pressure For example the viscosity of N2 gas at 298 K approximately doubles in going from 100 kPa to about 5 x 104 kPa R1 In liquids the viscosity decreases with increasing temperature Since liquids are essentially incompressible the viscosity is not affected by pressure